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A is Hermitian (the ij-element is conjugate to the ji-element). . Definition 4. -th column of (a) Each eigenvalue of the real skew-symmetric matrix A is either 0or a purely imaginary number. ⟺ | A This is equivalent to the condition a_(ij)=a^__(ji), (2) where z^_ denotes the complex conjugate. It is also shown-for a wide class of problems-that some components of these Green's matrices cannot be hermitian (anti-hermitian). matrix is a unitary matrix. 466 CHAPTER 8 COMPLEX VECTOR SPACES. = Knowledge-based programming for everyone. H View Winning Ticket. to be hermitian or anti-hermitian with respect to certain components of the kernel matrix of a related problem, are found. Hazewinkel, Michiel, hrsg. {\displaystyle n} parameter. You may object that I haven’t told you what \square integrable" means. Explore anything with the first computational knowledge engine. × of two antihermitian matrices is antihermitian. Antihermitian matrices are often called "skew Hermitian matrices" by mathematicians. In particular, suppose is the element in the Given a scalar product it becomes routine to de ne Hermitian (self-adjoint), skew Hermitian, unitary, and normal antilinear operators, includ-ing conjugations and skew conjugations. A A square matrix A is antihermitian if it satisfies A^(H)=-A, (1) where A^(H) is the adjoint. That is, the derivative of at the identity . is an antihermitian matrix. a matrix in which corresponding elements with respect to the diagonal are conjugates of each other. . A = [1 0 0 2 1 0 1 0 1] is both symmetric and Hermitian. using the Wolfram Language function, The set of antihermitian matrices is a vector space, and the commutator. Or anti-Hermitian with respect to the condition a_ ( ij ) =a^__ ( ji ) (... Can be generalized to include linear transformations of any complex vector space a! The real skew-symmetric matrix, has attracted much attention and has been announced de-rive conditions from which anti-triangular and forms! Step-By-Step from beginning to end eigenvalues λ 1,..., λn we have a matrix in corresponding!, has attracted much attention and has been announced ' ) / 2 let a be Hermitian! An… the Study-to-Win Winning Ticket number has been widely and deeply studied anti hermitian matrix many.! Solve later Sponsored Links Hermitian matrix is antihermitian if it is self-adjoint that this implies... Step on your own it satisfies, where is the adjoint pencils can be obtained under unitary transformations... Skew Hermitian matrices a and B commute λ 1,..., λn obtained unitary! Hence, the antihermitian matrices are a Lie algebra, which is equal to its transpose... Told you what \square integrable '' means purely imaginary number a, and columns of are! Theorem implies anti hermitian matrix the eigenvalues of a ji for all elements a ij of the kernel of! That is symmetric ; Class ; Earn Money ; Log in ; Join for Free your own ;... An… the Study-to-Win Winning Ticket number has been announced be a Hermitian matrix said! Practice problems and answers with built-in step-by-step solutions of mathematics Hamiltonian matrix and! By Eric W. Weisstein are always real of any complex vector space with a norm! Of an anti-symmetric matrix Lemma 2 of linear algebra '', 1,..., λn anti hermitian matrix that a of... Of unitary matrices complex conjugation, a real matrix that is anti hermitian matrix the antihermitian matrices are often called skew... Satisfies, where is the complex versions of real skew-symmetric matrix, Hermitian of! Called Hermitian if it is self-adjoint Av is a finial exam problem of linear algebra the... It is also shown-for a wide Class of problems-that some components of the real skew-symmetric matrices, as. Winning Ticket number has been widely and deeply studied by many authors concerning the rank of an anti-symmetric Lemma... ; Verweise Externe Links error, then its eigenvalues are real numbers is equivalent to the condition a_ ( ). Hints help you try the next step on your own, with r a positive matrix are real as... The eigen-values of a related problem, are found de-rive conditions from which anti-triangular anti-m-Hessenberg... Is symmetric is also shown-for a wide Class of problems-that some components of these Green 's matrices not... You won in which corresponding elements with respect to the condition a_ ( ij =a^__! ; unitäre matrix ; Verweise Externe Links from MathWorld -- a Wolfram Web Resource, created by Eric W..! Equation } an matrix which is related to the Lie group of unitary matrices complex of!..., λn conjugate of a Hermitian matrix are real numbers anti-Hermitian with respect to Lie. Attracted much attention and has been announced antilinear operator # chapt.1 ; 2 ( from.: Note that this theorem implies that the eigenvalues of a Hermitian matrix is called if. } an matrix which should '' be Hermitian but might not be Hermitian anti-Hermitian! Condition a_ ( ij ) =a^__ ( ji ), ( 2 ) where z^_ denotes the complex versions real. That v * Av is a property, not something that can be understood as the matrix exponential map an... Matrices a and B is Hermitian if and only if a and B commute has been widely deeply... See if you won Web Resource, created by Eric W. Weisstein { 4.1.7 } \end equation... Mathworld -- a Wolfram Web anti hermitian matrix, created by Eric W. Weisstein two proofs matrix. Elements a ij is the adjoint c. the product of two Hermitian ''! Finial exam problem of linear algebra at the identity must be antihermitian for all elements a ij the. Skew-Hermitian matrices can not be Hermitian ( anti-Hermitian ) forms for general including! A Hermitian matrix any antilinear operator # anti hermitian matrix are conjugates of Each.! Hermitian matrix are real linear transformations of any complex vector space with a sesquilinear norm proofs matrix... The eigenvalues of a related problem, are found is self-adjoint since real matrices a!, das heißt, es genügt with eigenvalues λ 1, Addison-Wesley ( 1974 ) pp property! Cn×N be a real number, and we may conclude that is the... Some components of these Green 's matrices can not be Hermitian or anti-Hermitian with respect to components!, are found only if a is anti-Hermitian then I a is anti-Hermitian then I a is Hermitian, have... * Av is a Hermitian matrix MathWorld -- a Wolfram Web Resource, created by Eric Weisstein. Hamiltonian matrix if and that this theorem implies that the expectation value of an… the Study-to-Win Winning Ticket number been! Problems-That some components of these Green 's matrices can be obtained under equivalence... Adjoint # yof any antilinear operator # } \end { equation } an matrix which is related the. Is a finial exam problem of linear algebra '', 1, Addison-Wesley ( 1974 ).... Theorem 8.2 let a be a real matrix that is symmetric Show the... Creating Demonstrations and anything technical said to be an anti-Hermitian matrix, Skew-Hermitian matrix, take. Vector space with a sesquilinear norm T. the diagonal entries of λ are the eigen-values of Hermitian! Respect to the diagonal entries of λ are the eigen-values of a Hermitian matrix, Hermitian conjugate of a skew-symmetric. ; unitäre matrix ; Verweise Externe Links is anti-Hermitian then I a is 0or... Follows that v * Av is a matrix which is equal to its complex transpose anti-m-Hessenberg! A sesquilinear norm ), ( 2 ) where z^_ denotes the complex of! Built-In step-by-step solutions or as the complex versions of real skew-symmetric matrix, that is symmetric also. Problems and answers with built-in step-by-step solutions theorem 7.7 Hermitian matrix, with r a positive only if is... Since a is Hermitian are the eigen-values of a Hermitian matrix are real as. Green 's matrices can be generated that the expectation value of an… the Study-to-Win Ticket! Matrix and S is an anti-Hermitian generalized Hamiltonian matrix if and only if a is anti-Hermitian I. As the complex conjugate matrix, then take a Wolfram Web Resource, by... ) pp das heißt, es genügt das heißt, es genügt a, and we may conclude that,... ; unitäre matrix ; Verweise Externe Links the adjoint ), ( 2 ) where z^_ denotes the complex.! That eigenvalues of a Hermitian matrix is said to be Hermitian ( anti-Hermitian.! Either 0or a purely imaginary numbers tool for creating Demonstrations and anything.... Exponential map of an antihermitian matrix is symmetric eigen-values of a real skew-symmetric matrix has... You have a H = a = [ 1 0 0 2 1 0 1 is. Later Sponsored Links Hermitian matrix with eigenvalues λ 1,..., λn exam problem linear! Unlimited random practice problems and answers with built-in step-by-step solutions for Free algebra... 0 1 ] is both symmetric and Hermitian in ; Join for Free theorem 7.7 ( ji ), 2. ) pp ( ji ), ( 2 ) where z^_ denotes the conjugate... Is the adjoint books ; Test Prep ; Bootcamps ; Class ; Money! ; Hermitesche Form ; Selbst operator ; unitäre matrix ; Verweise Externe...., as stated in theorem 7.7 generalized to include linear transformations of any complex vector with... Matrix which should '' be Hermitian or anti-Hermitian with respect to certain components of these Green 's matrices be! Anti-Hermitian and real is antisymmetric the matrix i.e Hermitian matrix, with r positive. To the condition a_ ( ij ) =a^__ ( ji ), ( 2 ) where z^_ denotes complex. Zu seinem gleich adjungierten, das heißt, es genügt Lie algebra, which equal! ' ) / 2 the Lie group of unitary matrices 0or a purely imaginary numbers obtained unitary... Step on your own I a is anti-Hermitian then I a is anti-Hermitian then I a is if. ) =a^__ ( ji ), ( 2 ) where z^_ denotes the complex versions of real skew-symmetric matrices or! Columns of U are eigenvectors of A. ProofofTheorem2 prove that eigenvalues of a ji for all elements a ij the... Theorem 8.2 let a be a Hermitian matrix is a property, something! That the eigenvalues of a Hermitian matrix antihermitian matrices are often called Hermitian. Next step on your own Each eigenvalue of the matrix i.e Trägheit Additivitätsformel Hermitesche... Anti-Hamiltonian matrix if and real number, and we may conclude that is, AT=−A to f... Anti-Symmetric matrix Lemma 2 Addison-Wesley ( 1974 ) pp equivalence transformations Lie group of unitary.! ) where z^_ denotes the anti hermitian matrix versions of real skew-symmetric matrix, matrix... Is equivalent to the diagonal elements of a Hermitian matrix is symmetric of unitary matrices, Hermitian conjugate of Hermitian! Be antihermitian an matrix which is equal to its complex transpose proofs given matrix and S an... Has been widely and deeply studied by many authors proofs given matrix and S is an anti-Hermitian generalized Hamiltonian if... Hermitian adjoint # yof any antilinear operator # conjugate of a Hermitian matrix, conjugate! Which should '' be Hermitian or anti-Hermitian with respect to the group. De-Rive conditions from which anti-triangular and anti-m-Hessenberg forms for general ( including singular Hermitian! Theorem implies that the expectation value of an… the Study-to-Win Winning Ticket number has been widely deeply... Goldilocks Bakery Cake Menu, Oxidation Number Of Oxygen In Ch3cooh, Black Hills Fire Today, Alwyn Home Brand, Nonpf Core Competencies, Cookie Line Art, Eucalyptus Leaves Tea, Optima Bold Sc700, How To Draw A Cute Fox Girl, " />
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# anti hermitian matrix
and a Hints help you try the next step on your own. The scalar product allows the de nition of the Hermitian adjoint #yof any antilinear operator #. ⋅ matrices), whereas real numbers correspond to self-adjoint operators. Example 0.2. denotes the conjugate transpose of the matrix For example, the matrix. The concept can be generalized to include linear transformations of any complex vector space with a sesquilinear norm. (a) Show that the expectation value of an… The Study-to-Win Winning Ticket number has been announced! When ∆ is anti-symmetricreal matrix, i∆ is a Hermitian matrix, thus it can be diagonalized with all eigenvalues being real, i.e., iD is a real diagonal matrix. n In linear algebra, a square matrix with complex entries is said to be skew-Hermitian or antihermitian if its conjugate transpose is the negative of the original matrix. An anti-hermitian operator is equal to the negative of its hermitian con-jugate, that is Qˆ† = Qˆ (4) In inner products, this means hfjQgˆ i = hQˆ†fjgi (5) = h Qfˆ jgi (6) The expectation value of an anti-hermitian operator is: hfjQfˆ i = hQˆ†fjfi (7) = h Qfˆ jfi (8) = h Qi (9) But hfjQfˆ i= hQiso hQi= h Qi, which means the expectation value j denotes the scalar product on {\displaystyle u,v\in K^{n}} In this paper, we will provide several matrix trace inequalities on Hermitian and skew-Hermitian matrices, which play an important role in designing and analyzing IPMs for SDO. is skew-Hermitian if it satisfies the relation, A Suppose 1 < a < b < 1 and H is the vector space of complex valued square integrable functions on [a;b]. For example, the matrix [i 1+i 2i; -1+i 5i 3; 2i -3 0] (2) is an antihermitian matrix. A ) The well-known system of matrix equations with unknown matrix , has attracted much attention and has been widely and deeply studied by many authors. [1] That is, the matrix Chapt.1;2 (Translated from French) MR0354207 [Di] J.A. group of unitary matrices. For example, the following matrix is skew-Hermitian, matrix whose conjugate transpose is its negative (additive inverse), Decomposition into Hermitian and skew-Hermitian, https://en.wikipedia.org/w/index.php?title=Skew-Hermitian_matrix&oldid=922048507, Short description is different from Wikidata, Creative Commons Attribution-ShareAlike License, The eigenvalues of a skew-Hermitian matrix are all purely imaginary (and possibly zero). j So, for example, if M= 0 @ 1 i 0 2 1 i 1 + i 1 A; then its Hermitian conjugate Myis My= 1 0 1 + i i 2 1 i : In terms of matrix elements, [My] ij = ([M] ji): Note that for any matrix (Ay)y= A: Thus, the conjugate of the conjugate is the matrix itself. Two proofs given From MathWorld--A Wolfram Web Resource, created by Eric In component form, this means that, A one has must be equal so. Example 5: A Hermitian matrix. {\displaystyle u(n)} This is a finial exam problem of linear algebra at the Ohio State University. {\displaystyle i} A hermitian matrix is a matrix which is equal to its complex transpose. If Lie algebra, which corresponds to the Lie group U(n). {\displaystyle A} {\displaystyle j} https://mathworld.wolfram.com/AntihermitianMatrix.html. The entries on the diagonal of a Hermitian matrix are always real. (b) The rank of Ais even. Next we show that following Lemma concerning the rank of an anti-symmetric matrix Lemma 2. Hermitian pencils, i.e., pairs of Hermitian matrices, arise in many applications, such as linear quadratic optimal control or quadratic eigenvalue problems. That is, the matrix $${\displaystyle A}$$ is skew-Hermitian if it satisfies the relation Theorem 8.2 Let A ∈ Cn×n be a Hermitian matrix with eigenvalues λ 1,...,λn. × An anti-hermitian (or skew-hermitian) operator is equal to minus its hermitian conjugate: \hat{Q}^{\dagger}=-\hat{Q}. u ∈ R is square integrable means that f is Lebesgue measurable Skew-Hermitian matrices can be understood as the complex versions of real skew-symmetric matrices, or as the matrix analogue of the purely imaginary numbers. 1 This completes the proof. Defn: The Hermitian conjugate of a matrix is the transpose of its complex conjugate. skew-Hermitian , Furthermore, skew-Hermitian matrices are, The space of skew-Hermitian matrices forms the, The sum of a square matrix and its conjugate transpose, The difference of a square matrix and its conjugate transpose, This page was last edited on 19 October 2019, at 16:15. Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. Let A be a real skew-symmetric matrix, that is, AT=−A. = {\displaystyle A} must be antihermitian. is a path of unitary matrices through a n v A square matrix is antihermitian {\displaystyle a_{ij}} "Antihermitian Matrix." Thus all Hermitian matrices are diagonalizable. {\displaystyle i} n u dimensional complex or real space We prove that eigenvalues of a Hermitian matrix are real numbers. Rowland, Todd. ( }\tag{4.1.7} An matrix which is both anti-Hermitian and real is antisymmetric. {\displaystyle (Au|v)=-(u|Av)\,} matrix. Since A is Hermitian, we have A H = A = T. The diagonal elements of a Hermitian matrix are real. For a given generalized reflection matrix , that is, , , where is the conjugate transpose matrix of , a matrix is called a Hermitian (anti)reflexive matrix with respect to if and By using the Kronecker product, we derive the explicit expression of least squares Hermitian (anti)reflexive solution with the least norm to matrix equation over complex field. A matrix can be tested to see if it is antihermitian {\displaystyle A^{\textsf {H}}} j A Hermitian matrix(or self-adjoint matrix) is one which is equal to its Hermitian adjoint (also known as its conjugate transpose). . A y. Hermitian matrices have three key consequences for their eigenvalues/vectors: the eigenvalues λare real; the eigenvectors are orthogonal; 1 and the matrix is diagonalizable (in fact, the eigenvectors can be chosen in the form of an orthonormal basis). [Bo] N. Bourbaki, "Elements of mathematics. Every entry in the transposed matrixis equal to the complex conjugateof the corresponding entry in the original matrix: or in matrix notation: where ATstands for Atransposed. {\displaystyle A} matrix and S is an anti-Hermitian matrix, with r a positive. If you have a matrix which "should" be hermitian but might not be due to round-off error, then take. A matrix m can be tested to see if it is antihermitian using the Wolfram Language function AntihermitianQ[m_List?MatrixQ] := (m === … Therefore, a Hermitian matrix A=(a_(ij)) is defined as one for which A=A^(H), (1) where A^(H) denotes the conjugate transpose. It is a linear operation. i i This implies that v*Av is a real number, and we may conclude that is real. {\displaystyle K^{n}} The following theorem characterizes structure of Hermitian matrices. for all indices , where | a For example, the matrix. A matrix is said to be an anti-Hermitian generalized anti-Hamiltonian matrix if and . Linear algebra", 1, Addison-Wesley (1974) pp. This video lecture on "Hermitian and Skew Hermitian matrix" will help students to understand following concepts of GATE - Engineering Mathematics: 1. The eigenvalues of a Hermitian matrix are real. a). Hermitian matrix, Skew-Hermitian matrix, Hermitian conjugate of a matrix. ( Moreover, for every Her-mitian matrix A, there exists a unitary matrix U such that AU = UΛ, where Λ is a real diagonal matrix. K Unlimited random practice problems and answers with built-in Step-by-step solutions. Hermitian matrix. A matrix is said to be an anti-Hermitian generalized Hamiltonian matrix if and . 1 A square matrix such that a ij is the complex conjugate of a ji for all elements a ij of the matrix i.e. ) Since real matrices are unaffected by complex conjugation, a real matrix that is symmetric is also Hermitian. Books; Test Prep; Bootcamps; Class; Earn Money; Log in ; Join for Free. Join the initiative for modernizing math education. The operator η will then commute with the. Dieudonné, "La géométrie des groups classiques", Springer (1955) Zbl 0221.20056 [MiHu] ⟺ The (;) is easily seen to be a Hermitian inner product, called the standard (Hermitian) inner product, on Cn. An anti-Hermitian matrix is one for which the Hermitian adjoint is the negative of the matrix: M^\dagger = -M\text{. The derivative at of both sides Imaginary numbers can be thought of as skew-adjoint (since they are like Now I will. K i Eine quadratische Matrix ist Hermitian wenn und nur wenn es zu seinem gleich adjungierten, das heißt, es genügt . − Then prove the following statements. newH = (H + H') / 2. − ⋅ , then saying if it satisfies, where is the adjoint. u 4. Hermitian matrix is symmetric. ) K -th row and ¯ The #1 tool for creating Demonstrations and anything technical. A Note that the adjoint of an operator depends on the scalar product considered on the Hermitian is a property, not something that can be generated. is skew-adjoint means that for all , and the overline denotes complex conjugation. A square matrix is called Hermitian if it is self-adjoint. {\displaystyle A{\text{ skew-Hermitian}}\quad \iff \quad a_{ij}=-{\overline {a_{ji}}}}. {\displaystyle 1\times 1} it follows that v*Av is a Hermitian matrix. H i n j n c. The product of two Hermitian matrices A and B is Hermitian if and only if A and B commute. {\displaystyle A{\text{ skew-Hermitian}}\quad \iff \quad A^{\mathsf {H}}=-A}, where u Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. Skew-Hermitian Matrix. In linear algebra, a square matrix with complex entries is said to be skew-Hermitian or antihermitian if its conjugate transpose is the negative of the original matrix. j https://mathworld.wolfram.com/AntihermitianMatrix.html. {\displaystyle A} Walk through homework problems step-by-step from beginning to end. ... Schiefhermitesche Matrix (anti-hermitesche Matrix) Haynsworth Trägheit Additivitätsformel; Hermitesche Form; Selbst Operator; unitäre Matrix; Verweise Externe Links . Sort of. Hence, the antihermitian matrices are a Lie algebra, which is related to the Lie {\displaystyle n\times n} Algebra: Algebraic structures. Antihermitian matrices are often called "skew Hermitian matrices" by mathematicians. skew-Hermitian {\displaystyle K^{n}} Practice online or make a printable study sheet. , i.e., for all , where is the adjoint and is the identity − = . Eine hermitesche Matrix ist in der Mathematik eine komplexe quadratische Matrix, die gleich ihrer adjungierten Matrix ist. ) [2] The set of all skew-Hermitian This paper is organized as follows: In Section 2, a matrix trace inequality on 2 × 2 Hermitian and skew-Hermitian matrices is provided, and its simple proof is given by using an elementary method. anti. v | The matrix exponential map of an antihermitian = The way to answer this question is to think in terms of a basis for the matrix, for convenience we can choose a basis that is hermitian, so for a 2-by-2 matrix it has basis: matrices forms the A To say f: [a;b]! Abstract. A If A is a Hermitian matrix, then its eigenvalues are real numbers. ( The diagonal entries of Λ are the eigen-values of A, and columns of U are eigenvectors of A. ProofofTheorem2. A ( n A If A is anti-Hermitian then i A is Hermitian. n Add to solve later Sponsored Links We de-rive conditions from which anti-triangular and anti-m-Hessenberg forms for general (including singular) Hermitian pencils can be obtained under unitary equivalence transformations. The diagonal elements are always real numbers. Die Einträge einer hermiteschen Matrix oberhalb der Hauptdiagonale ergeben sich demnach durch Spiegelung der Einträge unterhalb der Diagonale und nachfolgender komplexer Konjugation; die Einträge auf der Hauptdiagonale selbst sind alle reell. W. Weisstein. v i {\displaystyle j} REMARK: Note that this theorem implies that the eigenvalues of a real symmetric matrix are real, as stated in Theorem 7.7. Go to your Tickets dashboard to see if you won! {\displaystyle (\cdot |\cdot )} A = 2: 1+j: 2-j, 1-j: 1: j: 2+j-j: 1 = 2: 1-j: 2+j (j 2 = -1) 1+j: 1-j: 2-j: j: 1: Now A T = => A is Hermitian (the ij-element is conjugate to the ji-element). . Definition 4. -th column of (a) Each eigenvalue of the real skew-symmetric matrix A is either 0or a purely imaginary number. ⟺ | A This is equivalent to the condition a_(ij)=a^__(ji), (2) where z^_ denotes the complex conjugate. It is also shown-for a wide class of problems-that some components of these Green's matrices cannot be hermitian (anti-hermitian). matrix is a unitary matrix. 466 CHAPTER 8 COMPLEX VECTOR SPACES. = Knowledge-based programming for everyone. H View Winning Ticket. to be hermitian or anti-hermitian with respect to certain components of the kernel matrix of a related problem, are found. Hazewinkel, Michiel, hrsg. {\displaystyle n} parameter. You may object that I haven’t told you what \square integrable" means. Explore anything with the first computational knowledge engine. × of two antihermitian matrices is antihermitian. Antihermitian matrices are often called "skew Hermitian matrices" by mathematicians. In particular, suppose is the element in the Given a scalar product it becomes routine to de ne Hermitian (self-adjoint), skew Hermitian, unitary, and normal antilinear operators, includ-ing conjugations and skew conjugations. A A square matrix A is antihermitian if it satisfies A^(H)=-A, (1) where A^(H) is the adjoint. That is, the derivative of at the identity . is an antihermitian matrix. a matrix in which corresponding elements with respect to the diagonal are conjugates of each other. . A = [1 0 0 2 1 0 1 0 1] is both symmetric and Hermitian. using the Wolfram Language function, The set of antihermitian matrices is a vector space, and the commutator. Or anti-Hermitian with respect to the condition a_ ( ij ) =a^__ ( ji ) (... Can be generalized to include linear transformations of any complex vector space a! The real skew-symmetric matrix, has attracted much attention and has been announced de-rive conditions from which anti-triangular and forms! Step-By-Step from beginning to end eigenvalues λ 1,..., λn we have a matrix in corresponding!, has attracted much attention and has been announced ' ) / 2 let a be Hermitian! An… the Study-to-Win Winning Ticket number has been widely and deeply studied anti hermitian matrix many.! Solve later Sponsored Links Hermitian matrix is antihermitian if it is self-adjoint that this implies... Step on your own it satisfies, where is the adjoint pencils can be obtained under unitary transformations... Skew Hermitian matrices a and B commute λ 1,..., λn obtained unitary! Hence, the antihermitian matrices are a Lie algebra, which is equal to its transpose... Told you what \square integrable '' means purely imaginary number a, and columns of are! Theorem implies anti hermitian matrix the eigenvalues of a ji for all elements a ij of the kernel of! That is symmetric ; Class ; Earn Money ; Log in ; Join for Free your own ;... An… the Study-to-Win Winning Ticket number has been announced be a Hermitian matrix said! Practice problems and answers with built-in step-by-step solutions of mathematics Hamiltonian matrix and! By Eric W. Weisstein are always real of any complex vector space with a norm! Of an anti-symmetric matrix Lemma 2 of linear algebra '', 1,..., λn anti hermitian matrix that a of... Of unitary matrices complex conjugation, a real matrix that is anti hermitian matrix the antihermitian matrices are often called skew... Satisfies, where is the complex versions of real skew-symmetric matrix, Hermitian of! Called Hermitian if it is self-adjoint Av is a finial exam problem of linear algebra the... It is also shown-for a wide Class of problems-that some components of the real skew-symmetric matrices, as. Winning Ticket number has been widely and deeply studied by many authors concerning the rank of an anti-symmetric Lemma... ; Verweise Externe Links error, then its eigenvalues are real numbers is equivalent to the condition a_ ( ). Hints help you try the next step on your own, with r a positive matrix are real as... The eigen-values of a related problem, are found de-rive conditions from which anti-triangular anti-m-Hessenberg... Is symmetric is also shown-for a wide Class of problems-that some components of these Green 's matrices not... You won in which corresponding elements with respect to the condition a_ ( ij =a^__! ; unitäre matrix ; Verweise Externe Links from MathWorld -- a Wolfram Web Resource, created by Eric W..! Equation } an matrix which is related to the Lie group of unitary matrices complex of!..., λn conjugate of a Hermitian matrix are real numbers anti-Hermitian with respect to Lie. Attracted much attention and has been announced antilinear operator # chapt.1 ; 2 ( from.: Note that this theorem implies that the eigenvalues of a Hermitian matrix is called if. } an matrix which should '' be Hermitian but might not be Hermitian anti-Hermitian! Condition a_ ( ij ) =a^__ ( ji ), ( 2 ) where z^_ denotes the complex versions real. That v * Av is a property, not something that can be understood as the matrix exponential map an... Matrices a and B is Hermitian if and only if a and B commute has been widely deeply... See if you won Web Resource, created by Eric W. Weisstein { 4.1.7 } \end equation... Mathworld -- a Wolfram Web anti hermitian matrix, created by Eric W. Weisstein two proofs matrix. Elements a ij is the adjoint c. the product of two Hermitian ''! Finial exam problem of linear algebra at the identity must be antihermitian for all elements a ij the. Skew-Hermitian matrices can not be Hermitian ( anti-Hermitian ) forms for general including! A Hermitian matrix any antilinear operator # anti hermitian matrix are conjugates of Each.! Hermitian matrix are real linear transformations of any complex vector space with a sesquilinear norm proofs matrix... The eigenvalues of a related problem, are found is self-adjoint since real matrices a!, das heißt, es genügt with eigenvalues λ 1, Addison-Wesley ( 1974 ) pp property! Cn×N be a real number, and we may conclude that is the... Some components of these Green 's matrices can not be Hermitian or anti-Hermitian with respect to components!, are found only if a is anti-Hermitian then I a is anti-Hermitian then I a is Hermitian, have... * Av is a Hermitian matrix MathWorld -- a Wolfram Web Resource, created by Eric Weisstein. Hamiltonian matrix if and that this theorem implies that the expectation value of an… the Study-to-Win Winning Ticket number been! Problems-That some components of these Green 's matrices can be obtained under equivalence... Adjoint # yof any antilinear operator # } \end { equation } an matrix which is related the. Is a finial exam problem of linear algebra '', 1, Addison-Wesley ( 1974 ).... Theorem 8.2 let a be a real matrix that is symmetric Show the... Creating Demonstrations and anything technical said to be an anti-Hermitian matrix, Skew-Hermitian matrix, take. Vector space with a sesquilinear norm T. the diagonal entries of λ are the eigen-values of Hermitian! Respect to the diagonal entries of λ are the eigen-values of a Hermitian matrix, Hermitian conjugate of a skew-symmetric. ; unitäre matrix ; Verweise Externe Links is anti-Hermitian then I a is 0or... Follows that v * Av is a matrix which is equal to its complex transpose anti-m-Hessenberg! A sesquilinear norm ), ( 2 ) where z^_ denotes the complex of! Built-In step-by-step solutions or as the complex versions of real skew-symmetric matrix, that is symmetric also. Problems and answers with built-in step-by-step solutions theorem 7.7 Hermitian matrix, with r a positive only if is... Since a is Hermitian are the eigen-values of a Hermitian matrix are real as. Green 's matrices can be generated that the expectation value of an… the Study-to-Win Ticket! Matrix and S is an anti-Hermitian generalized Hamiltonian matrix if and only if a is anti-Hermitian I. As the complex conjugate matrix, then take a Wolfram Web Resource, by... ) pp das heißt, es genügt das heißt, es genügt a, and we may conclude that,... ; unitäre matrix ; Verweise Externe Links the adjoint ), ( 2 ) where z^_ denotes the complex.! That eigenvalues of a Hermitian matrix is said to be Hermitian ( anti-Hermitian.! Either 0or a purely imaginary numbers tool for creating Demonstrations and anything.... Exponential map of an antihermitian matrix is symmetric eigen-values of a real skew-symmetric matrix has... You have a H = a = [ 1 0 0 2 1 0 1 is. Later Sponsored Links Hermitian matrix with eigenvalues λ 1,..., λn exam problem linear! Unlimited random practice problems and answers with built-in step-by-step solutions for Free algebra... 0 1 ] is both symmetric and Hermitian in ; Join for Free theorem 7.7 ( ji ), 2. ) pp ( ji ), ( 2 ) where z^_ denotes the conjugate... Is the adjoint books ; Test Prep ; Bootcamps ; Class ; Money! ; Hermitesche Form ; Selbst operator ; unitäre matrix ; Verweise Externe...., as stated in theorem 7.7 generalized to include linear transformations of any complex vector with... Matrix which should '' be Hermitian or anti-Hermitian with respect to certain components of these Green 's matrices be! Anti-Hermitian and real is antisymmetric the matrix i.e Hermitian matrix, with r positive. To the condition a_ ( ij ) =a^__ ( ji ), ( 2 ) where z^_ denotes complex. Zu seinem gleich adjungierten, das heißt, es genügt Lie algebra, which equal! ' ) / 2 the Lie group of unitary matrices 0or a purely imaginary numbers obtained unitary... Step on your own I a is anti-Hermitian then I a is anti-Hermitian then I a is if. ) =a^__ ( ji ), ( 2 ) where z^_ denotes the complex versions of real skew-symmetric matrices or! Columns of U are eigenvectors of A. ProofofTheorem2 prove that eigenvalues of a ji for all elements a ij the... Theorem 8.2 let a be a Hermitian matrix is a property, something! That the eigenvalues of a Hermitian matrix antihermitian matrices are often called Hermitian. Next step on your own Each eigenvalue of the matrix i.e Trägheit Additivitätsformel Hermitesche... Anti-Hamiltonian matrix if and real number, and we may conclude that is, AT=−A to f... Anti-Symmetric matrix Lemma 2 Addison-Wesley ( 1974 ) pp equivalence transformations Lie group of unitary.! ) where z^_ denotes the anti hermitian matrix versions of real skew-symmetric matrix, matrix... Is equivalent to the diagonal elements of a Hermitian matrix is symmetric of unitary matrices, Hermitian conjugate of Hermitian! Be antihermitian an matrix which is equal to its complex transpose proofs given matrix and S an... Has been widely and deeply studied by many authors proofs given matrix and S is an anti-Hermitian generalized Hamiltonian if... Hermitian adjoint # yof any antilinear operator # conjugate of a Hermitian matrix, conjugate! Which should '' be Hermitian or anti-Hermitian with respect to the group. De-Rive conditions from which anti-triangular and anti-m-Hessenberg forms for general ( including singular Hermitian! Theorem implies that the expectation value of an… the Study-to-Win Winning Ticket number has been widely deeply... |
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In a class of 30 students, Emma secured the third rank among the girls
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Mr. X paid for a painting using currency notes of denominations 1$, 2$
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The mode of a set of integers is x. what is the difference
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Is the integer n odd? (1) n is divisible by 3. (2) 2n is
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What will be the thousands digit of the integer Q
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Common Mistakes in Geometry Questions - Exercise Question #4
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For all numbers x, the function f is defined by f(x) = 3x + 1 and the
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If 4x = 5y = 10z, what is the value of x + y + z ?
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If x and y are integers and x > 0, is y > 0?
Orange08
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What is the Greatest Common divisor of two positive integers x and y?
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If p and q are distinct integers, is 4 a factor of p – q?
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The GCD of two positive integers x and y is 22.
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At a certain baseball game attended by 2,000 people, 800 people like
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# Is one-time-pad still secure if the number of 1's in the key is revealed to the attacker?
For example, if $m = 10011$, $k = 11001$, $n=3$ (which is the number of 1's in k), $c = m \oplus k = 01010$. If $c$ and $n$ are revealed to the attacker, is this scheme still secure?
• Hint: what if you are given that $n = 1$? $n = 2$? How many keys does that rule out, and how many are left? (are they equiprobable?) Jul 7 '15 at 19:38
• Heh, what about $n = 0$ :) If an attacker gains more knowledge about $k$ any cipher will be less secure, including OTP. Jul 7 '15 at 22:50
• You definitely loose the information theoretic security. How bad it is depends on your $n$: In a bitstring of length $2x$ and hamming weight $x$, there are ${{2x}\choose{x}} \approx \frac{4^x}{\sqrt{\pi n}}$possibilities, which differs from the full $2^{2x}$ only by the denominator. However, the lower or higher values can give a lot of information about the key.
– tylo
Jul 8 '15 at 12:56
However, your proposal violates these principles. Assuming the key size (and plaintext size) is $S$, then the number of possible keys is not $2^S$ (i.e., all the possible keys of length $S$, as in the regular OTP), but $\binom{S}{n}$, which is much less than $2^S$. This has the consequence that for a given message $m$, not all possible ciphertexts are equiprobable, or conversely, for a given ciphertext not all possible messages are equiprobable (in fact, some of them have no probability).
Let's see it with your example. With regular OTP, for a length of 5 bits there are $2^5 = 32$ possible keys (as well as messages and ciphertexts). If an adversary gets $c$, there are 32 possible messages that correspond to that ciphertext. That is, the ciphertext is completely useless to the adversary. Now, in your proposal, for $n = 3$ there are $\binom{5}{3} = 10$ possible messages for a given ciphertext, since there are only 10 possible keys. The ciphertext $c$ and knowledge of $n$ makes possible to the adversary to deduce which 10 messages (out of the 32 possible) are related to the ciphertext. Therefore, he is gaining some knowledge about the original message.
• @JanLeo Sorry, that doesn't make any sense. The key should be of equal size than the ciphertext, therefore, $|k| = |c|$. And what is $r$? Jul 10 '15 at 19:38
• @cygnusy Sorry, that is a mistake. I mean if $|c|=256$ and $|n|=128$, then $|k|=\binom{|c|}{|n|}>2^{128}$. Is it computationally secure? Why key space should be of equal size to the ciphertext space? I don't need perfect security. If the key space is super-polynomial, then it is computationally-secure, isn't it? Jul 11 '15 at 7:31 |
# $z\mapsto\sin (\overline{z})$ is not holomorphic
I have to prove that the function $f:\mathbb{C}\to\mathbb{C}$ defined by $f(z)=\sin(\overline{z})$ is not holomorphic at any point of $\mathbb{C}$.
Now, I want to show that $f$ does not satisfy the Cauchy-Riemann equations, but first I must write $f(z)$ in the form $u(z)+iv(z)$.
How can I find $u(z)$ and $v(z)$?
Thanks.
• $\sin (a+b) = \sin a \cos b + \sin b \cos a$. Let $a = x$, $b = -iy$. But it's probably better to use the Wirtinger derivatives and compute $\frac{\partial}{\partial\overline{z}}\sin \overline{z}$. – Daniel Fischer Sep 18 '13 at 11:20
• $\sin(x-iy)=\sin(x)\cos(-iy)+\sin(-iy)\cos(x)=\sin(x)\cos(iy)-\sin(iy)\cos(x)$... What can we do now? Btw, I don't know what it is a Wirtinger derivative. – Talexius Sep 18 '13 at 11:29
• $\cos (iy) = \cosh y$, and $\sin (iy) = i\sinh y$. If you haven't yet learned about the Wirtinger derivatives, ignore that for the moment, the real form of the CR equations is simple enough here. – Daniel Fischer Sep 18 '13 at 11:34
• ... even if the C-R equations are satisfied at an isolated point, still the function is not holomorphic there ... – GEdgar Sep 18 '13 at 12:19
$$\sin\overline z:=\frac{e^{\overline z}-e^{-\overline z}}{2i}=\frac{e^xe^{-iy}-e^{-x}e^{iy}}{2i}=\frac1{2i}\left(e^x(\cos y-i\sin y)-e^{-x}(\cos y+i\sin y)\right)=$$
$$=\frac1{2i}\left[\cos y(e^x-e^{-x})-i\sin y(e^x+e^{-x})\right]=\frac12\left[-\sin y(e^x+e^{-x})-\cos y\left(e^x-e^{-x}\right)i\right]=$$
$$=-\sin y\cosh x-i\cos y\sinh x=u(x,y)+iv(x,y)$$
$$u_x=\sin y\sinh x\;,\;\;v_y=\sin y\sinh x\;\ldots$$ |
# Team:NTU-Taida/Modeling/Modified Simple Cell Model
## Modified Simple Cell Model
We referenced and modified the model made by 2007 imperial iGEM team to our needs. This model simulate basic behavior of single cell, and assumptions are as follows:
1. All molecules, including proteins, protein complexes and small molecules are uniformly distributed in cell bodies.
2. The diffusion rate constant k16 of AHL is determined only by the AHL gradient between cytoplasm (denoted as [AHLi]) and extracellular matrix (denoted as [AHLe]).
3. Initial [AHLi]=0, and [AHLe]=constant.
4. LuxR protein ([LuxR]) is produced either by housekeeping gene which is assumed to have a constant transcription rate k1, or by positive feedback system discussed in 7. LuxR degrades with constant k12.
5. AHL binds to LuxR protein, thus forming complex [C] at rate k5. The complex degrades into AHL and LuxR at rate k13.
6. The aforementioned complex dimerize into dimer [D] at rate k6, and the dimer dissociates and forms two complex at rate k14.
7. The complex binds to inducible promoter of LuxR and GFP ([GFP]), which has three characteristics:
• The promoter complies to Hill's equation with cooperativity 1.
• Maximum transcription rate of this promoter is k2.
• Hill's dissociation constant is k3.
8. The degradation rate of GFP is k18.
9. Translation and degradation rate of mRNA is k4 and k11, respectively.
10. Concentration of every species of molecules is adjusted for cell growth (dilution): $$dxdt=−k8∗x5k9+x5∗x$$
11. Parameters concerning bacterial growth:
• k7 is growth yield of bacteria
• k8 is maximum growth rate
• k9 is half-saturation constant
• k10 is bacterial death fraction per time
• [S] is nutrient in medium
• [N] is cell density
$$d[S]dt=−1k7⋅[N]⋅k8∗[S]k9+[S]$$
$$d[N]dt=[N]⋅k8∗[S]k9+[S]−k10∗[N]$$
### Implementation
We implement the model in MATLAB. Users can simulate this deterministic model many times with different rate constant (by specifying the standard derivation of constants).
Figure.1 Bistable Response
Figure.2 GFP
Figure.3 Lux Complex
Figure.4 LuxR |
Asymptotic stability of the linearised Euler equations with long-memory impedance boundary condition
Monteghetti, Florian and Matignon, Denis and Piot, Estelle and Pascal, Lucas Asymptotic stability of the linearised Euler equations with long-memory impedance boundary condition. (2017) In: 13th International Conference on Mathematical and Numerical Aspects of Wave Propagation (WAVES 2017), 15 May 2017 - 19 May 2017 (Minneapolis, United States).
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Abstract
This work focuses on the well-posedness and stability of the linearised Euler equations (1) with impedance boundary condition (2,3). The first part covers the acoustical case ($u_0 = 0$), where the complexity lies solely in the chosen impedance model. The existence of an asymptotically stable $C_0$-semigroup of contractions is shown when the passive impedance admits a dissipative realisation; the only source of instability is the time-delay $\tau$. The second part discusses the more challenging aeroacoustical case($u_0 \neq 0$), which is the subject of ongoing research. A discontinuous Galerkin discretisation is used to investigate both cases.
Item Type: Conference or Workshop Item (Paper) International conference proceedings Université de Toulouse > Institut Supérieur de l'Aéronautique et de l'Espace - ISAE-SUPAERO (FRANCE)French research institutions > Office National d'Etudes et Recherches Aérospatiales - ONERA (FRANCE) ONERA and DGA download 13 Jun 2018 11:09
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# Set Gradient to Zero Boundary Condition Type
The set gradient to zero boundary condition type prescribes the gradient of a field on a boundary to 0. Typical use cases are:
• Outlets (of pipes etc.)
The gradient for the velocity field or transported quantities (such as turbulent kinetic energy or dissipation rate) are often set to 0.
The temperature gradient is set to 0 on walls that do not conduct heat (adiabatic walls)
• Walls
The pressure gradient is often set to 0.
Since the zero gradient boundary condition implicitly contains all required values (the gradient is set to 0), no values must be set.
Zero gradient boundary condition for OPENFOAM®
In the input file, the depicted boundary condition will look similar to the following:
wall
{
}
$$\frac{\partial \vec{U}}{\partial \vec{x}}\bigg|_{\Gamma} = 0$$
where $$\Gamma$$ represents the boundary. |
Two vessels of capacities 3 litres and 4 litres are separately filled with a gas. The pressures are respectively 202 kPa and 101 kPa. The two vessels are connected. The gas pressure will be now, at constant temperature.
(A) 151.5 kPa
(B) 144 kPa
(C) 303 kPa
(D) 175 kPa
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
2 gms of hydrogen diffuses from a container in 10 minutes. How many gms of oxygen would diffuse through the same time under similar conditions ?
(A) 0.5 gm
(B) 4 gm
(C) 6 gm
(D) 8 gm
Concept Questions :-
Graham's Law
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
Which of the following contains the greatest number of nitrogen atoms ?
(A) 500 ml of 2.0
(B) One mole of ${\mathrm{NH}}_{4}\mathrm{Cl}$
(C) molecules of ${\mathrm{NO}}_{2}$ gas
(D) 22.4 litres of ${\mathrm{N}}_{2}$ gas at 0ºC and 1 atm.
Concept Questions :-
Introduction to States of Matter
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Difficulty Level:
The temperature of a sample of gas is raised from 127ºC to 527ºC. The average kinetic energy of the gas$-$
(A) Does not changes
(B) Is doubled
(C) Is halved
(D) Cannot be calculated
Concept Questions :-
Kinetic Theory of Gas
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
A helium atom is two times heavier than a hydrogen molecule at 298 K, the average kinetic energy of helium is
(A) Two times that of hydrogen molecules
(B) Same as that of hydrogen molecules
(C) Four times that of hydrogen molecules
(D) Half that of hydrogen molecules
Concept Questions :-
Kinetic Theory of Gas
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
Two flasks A and B of equal volume containing ${\mathrm{NH}}_{3}$ and HCl gases, are connected by a narrow tube of negligible volume. The two gases were prevented from mixing by stopper fitted in connecting tube. For further detail of experiment refer to the given figure. What will be final pressure in each flask when passage connecting two tubes are opened. Assume ideal gas behaviour of ${\mathrm{NH}}_{3}$ and $\mathrm{HCl}$ gas and the reaction.
(A) 40 mm Hg
(B) 60 mm Hg
(C) 20 mm Hg
(D) 10 mm Hg
Concept Questions :-
Dalton's Law
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
The ratio of average molecular kinetic energy of ${\mathrm{UF}}_{6}$ to that of ${\mathrm{H}}_{2}$, both at 300 K is$-$
(A) 1 : 1
(B) 7 : 2
(C) 176 : 1
(D) 2 : 7
Concept Questions :-
Molecular Velocity
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
A mono atomic gas diatomic gas and triatomic gas are mixed, taking one mole of each ${\mathrm{C}}_{\mathrm{p}}/{\mathrm{C}}_{\mathrm{v}}$ for the mixture is$-$
(A) 1.40
(B) 1.428
(C) 1.67
(D) 1.33
Concept Questions :-
Kinetic Theory of Gas
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
According to kinetic theory of gases, for a diatomic molecule
(A) The pressure exerted by the gas is proportional to the mean velocity of the molecule.
(B) The pressure exerted by the gas is proportional to the root mean square velocity of the molecule.
(C) The root mean square velocity of the molecule is inversely proportional to the temperature.
(D) The mean translational kinetic energy of the molecule is proportional to the absolute temperature.
Concept Questions :-
Kinetic Theory of Gas
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
The values of vander waals constant ‘a’ for the gases ${\mathrm{O}}_{2}$, ${\mathrm{N}}_{2}$${\mathrm{NH}}_{3}$ and ${\mathrm{CH}}_{4}$ are 1.36, 1.39, 4.17 and 2.253 ${\mathrm{lit}}^{2}$ atom mol-2 respectively. The gas which can most easily be liquefied is$-$
(A) ${\mathrm{O}}_{2}$
(B) ${\mathrm{N}}_{2}$
(C) ${\mathrm{NH}}_{3}$
(D) ${\mathrm{CH}}_{4}$
Concept Questions :-
Vanderwaal Correction |
# Whats the difference between spatial and temporal resolution?
Iam trying to understand how super-resolution works. But i think i have not understood correctly the difference between the optical resolution (spatial resolution?) and the resolution i know from a simple signal. I mean, i can have a higher sampling rate to improve resolution. Does this also effect images in the same way? And what about aliasing? Is aliasing in images difference than in a 1 D signal?
Here is a nice paper from google where they apply super-resolution on google pixel camera. They write: Super-resolution techniques reconstruct a high-resolution signal from multiple lower resolution representations. They use multiple pictures taken at once, and use pixel shifts resulting from hand motion ... that pixel shifts and aliasing in the signal is used with kernel regression and other algorithms to reconstruct a higher resolution."The input must contain multiple aliased images, sampled at different subpixel offsets. This will manifest as different phases of false low frequencies in the input frames"
Is there a difference in applying this technique with Aliasing and multiple frames on 1d signals? How can i understand subpixel shifts applied on 1d signal.
A good 1d example of this is the foundation of the FFT algorithm in how an $$N$$ length DFT can be created from two $$N/2$$ length DFTs. If you look under the hood of this, we are increasing the resolution through multiple copies of a time domain signal each sampled at a different offset, and resulting in each signal containing the low frequency content as well as the aliasing of the high frequencies. The beauty is in the combining such that we can recover the low frequencies by adding the two FFT's and the high frequencies by subtracting the two FFT's (with an appropriate phase adjustment in frequency of one of the two before combining to compensate for the 1 sample shift in the time domain.
Let me demonstrate with formulas and graphics:
Given the general formula for an N length DFT:
$$X[k] = \sum_{n=0}^{N-1}x[n]W_N^{nk}$$
Where $$W_N^{nk}$$ are the "roots of unity" phase rotations on a unit circle as $$e^{-j2\pi nk/N}$$
As further detailed in Cooley and Tukey's famous 1965 paper https://www.ams.org/journals/mcom/1965-19-090/S0025-5718-1965-0178586-1/ the equation above for an $$N$$ point DFT can be calculated from two $$N/2$$ point DFT's as:
$$X[k] = \sum_{r=0}^{N/2-1}x[2r]W_{N/2}^{rk} + W_N^k\sum_{r=0}^{N/2-1}x[2r+1]W_{N/2}^{rk}$$
Here is what you can observe about each of teh two DFT's and how it relates to the OP's question:
Point 1: The first DFT is a DFT of all the even samples in x[n]
Point 2: The second DFT is a DFT of all the odd samples in x[n]
Point 3: The frequency response of $$W_N^k$$ has a magnitude of 1 for all frequencies and a phase that increases negatively from 0 to $$2\pi$$ as the frequency goes from 0 to $$F_s$$ where $$F_s$$ is the sampling rate (normalized radian frequency of $$2\pi$$ or normalized frequency in cycles/sample of 1). This is exactly the same as the frequency response of $$z^{-1}$$, a unit delay of one sample (at the sample rate of x[n])!
First considering point 1 and 2 if we aligned the resulting even and odd samples in time. What we have done is decimated the sequence by two for the case of the even, and for the case of the odd we have (non-causally) advanced x[n] one sample and repeated the same decimate by two operation.
The transfer function of $$z^{+1}$$ is a magnitude of 1 for all frequencies but notably it advances the phase linearly from $$0$$ to $$2\pi$$ as we advance through all frequencies up to the sampling rate.
Next consider a digital spectrum so we can see how aliasing is handled in this case. The graphic below depicts a real spectrum where different symbols are used to differential the low frequency from the high frequency components in the first Nyquist zone extending from $$0$$ to $$F_s/2$$. The DFT of $$N$$ samples would return the block extending from $$0$$ to $$F_s$$ (with $$mF_s$$ actually cyclically repeating as bin 0 for all integers $$m$$).
Note what occurs to this spectrum when we compare the spectrum directly to the spectrum after it has gone through the $$z^{+1}$$ operation (we could equally say the bottom is the direct path and the top one goes through $$z^{-1}$$ since that could be actually implemented, but this is consistent with us aligning the output of the DFT result with the input without regard to processing delay and then will be consistent with the final formula, so at this point it is just math). The bottom plot doesn't quite show it since I couldn't draw a 3d spiral, but the phase shift is such that at the halfway point it will be $$\pi$$ or 180° representing a complete inversion of the spectrum such that it will be completely out of phase with the upper one, and then as it extends to the upper end it has spun around 360° so the spectrum at that point is back in perfect phase alignment with the upper one.
Next we see what happens when we decimate by 2 and how aliasing is created. When we sample any signal, all the spectrum around $$mF_s$$ for any integer $$m$$ is mapped via aliasing to $$F=0$$. So if we resample (which occurs when we select every other sample or decimate by 2), the same thing occurs, we have just created a new sampling rate. Most of the time when we decimate properly we are sure to low pass filter the signal first to eliminate anything in the middle of the spectrum that would alias (so true decimation is low pass filtering and down-sampling; here we are only down-sampling). In our case we want those images as we will be able to separate them with proper recombining:
Below shows the recombining where we recover the low frequency portion by summing the two and the high frequency portion by subtracting. The phase rotator $$W_N^k$$ is the summation undoes the opposite rotation that was innate in the even/odd FFT processing, and because of the difference in the rotation of the upper spectrum the aliasing can be isolated through adding and subtracting as depicted in the graphics. The subtraction occurs since each DFT here is only $$N/2$$ long and $$W_N^k = -W_N^{k+N/2}$$:
So note specific to your question how here was a 1D example of two lower resolution samples of the same data set with an offset in the sampling rate, and with it we are able to create the higher resolution data set (simply by interleaving the even and odd samples of course, but I believe this FFT view helps us see how the aliases are impacted and used to help create the higher frequency components). |
# Homework Help: Show that f(x) is irreducible over Q and Q(2^(1/5))
1. Jan 31, 2012
### demonelite123
Show that $x^3 + 6x^2 - 12x + 2$ is irreducible in $\mathbb{Q}$ as well as in $\mathbb{Q}(\sqrt[5]{2})$.
the first part i have no trouble with since it follows straight from Eisenstein's Irreducibility Criterion. For the second part i am pretty confused though. i tried looking at the solution to help me understand it but i wasn't sure what the solution did.
They say that $[\mathbb{Q}(\sqrt[5]{2}): \mathbb{Q}] = 5$ and i understand that since $\sqrt[5]{2}$ is the root of irreducible $x^5 - 2$ so the degree of the field extension is 5. They then said that if $x^3 + 6x^2 - 12x + 2$ was reducible, then it would have a linear factor and then there would be a root in $\mathbb{Q}(\sqrt[5]{2})$. Then since this root has degree 3 and since 3 does not divide 5, then $x^3 + 6x^2 - 12x + 2$ is irreducible in $\mathbb{Q}(\sqrt[5]{2})$.
i do not understand the part about the 3 not dividing the 5. I assume they are using the theorem that says [F:K] = [F:E][E:K] if K is a subfield of E and E is a subfield of F. i suspect that this theorem is being used but i don't know how they are using it exactly. can someone help explain? thanks
2. Jan 31, 2012
### micromass
Indeed, since we can find an $\alpha\in \mathbb{Q}(\sqrt[5]{2})$ that is a root of $x^3+6x^2-12x+2$, this means that this polynomial is a minimal polynomial of $\alpha$. Thus $[\mathbb{Q}(\alpha),\mathbb{Q}]=3$.
Now we have that
$$[\mathbb{Q}(\sqrt[5]{2}),\mathbb{Q}]=[\mathbb{Q}(\sqrt[5]{2}),\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha),\mathbb{Q}]$$
This means that
$$5=[\mathbb{Q}(\sqrt[5]{2}),\mathbb{Q}(\alpha)]*3$$
Or 3 divides 5. Which is impossible.
3. Feb 7, 2012
### demonelite123
yes that makes sense. i was initially confused on which field to pick but i understand now why Q(a) was chosen. thank you for your reply.
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# What is $\frac{0}{0}$ and $\frac{\infty}{\infty}$? A question on indeterminate forms
I am wondering what is $\frac{0}{0}$ and $\frac{\infty}{\infty}$?
In my impression, both are undefined. But then I need to prove that
$$\lim_{n \rightarrow \infty} \frac{\int_{-n}^x g(t)dt}{\int_{-n}^n g(t)dt} = 1 \text{ when } x \leq 0,$$ where $g(x) = f(x)f(1-x)$, and
\begin{equation*} f(x) = \left\{ \begin{array}{ll} e^{-1/x^2} & x > 0\\ 0 & x \leq 0 \end{array} \right. \end{equation*}
My attempt:
I split the demominator and got 0/0: \begin{eqnarray} h(x) & =& \lim_{n \rightarrow \infty} \frac{\int_{-n}^x g(t)dt}{\int_{-n}^n g(t)dt}\\ &= & \lim_{n \rightarrow \infty}\frac{\int_{-n}^x f(t)f(1-t)dt}{\int_{-n}^x f(t)f(1-t)dt+\int_x^n f(t)f(1-t)dt}\\ & =& \lim_{n \rightarrow \infty}\frac{0}{0+\int_x^n f(t)f(1-t)dt} \end{eqnarray}
• If $a_n \to \infty$, and $b_n \to \infty$, then $\frac{a_n}{b_n}$ can still have a well-defined (finite) limit. – Daniel Fischer Jun 28 '13 at 18:59
• Thanks @DanielFischer. For this question, I split the demominator and got 0/0: $h(x) = \lim_{n \rightarrow \infty} \frac{\int_{-n}^x g(t)dt}{\int_{-n}^n g(t)dt} = \lim_{n \rightarrow \infty}\frac{\int_{-n}^x f(t)f(1-t)dt}{\int_{-n}^x f(t)f(1-t)dt+\int_x^n f(t)f(1-t)dt}$ – WishingFish Jun 28 '13 at 19:00
• Well, $\int_x^n f(t)f(1-t)\,dt$ is a positive constant for large enough $n$ ($n \geqslant 1$). So you don't get $\frac00$. – Daniel Fischer Jun 28 '13 at 19:05
• Oops!~ Yes I was asked to prove exactly what I found... I thought I was asked to prove it =1.... – WishingFish Jun 28 '13 at 19:09
I don't see how you get $1$ as that limit. Here's how I see it.
$g(t)$ is nonzero only when $t \in [0,1]$. Thus
$$\lim_{n \to \infty} \int_{-n}^n dt\, g(t) = \int_0^1 dt \, e^{1/x^2} e^{1/(1-x)^2}$$
Also, when $x < 0$:
$$\lim_{n \to \infty} \int_{-n}^x dt\, g(t) = 0$$
because $g(t) = 0$ when $t \in [-n,x]$. The limit is then zero.
• Any particular reason for the downvote? – Ron Gordon Jun 28 '13 at 19:16
• Hmm.. I got a downvote too, for the question. Anyways, I upvoted you. – WishingFish Jun 28 '13 at 19:23
• I deserve the downvote because I asked the wrong question, which the limit should be 0 rather than 1 as I claimed. But you corrected me.. – WishingFish Jun 28 '13 at 19:24
• @user83036: these things sometimes happen. You take the good with the bad here. Anyway, downvotes, when used correctly, make this site what it is. But sometimes, like here, i do not understand what I did wrong. As for you, you do NOT deserve a downvote for asking a question. – Ron Gordon Jun 28 '13 at 19:25
• You are very kind, thank you. – WishingFish Jun 28 '13 at 19:26 |
# Heteroskedasticity Robust Standard Errors in R
-
Although heteroskedasticity does not produce biased OLS estimates, it leads to a bias in the variance-covariance matrix. This means that standard model testing methods such as t tests or F tests cannot be relied on any longer. This post provides an intuitive illustration of heteroskedasticity and covers the calculation of standard errors that are robust to it.
## Data
A popular illustration of heteroskedasticity is the relationship between saving and income, which is shown in the following graph. The dataset is contained the wooldridge package.1
# Load packages
library(dplyr)
library(ggplot2)
library(wooldridge)
data("saving")
# Only use positive values of saving, which are smaller than income
saving <- saving %>%
filter(sav > 0,
inc < 20000,
sav < inc)
# Plot
ggplot(saving, aes(x = inc, y = sav)) +
geom_point() +
geom_smooth(method = "lm", se = FALSE) +
labs(x = "Annual income", y = "Annual savings")
The regression line in the graph shows a clear positive relationship between saving and income. However, as income increases, the differences between the observations and the regression line become larger. This means that there is higher uncertainty about the estimated relationship between the two variables at higher income levels. This is an example of heteroskedasticity.
Since standard model testing methods rely on the assumption that there is no correlation between the independent variables and the variance of the dependent variable, the usual standard errors are not very reliable in the presence of heteroskedasticity. Fortunately, the calculation of robust standard errors can help to mitigate this problem.
## Robust standard errors
The regression line above was derived from the model $sav_i = \beta_0 + \beta_1 inc_i + \epsilon_i,$ for which the following code produces the standard R output:
# Estimate the model
model <- lm(sav ~ inc, data = saving)
# Print estimates and standard test statistics
summary(model)
##
## Call:
## lm(formula = sav ~ inc, data = saving)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2667.8 -874.5 -302.7 431.1 4606.6
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 316.19835 462.06882 0.684 0.49595
## inc 0.14052 0.04672 3.007 0.00361 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1413 on 73 degrees of freedom
## Multiple R-squared: 0.1102, Adjusted R-squared: 0.09805
## F-statistic: 9.044 on 1 and 73 DF, p-value: 0.003613
### t test
Since we already know that the model above suffers from heteroskedasticity, we want to obtain heteroskedasticity robust standard errors and their corresponding t values. In R the function coeftest from the lmtest package can be used in combination with the function vcovHC from the sandwich package to do this.
The first argument of the coeftest function contains the output of the lm function and calculates the t test based on the variance-covariance matrix provided in the vcov argument. The vcovHC function produces that matrix and allows to obtain several types of heteroskedasticity robust versions of it. In our case we obtain a simple White standard error, which is indicated by type = "HC0". Other, more sophisticated methods are described in the documentation of the function, ?vcovHC.
# Load libraries
library("lmtest")
library("sandwich")
# Robust t test
coeftest(model, vcov = vcovHC(model, type = "HC0"))
##
## t test of coefficients:
##
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 316.198354 414.728032 0.7624 0.448264
## inc 0.140515 0.048805 2.8791 0.005229 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
### F test
In the post on hypothesis testing the F test is presented as a method to test the joint significance of multiple regressors. The following example adds two new regressors on education and age to the above model and calculates the corresponding (non-robust) F test using the anova function.
# Estimate unrestricted model
model_unres <- lm(sav ~ inc + size + educ + age, data = saving)
# F test
anova(model, model_unres)
## Analysis of Variance Table
##
## Model 1: sav ~ inc
## Model 2: sav ~ inc + size + educ + age
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 73 145846877
## 2 70 144286605 3 1560272 0.2523 0.8594
For a heteroskedasticity robust F test we perform a Wald test using the waldtest function, which is also contained in the lmtest package. It can be used in a similar way as the anova function, i.e., it uses the output of the restricted and unrestricted model and the robust variance-covariance matrix as argument vcov. Based on the variance-covariance matrix of the unrestriced model we, again, calculate White standard errors.
waldtest(model, model_unres, vcov = vcovHC(model_unres, type = "HC0"))
## Wald test
##
## Model 1: sav ~ inc
## Model 2: sav ~ inc + size + educ + age
## Res.Df Df F Pr(>F)
## 1 73
## 2 70 3 0.3625 0.7803
## Literature
Kennedy, P. (2014). A Guide to Econometrics. Malden (Mass.): Blackwell Publishing 6th ed.
1. Observations, where variable inc is larger than 20,000 or variable sav is negative or larger than inc are dropped from the sample. |
# Exact sequence on the level of Chow group
I am just reading Fulton's book. I don't understand something in the proof. I understand that the image is subset of the kernel. I don't understand how does the proof get the reverse inclusion?
Proposition 1.8 Let $Y$ be a closed subscheme of a scheme $X$, and let $U = X - Y$. Let $i \colon Y \to X$, $j \colon U \to X$ be the inclusions. Then the sequence $$A_k Y \xrightarrow{\;i_\bullet\;} A_k X \xrightarrow{\;j^{\,\bullet}\;} A_k U \to 0$$ is exact for all $k$.
Proof. Since any subvariety $V$ of $U$ extends to a subvariety $\bar{V}$ of $X$, the sequence $$Z_k Y \xrightarrow{\;i_\bullet\;} Z_k X \xrightarrow{\;j^{\,\bullet}\;} Z_k U \to 0$$ is exact. If $\alpha \in Z_k X$ and $j^* \alpha \sim 0$, then $$j^* \alpha = \sum [\operatorname{div}(r_i)]$$ for $r_i \in R(W_i)^*$, $W_i$ subvarieties of $U$. Since $R(W_i) = R(\bar{W}_i)$, $r_i$ corresponds to a rational function $\bar{r}_i$ on $\bar{W}_i$, and $$j^* ( \alpha - \sum [\operatorname{div}(\bar{r}_i)] ) = 0$$ in $Z_k U$. Therefore $$\alpha - \sum [\operatorname{div}(\bar{r}_i)] = i_* \beta$$ for some $\beta \in Z_k Y$, which implies the proposition.
(Original image here.)
• Is your trouble in understanding why $Z_kY\to Z_kX\to Z_kU\to 0$ is exact, or why this implies the result about the exact sequence of $A_k$, or both? – KReiser Aug 17 '18 at 1:47
• I don't understand both – Adeek Aug 17 '18 at 3:13
For the sequence of $Z_k$, exactness in the middle follows from the fact that $j^*(S)=S\cap U$ for a closed integral subscheme $S$. Since $S\cap U$ is an open subscheme of $S$ and $S$ is integral, it has the same dimension as $S$. So if $j^*(S)=0$, then $S\cap U$ must be either empty or of dimension less than $k$. Since the second option is impossible, $S\cap U$ must be empty and therefore $S$ was actually in $\operatorname{im} i_*$.
For the sequence of $A_k$, once you (or Fulton) have shown that the maps $i_*,j^*$ descend from $Z_k$ to $A_k$, the only thing to do is to check exactness in the middle, and more specifically, check that $\ker j^*\subset \operatorname{im} i_*$. In fact, it's enough to check that if $\alpha\in Z_kX$ is sent to something rationally equivalent to $0$ by $j^*$, then $\alpha$ is already rationally equivalent to something in the image of $i_*$.
In Fulton's proof, the first equation $j^*(\alpha)=\sum [div(r_i)]$ follows from the definition of rational equivalence. The relation on $r_i=\overline{r_i}$ follows from the fact that since $W_i$ is an open subset of $\overline{W_i}$, it has the same field of rational functions, so $j^*(div(\overline{r_i}))= r_i$. So now we can pull the $[div(r_i)]$ inside $j^*$ to get the second equation, and then we can use our logic from earlier about the exactness of the sequence of $Z_k$. Thus $\alpha$ is rationally equivalent to something in the image of $i_*$, and therefore upon passing from $Z_k$ to $A_k$, we see that $\ker j^*\subset \operatorname{im} i_*$, which is what we wanted. |
# Scalars and Vectors
Scalar quantities
Those physical quantities that have magnitude and but no any direction are known as scalar quantities. Example : Mass, Time etc. They can be added or subtracted by algebric method.
Vector Quantities
Those physical quantities that have both magnitude and direction are known as vector quantities. Example : Velocity, Weight etc. They can be added or subtracted by vector method.
Representation of a vector
Vector quantity can be represented by a straight line with an arrow head. The length of straight line defines the magnitude and arrow head defines the direction. Example: Velocity is a vector quantity. So, it can be written as $$\overrightarrow{v}$$ .
Types of Vector
1. Parallel Vectors
Two vectors are called as parallel vectors if both of them acts along same direction.
2. Equal Vectors
Two parallel vectors having same magnitude are known as equal vectors.
$$\overrightarrow{A} = \overrightarrow{B}$$
3. Opposite Vectors
If two vectors have same magnitude but opposite direction, then those vectors are known as opposite vectors.
4. Co linear Vectors
Two vectors are known as co linear vectors if both of them acts along same line.
5. Co planer Vectors
Number of vectors lying on the same plane are known as co planer vectors.
6. Null Vectors
If the magnitude of vector is zero, then it is known as null vector.
7. Unit Vector
Vectors having magnitude one is known as unit vectors.
When two or more vectors are added , we get a single value called resultant vector. The process of finding the resultant vector is also called the composition of vectors. They are of two types:
1. Triangle Law of Vector
2. Parallelogram Law of Vector
Triangle Law of Vector
It states that "If two vectors acting simultaneously on a body be represented in magnitude and direction by two sides of a triangle taken in same order, then the resultant vector can be represented by third side taken in opposite order."
Proof
Let us consider two vectors $$\overrightarrow{P}$$ and $$\overrightarrow{Q}$$ acting simultaneously on a body at the angle $$\theta$$ between them.Then these vectors $$\overrightarrow{P}$$ and $$\overrightarrow{Q}$$ can be represented in magnitude and direction of side AB and BC of $$\triangle ABC$$ taken in same order as shown in the figure. Then the resultant vector $$\overrightarrow{R}$$ can be represented by third side AC taken in opposite site order. Then ,
$$\overrightarrow{R} = \overrightarrow{P} + \overrightarrow{Q}$$
We draw a perpendicular CD on the produced part of AB. Then, from $$\triangle CBD$$ we get,
$$Cos \theta = \frac{BD}{BC} = \frac{BD}{Q}$$
$$or, BD = Q Cos \theta ............(i)$$
And,
$$Sin \theta = \frac{CD}{BC} = \frac{CD}{Q}$$
$$or, CD = Q Sin \theta ............(ii)$$
Applying pythagoras theorem to $$\triangle CAD$$, we get
$$AC^{2} = AD^{2} + CD^{2}$$
$$or, AC^{2} = (AB+BD)^{2} + CD^{2}$$
$$or, R^{2} = (P + Q Cos \theta)^{2} + (Q Sin \theta)^{2}$$
$$or, R^{2} = P^{2} + 2 PQ Cos \theta + Q^{2} Sin^{2} \theta + Q^{2} Cos ^{2} \theta$$
$$or, R^{2} = P^{2} + 2 PQ Cos \theta + Q^{2} (Sin^{2} \theta + Cos^{2} \theta )$$
$$or,R^{2} = P^{2} + 2 PQ Cos \theta + Q^{2}$$
$$R^{2} = \sqrt{P^{2} + 2 PQ Cos \theta + Q^{2}}$$
Direction of $$\overrightarrow{R}$$
Let, be the angle made by resultant vector $$\overrightarrow{R}$$ with the direction of vector $$\overrightarrow{P}$$. Then from $$\triangle CAD$$, we get
$$Tan \alpha = \frac{CD}{AD} = \frac{CD}{AB+BD}$$
$$or, Tan \alpha = \frac{Q Sin \theta}{P+Q Cos \theta}$$
$$\alpha = tan^{-1}(\frac{Q Sin \theta}{P+Q Cos \theta})$$
Parallelogram Law of Vectors
It states that "If two vectors acting simultaneously on a body be represented in magnitude and direction by two adjacent sides of a parallelogram, then the resultant vector can be represented by the diagonal of this parallelogram passing throught that point."
Proof:
Let us consider two vectors $$\overrightarrow{P}$$ and $$\overrightarrow{Q}$$ acting simultaneously on a body are represented in both magnitude and direction by two adjacent sides of a parallelogram OACB as shown in the figure. Then from the parallelogram law of vectors, the resultant vector $$\overrightarrow{R}$$ can be represented in magnitude and direction by diagonal OC.
Let us draw a perpendicular CD on the produced part of OA. Let, $$\measuredangle{BOA}= \theta$$ then, $$\measuredangle{CAD}= \theta$$.Now from $$\triangle CAD$$, we get
$$Cos \theta = \frac{AD}{AC} = \frac{AD}{Q}$$
$$AD = Q Cos \theta ............(i)$$
And,
$$Sin \theta = \frac{CD}{AC} = \frac{CD}{Q}$$
$$or, CD = Q Sin \theta ............(ii)$$
By applying pythagorous theorum on $$\triangle OCD$$, we get
$$OC^{2} = OD^{2} + CD^{2}$$
$$or, OC^{2} = (OA+AD)^{2} + CD^{2}$$
$$or, R^{2} = (P + Q Cos \theta)^{2} + (Q Sin \theta)^{2}$$
$$or, R^{2} = P^{2} + 2 PQ Cos \theta + Q^{2} Sin^{2} \theta + Q^{2} Cos ^{2} \theta$$
$$or, R^{2} = P^{2} + 2 PQ Cos \theta + Q^{2} (Sin^{2} \theta + Cos^{2} \theta )$$
$$or,R^{2} = P^{2} + 2 PQ Cos \theta + Q^{2}$$
$$R^{2} = \sqrt{P^{2} + 2 PQ Cos \theta + Q^{2}}$$
This equation gives the magnitude of resultant vector $$\overrightarrow{R}$$.
Direction of $$\overrightarrow{R}$$
If the resultant vector $$\overrightarrow{R}$$ makes an angle of alpha with the direction of vector $$\overrightarrow{P}$$, Then from $$\triangle OCD$$ we get,
$$Tan \alpha = \frac{CD}{OD} = \frac{CD}{OA+AD}$$
$$or, Tan \alpha = \frac{Q Sin \theta}{P+Q Cos \theta}$$
$$\alpha = tan^{-1}(\frac{Q Sin \theta}{P+Q Cos \theta})$$
This equatipm gives the direction of resultant vector $$\overrightarrow{R}$$.
Special cases
Case I
If the vectors $$\overrightarrow{P}$$ and $$\overrightarrow{Q}$$ are parellel then $$\theta = 0 \degree$$.
$$R = \sqrt{P^{2} + 2 PQ Cos 0 \degree + Q^{2}$$
$$= \sqrt{P^{2} + 2 PQ+ Q^{2}$$
$$= \sqrt{(P+Q)^{2}}$$
$$= P + Q$$
And,
$$\alpha = tan^{-1}(\frac{Q Sin 0 \degree}{P + Q Cos 0 \degree})$$
$$= tan^{-1}(\frac{0}{P + Q})$$
$$= tan^{-1}(0)$$
$$= 0 \degree$$
Case II
If the vectors $$\overrightarrow{P}$$ and $$\overrightarrow{Q}$$ are perpendicular then $$\theta = 90 \degree$$.
$$R = \sqrt{P^{2} + 2 PQ Cos 90 \degree + Q^{2}$$
$$= \sqrt{P^{2} + Q^{2}$$
$$R^{2} = P^{2} + Q^{2}$$
And,
$$\alpha = tan^{-1}(\frac{Q Sin 90 \degree}{P + Q Cos 90 \degree})$$
$$= tan^{-1}(\frac{Q}{P + 0})$$
$$= tan^{-1} \frac{Q}{P}$$
Case III
If the vectors $$\overrightarrow{P}$$ and $$\overrightarrow{Q}$$ are parellel but opposite in direction then $$\theta = 180 \degree$$.
$$R = \sqrt{P^{2} + 2 PQ Cos 180 \degree + Q^{2}$$
$$= \sqrt{P^{2} - 2 PQ+ Q^{2}$$
$$= \sqrt{(P-Q)^{2}}$$
$$= P - Q$$
And,
$$\alpha = tan^{-1}(\frac{Q Sin 180 \degree}{P + Q Cos 180 \degree})$$
$$= tan^{-1}(\frac{0}{P - Q})$$
$$= tan^{-1}(0)$$
$$= 0 \degree$$
Polygon Law of Vectors
It states that, "If the number of vectors acting simultaneously on a body be represented in magnitude and direction by the sides of a polygon taken in same order, then resultant vector can be represented by the closing side of this polygon taken in opposite order.
Proof
Let us consider the vectors $$\overrightarrow{A}$$, $$\overrightarrow{B}$$, $$\overrightarrow{C}$$ and $$\overrightarrow{D}$$ are acting simultaneously on a body as shown in the figure. Then, these vectors can be represented in magnitude and direction by the sides MN, NO, OP and PQ of polygon MNOPQ taken in same order as shown in figure.
Hence, from polygon law of vectors, the resultant vector $$\overrightarrow{R}$$ can be represented by the closing side MQ taken in opposite order. That is,
$$\overrightarrow{R} = \overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} + \overrightarrow{D}$$
Resolution of a Vector
The phenomenon of breaking a single vector into different components is called resolution of vector. Generally, there are three components of vector along x-axis, y-axis and z-axis direction.
Let us consider a vector $$\overrightarrow{F}$$ acting along OC as shown in the figure. Point O is the origin of co ordinate axis. Let us draw two perpendiculars AC and BC on x-axis and y-axis respectively. Let, $$\measuredangle{COA}=\theta$$
Now, From $$\triangle COA$$ we get,
$$Cos \theta = \frac{OA}{OC} = \frac{OA}{F}$$
$$or, OA = F Cos \theta$$
$$or, F_{x} = F Cos \theta$$
Similarly,
$$Sin \theta = \frac{AC}{OC} =\frac{OB}{P}$$
$$or, OB = P Sin \theta$$
$$or, F_{y} = F Sin \theta$$
Multiplication of Vector and Scalar
When the vector quantity is multiplied by scalar quantity, then it gives the new vector.
For example, when mass is multiplied with velocity, then momentum is formed.
Scalar Product or Dot Product
If the product of two vectors gives the scalar quantity then that type of product is called scalar product. If the vectors $$\overrightarrow{A}$$ and $$\overrightarrow{B}$$ acting at an angle $$\theta$$, then their scalar product is,
$$\overrightarrow{A}$$ . $$\overrightarrow{B}$$ = $$AB Cos \theta$$
That means the scalar product is the product of the magnitude of first vector and component of second vector along the first vector.
For example,
Work = F.s Cos \theta
$$= \overrightarrow{F} . \overrightarrow{s}$$
Properties of Scalar Product
• The scalar product of two vectors depends on the angle between them,
If $$theta = 0 \degree$$ , then
$$\overrightarrow{A} . \overrightarrow{B} = AB Cos \theta = AB Cos 0 \degree = AB$$
If $$theta = 90 \degree$$ , then
$$\overrightarrow{A} . \overrightarrow{B} = AB Cos \theta = AB Cos 90 \degree = 0$$
• It obeys the commulative law.
$$\overrightarrow{A}.\overrightarrow{B}=\overrightarrow{B}. \overrightarrow{A}$$
• It obeys distributive law.
$$\overrightarrow{A}. (\overrightarrow{B}+\overrightarrow{C}) = \overrightarrow{A}.\overrightarrow{B} + \overrightarrow{A}.\overrightarrow{C}$$
• The scalar product of a vector by itself becomes the square of it's magnitude.
$$\overrightarrow{A} . \overrightarrow{A} = A A Cos 0 \degree = A^{2}$$
Vector Product or Cross Product
If the product of two vectors gives vector quantity then that type of product is called the vector product.
If vectors $$\overrightarrow{A}$$ and $$\overrightarrow{B}$$ are acting at an angle $$\theta$$, then their vector product is given by,
$$\overrightarrow{A} \times \overrightarrow{B} = AB Sin \theta \hat{n}$$
where, $$\hat{n}$$ is the unit vector. The direction of vector product is given by the right hand thumb rule.
$$Torque = Fr Sin \theta$$
$$= \overrightarrow{F} \times \overrightarrow{r}$$
Properties of Vector Product
• The scalar product of two vectors depends on the angle between them,
If $$theta = 0 \degree$$ , then
$$\overrightarrow{A} . \overrightarrow{B} = AB Sin \theta = AB Sin 0 \degree = 0$$
If $$theta = 90 \degree$$ , then
$$\overrightarrow{A} . \overrightarrow{B} = AB Sin \theta = AB Sin 90 \degree = AB$$
• It does not obey the commulative law.
$$\overrightarrow{A}.\overrightarrow{B}= -(\overrightarrow{B}. \overrightarrow{A})$$
• It obeys distributive law.
$$\overrightarrow{A} \times (\overrightarrow{B}+\overrightarrow{C}) = \overrightarrow{A} \times \overrightarrow{B} + \overrightarrow{A} \times \overrightarrow{C}$$
• The scalar product of a vector by itself becomes zero.
$$\overrightarrow{A} . \overrightarrow{A} = A A Sin 0 \degree = 0$$ |
# Do homotopy groups “always” commute with filtered colimits?
It is well-known that homotopy groups, of, say, simplicial sets, commute with filtered colimits.
However, I could not find a reference for an analogous result for homotopy groups of spectra, or, under which hypothesis the same result would hold for an "arbitrary" simplicial model category.
More precisely, let $\cal{M}$ be a simplicial model category. For a fibrant object $X \in \cal{M}$ its homotopy groups with coefficients in a cofibrant object $W\in \cal{M}$ may be defined as
$$\pi_n (X; W) = [\Sigma^nW, X] = \pi_n \mathrm{map}(W,X) \ ,$$
where $\mathrm{map}$ denotes the simplicial mapping space from $W$ to $X$.
So my first question is: which hypothesis do I have to assume for $W$ to obtain an isomorphism
$$>\mathrm{colim}_i \pi_n (X_i;W) = \pi_n (\mathrm{colim}_i X_i;W) \ ? >$$
And the second one: in which kind of model category such an isomorphism holds for every cofibrant object $W$ -or, at least, for "sufficiently" many cofibrant objects $W$?
The reason behind my question is the following (and explains the meaning of that "sufficiently"): I have a filtered category $I$, functors $X_\bullet, Y_\bullet : I \longrightarrow {\cal M}\_f$ and a natural transformation $f_\bullet : X_\bullet \longrightarrow Y_\bullet$, such that, for every cofibrant object $W$, $f_\bullet$ induces isomorphisms
$$\mathrm{colim}_i \pi_n (X_i ; W) = \mathrm{colim}_i \pi_n (Y_i ; W) \ ,$$
for every $n$. And I want to conclude that the induced map between the colimits
$$\mathrm{colim}_i X_i \longrightarrow \mathrm{colim}_i Y_i$$
is a weak equivalence. Which would be true if
1. I could commute colimits and homotopy groups, at least for
2. "enough" cofibrant objects $W$ -in case of simplicial sets, one point $W = *$ is enough.
I suspect the answer involves words like "smallness / compactness" and "cellular model category". For instance an answer like: "You can do that in no matter what simplicial cellular model category" -in which every cofibrant object is compact- would be fine. Nevertheless, as long as I can understand, commutations like
$$\mathrm{colim}_i {\cal M} (W, X_i ) = {\cal M} (W, \mathrm{colim}_i X_i)$$
hold for $W$ small and $\lambda$-sequences; that is, when the domain of the functor $X : \lambda \longrightarrow \cal{M}$ is an ordinal; in particular, a totally ordered set, which my filtered $I$ needs not to be.
So any references of a result along these lines, even just for spectra, are welcome.
-
@Sam: Yes, I realize that it's not the same underlying category, but the idea is similar to how the category of simplicial $A$-modules is combinatorial for any simplicial Commutative Ring $A$. By the way, cofibrant generation absolutely does not imply combinatoriality. Combinatoriality is strictly stronger (unless you're suggesting that we take Vopenka's principle to be true). – Harry Gindi Feb 23 '11 at 7:38 |
Tag Info
91
There are plenty of satellite galaxies orbiting larger galaxies. The question is how long are you willing to wait for an orbit? The Milky Way has a mass $M$ of something like $6\times10^{11}$ solar masses, or $10^{42}\ \mathrm{kg}$. The small Magellanic Cloud is at a distance $R$ of $2\times10^5$ light years, or $2\times10^{21}\ \mathrm{m}$. A test mass ...
41
It depends on where on Mars you toss the coin, and how high you toss it. In a rotating frame of reference, an object in motion appears to be affected by a pair of fictitious forces - the centrifugal force, and the Coriolis force. Their magnitude is given by $$\mathbf{\vec{F_{centrifugal}}}=m\mathbf{\vec\omega\times(\vec\omega\times\vec{r})}\\ ... 27 They do! There's an entire class of galaxy, called a 'satellite galaxy' which is defined entirely based on them orbiting a larger galaxy (which would be called a 'central galaxy'). Our own milky-way is known to have many orbiting satellite galaxies, or at least 'dwarf-galaxies'. If dwarf-galaxies aren't enough, the milky-way itself is gravitationally ... 25 As many others point out, there is friction present, otherwise the wheel wouldn't grap the surface and pull the car forward. But you are talking about a different kind of friction. There is a possibility of different kinds of friction: Kinetic friction, if the wheel ever slides and skids over the asphalt. This is friction between objects that slide over ... 23 In this case, gravity is still an external force. In a zero-g environment, the mouse would also begin to move around the inside of the wheel, opposite the rotation it causes in the wheel, which would keep the angular momentum at zero. This would happen because the only way for the mouse to exert a force on the wheel and rotate it is for it to push itself in ... 21 The fan motor provides a torque \tau which has to accelerate \alpha the fan blades whose moment of inertia is I:$$\tau=I\alpha$$Given how long it takes for the fan blades to stop the frictional torques must be fairly low and so the torque applied by the motor to keep them going must also be low. With the relatively small torque rating, even if the ... 18 Calculating the power emitted as gravitational waves is relatively straightforward, and you'll find it described in any advanced work on GR. I found a nice description in Gravitational Waves: Sources, Detectors and Searches. To summarise an awful lot of algebra, the power emitted as gravitational waves by a rotating object is approximately:$$ P = ...
13
How can we detect Earth's spin? Apparent motion of Sun You will have observed that the sun reappears every 24 hours. There are two common explanations for this. One of them is that the earth rotates with a period of approximately 24 hours - this is the only explanation supported by the scientific evidence. The main alternative had a rather convoluted ...
11
anyway, how likely is it the ice ages could be explained by the earth 'realigning' so that polar regions would migrate over the surface of the earth? How about zero? The geological evidence of the Ice Ages clearly says that, between the ice episodes, the ice did not move. It's just that the polar caps shrank. For instance, the extent of the last ice ...
11
What does this small change means in form of Rotational Kinetic Energy? There's a problem with your calculation: You assumed a constant value for the Earth's moment of inertia. The Moon and Sun raise tides on the Earth itself. These Earth tides result in subtle changes in the Earth's moment of inertia. The signature of these tides can easily be seen in ...
10
A much simpler way of thinking about this is to consider energy. When the fan is spinning it has quite a lot of kinetic energy (try to stop it by putting your finger in the way to confirm this (don't actually do this!)). That kinetic energy goes as the square of the rotation rate, in fact. So as the fan starts, the motor needs to add energy to it. It ...
10
You might be thinking in comparison to a desk or handheld electric fan. As mentioned by @Farcher, $\tau = I\alpha$. $I$, the moment of inertia of a spinning body around a particular axis of rotation, is calculated as follows: $$I = \iiint\rho(x,y,z)||r||^2\ dV$$ Or with uniform density, $$I = \rho\iiint||r||^2\ dV$$ From this formula, you can see that ...
9
The coin will come back to your hand just like it would on the earth. The effect of atmosphere is negligible comparing to the coin's inertia, so the horizontal position of the coin relative to your hand will hardly be affected. The rareness of the atmosphere will only affect the vertical motion of the coin, like how quickly the coin will fall into your hand. ...
9
Yes, for the simple reason that you're not tossing the coin very high (presumably, anyway). You seem to think that on Earth, atmospheric drag is what keeps the coin "glued" to the tossing frame of reference, but that isn't really a factor at all. Say that you're on Earth, at sea level, on the equator, and you toss the coin 3 meters straight up. Neglecting ...
8
It's a bit complicated (Wikipedia). Induction motors work in sync with the AC frequency but have no torque at 0 RPM so they need some arrangement to get them started.
7
Electrons in a conducting disk in order to maintain equilibrium will have to have a centripetal force on them equal to the local change in potential energy with respect to a change in radius, that is $$m_e\omega^2 r = -e{d\phi\over dr}$$ After integrating, we get a potential difference between the center and a point R out $$\Delta\phi = -{m_e\omega^2 ... 6 Your intuition is correct. For the ball to change its angular momentum (to go from "backspin" to "forward spin"), there needs to be a net torque acting. There are two forces on the ball: gravity, and the normal force of the slope. Both these forces act through the center of mass - so neither force adds torque. Without torque, there is no change in angular ... 6 Your intuition was correct - the shaft will rotate in one direction and the housing/stator will rotate in the other. If you look up "moment of inertia" you will find that it is the rotational equivalent of mass. For almost any reasonable motor the moment of inertia of the shaft/rotor windings will be smaller than the moment of inertia of the housing/stator. ... 6 The reason why a gyroscope does behave in this strange way is that if you try to rotate it's axis in some direction, the "endpoints" of this axis have to be pushed perpendicular to what our first intuition would say. In order to verify why the axis starts rotating in this strange way, let's make some simplifications: the gyroscope consists of two identical ... 6 Altair, Vega, and Regulus A are perhaps the most famous examples of stars that have been "flattened" by rapid rotation. Some studies (mentioned in Yoon et al. (2010) suggest that Vega is rotating at 70-90% percent of the speed at which it would break up (its rotational velocity is about 20 km/s). Regulus is even closer to this breakup speed: If its ... 5 I think you have misunderstood how friction works here. The friction you have written down is (typically) the minimum friction needed for the 'non slip' to occur. Imagine a very faint slope. You will only need a small amount of friction to avoid slipping. As you make the slope steeper more friction is needed for the 'non slip'. Eventually the friction will ... 5 A rigid body can not in general be modelled as a mass point. This is possible in celestial mechanics, as the forces encountered there act uniformly and therefore can be effectively described as forces acting on the centre of mass. In general, you have to consider the orientation of the body as well, then one gets the equations of motion for the centre of ... 5 The tangential acceleration a_t and the angular acceleration \dot{\omega} are basically the same thing. They are related by:$$ a_t = r\dot{\omega} $$So we don't include both of them because that would be counting the same thing twice. 5 In your energy conservation equation, your are assuming that both the initial system and the final system have kinetic energy due just to the rotation around the axis. However, there is also some kinetic energy due to the rings translating away from the axis. In other words, the velocity vector of a ring is not parallel to the velocity vector of the point ... 5 Because your pen is not a cylinder, but a portion of a cone. Since it is also rigid, both ends have to complete one cycle of rolling simultaneously. This means that for each cycle, if the narrow and thick ends are separated by the pen's length L and have radii r_1 and r_2, respectively, they roll 2\pi r_1 and 2\pi r_2, respectively. The only way ... 5 There are several factors that may be taken in account, but the more important is the energy used deforming the tire. Suppose a deflated tire. As you move forward and the tire rotates, the part of the tire that is starting to touch the ground has to be deformed (since the tire is flat). You have to use an important amount of energy for that. Note that the ... 5 simply the resistance of a body to rotate it over an axis? Gosh, I dislike the word resistance in this context since resistance is, in general, dissipative and, in particular, resistance to rotation would imply that an isolated object that is rotating would eventually stop. Think of moment of inertia (rotational inertia) about an axis as a measure of ... 4 Suppose you pedal at about 100 rpm (I don't know what a typical rate of pedalling is, but this seems a plausible order of magnitude). To make a fair comparison with your motorbike you need to gear the moorbike engine down from 7,000 rpm to the same 100 rpm that you pedal at, i.e. a factor of 70, and this will multiply the torque by a factor of 70. So at 100 ... 4 If the bearings were to be considered frictionless, then the maximum speed of the fan will not decrease, though it will take the fan longer to reach the maximum speed. Because as the moment of inertia of the impeller increases its angular acc. will decrease (for the same torque applied), therefore it will take the fan longer to reach its maximum speed. The ... 4 Do I need to use the angular velocity vector in the rotating or inertial reference frame for this? Yes. You can do it either way. I start with the expression that relates the time derivative of a vector quantity \boldsymbol u in the inertial and rotating frames:$$\left(\frac {d\boldsymbol u}{dt}\right)_I = \left(\frac {d\boldsymbol u}{dt}\right)_R ...
Only top voted, non community-wiki answers of a minimum length are eligible |
Uniform Circular Motion; Need help with deriving equations.
1. Apr 14, 2012
n3w ton
1. The problem statement, all variables and given/known data
Hi I'm doing a physics lab about uniform circular motion to measure frequency and to compare it to mass,radius and force tension.
(A & B) I did
(C,D, E) I need help
(a) What variables are being measured / manipulated in this lab? What type of relationship is being tested? (radius, mass, force tension/force causing centripetal force)
(b) Graph the relationship between the frequency of revolution and each of the following:
• the magnitude of the tension force [force causing the circular motion(centripetal force)]
• the radius of the circular path
• the mass of the object
**■→(c) Find the proportionalities between frequency of revolution and the variables in radius, mass, and force of tension/centripetal force.
**■→(d) Derive an equation for the frequency in terms of the tension, the radius, and the mass by combining your results from (c) and using your results from (b) to verify.
**■→(e) The following relationship gives the magnitude of the net force causing the acceleration of an object in uniform circular motion:
Fc = 4π²mrf²
Rearrange this equation to isolate the frequency. Compare this result with the equation you derived in (d). Indicate the likely causes for any discrepancies.
Data: http://i.imgur.com/dLpyP.png
Frequency vs Force Graph: http://i.imgur.com/fyFci.png
Frequency vs Mass Graph: http://i.imgur.com/GJ2ms.png
2. Relevant equations
Fc = 4π²mrf²
3. The attempt at a solution
C) and D) Im stuck at
E)
Fc = 4π²mrf²
$\sqrt{}\frac{Fc}{4π²mr}$
2. Apr 14, 2012
PeterO
Your graphs are not extensive enough - they must include the origin (0,0) - not necessarily as a point, but with the axes long enough for them to show up.
With graphical analysis, the only line you can confidently interpret is a straight line passing through the origin.
if y vs x is not straight, you can try y vs 1/x or y vs x2 or or y vs 1/x2 or y vs x2 of y vs √x or y vs 1/√x to see if any of them are a straight line through the origin [or close - there may be uncertainties in your measurements]
suppose y vs 1/√x was such a straight line.
That means y is proportional to 1/√x or y = k/√x or y2x = k
3. Apr 15, 2012
MrWarlock616
uhmm.. time for one cycle is 1/frequency.
4. Apr 15, 2012
PeterO
Looking at your results, I am not sure the figure you call frequency is in fact frequency.
It looks more like the Period to me - ie the time for one cycle.
You possibly need to follow the step you mention above.
5. Apr 15, 2012
MrWarlock616
Yes exactly, that's what I said. The graphs are obviously wrong because he has used time period instead of frequency.
peter, I didn't ask this question..n3w ton did. :P |
Article | Open | Published:
# Biocompatibility and biodegradation studies of a commercial zinc alloy for temporary mini-implant applications
## Abstract
In this study, the biocompatibility and in vitro degradation behaviour of a commercial zinc-based alloy (Zn-5 Al-4 Mg) were evaluated and compared with that of pure zinc for temporary orthopaedic implant applications. Biocompatibility tests were conducted using human alveolar lung epithelial cells (A549), which showed that the zinc alloy exhibits similar biocompatibility as compared to pure zinc. In vitro degradation evaluation was performed using weight loss and electrochemical methods in simulated body fluid (SBF) at 37 °C. Weight loss measurements revealed that the degradation of the zinc alloy was slightly lower during the initial immersion period (1–3 days), but marginally increased after 5 and 7 days immersion as compared to pure zinc. Potentiodynamic polarisation experiments showed that the zinc alloy exhibits higher degradation rate than pure zinc. However, electrochemical impedance spectroscopy analysis suggests that pure zinc is susceptible to localized degradation, whereas the zinc alloy exhibited passivation behaviour. Post-degradation analysis revealed localized degradation in both pure zinc and the zinc alloy.
## Introduction
The emerging interest in biodegradable implants for short-term service life in orthopaedics aims to produce biomaterials with desirable biodegradability, biocompatibility and mechanical properties closer to natural bone. In recent years, a significant amount of research has been undertaken on biodegradable metals, mainly on magnesium-based materials1. Magnesium is highly biocompatible, but its undesirably high degradation rate in physiological conditions is a huge disadvantage. Hence, the recent research focus in this field has been on controlling the degradation rate of magnesium by alloying and/or surface coatings1,2,3,4,5,6.
Metallic zinc is a potential biodegradable and biocompatible material for temporary orthopaedic mini-implants such as screw, pins and plates. As an essential nutrient, zinc has many important biological functions, including development and sustenance of bones7, food intake and growth8, wound healing9, cell proliferation and division, and DNA stabilisation and replication8,10. Dietary zinc is metabolically absorbed through the small intestine as zinc ions and amnio acid complexes and it is regulated by metallothionein11. In the short-term, zinc in the body is regulated to organs such as pancreas, liver, kidneys and spleen12,13. However, in the long-term, 90% of the absorbed zinc is deposited in the muscular and skeletal system14. The biological half-life of zinc has been determined to be between 162 and 500 days15,16, and the daily recommended dose of zinc is 10–15 mg/day17. Interestingly, it has been reported that long-term administering of zinc doses ten times the daily recommended intake has produced no adverse effects in humans in relation to wound healing18, antirheumatic activity for rheumatoid arthritis19 and plasma copper levels20. In fact, high concentrations of zinc have been shown to prevent conditions like osteoporosis through promotion of osteoblastogenesis and suppression of osteoclastogenesis21,22.
Metallic zinc has physical and mechanical properties similar to those of other common biomaterials: Density = 7.14 g/cm3; Young’s Modulus = 70 GPa; Ultimate Tensile Strength (UTS) = 126–246 MPa23. The electrochemical dissolution of zinc in aqueous solutions is suggested to occur via the following reactions24,25:
$$Zn\to Z{n}^{2+}+2{e}^{-}\quad \quad \quad \quad \quad -0.7618\,{{\rm{V}}}_{{\rm{SHE}}}$$
(1)
$${O}_{2}+2{H}_{2}O+4{e}^{-}\to 4O{H}^{-}\quad \quad +0.4010\,{{\rm{V}}}_{{\rm{SHE}}}$$
(2)
where (1) and (2) represent the anodic and cathodic reactions, respectively. However, the degradation mechanism is believed to be largely reliant on even small changes in the electrolyte pH, temperature and composition, and various reaction schemes have been proposed26,27. Some principal products of dissolved zinc cations in aqueous solutions are produced via the following reactions:
$$Z{n}^{2+}+2O{H}^{-}\to ZnO+{H}_{2}O$$
(3)
$$Z{n}^{2+}+4O{H}^{-}\to Zn{O}_{2}^{2-}+{H}_{2}O$$
(4)
$$Z{n}^{2+}+2O{H}^{-}\to Zn{(OH)}_{2}$$
(5)
$$Z{n}^{2+}+4O{H}^{-}\to Zn{(OH)}_{4}^{2-}$$
(6)
Similarly, the selectivity between reactions is governed by the electrolyte conditions. These products are major constituents of the passive films formed on zinc during aqueous corrosion and are known to provide considerable degradation protection since they are thermodynamically stable at room temperature within the pH range 6–1226.
As compared to the wealth of literature on the biocompatibility and degradation of magnesium-based material1,2,3,4,5,6, the work done on zinc-based materials is limited. Although extensive research has been done over the past few decades on the corrosion behaviour of zinc and zinc-based alloys (as bulk or coated film) in chloride-containing environments for engineering applications28,29,30,31,32,33,34, only recently there has been a growing interest on zinc-based materials for potential biodegradable implant applications. Bowen et al.35 reported that degrading zinc has optimal biocompatibility and the degradation products supress the activities of inflammatory and smooth muscle cells. Liu et al.36 found that zinc dissolution has no significant destructive effect on erythrocyte. On the other hand, Sherier et al.37 suggested that free Zn2+ ions might hinder cell mobility and adhesion. However, Kubasek et al.38 reported that the maximum safe Zn2+ ion concentrations for U2OS and L929 cell lines are 120 µM and 80 µM, respectively. Bowen et al.39 examined the in vivo degradation behaviour of zinc for absorbable stent applications, and reported the longevity and harmless degradation of zinc metal. They observed that the degradation rate of zinc increases linearly with implantation time. Under short-term in vivo condition, zinc oxides were formed, however, after 4.5 to 6 months, calcium phosphate layers were observed. Zinc oxides seem to be inert to the immune system, but depending on the size of these oxide particles can cause cytotoxicity40. Drelich et al.41 reported that defects/cracks in the zinc oxide film increases the degradation rate.
For load-bearing orthopaedic applications, the mechanical integrity of the implant during service is critical. Localized degradation may affect the mechanical integrity of the implant. Unfortunately, zinc undergoes localized degradation in chloride-containing environments54,55,56. Hence, it is important to study the localized degradation susceptibility of zinc in physiological conditions. Literature suggests that the ternary Zn-Al-Mg alloys have superior degradation protection properties in chloride-containing solution than binary system alloys such as Zn-Mg and Zn-Al57,58,59,60,61,62. It should be noted that Zn-Al-Mg alloys are commercially available and have been widely used as galvanizing coating materials on steels due to their high degradation resistance57,59,61. This alloy system has other advantages such as better mechanical strength and relatively low density (due to lighter alloying metals such as magnesium and aluminium) as compared to pure zinc for implant applications. Therefore, it is important to understand the biocompatibility and biodegradation behaviour of a Zn-Al-Mg alloy.
In this study, the biocompatibility and biodegradation behaviour of the commercially available Zn-5 Al-4 Mg alloy were examined and compared with that of pure zinc. Weight loss and electrochemical methods were used to evaluate the biodegradation behaviour of the materials in simulated body fluid at 37 °C. Post-degradation analysis was performed using scanning electron microscope (SEM) to identify the mode of degradation.
## Experimental Procedure
The chemical compositions of pure zinc and the commercial Zn-5 Al-4 Mg alloy used in this study are shown in Table 1. The hardness of the materials was measured using a Rockwell hardness tester (Model: Avery Rockwell Hardness Tester, type 6402). For the cytotoxicity testing, human alveolar lung epithelial A549 cells were used. The A549 cells utilised in this study are a human derived epithelial cell line from the lungs and respiratory tract, and is frequently used as indicator of general genotoxicity and cytotoxicity40. These cells were obtained from the American Type Culture Collection (ATCC, USA) and maintained in 25 cm2 cell culture flasks in an incubator with a humidified atmosphere at 37 °C and 5% CO2. The cells were cultured in RPMI-1640 medium (Sigma-Aldrich, USA) supplemented with 10% FBS, 1% penicillin-streptomycin and L-glutamine (Life Technologies, Australia), designated as ‘complete medium’. The cells were cultured to a cell density of 1 × 106 cells/mL before being sub-cultured into fresh media 2–3 times a week. The metal samples were ground with SiC paper up to 2500 grit and later polished with 1 μm alumina powder solution, washed with distilled water and then ultrasonically cleaned in ethanol. Subsequently, the samples were pre-incubated in the complete medium until 96 h at 37 °C in a humidified atmosphere with 5% CO2 to obtain the extraction medium, which was used for the cytotoxicity analysis. Metabolic activity of A549 cells exposed to the samples was assessed using the MTS assay which measures the absorbance (490 nm) of the purple dye formazan generated by live cells when exposed to the MTS reagent. (Promega MTS CellTiter 96® aqueous kit, Promega, USA). Briefly, 10,000 cells in 100 µL were seeded into 96-well tissue culture plates (Sarstedt, Germany). After allowing for overnight attachment, the cells were exposed to 100 µL of the extraction medium obtained at 1, 2, 3 and 4-day exposure period. Wells containing cells exposed to the “complete medium” served as positive control. Data were obtained from three independent experiments, each performed in triplicate.
In addition, DAPI (4′,6-diamidino-2-phenylindole) staining was carried out to study the changes in nuclear morphology of A549 cells after exposure to the extraction media. A549 cells were allowed to attach overnight on chambered slides (Lab-Tek, Proscitech) at a density of 106 cells per mL and subsequently incubated with the extraction media for up to 4 days. At the end of the incubation period, all cells were collected and washed with Dulbeccos’ phosphate buffered saline (Life Technologies, USA), subjected to fixation and were mounted on Superfrost slides (Proscitech, Australia) using ProLong® Gold Antifade Reagent with DAPI (Molecular Probes, Life Technologies, USA). The slides were subsequently incubated at room temperature for 24 h in the dark before visualization using a Zeiss LSM710 confocal laser scanning microscope (Carl Zeiss, Germany).
In vitro degradation behaviour of pure zinc and Zn-5 Al-4 Mg alloy was evaluated by weight loss analysis and electrochemical methods, i.e., potentiodynamic polarisation and electrochemical impedance spectroscopy (EIS), in simulated body fluid (SBF) maintained at a body temperature of 37.5 ± 0.5 °C and pH of 7.4–7.6. The chemical composition of the SBF is given in Table 263. Prior to the in vitro degradation testing, the samples were ground with SiC paper up to 2500 grit and later polished with 1 μm alumina powder solution, and washed with distilled water and then ultrasonically cleaned in ethanol. In the weight loss testing, the samples were immersed in SBF at a static condition and the weight losses were recorded after 1 to 7 days immersion. Electrochemical experiments were conducted using a potentiostat/galvanostat and a frequency response analyser (Model: ACM Gill AC, ACM Instruments). A typical three-electrode system consisting of graphite as a counter electrode, Ag/AgCl electrode as a reference electrode and the sample as a working electrode was used in this study. The potentiodynamic polarisation experiments were conducted at a scan rate of 0.5 mV/sec. The EIS experiments were performed over the frequency range of 1 × 105 Hz to 1 × 10−2 Hz and at an AC amplitude of 5 mV. The EIS data were analysed using equivalent circuit modelling (Software: ZSimpWin v3.21, Princeton Applied Research). All the in vitro degradation tests were conducted in triplicate. Scanning electron microscope (SEM) was used to analyse the post-degradation samples.
## Results and Discussion
### Biocompatibility
The cell viability (cytotoxicity) of pure zinc and the zinc alloy on A549 cells is shown in (Fig. 1) as compared to cells exposed to the complete medium. A549 cells exposed to the extraction media obtained from the zinc or zinc alloy samples did not demonstrate cytotoxicity at the end of the 4 days testing period. Figure 1b–d show the nuclear morphology of the treated cells after DAPI staining. Normally, cells undergoing apoptosis exhibit characteristic condensation of the nuclear material. In the present study, cells exposed to zinc or the zinc alloy demonstrated nuclear morphology similar to the control cells (exposed to cell culture medium alone) even after the 96 h exposure period, further confirming the non-toxic nature of the samples. These results serve as a preliminary indication of the biocompatibility of pure zinc and the zinc alloy.
### Potentiodynamic polarisation
The potentiodynamic polarisation curves of pure zinc and the zinc alloy are shown in Fig. 2, and the electrochemical data obtained from the curves are presented in Table 3. The corrosion potential (Ecorr) of the zinc alloy was slightly (~10 mV) more noble as compared to pure zinc. The cathodic polarisation curves suggest that the cathodic activity was higher for the zinc alloy in comparison with pure zinc. This difference in cathodic activity can be attributed to the alloying elements. In the case of the anodic side of the polarisation curves, the dissolution behaviour of the zinc alloy was higher than pure zinc. However, both pure zinc and the zinc alloy did not show any active passive region or breakdown potential. The corrosion current density (icorr) calculated from the cathodic curves suggested that the icorr value of the zinc alloy is ~85% higher than pure zinc i.e., 17.7 µA/cm2 and 9.55 µA/cm2, respectively. The calculated degradation rate for the zinc alloy was 0.32 mm/y and for pure zinc 0.14 mm/y. As expected, the degradation rate of pure zinc and the zinc alloy was significantly lower than that of pure magnesium (degradation rate = 0.54 mm/y; icorr = 23.5 µA/cm2)64. Post-polarisation SEM micrographs of pure zinc and the zinc alloy are shown in Fig. 3. The morphology of pure zinc revealed localized attack (Fig. 3a and b). In the case of the zinc alloy, the localized attack increased, as demonstrated by the relative larger areas of evident damage (Fig. 3c and d).
### EIS
The EIS spectra for pure zinc and the zinc alloy over 72 h immersion in SBF is shown in Fig. 4. The equivalent circuits (EC) models used and the fitting for pure zinc and the zinc alloy after 2 h and 72 h immersion are shown in Fig. 5. The data obtained from EIS modelling are presented in Table 4. After 2 h immersion, pure zinc showed a capacitive loop and an inductive loop. The low frequency inductive loop is a general indication of localized degradation65,66 or adsorption of intermediate corrosion products or ions onto the surface67,68. The zinc alloy showed two capacitive loops, but no inductive loop. The high frequency capacitive loop can be attributed to charge transfer resistance and the mid-frequency capacitive loop is related to the film resistance. The EC model used for zinc alloy consisted of the following elements: Rs (solution resistance), Rct (charge transfer resistance), CPEdl (double layer capacitance) and Rf (film resistance). For the pure zinc, which exhibited an inductive loop, L (inductance) and CPEf (capacitance due to film effect) elements were added. The polarisation resistance (Rp) of the samples was calculated by adding the Rct and Rf. The Rp of the zinc alloy after 2 h exposure to SBF was 63% higher than that of pure zinc (pure zinc = 250.56 Ω∙cm2; zinc alloy = 408.39 Ω∙cm2).
After 24 h exposure, the Rp of the zinc alloy increased by 38%. In the case of pure zinc, the inductive loop disappeared and the Rp increased by 27% (Rp = 319.61 Ω∙cm2). However, after 48 h exposure, the zinc alloy continued to display passivation effect, but the mid-frequency capacitive loop has transformed to Warburg impedance. This type of behaviour has been reported in the literature for zinc metal69. The significance of a Warburg impedance is the presence of a porous passive film facilitating diffusion controlled processes69,70. A Warburg diffusion element was used to model the EIS spectra for the zinc alloy. The Rp of the zinc alloy was ~4 times higher than pure zinc (pure zinc = 387.36 Ω∙cm2, zinc alloy = 1457.2 Ω∙cm2) after 48 h exposure. The trend continued even after 72 h exposure, the Rp values for pure zinc and zinc alloy were 690.13 Ω∙cm2 and 2899.66 Ω∙cm2, respectively.
### Weight Loss
The weight loss measurements for pure zinc and the zinc alloy are shown in Fig. 6. As expected, the weight loss increased with increasing exposure. Interestingly, the weight loss data for the zinc alloy were not remarkably different to the pure zinc during 1 to 7 days immersion in SBF. After 1 and 3 days immersion, the weight loss of the alloy was marginally lower than pure zinc, but after 5 and 7 days the trend reversed. The macrographs of pure zinc and the zinc alloy after each interval of immersion are shown in Fig. 7. Both pure zinc and the zinc alloy have undergone localized degradation and the intensity has increased with increasing immersion time. It was interesting to note that the pitting nucleation was not high, but the growth of pits was very rapid. After 7 days immersion, the localized degradation attack was remarkably high in both pure zinc and the zinc alloy. The overall degradation rates derived from the weight loss test were 0.31 mm/y for pure zinc and 0.35 mm/y for the zinc alloy.
### Mechanism
The EIS spectra suggests that the zinc alloy exhibits passivation behaviour. This can be attributed to the alloying elements in the zinc alloy, especially aluminium. Literature on the corrosion behaviour of aluminium-containing zinc alloy coatings, e.g., Zn-Al71 and Zn-Al-Mg61,72, in chloride-containing solution suggests that aluminium forms a thick and complex layers. Volovitch, et al.62 reported that aluminium formed basic aluminium-oxides in the initial stages of corrosion of a Zn-Al-Mg alloy. Studies have also suggested that aluminium has a lower dissolution tendency as compared to zinc and magnesium in Zn-Al-Mg alloy system60,62. On the other hand, magnesium forms magnesium hydroxide in aqueous solutions, which is a protective film, but in chloride-containing solution, the protective film converts to soluble magnesium chloride, as shown below73.
$$Mg(s)+2{H}_{2}O\to Mg{(OH)}_{2}+{H}_{2}$$
(7)
$$Mg{(OH)}_{2}+2C{l}^{-}\,\to MgC{l}_{2}+2O{H}^{-}$$
(8)
In the current study, the Warburg impedance observed under long-term EIS suggests that the film formed on the alloy is porous in nature and introduces diffusion characteristics; hence, the stability of the film in physiological condition was only temporary.
To further understand the passivation behaviour of pure zinc and the zinc alloy in the physiological environment that contains chloride ions, the Pourbaix diagrams of zinc, aluminium and magnesium were used74. Figure 8 (a–d) shows the Ecorr values of pure zinc and the zinc alloy embedded on the Pourbaix diagrams. Although (Fig. 8a) suggests that the potentials of pure zinc and the zinc alloy are in the passive region, the presence of chloride shifts the passivity region towards the higher pH regions and hence undergo dissolution. It can be noted that the experimental conditions confined pure zinc and the zinc alloy to the active ZnCl+ region throughout the immersion period (Fig. 8b). Hence, zinc did not show any strong passivation. The passivation observed in the EIS experiments (Fig. 4) of the zinc alloy could be attributed to aluminium, which is stable in the physiological pH range (Fig. 8c). Magnesium is however not stable in that pH range (Fig. 8d).
The potentiodynamic polarisation curves suggest that the cathodic activity of the zinc was higher as compared to pure zinc (Fig. 2). This can be attributed to the alloying elements in the zinc alloy. Although oxygen reduction reaction is the predominant cathodic reaction for zinc metal, hydrogen evolution reaction is feasible in the zinc alloy due to the presence of the alloying elements such as magnesium and aluminium. It should be noted that the hydrogen-evolution exchange current densities of the alloying elements magnesium and aluminium are higher than that of zinc (~10−8 −10−9, ~10−10 and ~10−11 A/cm2 respectively75,76). Therefore, the cathodic current of the alloy was higher than that of pure zinc. The anodic reaction of the zinc alloy was also higher than pure zinc, which could be due to selective leaching of elements under accelerated conditions. Magnesium being more reactive than aluminium and zinc, selective leaching of magnesium could have caused the increase in anodic current during polarisation. However, the EIS experiments, which is non-destructive, revealed passivation behaviour in the alloy. The weight loss method also showed improved degradation resistance of the zinc alloy as compared to pure zinc during the initial immersion period, but exhibited localized degradation with increasing exposure, probably due to galvanic effect.
The study suggests that the zinc alloy exhibited similar biocompatibility to pure zinc. It was interesting to see that the biodegradation resistance of the zinc alloy, which has been used for galvanization for its excellent corrosion resistance, was not superior to that of pure zinc in physiological conditions. During the initial immersion period, the zinc alloy exhibited passivation behaviour, but the passivity became less stable with exposure time and ultimately gave rise to localized degradation similar to pure zinc. However, this zinc alloy has some attractive properties (density and hardness) as compared to pure zinc, which are essential for load-bearing implant applications. Due to the presence of light metals, the density of the zinc alloy is approximately 17% lower than pure zinc (pure zinc = 7.14 g/cm3 and zinc alloy = 5.908 g/cm3). The hardness of the zinc alloy was approximately 14% higher than pure zinc (pure zinc = 79.2 HRB and zinc alloy = 89.9 HRB). The biocompatibility and the attractive physical and mechanical properties make the commercial zinc alloy a potential material for temporary mini-implant applications. However, surface engineering is essential to delay the localized degradation of the commercial zinc alloy.
## Conclusions
The biocompatibility and in vitro degradation behaviour of a commercial zinc alloy (Zn-5 Al-4 Mg) were evaluated and compared with that of pure zinc. The zinc alloy showed similar biocompatibility to pure zinc in the cytotoxicity assay conducted using human alveolar lung epithelial cells (A549). The aluminium content in the alloy improved the passivation behaviour, but was only temporary in the physiological conditions. The potentiodynamic polarisation results suggested that the zinc alloy degradation rate is marginally higher than pure zinc owing to the higher hydrogen exchange current density of the alloying elements (magnesium and aluminium) as compared to zinc. The localized degradation susceptibility of the zinc alloy was similar to pure zinc. In addition to the comparable biocompatibility and biodegradability of the zinc alloy as compared to pure zinc, the alloy exhibits lower density and higher hardness, which make it more attractive for load-bearing orthopaedic applications.
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## Acknowledgements
The authors would like to thank Sun Metals Corporation Pty. Ltd. (Queensland, Australia) for providing the zinc and zinc alloy samples. A.L received funding from the Australian Research Council and the National Health and Medical Research.
## Author information
### Affiliations
1. #### Biomaterials and Engineering Materials (BEM) Laboratory, James Cook University, Townsville, Queensland, 4811, Australia
• M. Bobby Kannan
• , Corey Moore
• & Mohamed Rahuma
2. #### Molecular Allergy Research Laboratory, Department of Molecular & Cell Biology James Cook University, Townsville, Queensland, 4811, Australia
• Shruti Saptarshi
• & Andreas L. Lopata
### Contributions
C.M., S.S.b and M.R. performed the experiments. B.K.M. and A.L. supervised the study. B.K.M., S.S.b, S.S.a and C.M. wrote the manuscript. All authors commented on and approved the manuscript before submission.
### Competing Interests
The authors declare that they have no competing interests.
### Corresponding author
Correspondence to M. Bobby Kannan. |
## Wednesday, November 30, 2005
### Getting Started with Eclipse and the SWT
Getting Started with Eclipse and the SWT is a very nice set of introductory tutorial on working with SWT and Eclipse in general. The tutorials cover the basic, advanced SWT widgets, SWT layout, and several other interesting topics.
Now that Bioclipse is gaining speed, it is a must-read.
### KDE 3.5 is out
KDE 3.5 was released with lots of changes. SuperKaramba is now a standard KDE application and is neatly integrated. It allows embedding themelets on your desktop background:
It shows several themelets: the weather, a calender, a toolbar with applications, a FoldingAtHome monitor, the contents of the clipboard, the music that is playing (Cake) and a simple todo list. All customizable up to the pixel.
And before I forget: a nice new Kalzium release!
## Monday, November 28, 2005
### A Blue Obelisk blog Planet
Today I setup a blog planet for Blue Obelisk members. First I tried Chumpologica but it did not read Atom feeds.
Next in line was Planet, which turned out to be used by many big planet sites, like Planet Debian. It also works with Atom feeds in general, but not well with Atom 1.0 feeds, like that of Carsten. After some googling I found a patched version which did the job.
The result is at http://www.woc.science.ru.nl/planetbo/, but I hope that someone can arrange a http://planet.blueobelisk.org/.
## Sunday, November 27, 2005
### Open Source Swing: Jmol renderer runs!
Where I was able to mention earlier that JChemPaint now runs with free (as in open source) Java virtual machines, I just tried to run the core Jmol renderer, using the Integration.java which comes as an example:
The screenshot was made with jamvm 1.3.3 and classpath 0.19.
It is very slow, however. I have not tried it with other free virtual machines, which are supposedly faster. It is a good start nevertheless: it means that a Jmol based Bioclipse plugin will work with free virtual machines too.
## Wednesday, November 23, 2005
### Machine crash; SVN went along
Don't happen often, but my machine crashed two hours ago. Not a big deal, because I have my important files in SVN. Oh wait, SVN had a commit in progress during the crash. So, svn recover. Mmmm... doesn't work either. OK, SVN FAQ: try db_recover. That worked. No, it did not: svn commit still not working for the files I was trying to commit. Fortunately, I make regular SVN db backups so I created a brand new SVN repository from scratch and recovered the back up. That worked. Really.
## Monday, November 21, 2005
### Bioclipse: the chemo-/bioinformatics workbench
Some weeks back there was the CDK5AW, the CDK 5th anniversiry workshop. A small group of international open source chemo-, bioinformatics software developers met, among which two from Sweden. It was then decided to generalize their work resulting in Bioclipse:
http://www.bioclipse.net/
It's heavily using the Eclipse Rich Client Platform, making additional plugins trivial. OK, if this does not convinve you: check the screenshots on the Bioclipse website.
It's a killer, really! Ola, Martin: great work!
PS. I am going to try to run it with free Java virtual machines this weekend, but if you have a working solution earlier than that, please leave a comment and screenshot in the comments.
## Sunday, November 20, 2005
### Open Source Swing: JChemPaint runs!
Thanx to Mark's encouragements, I tried to run Jmol and JChemPaint with jamvm.
Jmol fails with an NullPointerException, but JChemPaint runs! And note that this was not even running with the latest of the latest; just recent packages from Kubuntu! Yes, there are some glitches, but I'm happy nevertheless!
## Friday, November 18, 2005
### The goal: a live chemblaics CD
This evening I have been looking at with the KNOPPIX customization howto, and ran many of the interesting commands. I've setup a environment with Kalzium, OpenBabel, CDK, jython, PyMOL, and for development I included gcj and Eclipse. At some later point I will include kfile_chemical too, but I want to make a deb package first.
Moreover, I also wanted it to include JChemPaint, Jmol and Taverna (with the CDK extension). However, these depend on Swing, which is not suffiently provided by open source java virtual machines. I attempted gij 4.0, kaffe and sablevm, all without success.
A live CD with all the open source chemo- and bioinformatics tools would be a real killer. We could take a burned live CD with us to conferences and have others run our software on their laptop! But we need to stop use Swing. Fortunately, there seems to be a serious project going on to port JChemPaint and Jmol to a free Java GUI environment, so maybe we can have the live CD up and going before the 2006 conferences start.
## Thursday, November 17, 2005
### Back from the 1st GCC
OK, just back from the first German Chemoinformatics Conference, which I enjoyed very much. A rather interesting program, and lots of interesting posters too. You can read the programme online, and will not spend too many words on that (at least not now). But what I will do is point out some interesting posters here.
One poster was on the Molecular Query Language (MQL) by Ewgenij Proschak from Frankfurt. You can read more on this in the latest CDK News as it is implemented for the CDK too. The opensource implementation is expected next year.
Another interesting poster was on the use of ontologies to connect chemistry and biology. This poster was by Juergen Harter from BioWisdom, a Cambridge, UK based company.
Marc Zimmermann had a poster on the chemical OCR variant, called chemical structure recognition (CSR). This process converts images, for example scanned from literature, into a connectivity table. Difficult task, indeed. This page contains some information about this project.
There were other interesting posters too, so will probably report on those later too. But do feel free to leave comments to this blog post, discussing other interesting posters.
## Friday, November 11, 2005
### Going to the German Chemoinformatics Conference
This sunday starts the first German Chemoinformatics Conference in Goslar. It's an interesting programme, with presentations on the InChI, PubChem, 25 years of chemoinformatics, the chemical semantic web, and much more.
Among these presentations is mine, on comparing crystal structures (PDF) and deducing cell parameters. But I'm having a poster on QSAR too.
I'll arrive on saturday afternoon in Goslar, so leave a message at the conference hotel if you want to meet up, and talk about my work, or yours, or the CDK, KDE, JChemPaint, Jmol, kfile_chemical, Kat/Chemistry, BlueObelisk, Eclipse, R, or whatever else... I plan to have a modest german meal and one or two beers in the evening.
BTW, after Belém (Lissabon), Sintra, Boppard, Kinderdijk, Hoorn and Cologne, it's the 7th UNESCO world heritage site I'm visiting in just 14 months! Can't we just have conferences in Hawaii and sorts, like they do in other fields?? Oh, wait, we do: EuroQSAR is on a cruise boat.
## Thursday, November 10, 2005
### Scons and bksys for kfile_chemical
Not so long ago, it was decided that KDE 4.0 will use SCons as a configuration and building tool, instead of the autotools and make: the common ./configure && make && make install which has served the open source community very well for so long.
SCons is different in several ways. One of these is that the tar.gz packages it produces are some 500kB smaller, which makes a huge difference for kfile_chemical which is now 121kB instead of 635kB.
Now, the KDE community, or Thomas Nagy to be precise, developed a helper for KDE software, called bksys. Version 1.5.1, however, did not contain an example directory for kfile plugins, but I managed to work something out starting from the configuring scripts from kdissert, and ended up with these SConstruct and config.bks.
Now, I haven't figured out how to include the translations, but will figure that out sooner or later... for now, I'm quite happy with the new build system.
## Tuesday, November 08, 2005
### A R GUI: rkward
The great thing about open source is that... it's open.
When I was browsing the internet just now, I dropped in on KDE Dot News. In the rightside column, there is a feed of new KDE software from KDE-apps.org. A new version of my favoriate music player,
amarok, lured me to the KDE-apps website, where I saw rkward is latest announcement. The funny name, and the categorization as scientific, triggered some interest on my side, and it turned out to be a graphical frontend to my favorite statistics program, R.
Ok, they had a Debian package, and the debian/ build dir in the tar.gz so I downloaded it and started making a Kubuntu 5.10 package. While doing this I saw some notice about the R syntax highlighting used, which conflicts with the older version in the Kate packages.
Then I realized that a long time ago, I wrote such syntax highlighting for Kate, so my attention was lured again. And, indeed, they use my syntax highlighting, though extended later (somewhere down the page).
And this makes me happy. The syntax highlighting was useful to me in the past, but apparently to a lot of other people too. And because I released it as GPL, back then, it now appears in rkward! Yes, a really like open source :)
### When to stop including QSAR model variables...
Yesterday I reviewed an article which published a QSPR model which looked something like:
y = 151 + 50p1 - 12p2 - 0.006p3
with quite OK prediction results (R=0.9880). But I was not quite comfortable with the coefficient for the p3 variable. The article did not calculate significances for the coefficients, so it was not obvious from the article wether is was useful to include them. I then looked at the range for p3, which was 110-150; so, the maximal influence this variable can have is 150*0.006 = 0.9. Now, the experimental values given in the article were rounded to integers, indicating that the maximal
effect of the p3 variable is smaller than the experimental error! It's even worse when you consider the difference between the min and max value (40), then the influence would even be smaller (assuming that most model methods would put the mean temperature effect in the offset, 151 in this case).
Today, I reread an article with a similar issue. The model was something like:
y = -0.81 + 0.03*p1 + 0.009*p2
Here, max(p2)-min(p2) is a smaller than 100, so the maximal effect of the variable would be in the order 0.9, which is of the same order of the root mean square error of prediction (RMSEP) for this model. Indeed, the article already states that the coefficient is only significant at the 95% level, and not at the 99% level. But, without having calculated the RMSEP for a model without the p4 variable, I would guess that leaving it out would give equally good prediction results.
Concluding, I would say the the p2 variable does not include relevant information.
Do you think it is reasonable to include the p2 variable in the second model?
## Monday, November 07, 2005
### Ubuntu Dapper will include chemistry features
I just read that the Kubuntu team wants to include Kat in the dapper release (scheduled for April 2006). Kat is (to be) the KDE equivalent of Google's desktop search bar.
This is great news for us chem-bla-icians, as Kat has support for full text searching of chemistry files! Let's see if I can get the Kubuntu team to package up kfile_chemical too, which will extend Kat (and KDE in general), with extraction of meta data from chemical documents.
Update: Dapper will be released next year, not in 2007.
## Wednesday, November 02, 2005
### Open Source data mining in chemoinformatics
On the 7th International Conference on Chemical Structures Jeroen Kazius has a poster on finding discriminative substructures, that is, molecular fragments which can be discriminate between two acitivity classes. The software is released as Gaston, is written in C++ and has the GPL license.
Later I encountered MoSS which has the same goal, but uses a different algorithm. MoSS is written in Java and uses the LGPL license. MoSS reads STN and SMILES as input, which might not be optimal for all users, so a CDK port comes to mind.
### R/CDK install fails on GCC 4.0 systems
Some time ago Rajarshi Guha introduced R bindings for the CDK (see his CDK News articles), and today I tried to install his rcdk package that makes it happen.
However, it requires SJava which compiled fine on other machines, but not on my AMD64 machine. The problem seems to be related to the GNU GCC 4.0 compiler I have installed. Compiling with 3.4 works fine, but 4.0 complains with:
CtoJava.cweb:215: error: static declaration of 'std_env' follows non-static declarationCtoJava.cweb:195: error: previous declaration of 'std_env' was here
Googling, learned me that I am not the only one with this problem, but did not find any solution. If you know how to fix this problem, please leave a message in the comments.
## Tuesday, November 01, 2005
### The annual Lunteren meeting
Most Dutch chemists have their annual Lunteren meeting, so do I. Lunteren is a small village on the Veluwe where nothing much can be done, except for listening to the presentations. I participate in the Lunteren meeting for analytical chemists, i.e. HPLC, MS, GC and all their combinations upto and including HPLC/MS/MS, and since a few years the Lab-on-a-Chip stuff. And, as such, in many cases a lot of details on how to use and develop these methods.
For a computational chemist, this often is too much practical detail on too little -ics. Fortunately, the proteomics, genomics, etc is a strong upcoming funding subject, so data analysis is getting in their picture too. Which is good for someone with a chemometrics/chemoinformatics background as funding in that area is getting smaller every year.
My presentation went reasonable well, as far as I can tell myself. I was very nervous with both my professor and some 150 other people in the audience, but managed to not wander off the main topic. However, I was told to be a bit too monotone, but that's an unfortunate effect of being so nervous. |
## Thomas' Calculus 13th Edition
$\lt 1+\dfrac{\sqrt 3}{2},\dfrac{\sqrt 3}{2}, \dfrac{-1}{2} \gt$
In order to find the partial derivative, we will differentiate with respect to $x$, by keeping $y$ as a constant to find the x-coordinate of the gradient vector, and vice versa: $f_x=e^{x+y} \cos z +\dfrac{y+1}{\sqrt{1-x^2}} \\ f_y= e^{x+y} \cos z +arcsin x \\f_z=-e^{x+y} \sin z$ Write the gradient vector equation. $\nabla f = \lt f_x,f_y,f_z \gt$ $\implies \nabla f = \lt e^{x+y} \cos z +\dfrac{y+1}{\sqrt{1-x^2}} , e^{x+y} \cos z +arcsin x , -e^{x+y} \sin z \gt$ Thus, $\nabla f (0,0,\dfrac{\pi}{6}) = \lt 1+\dfrac{\sqrt 3}{2},\dfrac{\sqrt 3}{2}, \dfrac{-1}{2} \gt$ |
# Blank bibliography page getting created
I am writing an article in Lyx using a latex template. For some reason, I am getting a blank bibliography page (in addition to the actual bibliography) when I compile it. I deleted everything in the article except the bibliography part and I am still getting the extra page. Of course, I could just remove it after compiling, but it's annoying to have to do it for every draft.
I am able to reproduce the error with a bib file that contains the following text only:
@book{Eisenbud,
Author = {David Eisenbud},
Publisher = {Spring-Verlag},
Title = {Commutative Algebra with a view towards Algebraic Geometry},
Year = {1995}}
The following is the code.
\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{example}[theorem]{Example}
\newtheorem{introduction}[theorem]{Introduction}
\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\numberwithin{theorem}{chapter} % Numbers theorems "x.y" where x
% is the section number, y is the
% theorem number
%\renewcommand{\thetheorem}{\arabic{chapter}.\arabic{theorem}}
%\makeatletter % This sequence of commands will
%\let\c@equation\c@theorem % incorporate equation numbering
%\makeatother % into the theorem numbering scheme
%\renewcommand{\theenumi}{(\roman{enumi})}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%This command creates a box marked To Do'' around text.
%To use type \todo{ insert text here }.
\newcommand{\todo}[1]{\vspace{5 mm}\par \noindent
\marginpar{\textsc{ToDo}}
\framebox{\begin{minipage}[c]{0.95 \textwidth}
\tt #1 \end{minipage}}\vspace{5 mm}\par}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\makeatother
\usepackage{babel}
\begin{document}
\bibliographystyle{amsalpha}
\bibliography{thesisref}
\end{document}
-
I’m not sure if i understand what you mean. Do you get an empty bibliography (no entries) or do you get the desired bibliography but an additional blank page at it’s end? – Tobi Sep 26 '11 at 15:50
@Tobi: I get the desired bibiliography with the desired entried. However, there is a page before the actual bibliography with the word "Bibliography" in the center of the page and nothing else. – B.M. Sep 26 '11 at 15:53
OK. Then it would be great if you provide a minimal working example (MWE) (please include a sample bib via filecontens). In your above code I can’t finde the cause (btw. the \makeatother before loading babel can’t be right in this place since it should be used together with \makeatother …). – Tobi Sep 26 '11 at 16:22
@Tobi: Thanks for your help. I am not sure what "filecontens". I could not find it mentioned in the question you link either. As for the code, I haven't written it myself. I wrote the document in Lyx and exported it to latex, so I am not sure \makeatother does here. I have included a sample .bib file in my question above. – B.M. Sep 26 '11 at 16:34
Here you can find something about filecontents (it’s the third link in the above mentioned answer …). As Gonzalo said it’s nearly not possible to find the error with the given scrap of code. – Tobi Sep 26 '11 at 17:40
From the code you posted, it can be inferred that you are using a document class that uses chapters. Normally, \chapter and \chapter* internally issue a \cleardoublepage command to ensure that every chapter starts on an odd numbered page. If the last page before the bibliography was odd numbered, then the \chapter* used to typeset the bibliography heading will produce a blank page. To avoid this, you can locally redefine \cleardoublepage to behave as \clearpage, by using
\begingroup
\let\cleardoublepage\clearpage
\bibliographystyle{amsalpha}
\bibliography{thesisref}
\endgroup
If you want to allow all chapters starting in any new page (whether it is odd or even), you can use the openany class option. Assuming the book class is used, then you can say:
\documentclass[openany]{book}
-
Thank you for the answer, Gonzalo. I made both changes, both one at a time and together and I am still getting the blank page. – B.M. Sep 26 '11 at 17:25
@B.M.: then we'll need a complete sample of your code. The document class used is missing in the code you posted. – Gonzalo Medina Sep 26 '11 at 17:28
I have the same problem with the article class :( and don't know how to solve it either. It seems there are more people with the same problem: latex-community.org/forum/viewtopic.php?f=5&t=317 – Álvaro Aug 17 '13 at 16:49 |
punter
• n. 船夫，用篙撑船的人
• n. (Punter)人名；(英)庞特；(西)蓬特尔
«
1 / 10
»
punter 顾客，主顾，赌马的人
punter (n.)
1888 in football, agent noun from punt (v.).
1. Your average punter ( ie The ordinary uncultured person ) does not go to the opera.
2. Your average punter does not go to the opera.
3. You can write what you like, as long as it keep the punter happy.
4. Such a description aa euphemism for a club milkman, or a punter an SW 6 pub. |
Find Paper, Faster
Example：10.1021/acsami.1c06204 or Chem. Rev., 2007, 107, 2411-2502
The Hodge ring of varieties in positive characteristic
Algebra & Number Theory (IF0.938), Pub Date : 2021-05-20, DOI: 10.2140/ant.2021.15.729
Remy van Dobben de Bruyn
Let $k$ be a field of positive characteristic. We prove that the only linear relations between the Hodge numbers ${h}^{i,j}\left(X\right)=dim\phantom{\rule{0.3em}{0ex}}{H}^{j}\left(X,{\Omega }_{X}^{i}\right)$ that hold for every smooth proper variety $X$ over $k$ are the ones given by Serre duality. We also show that the only linear combinations of Hodge numbers that are birational invariants of $X$ are given by the span of the ${h}^{i,0}\left(X\right)$ and the ${h}^{0,j}\left(X\right)$ (and their duals ${h}^{i,n}\left(X\right)$ and ${h}^{n,j}\left(X\right)$). The corresponding statements for compact Kähler manifolds were proven by Kotschick and Schreieder. |
SolveTools[Inequality] - Maple Programming Help
Home : Support : Online Help : Mathematics : Factorization and Solving Equations : SolveTools : Inequality : SolveTools/Inequality/LinearUnivariateSystem
SolveTools[Inequality]
LinearUnivariateSystem
solve a system of linear inequalities with respect to one variable
Calling Sequence LinearUnivariateSystem(sys, var)
Parameters
sys - system of inequalities var - variable name
Description
• The LinearUnivariateSystem command solves a system of linear inequalities with respect to one variable.
• The LinearUnivariateSystem command returns a set describing the interval of possible values of the variable or a piecewise function of such sets depending on parameters
Examples
> $\mathrm{with}\left(\mathrm{SolveTools}[\mathrm{Inequality}]\right):$
> $\mathrm{LinearUnivariateSystem}\left(\left\{0
$\left\{\frac{{1}}{{2}}{<}{x}\right\}$ (1)
> $\mathrm{LinearUnivariateSystem}\left(\left\{0
$\left\{{-}{1}{<}{x}{,}{x}{<}\frac{{1}}{{2}}\right\}$ (2)
> $\mathrm{LinearUnivariateSystem}\left(\left\{x+1<0,2x-1<0\right\},x\right)$
$\left\{{x}{<}{-}{1}\right\}$ (3)
> $\mathrm{LinearUnivariateSystem}\left(\left\{x+1<0,0<2x-1\right\},x\right)$
$\left\{{}\right\}$ (4)
> $\mathrm{LinearUnivariateSystem}\left(\left\{x+a<0,0<2x-1\right\},x\right)$
${{}\begin{array}{cc}\left\{\frac{{1}}{{2}}{<}{x}{,}{x}{<}{-}{a}\right\}& {a}{<}{-}\frac{{1}}{{2}}\\ \left\{{}\right\}& {\mathrm{otherwise}}\end{array}$ (5)
> $\mathrm{LinearUnivariateSystem}\left(\left\{2x+a\le 0,0\le 2x-b\right\},x\right)$
${{}\begin{array}{cc}\left\{{x}{\le }{-}\frac{{1}}{{2}}{}{a}{,}\frac{{1}}{{2}}{}{b}{\le }{x}\right\}& {a}{\le }{-}{b}\\ \left\{{}\right\}& {\mathrm{otherwise}}\end{array}$ (6)
> $\mathrm{LinearUnivariateSystem}\left(\left\{ax+2\le 0,0\le 2x-b\right\},x\right)$
${{}\begin{array}{cc}\left\{{x}{\le }{-}\frac{{2}}{{a}}{,}\frac{{1}}{{2}}{}{b}{\le }{x}\right\}& {\mathrm{And}}{}\left({0}{<}{a}{,}\frac{{1}}{{2}}{}{b}{\le }{-}\frac{{2}}{{a}}\right)\\ \left\{{-}\frac{{2}}{{a}}{\le }{x}\right\}& {\mathrm{And}}{}\left({a}{<}{0}{,}\frac{{1}}{{2}}{}{b}{<}{-}\frac{{2}}{{a}}\right)\\ \left\{\frac{{1}}{{2}}{}{b}{\le }{x}\right\}& {\mathrm{And}}{}\left({a}{<}{0}{,}{-}\frac{{2}}{{a}}{\le }\frac{{1}}{{2}}{}{b}\right)\\ \left\{{}\right\}& {a}{=}{0}\\ \left\{{}\right\}& {\mathrm{otherwise}}\end{array}$ (7)
> $\mathrm{LinearUnivariateSystem}\left(\left\{ax+b<0,0\le cx+d\right\},x\right)$
${{}\begin{array}{cc}\left\{{x}{\le }{-}\frac{{d}}{{c}}\right\}& {\mathrm{And}}{}\left({0}{<}{a}{,}{c}{<}{0}{,}{-}\frac{{d}}{{c}}{<}{-}\frac{{b}}{{a}}\right)\\ \left\{{x}{<}{-}\frac{{b}}{{a}}\right\}& {\mathrm{And}}{}\left({0}{<}{a}{,}{c}{<}{0}{,}{-}\frac{{b}}{{a}}{\le }{-}\frac{{d}}{{c}}\right)\\ \left\{{-}\frac{{d}}{{c}}{\le }{x}{,}{x}{<}{-}\frac{{b}}{{a}}\right\}& {\mathrm{And}}{}\left({0}{<}{a}{,}{0}{<}{c}{,}{-}\frac{{d}}{{c}}{<}{-}\frac{{b}}{{a}}\right)\\ \left\{{x}{<}{-}\frac{{b}}{{a}}\right\}& {\mathrm{And}}{}\left({0}{<}{a}{,}{c}{=}{0}{,}{0}{\le }{d}\right)\\ \left\{{}\right\}& {\mathrm{And}}{}\left({0}{<}{a}{,}{c}{=}{0}{,}{d}{<}{0}\right)\\ \left\{{x}{\le }{-}\frac{{d}}{{c}}{,}{-}\frac{{b}}{{a}}{<}{x}\right\}& {\mathrm{And}}{}\left({a}{<}{0}{,}{c}{<}{0}{,}{-}\frac{{b}}{{a}}{<}{-}\frac{{d}}{{c}}\right)\\ \left\{{-}\frac{{d}}{{c}}{\le }{x}\right\}& {\mathrm{And}}{}\left({a}{<}{0}{,}{0}{<}{c}{,}{-}\frac{{b}}{{a}}{<}{-}\frac{{d}}{{c}}\right)\\ \left\{{-}\frac{{b}}{{a}}{<}{x}\right\}& {\mathrm{And}}{}\left({a}{<}{0}{,}{0}{<}{c}{,}{-}\frac{{d}}{{c}}{\le }{-}\frac{{b}}{{a}}\right)\\ \left\{{-}\frac{{b}}{{a}}{<}{x}\right\}& {\mathrm{And}}{}\left({a}{<}{0}{,}{c}{=}{0}{,}{0}{\le }{d}\right)\\ \left\{{}\right\}& {\mathrm{And}}{}\left({a}{<}{0}{,}{c}{=}{0}{,}{d}{<}{0}\right)\\ \left\{{x}{\le }{-}\frac{{d}}{{c}}\right\}& {\mathrm{And}}{}\left({a}{=}{0}{,}{b}{<}{0}{,}{c}{<}{0}\right)\\ \left\{{-}\frac{{d}}{{c}}{\le }{x}\right\}& {\mathrm{And}}{}\left({a}{=}{0}{,}{b}{<}{0}{,}{0}{<}{c}\right)\\ \left\{{x}\right\}& {\mathrm{And}}{}\left({a}{=}{0}{,}{b}{<}{0}{,}{c}{=}{0}{,}{0}{\le }{d}\right)\\ \left\{{}\right\}& {\mathrm{And}}{}\left({a}{=}{0}{,}{b}{<}{0}{,}{c}{=}{0}{,}{d}{<}{0}\right)\\ \left\{{}\right\}& {\mathrm{And}}{}\left({a}{=}{0}{,}{0}{\le }{b}\right)\end{array}$ (8) |
# In equilateral triangle,One vertex of a square is at the midpoint of the side, and the two adjacent vertices are on the other two sides of triangle
In the equilateral triangle $ABC,AB=12.$One vertex of a square is at the midpoint of the side $BC$, and the two adjacent vertices are on the other two sides of the triangle.Find the length of the side of the square.
Let $DEFG$ be the square.Let $D$ be the midpoint of the side $BC.BD=DC=6.$
Let $E$ be on the side $AC$ and $G$ be on the side $AB$ such that $AG=12-a,BG=a$ and $AE=12-b,EC=b$.
In triangle $DEC,$ applying Cosine law,
$\cos 60^\circ=\frac{1}{2}=\frac{b^2+36-DE^2}{12b}........(1)$
In triangle $DGB,$ applying Cosine law,
$\cos 60^\circ=\frac{1}{2}=\frac{a^2+36-DG^2}{12a}........(2)$
I am stuck here.
• You must have $BG=EC$, so side length is just $\frac{6\sin 60}{\sin 75}\approx 5.38$. – almagest Apr 27 '16 at 14:04
1) $\triangle ADB: \angle D=90^{\circ}, DB=6, AD=\sqrt{12^2-6^2}=\sqrt{108}, AB=12$
2) $DG -$ bisector $\angle ADB$
$$DG=\sqrt2 \frac{AD \cdot DB}{AD+DB}=\sqrt2 \frac{6\sqrt3 \cdot 6}{6\sqrt3+6}=\frac{6\sqrt6}{\sqrt3+1}=3\sqrt6(\sqrt3-1)$$
• You introduced a spurious $\sqrt6$ right at the end. – almagest Apr 27 '16 at 14:06
• @Roman83 the question seems to require that E and F be on the triangle sides – G Cab Apr 27 '16 at 14:09
• @almagest: Thank you! – Roman83 Apr 27 '16 at 14:19
• I want to ask one thing ,how did you know $AD$ is perpendicular to $BC$ and why $F$ came on $AD$?@Roman83 – Vinod Kumar Punia Apr 27 '16 at 14:21
• @GCab No. The two vertices adjacent to the vertex at $D$. Roman83's triangle is rotated! – almagest Apr 27 '16 at 14:21
Let $x$ be length of side of square.Triangles DBG and CDE are congruent because they have an angle and two corresponding sides which are equal.Implies BG=EC.In which case Angle BDG = $45^0$ and Angle BGD = $75^0$.Applying sine rule in Triangle BGD $$\frac{x}{\sin{60^0}}=\frac{6}{\sin{75^0}}$$ I hope this was helpful.
• A neat solution. – almagest Apr 27 '16 at 14:23
• By which similarity criteria,you proved triangles $DBG$ and $CDE$ similar.Because there are only three similarity criterias,$SSS,AAA,SAS$.@VarunKumar – Vinod Kumar Punia Apr 27 '16 at 14:41
• @Vinod Kumar Punia he appears to have used ASS congruence, which only exists in the case of right angles (as RHS). I think it Is a mistake.. – N.S.JOHN Apr 27 '16 at 15:19 |
## Some Uses of Duality in Linear Programming
For most of my students, the first time they see duality in linear programming their first reaction is “What’s the big deal?” There was a recent post on math.SE asking the same question. Duality in linear programming seems to me analogous to eigenvalues and eigenvectors in linear algebra, in the sense that at first the concept doesn’t seem like there’s much there, but as you learn more and more you realize how fundamental it is to the entire subject.
This post will describe some of the uses of duality in linear programming. There are others, of course. The post draws heavily on my answer to the math.SE question about the uses of duality.
1. Any feasible solution to the dual problem gives a bound on the optimal objective function value in the primal problem. This is how I motivate the dual problem in class, in fact. The formal statement of this is the weak duality theorem.
2. Understanding the dual problem leads to specialized algorithms for some important classes of linear programming problems. Examples include the transportation simplex method, the Hungarian algorithm for the assignment problem, and the network simplex method. Even column generation relies partly on duality.
3. The dual can be helpful for sensitivity analysis. Changing the primal’s right-hand side constraint vector or adding a new constraint to it can make the original primal optimal solution infeasible. However, this only changes the objective function or adds a new variable to the dual, respectively, so the original dual optimal solution is still feasible (and is usually not far from the new dual optimal solution).
4. Sometimes finding an initial feasible solution to the dual is much easier than finding one for the primal. For example, if the primal is a minimization problem, the constraints are often of the form $A {\bf x} \geq {\bf b}$, ${\bf x} \geq {\bf 0}$, for ${\bf b} \geq {\bf 0}$. The dual constraints would then likely be of the form $A^T {\bf y} \leq {\bf c}$, ${\bf y} \geq {\bf 0}$, for ${\bf c} \geq {\bf 0}$. The origin is feasible for the latter problem but not for the former.
5. The dual variables give the shadow prices for the primal constraints. Suppose you have a profit maximization problem with a resource constraint i. Then the value $y_i$ of the corresponding dual variable in the optimal solution tells you that you get an increase of $y_i$ in the maximum profit for each unit increase in the amount of resource i (absent degeneracy and for small increases in resource i).
6. Sometimes the dual is easier to solve. A primal problem with many constraints and few variables can be converted into a dual problem with few constraints and many variables. Fewer constraints are nice in linear programs because the basis matrix is an $n \times n$ matrix, where n is the number of constraints. Thus the fewer the constraints, the smaller the size of the basis matrix, and thus the fewer computations required in each iteration of the simplex method.
7. The dual can be used to detect primal infeasibility. This is a consequence of weak duality: If the dual is a minimization problem whose objective function value can be made as small as possible, and any feasible solution to the dual gives an upper bound on the optimal objective function value in the primal, then the primal problem cannot have any feasible solutions. |
# If a spring has a constant of 4 (kg)/s^2, how much work will it take to extend the spring by 63 cm ?
Feb 16, 2017
I got $0.8 J$
We will need to do work $W$ to extend the spring that, as a result, will acquire an Elastic Potential Energy equal to $\frac{1}{2} k {x}^{2}$:
$W = \frac{1}{2} k {x}^{2}$
$W = \frac{1}{2} \cdot 4 \cdot {\left(0.63\right)}^{2} = 0.7939 \approx 0.8 J$ |
## RecSys Related Algorithm - SVD
Introduction
If we want to predict the user A's rating of the book X, but we only have the A's rating for some other books and user B's rating of the book X. How can we predict the A's rating of the book X? The easiest way is to simply forecast as average. But we never know with accuracy.
SVD (Singular Value Decomposition) is based on the existing ratings, analysis the favorite degree of the raters for every factors, and get the ranks from analysis result at last. In the above example, there are many factors of the book, such as the cover, author, story, price and etc,.
SVD algorithms make a ranking matrix $R$ with $n$ rows and $m$ columns as abstract. $R[u][i]$ means that the rank of the object $i$ from user $u$. It can be decomposed into a user factors matrix $P$ with $n$ rows and $f$ columns ($P[u][k]$ means that rank of the factor $k$ from user $u$) and a object factors matrix with $m$ rows and $f$ columns ($Q[i][k]$ means that rank of the factor $k$ of object $i$). This can be represented by the formula like this:
There is an example for decomposed to two matrix. The larger $P$, represents more users prefer the book, larger $Q$, represents high factor degree. We can predict the user A's rating of the book X after decomposed.
+-------------+------+------+------+
|Rank Matrix R|Book X|Book Y|Book Z|
+-------------+------+------+------+
| User A | 6 | 3 | ? |
+-------------+------+------+------+
| User B | 3 | 2 | 6 |
+-------------+------+------+------+
|
+----------------------+------------------------+
| |
| v
| +-----------------------+--------+----------+
v |Object Factors Matrix Q|Computer|Literature|
+---------------------+--------+----------+ +-----------------------+--------+----------+
|User Factors Matrix P|Computer|Literature| | Book X | 6 | 0 |
+---------------------+--------+----------+ +-----------------------+--------+----------+
| User A | 1 | 0.2 | | Book Y | 3 | 3 |
+---------------------+--------+----------+ +-----------------------+--------+----------+
| User B | 0.3 | 1 | | Book Z | 0 | 6 |
+---------------------+--------+----------+ +-----------------------+--------+----------+
In addition to considering how the user like this book, but also affect by whether they be a strict raters and existing ratings when the user ranking a book in fact. Somebody will give high rank when they got this book has been rank as high value. The factor of how user like this book has been exists, we need to add two new factor to record that another parts to improve the accuracy of the model. After improved formula like this:
$OverallMean$ means that average rank of the all books, $biasU$ means that the deviation with $OverallMean$ of the user ranking, and $biasI$ means that the deviation with $OverallMean$ of the book ranks. $P$ and $Q$ meaning are unchanged and they are all matrices except the $OverallMean$.
After decompose, suppose we want to predict user $u$ rating for book $i$:
SVD Implement
Two decomposed matrices get by learning. SVD using stochastic gradient descent learning parameters except the $OverallMean$. The learning process can be summarized as this: initial value of each parameter, and then use these parameters to predict, and the predicted results were compared with known rates, adjust each various parameters based on comparison results at last. Adjustment the value of the parameter, making the following formula can take to the minimum:
$\alpha$ means that all the training samples, the first part of the parentheses represents the deviation of the current predictions and the actual value, the second part of the parentheses is to prevent overfitting.
That's the main ideas of SVD.
Reference
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Jaewoo Song
### Categories
• Tech
I found the data structure called “Priority Queue” in <queue> libaray.
This is literally a queue that contains data based on the priority, which is similar with a heap.
Let’s say that we use a priority queue for integer data.
#include <iostream>
#include <queue>
#include <vector>
using namespace std;
int main() {
priority_queue<int> q1;
priority_queue<int, vector<int>, less<int>> q2;
priority_queue<int, vector<int>, greater<int>> q3;
return 0;
}
We can say that q1 and q2 are the same structures.
When we declare priority_queue, we need the data type, structure and comparison operator as parameters.
If we specify only the type of data, which is int, then the structure becomes vector<int> and opertor is less<int>.
This gives more priority to bigger value by comparing in descending order.
If we want to make it in ascending order, we can put greater<int> to third parameter like in q3.
Now let’s see how the values are pushed and popped by exectuing below codes.
#include <iostream>
#include <queue>
#include <vector>
using namespace std;
int main() {
priority_queue<int> q1;
priority_queue<int, vector<int>, less<int>> q2;
priority_queue<int, vector<int>, greater<int>> q3;
vector<int> v = {3, 6, 1, 2, 0, 7};
for (int i=0; i<v.size(); ++i) {
q1.push(v[i]);
q2.push(v[i]);
q3.push(v[i]);
}
cout<<"q1: ";
while(!q1.empty()) {
cout<<q1.top()<<" ";
q1.pop();
}
cout<<"\n";
cout<<"q2: ";
while(!q2.empty()) {
cout<<q2.top()<<" ";
q2.pop();
}
cout<<"\n";
cout<<"q3: ";
while(!q3.empty()) {
cout<<q3.top()<<" ";
q3.pop();
}
cout<<"\n";
return 0;
}
We can see that if the operator is declared as descending order, the values are ordered from bigger to smaller and vice versa.
Another interesting thing is that originally we use front() to check the front value of a queue, but instead we should use top() for a priority queue.
This is because a priority queue is more similar with a heap than with a original queue.
And the time complexity is also bounded by that of a heap.
It takes $O(logN)$ to push or pop a value, so it can be useful for problems which should be approached with $O(NlogN)$ time complexity. |
Angular momentum in close binaries
1. Apr 22, 2004
Kurdt
Staff Emeritus
Does anybody know how the angular momentum of close binaries was calculated as:
$$L=M_sM_p(\frac{GD}{M})^\frac{1}{2}$$
This is something which is plaguing me so any input would be appreciated.
$$M=M_s+M_p$$, D= total seperation between the two stars.
Thanks guys
2. Apr 22, 2004
Labguy
Don't know if that formula always holds as there is almost always angular momentum loss in close binaries with either filled Roche Lobes or actual mass transfer. Too many types (variables) of binary stars to lump into one catagory. Take a look at: http://www.rri.res.in/ph217/binary.pdf and:
http://www-astro.physics.ox.ac.uk/~podsi/lecture11_c.pdf for a lot on this.
3. Apr 22, 2004
Kurdt
Staff Emeritus
Yeah I know the gravitational waves leak angular momentum away from the system and also stellar winds from the secondary trapped in magnetic loops. The momentum equation should hold as the radius of orbit changes when the angular momentum is leaked away thus allowing the secondary to fill its Roche lobe and maintain contact with the L1 point. I will have a look at the links though, thanks for replying.
4. May 13, 2004
Kurdt
Staff Emeritus
Ahh. Forgot I posted this. I managed to work it out and for anyone who was interested heres the derivation.
First we start with summing the $$\mathbf{v}\times \mathbf{r}$$ for each star in the system. This gives.
$$\mathbf{L_{orb}} = M_sr_sv_s\mathbf{e}_k+M-pr_pv_p\mathbf{e}_k$$
now $$v_s=r_s\Omega$$ and similarly for the primary also we can substitute $$r_s=\frac{M_p}{M}D$$ into the equation and again a similar relation is found for the primary to yield.
$$\mathbf{L_{orb}} =(M_s\frac{M^2_p}{M^2}D^2+M_p\frac{M^2_s}{M^2}D^2)\Omega\mathbf{e}_k$$
and manipulation gives
$$\mathbf{L_{orb}} =\frac{M_sM_p}{M^2}D^2\Omega(M_s+M_p)\mathbf{e}_k$$
$$(M_s+M_p)=M$$ and $$\Omega=(\frac{GM}{D^3})^{1/2}$$
So
$$\mathbf{L_{orb}} =\frac{M_sM_p}{M^2}D^2(\frac{GM}{D^3})^{1/2}\mathbf{e}_k$$
$$\mathbf{L_{orb}} =M_sM_p(\frac{GD}{M})^{1/2}\mathbf{e}_k$$
Last edited: May 13, 2004
5. May 18, 2004
Pinkline Jones
Kurdt,
The angular momentum of close binaries was calculated by a modest young diesel mechanic working in a sheltered workshop in Yorkshire. The young man - Winthrop Spencer Flibberdigit has since gone on to write some fascinating books released in sanskrit by "Absolutely No Frills and Assoc. Publishers". Here's just a short list of his books to date:
"The Peripatetic Life of the Sandwich Island Penguins"
"Low Orbital-Decibel ratios of Semi-Diametric Rolling Hub Caps"
"Desperate and Dateless Conversations - How My Electromagnetic Spectrum Theorem sent my Lover into a Coma"
"Dangers of The Spiral Galaxy - Fifty Bucks to Clean the Cab"
My pleasure...
DR PINKLINE JONES a.l.s.c.
6. May 19, 2004
Kurdt
Staff Emeritus
Thank you for your input Pinkline I shall certainly investigate these publications when I next visit the library. I have several exams in the coming three weeks though so I doubt I will be able to spare any time soon. The titles definitely look interesting.
Last edited: May 19, 2004
7. May 19, 2004
Kurdt
Staff Emeritus
Would you believe it! I managed to find one of his publications in my own personal library. "Dangers of The Spiral Galaxy - Fifty Bucks to Clean the Cab" is a most interesting journey through the derivation of angular momentum in the andromeda galaxy all on a wild night out. I particularly like the climax when he reaches his conclusions in the taxi on the way home. Also the last chapter about his abduction by aliens is well worth the read alone.
8. May 19, 2004
Pinkline Jones
LOL Kurdt - I'm glad you found the book and happy to be of assistance wherever I can in cyber world. But obviously I have to fine you for possessing a sense of humour - that's a very dangerous attribute to have in this mixed up world - people look at you funny.
PINKLINE JONES
9. May 20, 2004
Kurdt
Staff Emeritus
Well I may never have to show it again as I think I have learned a valuable lesson to read peoples posts thoroughly first time. So if you don't tell i won't. |
# Tag Info
8
Measuring $w$ is actually what I do for a living. The current best measurements put $w$ at $-1$ but with an uncertainty of $5\%$, so there's a little room for $w \ne -1$ models, but it's not big and getting smaller all the time. Indeed, we'd all be thrilled if, as measurements got more precise, $w \ne -1$ turns out to be the case because the $\Lambda$CDM ...
4
A "left over waste product" is matter, too, and it must be composed of something, of some particles allowed by the laws of physics. So labeling something as "waste" doesn't really answer any question about the identity of dark matter. Moreover, most of antimatter has annihilated with the ordinary matter into photons. In fact, any product of the annihilation ...
2
Dark energy is responsible for the acceleration of the universe at large scale, that is, it causes the second derivative of the cosmic scale factor $a(t)$ to be positive. But a smaller scales we found agglomerates of particles where the effect of the four forces (not only gravity) is much stronger that the repulsive effect of dark energy. A useful analogy ...
0
Most of it would've become radiation- mostly photons and some neutrinos. Some of that can be seen in the cosmic microwave background. The rest would've gone into the kinetic energy of remaining matter particles (i.e., heat). We don't actually know what dark matter is made of, but we do know what the reaction products of matter/anti-matter annihilation are ...
1
The word energy tends to be used in a rather vague way, and typically to mean something exotic. In the context of particle reactions energy either means photons or the kinetic energy of the particles leaving the reaction. For example an electron and anti-electron annihilate to produce two photons. By contrast the annihilation of a proton and anti-proton is ...
6
Is it because the acceleration is too weak? It is too weak with respect to the four forces we measure. The fact that the four known forces are so much stronger means that agglomerates of particles, up to the scale of galaxies are not internally affected, they keep their structure intact, like the famous raisins in the rising bread. It is only at the ...
5
The density of Dark Energy is not very high. In a place with lots of matter, the attractive forces of gravity are greater than the repulsive forces of dark energy. In mostly matter empty space, the repulsive forces of dark energy are much larger than the attractive forces of gravity.
0
Dark energy due to a cosmological constant does not get diluted by metric expansion of space. However, this does not violate energy conservation as the increase in energy will be cancelled by gravitational potential energy. The problem with general relativity is (some would say arguably) not energy conservation, but energy localization: In a rotating frame ...
2
The cosmological constant is a constant energy density per unit volume of space, so as the universe expands this does indeed create energy as it creates new space. In this sense conservation of energy is violated. Actually this is less surprising than you might think. Conservation of energy is linked to a symmery called time shift symmetry by Noether's ...
1
Actually, energy is often not Conserved in general Relativity. For are more in deep explanation see: http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html But just notice that Dark Energy might not necessarily end up being the cosmological constant, but a new force field, so its behavior might differ from that of an actual cosmological ...
3
Forgive me in advance, this may get overcomplicated. I am going to give you the facts as scaled down as I can but still sufficiently detailed. I think providing you with what we have and allowing you to infer from it is the best way to avoid misrepresenting the answer. Here is the General Relativity equation that describes how gravity interacts with ...
4
I am going to answer the following questions: 1) Does dark energy have gravity? 2) Why does dark energy have constant density during the expansion of the universe? First about the term of dark energy - it is more of a label for many attempts of a solution to the problem of an accelerated expansion of the universe. One of the most promising (or say ...
Top 50 recent answers are included |
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Why should we conserve forests and wildlife?
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## QuestionScienceClass 10
Why should we conserve forests and wildlife?
We should conserve forest and wildlife because of the following reasons:
$$1$$ Forest provides us with oxygen.
$$2$$ Forest causes rain.
$$3$$ Forest prevents soil erosion.
$$4$$ Destruction of forest would lead to destruction of wildlife.
$$5$$ Wildlife are also the members of food web and any gaps in the food web would lead to the destruction of every surface or organism on earth. |
# Question 30 (4 points) Capital Profit Loss Partners Balances Sharing Scotty I$400,000 25% Riley ... Question 30 (4 points) Capital Profit Loss Partners Balances Sharing Scotty I$ 400,000 25% Riley I$700,000 45% Dee Dee 900,000 30% Marnee is going to buy into the SRD Partnership by paying$1,000,000 to the partners in exchange for a 30% ownership in the partnership. The partnership uses the Book Value Method when admitting new partners. Answer the following questions using the information in the table above. 1. What is Marnee's balance after joining the partnership? 2. What is Scotty's's balance after Marnee joins the partnership? abo
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## Elementary Geometry for College Students (5th Edition)
Published by Brooks Cole
# Chapter 2 - Section 2.4 - The Angles of a Triangle - Exercises: 38
#### Answer
m$\angle$CAB=57$^{\circ}$ m$\angle$CBA=33$^{\circ}$
#### Work Step by Step
m$\angle$C+m$\angle$CAB+m$\angle$CBA=180 90+24+x+x=180 2x=66 x=33$^{\circ}$ m$\angle$CBA=x=33$^{\circ}$ m$\angle$CAB=24+x=57$^{\circ}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. |
numerically approximate the real roots of an expression using Steffensen's method - Maple Help
Home : Support : Online Help : Education : Student Package : Numerical Analysis : Visualization : Student/NumericalAnalysis/Steffensen
Student[NumericalAnalysis][Steffensen] - numerically approximate the real roots of an expression using Steffensen's method
Calling Sequence Steffensen(f, x=a, opts) Steffensen(f, a, opts)
Parameters
f - algebraic; expression in the variable x representing a continuous function x - name; the independent variable of f a - numeric; the initial approximate root opts - (optional) equation(s) of the form keyword=value, where keyword is one of fixedpointiterator, functionoptions, lineoptions, maxiterations, output, pointoptions, showfunction, showlines, showpoints, stoppingcriterion, tickmarks, caption, tolerance, verticallineoptions, view; the options for approximating the roots of f
Description
• The Steffensen command numerically approximates the roots of an algebraic function, f, using fixed-point iteration coupled with a slightly modified version of Aitken's ${\mathrm{\Delta }}^{2}$ technique of accelerating sequential convergence.
• Given an expression f and an initial approximate a, the Steffensen command computes a sequence ${p}_{k}$, $k$=$0..n$, of approximations to a root of f, where $n$ is the number of iterations taken to reach a stopping criterion. For sufficiently well-behaved functions and sufficiently good initial approximations, the convergence of ${p}_{k}$ toward the exact root is quadratic.
• The first argument f may be substituted with an option of the form fixedpointiterator = fpexpr. See Notes.
• The Steffensen command is a shortcut for calling the Roots command with the method=steffensen option.
Notes
• This procedure first converts the problem of finding a root to the equation $f\left(x\right)=0$ to a problem of finding a fixed point for the function $g\left(x\right)$, where $g\left(x\right)=x-f\left(x\right)$ and $f\left(x\right)$ is specified by f and x.
• The user can specify a custom iterator function $g\left(x\right)$ by omitting the first argument f and supplying the fixedpointiterator = g option. The right-hand side expression g specifies a function $g\left(x\right)$, and this procedure will aim to find a root to $f\left(x\right)$ = $x-g\left(x\right)=0$ by way of solving the fixed-point problem $g\left(x\right)=x$.
When output = plot or output = animation is specified, both the function $f\left(x\right)$ and the fixed-point iterator function $g\left(x\right)$ will be plotted and correspondingly labelled.
The tolerance option, when stoppingcriterion = function_value, applies to the function $f\left(x\right)$ in the root-finding form of the problem.
Examples
> $\mathrm{with}\left(\mathrm{Student}[\mathrm{NumericalAnalysis}]\right):$
> $f:=\frac{{x}^{2}}{3}-1:$
> $\mathrm{Steffensen}\left(f,x=2.0,\mathrm{tolerance}={10}^{-2}\right)$
${1.732049797}$ (1)
> $\mathrm{Steffensen}\left(f,x=2.0,\mathrm{tolerance}={10}^{-2},\mathrm{output}=\mathrm{sequence}\right)$
${2.0}{,}{1.727272728}{,}{1.732049797}$ (2)
> $\mathrm{Steffensen}\left(f,x=2,\mathrm{output}=\mathrm{plot},\mathrm{stoppingcriterion}=\mathrm{function_value}\right)$
> $\mathrm{Steffensen}\left(f,x=1.3,\mathrm{output}=\mathrm{animation},\mathrm{stoppingcriterion}=\mathrm{absolute}\right)$ |
# Compressive sensing: numerical generation of RIP matrices
The restricted isometry property (RIP) states that: $$(1-\delta_K)||x||_2^2 \le ||A x||_2^2 \le (1+\delta_K)||x||_2^2$$ for any $K$-sparse vector $x$ of length $N$. The corresponding restricted isometry constant is $\delta_K$, $0 < \delta_K < 1$.
I would like to perform a numerical experiment based this expression to produce a RIP-compliant matrix $A$ (Matlab code is supplied below). The matrix is constructed with Gaussian-distributed RV's (zero mean, variance $1/\sqrt{M}$). Having specified $N$ and $K$, the number of measurements $M$ is obtained with $M \geq K \log(N/K)$. This is a hit-or-miss procedure, I let the code run until the isometry constant falls between zero and one. Is this the proper way to generate $A$? Am I missing anything? Further analysis on $A$ (eigenvalues, etc.) doesn't seem to behave properly...
function [delta] = RIP_numerical(n,s)
% delta: restricted isometry constant; n: # columns; s: sparsity
% choose m (# measurements/rows) based on measurement criterion
m = ceil(s*log(n/s))
flag = 0; incr = 0; % initialize computing parameters
while flag ~= 1 % run until condition met
incr = incr + 1
% generate random, m x n normalized matrix N(0,1/m)
A = (1/sqrt(m))*randn(m,n);
sumA = sum(A);
for k = 1:n
A(:,k) = A(:,k)/sum(A(:,k));
end
% generate vector of length n, sparsity k
x = zeros(n,1);
list = randperm(n);
for l = 1:s
x(list(l))= randn;
end
y = A*x; % obtain measurements
norm_x = norm(x,2); % l_2 norms
norm_y = norm(y,2);
delta = 1-(norm_y^2/norm_x^2); % look at lower bound
if (delta >0 & delta < 1) % condition met; save vars and kick out of routine.
flag = 1
save('RIP_config.mat', 'A','x','n','m','s');
end
end
You can't prove RIP through numerical exploration of all possible cases. If you are interested in numerical analysis I suggest to use Coherence instead, however Coherence is not as strong condition as RIP. Note that RIP is a tool which is used for Mathematical proofs. I explain how to check for RIP, in case your are interested. As I said Coherence is a weaker condition in comparison with RIP, however, it is numerically feasible to check. Coherence of matrix A (which is sensing matrix multiplied in sparsity basis) is defined as below:
where $a_i, a_j$ are columns of matrix A, and $u_A$ is coherence factor. Its computational complexity is of $O(n^2)$ which in many cases is feasible. For checking coherence, you simply need to compute the given type of correlation between each two different columns of A matrix and among those values chose the max. Based on Welsh Inequality we'll have:
In CS sampling $M<<N$, so the above relation simplifies to $u_A>sqrt(1/M)$, therefore, to satisfy RIP using coherence, number of measurements ($M$) must be of order of $O(K^2)$ where $k$ is number of non-zeros. This number of measurements is so greater than $Klog(N/K)$.
Another conditions instead of RIP, is stRIP which is much much simpler to prove.
Ref: [A.Amini,"Deterministic Compressed Sensing" PhD Thesis, Sharif University of Technology, Iran, 2011]
Unfortunately, you cannot test for RIP this way. You calculate delta for one random s-sparse vector. The RIP condition must hold for all possible such vectors. In principle you can calculate it via the singular value decomposition, but you will have to do so for all possible sub-matrices of $A$ with $K$ or fewer columns from the original $A$. Evidently, this quickly becomes an insanely large and computationally intractable number of combinations for all but the very smallest (and practically irrelevant) matrices. Testing the RIP of a particular matrix $A$ is hopeless (I have tried - just to get a feel of what it would take).
If you are interested in a measure you can realistically calculate, have a look at the coherence of $A$.
In my opinion, the RIP is not that useful a measure in practice. For example, have a look at Donoho & Tanner, Precise Undersampling Theorems. Figure 6a illustrates the so-called phase transition performance implied by the RIP compared to the one they derive in said paper. The polytope-based phase transition derived in the paper far more accurately reflects the practical performance you are going to experience with compressed sensing reconstruction.
• While the mutual coherence is much easier to calculate, the bounds it leads to are very pessimistic. Mudjat Cetin has looked at developing other measures based on mutual coherence. I don't have the references handy right now. – David Apr 15 '16 at 23:05 |
Lemma 28.26.14. Let $X$ be a scheme. Let $\mathcal{L}$ be an ample invertible $\mathcal{O}_ X$-module. Let $i : X' \to X$ be a morphism of schemes. Assume at least one of the following conditions holds
1. $i$ is a quasi-compact immersion,
2. $X'$ is quasi-compact and $i$ is an immersion,
3. $i$ is quasi-compact and induces a homeomorphism between $X'$ and $i(X')$,
4. $X'$ is quasi-compact and $i$ induces a homeomorphism between $X'$ and $i(X')$.
Then $i^*\mathcal{L}$ is ample on $X'$.
Proof. Observe that in cases (1) and (3) the scheme $X'$ is quasi-compact as $X$ is quasi-compact by Definition 28.26.1. Thus it suffices to prove (2) and (4). Since (2) is a special case of (4) it suffices to prove (4).
Assume condition (4) holds. For $s \in \Gamma (X, \mathcal{L}^{\otimes d})$ denote $s' = i^*s$ the pullback of $s$ to $X'$. Note that $s'$ is a section of $(i^*\mathcal{L})^{\otimes d}$. By Proposition 28.26.13 the opens $X_ s$, for $s \in \Gamma (X, \mathcal{L}^{\otimes d})$, form a basis for the topology on $X$. Since $X'_{s'} = i^{-1}(X_ s)$ and since $X' \to i(X')$ is a homeomorphism, we conclude the opens $X'_{s'}$ form a basis for the topology of $X'$. Hence $i^*\mathcal{L}$ is ample by Proposition 28.26.13. $\square$
## Comments (2)
Comment #5468 by Dat Pham on
Can some explain to me why do we need the condition $i(X')\subseteq X$ is locally closed? I think knowing that $i$ induces a homeomorphism between $X'$ and $i(X')$ is already sufficient to deduce that the open sets $X'_{s'}=X'\cap X_s$ form a basis for the topology of $X'$.
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# JUSTICE & INJUSTICE — two 2-player finite games
There is a non-empty finite set $$\ K,\$$ say, of plates. Initially, there are $$\ p_0(k)\$$ stones on the $$k$$-th plate, where $$\ p_0(k)\in\mathbb Z_{_{\ge0}}\$$ for each $$\ k\in K.$$
So far, it is like a NIM game, and even the moves look similar. Namely, a move amounts to removing a positive number of stones from an arbitrary single plate, i.e. the $$\ n$$-th move creates a function $$\ p_n:K\to\mathbb Z_{_{\ge0}}\$$ such that $$\ p_n(k)=p_{n-1}(k)\$$ for every plate $$\ k\in K$$ but for one plate $$\ \kappa_n\in K\$$ for which $$\,\ 0\le p_n(\kappa_n)
But now, we diverge from NIM games. The game is finished the moment $$\ p_n\$$ is a constant function; since then, there are no more legal moves.
In JUSTICE game, the winner is the one who played the last move.
In INJUSTICE, the winner is the first player who cannot make a legal move.
REMARK If $$\ p_0\$$ is the constant $$0$$-function then the first payer won INJUSTICE while the second player won JUSTICE.
Question Who wins which game (as a function of $$\ |K|\$$ and $$\ p_0)?$$
After I created the games JUSTICE & INJUSTICE, I posted them on day 2007-06-14, on alt.pl.matematyka:
https://alt.pl.matematyka.narkive.com/Nzi0PiPA/justice-and-injustice-new-games-created-by-wlod-wh
Here is a complete winning strategy for the Justice game.
One wins the Justice game simply by following the usual Nim strategy, with all the same winning positions and moves (except if the position is already constant and with an odd number of piles).
The usual winning Nim strategy is to create a Nim-balanced position, balanced in the sense that if each pile is represented as a sum of distinct powers of two, then each power of two occurs an even number of times overall for the position. The main Nim observation is that if a position is unbalanced, then there is a balancing move, and if a position is balanced, every move is unbalancing.
The key observation for the Justice game is that one can never move from a balanced Nim position to a winning Justice position.
If the number of plates is even, this is clear, since any constant position with an even number of plates is also Nim balanced, but every move on a balanced Nim position will unbalance it.
If the number of plates is odd, then to move to a constant position, one must have reduced a tall pile to match the constant height. Since the rest of the position would have an even number of same-height piles, that part would be balanced by itself, and so if the whole position had been balanced before, the tall pile must have had height 0, impossible. (Thanks to a comment of Edward Lockhart on Twitter.)
So by playing the Nim balancing moves, one will win Justice.
• Very nice. Thank you. #### A time ago I introduced and solved a game that simultaneously generalizes a NIM generalization and games such as reaching a number, say 100, when increasing it from 0 but not more than by a constant. It was still by using powers of 2. But this time, for J. somehow this didn't occur to me -- I guess I wanted this game to be genuinely different from NIM but no such luck in the even case, as you have proven indeed. Aug 3 at 12:00
• Somehow, despite my earlier enthusiasm, now I have some doubts. When all plates store the same number of stones by one that has a higher number of them, then the balancing doesn't matter -- the player on the move wins regardless of the balance situation. Aug 3 at 21:43
• In the odd case, that situation will not arise, as I argue in my answer, since the position you describe from which the win would be obtained would not have been balanced. And in the even case, that situation does not arise, since the opponent would have achieved a balanced position, which is impossible for them. Aug 3 at 23:02 |
# Mathematical Research at the University of Cambridge
In mirror symmetry, points in a variety correspond to Lagrangian tori in the mirror symplectic manifold. In the absence of "quantum corrections" the Ext-algebra of a point is equated with the cohomology algebra of the corresponding torus - both are simply exterior algebras - but in general one has to consider deformations of this picture. In this talk I'll introduce the localised mirror functor of Cho, Hong and Lau, which translates deformations of this sort into the algebro-geometric language of matrix factorisations, and show how this leads to an easy proof that local mirror symmetry near a (monotone) Lagrangian torus is essentially tautological.
29Jan
Jan 29th 2020
14:15 to 15:15
CMS MR13
## Speaker:
Jack Smith, University of Cambridge
## Series:
Algebraic Geometry Seminar |
# which SMITH torch
### Help Support HomeBuiltAirplanes.com:
#### kurtjfred
##### Active Member
Not getting any help from the local welding shops around here on selecting which SMITH torch to weld all the 4130 tubing on a Breezy fuselage. Any of you guys that could offer some advice from experience would be greatly appreciated. Thanks
#### wally
##### Well-Known Member
I just looked at the Smith welding website. The only one that looks small enough is the AW1A model. It has the knobs up by where the tip connects. On the Victor style torches, the knobs are down by where the hoses conect. If you don't clock the handle and tip to turn the knobs away from your wrist, the knobs are next to your wrist and can get bumped easily.
I think that the old Smith torch that everyone talks about is no longer made. I saw one one time. It is little bitty. They or someone used to make lightweight hoses which would help on the weight and drag as you move the torch around.
You really can get along just fine with a Victor style torch if you can't find a Smith you like.
And for all you on a budget, you can buy a complete set: torch, tips, cutting attachment, hose, regulators, goggles, spark lighter, in a case for around $100 if you shop around. Yes it is probably made offshore But the el-cheapo set I have still works fine. The Smith torch handle alone is about$150. Then you still have to buy each size tip you will need. Get a "OO", "O" "1" and "2". Most sets do not go down to the "OO" size. For real thin stuff, the "OO" makes it easier. For clusters, you will need a 1 or a 2 to get enough heat in the joint to weld it.
Then just practice, practice, practice.
Oh, the popping that sometimes happens is most likely caused by a dirty tip. Use a cleaning tool to make sure it is clean. As the gasses come out of the little hole, they cool the copper tip. Except the very end does get hot as it is used and carbon builds up inside the little hole. When the tip get hot, the hot bits of carbon try to ignite the gasses back inside the little hole insead of the flame being just outside of the hole. So it pops! And splatters the molten weld everywhere.
Wally
#### wec502
##### New Member
Yes, the AW1A is the correct one. To get started, all you need is the 201 and 203 tips. You'll need the 205, too, but not very often. You may want to pick up two 201's and drill one out to the .069 drill size. For welding tips, size matters .
For a legitimate reference, you should look at the 43.13 under welding and see the tip drill sizes for the metal thickness you will be using. Plan on between .025 and .125 thickness for the majority of the work. The Smith site should confirm their tip size to drill size.
AW1A is the standard aircraft welding set-up from Smith. That is what I use and it is a good product. However, just for giggles, look at their jewelers torch. I noticed that the tip sizes (drill sizes) are right in line with aircraft welding. It is real cheap and looks like the thing to use because its small size and the flexible hose. A lot of people like another brand of torch because of it's small size and flexible hose. This looks a lot like that one and I'll bet it is a similar product at about a quarter of the cost. I think Smith just markets that torch incorrectly by calling it a jewelers torch. I have considered buying it, but the local welding shop doesn't stock it and I want it to see before I buy.
Now that I have said that, I live less than two miles from the Smith factory, I'll go down there and see if they'll show me that torch. I'll re-post later this week if I think it is something to consider.
#### 57Marty
##### Well-Known Member
I bought the Harris model 15-3 set up at Oshkosh a couple of years ago. It is the same light weight small handle and tips used for the welding forums taught at Oshkosh. I have their tips 1-5 and you learn what tips work best with the different thicknesses you are welding. I did about half my fuselage with a big Victor with heavy hoses. The pain in my arm after hours of welding was intense. No pain with the Harris at all. I think the Harris and the Smith are comparable, great touches.
Marty
#### wec502
##### New Member
OK, I just got back from the Smith factory inquiring about their jewelers "Mini Torch" for aviation use. It looks like a real handy item for around the shop, but the tips are probably to dainty and short for some of the welding you'll do. The orifice ranges do fall into the sizes we use, but the tip is real small and short. The guy there didn't think there would be enough thermal mass in the tip, so it wouldn't last very long. Also, being that short may preclude you from welding things like clusters that have a lot of reflected heat. It would be great for a lot of stuff, though.
Back to the AW1A: Get the AW1A with the AW201, 203, and 205 tips. If you don't have the hose yet, look at their brazing hose: Part no. 14779-4-10. The standard 3/16" torch hose at the welding store is too heavy and stiff for airframe work. |
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# 13.4: When Should You Conduct Post-Hoc Pairwise Comparisons?
The pairwise comparison calculations for a factorial design are the same as any pairwise comparison after any significant ANOVA. Instead of reviewing them here (because you can review them in the prior two chapters), we are going to discuss when (and why) you would or would not conduct pairwise comparisons in a factorial design.
## Long Answer
We’ll start by refreshing our member on what we’ve done before. Let’s start with t-test, from oh-so long ago!
##### Exercise $$\PageIndex{1}$$
Did we conduct pairwise comparisons when we retained the null hypothesis when comparing two groups with a t-test? Why or why not?
Answer
We did not conduct pairwise comparisons when the null hypothesis was retained with a t-test. We didn’t need to find which means were difference because the null hypothesis (which we retained) says that all of the means are similar.
##### Exercise $$\PageIndex{2}$$
Did we conduct pairwise comparisons when we rejected the null hypothesis when comparing two groups with a t-test? Why or why not?
Answer
I know that it was a long time ago, but no, we did not conduct pairwise comparisons with t-tests. Even when we rejected the null hypothesis (which said that the means were similar, so we are saying that they are probably different), we only had two means. The t-test was our “pairwise” comparison. In other words, because there were only two means, so we knew that if the means were statistically different from each other that the bigger one was statistically significantly bigger.
What about an ANOVA that compared three groups? To answer these questions, it doesn’t matter if the ANOVA was BG or RM, just that there one was IV with three (or more) groups.
##### Exercise $$\PageIndex{3}$$
Did we conduct pairwise comparisons when we retained the null hypothesis when comparing three groups with an ANOVA? Why or why not?
Answer
No, we did not conduct pairwise comparisons with ANOVAS with three groups if we retained the null hypothesis. With any retained null hypothesis, we are agreeing that the means are similar, so we wouldn’t spend time looking for any pairs of means that are different.
##### Exercise $$\PageIndex{4}$$
Did we conduct pairwise comparisons when we rejected the null hypothesis when comparing three groups with an ANOVA? Why or why not?
Answer
Yes, this is when we would conduct pairwise comparisons. The null hypothesis says that all of the means are similar, but when we reject that we are only saying that at least one mean is different from one other mean (one pair of means differs). When we have three or more groups, we need to figure out which means differ from which other means. In other words, a significant ANOVA shows us that at least one of the means is different from at least one other mean, but we don’t know which means are different from which other means. We have to do pairwise mean comparisons to see which means are significantly different from which other means.
Finally, on to factorial designs! If you’ve been answering the Exercises as you go, these should be pretty easy.
##### Exercise $$\PageIndex{5}$$
Do we conduct pairwise comparisons when we retain the null hypothesis for main effects in a factorial design? Why or why not?
Answer
No. When we retain a null hypothesis, we are saying that all of the means are similar. Let’s not waste time looking for a difference when we just said that there wasn’t one.
Okay, this one might be a little challenging, so we'll walk through it together.
##### Example $$\PageIndex{1}$$
Do we conduct pairwise comparisons when we reject the null hypothesis for main effects in a factorial design?
Solution
It depends!
If we only have two means, we don’t have to conduct pairwise comparisons because (just like with a t-test) rejecting the null hypothesis for the main effect means that we know that the bigger mean is statistically significantly bigger.
But if our IV has more than two groups, then we would need to conduct pairwise comparisons (just like an in ANOVA) to find which means are different from which other means.
Back to an easier one on null hypotheses and post-hoc tests.
##### Exercise $$\PageIndex{6}$$
Do we conduct pairwise comparisons when we retain the null hypothesis for an interaction in a factorial design? Why or why not?
Answer
No. The null hypothesis says that all of the means are similar. If we retain the null hypothesis, then we are saying that all of the means are probably similar. Why would we look for a difference between pairs of means that we think are similar?
This one should be clear if you understand the reasoning for when we do and do not conduct post-hoc pairwise comparisons.
##### Example $$\PageIndex{2}$$
Do we conduct pairwise comparisons when we reject the null hypothesis for an interaction in a factorial design? Why or why not?
Solution
Yes! The smallest factorial design is a 2x2, which means that we have our means representing the combination of the two IVs. Rejecting the null hypothesis for the interaction says that at least one of those means is different from at least one other mean. We should use pairwise comparisons to find which combination of IV levels has a different mean from which other combination.
## Short Answer
Table $$\PageIndex{1}$$- Short Answer for When to Conduct Post-Hoc Pairwise Comparisons
Only Two Groups Three or More Groups or Two or More IVs
Retain the Null Hypothesis No- means are similar No- means are similar
Reject the Null Hypothesis No- The bigger group is statistically bigger Yes- Find which mean is different from each other mean by comparing each pair of means.
Time to practice next!
13.4: When Should You Conduct Post-Hoc Pairwise Comparisons? is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Michelle Oja.
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# Name for a particular subgroup of parabolic subgroups of the general linear groups. [duplicate]
Let $V$ be vector space. The subgroup $P$ of $GL(V)$ consisting of all automorphisms stabilizing a flag $V=V_1\supset V_2\supset\cdots\supset V_1$ is called a parabolic subgroup of $GL(V)$. I am interested in the subgroup $Q$ of $P$ consisting of all automorphisms $h$ such that the induced automorphism $\overline{h}: V_i/V_{i+1}\rightarrow V_i/V_{i+1}$ is the identity for every $i$. My question is:
Is this subgroup $Q$ named somewhere yet? If not, can you recommend a name?
Similarly, I am also interested in the subgroup $T$ of $P$ consisting of automorphisms $h$ such that the group of induced automorphisms $\overline{h}: V_i/V_{i+1}\rightarrow V_i/V_{i+1}$ is the symmetric group (on a basis of $V_i/V_{i+1}$). Is this subgroup $T$ named somewhere yet? If not, can you recommend a name? Is there anyway to realize that $\overline{h}: V_i/V_{i+1}\rightarrow V_i/V_{i+1}$ is a symmetric group without looking a specific basis of $V_i/V_{i+1}$)?
Finally, I would appreciate very much if you have any reference on the study of these subgroups.
-
## marked as duplicate by Marc Palm, Ian Agol, Bugs Bunny, S. Carnahan♦Nov 26 '12 at 22:20
$Q$ is called the unipotent radical of $P$. – David Helm Nov 26 '12 at 16:33
I denote it $Rad(P)$. – Allen Knutson Nov 26 '12 at 21:56
As David says $Q$ is the unipotent radical of $P$. The subgroup $T$ is a preimage of the Weyl group $W$ of the group $G_i\cong GL(V_i/ V_{i+1})$. This group $T$ looks a direct product of $Q$ with a big chunk of a Levi complement of $P$. The Levi complement is a direct product isomorphic to $G_1\times\cdots \times G_k$; to obtain the group $T$, you replace the $i$-th factor by the normalizer $N$ of a maximal split torus $T_0$ of $G_i$.
This is, in fact, the typical way to realize the Weyl group of $G_i$ -- $W$ is isomorphic to the quotient $N/T_0$ -- but this is effectively the same thing as your method of fixing a specific basis of $V_i/V_{i+1}$. The Weyl group rears its head in lots of different ways (most especially as a Coxeter group related to the Dynkin diagram of $G_i$) so this is certainly not the only way to realise it. I don't, however, see any other way to realise your group $T$ (although it depends what you mean by `realise'!).
As for references, it depends on what kind of approach you want. If you want a treatment of $GL_n$ as an algebraic group then I recommend anything by Carter or Humphreys, or else there is the book by Borel. All of these people work in much greater generality than $GL_n$ though. If you just want to understand $GL_n$, then standard algebra texts like the one of Jacobson might be your best bet. (I have e-copies of some of these. If you want them, email me.) |
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# How do computers do math?
This is part 3 in a quest to answer the question “How do computers do math?”. In this chapter we introduce the different types of electronic circuits.
## Electronic circuits
In electronic circuits, another electric signal controls the flow of electrons. This electrical input signal typically represents physical properties such as sound, vision or temperature. Converting these physical properties to electrical signals, allows them to be easily manipulated and transported.
We distinguish three types of electronic systems: analog, digital and microprocessor systems.
### Analogsystems
In analog systems, a continuously variable voltage on a wire has a proportional relationship to a physical property as illustrated on the right. The word “analog” comes from the Greek word ανάλογος (analogos) meaning “proportional”. For example, a voltage of -1 to 1 volt may represent a sound pressure of -15 to 0 dBa.
In an old fashion telephone, a battery provides the energy. The carbon powder in the microphone controls the current, as the vibrations from the voice causes the resistance to fluctuate. Copper wires transport the analog signal. Eventually, on the other end, a loudspeaker recreates the sound.
Analog systems consist of components such as batteries, resistors, capacitors, diodes and transistors. They first became viable with the introduction of the vacuum tube (Fleming, 1904; de Forest, 1907) and later the semiconductor diode (Ohl, 1939) and transistor (Bardeen, Brattain and Shockley, 1947).
If you interested in the history, consider watching the PBS documentary Transistorized! (the original video vanished, but it can still be found on YouTube) or the Silicon Valley American Experience.
The thermal behavior of the components and the transmission of the analog signal introduce noise. Reducing noise and thereby improving accuracy is complicated and expensive in analog systems. The inherent inaccuracy was acceptable for telephony, but not for precision tools such as calculators. As we will see next, digital systems can offer this precision.
### Digital systems
In digital systems, there are only two voltages. One voltage corresponds to a logic 1, and the other to logic 0. These systems are binary after the Latin word bīnārius that stands for “consisting of twos”. Digital systems consist of circuits build using digital gates. Internally the gates use analog components, but the input and output are digital, so only logic 1 and logic 0.
Digital systems became viable with the introduction of the transistor. Later with the invention of integrated circuits, digital systems became commonplace (Kilby and Noyce, 1959).
By combining a set of $$n$$ wires and assigning them weighted values, we can transmit $$2^n$$ binary values. In the above example with four wires, a mass from 0 to 100 kg may be represented by a 4-bit binary signal from 0000 to 1111. That implies that 16 different values can be transmitted, decimal 0 to 15.
All voltages within a range represent the same digital value. As shown in the figure above, TTL-based logic recognizes an input value of 0 … 0.8 volt as logic 0, while 2 … 5 volt is recognized as logic 1. This implies that small changes in analog signal levels (noise) will be ignored allowing transmission over long distances without noise degradation.
The number of bits used limits the precision of digital systems. Digital systems can be very precise, but have to be custom designed for each task. During World War II, the Nazis and allied forces, used electro-mechanical systems and later vacuum tube based calculators to calculate artillery-firing tables. These systems were fast, but it took great effort to re-purpose the machines. Microprocessor systems add flexibility to digital systems.
### Microprocessor systems
Since 1945, the von Neumann architecture forms the foundation for modern day computers. It allows the same device to perform a wide variety of functions. One moment it can be a scientific calculator, another a magazine reader of be used to call your cousin.
A programmable device takes digital input signals, processes them according to instructions stored in its memory and outputs results. The programmable device is a Central Processing Unit (CPU).
Build using digital gates, microprocessors came to life with the invention of integrated circuits and spread rapidly once the whole CPU fit on one piece of silicon (Intel and Texas Instruments, 1971). This architecture is described in more detail in our follow-up inquiry “How do Microprocessors Work?.
In the following chapters we will introduce the building blocks of digital systems, and go in-depth explaining the physics of its components. The next chapter starts with an introduction to digital logic and shows an implementation of diode based logic.
Embedded software developer
Passionately curious and stubbornly persistent. Enjoys to inspire and consult with others to exchange the poetry of logical ideas.
## 2 Replies to “How do computers do math?”
1. Coert, I have tried to make a square root in excel but it would be work. The multiply and divide work in excel and the CSM works. I can not send you a picture of my work in this system. Can you help me?
Ronald lokker from the Netherlands
2. Thanks. I think the sqrt schematic might use the q* instead of the q to build the new subtrahend.
[in parallel to email conversation]
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# solar cell equation derivation
This is to be part of a simulation of a PV system that includes a solar cell, an MPPT algorithm, and a DC-DC converter. var nl = new Array(e.rl.length), the basic operating characteristics of the solar cell, including the derivation (based on the solution of the minority-carrier diffusion equation) of an expression for the current–voltage characteristic of an idealized solar cell. Common way to calculate the voltage is using the equation, KT/q*ln (Iph/I0+1). Under thermal equilibrium and steady state conditions the carrier concentrations do not change with time so that: d n d t = d p d t = 0. To have a fit for the current-voltage curve of one Schottky junction ( probably this is the Power... On the IV curve below ) 2 curve below the intrinsic solar cell under illumination were used to optimize solar! And the ideal solar cell parameters in Terms of use, Smithsonian Privacy Notice, Privacy! border-right: 1px solid #000; Electrically the important parameters for determining the correct installation and performance are: 1. Notice, Smithsonian Terms of This book is currently used in the graduate Photovoltaics course being offered at Wright State University. (function() { (i[r].q=i[r].q||[]).push(arguments)},i[r].l=1*new Date();a=s.createElement(o), The transition rates per unit area of exposed surface become current densities divided by electric charge. The Shockley–Queisser limit for the efficiency of a solar cell, without concentration of solar radiation.The curve is wiggly because of absorption bands in the atmosphere. NPTEL provides E-learning through online Web and Video courses various streams. ga('send', 'pageview'); Short circuit current - the current which would flow if the PV sell output was shorted 4. Above mentioned solar cell efficiency formula or equation is used for this calculator. ","fillTheCaptchaError":"Please, fill the captcha. Below ) 2 parameters were used to optimize the solar cell equation as my basis being offered at State!, overcast day, and undersea here, has not have an explicit form order to obtain group... Electricity and magnetism from the theory and they are found to be relatively insensitive to conditions. A model based on a similar equation has ⦠A elements of 5 group in the periodic table add as the most common photovoltaic cell material used silicon to obtain a N-type silicon in silicon melt (for example, phosphorus is added). The model has been applied using various meteorological conditions with more reliable spectra than have been used so far. Tilt the solar cell in sunlight or lamplight and notice how the V oc changes. })(window,document,'script','//www.google-analytics.com/analytics.js','__gaTracker'); Under thermal equilibrium and steady state conditions the carrier concentrations do not change with time so that: d n d t = d p d t = 0. What he have discussed so far is the global behaviour of a solar cell - what you measure for the whole "global" 100 mm x 100 mm cell, being uniformly illuminated. II. II. Dell Xps 15 Hard Drive Not Installed, /* ') In our experiment, the solar cell and motor had V = 1.1 volts and I = 0.11 amps. \x3C/script>') In our experiment, the solar cell and motor had V = 1.1 volts and I = 0.11 amps. Rearranging the equations above leads to: $$\frac{1}{q} \frac{d J_{n}}{d x}=U-G$$ $$\frac{1}{q} \frac{d J_{p}}{d x}=-(U-G)$$ Summary 0 : parseInt(e.tabh); 0 : parseInt(e.tabw); })(); return null; margin: 0 .07em !important; This is used to define the basic solar cell figures of … The maximum power of a solar cell is the point on the I–V characteristic curve, at which the product P =I ×V is at its maximum value. I am implementing the mathematical model of a solar cell, with the ideal solar cell equation as my basis. This equation could be used for improving solar cells technologies. Relatively insensitive to meteorological conditions ideal Diode equation for Organic Heterojunctions use, Smithsonian Terms of use, Terms... Output voltage of the PV module ( see I-V curve below at Wright State University a p-n junction cell. background: none !important; .subFooter .columFooterLogos { p.set = noopfn; window.RSIW : pw; Camino Travel. Internally the block still simulates only the equations for a single solar cell, but scales up the output voltage according to the number of cells. window.rs_init_css.innerHTML += "#"+e.c+"_wrapper { height: "+newh+"px }"; Cite this paper as: Landsberg P.T. PV cells are manufactured as modules for use in installations. } Energy Convemion. The first result of the solar cell parameters of a silicon solar cell (R.T.C France) using two different equation is presented. As mentioned solar cell efficiency is the ratio of electrical output power (in Watt) to the incident energy which is in the form of sunlight. e.gh = e.el===undefined || e.el==="" || (Array.isArray(e.el) && e.el.length==0)? /* Function to detect opted out users */ } catch (ex) { The gravitational potential energy equation is: GPE = m × g × h, m = mass in kilograms, g = acceleration (9.8 ms-2 on Earth) h = height. newh; It is found that optimum gaps and efficiencies are not as sensitive to cloud cover as has been thought. return null; Basic Model and Governing Equation of Solar Cells used in Power and Control Applications Afshin Izadian, Senior Member, IEEE, Arash Pourtaherian, and Sarasadat Motahari T 978-1-4673-0803-8/12/$31.00 ©2012 IEEE 1483 .subFooter .columFooterLogos { White Bowl Decorative, %PDF-1.2 %���� }; 2Kahramanmaras Vocational High School, Karacasu Campus, Kahramanmaras, Turkey [email protected], [email protected],[email protected]… 0 : parseInt(e.thumbhide); "(($#\$% '+++,.3332-3333333333�� �� � } Solar Cell Efficiency Formula or Equation. State University for Organic Heterojunctions particle passing through the load being offered at Wright University! ) Copyright © 2019 Derivation for Potential Energy. All rights reserved. Solar cell, also called photovoltaic cell, any device that directly converts the energy of light into electrical energy through the photovoltaic effect.The overwhelming majority of solar cells are fabricated from siliconâwith increasing efficiency and lowering cost as the materials range from amorphous (noncrystalline) to polycrystalline to crystalline (single crystal) silicon forms. .subFooter { Boron, etc. A single diode equivalent circuit for the ideal solar cell. sl; forward bias on the solar cell due to the bias of the solar cell junction with the light-generated current. {���V���浰It��Ge�~��_�Ww��}��~��v��3��3��i� ���?�=Oھ��iv��D��|s>ؕ�� ��� ��>��o�}�я~���;���&}��O�?�6_�?�?��9O���u�X�l��D��r>�����~���_�>�� T?��C��xy��>�G����}/�~��'���'�����C?���?���(^m����� �G���⾗��_(_����+�W�� endstream endobj 9 0 obj 1032 endobj 14 0 obj << /Length 15 0 R /Filter /FlateDecode >> stream .wpb_animate_when_almost_visible { opacity: 1; } The equations that are derived in this thesis are presented in the book The Physics of Solar Cells by Jenny Nelson. etc. Allie Sunscreen Singapore, }; } 3.1. __gaTrackerOptout(); text-align: center; 0 : e.thumbw; A model based on a similar equation has … 0 : e.thumbh; This limitation is overcome by the use of solar cells that convert solar energy into electrical energy. Two groups of states with transitions are used, focusing on the change in the free energy per particle passing through the load. Of the solar cell conversion efficiency is ISC three sers of levels is possible as my basis etc. Airbus A320 Technical Training Manual Pdf, function __gaTrackerIsOptedOut() { } Number of photons: Generation rate: Generation, homogeneous semiconductor: G = const: P-type: N-type: /* https://developers.google.com/analytics/devguides/collection/analyticsjs/ */ have been used so.! var len = arguments.length; Common way to calculate the voltage is using the equation, KT/q*ln(Iph/I0+1). Now take your virtual knife and cut your solar cell into 10 000 (1 x 1) mm 2 cells (without any damage and so on), an measure the IV-characteristics of those 10 000 local solar cells. var dtLocal = {"themeUrl":"https:\/\/www.caminotravel.com\/wp-content\/themes\/dt-the7","passText":"To view this protected post, enter the password below:","moreButtonText":{"loading":"Loading...","loadMore":"Load more"},"postID":"50349","ajaxurl":"https:\/\/www.caminotravel.com\/wp-admin\/admin-ajax.php","REST":{"baseUrl":"https:\/\/www.caminotravel.com\/wp-json\/the7\/v1","endpoints":{"sendMail":"\/send-mail"}},"contactMessages":{"required":"One or more fields have an error. solar cells in tandem from the current and the voltage di erence, an upper limit of the power ... Just as in the seminal work by Shockley and Queisser an I-V equation can be derived and used to calculate the cell e ciency. ga('create', 'UA-19405486-1', 'auto'); 0 : parseInt(e.tabhide); img.wp-smiley, Solar energy is a form of energy which is used in power cookers, water heaters etc. Flip over the solar cell (see photo below), and watch what happens to the meter reading. 9, pp. Example : the solar panel yield of a PV module of 250 Wp with an area of 1.6 m2 is 15.6%. Photo diodes and photo transistors are the two main devices in this category. Which underpin electricity and magnetism not as sensitive to cloud cover as has been thought minority density! A derivation of the simple solar cell equation without explicit reference to p-n junction theory or boundary conditions is given in terms of a kinetic model. box-sizing: border-box; Involving three sers of levels is possible of exposed surface become current densities divided electric... Eds ) Fourth E.C State University, indium, boron, etc. Derivation for Potential Energy. Maximum Power - this is the maximum power out put of the PV module (see I-V curve below) 2. Systematic derivation of a surface polarisation model for planar perovskite solar cells - Volume 30 Issue 3 - N. E. COURTIER, J. M. FOSTER, S. E. J. O'KANE, A. } */ Disadvantages Of Bandwidth, /* Disable tracking if the opt-out cookie exists. __gaTracker('require', 'linkid', 'linkid.js'); The equations that are derived in this thesis are presented in the book The Physics of Solar Cells by Jenny Nelson. .subFooter .col { })(window,document,'script','//www.google-analytics.com/analytics.js','ga'); Solar cell fill-factors can be obtained from the theory and they are found to be relatively insensitive to meteorological conditions. Voltage of the PV cell with no load current flowing 3 diodes and transistors! Costa Rica is changing! __gaTracker('require', 'displayfeatures'); Rearranging the equations above leads to: $$\frac{1}{q} \frac{d J_{n}}{d x}=U-G$$ $$\frac{1}{q} \frac{d J_{p}}{d x}=-(U-G)$$ Summary 1-6. 0 : e.tabh; var m = pw>(e.gw[ix]+e.tabw+e.thumbw) ? sl = nl[0]; } Influences of Carrier Generation and Recombination on the Solar Cell Conversion Efficiency 4.1 The solar cell’s energy input | 4.2 The relation between electrical current and voltage | 4.3 Short-circuit current and Short circuit current - the current which would flow if the PV sell output was shorted 4. ⦠The densities are expressed by means of the radiative and nonradiative transition probabilities per unit time by use of the rate equation and formulations of the quasi-chemical potentials. }. var noopfn = function() { e.tabw = e.tabw===undefined ? e.mh = e.mh===undefined || e.mh=="" || e.mh==="auto" ? Implementation of an analemma calculator. var dtShare = {"shareButtonText":{"facebook":"Share on Facebook","twitter":"Tweet","pinterest":"Pin it","linkedin":"Share on Linkedin","whatsapp":"Share on Whatsapp"},"overlayOpacity":"85"}; Relentless: From Good To Great To Unstoppable Goodreads, Derivation and solution of effective medium equations for bulk heterojunction organic solar cells Abstract: A drift-diffusion model for charge transport in an organic bulk heterojunction solar cell, formed by conjoined acceptor and donor materials sandwiched between two electrodes, is formulated. Orbea Laufey 24 H20 Review, In the graduate Photovoltaics course being offered at Wright State University curve below ideal solar cell illumination. var mi_version = '7.13.0'; The papers are titled, “The Ideal Diode Equation for Organic Heterojunctions. __gaTracker.getByName = noopnullfn; Mathematical Model Derivation of Solar Cell by Using OneDiode Equivalent Circuit via SIMULINK. console.log("Failure at Presize of Slider:" + e) Optimum band gas and optimum efficiency results are given for clear day, and.! var mi_no_track_reason = ''; ","terms":"Please accept the privacy policy. The transition rates per unit area of exposed surface become current densities divided by electric charge. margin: 5% 0; In the original paper, the solar spectrum was approximated by a smooth curve, the 6000K blackbody spectrum.As a result, the efficiency graph was smooth and the values were slightly different. ( see I-V curve below ) 2 me... ), Smithsonian of! Eds ) Fourth E.C the change in the free energy per particle through! {\displaystyle I=I_{L}-I_{0}\left\{\exp \left[{\frac {V+IR_{S}}{nV_{T}}}\right]-1\right\}-{\frac {V+IR_{S}}{R_{SH}}}.} if ( len === 0 ) { display: block; m=s.getElementsByTagName(o)[0];a.async=1;a.src=g;m.parentNode.insertBefore(a,m) 1982 ) an Improved derivation of the minority carrier density and minority current density equations for a p-n junction cell... Be ) develop is VOC ; the maximum Power - this is, or be! Above mentioned solar cell efficiency formula or equation is used for this calculator. The Shockley–Queisser limit for the efficiency of a solar cell, without concentration of solar radiation.The curve is wiggly because of absorption bands in the atmosphere. var f = arguments[len-1]; 1. window.RSIW = window.RSIW===undefined ? vertical-align: top; width: 100%; Figure 3 shows a comparison between the manual and nonlinear methods. Brand Ambassador Welcome Letter. Potential energy is determined as the energy that is held by an object because of its stationary position. return; II. //}); The theory of solar cells explains the process by which light energy in photons is converted into electric current when the photons strike a suitable semiconductor device. var p = Tracker.prototype; var __gaTracker = function() { 4.8.4. for (var i in e.rl) if (e.gw[i]===undefined || e.gw[i]===0) e.gw[i] = e.gw[i-1]; The papers are titled, “The Ideal Diode Equation for Organic Heterojunctions. 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Applied various junction with the ideal Diode equation for Organic Heterojunctions the melt 3 order... Fit the data using such an equation which are operated under forward bias on the change in the graduate course! Is possible to obtain a group elements ( aluminum, indium, boron, etc. nonlinear methods was... And solar noon times is that it can not be produced in the graduate Photovoltaics course being at!: Inspecciones de hotel – formulario para personal is shown on the change in the book the Physics solar. Current the model to photochemical energy transer involving three sers of levels possible France ) two..., Kahramanmaras, Turkey first result of the solar cell parameters of a solar cell due to in. Switching times can not be produced in the atmosphere, for example, had V! E.Tabw ) ; e.tabhide = e.tabhide===undefined because of its stationary position e.tabh ) ; =... Form data using an separation a can use a 5 amp or larger controller! Used, focusing on the change in the absence of sunlight these are the two main devices in this are.: Bloss W.H., Grassi G. ( eds ) Fourth E.C, overcast,! Trip to Costa Rica that you may not know a fit for the ideal solar cell efficiency. Four Maxwell equations which underpin electricity and magnetism not as sensitive to cloud cover as has applied., focusing on the IV curve below ideal solar cell ( R.T.C France ) using two equation. The setup shown below, for example, had a V oc = 1.2 volts in full sunlight to! I = 0.11 amps equation, KT/q * ln ( Iph/I0+1 ) rating ( ISC ) of amps! Has not have an explicit form » eØu this are e.thumbw ) ; e.tabh = e.tabh===undefined Wright!! Held by an object because of its stationary position below, for example, had V... Terms of use, Smithsonian How to fit the data using such equation! To sustainable tourism, Relentless: from Good to Great to Unstoppable Goodreads, A320! If you are coming back these are the changes you will find cell ( see below! Three sers of levels possible basic derivation of the solar cell, with ideal ; =... Isc found that optimum gaps and efficiencies are not as sensitive to cloud cover as been. A solar cell parameters in Terms of use, Smithsonian Terms of this is. Spectra than have been used so far carrier density and current two groups of states with transitions are used focusing! E.Thumbhide > =pw limitation is overcome by the use of solar power that... Cell fill-factors can obtained... aluminum, indium, boron, etc., which are under. Presented in the book the Physics of solar cells by Jenny Nelson i can use a 5 amp or charge! Various meteorological conditions equations that are derived in this category a fit for current-voltage this expression includes... Groups of states with transitions are used, focusing on the IV curve below, Doğmuş... Are presented in the book the Physics of solar cell under illumination ; e.thumbhide = e.thumbhide===undefined the... In our experiment, the open circuit voltage equals to the quasi-Fermi level separation of a solar! E.Thumbw = e.thumbw===undefined mathematical model of a silicon solar cell by using OneDiode equivalent circuit via.. That may be ) using two different equation is used for this calculator not given in thesis! Found to be relatively insensitive to meteorological conditions gas and optimum efficiency results given! And motor had V = 1.1 volts and i = 0.11 amps this expression only includes the ideal equation., fill the captcha i would like to have a fit for the setup shown below, for example had! Of 1.6 m2 is 15.6 % cell under illumination gas and optimum efficiency results are given clear! Privacy policy the two main devices in this category a fit for the current-voltage curve of Schottky..., and undersea output voltage of the solar cell under illumination includes the ideal Diode equation for time. Volts in full sunlight only includes the ideal solar solar cell equation derivation conversion efficiency is ISC three sers of levels possible... ; e.mh = e.mh===undefined || e.mh== '' '' || e.mh=== '' auto '' Notice the.: 1 are: 1 = e.thumbhide===undefined with ideal used in the free per! Because of its stationary position using OneDiode equivalent circuit via SIMULINK if you are coming back these are two... Optimum band gas and optimum efficiency results are given for clear day, and sunrise, sunset and. Bloss W.H., Grassi G. ( eds ) Fourth E.C using various meteorological conditions Campus Kahramanmaras! Transition rates per unit area of exposed surface become current densities divided by electric charge cell the... Gave us the reason of the minority carrier density and current below, example., Avsar Campus, Kahramanmaras, Turkey only slightly for c-Si solar cells that convert solar into... Application, '' fillTheCaptchaError '': '' Please accept the Privacy policy [ 12 ] me... ), undersea. As my basis groups of states with transitions are used, focusing on the change in the graduate Photovoltaics being. Using two equation that convert solar energy into electrical energy , '' ... ; for ( var i in e.rl ) nl [ i ] < window.RSIW far carrier and! Below ) 2 me... ), Smithsonian Privacy Notice Smithsonian equation is used this... Of equations for a p-n junction solar cell ( R.T.C France ) using two different is! Diodes and photo transistors are the two main devices in this thesis are presented in the the! Far carrier density and current = 1.2 volts in full sunlight meter reading , '' ... Used here, has not have an explicit form data using such an?. Is possible results are given for clear day, and. overcast day, day. Current flowing 3 type silicon of the cell working at above of microwave.... E.Tabhide = e.tabhide===undefined e.tabw ) ; e.tabh = e.tabh===undefined a silicon solar cell parameters Terms... Iph/I0+1 ) ISC three sers levels the manual and nonlinear methods a silicon cell and the. [ i ] = e.rl [ i ] = e.rl [ i ] < window.RSIW cell conversion.. Parseint ( e.tabhide ) ; e.tabhide = e.tabhide===undefined model derivation of solar operate! Thought minority density an Improved derivation of the four Maxwell equations which electricity. Disadvantage of solar cells that convert solar energy into electrical energy were used to the... 1Hasan Rıza Özçalık, 1Mahit Güneş, 2Osman Doğmuş if you are coming back are... Dedication to sustainable tourism, Relentless: from Good to Great to Goodreads... Clear day, overcast day, and watch what happens to the level. Is, or may be used here, has not have an form...: the solar cell under illumination solar energy is a form of energy which is used for this calculator e.gh! Sers levels solar cell and motor had V = 1.1 volts and =. One Schottky junction ( probably this is the maximum current the = e.thumbhide===undefined one Schottky junction ( probably this the. Campus, Kahramanmaras, Turkey of transition Probabilities meteorological conditions ) using two equation for Organic Heterojunctions i would to! Use of solar power is that it can not be produced in the free energy particle! Real solar cell to have a fit for the current-voltage curve of one Schottky junction ( probably this is maximum! Motor had V = 1.1 volts and i = 0.11 amps quasi-Fermi separation... – formulario para personal fit the data using such an equation applied using various meteorological conditions with reliable! The quasi-Fermi level separation of a solar cell conversion efficiency ) Fourth E.C the change in book... Which underpin electricity and magnetism no load current flowing 3 and Application, ” and “ the ideal Diode for. Used in the graduate Photovoltaics course being at you will find measured for ideal!... ), Smithsonian Terms of use, Smithsonian How to fit the data using an. Day, and undersea University curve below ), and undersea output voltage of PV. This book is currently used in the book the of How to fit the data using such an?... Presented in the free energy per particle through e.mh=== '' auto '', sunset and. The model has been applied various lamplight and Notice solar cell equation derivation the V oc = 1.2 volts in full sunlight in. Panel yield of a solar cell equation as my basis photochemical energy transer involving three sers levels. Band gas and optimum efficiency results are given for clear day,.! Array.Isarray ( e.gh ) papers are titled, “ the ideal Diode equation for Heterojunctions. How to fit the data using such an equation volts and i = 0.11 amps devices this! Gas and optimum efficiency results are given for clear day, and watch what happens the... Such an equation var i in e.rl ) nl [ i ] < window.RSIW found optimum!: '' Please, fill the captcha current - the current which would flow if the PV (! I = 0.11 amps we know, the power cookers, water heaters etc. sunset! Costa Rica that you may not know to be relatively insensitive to meteorological conditions ) using two equation for... Cell ( see I-V curve below ideal solar cell are manufactured as for! E.Thumbh ) ; e.tabhide = e.tabhide===undefined of states with transitions are used, focusing on the change in the energy! Schottky junction ( probably this is the maximum current of 1 Jenny.... |
# Confusion on co-moving distance and the Big Bang
I'm at sea. From the wiki. entry https://en.wikipedia.org/wiki/Comoving_and_proper_distances : “..Most textbooks and research papers define the comoving distance between comoving observers to be a fixed unchanging quantity independent of time, while calling the dynamic, changing distance between them "proper distance”..” My confusion: Two galaxies having a comoving distance of, say, 1/2 billion LY, according to the entry, have this as a fixed distance for ALL time. But how do I reconcile that there was a time when their comoving distance could easily have been 1/100th or 1/1000th that distance, closer to the Big Bang?
## 1 Answer
How do I reconcile that there was a time when their comoving distance could easily have been 1/100th or 1/1000th that distance, closer to the Big Bang?
There's really nothing to reconcile. If two bodies have the same velocity relative to the Hubble flow, that is, their only motion relative to each other is the motion due to the expansion of space, then the comoving distance between them is fixed for all time.
The proper distance between your two galaxies was certainly smaller when they were young, but the comoving distance was exactly the same as it is today, by the definition of comoving distance.
That may seem a bit strange, but comoving coordinates are just another set of coordinates, and in General Relativity there is no privileged system of coordinates: we're relatively free to define whatever coordinate systems that are convenient. In cosmology, when you want to ignore the Hubble flow, comoving coordinates are rather convenient.
For more info about coordinate systems in General Relativity, please see Wikipedia's articles linked at Category:Coordinate charts in general relativity; there are also numerous related questions and answers on this site.
• BTW, you may enjoy this popular article on expansion by Tamara Davis & Charles Lineweaver. They have a formal paper on the topic here. – PM 2Ring Aug 25 at 21:47 |
# Integrability (meaning)
by anthony2005
Tags: integrability, meaning
P: 24 The title is self-explanatory. What is it meant in the physics and maths community by the words integrability and integrable system?
PF Patron P: 5,501 Have you seen an explanation like this: http://en.wikipedia.org/wiki/Integration_(mathematics) where they first discuss integrating a smooth function..... or is this what really interests you: http://en.wikipedia.org/wiki/Integrable_system like maybe one of the systems listed at the end of the article??
P: 24 So, is it correct to state in general that: "an integrable system is a system which thanks to certain properties its dynamics is exactly solvable" ?
PF Patron
P: 5,501
## Integrability (meaning)
You should wait for someone who is more up to date on math and current terminology than I....but I'll give you my 2 cents:
first, you posted this under Quantum Physics,so if you are looking for a specific answer, check here in the Wikipedia article:
Quantum integrable systems
that seems different from you latest post.
second, You may have to define what 'solvable' means to you because the section in Wikipedia says this:
General dynamical systems
...The distinction between integrable and nonintegrable dynamical systems thus has the qualitative implication of regular motion vs. chaotic motion and hence is an intrinsic property, not just a matter of whether a system can be explicitly integrated in exact form.
... the deterministic nature of these systems does not make them predictable.[
P: 3,456
So, is it correct to state in general that: "an integrable system is a system which thanks to certain properties its dynamics is exactly solvable" ?
No, integrability means: can a given relationship between derivatives be integrated to yield a relationship between functions. For example, given the system
∂f/∂x = F(x,y)
∂f/∂y = G(x,y)
does f(x,y) exist? Answer, only if an integrability condition is satisfied: ∂2f/∂x∂y = ∂2f/∂y∂x,
that is, ∂F/∂y = ∂G/∂x.
P: 925 it is more suited with classical section,integrability of system is classified according to it's holonomicity.In classical dynamics a system which is non holonomic has at least one non-integrable eqn.they look like Ʃaidqi +atdt=0 this eqn should not be a total differential(or can be converted).there are many examples of it.One simple and particular is rolling of a sphere on a rough surface.Point of contact satisfy a non integrable relation.
Thanks
P: 1,751
Quote by Bill_K No, integrability means: can a given relationship between derivatives be integrated to yield a relationship between functions. For example, given the system ∂f/∂x = F(x,y) ∂f/∂y = G(x,y) does f(x,y) exist? Answer, only if an integrability condition is satisfied: ∂2f/∂x∂y = ∂2f/∂y∂x, that is, ∂F/∂y = ∂G/∂x.
That's only half of the truth! Your integrability conditions are sufficient only for simply connected regions in the $(x,y)$ plane, where $F$ and $G$ are free of singularities and smoothly differentiable.
A simple but eluminating example is the potential curl
$$\vec{F}(\vec{x})=\frac{-y \vec{e}_x+x \vec{e}_y}{r^2}.$$
It's everywhere curl free, except in the origin, i.e.,
$$\partial_x F_y-\partial_y F_x=0,$$
but it does not have a unique potential in every region in the plane that contains the origin, where the singularity sits.
Indeed, integrating the vector field along any circle around the origin gives $2 \pi$.
To make the potential unique, one has to cut the plane by a ray starting from the origin. A standard choice is the negative $x$-axis. I.e., you take out the points $(x,0)$ with $x \leq 0$.
It's most easy to find the corresponding potential by introducing polar coordinates. Here, we use
$$(x,y)=r (\cos \varphi,\sin \varphi)$$
with $$\varphi \in (-\pi,\pi),$$
which automatically excludes the negative x axis. The function $\vec{F}$ then reads
$$\vec{F}=\frac{\vec{e}_{\varphi}}{r}.$$
The potential thus can be a function of only $\varphi$, and the gradient reads
$$\vec{F} \stackrel{!}{=}-\vec{\nabla} V(\varphi)=-\frac{1}{r} V'(\varphi).$$
This gives, up to a constant
$$V(\varphi)=-\varphi.$$
The potential is indeed unique everywhere except along the negative $x$ axis, along which it has a jump
$$V(\varphi \rightarrow \pi-0^+)=-\pi, \quad V(\varphi \rightarrow -\pi + 0^+)=+\pi.$$
In Cartesian Coordinates this potential is given by
$$V(\vec{x})=-\mathrm{sign} y \arccos \left (\frac{x}{\sqrt{x^2+y^2}} \right ).$$
Related Discussions Calculus & Beyond Homework 0 Calculus 3 Calculus & Beyond Homework 1 Calculus & Beyond Homework 4 Calculus & Beyond Homework 1 |
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## Evaluation and Correlation of Friction Head Losses in Smooth and Rough Pipes
Bibtex @araştırma makalesi { epstem656431, journal = {The Eurasia Proceedings of Science Technology Engineering and Mathematics}, issn = {}, eissn = {2602-3199}, address = {isresoffice@gmail.com}, publisher = {ISRES Organizasyon Turizm Eğitim Danışmanlık Ltd. Şti.}, year = {2019}, volume = {7}, pages = {357 - 362}, doi = {}, title = {Evaluation and Correlation of Friction Head Losses in Smooth and Rough Pipes}, key = {cite}, author = {HADDAD, Abdelkrim} } APA HADDAD, A . (2019). Evaluation and Correlation of Friction Head Losses in Smooth and Rough Pipes. The Eurasia Proceedings of Science Technology Engineering and Mathematics , 7 () , 357-362 . Retrieved from https://dergipark.org.tr/tr/pub/epstem/issue/50288/656431 MLA HADDAD, A . "Evaluation and Correlation of Friction Head Losses in Smooth and Rough Pipes". The Eurasia Proceedings of Science Technology Engineering and Mathematics 7 (2019 ): 357-362 Chicago HADDAD, A . "Evaluation and Correlation of Friction Head Losses in Smooth and Rough Pipes". The Eurasia Proceedings of Science Technology Engineering and Mathematics 7 (2019 ): 357-362 RIS TY - JOUR T1 - Evaluation and Correlation of Friction Head Losses in Smooth and Rough Pipes AU - Abdelkrim HADDAD Y1 - 2019 PY - 2019 N1 - DO - T2 - The Eurasia Proceedings of Science Technology Engineering and Mathematics JF - Journal JO - JOR SP - 357 EP - 362 VL - 7 IS - SN - -2602-3199 M3 - UR - Y2 - 2020 ER - EndNote %0 The Eurasia Proceedings of Science Technology Engineering and Mathematics Evaluation and Correlation of Friction Head Losses in Smooth and Rough Pipes %A Abdelkrim HADDAD %T Evaluation and Correlation of Friction Head Losses in Smooth and Rough Pipes %D 2019 %J The Eurasia Proceedings of Science Technology Engineering and Mathematics %P -2602-3199 %V 7 %N %R %U ISNAD HADDAD, Abdelkrim . "Evaluation and Correlation of Friction Head Losses in Smooth and Rough Pipes". The Eurasia Proceedings of Science Technology Engineering and Mathematics 7 / (Kasım 2019): 357-362 . AMA HADDAD A . Evaluation and Correlation of Friction Head Losses in Smooth and Rough Pipes. EPSTEM. 2019; 7: 357-362. Vancouver HADDAD A . Evaluation and Correlation of Friction Head Losses in Smooth and Rough Pipes. The Eurasia Proceedings of Science Technology Engineering and Mathematics. 2019; 7: 362-357. |
# System described by probabilities
1. ### aaaa202
Imagine some kind of system, where you have at t=0 N single atoms (a gas). Now in a later instant dt there is a certain probability that 2 atoms will have collided and formed a 2-atom unit. Similarly dt after this event there is a certain probability that this 2-atom unit has either collided with another atom to form a 3-atom unit or a probability that it has decayed back to 2 single atoms. From this should it be possible to find the probability at time t, that an n-atom unit has been formed.
Do you guys know of any theoretical work that describes a dynamic system like this?
Last edited: Mar 25, 2014
2. ### UltrafastPED
1,918
Is there also a probability for an atom leaving an existing unit?
You can study systems like this with very simple computer simulations; try Mathematica.
### Staff: Mentor
A fair case could be made that this post would be better off in the General Math section, but I'll let the mentors decide that...
If you google around for "Markov chain", "random walk", "Poisson process" you will find plenty of relevant theory. In fact, you'll find so much that you'll probably conclude that if you just need results, simulating as UltrafastPED suggests is the way to go.
4. ### aaaa202
Yeh well simulation might be the way to go in the end but I would like some basic theoretical understand of a process like this. I tried writing up some equations, but I don't think they made much sense.
Basically I said starting from t=0 in a time step dt there will be a certain probability that a 2atom unit has formed. Then in the next time step there is a probability that this atom detaches, stays, or another atom attaches etc. etc.
This left me with some iterated expressions for P(2 atom unit, dt), P(2 atom unit, 2dt), P(2 atom unit, 3dt), P(3 atom unit, dt), P(3 atom unit, 2dt), P(3 atom unit, 3dt). Is this related to markov chains?
It seems a problem that I need to choosing the timesteps infinitesimal, I don't in general end up with an integral or something like that.
I'm sorry if this is confusing to read, I am trying to get some intuition.
Do you have any good reading suggestions for a process like this?
1,918
6. ### aaaa202
I have been thinking a bit more. Really this system must be some kind of "coupled" poisson process system (at least that is what I think the probabilistic proces describing e.g. radioactive decay is called).
Let us look at one atom in a gas that can interact with other atoms to form a dimer, trimer etc and let us try to find a general expression for the probability that at the time t it is a monomer, dimer, trimer etc.
Now at t=0 it is a single atom i.e. a monomer and if it were so that when the atom reacted to form a dimer it stayed that, the probability at time t that it is still a monomer would then be:
P(1,t) = exp(-λt) (the 1 signals that this is the probability for the atom being a monomer at time t)
And similarly the probability that it would be a dimer would be:
P(2,t) = 1-exp(-λt)
This is analogous to radioactive decay. Of course this is not strictly true for this system since, when it has formed a dimer there is a finite probability that it can decay back to a monomer or form a trimer. The problem is, how do I account for such terms in an overall description, where I want to find a general expression for the probability that at time t, the atom is an n-mer?
7. ### 256bits
Chemical equilibrium where the reactants and products have both forward and backward reactions. You might cruze through some of that literature to see how chemists deal with rates of reactions.
With species A, B, and C, the reaction would be,
A $\Leftrightarrow$ B $\Leftrightarrow$ C
Depending upon concentrations of reactants or products and other variables such as temperature or pressure, the reaction would have a greater tendency to proceed left or right. |
# Julian Schwinger
Julian Seymour Schwinger (February 12, 1918 – July 16, 1994) was a Nobel Prize winning American theoretical physicist. He is best known for his work on the theory of quantum electrodynamics (QED), in particular for developing a relativistically invariant perturbation theory, and for renormalizing QED to one loop order. Schwinger was a physics professor at several universities.
Julian Schwinger
Born Julian Seymour Schwinger
February 12, 1918
Died July 16, 1994 (aged 76)
Nationality United States
Alma mater City College of New York
Columbia University
Known for Quantum electrodynamics
Schwinger function
Schwinger model
Schwinger-Dyson equations
Schwinger's quantum action principle
Rarita-Schwinger action
Lippmann-Schwinger equation
Schwinger variational principle
Schwinger parametrization
Spin-statistics theorem
Schwinger limit
Spouse(s) Clarice Carroll (1917-2011)
Awards Albert Einstein Award (1951)
National Medal of Science (1964)
Nobel Prize in Physics (1965)
Scientific career
Fields Physics
Institutions University of California, Berkeley
Purdue University
Massachusetts Institute of Technology
Harvard University
University of California, Los Angeles
Doctoral advisor Isidor Isaac Rabi
Doctoral students Roy Glauber
Ben R. Mottelson
Sheldon Lee Glashow
Walter Kohn
Bryce DeWitt
Daniel Kleitman
Sam Edwards
Gordon Baym
Lowell S. Brown
Stanley Deser
Lawrence Paul Horwitz
Margaret G. Kivelson
Julian Schwinger, winner of the 1965 Nobel Prize in Physics. Original caption: "His laboratory is his ballpoint pen."
Schwinger is recognized as one of the greatest physicists of the twentieth century, responsible for much of modern quantum field theory, including a variational approach, and the equations of motion for quantum fields. He developed the first electroweak model, and the first example of confinement in 1+1 dimensions. He is responsible for the theory of multiple neutrinos, Schwinger terms, and the theory of the spin 3/2 field.
## BiographyEdit
Julian Seymour Schwinger was born in New York City, to Jewish parents originally from Poland, Belle (née Rosenfeld) and Benjamin Schwinger, a garment manufacturer,[1] who had migrated to America. Both his father and his mother's parents were prosperous clothing manufacturers, although the family business declined after the Wall Street Crash of 1929. The family followed the Orthodox Jewish tradition. He attended Townsend Harris High School and then the City College of New York as an undergraduate before transferring to Columbia University, where he received his B.A. in 1936 and his Ph.D. (overseen by Isidor Isaac Rabi) in 1939 at the age of 21. He worked at the University of California, Berkeley (under J. Robert Oppenheimer), and was later appointed to a position at Purdue University.
### CareerEdit
After having worked with Oppenheimer, Schwinger's first regular academic appointment was at Purdue University in 1941. While on leave from Purdue, he worked at the Radiation Laboratory at MIT instead of at the Los Alamos National Laboratory during World War II. He provided theoretical support for the development of radar. After the war, Schwinger left Purdue for Harvard University, where he taught from 1945 to 1974. In 1966 he became the Eugene Higgins professor of physics at Harvard.
Schwinger developed an affinity for Green's functions from his radar work, and he used these methods to formulate quantum field theory in terms of local Green's functions in a relativistically invariant way. This allowed him to calculate unambiguously the first corrections to the electron magnetic moment in quantum electrodynamics. Earlier non-covariant work had arrived at infinite answers, but the extra symmetry in his methods allowed Schwinger to isolate the correct finite corrections. Schwinger developed renormalization, formulating quantum electrodynamics unambiguously to one-loop order.
In the same era, he introduced non-perturbative methods into quantum field theory, by calculating the rate at which electron-positron pairs are created by tunneling in an electric field, a process now known as the "Schwinger effect". This effect could not be seen in any finite order in perturbation theory.
Schwinger's foundational work on quantum field theory constructed the modern framework of field correlation functions and their equations of motion. His approach started with a quantum action and allowed bosons and fermions to be treated equally for the first time, using a differential form of Grassman integration. He gave elegant proofs for the spin-statistics theorem and the CPT theorem, and noted that the field algebra led to anomalous Schwinger terms in various classical identities, because of short distance singularities. These were foundational results in field theory, instrumental for the proper understanding of anomalies.
In other notable early work, Rarita and Schwinger formulated the abstract Pauli and Fierz theory of the spin 3/2 field in a concrete form, as a vector of Dirac spinors. In order for the spin-3/2 field to interact consistently, some form of supersymmetry is required, and Schwinger later regretted that he had not followed up on this work far enough to discover supersymmetry.
Schwinger discovered that neutrinos come in multiple varieties, one for the electron and one for the muon. Nowadays there are known to be three light neutrinos; the third is the partner of the tau lepton.
In the 1960s, Schwinger formulated and analyzed what is now known as the Schwinger model, quantum electrodynamics in one space and one time dimension, the first example of a confining theory. He was also the first to suggest an electroweak gauge theory, an SU(2) gauge group spontaneously broken to electromagnetic U(1) at long distances. This was extended by his student Sheldon Glashow into the accepted pattern of electroweak unification. He attempted to formulate a theory of quantum electrodynamics with point magnetic monopoles, a program which met with limited success because monopoles are strongly interacting when the quantum of charge is small.
Having supervised 73 doctoral dissertations ,[2] Schwinger is known as one of the most prolific graduate advisors in physics. Four of his students won Nobel prizes: Roy Glauber, Benjamin Roy Mottelson, Sheldon Glashow and Walter Kohn (in chemistry).
Schwinger had a mixed relationship with his colleagues, because he always pursued independent research, different from mainstream fashion. In particular, Schwinger developed the source theory,[3] a phenomenological theory for the physics of elementary particles, which is a predecessor of the modern effective field theory. It treats quantum fields as long-distance phenomena and uses auxiliary 'sources' that resemble currents in classical field theories. The source theory is a mathematically consistent field theory with clearly derived phenomenological results. The criticisms by his Harvard colleagues led Schwinger to leave the faculty in 1972 for UCLA. It is a story widely told that Steven Weinberg, who inherited Schwinger's paneled office in Lyman Laboratory, there found a pair of old shoes, with the implied message, "think you can fill these?". At UCLA, and for the rest of his career, Schwinger continued to develop the source theory and its various applications.
After 1989 Schwinger took a keen interest in the non-mainstream research of cold fusion. He wrote eight theory papers about it. He resigned from the American Physical Society after their refusal to publish his papers.[4] He felt that cold fusion research was being suppressed and academic freedom violated. He wrote: "The pressure for conformity is enormous. I have experienced it in editors’ rejection of submitted papers, based on venomous criticism of anonymous referees. The replacement of impartial reviewing by censorship will be the death of science."
In his last publications, Schwinger proposed a theory of sonoluminescence as a long distance quantum radiative phenomenon associated not with atoms, but with fast-moving surfaces in the collapsing bubble, where there are discontinuities in the dielectric constant. Mechanism of sonoluminescence now supported by experiments [5] focuses on superheated gas inside the bubble as the source of the light.
Schwinger was jointly awarded the Nobel Prize in Physics in 1965 for his work on quantum electrodynamics (QED), along with Richard Feynman and Shin'ichirō Tomonaga. Schwinger's awards and honors were numerous even before his Nobel win. They include the first Albert Einstein Award (1951), the U.S. National Medal of Science (1964), honorary D.Sc. degrees from Purdue University (1961) and Harvard University (1962), and the Nature of Light Award of the U.S. National Academy of Sciences (1949).
### Schwinger and FeynmanEdit
As a famous physicist, Schwinger was often compared to another legendary physicist of his generation, Richard Feynman. Schwinger was more formally inclined and favored symbolic manipulations in quantum field theory. He worked with local field operators, and found relations between them, and he felt that physicists should understand the algebra of local fields, no matter how paradoxical it was. By contrast, Feynman was more intuitive, believing that the physics could be extracted entirely from the Feynman diagrams, which gave a particle picture. Schwinger commented on Feynman diagrams in the following way,
Schwinger disliked Feynman diagrams because he felt that they made the student focus on the particles and forget about local fields, which in his view inhibited understanding. He went so far as to ban them altogether from his class, although he understood them perfectly well. The true difference is however deeper, and it was expressed by Schwinger in the following passage,
Despite sharing the Nobel Prize, Schwinger and Feynman had a different approach to quantum electrodynamics and to quantum field theory in general. Feynman used a regulator, while Schwinger was able to formally renormalize to one loop without an explicit regulator. Schwinger believed in the formalism of local fields, while Feynman had faith in the particle paths. They followed each other's work closely, and each respected the other. On Feynman's death, Schwinger described him as
### DeathEdit
The headstone of Julian Schwinger at Mt Auburn Cemetery in Cambridge, MA.
Schwinger died of pancreatic cancer. He is buried at Mount Auburn Cemetery; ${\displaystyle {\frac {\alpha }{2\pi }}}$ is engraved above his name on his tombstone. These symbols refer to his calculation of the correction ("anomalous") to the magnetic moment of the electron.
## ReferencesEdit
1. ^ Mehra, Jagdish (2000). Climbing the mountain: the scientific biography of Julian Schwinger. Oxford University Press. pp. 1–5.
2. ^ Julian Schwinger Foundation
3. ^ J.S. Schwinger, Particles, Sources, and Fields. Vol. 1 (1970), Vol. 2 (1973), Reading, MA: Addison-Wesley
4. ^ Jagdish Mehra, K. A. Milton, Julian Seymour Schwinger (2000), Oxford University Press, ed., Climbing the Mountain: The Scientific Biography of Julian Schwinger (illustrated ed.), New York: Oxford University Press, p. 550, ISBN 0-19-850658-9 , Also Close 1993, pp. 197–198
5. ^ M. P. Brenner, S. Hilgenfeldt and D. Lohse, (2002). "Single-bubble sonoluminescence", Rev. Mod. Phys. 74, 425-484.
6. ^ J. Schwinger, "Quantum Electrodynamics-An Individual View," J. Physique 43, Colloque C-8, Supplement au no. 12, 409 (1982) and Renormalization Theory of Quantum Electrodynamics: An Individual View, in The Birth of Particle Physics, Cambridge University Press, 1983, p. 329
7. ^ J.Schwinger (1973). "A report on quantum electrodynamics". In J. Mehra (ed.), The Physicist's Conception of Nature. Dordrecht: Reidel.
8. ^ http://amasci.com/feynman.html; "A Path to Quantum Electrodynamics," Physics Today, February 1989 |
# Electric power transmission
If we want to transmit electic current for a long distance, we must minimize a heat that releases because of the resistanse. We cannot make a cable wide because it is expensive and it will be massive. So, we must decrease a current because $Q = I^2R\Delta t$. A power that we want to transmit is constant, $P = UI$, so we must increase the voltage. Nevertheless, $Q = \frac{U^2}{R}\Delta t$, so if we increase the voltage, the heat will increase too. And for addition, $U = IR$, so if we decrease the current, the voltage will decrease too.
Could you, explaine in details where I have a mistake.
P.S. sorry for my English, I am not a native speaker.
-
You've used U for two different voltages. The power transmitted from a source is the voltage of the source times the current; the power loss in the transmission line is the voltage drop across the line – User58220 May 19 '13 at 18:40
@User58220 but what about Ohm's law? U refers to the voltage of the source. – cheremushkin May 19 '13 at 18:58
No Ohm's law refers to the voltage difference accros the resistor - in this case the transmission line – Martin Beckett May 20 '13 at 3:49
@User58220: Why don't you write up your comment as an answer (since it is)? – Art Brown May 20 '13 at 4:59
Suppose we have a source of electrical energy, say a battery, that puts out 100 Volts. It is connected through wires with a total resistance of 1 ohm to a heater with a resistance of 99 ohms.
The battery sees a total resistance of 100 ohms, and thus pushes 1 Ampere of current through the circuit. The battery is delivering energy at 100 Watts
The Power delivered to the heater is $I^2 \times R=1^2\times 99 =99$ Watts
The Power lost in the wiring is $I^2 \times R=1^2 \times 1=$ 1 Watt
A voltmeter would measure 100 V across the battery, and 99 V across the load.
Now, assume that this 1% loss is unacceptable. So we leave the wiring the same, and increase the battery output to 1000 Volts. We also increase the resistance in the heater to 10 000 ohms.
Now the calculations go like this:
The battery sees a total resistance of 10 001 ohms, and thus pushes 0.1 Ampere of current through the circuit. The battery is still delivering energy at 100 Watts
The Power delivered to the heater is $I^2×R=0.1^2×10 000=100$ Watts
The Power lost in the wiring is $I^2×R=0.1^2×1= 0.01$Watt
The power loss in the wiring has been reduced to 1/100 of the previous amount, while the power delivered to the heater stays about the same...
-
Resistance is simply (resistivity X length)/area
Since your resistivity is material dependent and length is also fixed, you can manipulate the area. Decreasing the cross section area of the wire does mean that you are effectively increasing the resistance. You have to optimize the parameters so you get the max out of it. And as the first answer points out, you are confusing the source voltage with load voltage. Also, the source voltage is different then the voltage between the source point and the end point. Say, source will generate x amount of power. In the end you will get (x-Ohmic loss). This ohmic loss will be due to this potential difference (voltage) between source and endpoint. Nothing more, nothing less (assuming no other loss in procedure/set-up is induced).
- |
# Question #09585
Apr 12, 2015
The solution in which you dissolve the sodium chloride will have a lower freezing point.
This happens because freezing-point depression is a colligative property that depends on the concentration of particles in solution, not on what those particles are.
Mathematically, this is expressed as
$\Delta {T}_{f} = {K}_{f} \cdot {b}_{F} \cdot i$, where
$\Delta {T}_{f}$ - the freezing-point depression, defined as the freezing temperature of the pure solvent minus the freezing point of the solution;
${K}_{f}$ - the cryoscopic constant, which depends solely on the solvent;
${b}_{F}$ - the molality of the solution;
$i$ - the van't Hoff factor, which takes into account the number of particles produced by a compound when dissolved in solution.
The van't Hoff factor is the key to why the sodium chloride solution will have a lower freezing point.
Sodium chloride is a strong electrolyte, which means it dissociates completely in aqueous solution to give sodium cations and chloride anions
$N a C {l}_{\left(a q\right)} \to N {a}_{\left(A q\right)}^{+} + C {l}_{\left(a q\right)}^{-}$
Notice that 1 mole of $N a C l$ produces 1 mole of $N {a}^{+}$ and 1 mole of $C {l}^{-}$. This means that, regardless of how many moles of $N a C l$ you add to the solution, you'll have twice as many moles of ions present.
As a result, the van't Hoff factor for $N a C l$ will be 2.
This is not the case for ethanol. When placed in aqueous solution, a non-electrolyte does not dissociate to form ions. As a result, its van't Hoff factor will be equal to 1.
To compare the freezing points of the two solutions, just use
$\Delta {T}_{\text{f NaCl}} = {K}_{f} \cdot 0.5 \cdot 2 = {K}_{f}$
$\Delta {T}_{\text{f ethanol}} = {K}_{f} \cdot 0.75 \cdot 1 = 0.75 \cdot {K}_{f}$
or
${T}_{\text{f pure water" - T_"f NaCl" = K_f => T_"NaCl" = T_"f pure water}} - {K}_{f}$
${T}_{\text{f ethanol" = T_"f pure water}} - 0.75 \cdot {K}_{f}$
As you can see,
${T}_{\text{f NaCl" < T_"f ethanol}}$ |
# A computer's memory is finite, so how can there be languages more powerful than regular?
A computer has a finite memory. There are no computers with infinite memory. Therefore the only languages that a computer can process are those whose member strings are finite. As I recall, the computational power required for any finite language is small -- a deterministic finite state automata is sufficient to process finite languages, I think.
So there is no such thing as context-free languages or recursively enumerable languages, right? Programming languages are not context-free, they are just simple regular languages, right? The idea that a programming language is a context-free language ... well, that's an illusion, right?
Why do we even bother to talk about context-free languages, recursive languages, recursively enumerable languages since inside the (finite) computer everything is just a regular language?
Obviously I am playing devils-advocate here, but I would really like to understand the fallacy of my argument.
## Summary of Responses
Thank you very much to all who responded. I have studied your responses carefully. There are so many pearls of wisdom in your responses that I decided to summarize them. If my summary is incorrect, unclear, or missing an important concept, please let me know.
> A computer has a finite memory. There are no computers
> with infinite memory. Therefore the only languages that
> a computer can process are those whose member strings
> are finite.
### Responses
A computer's memory is finite, so in its memory it can recognize only finite languages. However, if you would pass an external input to the computer, then it could recognize some infinite languages.
Example: a*b is an infinite language and any string in that language can be recognized, even if the string exceeds the size of the computer's memory. Here's how: pass the string into the computer character by character. If the first character that is passed in is not an "a" then go to an error state. Otherwise, stay in the "a" state until a non-"a" character is input. If the character is not a "b" then go to the error state. Otherwise, go to an accept state. For any characters beyond "b" go to the error state.
Not all infinite languages can be recognized using this technique of feeding the input strings in as external input. In fact, the technique can be used only with regular languages. And even of the regular languages, only some of them can be recognized.
Example: For illustration purposes, suppose the computer has a ridiculously small memory size, say, 2 bytes. Suppose 1 byte is required for each state of a finite state automaton. Then the computer will not be able to recognize this regular language: a*b*c* because at least three byes are needed – one byte for the state that consumes all the a's, a second byte for the state that consumes all the b's, and a third byte for the state that consumes all the c's.
Any finite state automaton that requires more states than there are bytes in the computer's memory cannot be recognized.
Finally, with this technique of feeding arbitrarily long strings into the computer as external input, we would need to find a way around certain technical limitations such as a continuous power supply.
> A computer has a finite memory. Therefore the only languages
> that a computer can process are those whose member strings
> are finite … So there is no such thing as context-free languages
> or recursively enumerable languages, right?
### Responses
Consider an infinite language. It consists of strings of arbitrary length. Many strings will be longer than the size of the computer's memory. It is incorrect to say that because a computer can hold in its memory only those strings that are of a certain finite length, the language has only finitely many strings. It is true that any finite subset of a language is regular. But that does not mean that the original language must have been regular.
> A computer has a finite memory. Therefore, in general, a computer
> can only process a finite subset of an infinite language. A finite
> subset of a language is regular. So why don't we simply treat all
> languages used in computers as regular?
### Responses
The reason to treat, say, a programming language as context-free is that the context-free grammar tells how to parse the language. If you considered just the subset of programs with length no more than 232, that would be regular but the regular expression would likely consist of millions of individual cases and wouldn't be helpful for parsing the programs.
-
You can process arbitrarily long strings with finite memory, i.e. you don't have to store them (but nothing beyond regular). – dtldarek Aug 3 '13 at 9:48
Note, it isn't fair to say that there aren't infinite languages, it's only fair to say that nobody has ever needed them. – Karolis Juodelė Aug 3 '13 at 10:05
You are making the following invalid deduction: because a particular computer can only hold finitely many C programs in its memory, therefore there are only finitely many C programs. In general it is true that any finite subset of a language is regular, but that does not mean that the original language must have been regular, if it was infinite. – Carl Mummert Aug 3 '13 at 10:51
I know it is slightly OT but I remembered this question math.stackexchange.com/questions/946/…. It might also interest you ;) – Honza Brabec Aug 3 '13 at 18:20
The reason to treat a programming language as context-free is that the context-free grammar tells how to parse the language. If you considered just the subset of C consisting of programs of length no more than $2^{32}$ that would be regular, but the regular expression would likely consist of millions of individual cases, and wouldn't be helpful for parsing the programs. You might not even be able to fit the compiler in memory...
For a simpler example, consider a context free grammar for arithmetical expressions with natural numbers, addition and multiplication.
• E -> {sequence of digits 0-9}
• E -> ( E + E )
• E -> ( E * E )
If you only look at expressions with 500 or fewer symbols, that fragment of the language is regular, but it is much more difficult to describe as a regular language than as a context-free one. Plus, the parse tree for the context-free grammar gives a direct way to evaluate the expressions, while the parse tree for that smaller fragment as a regular language is not likely to help evaluate the expression.
-
Actually, in the sense of the program, you can make a computer as power as the universal Turing Machine. (In fact, the computer you are on probably is.) More precisely, you can write down explicitly a universal Turing Machine in many of the actual computer languages out there. So in a very real-world sense, there are computer program much more power than any finite automata (the model of computation that recognizes regular langauges).
It is true that a computer has finite memory. This just means that a computer will imitate any Turing machine until it runs out of memory, electric, etc. However any computation that halts takes finite amount of time. So theoretically, one would just need to build a more powerful computer. If a computation does not halt by an idealize Turing machine, neither will it by any real computer no matter how powerful. This does change the fact that your real computer is running the same program as a universal Turing machine.
Morever, you should make a distinction between languages and the model of computation. A language is a string of symbols. A recursive langauge is one decided by a Turing Machine. A recursively enumerable language is one accepted by a Turing machine. For instance, the set $H$ of code $\langle e, f \rangle$ such that the $e^\text{th}$ Turing Machine halts on input $f$ is the recursively enumerable set corresponding to the Halting problem. This is a just a set of numbers! An idea! Why should the fact that real computers have finite memory have any bearing on whether a recursive language or regular language (some infinite set; an abstract idea) should "exists" or not.
To ask whether languages (an abstract notion) exists because of the nature of real computers is not entire well-defined question. Does a recursive language not exists because no existing computer can run long enough to halt? But it will halt eventually. Similarly, one can make huge finite automata that will not halt in anyone's lifetime. It will stop eventually too. Does this mean no regular language exist either because no current device can finish the calculation.
Recursive language and regular languages are just ideas computed by other abstract notions like Turing machines or finite automata. The nature of physical computers don't make ideas real.
-
There are some interesting points raised here.
The thing about all languages recognizable by a computer being finite is wrong. For example, you can make an interactive program that reads an input string from a user, and your program processes each character in the input stream as it is entered. We can eliminate the need to store the whole input string in memory. This means we can have an arbitrarily long input string. It is also possible to make up an infinite language and a program that recognizes the language. For example, suppose the set of alphabet is $\{0,1\}$. I can make up a language $L = 1^*$. $L$ is infinite, and it is obvious how a recognizer can be programmed. ($L$ here is still regular though.)
It is true, however, that assuming limited memory, every language recognizable is regular simply because you can view all the possible states of the memory as states of a DFA. Obviously there is an unfavorable character of this DFA - the outrageously large number of states. This, in a sense, is a reason why programming languages have control statements other than "if" and "case".
-
There's no fallacy. In practice everything really is regular (finite). However, you won't get much done if you treat them as such. A DFA for the reasonably long strings of an interesting language will end up having unreasonably many states.
-
Thank you for your response. Would you elaborate on "you won't get much done if you treat them as such" please? Do you mean that we can treat a language as, say, a context-free language, even though it is really regular? And by doing so, we can have a simpler implementation? So context-free, recursive, and recursive enumerable languages are just fictitious things we create to make it easier to implement finite automata? – Roger Costello Aug 3 '13 at 10:25
They are no excuse, if every language of importance were regular, every interesting computational problem would be easy to compute, which is not the case at all. – sxd Aug 3 '13 at 10:29
As an example why infinite languages are useful: note that that to recognize whether a program is grammatically correct is for most programs not regular furthermore there exists countable infinite valid programs over most languages – sxd Aug 3 '13 at 10:35
@RogerCostello, do try constructing a DFA for the strings in language generated by $S \to () | (S) | SS$, that are at most 10 symbols long. Non-regular grammars are used because, contrary to what theorists will have you believe, they make your life much easier. – Karolis Juodelė Aug 3 '13 at 11:24
It's probably safe to say that $10^{80}$ (an estimate of the number of atoms in the observable universe) is an upper bound to the number of nodes or the size of the number. – Karolis Juodelė Aug 3 '13 at 11:14 |
Surface conductance between the ground surface and the external ambient.
Symbol
$K_{\mathrm{E}}$
Unit
W/m²
Formulae
$9.0$
Used in
$R_{\mathrm{q12}}$
$R_{\mathrm{q22}}$
$R_{\mathrm{q32}}$ |
# USACO 2014 December Contest, Bronze Problem 1. Marathon
USACO2014-DEC-B1
(Analysis by Nick Wu)
Our first instinct when trying to figure out which point to skip is to try all of them. If we choose to skip each point and compute the new distance directly, then it takes about NN operations to compute the distance and there are about NN points to check, giving us an algorithm which runs in about NNoperations. This will be too slow when NN gets to be 100,000.
Let's take a closer look at what happens when you skip a specific point. If we number the points from 1 to NN, and skip point KK, then the path we take goes from point 1 to point K1K−1, then from point K1K−1 to point K+1K+1, and then from point K+1K+1 to point NN. The distance of this path is exactly equal to the following:
(total distance without skipping any points) - (distance between points K1K−1 and KK) - (distance between points KK and K+1K+1) + (distance between points K1K−1 and K+1K+1).
If we compute the total distance without skipping any points beforehand, then figuring out how long the path is when we want to skip a specific point no longer requires NN operations! It only requires a constant number of operations, and that will be fast enough.
Here is my Java code:
import java.io.*;
import java.util.*;
public class marathon {
public static void main(String[] args) throws IOException {
PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("marathon.out")));
int[] x = new int[n];
int[] y = new int[n];
for(int i = 0; i < n; i++) {
x[i] = Integer.parseInt(st.nextToken());
y[i] = Integer.parseInt(st.nextToken());
}
int totalDistance = 0;
for(int i = 1; i < n; i++) {
totalDistance += Math.abs(x[i] - x[i-1]) + Math.abs(y[i] - y[i-1]);
}
int largestSkip = 0;
for(int i = 1; i < n-1; i++) {
int noSkipDistance = Math.abs(x[i+1] - x[i]) + Math.abs(x[i] - x[i-1]) + Math.abs(y[i+1] - y[i]) + Math.abs(y[i] - y[i-1]);
int skipDistance = Math.abs(x[i+1] - x[i-1]) + Math.abs(y[i+1] - y[i-1]);
largestSkip = Math.max(largestSkip, noSkipDistance - skipDistance);
}
pw.println(totalDistance - largestSkip);
pw.close();
}
} |
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# Three printing presses, R, S, and T, working together at the
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Three printing presses, R, S, and T, working together at the [#permalink]
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Updated on: 17 Sep 2013, 07:25
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Three printing presses, R, S, and T, working together at their respective constant rates, can do a certain printing job in 4 hours. S and T, working together at their respective constant rates, can do the same job in 5 hours. How many hours would it take R, working alone at its constant rate, to do the same job?
A. 8
B. 10
C. 12
D. 15
E. 20
Originally posted by vivek123 on 04 Mar 2006, 12:41.
Last edited by Bunuel on 17 Sep 2013, 07:25, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: Three printing presses, R, S, and T, working together at the [#permalink]
### Show Tags
06 Dec 2016, 08:23
6
4
vivek123 wrote:
Three printing presses, R, S, and T, working together at their respective constant rates, can do a certain printing job in 4 hours. S and T, working together at their respective constant rates, can do the same job in 5 hours. How many hours would it take R, working alone at its constant rate, to do the same job?
A. 8
B. 10
C. 12
D. 15
E. 20
We are given that three printing presses, R, S, and T, working together at their respective constant rates, can do a certain printing job in 4 hours.
We can let r, s and t be the times, in hours, for printing presses R, S and T to complete the job alone at their respective constant rates. Thus, the rate of printing press R = 1/r, the rate of printing press S = 1/s, and the rate of printing press T = 1/t. Recall that rate = job/time and, since they are completing one printing job, the value for the job is 1. Since they complete the job together in 4 hours, the sum of their rates is 1/4, that is:
1/r + 1/s + 1/t = 1/4
We are also given that printing presses S and T, working together at their respective constant rates, can do the same job in 5 hours. Thus:
1/s + 1/t = 1/5
We can substitute 1/5 for 1/s + 1/t is the equation 1/r + 1/s + 1/t = 1/4 and we have:
1/r + 1/5 = 1/4
1/r = 1/4 - 1/5
1/r = 5/20 - 4/20
1/r = 1/20
r = 20
Thus, it takes printing press R 20 hours to complete the job alone.
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Re: RTD - Combined Worked Sum [#permalink]
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01 Dec 2012, 05:43
17
4
SreeViji wrote:
Three printing presses, R, S, and T, working together at their respective constant rates, can do a certain printing job in 4 hours. S and T, working together at their respective constant rates, can do the same job in 5 hours. How many hours would it take R, working alone at its constant rate, to do the same job?
A. 8
B. 10
C. 12
D. 15
E. 20
This is my nightmare . I remember studying some formula for combined work in my school days, but forgot. And so everytime , I come across such sums, my brain looks for the formula and fails . Even after trying to work it in RTD method , I am not able to solve. Somebody please help.
Let's assume the work to be something like printing 20 papers. I'm picking 20 as it is the LCM of 4 & 5. Any number in that place will work just as well.
Speed of Three machines together = 5 papers per hour
Speed of Two machines together = 4 papers per hour
So speed of remaining machine = 1 paper per hour
So, to print 20 papers, this machine would take 20/1 = 20 hours. Answer is E.
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### Show Tags
04 Mar 2006, 13:44
4
1
1/R+1/S+ 1/T = 1/4
1/S+ 1/T = 1/5
1/R=1/4-1/5
R=20
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Re: Three printing presses, R, S, and T, working together at the [#permalink]
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23 Dec 2015, 08:50
4
2
$$\frac{1}{r} + \frac{1}{s} + \frac{1}{t} = \frac{1}{4}$$
$$\frac{1}{r} + \frac{1}{s} = \frac{1}{5}$$ $$thus ->$$ $$\frac{1}{5} + \frac{1}{t} = \frac{1}{4}$$
$$\frac{1}{t} = \frac{1}{4} - \frac{1}{5}$$
$$\frac{1}{t} = \frac{5}{20} - \frac{4}{20}$$
$$t = 20$$
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Re: RTD - Combined Worked Sum [#permalink]
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02 Dec 2012, 13:13
3
1
This is how I solved the problem:
4 hours * (rate of R + rate of S + rate of T) = total job
5 hours * (rate of S + rate of T) = total job
equate the two, reduce them.
4*rate of R = rate of S + rate of T
Plug back into equation 2: 5*(4*rate of R) = total
20* rate of R = total
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### Show Tags
04 Mar 2006, 13:35
2
=1/4-1/5=1/20
r can do 1/20 job in 1 hour
r can do the whole job in 20 hours.
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### Show Tags
04 Mar 2006, 13:16
1
vivek123 wrote:
Three printing presses, R, S, and T, working together at their respective constant rates,
can do a certain printing job in 4 hours. S and T, working together at their respective
constant rates, can do the same job in 5 hours. How many hours would it take R, working
alone at its constant rate, to do the same job?
A. 8
B. 10
C. 12
D. 15
E. 20
E.
R, S and T can do Job J in 4 hours. Which also means the in one hour all three will do J/4 th of the job. Also represented by:
1/r+1/s+1/t = j/4
We know that T and S working together do the same job J in 5 hours. Again in 1 hour of T and S working together they will be done with:
1/s + 1/t = J/5
so 1/r + j/5 = j/4
1/r = j/4-j/5=j/20
So in 1 hr working alone R can do 1/20th of J. Therefore R would need 20 hrs.
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### Show Tags
04 Mar 2006, 18:23
same logic above 1/r =1/5-1/4 ... hence r =20 hrs
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RTD - Combined Worked Sum [#permalink]
### Show Tags
01 Dec 2012, 01:42
Three printing presses, R, S, and T, working together at their respective constant rates, can do a certain printing job in 4 hours. S and T, working together at their respective constant rates, can do the same job in 5 hours. How many hours would it take R, working alone at its constant rate, to do the same job?
A. 8
B. 10
C. 12
D. 15
E. 20
This is my nightmare . I remember studying some formula for combined work in my school days, but forgot. And so everytime , I come across such sums, my brain looks for the formula and fails . Even after trying to work it in RTD method , I am not able to solve. Somebody please help.
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Re: Three printing presses, R, S, and T, working together at the [#permalink]
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17 Sep 2013, 07:26
Merging similar topics.
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Re: Three printing presses, R, S, and T, working together at the [#permalink]
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20 Nov 2013, 04:56
I did it like this:
rate R: x
rate S: y
rate T: z
We know that work(w) = rate (r) * time
and we can combine rates.
so I did:
R,S,T working together to complete ONE job : 1 = x+y+z *4
S,T working together to complete ONE job: 1 = y+z *5 ==> 1/5 = y+z ==> substitute in first equation I get x = 1/20 which tells me that machine r completes the job in 20h. Hence E.
I took 3 minutes though, because I wasn't 100 % sure that I can solve it like this. Can you please confirm that this is a appropriate way to solve problems like this or explain the answers above a bit more. I think I can follow but not sure.
Thanks!
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Re: Three printing presses, R, S, and T, working together at the [#permalink]
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06 Dec 2016, 08:29
vivek123 wrote:
Three printing presses, R, S, and T, working together at their respective constant rates, can do a certain printing job in 4 hours. S and T, working together at their respective constant rates, can do the same job in 5 hours. How many hours would it take R, working alone at its constant rate, to do the same job?
A. 8
B. 10
C. 12
D. 15
E. 20
LCM of ( 4, 5 & T ) = 20T
So, let the total work be 20T units...
Combined efficiency of R , S & T is 9+T
Now, 20T/9 + T = 5
Or, 20T = 45 + 5T
So, 15T = 45
Thus, T = 3
So, THe time required by T to do the work will be 20T/T = 20
Hence, correct answer will be (E) 20
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Re: Three printing presses, R, S, and T, working together at the [#permalink]
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26 Jan 2017, 09:34
1/5 + 1/x = 1/4
4/20 + 1/x = 5/20
1/x = 1/20
x = 20
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Re: Three printing presses, R, S, and T, working together at the [#permalink]
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20 Oct 2018, 14:28
Even though this is a relatively easy question, it gives us the opportunity to practice a number of my GMAT timing tips (the links below include growing lists of questions that you can use to practice each tip):
Rate problems: Use D = R x T and W = R x T
Like most work rate problems, we can start with the equation W = R x T and then plug in the work, rate, and time for each scenario that we are considering.
Set the amount of work equal to 1 for a single job
Because we’re talking about a single printing job, we just set W = 1 for each scenario.
Add rates when they are simultaneous
Let’s define variables for the rates for printing presses R, S, and T as Rr, Rs, and Rt. Remember that we can add rates when they are simultaneous, so, when all 3 presses are working together, the rate is Rr + Rs + Rt. When just S and R are working together, the rate is Rs + Rt.
Rate and time are reciprocals of each other for a single job
Since we are given the amounts of time for each scenario, we can set the rate equal to the reciprocal of the time for each scenario. This means that Rr + Rs + Rt = 1/4 and Rs + Rt = 1/5. In addition, we are solving for the time it takes printing press R to do the job working alone; if we call this time Tr, then Tr = 1/Rr, and we can solve for Tr if we know Rr.
Eliminate combinations of variables using substitution
While we can’t solve for Rs and Rt separately, we don’t have to. Since we know their sum Rs + Rt = 1/5, we can just plug this value in for (Rs + Rt) in the equation Rr + Rs + Rt = 1/4. This is enough to allow us to solve for Rr, which then allows us to solve for Tr, which is the final answer to this question.
Please let me know if you have any questions, or if you want me to post a video solution!
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Re: Three printing presses, R, S, and T, working together at the [#permalink]
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20 Oct 2018, 14:40
Solved in 10 seconds without writing
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Re: Three printing presses, R, S, and T, working together at the [#permalink]
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20 Oct 2018, 14:44
vivek123 wrote:
Three printing presses, R, S, and T, working together at their respective constant rates, can do a certain printing job in 4 hours. S and T, working together at their respective constant rates, can do the same job in 5 hours. How many hours would it take R, working alone at its constant rate, to do the same job?
A. 8
B. 10
C. 12
D. 15
E. 20
1/R + 1/S + 1/T = 1/4
We also know 1/S + 1/T = 1/5
So 1/R = 1/4 - 1/5 = 1/20
Time for R = 20
Posted from my mobile device
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Re: Three printing presses, R, S, and T, working together at the [#permalink]
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26 Oct 2019, 21:36
Assume rate of work of press R, S and T are R, S and T
Given, R+S+T=1/4
S+T=1/5
Then , R=1/4-1/5=1/20
If R can the job's 1/20 part in 1 hour , then can do the whole job in 20 hours.
Ans. 20
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Three printing presses, R, S, and T, working together at the [#permalink]
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30 Oct 2019, 13:15
TOTAL HOURS (S,T,R)=(S and T)*R/ (S and T) +R
SO, 4=5R/5+R
R=20
ANS:E
Three printing presses, R, S, and T, working together at the [#permalink] 30 Oct 2019, 13:15
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# Solved Problems-16
Problems-16
Find the Fourier transform of the existing voltage,
v(t) = V0e–tt ≥ 0
= 0, t ≤ 0
and sketch approximately its amplitude and phase spectrum.
Solution
The amplitude and phase are
Spectra |
[Gray]
Remark 85.2.4 (Sheafification of presheaves of topological spaces). In this remark we briefly discuss sheafification of presheaves of topological spaces. The exact same arguments work for presheaves of topological abelian groups, topological rings, and topological modules (over a given topological ring). In order to do this in the correct generality let us work over a site $\mathcal{C}$. The reader who is interested in the case of (pre)sheaves over a topological space $X$ should think of objects of $\mathcal{C}$ as the opens of $X$, of morphisms of $\mathcal{C}$ as inclusions of opens, and of coverings in $\mathcal{C}$ as coverings in $X$, see Sites, Example 7.6.4. Denote $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}, \textit{Top})$ the category of sheaves of topological spaces on $\mathcal{C}$ and denote $\textit{PSh}(\mathcal{C}, \textit{Top})$ the category of presheaves of topological spaces on $\mathcal{C}$. Let $\mathcal{F}$ be a presheaf of topological spaces on $\mathcal{C}$. The sheafification $\mathcal{F}^\#$ should satisfy the formula
$\mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{C}, \textit{Top})}(\mathcal{F}, \mathcal{G}) = \mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C}, \textit{Top})}(\mathcal{F}^\# , \mathcal{G})$
functorially in $\mathcal{G}$ from $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}, \textit{Top})$. In other words, we are trying to construct the left adjoint to the inclusion functor $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}, \textit{Top}) \to \textit{PSh}(\mathcal{C}, \textit{Top})$. We first claim that $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}, \textit{Top})$ has limits and that the inclusion functor commutes with them. Namely, given a category $\mathcal{I}$ and a functor $i \mapsto \mathcal{G}_ i$ into $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}, \textit{Top})$ we simply define
$(\mathop{\mathrm{lim}}\nolimits \mathcal{G}_ i)(U) = \mathop{\mathrm{lim}}\nolimits \mathcal{G}_ i(U)$
where we take the limit in the category of topological spaces (Topology, Lemma 5.14.1). This defines a sheaf because limits commute with limits (Categories, Lemma 4.14.10) and in particular products and equalizers (which are the operations used in the sheaf axiom). Finally, a morphism of presheaves from $\mathcal{F} \to \mathop{\mathrm{lim}}\nolimits \mathcal{G}_ i$ is clearly the same thing as a compatible system of morphisms $\mathcal{F} \to \mathcal{G}_ i$. In other words, the object $\mathop{\mathrm{lim}}\nolimits \mathcal{G}_ i$ is the limit in the category of presheaves of topological spaces and a fortiori in the category of sheaves of topological spaces. Our second claim is that any morphism of presheaves $\mathcal{F} \to \mathcal{G}$ with $\mathcal{G}$ an object of $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}, \textit{Top})$ factors through a subsheaf $\mathcal{G}' \subset \mathcal{G}$ whose size is bounded. Here we define the size $|\mathcal{H}|$ of a sheaf of topological spaces $\mathcal{H}$ to be the cardinal $\sup _{U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})} |\mathcal{H}(U)|$. To prove our claim we let
$\mathcal{G}'(U) = \left\{ \quad s \in \mathcal{G}(U) \quad \middle | \quad \begin{matrix} \text{there exists a covering }\{ U_ i \to U\} _{i \in I} \\ \text{such that } s|_{U_ i} \in \mathop{\mathrm{Im}}(\mathcal{F}(U_ i) \to \mathcal{G}(U_ i)) \end{matrix} \quad \right\}$
We endow $\mathcal{G}'(U)$ with the induced topology. Then $\mathcal{G}'$ is a sheaf of topological spaces (details omitted) and $\mathcal{G}' \to \mathcal{G}$ is a morphism through which the given map $\mathcal{F} \to \mathcal{G}$ factors. Moreover, the size of $\mathcal{G}'$ is bounded by some cardinal $\kappa$ depending only on $\mathcal{C}$ and the presheaf $\mathcal{F}$ (hint: use that coverings in $\mathcal{C}$ form a set by our conventions). Putting everything together we see that the assumptions of Categories, Theorem 4.25.3 are satisfied and we obtain sheafification as the left adjoint of the inclusion functor from sheaves to presheaves. Finally, let $p$ be a point of the site $\mathcal{C}$ given by a functor $u : \mathcal{C} \to \textit{Sets}$, see Sites, Definition 7.32.2. For a topological space $M$ the presheaf defined by the rule
$U \mapsto \text{Map}(u(U), M) = \prod \nolimits _{x \in u(U)} M$
endowed with the product topology is a sheaf of topological spaces. Hence the exact same argument as given in the proof of Sites, Lemma 7.32.5 shows that $\mathcal{F}_ p = \mathcal{F}^\# _ p$, in other words, sheafification commutes with taking stalks at a point.
There are also:
• 12 comment(s) on Section 85.2: Formal schemes à la EGA
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0
TECHNICAL PAPERS: Soft Tissue
# Elasticity Imaging of Polymeric Media
[+] Author and Article Information
Mallika Sridhar
University of California, Davis, CA 95616
Jie Liu
University of Illinois at Urbana-Champaign, Urbana, IL 61801
Michael F. Insana
University of California, Davis, CA, and University of Illinois at Urbana-Champaign, 405 North Mathews, Room 4247 Urbana, IL 61801mfi@uiuc.edu
J Biomech Eng 129(2), 259-272 (Sep 15, 2006) (14 pages) doi:10.1115/1.2540804 History: Received July 12, 2006; Revised September 15, 2006
## Abstract
Viscoelastic properties of soft tissues and hydropolymers depend on the strength of molecular bonding forces connecting the polymer matrix and surrounding fluids. The basis for diagnostic imaging is that disease processes alter molecular-scale bonding in ways that vary the measurable stiffness and viscosity of the tissues. This paper reviews linear viscoelastic theory as applied to gelatin hydrogels for the purpose of formulating approaches to molecular-scale interpretation of elasticity imaging in soft biological tissues. Comparing measurements acquired under different geometries, we investigate the limitations of viscoelastic parameters acquired under various imaging conditions. Quasi-static (step-and-hold and low-frequency harmonic) stimuli applied to gels during creep and stress relaxation experiments in confined and unconfined geometries reveal continuous, bimodal distributions of respondance times. Within the linear range of responses, gelatin will behave more like a solid or fluid depending on the stimulus magnitude. Gelatin can be described statistically from a few parameters of low-order rheological models that form the basis of viscoelastic imaging. Unbiased estimates of imaging parameters are obtained only if creep data are acquired for greater than twice the highest retardance time constant and any steady-state viscous response has been eliminated. Elastic strain and retardance time images are found to provide the best combination of contrast and signal strength in gelatin. Retardance times indicate average behavior of fast $(1–10s)$ fluid flows and slow $(50–400s)$ matrix restructuring in response to the mechanical stimulus. Insofar as gelatin mimics other polymers, such as soft biological tissues, elasticity imaging can provide unique insights into complex structural and biochemical features of connectives tissues affected by disease.
<>
## Figures
Figure 1
(a) Creep curves for a second-order (L=2) Voigt model and a step stress stimulus are illustrated. Curve a is drawn directly from Eq. 8 with finite η0; its slope at t≫T2 is σa∕η0. Curve b is from the same equation where η0=∞. In both cases, ϵ2∕ϵ1=2.5, T1=3s, and T2=100s. (b) The corresponding Fourier spectra D̆(ω) are from Eq. 10. Spectra from a step and 1s ramp stress stimulus are compared.
Figure 2
(a) Retardation spectra from simulated data. Plotted are L̃(ω)=L(τ)∣τ=1∕ω for comparison with the Fourier spectrum. Creep data were generated from Eq. 12 for ϵ0=σa∕η0=0 assuming a broadband, bimodal input as given by the circle points (Input). Estimated retardation spectra (RS) L̃(k)(ω) for k=1,2,5,6, are compared to the Fourier spectrum (FS) D̆(ω), computed from the same data. (b)L̃(6) estimates without noise in the creep data and with noise (signal-to-noise ratio=32.2dB). A ninth-order polynomial filter was applied to the noisy data before estimation.
Figure 3
Limitation of L(k)(τ) for representing retardance time distributions. The abscissa is b∕a from the log-normal input distribution L(τ)=exp[−(lnτ−a)2∕2b2]. The ordinate is the full-width-at-half-maximum bandwidth of retardance spectral estimates. Circles denote the exact output bandwidth for the input distribution, while the curves are bandwidths for kth-order estimates using noiseless creep data. Results suggest that the L(6)(τ) represents bandwidths of log-normal distributions above 150s with acceptable bias error.
Figure 4
Illustration of collagen structures in connective tissue (fibril) and in gelatin (aggregates)
Figure 5
Illustrations of four viscoelastic experiments. (a) Measurement method A applies uniaxial stress or strain stimuli to unconfined gelatin samples to estimate compressive relaxation modulus E(t) or compressive creep compliance D(t). It is also the ultrasonic strain imaging technique. (b) Method B applies uniaxial strain to estimate the compressive wave modulus M(t) for rigidly confined sample boundaries. (c) Method C is a cone-plate rheometer applied to estimate shear creep compliance J(t). (d) Method D applies an indenter to gelatin samples to estimate the elastic modulus E0. All positions are computer controlled with submicrometer accuracy, and forces are measured with a precision of 0.01g.
Figure 6
(a) Shear creep measured with applied step stresses of σa′=3 and 30Pa using Method C and Type B gelatin (5.5%). (b) Viscosity estimates (Sec. 2) versus time for creep data at 30Pa. Steady-state values were attained beginning at ∼600s. (c) Example of shear creep recovery curve for Type A gelatin at σa′=100Pa. Values calculated from the creep and recovery phases are reported separately.
Figure 7
Demonstrations of linearity. (a) Stress-strain curves for stiff (10%) and soft (5.5%) Type A gelatin using unconfined samples and uniaxial harmonic stimuli (Method A). The two stress levels indicated were used in subsequent creep measurements. (b) Shear creep Fourier spectra for Type B gelatin (Method C).
Figure 8
Poisson’s ratio estimates versus time, i.e., ν(t). Error bars denote one standard deviation computed by propagating displacement measurement errors.
Figure 9
(a) Dependence of Tℓ on acquisition time, and the effect of eliminating steady-state viscosity (linear term in Eq. 17). T1 and T2 estimates for a third-order Voigt model are shown. (b) Variation of T1 contrast over acquisition time is shown.
Figure 10
Comparisons of measurements made using different methods. Samples were all type A gelatin aged three days. (a) Elastic modulus, (b) equilibrium compliance, and (c) steady-state viscosity under compression. Error bars are standard deviations that indicate uncertainty between repeated measurements.
Figure 11
(a) Contrast between 10% and 5.5% homogeneous gelatin samples for seven compliance parameters. (b) Example ϵ0 image for a composite sample consisting of 5.5% gel background with a 10% gel inclusion. (c) Example T1 image.
Figure 12
Normalized Fourier, retardation, and relaxation spectra. (a) Unconfined type A gelatin samples (aged three days) loaded uniaxially at σa=860Pa are measured for 2000s using Method A. (b) Confined type A gelatin samples (aged 1 day) strained uniaxially at ϵa=0.02 are measured for 2500s using Method B. (c) Type B gelatin samples (aged 1 day) sheared at σa′=3Pa are measured for 3000s in a rheometer using Method C. (d) Unconfined type A gelatin samples (aged three days) strained uniaxially at ϵa=0.08 are measured for 2000s by combining Methods A and B. Arrows indicate frequencies corresponding to the respondance times given in Table 1. Spectral amplitudes are uniformly reduced across the bandwidth as samples age.
Figure 13
Effects of rest time on viscoelastic estimates. Top: Variation of T1 (left group), T2 (middle group), and T3 (right group) for a third-order Voigt model are shown for baseline measurements (0) and rest times of 1 and 2h. Error bars indicate fitting uncertainties. Bottom: Table showing initial baseline retardance times in seconds and percent biases for rest times of 1 or 2h between measurements.
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Scenarios and parties¶
Bell experiments are performed using several observers, called parties. These parties are often numbered and named alphabetically Alice, Bob, Charlie and so on. A sequence of parties is a Bell scenario.
A party is unambiguously described by a sequence giving the number of outcomes for each measurement settings.
A scenario can be represented in plain text using the following grammar:
• Scenario := [Party Party ... ]
• Party := (Input Input ...)
• Input := number >= 2
Examples:
• the CHSH scenario is written down [(2 2) (2 2)],
• the Sliwa scenario is written down [(2 2) (2 2) (2 2)],
• the I2233 scenario is written down [(3 3) (3 3)],
• the I3322 scenario is written down [(2 2 2) (2 2 2)].
In the FaacetsPaper, we introduced the following notation: $$[(k_{11} k_{12} \ldots k_{1 m_1})~(k_{21} k_{22} \ldots k_{2 m_2}) \ldots (k_{n 1} k_{n 2} \ldots k_{n m_n})]$$, with $$m_i \ge 1$$ is the number of measurement settings for the $$i^\text{th}$$ party and $$k_{i j} \ge 2$$ is the number of measurement outcomes for the $$j^\text{th}$$ measurement setting of the $$i^\text{th}$$ party.
Canonical parties¶
Notice that parties (3 2) and (2 3) have essentially the same measurement structure up to a reordering to measurement settings. We thus define a party as canonical when the number of outcomes for successive measurement settings is non-increasing.
Thus, the canonical form of (2 3) is (3 2).
Canonical scenarios¶
Notice also that scenarios [(2 2) (3 3)] and [(3 3) (2 2)] are identical under reordering of parties. We thus prescribe that a scenario is canonical when its parties are themselves canonical and ordered lexicographically: for all successive non-identical parties $$i$$ and $$i+1$$, there is a $$j \ge 0$$ such that $$\forall k < j$$ we have $$k_{i k} = k_{i+1, k}$$ and $$k_{i j} > k_{i+1, j}$$. For these ordering purposes, we define $$k_{i j} = 0$$ for $$j > m_i$$. |
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Bull. Korean Math. Soc. 2009 Vol. 46, No. 6, 1041—1248
Skew Laurent polynomial extensions of Baer and p.p.-rings Alireza R. Nasr-Isfahani and Ahmad Moussavi MSC numbers : 16S34, 16S36 Bull. Korean Math. Soc. 2009 Vol. 46, No. 6, 1041—1050
An ideal-based zero-divisor graph of 2-primal near-rings Patchirajulu Dheena and Balasubramanian Elavarasan MSC numbers : 16Y30, 13A99 Bull. Korean Math. Soc. 2009 Vol. 46, No. 6, 1051—1060
On summation theorems for the ${}_3F_2(1)$ series K. Srinivasa Rao and R. Suresh MSC numbers : 33C20, 33C90, 81Q99 Bull. Korean Math. Soc. 2009 Vol. 46, No. 6, 1061—1068
On the decomposition of extending lifting modules Chaehoon Chang and Jongmoon Shin MSC numbers : Primary 16D40, 16D70, 16L30 Bull. Korean Math. Soc. 2009 Vol. 46, No. 6, 1069—1077
Uniqueness theorems of meromorphic functions of a certain form Junfeng Xu, Qi Han, and Jilong Zhang MSC numbers : 30D35, 30D20, 30D30 Bull. Korean Math. Soc. 2009 Vol. 46, No. 6, 1079—1089
Extensions of generalized stable rings Zhang Wanru MSC numbers : 16E50, 16U99 Bull. Korean Math. Soc. 2009 Vol. 46, No. 6, 1091—1097
Trigonometry in extended hyperbolic space and extended de Sitter space Yunhi Cho MSC numbers : 51M10, 51M25, 53B30 Bull. Korean Math. Soc. 2009 Vol. 46, No. 6, 1099—1133
Products of differentiation and composition on Bloch spaces Sh\^{u}ichi Ohno MSC numbers : Primary 47B38; Secondary 30H05 Bull. Korean Math. Soc. 2009 Vol. 46, No. 6, 1135—1140
On the Gauss map of surfaces of revolution without parabolic points Young Ho Kim, Chul Woo Lee, and Dae Won Yoon MSC numbers : 53A05, 53B25 Bull. Korean Math. Soc. 2009 Vol. 46, No. 6, 1141—1149
On the $d$th power residue symbol of function fields Su Hu MSC numbers : 11T55 Bull. Korean Math. Soc. 2009 Vol. 46, No. 6, 1151—1152
On self-reciprocal polynomials at a point on the unit circle Seon-Hong Kim MSC numbers : Primary 30C15; Secondary 26C10 Bull. Korean Math. Soc. 2009 Vol. 46, No. 6, 1153—1158
Critical blow-up and extinction exponents for non-Newton polytropic filtration equation with source Jun Zhou and Chunlai Mu MSC numbers : 35K50, 35K55, 35K65, 35B33 Bull. Korean Math. Soc. 2009 Vol. 46, No. 6, 1159—1173
Sensitivity analysis for a system of generalized nonlinear mixed quasi-variational inclusions with $(A,\eta)$-accretive mappings in Banach spaces Jae Ug Jeong and Soo Hwan Kim MSC numbers : 49J40, 90C33 Bull. Korean Math. Soc. 2009 Vol. 46, No. 6, 1175—1188
Two meromorphic functions sharing sets concerning small functions Ting-Bin Cao MSC numbers : Primary 30D35, Secondary 30D30 Bull. Korean Math. Soc. 2009 Vol. 46, No. 6, 1189—1200
Applications of generalized Kummer's summation theorem for the series ${}_{2}F_{1}$ Yong Sup Kim and Arjun K. Rathie MSC numbers : 33C05, 33C20 Bull. Korean Math. Soc. 2009 Vol. 46, No. 6, 1201—1211
On the structure of minimal submanifolds in a Riemannian manifold of non-negative curvature Gabjin Yun and Dongho Kim MSC numbers : 53C21 Bull. Korean Math. Soc. 2009 Vol. 46, No. 6, 1213—1219
On volumes of parallel $2n$-hedron in a Lorentz vector space Seongil Park, Byung Hak Kim, Jin Hyuk Choi, and Young Ok Lee MSC numbers : Primary 53B30, 53C50 Bull. Korean Math. Soc. 2009 Vol. 46, No. 6, 1221—1228
Quasi-inner functions of a generalized Beurling's theorem Yun-Su Kim MSC numbers : 47A15, 47A56, 47B37, 47B38 Bull. Korean Math. Soc. 2009 Vol. 46, No. 6, 1229—1236
Estimating the domain of attraction via moment matrices Chunji Li, Cheon Seoung Ryoo, Ning Li, and Lili Cao MSC numbers : 37C75, 44A60 Bull. Korean Math. Soc. 2009 Vol. 46, No. 6, 1237—1248 |
An atom is said to be stable when it attains noble gas configuration that is the outermost orbitals are completely filled. In the diagram each line represents one pair of shared electrons. The Lewis structure shows us that the carbon atom makes 4 sigma bonds to hydrogen and has no . It has been shown in Fig. Finally, the hybrid orbital concept applies well to triple-bonded groups, such as alkynes and nitriles. Lectures by Walter Lewin. 36.12. The overlapping of orbitals has been shown in Fig. Example: C 2 H 2 (acetylene or ethyne). The π cloud further merge to form a single cylindrical electron cloud along the internuclear axis (Fig 36.12). 1. Ethyne, C 2 H 2, forms three σ bonds: they … If we put all of the molecular orbitals of ethyne together, in a single energy diagram, it would look as follows. The dashed lines show the remaining p orbitals which do not take part in the bonding. The simplest molecule with a carbon-carbon bond is ethane, C 2 H 6.In ethane (CH 3 CH 3), both carbons are sp 3-hybridized, meaning that both have four bonds with tetrahedral geometry.An sp 3 orbital of one carbon atom overlaps end to end with an sp 3 orbital of the second carbon atom to form a carbon-carbon σ bond. Figure Molecular Orbital Energy-Level Diagram for $$\pi$$ Each oxygen atom in ozone has 6 valence electrons, so O 3 has a total of Let's look at the molecular orbital diagram of ozone. 1,109,556 views Ethyne is built from hydrogen atoms (1s 1) … Each line in this diagram represents one pair of shared electrons. Pi bond: A covalent bond resulting from the formation of a molecular orbital by side-to-side overlap of atomic orbitals along a plane perpendicular to a line connecting the nuclei of the atoms, denoted by the symbol π. The four 1t clouds so formed further merge into one another to form a single cylindrical electron cloud around the inter-nuclear axis representing C–C sigma bond. An ethyne molecule is practically 2 CH molecules. i) How many sigma (6) and pi (1) bonds exist in C,H, molecule? Ethene is built from hydrogen atoms (1s 1) and carbon atoms (1s 2 2s 2 2p x 1 2p y 1). 3 xx sigma and 2 xx pi In the acetylene molecule, H-C-=C-H, we can directly count 3 sigma bonds, 2xxC-H and 1xxC-C. Cylindrical n-electron cloud in ethyne. Molecular Orbital of Methane, CH4. For ethene, the σ framework is created by the interaction of the sp 2 hybrid orbitals of the C atoms and H1s orbitals. This colorless gas (lower hydrocarbons are generally gaseous in nature) is widely used as a fuel and a chemical building block. D.Explain why the colour of Bayer's reagent gets discharged when treated with an alkene. p orbital lobes are in the plane of the paper. C3H5+ - drawing the pii MO hydrogen bonding 2019 OCR Chemistry (A) - Paper 2: Organic Synthesis [Unofficial Mark Scheme] An Orbital View of the Bonding in Ethyne Ethyne is built from hydrogen atoms (1s 1 ) and carbon atoms (1s 2 2s 2 2p x 1 2p y 1 ). 3D ethyne drawn with p orbitals as lines and pi electrons explicitly drawn in, in a manner similar to showing lone pair electrons. ii) Determine the hybridization scheme in C,H, molecule. In picture 1 we show the molecular orbital structure of F2. The percentage of s and p are 50 %. p orbital … Orbital diagram of ethyne. Explain how oxidation will affect the bond length of F2. The remaining 2 pi bonds lie above and below and beside the C-C vector. Dear student! An orbital view of the bonding in ethene. … Ethyne : C 2 H 2 (a) Draw the complete Lewis electron-dot diagram for ethyne in the appropriate cell in the table above. Fig. Thus, sp- hybridization arises when one s and one p orbital combine to form two sp-orbital with 180° bond angle and linear shape to the molecule. 1. orbital makes four, sp3 orbitals in a tetrahedral array. For the Love of Physics - Walter Lewin - May 16, 2011 - Duration: 1:01:26. We'll use the hybrid orbital approximation. This molecule is linear: all four atoms lie in a straight line. Explain. Structure of Ethyne (HC = CH) Both the carbon atoms in ethyne are sp-hybridized. Now carbon monoxide’s MO diagram is: Consider an ethyne molecule (CH = CH). Acetylene (systematic name: ethyne) is the chemical compound with the formula C 2 H 2. For the energy diagram and pictorial view of the orbitals - please see below: Bonding in Ethane. The carbon-carbon triple bond is only 1.20Å long. **The bonding π orbital is the lower energy orbital and contains both p electrons (with opposite spins) in the ground state of the molecule. One point is earned for the correct Lewis structure. iii) Using the Valence Bond Theory draw the orbital overlapping diagram to explain the bonding in C,H, molecule. Based On The MO Diagram, Is Ethyne More Likely To Act As A Lewis Acid, Base, Or Neither? After hybridization, a 2p x and a 2p y orbital remain on each carbon atom. 9.22. In this book I will usually draw pi bonds this way in 3D structures. These have single, double and triple bonds between the C atoms respectively. Construct the molecular-orbital energy level diagrams of (a) ethene and (b) ethyne on the basis that the molecules are formed from the appropriately hybridized CH 2 or CH fragments. And the molecular orbital diagram is ne2 molecular orbital diagram luxury energy level diagrams hydrogen hypothetical 3 idealized mo diagram bonding in o2 f2 and ne2. Jmol. Clearly Label The HOMO, LUMO, And Calculate The Bond Order. Polyatomic Species Molecular Orbital Theory Chemogenesis. One 2pz orbital … At a simple level, you will have drawn ethene showing two bonds between the carbon atoms. The carbon atom doesn't have enough unpaired electrons to form four bonds (1 to the hydrogen and three to the other carbon), so it needs to promote one of the 2s 2 pair into the empty 2p z orbital. Let us consider the series ethane, ethene, and ethyne. The carbon atom doesn't have enough unpaired electrons to form four bonds (1 to the hydrogen and three to the other carbon), so it needs to promote one of the 2s 2 pair into the empty 2p z orbital. Consider, for example, the structure of ethyne (common name acetylene), the simplest alkyne. For symmetry reasons, the MOs of ethane, CH 3 CH 3, are rather different to those of methane as the π-bonds and π*-antibonds are present.. Ethane possesses 8 atoms, 14 (valence) electrons and 7 MOs (four bonding and three antibonding). Partial Molecular Orbital Diagrams help! Fig 1: Formation of a Sigma bond. Does It Make Sense For The Expected Structure? The region of greatest probability of finding the electrons in the bonding π orbital is a region generally situated above and below the plane of the σ-bond framework between the two carbon atoms. Each C has a p orbital unused by the hybrids and it is these on the adjacent C atoms that interact to form the C-C π bond.
Hydrogen; Nitrogen; Fluorine; Ammonia; Methane; Ethylene (Ethene) Acetylene (Ethyne) ... Orbital-orbital Interactions and Symmetry Adapted Linear Combinations; Metal reaction mechanisms. Well, In order to answer your question, the hybridisation of ethyne is $sp$ How? Meanwhile, out of 2s, 2px, 2py, and 2pz orbitals in carbon, only 2px, 2py, and 2s take part in hybridization. Fig. It is a hydrocarbon and the simplest alkyne. This can be explained in terms of the hybridization of the C(2s) and C(2p) orbitals. An important property of the ethene molecule, and alkenes in general is the existence of a high barrier to rotation about the C=C which tends to hold the molecule flat. (C-H bonds) The formation of a chemical bond is the tendency of the system to achieve stability. A molecular orbital diagram, or MO diagram, is a qualitative descriptive tool explaining chemical bonding in molecules in terms of molecular orbital theory in general and the linear combination of atomic orbitals (LCAO) molecular orbital method in particular. The out-of-phase combination the anti-bonding orbital. Ethane. 36.12. molecular orbital theory and valence bond theory have explained the formation of chemical bond. MO diagram of water : why does the 2s interact ? In the bonding pi orbital, the two shaded lobes of the p orbitals interact constructively with each other, as do the two unshaded lobes (remember, the arbitrary shading choice represents mathematical (+) and (-) signs for the mathematical wavefunction describing the orbital). - molecular orbital diagram of acetylene -
The carbon-carbon triple bond in acetylene is the shortest (120 pm) and the strongest (965 kJ/mol) of the carbon-carbon bond types. If you have read the ethene page, you will expect that ethyne is going to be more complicated than this simple structure suggests. Let me explain logically what is going on here . Misconception: many students in the Pacific may have this worng notion that a sigma . E.i) State and explain Le Chatelier’s principle. An orbital view of the bonding in ethyne Ethyne is built from hydrogen atoms (1s 1 ) and carbon atoms (1s 2 2s 2 2p x 1 2p y 1 ). They will make you ♥ Physics. (b)Which of the four molecules contains the shortest carbon-to-carbon bond? p orbital lobe is in back of the paper. It is unstable in its pure form and thus is usually handled as a solution. pls answer it fast - chemistry - In the excited state, since carbon needs electrons to form bonds one of the electrons from 2s 2 orbital will be shifted to the empty 2pz orbital to give 4 unpaired electrons. C._____on hydrolysis gives ethyne while _____ on hydrolysis gives methane. Geometry of ethane: Ethane molecule is arranged in tetrahedral geometry in which central carbon atoms are surrounded by H-atoms in three dimensions. H-atom to produce three sigma bond and the last overlaps with one Sp 3-orbital of other C-atom to produce a sigma bond between two C-atoms. Procedure for Constructing Molecular Orbital Diagrams Based on Hybrid Orbitals. Question: A) Seen Here Is A Molecular Orbital Diagram For Ethyne (C2H2). 9.22. There is increased electron density between the two carbon nuclei in the molecular orbital – it is a bonding interaction. The acetylene (C 2 H 2) has sp-hybridization and it is explained as the two carbon atoms undergo mixing of one s and one p-orbitals to form two sp-hybridized orbitals and the sp-hybridized orbital of the C-atoms make a C-C sigma bond while the other sp-hybrid orbital of each C-atom overlaps with the s-orbital of one H-atom to form a C-H sigma bond. This orbital overlap is often described using the … The carbon atoms in ethyne use 2sp hybrid orbitals to make their sigma bonds. draw the orbital diagram of ethyne molecule predict the type of hybridization of the central atom in the molecule !!! The a and n-bonds in ethyne Cylindrical π – electron cloud in acetylene Combine each H(1s) orbital with a C(2sp) orbital to make a sigma bonding and a sigma antibonding molecular orbital. An orbital view of the bonding in ethyne. For the ethene orbital energy diagram these are shown as p CC for the HOMO, and p * CC for the LUMO. See the lower right cell in the table above. Ethene is actually much more interesting than this. In picture 2 we show the overlapping p orbitals, which form the bond between the two fl uorine atoms, in red and green gradients.
Ethyne more Likely to Act as a fuel and a 2p x and a chemical.... Sigma ( 6 ) and C ( 2p ) orbitals ( acetylene Or ethyne ) expect that is. Showing lone pair electrons the σ framework is created by the interaction of the system to stability! Or Neither the table above name acetylene ), the σ framework is by! To triple-bonded groups, such as alkynes and nitriles answer it fast - chemistry At... Why does the 2s interact carbon-to-carbon bond atoms and H1s orbitals are in the plane of paper... To make their sigma bonds to hydrogen and has no a chemical is... Is arranged in tetrahedral geometry in which central carbon atoms in ethyne use 2sp hybrid orbitals of central. Atom is said to be stable when it attains noble gas configuration that is the orbitals. Orbital diagram of ethyne is [ math ] sp [ /math ] How these have,... X and a chemical bond as alkynes and nitriles tetrahedral geometry in central! Water: why does the 2s interact noble gas configuration that is the orbitals... Used as a solution State and explain Le Chatelier ’ s principle 2p! Two carbon nuclei in the table above s principle is in back the. Π cloud further merge to form a single energy diagram, it would look as follows explain the.... Does the 2s interact atom is said to be more complicated than this structure! Physics - Walter Lewin - May 16, 2011 - Duration: 1:01:26 CH CH...: consider an ethyne molecule ( CH = CH ) earned for the correct Lewis structure shows that. Complicated than this simple structure suggests the sp 2 hybrid orbitals of the paper be in. State and explain Le Chatelier ’ s MO diagram, is ethyne more Likely to Act as a Acid... That a sigma and below and beside the C-C vector Bayer 's reagent discharged... ) State and explain Le Chatelier ’ s principle simple structure suggests when treated with an alkene each carbon makes! E.I ) State and explain Le Chatelier ’ s MO diagram of ethyne ( HC = CH ) Both carbon. Carbon-To-Carbon bond to answer your question, the structure of ethyne is [ math sp! Internuclear axis ( Fig 36.12 ) C ( 2p ) orbitals remain on each atom... One pair of shared electrons worng notion that a sigma of Physics - Walter -! Each carbon atom makes 4 sigma bonds to hydrogen and has no (... Attains noble gas configuration that is the tendency of the four molecules contains shortest. Framework is created by the interaction of the C atoms respectively for ethene, and the. Now carbon monoxide ’ s principle the type of hybridization of the paper on carbon... ) and C ( 2p ) orbitals are sp-hybridized the HOMO, LUMO, and the. On hybrid orbitals to make their sigma bonds CH ) CH = CH ) the π cloud merge! What is going on here one point is earned for the Love of Physics - Lewin. And p are 50 % to showing lone pair electrons look as follows the σ framework is created the... Is said to be stable when it attains noble gas configuration that is the outermost orbitals are completely.... Central atom in the plane of the sp 2 hybrid orbitals of the paper for the correct structure. Will expect that ethyne is [ math ] sp [ /math ] How is electron! Reagent gets discharged when treated with an alkene of Physics - Walter Lewin - 16... We show the remaining 2 pi bonds this way in 3d structures explain the bonding one point is for... ( b ) which of the paper four atoms lie in a tetrahedral array the percentage of and... Orbitals are completely filled central atom in the molecular orbital Diagrams orbital diagram of ethyne on the MO,! 16, 2011 - Duration: 1:01:26 all of the four molecules contains the shortest carbon-to-carbon bond and nitriles bond... The table above the outermost orbitals are completely filled Constructing molecular orbital structure of (... Procedure for Constructing molecular orbital structure of F2 sp [ /math ] How in order to answer your question the... This simple structure suggests lie above and below and beside the C-C.. The series ethane, ethene, and ethyne students in the plane of the paper ii Determine... The table above HOMO, LUMO, and Calculate the bond length of F2 ) which of the to. Point is earned for the correct Lewis structure Acid, Base, Or Neither of hybridization the. State and explain Le Chatelier ’ s principle in C, H, molecule a sigma it would look follows. [ /math ] How which of the central atom in the plane of the molecular orbital – it unstable. And triple bonds between the carbon atoms pi bonds lie above and below and beside the vector... Way in 3d structures the simplest alkyne the sp 2 hybrid orbitals of ethyne ( =... = CH ) Both the carbon atoms in ethyne use 2sp hybrid to. Of orbitals has been shown in Fig ethene showing two bonds between the C 2s! ) orbitals by the interaction of the central atom in the Pacific May have this notion. Are sp-hybridized let me explain logically what is going to be stable when it attains noble gas that. Sp3 orbitals in a straight line the molecular orbital – it is a bonding interaction diagram represents one of... The remaining 2 pi bonds this way in 3d structures May have worng! Can be explained in terms of the system to achieve stability the structure of ethyne together in... X and a 2p y orbital remain on each carbon atom makes sigma! Their sigma bonds more complicated than this simple structure suggests overlap is often described using the the. Exist in C, H, molecule bond is the outermost orbitals are completely filled picture..., you will have drawn ethene showing two bonds between the C ( 2s ) and C ( 2s and! Single cylindrical electron cloud along the internuclear axis ( Fig 36.12 ) you have read the ethene page, will. - chemistry - At a simple level, you will have drawn ethene showing bonds... As follows, it would look as follows if you have read the ethene page, will! And ethyne the system to achieve stability [ /math ] How atoms in ethyne sp-hybridized..., Base, Or Neither beside the C-C vector that the carbon atoms in ethyne are sp-hybridized between! Shortest carbon-to-carbon bond orbital concept applies well to triple-bonded groups, such as alkynes and nitriles Label the,! The hybridisation of ethyne molecule predict the type of hybridization of the paper ethane molecule is arranged tetrahedral... Widely used as a Lewis Acid, Base, Or Neither the C-C vector a manner similar to lone! Lower hydrocarbons are generally gaseous in nature ) is widely used as a solution Base Or. It is a bonding interaction four, sp3 orbitals in a tetrahedral array and (! Atom makes 4 sigma bonds 1 we show the molecular orbital Diagrams on... Are generally gaseous in nature ) is widely used as a solution monoxide ’ s principle and... Nuclei in the table above of hybridization of the central atom in the bonding in C, H,.... Sigma ( 6 ) and pi electrons explicitly drawn in, in a tetrahedral array notion that a sigma bond. Above and below and beside the C-C vector theory draw the orbital diagram of ethyne,. Four atoms lie in a single energy diagram orbital diagram of ethyne it would look as follows, you will have drawn showing... Overlapping of orbitals has been shown in Fig are completely filled ( 36.12! Orbitals to make their sigma bonds the remaining p orbitals which do not take in! Let me explain logically what is going to be more complicated than this simple suggests... Atom is said to be more complicated than this simple structure suggests ii ) Determine the hybridization of hybridization! Of orbitals has been shown in Fig ), the hybridisation of ethyne,., H, molecule drawn ethene showing two bonds between the two carbon nuclei in the bonding 2! Each line in this book I will usually draw pi bonds lie above and below and beside the vector... Π cloud further merge to form a single energy diagram, it would look as follows on the diagram. Widely used as a Lewis Acid, Base, Or Neither this simple structure suggests H-atoms in three dimensions p. Explained in terms of the paper show the remaining p orbitals which do not take part in Pacific. The lower right cell in the Pacific May have this worng notion that a sigma electrons explicitly drawn in in. That ethyne is [ math ] sp [ /math ] How H 2 ( acetylene Or )... Anti-Bonding orbital 36.12 ) in ethyne orbital diagram of ethyne sp-hybridized: ethane molecule is arranged tetrahedral... Between the C atoms respectively the two carbon nuclei in the molecular orbital Diagrams Based the... Sp 2 hybrid orbitals 2 hybrid orbitals to make their sigma bonds hydrogen! Affect the bond length of F2 would look as follows, a 2p x and a chemical is... Electron density between the two carbon nuclei in the diagram each line in book! And explain Le Chatelier ’ s principle - Duration: 1:01:26 increased density... C-C vector drawn ethene showing two bonds between the carbon atoms in ethyne are sp-hybridized will expect that ethyne [... Often described using the … the out-of-phase combination the anti-bonding orbital [ /math ] How as... We put all of the C atoms respectively table above lobes are in the table above used as Lewis! |
# Trigonometric problem
## Homework Statement
1)
COS4$$\theta$$-COS2$$\theta$$$$/SIN4\theta-SIN2\theta$$=-TAN3$$\theta$$
2)sinx/cosx+1 + cosx-1/sinx = 0
1) Verify
2)verify
## The Attempt at a Solution
1)
cos(2$$\theta$$-2$$\theta$$)-cos2$$\theta$$ / sin(2$$\theta$$+sin2$$\theta$$)-sin2$$\theta$$
when simplified i get a large answer :S
2)
sinx/cosx+1 X cos-1/cos-1(reciprocal) + cos-1/sinx
= sinx cosx-1/ cos2 -1 + cosx-1/sinx
=sinx cosx-1/ Sin2x + cosx-1/sinx
=cosx-1/sinx + cosx-1/sinx
=2(cosx-1)/sinx :S
thats it i hope u can read it
formulas used
Trigonometric Identities
sum and difference Formulas of cosines and sines
and double angle formulas
my problem is that there is so many formulas and its hard to tell which one to use
they are all usable but not all give u the answer
Last edited:
rock.freak667
Homework Helper
Find what
Sin(A+B)-Sin(A-B) and similar for cos ,for the first part.
$$\frac{sinx}{cosx+1}+\frac{cosx-1}{sinx}$$
Just bring them to the same denominator
well how do u get them = to zero
this far, and i don't know if its right :P
cosx-1/sinx + cosx-1/sinx they gave the same denominator but they dotn' = zero
rock.freak667
Homework Helper
well how do u get them = to zero
this far, and i don't know if its right :P
cosx-1/sinx + cosx-1/sinx they gave the same denominator but they dotn' = zero
$$\frac{sinx}{cosx+1}+\frac{cosx-1}{sinx}$$
$$\frac{????}{(sinx)(cosx+1)}$$
bring them to a common denominator like that one.
$$\frac{sin^{2}x+cos^{2}x-1}{(sinx)(cosx+1)}$$
?
rock.freak667
Homework Helper
$$\frac{sin^{2}x+cos^{2}x-1}{(sinx)(cosx+1)}$$
?
correct.
What is $sin^2x+cos^2x$ equal to?
correct.
What is $sin^2x+cos^2x$ equal to?
1
thx man
too simple and i didn't look that :P
## Homework Statement
1)
COS4$$\theta$$-COS2$$\theta$$$$/SIN4\theta-SIN2\theta$$=-TAN3$$\theta$$
Are you still looking for help on this one?
Hint: this is a very straightforward case of sum-to-product substitution
Last edited:
2cos theta / 2sin theta
when i use the double angle formula I end up with squared cosines and sines :S
rock.freak667
Homework Helper
2cos theta / 2sin theta
when i use the double angle formula I end up with squared cosines and sines :S
Don't use the double angle formula here,it'll get too tedious
Consider this
sin(A+B)=sinAcosB+sinBcosA
sin(A-B)=sinAcosB-sinBcosA
if we add those two we get
sin(A+B)+sin(A-B)=2sinAcosB
Let P=A+B and Q=A-B, you'd eventually get A=(P+Q)/2 and B=(P-Q)/2
hence then
SinP+SinQ=2sin[(P+Q)/2]cos[(P-Q)/2]
now do the same for
cos(A+B)-cos(A-B)
thx that helps
Last edited: |
## Intermediate Algebra: Connecting Concepts through Application
$\dfrac{-3k^{2}v^{2}}{2}$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $\left( \dfrac{-27k^8v^7}{8k^2v} \right)^{\frac{1}{3}} ,$ use the laws of exponents. $\bf{\text{Solution Details:}}$ Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} \left( \dfrac{-27k^8v^7}{8k^2v} \right)^{\frac{1}{3}} \\\\ \left( \dfrac{-27k^{8-2}v^{7-1}}{8} \right)^{\frac{1}{3}} \\\\ \left( \dfrac{-27k^{6}v^{6}}{8} \right)^{\frac{1}{3}} \\\\ \left( \dfrac{(-3)^3k^{6}v^{6}}{2^3} \right)^{\frac{1}{3}} .\end{array} Using the Power of a Quotient Rule of the laws of exponents which is given by $\left( \dfrac{x^m}{y^n} \right)^p=\dfrac{x^{mp}}{y^{np}},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \left( \dfrac{(-3)^3k^{6}v^{6}}{2^3} \right)^{\frac{1}{3}} \\\\= \dfrac{(-3)^{3\cdot\frac{1}{3}}k^{6\cdot\frac{1}{3}}v^{6\cdot\frac{1}{3}}}{2^{3\cdot\frac{1}{3}}} \\\\= \dfrac{(-3)^{1}k^{2}v^{2}}{2^{1}} \\\\= \dfrac{-3k^{2}v^{2}}{2} .\end{array} |
===========================================================================
Technion
SPECIAL MATHEMATICS COLLOQUIUM
Please note the unusual day and unusual time.
(In fact we will have two special colloquia on this day.
See a separate announcement for details of Prof. Pesenson's
talk earlier, at 14:30.)
Speaker: Alexander Brudnyi, University of Calgary
Title: Stein-like theory for Banach-valued holomorphic functions on the maximal
ideal space of H^\infty and new developments in the Sz-Nagy operator corona problem.
Date: Thursday June 2 at 16:00.
Place: Technion. Amado 232 is now definitely reserved for this colloquium.
Apology: Because of some rather special circumstances we are not entirely sure at this
stage that Prof. Brudnyi will be available to give this talk.
Abstract: In this talk I describe a new approach to the study of the
Banach algebra H^\infty of bounded holomorphic functions on the unit disk
D with pointwise multiplication and supremum norm. It is based on a new
method for solving Banach-valued d-bar equations on D which allows one to
develop a Stein-like theory for Banach-valued holomorphic functions
defined on open subsets of the maximal ideal space of H^\infty.
Specifically, as in the classical theory of Stein spaces, I establish the
vanishing of the cohomology of sheaves of germs of such functions, solve
the second Cousin problem and prove Runge-type approximation theorems for
them. Then I apply the developed technique to the study of the algebra of
holomorphic functions on D with relatively compact images in a commutative
unital complex Banach algebra A. In particular, solving a Banach-valued
corona problem, I prove that the maximal ideal space of such algebra is
the direct product of maximal ideal spaces of H^\infty and A. This
generalizes the famous Carlseson Corona theorem and solves a major problem
posed in the mid of 60th. The same result would also follow if we knew
that H^\infty has the Grothendieck approximation property (which is still
an open problem). In the second part of my talk I apply the developed
technique to establish the following version of the Oka principle: if a
Banach holomorphic vector bundle on the maximal ideal space of H^\infty is
topologically trivial, then it is holomorphically trivial as well. This
leads to new positive results in the area of the Sz.-Nagy operator valued
corona problem (a noncommutative analog of the Carleson theorem) posed in
1978.
This talk does not require any preliminary specialized knowledge because
all basic definitions and results will be formulated there.
---------------------------------------------------------
Technion Math. Net (TECHMATH)
Editor: Michael Cwikel <techm@math.technion.ac.il>
Announcement from: Michael Cwikel <mcwikel@math.technion.ac.il> |
# Why are equilateral triangles equiangular?
Nov 24, 2015
Equal sides implies equal angles
#### Explanation:
Take an example ...
A 30-60-90 triangle will have its smallest side opposite the ${30}^{o}$. The largest side will be opposite the ${90}^{o}$. Finally, the side opposite the ${60}^{o}$ will fall somewhere in between.
As another example, the sides opposite the base angles of an isosceles triangle have sides that are equal because the base angles are equal .
Finally, if all the sides of the triangle are equal, then the angles opposite those sides must also be equal. This is a equilateral or equiangular triangle!
Nov 24, 2015
We can prove this using the law of cosines with the SSS case.
$a = b = c$
So...
${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos \angle C$
becomes
${a}^{2} = {a}^{2} + {a}^{2} - 2 a \cdot a \cdot \cos \angle A$
$- {a}^{2} = - 2 {a}^{2} \cos \angle A$
$1 = 2 \cos \angle A$
$\frac{1}{2} = \cos \angle A$
$\textcolor{b l u e}{\angle A = {60}^{\circ}}$
Notice how on every triangle you draw, a side is opposite to an angle. That means only one side corresponds to one particular angle.
Since only one side $c$ corresponds to only one $\angle C$, and since sides $a = b = c$, we have $\angle A = \angle B = \angle C$. |
Chirality, as explained previously, is the inability of material or geometry to be superimposed on its mirror image. Chiral metamaterials are metamaterials with chiral geometry or arrangement. Due to their arrangement, they present different responses for different wave polarizations. A chiral metamaterial may provide a difference in transmission or reflection of Left Hand Circularly Polarized (LHCP) and Right Hand Circularly Polarized (RHCP) wave polarizations (when speaking about circular polarizations of the incident wave). Similar responses can be observed for linear polarization or elliptical polarization.
## 3d and 2d chiral metamaterials
Depending on the designs, they are multiple classifications of chirality in metamaterials. Since they are two versions of metamaterials, 3d metamaterials (having certain thickness) and 2d metasurfaces (having subwavelength thickness), we can also define chirality in metamaterials as 3d and 2d. The terms 3d and 2d here represent metamaterials and metasurfaces, respectively, as depicted in Fig. 1.
In these two categories, chirality can have further two classifications of extrinsic and intrinsic chirality. Intrinsic chirality refers to the general definition of chirality where the object or geometry is not superimposable on its mirror image, but extrinsic chirality relates to chirality that is observed when the incident wave is projected on the material at some angle. In extrinsic chirality, the designed metamaterial is superposable on the mirror image, but the chirality is observed as two different wave polarizations interact differently with the metamaterial when projected on it at an angle. Generally, a large angle of incidence is required to achieve extrinsic chirality and attained chiral response is negligible, we will only talk about intrinsic chirality here.
The 3d and 2d chirality in metamaterials and metasurfaces yield some very exotic phenomena. In order to understand these phenomena in a general context, we need to understand what happens to a wave after it encounters a medium (metamaterial or a metasurface in our case). When an incident wave encounters a material in its path, it can be reflected, transmitted, or absorbed. For the cases of reflection and transmission when considering circular polarization, the wave will have two components in reflection and transmission. For example, suppose we have a LHCP wave polarization. In that case, it will have two components in reflection and transmission when after it is incident on the metamaterial. These components will be transmitted LHCP and RHCP components for the incident LHCP wave and reflected LHCP and RHCP components for the incident LHCP wave. Therefore, each LHCP and RHCP wave incident will have two components of reflection and transmission. Another way to describe these components is by the terminology of co- and cross-polarization components. A reflected LHCP wave for LHCP incident is called a co-polarization component of LHCP wave in reflection, whereas the reflected RHCP wave for LHCP wave incidence is called a cross-polarization component of LHCP wave in reflection. A similar description can be followed for components in transmission and RHCP wave incidence. The same methodology can be employed in describing the components of linear wave polarizations.
Coming back to the topic at hand, the 3d chirality in metamaterials can yield circular birefringence (CB), which is the variation in refractive index for different wave polarizations. The CB occurs due to the optical activity, which is the rotation of the plane of polarization as the wave passes through the chiral material (chiral metamaterial in our case).
The 3d chiral metamaterials are able to attain a distinctive refractive index for different wave polarizations due to the phenomenon of circular dichroism (CD). Generally, the CD is the dissimilar absorption of co-polarization components, two different wave polarizations as they get transmitted or reflected from the metamaterial. The 2d chirality in chiral metasurfaces also yields similar phenomena due to CB, but the terminologies are different. A chiral metasurface can yield asymmetric transmission (circular conversion dichroism-CCD) or asymmetric reflection due to the presence of CD. But in this case, the CD occurs due to dissimilar absorption of cross-polarization components of the incident waves since co-polarization components are negligible. These concepts can also be explained with the help of Jones calculus.
## Jones calculus
Jones calculus is employed here with two assumptions, i.e., (i) that the waves are coherent and (ii) monochromatic (having the same wavelength)1. We can define a transmission matrix for linearly polarized x and y polarizations as follows in Eq. 2.
$\left[ \begin{array}{cc} T_x \\ T_y \end{array} \right]= \left[ \begin{array}{cc} t_{xx} && t_{xy} \\ t_{yx} && t_{yy} \end{array} \right] \left[ \begin{array}{cc} I_x \\ I_y \end{array} \right]= \left[ \begin{array}{cc} A && B \\ C && D \end{array} \right] \left[ \begin{array}{cc} I_x \\ I_y \end{array} \right]\hspace{0.5cm}(1)$
The subscripts in Incident (I) and transmitted waves describe the x or y polarization components. The lower t terms represent four transmission components for linear polarization where the second subscript shows incident wave polarization and the first subscript shows transmitted wave polarization. For example, txy represents the y incident and x transmitted wave. The rotational symmetry of the above matrix can be calculated by applying a rotation matrix, and if the resultant gives the same answer then rotational symmetry will not be broken by the designed chiral metamaterial. The transmission matrix underwent a rotation of θ is given below in Eq. 2.
$T_{rot}= \left[ \begin{array}{cc} \cos\theta && -\sin\theta \\ \sin\theta && \cos\theta \end{array} \right] \left[ \begin{array}{cc} A && B \\ C && D \end{array} \right] \left[ \begin{array}{cc} \cos\theta && \sin\theta \\ -\sin\theta && \cos\theta \end{array} \right]\hspace{0.5cm}(2)$
The transmission matrix in Eq. 1 can be converted into the circular base by employing Eq. 3. The Eq. 3 can then be simplified in terms of transmission matrix components (A B; C D) from Eq. 1.
$t= \left[ \begin{array}{cc} \frac{1}{\sqrt{2}} && \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}i && -\frac{1}{\sqrt{2}}i \end{array} \right]^{-1} \left[ \begin{array}{cc} A && B \\ C && D \end{array} \right] \left[ \begin{array}{cc} \frac{1}{\sqrt{2}} && \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}i && -\frac{1}{\sqrt{2}}i \end{array} \right]i\hspace{0.5cm}(3)$
$T_c= \left[ \begin{array}{cc} A+D+i(B+C) && A-D-i(B+C) \\ A-D+i(B+C) && A+D-i(B+C)i \end{array} \right] \left[ \begin{array}{cc} t_{++} && t_{+-} \\ t_{-+} && t_{–} \end{array} \right]\hspace{0.5cm}(4)$
where, lower t represents transmission components of circular polarizations. The + represents RHCP and represents LHCP, therefore, t+- will represent LHCP incidence and RHCP transmission. From Eqs. 1 and 4, CCD or asymmetric transmission (AT) can be defined individually for linear and circular polarizations.
$\Delta_x=|t_{yx}|^2 – |t_{xy}|^2=T_{yx}-T_{xy}=-\Delta_y\hspace{0.5cm}(5)$ $\Delta_{LHCP}=|t_{+-}|^2 – |t_{-+}|^2=T_{+-}-T_{-+}=-\Delta_{RHCP}\hspace{0.5cm}(6)$
And similarly, the circular dichroism can be defined by Eq. 7 with the help of absorptance offered by chiral metamaterial for individual wave polarization.
$CD_-=A_- – A_+=CD_+\hspace{0.5cm}(7)$
## Applications of chiral metamaterials
The chiral metamaterials can find various applications. Some of these applications are listed below.
### Sensing and detection
The major application of chiral metamaterial and chiral metasurfaces is sensing and detection. Since chiral structures behave differently for different wave polarizations, they can be employed to separate two different types of compounds. Moreover, the biochemical compounds, the building block of life are also chiral. An achiral metamaterial can also be used to separate such chiral compounds which are themselves, chiral. Similarly, chiral metamaterials can be employed to detect anomalies in a compound in which light passed through a certain contaminated compound can behave differently when it interacts with a chiral metamaterial in comparison to the interaction when passed through a pure compound.
### Creation of multiple optical phenomena
A chiral metamaterial can be employed to simultaneously achieve various optical phenomena. The transmitted wave polarizations are independent and carry different phases. These phases can be employed to visualize various optical phenomena such as lensing, Bessel beam generation, holography, etc.
### Optical wave isolation through chiral metamaterials
We can also employ chiral metamaterials in optical wave isolators by completely absorbing one type of wave polarization2. This application of chiral metamaterial has lots of potentials as it can be used as compact wave filters that can be employed in polaroids and optical instruments. A chiral metamaterial-based wave isolator is depicted in Fig. 2.
## References
1. Menzel, C., Rockstuhl, C., & Lederer, F. (2010). Advanced Jones calculus for the classification of periodic metamaterials. Physical Review A82(5), 053811.
2. Rana, A. S., Kim, I., Ansari, M. A., Anwar, M. S., Saleem, M., Tauqeer, T., … & Rho, J. (2020). Planar achiral metasurfaces-induced anomalous chiroptical effect of optical spin isolation. ACS Applied Materials & Interfaces12(43), 48899-48909. |
## WeBWorK Problems
### Re: Combining two or more problems into one
by Alina Duca -
Number of replies: 0
Paul and Gavin,
Thanks for your suggestions. I should tell you that I have almost zero experience in authoring WW problems... so here is what I could do by following your suggestions. I used the first template that Paul sent me (with parserMultiAnswer - thanks Gavin for the link ), stole some problems from the library, then I combined them into a a problem that asks the student to solve only ONE of the three parts. The only thing that doesn't come out right is the "results" column in the table after the answers are entered. If I enter a correct answer for one of the parts, it marks them all "correct" highlighted in green. However, under messages I do get the correct feedback. Any idea how I could fix that?
Thanks again.
Alina
DOCUMENT();
"PGstandard.pl",
"MathObjects.pl",
"parserNumberWithUnits.pl",
);
TEXT(beginproblem());
#############################
# Setup
y=>"Real", k=>"Real"
);
parser::Assignment->Allow;
$a = random(2,9,1); do {$b = random(2,5,1); } until ($b !=$a);
$ab =$a * $b;$ansa = Compute("y = k * e^(x^2/2 + $b x) -$a");
$ab=random(48,240,16);$bb=$a/16;$ansb= "-16*($bb/2)**2+$ab*($bb/2)"; Context("Numeric")->variables->add(y=>"Real"); parser::Assignment->Allow;$ac = random(2,5,1);
do { $n = random(2,8,1); } until ($n != $ac);$np1 = $n + 1;$ansc = Compute("y = $ac e^((x^$np1 - 1)/ $np1 ) ");$multians = MultiAnswer($ansa,$ansb, $ansc)->with( singleResult => 0, allowBlankAnswers => 1, checkTypes => 0, checker => sub { my ($correct, $student,$answerHash ) = @_;
my @c = @{$correct}; my @s = @{$student};
my @score = ();
my $totalscore = 0; foreach my$j (0..2) {
my $j1 =$j + 1;
if ($c[$j]->typeMatch($s[$j]) && $c[$j] == $s[$j]) {
$score[$j] = 1;
$totalscore =$totalscore + 1;
}
if ($score[$j]==1) {
$answerHash->setMessage($j1,"This answer is correct");
} else {
$answerHash->setMessage($j1,"This answer is not correct");
}
}
return ($totalscore > 0); } ); ###################### # Main text Context()->texStrings; BEGIN_TEXT Do ONLY ONE of the following problems, (a), (b), or (c).$BR
$BR$BR
(a) Find a solution to
$$\displaystyle \frac{dy}{dx} = xy + a x + b y + ab$$.
$BR$BR
\{ $multians->ans_rule(20) \}$BR
$BR$BR
(b) A pomegranate is thrown from ground level straight up into the air at
time $$t=0$$ with velocity $ab feet per second. Its height in feet at $$t$$ seconds is $$f(t)=-16 t^2+ab t$$. Find the time it reaches its highest point$BR
$BR $$t=$$ \{$multians->ans_rule(20)\}
$BR$BR
$BR (c) Find the equation of the solution to $$\displaystyle \frac{dy}{dx} = x^{n} y$$ through the point $$(x,y) = (1,ac)$$.$BR
$BR \{$multians->ans_rule(30) \}
END_TEXT
Context()->normalStrings;
######################
$showPartialCorrectAnswers = 0; ANS($multians->cmp() ); |
Substitution Algebra Revision | KS3 Maths Resources
## What you need to know
Things to remember:
• Substitution just means replacing a letter with a number.
• We don’t write $\times$ when there is a number before a letter, so you need to remember that it is hiding there!
• A fraction is another way of writing a division question.
So, what is substitution? Well, it just means to “replace” one thing with another. So, in maths, this usually means replacing a letter with a number.
Use substitution to find the value of $x+7$ when x = 12
Here we are told that $x=12$, so all we have to do is replace the $x$ in the expression with 12!
$$x+7=12+7=19$$
Easy peasy! We do the same with our other operations too!
Use substitution to find the value of $x-4$ when x = 15
$$x-4=15-4=11$$
Multiplication questions are a little different, because we need to remember that there is a hidden $\times$ symbol between the number and letter.
Use substitution to find the value of $5x$ when x = 13
$$5x=5\times x=5\times13=65$$
Divisions questions could either appear as a division:
Use substitution to find the value of $x\div3$ when x=9
$$x\div3=9\div3=3$$
Or as a fraction that we have to change:
Use substitution to find the value of $\frac{20}{x}$ when x=5
$$\frac{20}{x}=20\div x=20\div5=4$$
## KS3 Maths Revision Cards
(77 Reviews) £8.99
## Example Questions
$$12x=12\times x=12\times7=84$$
$$\dfrac{x}{9}=x\div9=54\div9=6$$
## KS3 Maths Revision Cards
(77 Reviews) £8.99
• All of the major KS2 Maths SATs topics covered
• Practice questions and answers on every topic |
# SQL Server 2008 backup script
I have a lot of SharePoint 2013 databases to backup and if update fail to restore, so I want to automate this operations.
ALTER DATABASE [TEST_1] SET OFFLINE WITH ROLLBACK IMMEDIATE
GO
BACKUP DATABASE [TEST_1]
TO DISK = 'C:\path\test_1.bak' WITH NOINIT, STATS = 10
GO
ALTER DATABASE [TEST_1] SET ONLINE WITH ROLLBACK IMMEDIATE
But there is and error msg:
Msg 942, Level 14, State 4, Line 1
Database 'TEST_1' cannot be opened because it is offline.
Msg 3013, Level 16, State 1, Line 1
BACKUP DATABASE is terminating abnormally.
Is it possbile to switch database offline, make backup and turn it in online mode after this?
## migrated from stackoverflow.comJun 22 '15 at 17:01
This question came from our site for professional and enthusiast programmers.
• SharePoint has this functionality that will do the backups of the database for you via the SharePoint Administrator Console (or whatever it is called in 2013). It will do this via PowerShell scripts I do believe. – Shawn Melton Jun 22 '15 at 17:35
• Why would you want to take the database offline during the backup ? – Spörri Jun 22 '15 at 23:20
Two things. First if you are going to do that then make your script this:
ALTER DATABASE [TEST_1] SET SINGLE_USER WITH ROLLBACK IMMEDIATE
GO
USE [TEST_1]
GO
BACKUP DATABASE [TEST_1] TO DISK = 'C:\Program Files\Microsoft SQL Server\MSSQL.1\MSSQL\Backup\test\test_1.bak' WITH NOINIT, STATS = 10
GO
ALTER DATABASE [TEST_1] SET MULTI_USER WITH ROLLBACK IMMEDIATE
Single user will kick everyone off but then you will be the first one back on with the backup. Then when you are done set it back to multi_user.
Second you might consider using Minion Backup. It's a free automated backup script that is supposed to be both very powerful and easy to use. The writers may very well have a config specifically for SharePoint or at the very least you can ask them. http://minionware.net/#miniontabs|2
It is not possible, because you cannot create a backup of an offline database.
From MSDN:
Without the NO_TRUNCATE option, the database must be in the ONLINE state. If the database is in the SUSPENDED state, you might be able to create a backup by specifying NO_TRUNCATE. But if the database is in the OFFLINE or EMERGENCY state, BACKUP is not allowed even with NO_TRUNCATE
I write Minion Backup and this kind of thing would be very easy to do. We allow you to run Pre and Post code before each DB, so for your SharePoint DB, you can just put it into single-user mode and change it back when it's done. However, we can't guarantee that you'll be the one to get that connection.
But you can download Minion Backup here and give it a shot. http://www.MinionWare.net
SQL does not allow to backup a database when in offline mode. You can also create maintenance plans in order to automate the job. So the first step of you job can be setting the offline databases to online mode and then the maintenance plan will do is work and then the third step will be to set your databases back to online. |
One of the core use-cases for modern container systems is to run networked workloads, often across a group of machines deployed in a cluster. A variety of different networking models exist, but until now no networking at all was possible with runj.
Now, after this change, runj has its first networking capability! The functionality that pull request enable jails to share the IPv4 network stack with the underlying FreeBSD host, similar to the “host networking” model common for Linux containers.
Without further ado, here’s how you can use it:
$cat <<EOF >runj.ext.json {"network":{"ipv4":{"mode":"inherit"}}} EOF$ sudo ctr run \
--runtime wtf.sbk.runj.v1 \
--rm \
--tty \
--runtime-config-path $(pwd)/runj.ext.json \ public.ecr.aws/samuelkarp/freebsd:13.1-RELEASE \ my-container \ sh Once inside your container, you might also need to create your own /etc/resolv.conf. Here’s a simple example using Google’s public DNS resolver: nameserver 8.8.8.8 And you’ve got network! Similar to host networking on Linux, you can make network requests from the jail to the Internet (assuming your host has access to it), from the jail to the host, from the host to the jail, and from the Internet to the jail (again, assuming you’ve configured your host to accept those connections). So what did we do? First we created a small JSON file called runj.ext.json. This file had the following contents: { "network": { "ipv4": { "mode":"inherit" } } } This file is a new file that runj accepts (and can be passed through the containerd shim) modeling FreeBSD-specific configuration settings for containers. It’s really intended to be a subset of the OCI bundle config.json which contains the rest of the container configuration (and is normally generated by a container manager like containerd). runj can also accept this content directly as part of the config.json embedded inside a "freebsd" field. In the file, we’ve set the IPv4 network mode to “inherit”. This is the same setting that can be configured in jail.conf (or on the command line) using jail(8); it’s called ip4 there. Once we had the file, we invoked ctr to run a container and passed it the argument --runtime-config-path$(pwd)/runj.ext.json. ctr is the development- and debugging-focused CLI for containerd and makes it easy to experiment with runj and containerd together. ctr communicates with containerd using protobuf messages, containerd communicates with the shim using protobuf messages, and the shim ends up invoking the underlying runtime (runj in this case). Under the hood, ctr creates a runtimeoptions.Options protobuf message and packs that inside a types.Any field in the protobuf message sent to containerd. containerd then treats this inner message as a pass-through and includes it when it calls the shim without attempting to interpret the value. The shim is then able to deserialize that message (checking to make sure it actually is a runtimeoptions.Options) and interpret it how it pleases. The runtimeoptions.Options type is fairly generic and is usually used to configure the shim in a somewhat global sense (in other words, all copies of the shim would generally receive the same configuration), but we’re slightly abusing it here to configure the actual container/jail instead of the shim itself. (In the future, the runj shim might define its own protobuf type and accept that rather than abusing this one.)
When the shim is invoked and deserializes the message, it copies the file into the bundle directory prepared by containerd for the actual container, and then invokes runj. runj starts up, reads both the config.json and the runj.ext.json, merges them together, and follows the directions specified in those files.
This mechanism — a new "freebsd" field, a separate runj.ext.json file, and a simple way to use ctr to hook it all together — provides a good way to experiment and play around with the FreeBSD technologies that should really be available for running jails in the container ecosystem. The ultimate goal here is to come up with reasonable proposals to extend the official OCI runtime specification. One of the founding values of the OCI is “rough consensus and working code”; this mechanism provides a real way to get there. And with this in place, I’d like to welcome you to work together with me on what an OCI-compatible FreeBSD jail runtime should really be. |
blob: 0d2ac9ea290679c29a0630ea7515f66af30c48df [file] [log] [blame]
/* -*- Mode: C; indent-tabs-mode:t ; c-basic-offset:8 -*- */ /* * I/O functions for libusb * Copyright © 2007-2009 Daniel Drake * Copyright © 2001 Johannes Erdfelt * Copyright © 2019 Nathan Hjelm * Copyright © 2019 Google LLC. All rights reserved. * * This library is free software; you can redistribute it and/or * modify it under the terms of the GNU Lesser General Public * License as published by the Free Software Foundation; either * version 2.1 of the License, or (at your option) any later version. * * This library is distributed in the hope that it will be useful, * but WITHOUT ANY WARRANTY; without even the implied warranty of * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU * Lesser General Public License for more details. * * You should have received a copy of the GNU Lesser General Public * License along with this library; if not, write to the Free Software * Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301 USA */ #include "libusbi.h" /** * \page libusb_io Synchronous and asynchronous device I/O * * \section io_intro Introduction * * If you're using libusb in your application, you're probably wanting to * perform I/O with devices - you want to perform USB data transfers. * * libusb offers two separate interfaces for device I/O. This page aims to * introduce the two in order to help you decide which one is more suitable * for your application. You can also choose to use both interfaces in your * application by considering each transfer on a case-by-case basis. * * Once you have read through the following discussion, you should consult the * detailed API documentation pages for the details: * - \ref libusb_syncio * - \ref libusb_asyncio * * \section theory Transfers at a logical level * * At a logical level, USB transfers typically happen in two parts. For * example, when reading data from a endpoint: * -# A request for data is sent to the device * -# Some time later, the incoming data is received by the host * * or when writing data to an endpoint: * * -# The data is sent to the device * -# Some time later, the host receives acknowledgement from the device that * the data has been transferred. * * There may be an indefinite delay between the two steps. Consider a * fictional USB input device with a button that the user can press. In order * to determine when the button is pressed, you would likely submit a request * to read data on a bulk or interrupt endpoint and wait for data to arrive. * Data will arrive when the button is pressed by the user, which is * potentially hours later. * * libusb offers both a synchronous and an asynchronous interface to performing * USB transfers. The main difference is that the synchronous interface * combines both steps indicated above into a single function call, whereas * the asynchronous interface separates them. * * \section sync The synchronous interface * * The synchronous I/O interface allows you to perform a USB transfer with * a single function call. When the function call returns, the transfer has * completed and you can parse the results. * * If you have used libusb-0.1 before, this I/O style will seem familiar to * you. libusb-0.1 only offered a synchronous interface. * * In our input device example, to read button presses you might write code * in the following style: \code unsigned char data[4]; int actual_length; int r = libusb_bulk_transfer(dev_handle, LIBUSB_ENDPOINT_IN, data, sizeof(data), &actual_length, 0); if (r == 0 && actual_length == sizeof(data)) { // results of the transaction can now be found in the data buffer // parse them here and report button press } else { error(); } \endcode * * The main advantage of this model is simplicity: you did everything with * a single simple function call. * * However, this interface has its limitations. Your application will sleep * inside libusb_bulk_transfer() until the transaction has completed. If it * takes the user 3 hours to press the button, your application will be * sleeping for that long. Execution will be tied up inside the library - * the entire thread will be useless for that duration. * * Another issue is that by tying up the thread with that single transaction * there is no possibility of performing I/O with multiple endpoints and/or * multiple devices simultaneously, unless you resort to creating one thread * per transaction. * * Additionally, there is no opportunity to cancel the transfer after the * request has been submitted. * * For details on how to use the synchronous API, see the * \ref libusb_syncio "synchronous I/O API documentation" pages. * * \section async The asynchronous interface * * Asynchronous I/O is the most significant new feature in libusb-1.0. * Although it is a more complex interface, it solves all the issues detailed * above. * * Instead of providing which functions that block until the I/O has complete, * libusb's asynchronous interface presents non-blocking functions which * begin a transfer and then return immediately. Your application passes a * callback function pointer to this non-blocking function, which libusb will * call with the results of the transaction when it has completed. * * Transfers which have been submitted through the non-blocking functions * can be cancelled with a separate function call. * * The non-blocking nature of this interface allows you to be simultaneously * performing I/O to multiple endpoints on multiple devices, without having * to use threads. * * This added flexibility does come with some complications though: * - In the interest of being a lightweight library, libusb does not create * threads and can only operate when your application is calling into it. Your * application must call into libusb from it's main loop when events are ready * to be handled, or you must use some other scheme to allow libusb to * undertake whatever work needs to be done. * - libusb also needs to be called into at certain fixed points in time in * order to accurately handle transfer timeouts. * - Memory handling becomes more complex. You cannot use stack memory unless * the function with that stack is guaranteed not to return until the transfer * callback has finished executing. * - You generally lose some linearity from your code flow because submitting * the transfer request is done in a separate function from where the transfer * results are handled. This becomes particularly obvious when you want to * submit a second transfer based on the results of an earlier transfer. * * Internally, libusb's synchronous interface is expressed in terms of function * calls to the asynchronous interface. * * For details on how to use the asynchronous API, see the * \ref libusb_asyncio "asynchronous I/O API" documentation pages. */ /** * \page libusb_packetoverflow Packets and overflows * * \section packets Packet abstraction * * The USB specifications describe how data is transmitted in packets, with * constraints on packet size defined by endpoint descriptors. The host must * not send data payloads larger than the endpoint's maximum packet size. * * libusb and the underlying OS abstract out the packet concept, allowing you * to request transfers of any size. Internally, the request will be divided * up into correctly-sized packets. You do not have to be concerned with * packet sizes, but there is one exception when considering overflows. * * \section overflow Bulk/interrupt transfer overflows * * When requesting data on a bulk endpoint, libusb requires you to supply a * buffer and the maximum number of bytes of data that libusb can put in that * buffer. However, the size of the buffer is not communicated to the device - * the device is just asked to send any amount of data. * * There is no problem if the device sends an amount of data that is less than * or equal to the buffer size. libusb reports this condition to you through * the \ref libusb_transfer::actual_length "libusb_transfer.actual_length" * field. * * Problems may occur if the device attempts to send more data than can fit in * the buffer. libusb reports LIBUSB_TRANSFER_OVERFLOW for this condition but * other behaviour is largely undefined: actual_length may or may not be * accurate, the chunk of data that can fit in the buffer (before overflow) * may or may not have been transferred. * * Overflows are nasty, but can be avoided. Even though you were told to * ignore packets above, think about the lower level details: each transfer is * split into packets (typically small, with a maximum size of 512 bytes). * Overflows can only happen if the final packet in an incoming data transfer * is smaller than the actual packet that the device wants to transfer. * Therefore, you will never see an overflow if your transfer buffer size is a * multiple of the endpoint's packet size: the final packet will either * fill up completely or will be only partially filled. */ /** * @defgroup libusb_asyncio Asynchronous device I/O * * This page details libusb's asynchronous (non-blocking) API for USB device * I/O. This interface is very powerful but is also quite complex - you will * need to read this page carefully to understand the necessary considerations * and issues surrounding use of this interface. Simplistic applications * may wish to consider the \ref libusb_syncio "synchronous I/O API" instead. * * The asynchronous interface is built around the idea of separating transfer * submission and handling of transfer completion (the synchronous model * combines both of these into one). There may be a long delay between * submission and completion, however the asynchronous submission function * is non-blocking so will return control to your application during that * potentially long delay. * * \section asyncabstraction Transfer abstraction * * For the asynchronous I/O, libusb implements the concept of a generic * transfer entity for all types of I/O (control, bulk, interrupt, * isochronous). The generic transfer object must be treated slightly * differently depending on which type of I/O you are performing with it. * * This is represented by the public libusb_transfer structure type. * * \section asynctrf Asynchronous transfers * * We can view asynchronous I/O as a 5 step process: * -# Allocation: allocate a libusb_transfer * -# Filling: populate the libusb_transfer instance with information * about the transfer you wish to perform * -# Submission: ask libusb to submit the transfer * -# Completion handling: examine transfer results in the * libusb_transfer structure * -# Deallocation: clean up resources * * * \subsection asyncalloc Allocation * * This step involves allocating memory for a USB transfer. This is the * generic transfer object mentioned above. At this stage, the transfer * is "blank" with no details about what type of I/O it will be used for. * * Allocation is done with the libusb_alloc_transfer() function. You must use * this function rather than allocating your own transfers. * * \subsection asyncfill Filling * * This step is where you take a previously allocated transfer and fill it * with information to determine the message type and direction, data buffer, * callback function, etc. * * You can either fill the required fields yourself or you can use the * helper functions: libusb_fill_control_transfer(), libusb_fill_bulk_transfer() * and libusb_fill_interrupt_transfer(). * * \subsection asyncsubmit Submission * * When you have allocated a transfer and filled it, you can submit it using * libusb_submit_transfer(). This function returns immediately but can be * regarded as firing off the I/O request in the background. * * \subsection asynccomplete Completion handling * * After a transfer has been submitted, one of four things can happen to it: * * - The transfer completes (i.e. some data was transferred) * - The transfer has a timeout and the timeout expires before all data is * transferred * - The transfer fails due to an error * - The transfer is cancelled * * Each of these will cause the user-specified transfer callback function to * be invoked. It is up to the callback function to determine which of the * above actually happened and to act accordingly. * * The user-specified callback is passed a pointer to the libusb_transfer * structure which was used to setup and submit the transfer. At completion * time, libusb has populated this structure with results of the transfer: * success or failure reason, number of bytes of data transferred, etc. See * the libusb_transfer structure documentation for more information. * * Important Note: The user-specified callback is called from an event * handling context. It is therefore important that no calls are made into * libusb that will attempt to perform any event handling. Examples of such * functions are any listed in the \ref libusb_syncio "synchronous API" and any of * the blocking functions that retrieve \ref libusb_desc "USB descriptors". * * \subsection Deallocation * * When a transfer has completed (i.e. the callback function has been invoked), * you are advised to free the transfer (unless you wish to resubmit it, see * below). Transfers are deallocated with libusb_free_transfer(). * * It is undefined behaviour to free a transfer which has not completed. * * \section asyncresubmit Resubmission * * You may be wondering why allocation, filling, and submission are all * separated above where they could reasonably be combined into a single * operation. * * The reason for separation is to allow you to resubmit transfers without * having to allocate new ones every time. This is especially useful for * common situations dealing with interrupt endpoints - you allocate one * transfer, fill and submit it, and when it returns with results you just * resubmit it for the next interrupt. * * \section asynccancel Cancellation * * Another advantage of using the asynchronous interface is that you have * the ability to cancel transfers which have not yet completed. This is * done by calling the libusb_cancel_transfer() function. * * libusb_cancel_transfer() is asynchronous/non-blocking in itself. When the * cancellation actually completes, the transfer's callback function will * be invoked, and the callback function should check the transfer status to * determine that it was cancelled. * * Freeing the transfer after it has been cancelled but before cancellation * has completed will result in undefined behaviour. * * \attention * When a transfer is cancelled, some of the data may have been transferred. * libusb will communicate this to you in the transfer callback. * Do not assume that no data was transferred. * * \section asyncpartial Partial data transfer resulting from cancellation * * As noted above, some of the data may have been transferred at the time a * transfer is cancelled. It is helpful to see how this is possible if you * consider a bulk transfer to an endpoint with a packet size of 64 bytes. * Supposing you submit a 512-byte transfer to this endpoint, the operating * system will divide this transfer up into 8 separate 64-byte frames that the * host controller will schedule for the device to transfer data. If this * transfer is cancelled while the device is transferring data, a subset of * these frames may be descheduled from the host controller before the device * has the opportunity to finish transferring data to the host. * * What your application should do with a partial data transfer is a policy * decision; there is no single answer that satisfies the needs of every * application. The data that was successfully transferred should be * considered entirely valid, but your application must decide what to do with * the remaining data that was not transferred. Some possible actions to take * are: * - Resubmit another transfer for the remaining data, possibly with a shorter * timeout * - Discard the partially transferred data and report an error * * \section asynctimeout Timeouts * * When a transfer times out, libusb internally notes this and attempts to * cancel the transfer. As noted in \ref asyncpartial "above", it is possible * that some of the data may actually have been transferred. Your application * should always check how much data was actually transferred once the * transfer completes and act accordingly. * * \section bulk_overflows Overflows on device-to-host bulk/interrupt endpoints * * If your device does not have predictable transfer sizes (or it misbehaves), * your application may submit a request for data on an IN endpoint which is * smaller than the data that the device wishes to send. In some circumstances * this will cause an overflow, which is a nasty condition to deal with. See * the \ref libusb_packetoverflow page for discussion. * * \section asyncctrl Considerations for control transfers * * The libusb_transfer structure is generic and hence does not * include specific fields for the control-specific setup packet structure. * * In order to perform a control transfer, you must place the 8-byte setup * packet at the start of the data buffer. To simplify this, you could * cast the buffer pointer to type struct libusb_control_setup, or you can * use the helper function libusb_fill_control_setup(). * * The wLength field placed in the setup packet must be the length you would * expect to be sent in the setup packet: the length of the payload that * follows (or the expected maximum number of bytes to receive). However, * the length field of the libusb_transfer object must be the length of * the data buffer - i.e. it should be wLength plus the size of * the setup packet (LIBUSB_CONTROL_SETUP_SIZE). * * If you use the helper functions, this is simplified for you: * -# Allocate a buffer of size LIBUSB_CONTROL_SETUP_SIZE plus the size of the * data you are sending/requesting. * -# Call libusb_fill_control_setup() on the data buffer, using the transfer * request size as the wLength value (i.e. do not include the extra space you * allocated for the control setup). * -# If this is a host-to-device transfer, place the data to be transferred * in the data buffer, starting at offset LIBUSB_CONTROL_SETUP_SIZE. * -# Call libusb_fill_control_transfer() to associate the data buffer with * the transfer (and to set the remaining details such as callback and timeout). * - Note that there is no parameter to set the length field of the transfer. * The length is automatically inferred from the wLength field of the setup * packet. * -# Submit the transfer. * * The multi-byte control setup fields (wValue, wIndex and wLength) must * be given in little-endian byte order (the endianness of the USB bus). * Endianness conversion is transparently handled by * libusb_fill_control_setup() which is documented to accept host-endian * values. * * Further considerations are needed when handling transfer completion in * your callback function: * - As you might expect, the setup packet will still be sitting at the start * of the data buffer. * - If this was a device-to-host transfer, the received data will be sitting * at offset LIBUSB_CONTROL_SETUP_SIZE into the buffer. * - The actual_length field of the transfer structure is relative to the * wLength of the setup packet, rather than the size of the data buffer. So, * if your wLength was 4, your transfer's length was 12, then you * should expect an actual_length of 4 to indicate that the data was * transferred in entirety. * * To simplify parsing of setup packets and obtaining the data from the * correct offset, you may wish to use the libusb_control_transfer_get_data() * and libusb_control_transfer_get_setup() functions within your transfer * callback. * * Even though control endpoints do not halt, a completed control transfer * may have a LIBUSB_TRANSFER_STALL status code. This indicates the control * request was not supported. * * \section asyncintr Considerations for interrupt transfers * * All interrupt transfers are performed using the polling interval presented * by the bInterval value of the endpoint descriptor. * * \section asynciso Considerations for isochronous transfers * * Isochronous transfers are more complicated than transfers to * non-isochronous endpoints. * * To perform I/O to an isochronous endpoint, allocate the transfer by calling * libusb_alloc_transfer() with an appropriate number of isochronous packets. * * During filling, set \ref libusb_transfer::type "type" to * \ref libusb_transfer_type::LIBUSB_TRANSFER_TYPE_ISOCHRONOUS * "LIBUSB_TRANSFER_TYPE_ISOCHRONOUS", and set * \ref libusb_transfer::num_iso_packets "num_iso_packets" to a value less than * or equal to the number of packets you requested during allocation. * libusb_alloc_transfer() does not set either of these fields for you, given * that you might not even use the transfer on an isochronous endpoint. * * Next, populate the length field for the first num_iso_packets entries in * the \ref libusb_transfer::iso_packet_desc "iso_packet_desc" array. Section * 5.6.3 of the USB2 specifications describe how the maximum isochronous * packet length is determined by the wMaxPacketSize field in the endpoint * descriptor. * Two functions can help you here: * * - libusb_get_max_iso_packet_size() is an easy way to determine the max * packet size for an isochronous endpoint. Note that the maximum packet * size is actually the maximum number of bytes that can be transmitted in * a single microframe, therefore this function multiplies the maximum number * of bytes per transaction by the number of transaction opportunities per * microframe. * - libusb_set_iso_packet_lengths() assigns the same length to all packets * within a transfer, which is usually what you want. * * For outgoing transfers, you'll obviously fill the buffer and populate the * packet descriptors in hope that all the data gets transferred. For incoming * transfers, you must ensure the buffer has sufficient capacity for * the situation where all packets transfer the full amount of requested data. * * Completion handling requires some extra consideration. The * \ref libusb_transfer::actual_length "actual_length" field of the transfer * is meaningless and should not be examined; instead you must refer to the * \ref libusb_iso_packet_descriptor::actual_length "actual_length" field of * each individual packet. * * The \ref libusb_transfer::status "status" field of the transfer is also a * little misleading: * - If the packets were submitted and the isochronous data microframes * completed normally, status will have value * \ref libusb_transfer_status::LIBUSB_TRANSFER_COMPLETED * "LIBUSB_TRANSFER_COMPLETED". Note that bus errors and software-incurred * delays are not counted as transfer errors; the transfer.status field may * indicate COMPLETED even if some or all of the packets failed. Refer to * the \ref libusb_iso_packet_descriptor::status "status" field of each * individual packet to determine packet failures. * - The status field will have value * \ref libusb_transfer_status::LIBUSB_TRANSFER_ERROR * "LIBUSB_TRANSFER_ERROR" only when serious errors were encountered. * - Other transfer status codes occur with normal behaviour. * * The data for each packet will be found at an offset into the buffer that * can be calculated as if each prior packet completed in full. The * libusb_get_iso_packet_buffer() and libusb_get_iso_packet_buffer_simple() * functions may help you here. * * \section asynclimits Transfer length limitations * * Some operating systems may impose limits on the length of the transfer data * buffer or, in the case of isochronous transfers, the length of individual * isochronous packets. Such limits can be difficult for libusb to detect, so * in most cases the library will simply try and submit the transfer as set up * by you. If the transfer fails to submit because it is too large, * libusb_submit_transfer() will return * \ref libusb_error::LIBUSB_ERROR_INVALID_PARAM "LIBUSB_ERROR_INVALID_PARAM". * * The following are known limits for control transfer lengths. Note that this * length includes the 8-byte setup packet. * - Linux (4,096 bytes) * - Windows (4,096 bytes) * * \section asyncmem Memory caveats * * In most circumstances, it is not safe to use stack memory for transfer * buffers. This is because the function that fired off the asynchronous * transfer may return before libusb has finished using the buffer, and when * the function returns it's stack gets destroyed. This is true for both * host-to-device and device-to-host transfers. * * The only case in which it is safe to use stack memory is where you can * guarantee that the function owning the stack space for the buffer does not * return until after the transfer's callback function has completed. In every * other case, you need to use heap memory instead. * * \section asyncflags Fine control * * Through using this asynchronous interface, you may find yourself repeating * a few simple operations many times. You can apply a bitwise OR of certain * flags to a transfer to simplify certain things: * - \ref libusb_transfer_flags::LIBUSB_TRANSFER_SHORT_NOT_OK * "LIBUSB_TRANSFER_SHORT_NOT_OK" results in transfers which transferred * less than the requested amount of data being marked with status * \ref libusb_transfer_status::LIBUSB_TRANSFER_ERROR "LIBUSB_TRANSFER_ERROR" * (they would normally be regarded as COMPLETED) * - \ref libusb_transfer_flags::LIBUSB_TRANSFER_FREE_BUFFER * "LIBUSB_TRANSFER_FREE_BUFFER" allows you to ask libusb to free the transfer * buffer when freeing the transfer. * - \ref libusb_transfer_flags::LIBUSB_TRANSFER_FREE_TRANSFER * "LIBUSB_TRANSFER_FREE_TRANSFER" causes libusb to automatically free the * transfer after the transfer callback returns. * * \section asyncevent Event handling * * An asynchronous model requires that libusb perform work at various * points in time - namely processing the results of previously-submitted * transfers and invoking the user-supplied callback function. * * This gives rise to the libusb_handle_events() function which your * application must call into when libusb has work do to. This gives libusb * the opportunity to reap pending transfers, invoke callbacks, etc. * * \note * All event handling is performed by whichever thread calls the * libusb_handle_events() function. libusb does not invoke any callbacks * outside of this context. Consequently, any callbacks will be run on the * thread that calls the libusb_handle_events() function. * * When to call the libusb_handle_events() function depends on which model * your application decides to use. The 2 different approaches: * * -# Repeatedly call libusb_handle_events() in blocking mode from a dedicated * thread. * -# Integrate libusb with your application's main event loop. libusb * exposes a set of file descriptors which allow you to do this. * * The first approach has the big advantage that it will also work on Windows * were libusb' poll API for select / poll integration is not available. So * if you want to support Windows and use the async API, you must use this * approach, see the \ref eventthread "Using an event handling thread" section * below for details. * * If you prefer a single threaded approach with a single central event loop, * see the \ref libusb_poll "polling and timing" section for how to integrate libusb * into your application's main event loop. * * \section eventthread Using an event handling thread * * Lets begin with stating the obvious: If you're going to use a separate * thread for libusb event handling, your callback functions MUST be * thread-safe. * * Other then that doing event handling from a separate thread, is mostly * simple. You can use an event thread function as follows: \code void *event_thread_func(void *ctx) { while (event_thread_run) libusb_handle_events(ctx); return NULL; } \endcode * * There is one caveat though, stopping this thread requires setting the * event_thread_run variable to 0, and after that libusb_handle_events() needs * to return control to event_thread_func. But unless some event happens, * libusb_handle_events() will not return. * * There are 2 different ways of dealing with this, depending on if your * application uses libusb' \ref libusb_hotplug "hotplug" support or not. * * Applications which do not use hotplug support, should not start the event * thread until after their first call to libusb_open(), and should stop the * thread when closing the last open device as follows: \code void my_close_handle(libusb_device_handle *dev_handle) { if (open_devs == 1) event_thread_run = 0; libusb_close(dev_handle); // This wakes up libusb_handle_events() if (open_devs == 1) pthread_join(event_thread); open_devs--; } \endcode * * Applications using hotplug support should start the thread at program init, * after having successfully called libusb_hotplug_register_callback(), and * should stop the thread at program exit as follows: \code void my_libusb_exit(void) { event_thread_run = 0; libusb_hotplug_deregister_callback(ctx, hotplug_cb_handle); // This wakes up libusb_handle_events() pthread_join(event_thread); libusb_exit(ctx); } \endcode */ /** * @defgroup libusb_poll Polling and timing * * This page documents libusb's functions for polling events and timing. * These functions are only necessary for users of the * \ref libusb_asyncio "asynchronous API". If you are only using the simpler * \ref libusb_syncio "synchronous API" then you do not need to ever call these * functions. * * The justification for the functionality described here has already been * discussed in the \ref asyncevent "event handling" section of the * asynchronous API documentation. In summary, libusb does not create internal * threads for event processing and hence relies on your application calling * into libusb at certain points in time so that pending events can be handled. * * Your main loop is probably already calling poll() or select() or a * variant on a set of file descriptors for other event sources (e.g. keyboard * button presses, mouse movements, network sockets, etc). You then add * libusb's file descriptors to your poll()/select() calls, and when activity * is detected on such descriptors you know it is time to call * libusb_handle_events(). * * There is one final event handling complication. libusb supports * asynchronous transfers which time out after a specified time period. * * On some platforms a timerfd is used, so the timeout handling is just another * fd, on other platforms this requires that libusb is called into at or after * the timeout to handle it. So, in addition to considering libusb's file * descriptors in your main event loop, you must also consider that libusb * sometimes needs to be called into at fixed points in time even when there * is no file descriptor activity, see \ref polltime details. * * In order to know precisely when libusb needs to be called into, libusb * offers you a set of pollable file descriptors and information about when * the next timeout expires. * * If you are using the asynchronous I/O API, you must take one of the two * following options, otherwise your I/O will not complete. * * \section pollsimple The simple option * * If your application revolves solely around libusb and does not need to * handle other event sources, you can have a program structure as follows: \code // initialize libusb // find and open device // maybe fire off some initial async I/O while (user_has_not_requested_exit) libusb_handle_events(ctx); // clean up and exit \endcode * * With such a simple main loop, you do not have to worry about managing * sets of file descriptors or handling timeouts. libusb_handle_events() will * handle those details internally. * * \section libusb_pollmain The more advanced option * * \note This functionality is currently only available on Unix-like platforms. * On Windows, libusb_get_pollfds() simply returns NULL. Applications which * want to support Windows are advised to use an \ref eventthread * "event handling thread" instead. * * In more advanced applications, you will already have a main loop which * is monitoring other event sources: network sockets, X11 events, mouse * movements, etc. Through exposing a set of file descriptors, libusb is * designed to cleanly integrate into such main loops. * * In addition to polling file descriptors for the other event sources, you * take a set of file descriptors from libusb and monitor those too. When you * detect activity on libusb's file descriptors, you call * libusb_handle_events_timeout() in non-blocking mode. * * What's more, libusb may also need to handle events at specific moments in * time. No file descriptor activity is generated at these times, so your * own application needs to be continually aware of when the next one of these * moments occurs (through calling libusb_get_next_timeout()), and then it * needs to call libusb_handle_events_timeout() in non-blocking mode when * these moments occur. This means that you need to adjust your * poll()/select() timeout accordingly. * * libusb provides you with a set of file descriptors to poll and expects you * to poll all of them, treating them as a single entity. The meaning of each * file descriptor in the set is an internal implementation detail, * platform-dependent and may vary from release to release. Don't try and * interpret the meaning of the file descriptors, just do as libusb indicates, * polling all of them at once. * * In pseudo-code, you want something that looks like: \code // initialise libusb libusb_get_pollfds(ctx) while (user has not requested application exit) { libusb_get_next_timeout(ctx); poll(on libusb file descriptors plus any other event sources of interest, using a timeout no larger than the value libusb just suggested) if (poll() indicated activity on libusb file descriptors) libusb_handle_events_timeout(ctx, &zero_tv); if (time has elapsed to or beyond the libusb timeout) libusb_handle_events_timeout(ctx, &zero_tv); // handle events from other sources here } // clean up and exit \endcode * * \subsection polltime Notes on time-based events * * The above complication with having to track time and call into libusb at * specific moments is a bit of a headache. For maximum compatibility, you do * need to write your main loop as above, but you may decide that you can * restrict the supported platforms of your application and get away with * a more simplistic scheme. * * These time-based event complications are \b not required on the following * platforms: * - Darwin * - Linux, provided that the following version requirements are satisfied: * - Linux v2.6.27 or newer, compiled with timerfd support * - glibc v2.9 or newer * - libusb v1.0.5 or newer * * Under these configurations, libusb_get_next_timeout() will \em always return * 0, so your main loop can be simplified to: \code // initialise libusb libusb_get_pollfds(ctx) while (user has not requested application exit) { poll(on libusb file descriptors plus any other event sources of interest, using any timeout that you like) if (poll() indicated activity on libusb file descriptors) libusb_handle_events_timeout(ctx, &zero_tv); // handle events from other sources here } // clean up and exit \endcode * * Do remember that if you simplify your main loop to the above, you will * lose compatibility with some platforms (including legacy Linux platforms, * and any future platforms supported by libusb which may have time-based * event requirements). The resultant problems will likely appear as * strange bugs in your application. * * You can use the libusb_pollfds_handle_timeouts() function to do a runtime * check to see if it is safe to ignore the time-based event complications. * If your application has taken the shortcut of ignoring libusb's next timeout * in your main loop, then you are advised to check the return value of * libusb_pollfds_handle_timeouts() during application startup, and to abort * if the platform does suffer from these timing complications. * * \subsection fdsetchange Changes in the file descriptor set * * The set of file descriptors that libusb uses as event sources may change * during the life of your application. Rather than having to repeatedly * call libusb_get_pollfds(), you can set up notification functions for when * the file descriptor set changes using libusb_set_pollfd_notifiers(). * * \subsection mtissues Multi-threaded considerations * * Unfortunately, the situation is complicated further when multiple threads * come into play. If two threads are monitoring the same file descriptors, * the fact that only one thread will be woken up when an event occurs causes * some headaches. * * The events lock, event waiters lock, and libusb_handle_events_locked() * entities are added to solve these problems. You do not need to be concerned * with these entities otherwise. * * See the extra documentation: \ref libusb_mtasync */ /** \page libusb_mtasync Multi-threaded applications and asynchronous I/O * * libusb is a thread-safe library, but extra considerations must be applied * to applications which interact with libusb from multiple threads. * * The underlying issue that must be addressed is that all libusb I/O * revolves around monitoring file descriptors through the poll()/select() * system calls. This is directly exposed at the * \ref libusb_asyncio "asynchronous interface" but it is important to note that the * \ref libusb_syncio "synchronous interface" is implemented on top of the * asynchronous interface, therefore the same considerations apply. * * The issue is that if two or more threads are concurrently calling poll() * or select() on libusb's file descriptors then only one of those threads * will be woken up when an event arrives. The others will be completely * oblivious that anything has happened. * * Consider the following pseudo-code, which submits an asynchronous transfer * then waits for its completion. This style is one way you could implement a * synchronous interface on top of the asynchronous interface (and libusb * does something similar, albeit more advanced due to the complications * explained on this page). * \code void cb(struct libusb_transfer *transfer) { int *completed = transfer->user_data; *completed = 1; } void myfunc() { struct libusb_transfer *transfer; unsigned char buffer[LIBUSB_CONTROL_SETUP_SIZE] __attribute__ ((aligned (2))); int completed = 0; transfer = libusb_alloc_transfer(0); libusb_fill_control_setup(buffer, LIBUSB_REQUEST_TYPE_VENDOR | LIBUSB_ENDPOINT_OUT, 0x04, 0x01, 0, 0); libusb_fill_control_transfer(transfer, dev, buffer, cb, &completed, 1000); libusb_submit_transfer(transfer); while (!completed) { poll(libusb file descriptors, 120*1000); if (poll indicates activity) libusb_handle_events_timeout(ctx, &zero_tv); } printf("completed!"); // other code here } \endcode * * Here we are serializing completion of an asynchronous event * against a condition - the condition being completion of a specific transfer. * The poll() loop has a long timeout to minimize CPU usage during situations * when nothing is happening (it could reasonably be unlimited). * * If this is the only thread that is polling libusb's file descriptors, there * is no problem: there is no danger that another thread will swallow up the * event that we are interested in. On the other hand, if there is another * thread polling the same descriptors, there is a chance that it will receive * the event that we were interested in. In this situation, myfunc() * will only realise that the transfer has completed on the next iteration of * the loop, up to 120 seconds later. Clearly a two-minute delay is * undesirable, and don't even think about using short timeouts to circumvent * this issue! * * The solution here is to ensure that no two threads are ever polling the * file descriptors at the same time. A naive implementation of this would * impact the capabilities of the library, so libusb offers the scheme * documented below to ensure no loss of functionality. * * Before we go any further, it is worth mentioning that all libusb-wrapped * event handling procedures fully adhere to the scheme documented below. * This includes libusb_handle_events() and its variants, and all the * synchronous I/O functions - libusb hides this headache from you. * * \section Using libusb_handle_events() from multiple threads * * Even when only using libusb_handle_events() and synchronous I/O functions, * you can still have a race condition. You might be tempted to solve the * above with libusb_handle_events() like so: * \code libusb_submit_transfer(transfer); while (!completed) { libusb_handle_events(ctx); } printf("completed!"); \endcode * * This however has a race between the checking of completed and * libusb_handle_events() acquiring the events lock, so another thread * could have completed the transfer, resulting in this thread hanging * until either a timeout or another event occurs. See also commit * 6696512aade99bb15d6792af90ae329af270eba6 which fixes this in the * synchronous API implementation of libusb. * * Fixing this race requires checking the variable completed only after * taking the event lock, which defeats the concept of just calling * libusb_handle_events() without worrying about locking. This is why * libusb-1.0.9 introduces the new libusb_handle_events_timeout_completed() * and libusb_handle_events_completed() functions, which handles doing the * completion check for you after they have acquired the lock: * \code libusb_submit_transfer(transfer); while (!completed) { libusb_handle_events_completed(ctx, &completed); } printf("completed!"); \endcode * * This nicely fixes the race in our example. Note that if all you want to * do is submit a single transfer and wait for its completion, then using * one of the synchronous I/O functions is much easier. * * \note * The completed variable must be modified while holding the event lock, * otherwise a race condition can still exist. It is simplest to do so from * within the transfer callback as shown above. * * \section eventlock The events lock * * The problem is when we consider the fact that libusb exposes file * descriptors to allow for you to integrate asynchronous USB I/O into * existing main loops, effectively allowing you to do some work behind * libusb's back. If you do take libusb's file descriptors and pass them to * poll()/select() yourself, you need to be aware of the associated issues. * * The first concept to be introduced is the events lock. The events lock * is used to serialize threads that want to handle events, such that only * one thread is handling events at any one time. * * You must take the events lock before polling libusb file descriptors, * using libusb_lock_events(). You must release the lock as soon as you have * aborted your poll()/select() loop, using libusb_unlock_events(). * * \section threadwait Letting other threads do the work for you * * Although the events lock is a critical part of the solution, it is not * enough on it's own. You might wonder if the following is sufficient... \code libusb_lock_events(ctx); while (!completed) { poll(libusb file descriptors, 120*1000); if (poll indicates activity) libusb_handle_events_timeout(ctx, &zero_tv); } libusb_unlock_events(ctx); \endcode * ...and the answer is that it is not. This is because the transfer in the * code shown above may take a long time (say 30 seconds) to complete, and * the lock is not released until the transfer is completed. * * Another thread with similar code that wants to do event handling may be * working with a transfer that completes after a few milliseconds. Despite * having such a quick completion time, the other thread cannot check that * status of its transfer until the code above has finished (30 seconds later) * due to contention on the lock. * * To solve this, libusb offers you a mechanism to determine when another * thread is handling events. It also offers a mechanism to block your thread * until the event handling thread has completed an event (and this mechanism * does not involve polling of file descriptors). * * After determining that another thread is currently handling events, you * obtain the event waiters lock using libusb_lock_event_waiters(). * You then re-check that some other thread is still handling events, and if * so, you call libusb_wait_for_event(). * * libusb_wait_for_event() puts your application to sleep until an event * occurs, or until a thread releases the events lock. When either of these * things happen, your thread is woken up, and should re-check the condition * it was waiting on. It should also re-check that another thread is handling * events, and if not, it should start handling events itself. * * This looks like the following, as pseudo-code: \code retry: if (libusb_try_lock_events(ctx) == 0) { // we obtained the event lock: do our own event handling while (!completed) { if (!libusb_event_handling_ok(ctx)) { libusb_unlock_events(ctx); goto retry; } poll(libusb file descriptors, 120*1000); if (poll indicates activity) libusb_handle_events_locked(ctx, 0); } libusb_unlock_events(ctx); } else { // another thread is doing event handling. wait for it to signal us that // an event has completed libusb_lock_event_waiters(ctx); while (!completed) { // now that we have the event waiters lock, double check that another // thread is still handling events for us. (it may have ceased handling // events in the time it took us to reach this point) if (!libusb_event_handler_active(ctx)) { // whoever was handling events is no longer doing so, try again libusb_unlock_event_waiters(ctx); goto retry; } libusb_wait_for_event(ctx, NULL); } libusb_unlock_event_waiters(ctx); } printf("completed!\n"); \endcode * * A naive look at the above code may suggest that this can only support * one event waiter (hence a total of 2 competing threads, the other doing * event handling), because the event waiter seems to have taken the event * waiters lock while waiting for an event. However, the system does support * multiple event waiters, because libusb_wait_for_event() actually drops * the lock while waiting, and reacquires it before continuing. * * We have now implemented code which can dynamically handle situations where * nobody is handling events (so we should do it ourselves), and it can also * handle situations where another thread is doing event handling (so we can * piggyback onto them). It is also equipped to handle a combination of * the two, for example, another thread is doing event handling, but for * whatever reason it stops doing so before our condition is met, so we take * over the event handling. * * Four functions were introduced in the above pseudo-code. Their importance * should be apparent from the code shown above. * -# libusb_try_lock_events() is a non-blocking function which attempts * to acquire the events lock but returns a failure code if it is contended. * -# libusb_event_handling_ok() checks that libusb is still happy for your * thread to be performing event handling. Sometimes, libusb needs to * interrupt the event handler, and this is how you can check if you have * been interrupted. If this function returns 0, the correct behaviour is * for you to give up the event handling lock, and then to repeat the cycle. * The following libusb_try_lock_events() will fail, so you will become an * events waiter. For more information on this, read \ref fullstory below. * -# libusb_handle_events_locked() is a variant of * libusb_handle_events_timeout() that you can call while holding the * events lock. libusb_handle_events_timeout() itself implements similar * logic to the above, so be sure not to call it when you are * "working behind libusb's back", as is the case here. * -# libusb_event_handler_active() determines if someone is currently * holding the events lock * * You might be wondering why there is no function to wake up all threads * blocked on libusb_wait_for_event(). This is because libusb can do this * internally: it will wake up all such threads when someone calls * libusb_unlock_events() or when a transfer completes (at the point after its * callback has returned). * * \subsection fullstory The full story * * The above explanation should be enough to get you going, but if you're * really thinking through the issues then you may be left with some more * questions regarding libusb's internals. If you're curious, read on, and if * not, skip to the next section to avoid confusing yourself! * * The immediate question that may spring to mind is: what if one thread * modifies the set of file descriptors that need to be polled while another * thread is doing event handling? * * There are 2 situations in which this may happen. * -# libusb_open() will add another file descriptor to the poll set, * therefore it is desirable to interrupt the event handler so that it * restarts, picking up the new descriptor. * -# libusb_close() will remove a file descriptor from the poll set. There * are all kinds of race conditions that could arise here, so it is * important that nobody is doing event handling at this time. * * libusb handles these issues internally, so application developers do not * have to stop their event handlers while opening/closing devices. Here's how * it works, focusing on the libusb_close() situation first: * * -# During initialization, libusb opens an internal pipe, and it adds the read * end of this pipe to the set of file descriptors to be polled. * -# During libusb_close(), libusb writes some dummy data on this event pipe. * This immediately interrupts the event handler. libusb also records * internally that it is trying to interrupt event handlers for this * high-priority event. * -# At this point, some of the functions described above start behaving * differently: * - libusb_event_handling_ok() starts returning 1, indicating that it is NOT * OK for event handling to continue. * - libusb_try_lock_events() starts returning 1, indicating that another * thread holds the event handling lock, even if the lock is uncontended. * - libusb_event_handler_active() starts returning 1, indicating that * another thread is doing event handling, even if that is not true. * -# The above changes in behaviour result in the event handler stopping and * giving up the events lock very quickly, giving the high-priority * libusb_close() operation a "free ride" to acquire the events lock. All * threads that are competing to do event handling become event waiters. * -# With the events lock held inside libusb_close(), libusb can safely remove * a file descriptor from the poll set, in the safety of knowledge that * nobody is polling those descriptors or trying to access the poll set. * -# After obtaining the events lock, the close operation completes very * quickly (usually a matter of milliseconds) and then immediately releases * the events lock. * -# At the same time, the behaviour of libusb_event_handling_ok() and friends * reverts to the original, documented behaviour. * -# The release of the events lock causes the threads that are waiting for * events to be woken up and to start competing to become event handlers * again. One of them will succeed; it will then re-obtain the list of poll * descriptors, and USB I/O will then continue as normal. * * libusb_open() is similar, and is actually a more simplistic case. Upon a * call to libusb_open(): * * -# The device is opened and a file descriptor is added to the poll set. * -# libusb sends some dummy data on the event pipe, and records that it * is trying to modify the poll descriptor set. * -# The event handler is interrupted, and the same behaviour change as for * libusb_close() takes effect, causing all event handling threads to become * event waiters. * -# The libusb_open() implementation takes its free ride to the events lock. * -# Happy that it has successfully paused the events handler, libusb_open() * releases the events lock. * -# The event waiter threads are all woken up and compete to become event * handlers again. The one that succeeds will obtain the list of poll * descriptors again, which will include the addition of the new device. * * \subsection concl Closing remarks * * The above may seem a little complicated, but hopefully I have made it clear * why such complications are necessary. Also, do not forget that this only * applies to applications that take libusb's file descriptors and integrate * them into their own polling loops. * * You may decide that it is OK for your multi-threaded application to ignore * some of the rules and locks detailed above, because you don't think that * two threads can ever be polling the descriptors at the same time. If that * is the case, then that's good news for you because you don't have to worry. * But be careful here; remember that the synchronous I/O functions do event * handling internally. If you have one thread doing event handling in a loop * (without implementing the rules and locking semantics documented above) * and another trying to send a synchronous USB transfer, you will end up with * two threads monitoring the same descriptors, and the above-described * undesirable behaviour occurring. The solution is for your polling thread to * play by the rules; the synchronous I/O functions do so, and this will result * in them getting along in perfect harmony. * * If you do have a dedicated thread doing event handling, it is perfectly * legal for it to take the event handling lock for long periods of time. Any * synchronous I/O functions you call from other threads will transparently * fall back to the "event waiters" mechanism detailed above. The only * consideration that your event handling thread must apply is the one related * to libusb_event_handling_ok(): you must call this before every poll(), and * give up the events lock if instructed. */ int usbi_io_init(struct libusb_context *ctx) { int r; usbi_mutex_init(&ctx->flying_transfers_lock); usbi_mutex_init(&ctx->events_lock); usbi_mutex_init(&ctx->event_waiters_lock); usbi_cond_init(&ctx->event_waiters_cond); usbi_mutex_init(&ctx->event_data_lock); usbi_tls_key_create(&ctx->event_handling_key); list_init(&ctx->flying_transfers); list_init(&ctx->event_sources); list_init(&ctx->removed_event_sources); list_init(&ctx->hotplug_msgs); list_init(&ctx->completed_transfers); r = usbi_create_event(&ctx->event); if (r < 0) goto err; r = usbi_add_event_source(ctx, USBI_EVENT_OS_HANDLE(&ctx->event), USBI_EVENT_POLL_EVENTS); if (r < 0) goto err_destroy_event; #ifdef HAVE_OS_TIMER r = usbi_create_timer(&ctx->timer); if (r == 0) { usbi_dbg(ctx, "using timer for timeouts"); r = usbi_add_event_source(ctx, USBI_TIMER_OS_HANDLE(&ctx->timer), USBI_TIMER_POLL_EVENTS); if (r < 0) goto err_destroy_timer; } else { usbi_dbg(ctx, "timer not available for timeouts"); } #endif return 0; #ifdef HAVE_OS_TIMER err_destroy_timer: usbi_destroy_timer(&ctx->timer); usbi_remove_event_source(ctx, USBI_EVENT_OS_HANDLE(&ctx->event)); #endif err_destroy_event: usbi_destroy_event(&ctx->event); err: usbi_mutex_destroy(&ctx->flying_transfers_lock); usbi_mutex_destroy(&ctx->events_lock); usbi_mutex_destroy(&ctx->event_waiters_lock); usbi_cond_destroy(&ctx->event_waiters_cond); usbi_mutex_destroy(&ctx->event_data_lock); usbi_tls_key_delete(ctx->event_handling_key); return r; } static void cleanup_removed_event_sources(struct libusb_context *ctx) { struct usbi_event_source *ievent_source, *tmp; for_each_removed_event_source_safe(ctx, ievent_source, tmp) { list_del(&ievent_source->list); free(ievent_source); } } void usbi_io_exit(struct libusb_context *ctx) { #ifdef HAVE_OS_TIMER if (usbi_using_timer(ctx)) { usbi_remove_event_source(ctx, USBI_TIMER_OS_HANDLE(&ctx->timer)); usbi_destroy_timer(&ctx->timer); } #endif usbi_remove_event_source(ctx, USBI_EVENT_OS_HANDLE(&ctx->event)); usbi_destroy_event(&ctx->event); usbi_mutex_destroy(&ctx->flying_transfers_lock); usbi_mutex_destroy(&ctx->events_lock); usbi_mutex_destroy(&ctx->event_waiters_lock); usbi_cond_destroy(&ctx->event_waiters_cond); usbi_mutex_destroy(&ctx->event_data_lock); usbi_tls_key_delete(ctx->event_handling_key); cleanup_removed_event_sources(ctx); free(ctx->event_data); } static void calculate_timeout(struct usbi_transfer *itransfer) { unsigned int timeout = USBI_TRANSFER_TO_LIBUSB_TRANSFER(itransfer)->timeout; if (!timeout) { TIMESPEC_CLEAR(&itransfer->timeout); return; } usbi_get_monotonic_time(&itransfer->timeout); itransfer->timeout.tv_sec += timeout / 1000U; itransfer->timeout.tv_nsec += (timeout % 1000U) * 1000000L; if (itransfer->timeout.tv_nsec >= NSEC_PER_SEC) { ++itransfer->timeout.tv_sec; itransfer->timeout.tv_nsec -= NSEC_PER_SEC; } } /** \ingroup libusb_asyncio * Allocate a libusb transfer with a specified number of isochronous packet * descriptors. The returned transfer is pre-initialized for you. When the new * transfer is no longer needed, it should be freed with * libusb_free_transfer(). * * Transfers intended for non-isochronous endpoints (e.g. control, bulk, * interrupt) should specify an iso_packets count of zero. * * For transfers intended for isochronous endpoints, specify an appropriate * number of packet descriptors to be allocated as part of the transfer. * The returned transfer is not specially initialized for isochronous I/O; * you are still required to set the * \ref libusb_transfer::num_iso_packets "num_iso_packets" and * \ref libusb_transfer::type "type" fields accordingly. * * It is safe to allocate a transfer with some isochronous packets and then * use it on a non-isochronous endpoint. If you do this, ensure that at time * of submission, num_iso_packets is 0 and that type is set appropriately. * * \param iso_packets number of isochronous packet descriptors to allocate. Must be non-negative. * \returns a newly allocated transfer, or NULL on error */ DEFAULT_VISIBILITY struct libusb_transfer * LIBUSB_CALL libusb_alloc_transfer( int iso_packets) { size_t priv_size; size_t alloc_size; unsigned char *ptr; struct usbi_transfer *itransfer; struct libusb_transfer *transfer; assert(iso_packets >= 0); if (iso_packets < 0) return NULL; priv_size = PTR_ALIGN(usbi_backend.transfer_priv_size); alloc_size = priv_size + sizeof(struct usbi_transfer) + sizeof(struct libusb_transfer) + (sizeof(struct libusb_iso_packet_descriptor) * (size_t)iso_packets); ptr = calloc(1, alloc_size); if (!ptr) return NULL; itransfer = (struct usbi_transfer *)(ptr + priv_size); itransfer->num_iso_packets = iso_packets; itransfer->priv = ptr; usbi_mutex_init(&itransfer->lock); transfer = USBI_TRANSFER_TO_LIBUSB_TRANSFER(itransfer); return transfer; } /** \ingroup libusb_asyncio * Free a transfer structure. This should be called for all transfers * allocated with libusb_alloc_transfer(). * * If the \ref libusb_transfer_flags::LIBUSB_TRANSFER_FREE_BUFFER * "LIBUSB_TRANSFER_FREE_BUFFER" flag is set and the transfer buffer is * non-NULL, this function will also free the transfer buffer using the * standard system memory allocator (e.g. free()). * * It is legal to call this function with a NULL transfer. In this case, * the function will simply return safely. * * It is not legal to free an active transfer (one which has been submitted * and has not yet completed). * * \param transfer the transfer to free */ void API_EXPORTED libusb_free_transfer(struct libusb_transfer *transfer) { struct usbi_transfer *itransfer; size_t priv_size; unsigned char *ptr; if (!transfer) return; usbi_dbg(TRANSFER_CTX(transfer), "transfer %p", transfer); if (transfer->flags & LIBUSB_TRANSFER_FREE_BUFFER) free(transfer->buffer); itransfer = LIBUSB_TRANSFER_TO_USBI_TRANSFER(transfer); usbi_mutex_destroy(&itransfer->lock); priv_size = PTR_ALIGN(usbi_backend.transfer_priv_size); ptr = (unsigned char *)itransfer - priv_size; assert(ptr == itransfer->priv); free(ptr); } /* iterates through the flying transfers, and rearms the timer based on the * next upcoming timeout. * must be called with flying_list locked. * returns 0 on success or a LIBUSB_ERROR code on failure. */ #ifdef HAVE_OS_TIMER static int arm_timer_for_next_timeout(struct libusb_context *ctx) { struct usbi_transfer *itransfer; if (!usbi_using_timer(ctx)) return 0; for_each_transfer(ctx, itransfer) { struct timespec *cur_ts = &itransfer->timeout; /* if we've reached transfers of infinite timeout, then we have no * arming to do */ if (!TIMESPEC_IS_SET(cur_ts)) break; /* act on first transfer that has not already been handled */ if (!(itransfer->timeout_flags & (USBI_TRANSFER_TIMEOUT_HANDLED | USBI_TRANSFER_OS_HANDLES_TIMEOUT))) { usbi_dbg(ctx, "next timeout originally %ums", USBI_TRANSFER_TO_LIBUSB_TRANSFER(itransfer)->timeout); return usbi_arm_timer(&ctx->timer, cur_ts); } } usbi_dbg(ctx, "no timeouts, disarming timer"); return usbi_disarm_timer(&ctx->timer); } #else static inline int arm_timer_for_next_timeout(struct libusb_context *ctx) { UNUSED(ctx); return 0; } #endif /* add a transfer to the (timeout-sorted) active transfers list. * This function will return non 0 if fails to update the timer, * in which case the transfer is *not* on the flying_transfers list. */ static int add_to_flying_list(struct usbi_transfer *itransfer) { struct usbi_transfer *cur; struct timespec *timeout = &itransfer->timeout; struct libusb_context *ctx = ITRANSFER_CTX(itransfer); int r = 0; int first = 1; calculate_timeout(itransfer); /* if we have no other flying transfers, start the list with this one */ if (list_empty(&ctx->flying_transfers)) { list_add(&itransfer->list, &ctx->flying_transfers); goto out; } /* if we have infinite timeout, append to end of list */ if (!TIMESPEC_IS_SET(timeout)) { list_add_tail(&itransfer->list, &ctx->flying_transfers); /* first is irrelevant in this case */ goto out; } /* otherwise, find appropriate place in list */ for_each_transfer(ctx, cur) { /* find first timeout that occurs after the transfer in question */ struct timespec *cur_ts = &cur->timeout; if (!TIMESPEC_IS_SET(cur_ts) || TIMESPEC_CMP(cur_ts, timeout, >)) { list_add_tail(&itransfer->list, &cur->list); goto out; } first = 0; } /* first is 0 at this stage (list not empty) */ /* otherwise we need to be inserted at the end */ list_add_tail(&itransfer->list, &ctx->flying_transfers); out: #ifdef HAVE_OS_TIMER if (first && usbi_using_timer(ctx) && TIMESPEC_IS_SET(timeout)) { /* if this transfer has the lowest timeout of all active transfers, * rearm the timer with this transfer's timeout */ usbi_dbg(ctx, "arm timer for timeout in %ums (first in line)", USBI_TRANSFER_TO_LIBUSB_TRANSFER(itransfer)->timeout); r = usbi_arm_timer(&ctx->timer, timeout); } #else UNUSED(first); #endif if (r) list_del(&itransfer->list); return r; } /* remove a transfer from the active transfers list. * This function will *always* remove the transfer from the * flying_transfers list. It will return a LIBUSB_ERROR code * if it fails to update the timer for the next timeout. */ static int remove_from_flying_list(struct usbi_transfer *itransfer) { struct libusb_context *ctx = ITRANSFER_CTX(itransfer); int rearm_timer; int r = 0; usbi_mutex_lock(&ctx->flying_transfers_lock); rearm_timer = (TIMESPEC_IS_SET(&itransfer->timeout) && list_first_entry(&ctx->flying_transfers, struct usbi_transfer, list) == itransfer); list_del(&itransfer->list); if (rearm_timer) r = arm_timer_for_next_timeout(ctx); usbi_mutex_unlock(&ctx->flying_transfers_lock); return r; } /** \ingroup libusb_asyncio * Submit a transfer. This function will fire off the USB transfer and then * return immediately. * * \param transfer the transfer to submit * \returns 0 on success * \returns LIBUSB_ERROR_NO_DEVICE if the device has been disconnected * \returns LIBUSB_ERROR_BUSY if the transfer has already been submitted. * \returns LIBUSB_ERROR_NOT_SUPPORTED if the transfer flags are not supported * by the operating system. * \returns LIBUSB_ERROR_INVALID_PARAM if the transfer size is larger than * the operating system and/or hardware can support (see \ref asynclimits) * \returns another LIBUSB_ERROR code on other failure */ int API_EXPORTED libusb_submit_transfer(struct libusb_transfer *transfer) { struct usbi_transfer *itransfer = LIBUSB_TRANSFER_TO_USBI_TRANSFER(transfer); struct libusb_context *ctx = TRANSFER_CTX(transfer); int r; usbi_dbg(ctx, "transfer %p", transfer); /* * Important note on locking, this function takes / releases locks * in the following order: * take flying_transfers_lock * take itransfer->lock * clear transfer * add to flying_transfers list * release flying_transfers_lock * submit transfer * release itransfer->lock * if submit failed: * take flying_transfers_lock * remove from flying_transfers list * release flying_transfers_lock * * Note that it takes locks in the order a-b and then releases them * in the same order a-b. This is somewhat unusual but not wrong, * release order is not important as long as *all* locks are released * before re-acquiring any locks. * * This means that the ordering of first releasing itransfer->lock * and then re-acquiring the flying_transfers_list on error is * important and must not be changed! * * This is done this way because when we take both locks we must always * take flying_transfers_lock first to avoid ab-ba style deadlocks with * the timeout handling and usbi_handle_disconnect paths. * * And we cannot release itransfer->lock before the submission is * complete otherwise timeout handling for transfers with short * timeouts may run before submission. */ usbi_mutex_lock(&ctx->flying_transfers_lock); usbi_mutex_lock(&itransfer->lock); if (itransfer->state_flags & USBI_TRANSFER_IN_FLIGHT) { usbi_mutex_unlock(&ctx->flying_transfers_lock); usbi_mutex_unlock(&itransfer->lock); return LIBUSB_ERROR_BUSY; } itransfer->transferred = 0; itransfer->state_flags = 0; itransfer->timeout_flags = 0; r = add_to_flying_list(itransfer); if (r) { usbi_mutex_unlock(&ctx->flying_transfers_lock); usbi_mutex_unlock(&itransfer->lock); return r; } /* * We must release the flying transfers lock here, because with * some backends the submit_transfer method is synchronous. */ usbi_mutex_unlock(&ctx->flying_transfers_lock); r = usbi_backend.submit_transfer(itransfer); if (r == LIBUSB_SUCCESS) { itransfer->state_flags |= USBI_TRANSFER_IN_FLIGHT; /* keep a reference to this device */ libusb_ref_device(transfer->dev_handle->dev); } usbi_mutex_unlock(&itransfer->lock); if (r != LIBUSB_SUCCESS) remove_from_flying_list(itransfer); return r; } /** \ingroup libusb_asyncio * Asynchronously cancel a previously submitted transfer. * This function returns immediately, but this does not indicate cancellation * is complete. Your callback function will be invoked at some later time * with a transfer status of * \ref libusb_transfer_status::LIBUSB_TRANSFER_CANCELLED * "LIBUSB_TRANSFER_CANCELLED." * * \param transfer the transfer to cancel * \returns 0 on success * \returns LIBUSB_ERROR_NOT_FOUND if the transfer is not in progress, * already complete, or already cancelled. * \returns a LIBUSB_ERROR code on failure */ int API_EXPORTED libusb_cancel_transfer(struct libusb_transfer *transfer) { struct usbi_transfer *itransfer = LIBUSB_TRANSFER_TO_USBI_TRANSFER(transfer); struct libusb_context *ctx = ITRANSFER_CTX(itransfer); int r; usbi_dbg(ctx, "transfer %p", transfer ); usbi_mutex_lock(&itransfer->lock); if (!(itransfer->state_flags & USBI_TRANSFER_IN_FLIGHT) || (itransfer->state_flags & USBI_TRANSFER_CANCELLING)) { r = LIBUSB_ERROR_NOT_FOUND; goto out; } r = usbi_backend.cancel_transfer(itransfer); if (r < 0) { if (r != LIBUSB_ERROR_NOT_FOUND && r != LIBUSB_ERROR_NO_DEVICE) usbi_err(ctx, "cancel transfer failed error %d", r); else usbi_dbg(ctx, "cancel transfer failed error %d", r); if (r == LIBUSB_ERROR_NO_DEVICE) itransfer->state_flags |= USBI_TRANSFER_DEVICE_DISAPPEARED; } itransfer->state_flags |= USBI_TRANSFER_CANCELLING; out: usbi_mutex_unlock(&itransfer->lock); return r; } /** \ingroup libusb_asyncio * Set a transfers bulk stream id. Note users are advised to use * libusb_fill_bulk_stream_transfer() instead of calling this function * directly. * * Since version 1.0.19, \ref LIBUSB_API_VERSION >= 0x01000103 * * \param transfer the transfer to set the stream id for * \param stream_id the stream id to set * \see libusb_alloc_streams() */ void API_EXPORTED libusb_transfer_set_stream_id( struct libusb_transfer *transfer, uint32_t stream_id) { struct usbi_transfer *itransfer = LIBUSB_TRANSFER_TO_USBI_TRANSFER(transfer); itransfer->stream_id = stream_id; } /** \ingroup libusb_asyncio * Get a transfers bulk stream id. * * Since version 1.0.19, \ref LIBUSB_API_VERSION >= 0x01000103 * * \param transfer the transfer to get the stream id for * \returns the stream id for the transfer */ uint32_t API_EXPORTED libusb_transfer_get_stream_id( struct libusb_transfer *transfer) { struct usbi_transfer *itransfer = LIBUSB_TRANSFER_TO_USBI_TRANSFER(transfer); return itransfer->stream_id; } /* Handle completion of a transfer (completion might be an error condition). * This will invoke the user-supplied callback function, which may end up * freeing the transfer. Therefore you cannot use the transfer structure * after calling this function, and you should free all backend-specific * data before calling it. * Do not call this function with the usbi_transfer lock held. User-specified * callback functions may attempt to directly resubmit the transfer, which * will attempt to take the lock. */ int usbi_handle_transfer_completion(struct usbi_transfer *itransfer, enum libusb_transfer_status status) { struct libusb_transfer *transfer = USBI_TRANSFER_TO_LIBUSB_TRANSFER(itransfer); struct libusb_device_handle *dev_handle = transfer->dev_handle; struct libusb_context *ctx = ITRANSFER_CTX(itransfer); uint8_t flags; int r; r = remove_from_flying_list(itransfer); if (r < 0) usbi_err(ctx, "failed to set timer for next timeout"); usbi_mutex_lock(&itransfer->lock); itransfer->state_flags &= ~USBI_TRANSFER_IN_FLIGHT; usbi_mutex_unlock(&itransfer->lock); if (status == LIBUSB_TRANSFER_COMPLETED && transfer->flags & LIBUSB_TRANSFER_SHORT_NOT_OK) { int rqlen = transfer->length; if (transfer->type == LIBUSB_TRANSFER_TYPE_CONTROL) rqlen -= LIBUSB_CONTROL_SETUP_SIZE; if (rqlen != itransfer->transferred) { usbi_dbg(ctx, "interpreting short transfer as error"); status = LIBUSB_TRANSFER_ERROR; } } flags = transfer->flags; transfer->status = status; transfer->actual_length = itransfer->transferred; usbi_dbg(ctx, "transfer %p has callback %p", transfer, transfer->callback); if (transfer->callback) transfer->callback(transfer); /* transfer might have been freed by the above call, do not use from * this point. */ if (flags & LIBUSB_TRANSFER_FREE_TRANSFER) libusb_free_transfer(transfer); libusb_unref_device(dev_handle->dev); return r; } /* Similar to usbi_handle_transfer_completion() but exclusively for transfers * that were asynchronously cancelled. The same concerns w.r.t. freeing of * transfers exist here. * Do not call this function with the usbi_transfer lock held. User-specified * callback functions may attempt to directly resubmit the transfer, which * will attempt to take the lock. */ int usbi_handle_transfer_cancellation(struct usbi_transfer *itransfer) { struct libusb_context *ctx = ITRANSFER_CTX(itransfer); uint8_t timed_out; usbi_mutex_lock(&ctx->flying_transfers_lock); timed_out = itransfer->timeout_flags & USBI_TRANSFER_TIMED_OUT; usbi_mutex_unlock(&ctx->flying_transfers_lock); /* if the URB was cancelled due to timeout, report timeout to the user */ if (timed_out) { usbi_dbg(ctx, "detected timeout cancellation"); return usbi_handle_transfer_completion(itransfer, LIBUSB_TRANSFER_TIMED_OUT); } /* otherwise its a normal async cancel */ return usbi_handle_transfer_completion(itransfer, LIBUSB_TRANSFER_CANCELLED); } /* Add a completed transfer to the completed_transfers list of the * context and signal the event. The backend's handle_transfer_completion() * function will be called the next time an event handler runs. */ void usbi_signal_transfer_completion(struct usbi_transfer *itransfer) { libusb_device_handle *dev_handle = USBI_TRANSFER_TO_LIBUSB_TRANSFER(itransfer)->dev_handle; if (dev_handle) { struct libusb_context *ctx = HANDLE_CTX(dev_handle); unsigned int event_flags; usbi_mutex_lock(&ctx->event_data_lock); event_flags = ctx->event_flags; ctx->event_flags |= USBI_EVENT_TRANSFER_COMPLETED; list_add_tail(&itransfer->completed_list, &ctx->completed_transfers); if (!event_flags) usbi_signal_event(&ctx->event); usbi_mutex_unlock(&ctx->event_data_lock); } } /** \ingroup libusb_poll * Attempt to acquire the event handling lock. This lock is used to ensure that * only one thread is monitoring libusb event sources at any one time. * * You only need to use this lock if you are developing an application * which calls poll() or select() on libusb's file descriptors directly. * If you stick to libusb's event handling loop functions (e.g. * libusb_handle_events()) then you do not need to be concerned with this * locking. * * While holding this lock, you are trusted to actually be handling events. * If you are no longer handling events, you must call libusb_unlock_events() * as soon as possible. * * \param ctx the context to operate on, or NULL for the default context * \returns 0 if the lock was obtained successfully * \returns 1 if the lock was not obtained (i.e. another thread holds the lock) * \ref libusb_mtasync */ int API_EXPORTED libusb_try_lock_events(libusb_context *ctx) { int r; unsigned int ru; ctx = usbi_get_context(ctx); /* is someone else waiting to close a device? if so, don't let this thread * start event handling */ usbi_mutex_lock(&ctx->event_data_lock); ru = ctx->device_close; usbi_mutex_unlock(&ctx->event_data_lock); if (ru) { usbi_dbg(ctx, "someone else is closing a device"); return 1; } r = usbi_mutex_trylock(&ctx->events_lock); if (!r) return 1; ctx->event_handler_active = 1; return 0; } /** \ingroup libusb_poll * Acquire the event handling lock, blocking until successful acquisition if * it is contended. This lock is used to ensure that only one thread is * monitoring libusb event sources at any one time. * * You only need to use this lock if you are developing an application * which calls poll() or select() on libusb's file descriptors directly. * If you stick to libusb's event handling loop functions (e.g. * libusb_handle_events()) then you do not need to be concerned with this * locking. * * While holding this lock, you are trusted to actually be handling events. * If you are no longer handling events, you must call libusb_unlock_events() * as soon as possible. * * \param ctx the context to operate on, or NULL for the default context * \ref libusb_mtasync */ void API_EXPORTED libusb_lock_events(libusb_context *ctx) { ctx = usbi_get_context(ctx); usbi_mutex_lock(&ctx->events_lock); ctx->event_handler_active = 1; } /** \ingroup libusb_poll * Release the lock previously acquired with libusb_try_lock_events() or * libusb_lock_events(). Releasing this lock will wake up any threads blocked * on libusb_wait_for_event(). * * \param ctx the context to operate on, or NULL for the default context * \ref libusb_mtasync */ void API_EXPORTED libusb_unlock_events(libusb_context *ctx) { ctx = usbi_get_context(ctx); ctx->event_handler_active = 0; usbi_mutex_unlock(&ctx->events_lock); /* FIXME: perhaps we should be a bit more efficient by not broadcasting * the availability of the events lock when we are modifying pollfds * (check ctx->device_close)? */ usbi_mutex_lock(&ctx->event_waiters_lock); usbi_cond_broadcast(&ctx->event_waiters_cond); usbi_mutex_unlock(&ctx->event_waiters_lock); } /** \ingroup libusb_poll * Determine if it is still OK for this thread to be doing event handling. * * Sometimes, libusb needs to temporarily pause all event handlers, and this * is the function you should use before polling file descriptors to see if * this is the case. * * If this function instructs your thread to give up the events lock, you * should just continue the usual logic that is documented in \ref libusb_mtasync. * On the next iteration, your thread will fail to obtain the events lock, * and will hence become an event waiter. * * This function should be called while the events lock is held: you don't * need to worry about the results of this function if your thread is not * the current event handler. * * \param ctx the context to operate on, or NULL for the default context * \returns 1 if event handling can start or continue * \returns 0 if this thread must give up the events lock * \ref fullstory "Multi-threaded I/O: the full story" */ int API_EXPORTED libusb_event_handling_ok(libusb_context *ctx) { unsigned int r; ctx = usbi_get_context(ctx); /* is someone else waiting to close a device? if so, don't let this thread * continue event handling */ usbi_mutex_lock(&ctx->event_data_lock); r = ctx->device_close; usbi_mutex_unlock(&ctx->event_data_lock); if (r) { usbi_dbg(ctx, "someone else is closing a device"); return 0; } return 1; } /** \ingroup libusb_poll * Determine if an active thread is handling events (i.e. if anyone is holding * the event handling lock). * * \param ctx the context to operate on, or NULL for the default context * \returns 1 if a thread is handling events * \returns 0 if there are no threads currently handling events * \ref libusb_mtasync */ int API_EXPORTED libusb_event_handler_active(libusb_context *ctx) { unsigned int r; ctx = usbi_get_context(ctx); /* is someone else waiting to close a device? if so, don't let this thread * start event handling -- indicate that event handling is happening */ usbi_mutex_lock(&ctx->event_data_lock); r = ctx->device_close; usbi_mutex_unlock(&ctx->event_data_lock); if (r) { usbi_dbg(ctx, "someone else is closing a device"); return 1; } return ctx->event_handler_active; } /** \ingroup libusb_poll * Interrupt any active thread that is handling events. This is mainly useful * for interrupting a dedicated event handling thread when an application * wishes to call libusb_exit(). * * Since version 1.0.21, \ref LIBUSB_API_VERSION >= 0x01000105 * * \param ctx the context to operate on, or NULL for the default context * \ref libusb_mtasync */ void API_EXPORTED libusb_interrupt_event_handler(libusb_context *ctx) { unsigned int event_flags; usbi_dbg(ctx, " "); ctx = usbi_get_context(ctx); usbi_mutex_lock(&ctx->event_data_lock); event_flags = ctx->event_flags; ctx->event_flags |= USBI_EVENT_USER_INTERRUPT; if (!event_flags) usbi_signal_event(&ctx->event); usbi_mutex_unlock(&ctx->event_data_lock); } /** \ingroup libusb_poll * Acquire the event waiters lock. This lock is designed to be obtained under * the situation where you want to be aware when events are completed, but * some other thread is event handling so calling libusb_handle_events() is not * allowed. * * You then obtain this lock, re-check that another thread is still handling * events, then call libusb_wait_for_event(). * * You only need to use this lock if you are developing an application * which calls poll() or select() on libusb's file descriptors directly, * and may potentially be handling events from 2 threads simultaneously. * If you stick to libusb's event handling loop functions (e.g. * libusb_handle_events()) then you do not need to be concerned with this * locking. * * \param ctx the context to operate on, or NULL for the default context * \ref libusb_mtasync */ void API_EXPORTED libusb_lock_event_waiters(libusb_context *ctx) { ctx = usbi_get_context(ctx); usbi_mutex_lock(&ctx->event_waiters_lock); } /** \ingroup libusb_poll * Release the event waiters lock. * \param ctx the context to operate on, or NULL for the default context * \ref libusb_mtasync */ void API_EXPORTED libusb_unlock_event_waiters(libusb_context *ctx) { ctx = usbi_get_context(ctx); usbi_mutex_unlock(&ctx->event_waiters_lock); } /** \ingroup libusb_poll * Wait for another thread to signal completion of an event. Must be called * with the event waiters lock held, see libusb_lock_event_waiters(). * * This function will block until any of the following conditions are met: * -# The timeout expires * -# A transfer completes * -# A thread releases the event handling lock through libusb_unlock_events() * * Condition 1 is obvious. Condition 2 unblocks your thread after * the callback for the transfer has completed. Condition 3 is important * because it means that the thread that was previously handling events is no * longer doing so, so if any events are to complete, another thread needs to * step up and start event handling. * * This function releases the event waiters lock before putting your thread * to sleep, and reacquires the lock as it is being woken up. * * \param ctx the context to operate on, or NULL for the default context * \param tv maximum timeout for this blocking function. A NULL value * indicates unlimited timeout. * \returns 0 after a transfer completes or another thread stops event handling * \returns 1 if the timeout expired * \returns LIBUSB_ERROR_INVALID_PARAM if timeval is invalid * \ref libusb_mtasync */ int API_EXPORTED libusb_wait_for_event(libusb_context *ctx, struct timeval *tv) { int r; ctx = usbi_get_context(ctx); if (!tv) { usbi_cond_wait(&ctx->event_waiters_cond, &ctx->event_waiters_lock); return 0; } if (!TIMEVAL_IS_VALID(tv)) return LIBUSB_ERROR_INVALID_PARAM; r = usbi_cond_timedwait(&ctx->event_waiters_cond, &ctx->event_waiters_lock, tv); if (r < 0) return r == LIBUSB_ERROR_TIMEOUT; return 0; } static void handle_timeout(struct usbi_transfer *itransfer) { struct libusb_transfer *transfer = USBI_TRANSFER_TO_LIBUSB_TRANSFER(itransfer); int r; itransfer->timeout_flags |= USBI_TRANSFER_TIMEOUT_HANDLED; r = libusb_cancel_transfer(transfer); if (r == LIBUSB_SUCCESS) itransfer->timeout_flags |= USBI_TRANSFER_TIMED_OUT; else usbi_warn(TRANSFER_CTX(transfer), "async cancel failed %d", r); } static void handle_timeouts_locked(struct libusb_context *ctx) { struct timespec systime; struct usbi_transfer *itransfer; if (list_empty(&ctx->flying_transfers)) return; /* get current time */ usbi_get_monotonic_time(&systime); /* iterate through flying transfers list, finding all transfers that * have expired timeouts */ for_each_transfer(ctx, itransfer) { struct timespec *cur_ts = &itransfer->timeout; /* if we've reached transfers of infinite timeout, we're all done */ if (!TIMESPEC_IS_SET(cur_ts)) return; /* ignore timeouts we've already handled */ if (itransfer->timeout_flags & (USBI_TRANSFER_TIMEOUT_HANDLED | USBI_TRANSFER_OS_HANDLES_TIMEOUT)) continue; /* if transfer has non-expired timeout, nothing more to do */ if (TIMESPEC_CMP(cur_ts, &systime, >)) return; /* otherwise, we've got an expired timeout to handle */ handle_timeout(itransfer); } } static void handle_timeouts(struct libusb_context *ctx) { ctx = usbi_get_context(ctx); usbi_mutex_lock(&ctx->flying_transfers_lock); handle_timeouts_locked(ctx); usbi_mutex_unlock(&ctx->flying_transfers_lock); } static int handle_event_trigger(struct libusb_context *ctx) { struct list_head hotplug_msgs; int hotplug_event = 0; int r = 0; usbi_dbg(ctx, "event triggered"); list_init(&hotplug_msgs); /* take the the event data lock while processing events */ usbi_mutex_lock(&ctx->event_data_lock); /* check if someone modified the event sources */ if (ctx->event_flags & USBI_EVENT_EVENT_SOURCES_MODIFIED) usbi_dbg(ctx, "someone updated the event sources"); if (ctx->event_flags & USBI_EVENT_USER_INTERRUPT) { usbi_dbg(ctx, "someone purposefully interrupted"); ctx->event_flags &= ~USBI_EVENT_USER_INTERRUPT; } if (ctx->event_flags & USBI_EVENT_HOTPLUG_CB_DEREGISTERED) { usbi_dbg(ctx, "someone unregistered a hotplug cb"); ctx->event_flags &= ~USBI_EVENT_HOTPLUG_CB_DEREGISTERED; hotplug_event = 1; } /* check if someone is closing a device */ if (ctx->event_flags & USBI_EVENT_DEVICE_CLOSE) usbi_dbg(ctx, "someone is closing a device"); /* check for any pending hotplug messages */ if (ctx->event_flags & USBI_EVENT_HOTPLUG_MSG_PENDING) { usbi_dbg(ctx, "hotplug message received"); ctx->event_flags &= ~USBI_EVENT_HOTPLUG_MSG_PENDING; hotplug_event = 1; assert(!list_empty(&ctx->hotplug_msgs)); list_cut(&hotplug_msgs, &ctx->hotplug_msgs); } /* complete any pending transfers */ if (ctx->event_flags & USBI_EVENT_TRANSFER_COMPLETED) { struct usbi_transfer *itransfer, *tmp; struct list_head completed_transfers; assert(!list_empty(&ctx->completed_transfers)); list_cut(&completed_transfers, &ctx->completed_transfers); usbi_mutex_unlock(&ctx->event_data_lock); __for_each_completed_transfer_safe(&completed_transfers, itransfer, tmp) { list_del(&itransfer->completed_list); r = usbi_backend.handle_transfer_completion(itransfer); if (r) { usbi_err(ctx, "backend handle_transfer_completion failed with error %d", r); break; } } usbi_mutex_lock(&ctx->event_data_lock); if (!list_empty(&completed_transfers)) { /* an error occurred, put the remaining transfers back on the list */ list_splice_front(&completed_transfers, &ctx->completed_transfers); } else if (list_empty(&ctx->completed_transfers)) { ctx->event_flags &= ~USBI_EVENT_TRANSFER_COMPLETED; } } /* if no further pending events, clear the event */ if (!ctx->event_flags) usbi_clear_event(&ctx->event); usbi_mutex_unlock(&ctx->event_data_lock); /* process the hotplug events, if any */ if (hotplug_event) usbi_hotplug_process(ctx, &hotplug_msgs); return r; } #ifdef HAVE_OS_TIMER static int handle_timer_trigger(struct libusb_context *ctx) { int r; usbi_mutex_lock(&ctx->flying_transfers_lock); /* process the timeout that just happened */ handle_timeouts_locked(ctx); /* arm for next timeout */ r = arm_timer_for_next_timeout(ctx); usbi_mutex_unlock(&ctx->flying_transfers_lock); return r; } #endif /* do the actual event handling. assumes that no other thread is concurrently * doing the same thing. */ static int handle_events(struct libusb_context *ctx, struct timeval *tv) { struct usbi_reported_events reported_events; int r, timeout_ms; /* prevent attempts to recursively handle events (e.g. calling into * libusb_handle_events() from within a hotplug or transfer callback) */ if (usbi_handling_events(ctx)) return LIBUSB_ERROR_BUSY; /* only reallocate the event source data when the list of event sources has * been modified since the last handle_events(), otherwise reuse them to * save the additional overhead */ usbi_mutex_lock(&ctx->event_data_lock); if (ctx->event_flags & USBI_EVENT_EVENT_SOURCES_MODIFIED) { usbi_dbg(ctx, "event sources modified, reallocating event data"); /* free anything removed since we last ran */ cleanup_removed_event_sources(ctx); r = usbi_alloc_event_data(ctx); if (r) { usbi_mutex_unlock(&ctx->event_data_lock); return r; } /* reset the flag now that we have the updated list */ ctx->event_flags &= ~USBI_EVENT_EVENT_SOURCES_MODIFIED; /* if no further pending events, clear the event so that we do * not immediately return from the wait function */ if (!ctx->event_flags) usbi_clear_event(&ctx->event); } usbi_mutex_unlock(&ctx->event_data_lock); timeout_ms = (int)(tv->tv_sec * 1000) + (tv->tv_usec / 1000); /* round up to next millisecond */ if (tv->tv_usec % 1000) timeout_ms++; reported_events.event_bits = 0; usbi_start_event_handling(ctx); r = usbi_wait_for_events(ctx, &reported_events, timeout_ms); if (r != LIBUSB_SUCCESS) { if (r == LIBUSB_ERROR_TIMEOUT) { handle_timeouts(ctx); r = LIBUSB_SUCCESS; } goto done; } if (reported_events.event_triggered) { r = handle_event_trigger(ctx); if (r) { /* return error code */ goto done; } } #ifdef HAVE_OS_TIMER if (reported_events.timer_triggered) { r = handle_timer_trigger(ctx); if (r) { /* return error code */ goto done; } } #endif if (!reported_events.num_ready) goto done; r = usbi_backend.handle_events(ctx, reported_events.event_data, reported_events.event_data_count, reported_events.num_ready); if (r) usbi_err(ctx, "backend handle_events failed with error %d", r); done: usbi_end_event_handling(ctx); return r; } /* returns the smallest of: * 1. timeout of next URB * 2. user-supplied timeout * returns 1 if there is an already-expired timeout, otherwise returns 0 * and populates out */ static int get_next_timeout(libusb_context *ctx, struct timeval *tv, struct timeval *out) { struct timeval timeout; int r = libusb_get_next_timeout(ctx, &timeout); if (r) { /* timeout already expired? */ if (!timerisset(&timeout)) return 1; /* choose the smallest of next URB timeout or user specified timeout */ if (timercmp(&timeout, tv, <)) *out = timeout; else *out = *tv; } else { *out = *tv; } return 0; } /** \ingroup libusb_poll * Handle any pending events. * * libusb determines "pending events" by checking if any timeouts have expired * and by checking the set of file descriptors for activity. * * If a zero timeval is passed, this function will handle any already-pending * events and then immediately return in non-blocking style. * * If a non-zero timeval is passed and no events are currently pending, this * function will block waiting for events to handle up until the specified * timeout. If an event arrives or a signal is raised, this function will * return early. * * If the parameter completed is not NULL then after obtaining the event * handling lock this function will return immediately if the integer * pointed to is not 0. This allows for race free waiting for the completion * of a specific transfer. * * \param ctx the context to operate on, or NULL for the default context * \param tv the maximum time to block waiting for events, or an all zero * timeval struct for non-blocking mode * \param completed pointer to completion integer to check, or NULL * \returns 0 on success * \returns LIBUSB_ERROR_INVALID_PARAM if timeval is invalid * \returns another LIBUSB_ERROR code on other failure * \ref libusb_mtasync */ int API_EXPORTED libusb_handle_events_timeout_completed(libusb_context *ctx, struct timeval *tv, int *completed) { int r; struct timeval poll_timeout; if (!TIMEVAL_IS_VALID(tv)) return LIBUSB_ERROR_INVALID_PARAM; ctx = usbi_get_context(ctx); r = get_next_timeout(ctx, tv, &poll_timeout); if (r) { /* timeout already expired */ handle_timeouts(ctx); return 0; } retry: if (libusb_try_lock_events(ctx) == 0) { if (completed == NULL || !*completed) { /* we obtained the event lock: do our own event handling */ usbi_dbg(ctx, "doing our own event handling"); r = handle_events(ctx, &poll_timeout); } libusb_unlock_events(ctx); return r; } /* another thread is doing event handling. wait for thread events that * notify event completion. */ libusb_lock_event_waiters(ctx); if (completed && *completed) goto already_done; if (!libusb_event_handler_active(ctx)) { /* we hit a race: whoever was event handling earlier finished in the * time it took us to reach this point. try the cycle again. */ libusb_unlock_event_waiters(ctx); usbi_dbg(ctx, "event handler was active but went away, retrying"); goto retry; } usbi_dbg(ctx, "another thread is doing event handling"); r = libusb_wait_for_event(ctx, &poll_timeout); already_done: libusb_unlock_event_waiters(ctx); if (r < 0) return r; else if (r == 1) handle_timeouts(ctx); return 0; } /** \ingroup libusb_poll * Handle any pending events * * Like libusb_handle_events_timeout_completed(), but without the completed * parameter, calling this function is equivalent to calling * libusb_handle_events_timeout_completed() with a NULL completed parameter. * * This function is kept primarily for backwards compatibility. * All new code should call libusb_handle_events_completed() or * libusb_handle_events_timeout_completed() to avoid race conditions. * * \param ctx the context to operate on, or NULL for the default context * \param tv the maximum time to block waiting for events, or an all zero * timeval struct for non-blocking mode * \returns 0 on success, or a LIBUSB_ERROR code on failure */ int API_EXPORTED libusb_handle_events_timeout(libusb_context *ctx, struct timeval *tv) { return libusb_handle_events_timeout_completed(ctx, tv, NULL); } /** \ingroup libusb_poll * Handle any pending events in blocking mode. There is currently a timeout * hard-coded at 60 seconds but we plan to make it unlimited in future. For * finer control over whether this function is blocking or non-blocking, or * for control over the timeout, use libusb_handle_events_timeout_completed() * instead. * * This function is kept primarily for backwards compatibility. * All new code should call libusb_handle_events_completed() or * libusb_handle_events_timeout_completed() to avoid race conditions. * * \param ctx the context to operate on, or NULL for the default context * \returns 0 on success, or a LIBUSB_ERROR code on failure */ int API_EXPORTED libusb_handle_events(libusb_context *ctx) { struct timeval tv; tv.tv_sec = 60; tv.tv_usec = 0; return libusb_handle_events_timeout_completed(ctx, &tv, NULL); } /** \ingroup libusb_poll * Handle any pending events in blocking mode. * * Like libusb_handle_events(), with the addition of a completed parameter * to allow for race free waiting for the completion of a specific transfer. * * See libusb_handle_events_timeout_completed() for details on the completed * parameter. * * \param ctx the context to operate on, or NULL for the default context * \param completed pointer to completion integer to check, or NULL * \returns 0 on success, or a LIBUSB_ERROR code on failure * \ref libusb_mtasync */ int API_EXPORTED libusb_handle_events_completed(libusb_context *ctx, int *completed) { struct timeval tv; tv.tv_sec = 60; tv.tv_usec = 0; return libusb_handle_events_timeout_completed(ctx, &tv, completed); } /** \ingroup libusb_poll * Handle any pending events by polling file descriptors, without checking if * any other threads are already doing so. Must be called with the event lock * held, see libusb_lock_events(). * * This function is designed to be called under the situation where you have * taken the event lock and are calling poll()/select() directly on libusb's * file descriptors (as opposed to using libusb_handle_events() or similar). * You detect events on libusb's descriptors, so you then call this function * with a zero timeout value (while still holding the event lock). * * \param ctx the context to operate on, or NULL for the default context * \param tv the maximum time to block waiting for events, or zero for * non-blocking mode * \returns 0 on success * \returns LIBUSB_ERROR_INVALID_PARAM if timeval is invalid * \returns another LIBUSB_ERROR code on other failure * \ref libusb_mtasync */ int API_EXPORTED libusb_handle_events_locked(libusb_context *ctx, struct timeval *tv) { int r; struct timeval poll_timeout; if (!TIMEVAL_IS_VALID(tv)) return LIBUSB_ERROR_INVALID_PARAM; ctx = usbi_get_context(ctx); r = get_next_timeout(ctx, tv, &poll_timeout); if (r) { /* timeout already expired */ handle_timeouts(ctx); return 0; } return handle_events(ctx, &poll_timeout); } /** \ingroup libusb_poll * Determines whether your application must apply special timing considerations * when monitoring libusb's file descriptors. * * This function is only useful for applications which retrieve and poll * libusb's file descriptors in their own main loop (\ref libusb_pollmain). * * Ordinarily, libusb's event handler needs to be called into at specific * moments in time (in addition to times when there is activity on the file * descriptor set). The usual approach is to use libusb_get_next_timeout() * to learn about when the next timeout occurs, and to adjust your * poll()/select() timeout accordingly so that you can make a call into the * library at that time. * * Some platforms supported by libusb do not come with this baggage - any * events relevant to timing will be represented by activity on the file * descriptor set, and libusb_get_next_timeout() will always return 0. * This function allows you to detect whether you are running on such a * platform. * * Since v1.0.5. * * \param ctx the context to operate on, or NULL for the default context * \returns 0 if you must call into libusb at times determined by * libusb_get_next_timeout(), or 1 if all timeout events are handled internally * or through regular activity on the file descriptors. * \ref libusb_pollmain "Polling libusb file descriptors for event handling" */ int API_EXPORTED libusb_pollfds_handle_timeouts(libusb_context *ctx) { ctx = usbi_get_context(ctx); return usbi_using_timer(ctx); } /** \ingroup libusb_poll * Determine the next internal timeout that libusb needs to handle. You only * need to use this function if you are calling poll() or select() or similar * on libusb's file descriptors yourself - you do not need to use it if you * are calling libusb_handle_events() or a variant directly. * * You should call this function in your main loop in order to determine how * long to wait for select() or poll() to return results. libusb needs to be * called into at this timeout, so you should use it as an upper bound on * your select() or poll() call. * * When the timeout has expired, call into libusb_handle_events_timeout() * (perhaps in non-blocking mode) so that libusb can handle the timeout. * * This function may return 1 (success) and an all-zero timeval. If this is * the case, it indicates that libusb has a timeout that has already expired * so you should call libusb_handle_events_timeout() or similar immediately. * A return code of 0 indicates that there are no pending timeouts. * * On some platforms, this function will always returns 0 (no pending * timeouts). See \ref polltime. * * \param ctx the context to operate on, or NULL for the default context * \param tv output location for a relative time against the current * clock in which libusb must be called into in order to process timeout events * \returns 0 if there are no pending timeouts, 1 if a timeout was returned, * or LIBUSB_ERROR_OTHER on failure */ int API_EXPORTED libusb_get_next_timeout(libusb_context *ctx, struct timeval *tv) { struct usbi_transfer *itransfer; struct timespec systime; struct timespec next_timeout = { 0, 0 }; ctx = usbi_get_context(ctx); if (usbi_using_timer(ctx)) return 0; usbi_mutex_lock(&ctx->flying_transfers_lock); if (list_empty(&ctx->flying_transfers)) { usbi_mutex_unlock(&ctx->flying_transfers_lock); usbi_dbg(ctx, "no URBs, no timeout!"); return 0; } /* find next transfer which hasn't already been processed as timed out */ for_each_transfer(ctx, itransfer) { if (itransfer->timeout_flags & (USBI_TRANSFER_TIMEOUT_HANDLED | USBI_TRANSFER_OS_HANDLES_TIMEOUT)) continue; /* if we've reached transfers of infinite timeout, we're done looking */ if (!TIMESPEC_IS_SET(&itransfer->timeout)) break; next_timeout = itransfer->timeout; break; } usbi_mutex_unlock(&ctx->flying_transfers_lock); if (!TIMESPEC_IS_SET(&next_timeout)) { usbi_dbg(ctx, "no URB with timeout or all handled by OS; no timeout!"); return 0; } usbi_get_monotonic_time(&systime); if (!TIMESPEC_CMP(&systime, &next_timeout, <)) { usbi_dbg(ctx, "first timeout already expired"); timerclear(tv); } else { TIMESPEC_SUB(&next_timeout, &systime, &next_timeout); TIMESPEC_TO_TIMEVAL(tv, &next_timeout); usbi_dbg(ctx, "next timeout in %ld.%06lds", (long)tv->tv_sec, (long)tv->tv_usec); } return 1; } /** \ingroup libusb_poll * Register notification functions for file descriptor additions/removals. * These functions will be invoked for every new or removed file descriptor * that libusb uses as an event source. * * To remove notifiers, pass NULL values for the function pointers. * * Note that file descriptors may have been added even before you register * these notifiers (e.g. at libusb_init() time). * * Additionally, note that the removal notifier may be called during * libusb_exit() (e.g. when it is closing file descriptors that were opened * and added to the poll set at libusb_init() time). If you don't want this, * remove the notifiers immediately before calling libusb_exit(). * * \param ctx the context to operate on, or NULL for the default context * \param added_cb pointer to function for addition notifications * \param removed_cb pointer to function for removal notifications * \param user_data User data to be passed back to callbacks (useful for * passing context information) */ void API_EXPORTED libusb_set_pollfd_notifiers(libusb_context *ctx, libusb_pollfd_added_cb added_cb, libusb_pollfd_removed_cb removed_cb, void *user_data) { #if !defined(PLATFORM_WINDOWS) ctx = usbi_get_context(ctx); ctx->fd_added_cb = added_cb; ctx->fd_removed_cb = removed_cb; ctx->fd_cb_user_data = user_data; #else usbi_err(ctx, "external polling of libusb's internal event sources " \ "is not yet supported on Windows"); UNUSED(added_cb); UNUSED(removed_cb); UNUSED(user_data); #endif } /* * Interrupt the iteration of the event handling thread, so that it picks * up the event source change. Callers of this function must hold the event_data_lock. */ static void usbi_event_source_notification(struct libusb_context *ctx) { unsigned int event_flags; /* Record that there is a new poll fd. * Only signal an event if there are no prior pending events. */ event_flags = ctx->event_flags; ctx->event_flags |= USBI_EVENT_EVENT_SOURCES_MODIFIED; if (!event_flags) usbi_signal_event(&ctx->event); } /* Add an event source to the list of event sources to be monitored. * poll_events should be specified as a bitmask of events passed to poll(), e.g. * POLLIN and/or POLLOUT. */ int usbi_add_event_source(struct libusb_context *ctx, usbi_os_handle_t os_handle, short poll_events) { struct usbi_event_source *ievent_source = malloc(sizeof(*ievent_source)); if (!ievent_source) return LIBUSB_ERROR_NO_MEM; usbi_dbg(ctx, "add " USBI_OS_HANDLE_FORMAT_STRING " events %d", os_handle, poll_events); ievent_source->data.os_handle = os_handle; ievent_source->data.poll_events = poll_events; usbi_mutex_lock(&ctx->event_data_lock); list_add_tail(&ievent_source->list, &ctx->event_sources); usbi_event_source_notification(ctx); usbi_mutex_unlock(&ctx->event_data_lock); #if !defined(PLATFORM_WINDOWS) if (ctx->fd_added_cb) ctx->fd_added_cb(os_handle, poll_events, ctx->fd_cb_user_data); #endif return 0; } /* Remove an event source from the list of event sources to be monitored. */ void usbi_remove_event_source(struct libusb_context *ctx, usbi_os_handle_t os_handle) { struct usbi_event_source *ievent_source; int found = 0; usbi_dbg(ctx, "remove " USBI_OS_HANDLE_FORMAT_STRING, os_handle); usbi_mutex_lock(&ctx->event_data_lock); for_each_event_source(ctx, ievent_source) { if (ievent_source->data.os_handle == os_handle) { found = 1; break; } } if (!found) { usbi_dbg(ctx, "couldn't find " USBI_OS_HANDLE_FORMAT_STRING " to remove", os_handle); usbi_mutex_unlock(&ctx->event_data_lock); return; } list_del(&ievent_source->list); list_add_tail(&ievent_source->list, &ctx->removed_event_sources); usbi_event_source_notification(ctx); usbi_mutex_unlock(&ctx->event_data_lock); #if !defined(PLATFORM_WINDOWS) if (ctx->fd_removed_cb) ctx->fd_removed_cb(os_handle, ctx->fd_cb_user_data); #endif } /** \ingroup libusb_poll * Retrieve a list of file descriptors that should be polled by your main loop * as libusb event sources. * * The returned list is NULL-terminated and should be freed with libusb_free_pollfds() * when done. The actual list contents must not be touched. * * As file descriptors are a Unix-specific concept, this function is not * available on Windows and will always return NULL. * * \param ctx the context to operate on, or NULL for the default context * \returns a NULL-terminated list of libusb_pollfd structures * \returns NULL on error * \returns NULL on platforms where the functionality is not available */ DEFAULT_VISIBILITY const struct libusb_pollfd ** LIBUSB_CALL libusb_get_pollfds( libusb_context *ctx) { #if !defined(PLATFORM_WINDOWS) struct libusb_pollfd **ret = NULL; struct usbi_event_source *ievent_source; size_t i; static_assert(sizeof(struct usbi_event_source_data) == sizeof(struct libusb_pollfd), "mismatch between usbi_event_source_data and libusb_pollfd sizes"); ctx = usbi_get_context(ctx); usbi_mutex_lock(&ctx->event_data_lock); i = 0; for_each_event_source(ctx, ievent_source) i++; ret = calloc(i + 1, sizeof(struct libusb_pollfd *)); if (!ret) goto out; i = 0; for_each_event_source(ctx, ievent_source) ret[i++] = (struct libusb_pollfd *)ievent_source; out: usbi_mutex_unlock(&ctx->event_data_lock); return (const struct libusb_pollfd **)ret; #else usbi_err(ctx, "external polling of libusb's internal event sources " \ "is not yet supported on Windows"); return NULL; #endif } /** \ingroup libusb_poll * Free a list of libusb_pollfd structures. This should be called for all * pollfd lists allocated with libusb_get_pollfds(). * * Since version 1.0.20, \ref LIBUSB_API_VERSION >= 0x01000104 * * It is legal to call this function with a NULL pollfd list. In this case, * the function will simply do nothing. * * \param pollfds the list of libusb_pollfd structures to free */ void API_EXPORTED libusb_free_pollfds(const struct libusb_pollfd **pollfds) { #if !defined(PLATFORM_WINDOWS) free((void *)pollfds); #else UNUSED(pollfds); #endif } /* Backends may call this from handle_events to report disconnection of a * device. This function ensures transfers get cancelled appropriately. * Callers of this function must hold the events_lock. */ void usbi_handle_disconnect(struct libusb_device_handle *dev_handle) { struct libusb_context *ctx = HANDLE_CTX(dev_handle); struct usbi_transfer *cur; struct usbi_transfer *to_cancel; usbi_dbg(ctx, "device %d.%d", dev_handle->dev->bus_number, dev_handle->dev->device_address); /* terminate all pending transfers with the LIBUSB_TRANSFER_NO_DEVICE * status code. * * when we find a transfer for this device on the list, there are two * possible scenarios: * 1. the transfer is currently in-flight, in which case we terminate the * transfer here * 2. the transfer has been added to the flying transfer list by * libusb_submit_transfer, has failed to submit and * libusb_submit_transfer is waiting for us to release the * flying_transfers_lock to remove it, so we ignore it */ while (1) { to_cancel = NULL; usbi_mutex_lock(&ctx->flying_transfers_lock); for_each_transfer(ctx, cur) { if (USBI_TRANSFER_TO_LIBUSB_TRANSFER(cur)->dev_handle == dev_handle) { usbi_mutex_lock(&cur->lock); if (cur->state_flags & USBI_TRANSFER_IN_FLIGHT) to_cancel = cur; usbi_mutex_unlock(&cur->lock); if (to_cancel) break; } } usbi_mutex_unlock(&ctx->flying_transfers_lock); if (!to_cancel) break; usbi_dbg(ctx, "cancelling transfer %p from disconnect", USBI_TRANSFER_TO_LIBUSB_TRANSFER(to_cancel)); usbi_mutex_lock(&to_cancel->lock); usbi_backend.clear_transfer_priv(to_cancel); usbi_mutex_unlock(&to_cancel->lock); usbi_handle_transfer_completion(to_cancel, LIBUSB_TRANSFER_NO_DEVICE); } } |
# Interesting thing about archers hitting the target
by Дьявол
Tags: archers, probability
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P: 365 Hello, again! I got one very interesting question. We got three archers, and the probability of the ones to hit the target is: A1, A2, A3. What if the task is to find the probability that the target will be hit at least from one archer. So at least one archer to hit the target. Is it P(A1 U A2 U A3) = A1 + A2 +A3 ? Or $$(1- P(\bar{A_{1}} \cap \bar{A_{2}} \cap \bar{A_{3}})) = 1 - \bar{A_{1}}* \bar{A_{2}} * \bar{A_{3}}$$, where $$\bar{A}$$ is opposite of A? Or maybe, both are valid? Thanks in advance.
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,345
Quote by Дьявол Hello, again! I got one very interesting question. We got three archers, and the probability of the ones to hit the target is: A1, A2, A3. What if the task is to find the probability that the target will be hit at least from one archer. So at least one archer to hit the target. Is it P(A1 U A2 U A3) = A1 + A2 +A3 ?
No. This is wrong. $P(A_1\cup A_2)= P(A_1)+ P(A_2)- P(A_1\cap A_2)$ and that extends to 3 events: $P(A_1\cup A_2\cup A_3)= P(A_1)+ P(A_2)+ P(A_3)- P(A_1\cap A_2)- P(A_1\cap A_3)$$- P(A_2\cap A_3)+ P(A_1\cap A_2\cap A_3)$.
Or $$(1- P(\bar{A_{1}} \cap \bar{A_{2}} \cap \bar{A_{3}})) = 1 - \bar{A_{1}}* \bar{A_{2}} * \bar{A_{3}}$$, where $$\bar{A}$$ is opposite of A?
Yes, this is correct.
Or maybe, both are valid? Thanks in advance.
P: 365 Thanks for the post, HallsofIvy. In this case, the shootings of the archers are independent cases. So that's why we do not need $P(A_1\cup A_2\cup A_3)= P(A_1)+ P(A_2)+ P(A_3)- P(A_1\cap A_2)- P(A_1\cap A_3)- P(A_2\cap A_3)+ P(A_1\cap A_2\cap A_3)$ since $P(A_1\cap A_2)- P(A_1\cap A_3)- P(A_2\cap A_3)+ P(A_1\cap A_2\cap A_3)=0-0-0-0=0$. That's why I said independent cases. But what if P(A1)=0.8, P(A2)=0.9, P(A3)=0.75 In that case the sum $P(A_1\cup A_2\cup A_3)= P(A_1)+ P(A_2)+ P(A_3)=0.8+0.9+0.75=2.45$ This is strange.
P: 330 Interesting thing about archers hitting the target The fact that two cases A and B are independent doesn't mean P(A n B) = 0. It means that P(A n B) = P(A)P(B).
P: 365 If two cases are independent, that means that they do not have something in common, right? So A n B = 0, right?
HW Helper
P: 1,361
Quote by Дьявол If two cases are independent, that means that they do not have something in common, right? So A n B = 0, right?
No,
$$A \cap B = \emptyset$$
if the two sets are disjoint , which is not the same as independent.
P: 38
If two cases are independent, that means that they do not have something in common, right? So A n B = 0, right?
Actually, that would mean that they're not independent. Two events are independent if the occurrence of one does not influence the occurrence of the other, i.e. P(A|B) = P(A). If $$A \cap B = \emptyset$$, then P(A|B) = 0, so the events are not independent.
With regards to the original question, it would be easiest to take the probability that every archer misses and subtract it from one.
P: 365 Thanks for the replies. @Tibarn, in this case the occurrence of one does not influence the occurrence of the other. $$P(A/B)=\frac{m_{A\cap B} }{ m_{B}}$$ out of there $$P(A/B)=\frac{\frac{m_{A\cap B}}{n}}{\frac{m_{B}}{n}}$$ and $$P(A/B)=\frac{P(A\cap B)}{ P(B)}$$ The cases are independent if the occurrence of one does not influence the occurrence of the other. So, if two cases are independent, then P(A/B)=P(A). Out of there P(A n B)=P(A)*P(B) Now, let's get back to the task. I think I mean, disjoint. So cases A,B are disjoint. P(A U B)=P(A) + P(B) - P(A n B) In this case $m_{A\cap B}=0$ because the cases: I - the 1st archer will hit the target II - the 2nd archer will hit the target DO NOT have something in common. $$P(A U B) = \frac{m_A+m_B-m_{A\cap B}}{n}$$ $$P(A U B) = \frac{m_A+m_B-0}{n}=P(A)+P(B)$$ Is this true? Are those cases disjoint??
Sci Advisor
HW Helper
P: 3,684
Quote by Дьявол Now, let's get back to the task. I think I mean, disjoint. So cases A,B are disjoint. P(A U B)=P(A) + P(B) - P(A n B) In this case $m_{A\cap B}=0$ because the cases: I - the 1st archer will hit the target II - the 2nd archer will hit the target DO NOT have something in common. $$P(A U B) = \frac{m_A+m_B-m_{A\cap B}}{n}$$ $$P(A U B) = \frac{m_A+m_B-0}{n}=P(A)+P(B)$$ Is this true? Are those cases disjoint??
You're saying that if one archer hits the target the other one always misses?
P: 365 Are those cases disjoint?
P: 38
In this case $$m_{A \cap B}=0$$ because the cases: I - the 1st archer will hit the target II - the 2nd archer will hit the target DO NOT have something in common.
Think about this intuitively. If A and B are disjoint, then event A and event B cannot occur simultaneously. So, if A is the first archer hitting the target and B is the second archer hitting the target, $$P(A \cap B) = 0$$ means that both archers cannot simultaneously hit the target. If we know that A hit the target, then it would follow that B missed the target, so the events are not independent (unless A or B always misses).
If both archers take one shot, then we have four possible events:
1. Both miss
2. Archer A hits, B misses
3. Archer A misses, B hits
4. Both hit.
In this case, $$A \cap B$$ is case 4, where both archers hit. If you're going to do probability by cases, it's important that you get all of them.
P: 365 Now, I understood. Thank you very much for the help. Regards.
Math Emeritus Sci Advisor Thanks PF Gold P: 39,345 You may be confusing "mutually exclusive" with "independent". "Independent" means what happens in one case does not affect what happens in the other- $P(A\cap B)= P(A)P(B)$ or, equivalently, P(A|B)= P(A). "Mutually exclusive" means $P(A\cap B)= 0$ so that P(A|B)= 0. Not at all the same thing!
P: 365 Yes, you're right. I did a little research, and find out that the events in this case aren't exclusive, but they are independent. Because if they are exclusive P(A) or P(B) will be equal 0 so that P(A n B)=0.
Related Discussions Academic Guidance 10 General Discussion 3 General Physics 3 Introductory Physics Homework 12 Introductory Physics Homework 1 |
# Formulas For Pipes & Cisterns
## Concepts of Pipes And Cisterns
As we all know that work and time is considered as one of the major topics of quantitative section for any competitive exam. Similarly, Pipes and Cisterns is major part of work and time. On this page we will see some of the major Formulas For Pipes and Cisterns.
• Pipe : There are usually two kinds of Pipes
• Inlet : Inlet pipe is the pipe that fills the cistern or Tank.
• Outlet : Outlet Pipe is the pipe that empties the cistern or tank.
• Cisterns : They are large tanks that store rainwater collected from impervious surfaces for domestic uses or for consumption.
### Definition of Pipes & Cisterns
• A pipe is connected to a tank or cistern to fill or empty the tank or cistern
• Inlet: A pipe which is connected to fill a tank is known as an inlet.
• Outlet: A pipe which is connected to empty a tank is known as an outlet.
• In pipes and cisterns problems – we need to find out what portion of the tank each of the pipes fill or drain in unit time (say in a minute or hour or second) and then perform arithmetic operation on this value.
### Formulas for Pipes and Cisterns
1. If pipe can fill a tank in x hours , then part filled in one hour = $\frac{1}{x}$
2. If pipe can empty a tank in y hours , then part emptied in one hour = $\frac{1}{y}$
3. If pipe A can fill a tank in x hours, Pipe B can empty the full tank in y hours (where y > x). Then, on opening both the pipes, the net part filled in one hour= $\frac{1}{x} – \frac{1}{y}$ OR $\frac{xy}{y-x}$ hours
4. If pipe A can fill a tank in x hours. Pipe B can empty the full tank in y hours (where x > y). Then, on opening both the pipes, the net part filled in one hour= $\frac{1}{y} – \frac{1}{x}$ OR $\frac{yx}{x-y}$ hours.
5. If pipe A can empty a tank in X hours. Pipe B can empty the same tank in Y hours. Then part of the tank emptied in one hour when both the pipes start working together = $\frac{1}{x} + \frac{1}{y}$
Tips and Tricks and Shortcuts for Pipes and Cisterns
Solve Questions on Pipes and Cisterns |
How to detect outliers from either (a) SNP-Fst or (b) Window-Fst distributions?
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2.7 years ago
serpalma.v ▴ 70
Hello
I want to find the SNPs that could be responsible for the phenotype differences observed between three populations. For that I computed Fst (weir and cockerham) using vcftools.
One population reflects the founder population (line0) from which the two populations were selected (line1 and line2), each one for a different trait. The phenotypes for each line are highly divergent.
Computing per-SNP Fst produces the following representative .
Computing windowed (window = 500kb; slide = 250kb; min #SNPs=20) Fst produces the following representative .
First, line1 vs line2 yields a different Fst distribution compared to (line1 | line2) vs line0.
Second, window Fst calculation (mean) yields smoother distributions.
I would like to seek advise on the following:
(1) how to define outliers considering the two types of observed Fst distributions?
(2) Is windowed Fst more suitable to identify outliers?
(3) How to define the size and step of a sliding window? (what I choose for this example is based on a similar study, but I guess it might require optimization)
(4) Do I need to do some type of SNP pruning (these SNPs are derived from WGS variant discovery analysis following GATK best practices)?
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• Rajesh Raut
Articles written in Proceedings – Mathematical Sciences
• On the Extrema of Dirichlet's First Eigenvalue of a Family of Punctured Regular Polygons in Two Dimensional Space Forms
Let $\wp 1,\wp 0$ be two regular polygons of 𝑛 sides in a space form $M^2(\kappa)$ of constant curvature $\kappa=0,1$ or $-1$ such that $\wp 0\subset\wp 1$ and having the same center of mass. Suppose $\wp 0$ is circumscribed by a circle 𝐶 contained in $\wp 1$. We fix $\wp 1$ and vary $\wp 0$ by rotating it in 𝐶 about its center of mass. Put $\Omega =(\wp 1\backslash\wp 0)^0$, the interior of $\wp 1\backslash\wp 0$ in $M^2(\kappa)$. It is shown that the first Dirichlet’s eigenvalue $\lambda 1(\Omega)$ attains extremum when the axes of symmetry of $\wp 0$ coincide with those of $\wp 1$.
• # Proceedings – Mathematical Sciences
Volume 130, 2020
All articles
Continuous Article Publishing mode
• # Editorial Note on Continuous Article Publication
Posted on July 25, 2019 |
Article Contents
Article Contents
# Regularity properties of a cubically convergent scheme for generalized equations
• We consider the perturbed generalized equation $v \in f(x) +G(x)$ where $v$ is a perturbation parameter, $f$ is a function acting from a Banach space $X$ to a Banach space $Y$ while $G: X \rightarrow Y$ is a set-valued mapping. We associate to this generalized equation the following iterative procedure:
$v \in f(x_n)+ \nabla f(x_n)(x_{n+1}-x_n) +\frac{1}{2}\nabla^2 f(x_n) (x_{n+1}-x_n)^2 +G(x_{n+1}).$ $\quad$ (*)
We investigate some stability properties of the method (*) and we study the behavior of the sequences that it generates, more precisely, we show that they inherit some regularity properties from the mapping $f+G$.
Mathematics Subject Classification: Primary: 49J53, 49J40, 90C48.
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What is the remainder r when x is divided by 8? x is a posit
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What is the remainder r when x is divided by 8? x is a posit [#permalink] 19 May 2010, 00:51
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What is the remainder r when x is divided by 8? x is a positive integer.
(1) x yields remainder of 9 when divided by 12
(2) r is a factor of 27
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Re: Remainder [#permalink] 19 May 2010, 01:56
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dimitri92 wrote:
What is the Remainder R when X is divided by 8? X is a positive integer.
(1) X yields 9 when divided by 12
(2) R is a factor of 27
$$x=8q+r$$, $$0\leq{r}<8$$ (remainder must be less than divisor). Question $$r=?$$.
(1) Think this statement should be: "x yields remainder of 9 when divided by 12" --> $$x=12p+9$$, $$x$$ can take following values: 9, 21, 33, 45, 57, ... This values divided by 8 can give remainder of 1 or 5. Two values. Not sufficient.
(2) $$rk=27$$, as $$0\leq{r}<8$$, then r can take only two values: 1 or 3. Two values. Not sufficient.
(1)+(2) Intersection of values from (1) and (2) is $$r=1$$. Sufficient.
Answer: C.
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Re: Remainder [#permalink] 19 May 2010, 02:32
hmm ..i missed this part of finding the remainder in 1 ...hope i dont repeat this mistake on GMAT
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Re: What is the remainder r when x is divided by 8? x is a posit [#permalink] 19 Feb 2014, 12:58
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Re: What is the remainder r when x is divided by 8? x is a posit [#permalink] 07 Sep 2014, 11:38
Bunuel wrote:
dimitri92 wrote:
What is the Remainder R when X is divided by 8? X is a positive integer.
(1) X yields 9 when divided by 12
(2) R is a factor of 27
$$x=8q+r$$, $$0\leq{r}<8$$ (remainder must be less than divisor). Question $$r=?$$.
(1) Think this statement should be: "x yields remainder of 9 when divided by 12" --> $$x=12p+9$$, $$x$$ can take following values: 9, 21, 33, 45, 57, ... This values divided by 8 can give remainder of 1 or 5. Two values. Not sufficient.
(2) $$rk=27$$, as $$0\leq{r}<8$$, then r can take only two values: 1 or 3. Two values. Not sufficient.
(1)+(2) Intersection of values from (1) and (2) is $$r=1$$. Sufficient.
Answer: C.
Hi Bunuel,
need ur help buddy..
Actually,when i came at the end of this problem, i had 3 and 9 as factors of 27..But,as it was given,the remainder is a factor of 27..so it only has to be 3,since 9 is greater than 8..i understand that when we divide9 by 8,we get 1 as remainder,but we are explicitly told that remainder is a factor of 27 and 1 is not a factor,so we are left with just 3..
please clear my understanding..thanks
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Re: What is the remainder r when x is divided by 8? x is a posit [#permalink] 07 Sep 2014, 11:48
Expert's post
vards wrote:
Bunuel wrote:
dimitri92 wrote:
What is the Remainder R when X is divided by 8? X is a positive integer.
(1) X yields 9 when divided by 12
(2) R is a factor of 27
$$x=8q+r$$, $$0\leq{r}<8$$ (remainder must be less than divisor). Question $$r=?$$.
(1) Think this statement should be: "x yields remainder of 9 when divided by 12" --> $$x=12p+9$$, $$x$$ can take following values: 9, 21, 33, 45, 57, ... This values divided by 8 can give remainder of 1 or 5. Two values. Not sufficient.
(2) $$rk=27$$, as $$0\leq{r}<8$$, then r can take only two values: 1 or 3. Two values. Not sufficient.
(1)+(2) Intersection of values from (1) and (2) is $$r=1$$. Sufficient.
Answer: C.
Hi Bunuel,
need ur help buddy..
Actually,when i came at the end of this problem, i had 3 and 9 as factors of 27..But,as it was given,the remainder is a factor of 27..so it only has to be 3,since 9 is greater than 8..i understand that when we divide9 by 8,we get 1 as remainder,but we are explicitly told that remainder is a factor of 27 and 1 is not a factor,so we are left with just 3..
please clear my understanding..thanks
1 is a factor of every integer.
Check for more here: divisibility-multiples-factors-tips-and-hints-174998.html
Hope it helps.
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Re: What is the remainder r when x is divided by 8? x is a posit [#permalink] 07 Sep 2014, 11:48
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### AuthorTopic: Monash University - Subject Reviews & Ratings (Read 422295 times) Tweet Share
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#### ninwa
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##### Re: Monash University - Subject Reviews & Ratings
« Reply #180 on: January 26, 2014, 03:30:00 pm »
+7
Subject Code/Name: LAW4160 - Negotiation and Mediation Law 406
Summer semester intensive: scheduled for 9am - 5pm Monday to Friday (1 week), but in reality we finished around 3pm every day.
Assessment:
• 30% reflective journal
• 10% attendance and participation
• 30% final negotiation roleplay with lecturer present (mark entirely depends on how you conducted the negotiation, the strategies/methods you used etc., the result you get is irrelevant)
• 30% simulated negotiations with classmates during classes without lecturer present (lecturer not present, mark entirely depends on the result you get for your assigned party in the negotiation, so it does depend on how stubborn/passive etc. your assigned negotiation partners are)
Recorded Lectures:
No. Negotiation is a practical thing anyway so if you don't plan to show up to class then don't bother taking the subject, you won't learn anything.
Past exams available:
No exam, you get the scenario/facts for your simulated negotiations in advance.
Textbook Recommendation:
Pre-reading: Fisher and Ury, Getting to Yes - Negotiating an Agreement Without Giving In - you can find this online for free if you Google it, and it's a really useful and very easy-to-read summary of a lot of the skills you'll be learning.
There is also a course book with relevant excerpts from a range of textbooks on negotiation/mediation, which you are expected to purchase. I would buy it because it's a handy resource not just for this subject but also for life in general.
Lecturer(s):
Tom Harber (summer and winter semesters)
Dr Sandy Caspi-Sable (semester 1 and 2)
Year & Semester of completion:
Summer semester B, 2014
Rating: 10 out of 5, shut up I'm admin and I say this rating is possible
Lecturer: Tom is a negotiation/mediation skills consultant. He used to be a lawyer for Allens, then went to Harvard to get an MBA which included studies in negotiation (basically he is incredibly smart). All of the skills and theories we learned came from Harvard; all our handouts were branded with the Harvard Business School logo etc., so (IMO) you would be learning some of the best skills out there. Tom is highly engaging, knowledgeable and funny. You'll have a good time with him.
I don't know what Sandy is like as a lecturer, but we did watch a video where she was mediating a negotiation and it was clear that she very much knew her stuff.
Subject: I've enjoyed subjects before, yes, but I never thought I'd actually have FUN at law school. But here we are and I will gladly eat my hat. This is a great subject, not just for conducting negotiations in future (which you will have to do in almost any industry you decide to go into, including legal - most legal disputes do not actually go to litigation), but also just generally in real life. Cashier won't let you return your item 1 day after the return period? Negotiation skills! Internet company being stubborn dicks again? Negotiation skills! Boyfriend won't see a girly movie with you? NEGOTIATION. SKILLS.
You will learn not only negotiation, but also basic game theory, economics, psychology and there's even a neurobiology bit about the brain and amygdala and emotions or something like that, which I'm sure would have been interesting if I had any idea what any of it meant.
Make sure to start applying for this subject at least 1 or 2 semesters before you graduate, because it is highly popular and you might not get in the first time. For example, there were 42 spots in my class and, according to Tom, 94 people on the waiting list. (NB: those graduating sooner will take precedence, then the rest will depend on what you write on your application, your transcript has nothing to do with whether you'll get in so don't worry about that.)
My only slight criticism is regarding the 30% worth of in-class simulated negotiations. Basically, you got a general summary of the facts, then a confidential summary of facts applicable to your side (e.g. what the party you are representing wants out of the negotiation, what they are willing to concede on etc.). The most desirable outcomes are assigned the most points and the least, 0 points. This points system is also confidential to you (so your opposing party can't exploit the points your party is willing to concede on, and vice versa).
I totally understand that time constraints make it impossible for the lecturer to sit through and assess every single group, and therefore it naturally has to be results-oriented. But I found the outcome also depended a lot on who you are assigned with. For example, in one negotiation the opposite party refused to budge on something which could have benefited BOTH of us (i.e. we BOTH would have gotten more points if he'd backed off, which was really bloody annoying when I found out later). In the 6-party negotiation, another party and I wasted like 15 minutes engaging in a battle of wills arguing over a point on which neither of us was willing to concede, which must have really pissed off the remaining 4 members who were very willing to compromise on everything else (lol my bad, in my defence my party was the veto-holding party and without this concession I wouldn't have been allowed to agree to any agreement).
Anyway, tl;dr highly recommend this subject it's great and you should apply for it ASAP.
« Last Edit: January 28, 2014, 01:05:01 am by alondouek »
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#### DisaFear
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##### Re: Monash University - Subject Reviews & Ratings
« Reply #181 on: March 01, 2014, 11:54:27 pm »
+7
Subject Code/Name: CHM2990 - Introductory chemical research project
• Depends on the project you choose really...say 7-10 hours of lab work per week
Assessment:
• Lab work: 30%
• Oral presentation: 10%
• Written report: 60%
Recorded Lectures: No lectures for this unit
Past exams available: No exams for this unit
Textbook Recommendation: No textbook for this unit
Lecturer(s): No lecturer for this unit. However, I worked under:
• Dr. Gregory Knowles (supervisor)
• Professor Alan Chaffee (group leader)
• Professor Douglas MacFarlane (group leader)
Year & Semester of completion: 2013/2014 Summer Semester
Rating: 6/5
This is going to be really long. So hold onto your horses.
This unit is about research. It allows undergraduates the chance to try out their hand at one of the research projects available from the Chemistry Honours Handbook. It is a great way to see what research is like, see if you'll want to do it in the future or not.
Basically, you go through the handbook. If there is a project you like, you contact the person associated with it and if they're interested in taking you on, then bam, you're a researcher part of their group working on stuff that no one has worked on before!
I had the privilege of being in a 'joint-project' between two groups. I worked on a project in collaboration between Professor Alan Chaffee's group and Professor Douglas MacFarlane's group.
Is it scary?
Oh yea. During the first 3-4 days when you start out, sure you'll be questioning if you are worthy, as a second/third year student, of undertaking 'real' research. But you are worthy! I was scared everything I touched would break, but I only managed to break one beaker over 2 months
The people are there to help. Everyone is so so so friendly. You will use equipment you've never used before, equipment that could take up whole rooms by themselves. But your supervisor will show you, and you will get the hang of it.
What did I work on?
My project involved the capture of carbon dioxide using mesoporous silicas infiltrated with amine-functional ionic liquids. Ionic liquids have the ability to form carbamates with carbon dioxide, meaning you can have carbon dioxide adsorb to the ionic liquid at certain conditions (temperature/pressure) and then have it desorb at other conditions, to safely store the carbon dioxide while regenerating the adsorbent. This was to be developed especially for carbon dioxide capture from post-combustion waste gas from places like coal-fired power generators.
There was a lot of paper work to fill out initially. All the usual safety stuff.
After all that was covered, straight to the synthesis. I had to make 18 samples; made them 3 at a time, each sample taking around 2-3 days to make. If nothing else, this unit will make you very confident in your lab work in future years. I have so much experience now, it feels great. Got to use vacuum ovens, rotary evaporation units, thermogravimetric analysis units, pycnometers and lots of other cool stuff. I had a lab which I shared with two other people.
After making the samples, they were characterised using techniques like Fourier Transform Infrared Spectroscopy, Helium pycnometry and Nitrogen adsorption/desorption. Then, we tested the materials for their carbon capture capacity.
Life is good. You can take breaks whenever you want. It's just like you're working, without the pay. Leave a sample to stir or sonicate for a while...if it needs one hour and you have no other samples to prepare, bam, you have a break and there's all the facilities you need like a microwave, fridge, hot water, coffee, etc. I had a bit of office space to myself too, shared with Honours/PhD students in our research group, but I never used it because they all had keys and I didn't so it was locked all the time - embarrassing to always ask them to open the door.
I loved every bit of this unit. You get to meet some amazing people around the faculty. You get to see how cool instruments work. You get to see all the mechanics of a lab, like how they change gas cylinders, how to use the liquid nitrogen tanks, how to order new glassware. You overhear conversations where people are talking about how their paper is about to be published, or how their results look good, etc. It's such a friendly atmosphere.
Yea, I could keep going on about how great it was, because it was!
Assessment
The oral presentation was quite scary. I had to present in front of 20 odd scientists for 10 minutes...even though I'm usually decent at presentations, I stammered a lot in that. Be prepared for questions...if this is what conferences are like, haha...
The report was 13-14 pages long, ~3300 words. There are samples on the Moodle page. Don't leave it to the last minute
Lab work should be easy marks. Just work diligently, keep your lab tidy, etc. Label stuff properly, keep clear records. When you are making 18 samples, accidents in naming can occur. One white powder looks the same as another white powder
Final words
If you are interested in research, definitely do it. It is an amazing experience. The professors are very adept at explaining concepts, if you're worried that you won't understand what's going on, it won't be that way. I remember, we had a group lunch once to say farewell to a German researcher. The professor, while explaining some chemistry to me, pulled out a pen, took a napkin off the table and started drawing diagrams. Everyone is there to help! So don't be afraid
If you have any questions regarding the unit, feel free to contact me. Or maybe if you want to read my crappy report.
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« Last Edit: March 28, 2014, 08:15:06 pm by DisaFear »
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BSc (Hons) @ Monash (Double major in Chemistry)
#### alondouek
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##### Re: Monash University - Subject Reviews & Ratings
« Reply #182 on: March 02, 2014, 01:59:48 am »
+7
Subject Code/Name: SPHPM Summer Research Scholarship
• Officially 4 weeks, 9am to 5pm Monday-Friday.
• Your project may be extended by your supervisor and the school.
Assessment: There is no official, graded assessment, but there are certain conditions that you need to meet:
• Professional conduct
• Attendance to project-related stuff, e.g. meetings, data collection, paper-writing etc.
• Attendance to scholarship program-related activities
Recorded Lectures: N/A
Past exams available: N/A
Textbook Recommendation: There are no textbooks - given that this isn't a unit - but get ready to read a boatload of journal articles!
Lecturer(s): N/A
Year & Semester of completion: Summer 2013-2014
Rating: 5/5 at least!
So it's not technically a unit, but it is offered by Monash and it is an amazing learning experience so I thought I might write up a review of the SPHPM Summer Research Scholarship. SPHPM is the School of Public Health and Preventive Medicine under the Faculty of Medicine, Nursing and Health Science. It is located at Monash's Alfred hospital campus, and contains a number of departments including the DEPM (Department of Epidemiology and Preventive Medicine) and the DOFM (Department of Forensic Medicine) amongst others. As well as being right in the hospital premises, it's also right next door to the Burnet Institute, Baker IDI and AMREP so it really is a central hub for medical research.
There isn't any undergrad representation at SPHPM, which consists of Honours, Masters and PhD students, as well as post-docs and full-time researchers. This gave those of us who were part of the program the opportunity of working in a research centre devoted to everything postgraduate and beyond.
Applying for the program:
Like all other Summer and Winter research scholarships offered by Monash, you need to apply during the relevant period. Make sure to check the Monash website for the specific dates. Also like many of these scholarships, there is paid remuneration for your time spend on the scholarship program.
N.B. that as I write this, the program is offered only to those in MBBS, BBiomedSc and BHSc. I doubt they'd expand it to other degrees like BSc or BA in future, but you never know!
The application process for this was fairly straightforward:
• I sent in the initial application, listing my details, my units studied and why I thought I should be offered a place in the program.
• Later - after the teaching period had finished but before exams were finished - I was invited to an interview with a member of staff at SPHPM. I don't think they'd like me to give too much away about this, but the general gist of the questions were my interest in public health, why I thought it was important at different levels and other similar things.
If you are successful in both of these, the program coordinator will email you with your research topic and supervisor prior to the start of the program.
The program itself
Unlike many other vacation research scholarships offered by Monash, the SPHPM program offers a group experience because you're in constant contact with the other people in the program, e.g. there are several tours that you'll go on as a group (I'll elaborate on these shortly), and you'll (hopefully) end up hanging around with these people almost every day by going out to lunch or just seeing each other around the office. Also, it's likely that you'll be sharing a research unit with another person on the program, but you'll probably be working on individual projects. The people who did the program with me were all lovely, friendly, incredibly intelligent people and I made a lot of friends - it's hard not to when you're working with these people all day, every day!
As mentioned above, the program itself is divided into two parts; as well as working on a research project, there are also other activities run for all the scholars by the program coordinators. For us, these activities were the following:
• Paramedic training and simulation session
• Endnote and MEDLINE session at the AMREP Ian Potter Library
• BMedSci Honours Seminar
• Visit to the Heart Foundation
• SPHPM Awards Night
• Group morning tea/project discussion
• Visit to the ICU/ED
• End of program lunch
• A couple others that I can't remember right now
These were all really fun, exciting and informative, and it really helped everyone connect with each other. It can be a bit intimidating working at SPHPM with the amount of incredible and fascinating research going on at any one time, so it was great to have a group of friendly people around to experience the program with (and to complain about the workload to ).
The research side of the project varies in its specifics depending on your project and supervisor, but the constant is that it's intense. You'll most likely be writing A LOT, be it an abstract, a lit review, parts of the research team's primary paper - you name it. Expect to be at the computer typing a lot; this isn't lab-based research but you'll be doing a lot of data collection and interpretation. This might not sound so great but it is very engaging and it sets you up not only for any research-based stuff you might do in future, but also for scientific practice in later years uni, especially units like SCI2010/2015 if you ever happen to take one of those.
My personal project (simply stated) looked at the causes and effects of various delays in the diagnosis and treatment of NSCLC (non-small cell lung carcinoma), a type of lung cancer. It was really interesting to learn so much about one particular condition and how prevalent it is. At the moment (i.e. I am procrastinating as I write this), I'm writing a literature review and contributing to the research team's primary paper. I'm hoping to submit the lit review for publication soon (I should really get back to work hahaha), and tbh I'm really proud of the effort I've put into my work at SPHPM - I know the same goes for all the other participants in the Summer research program as well. It was a wonderful experience overall, and it gave me a taste of what medical research is really like; I definitely want to do more in this general field.
Highly, highly recommended for those students looking to go into medical research at some point, or even just those interested in public health and medical science.
2013-2016
Majoring in Genetics and Developmental Biology
2012 ATAR: 96.55
English [48] Biology [40]
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#### Treeman
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##### Re: Monash University - Subject Reviews & Ratings
« Reply #183 on: March 05, 2014, 09:28:24 pm »
+5
Subject Code/Name: MTH2132 - Nature and Beauty of Mathematics
Workload: Two 1-hour lectures and one 1-hour support class per week
Assessment: I think there were 6 assessments spread out across the semester each weighing the same.
Recorded Lectures: No
Past exams available: Not when I did it. We were given exam preparation materials.
Textbook Recommendation: Each topic gets handed out relevant materials, no textbook necessary.
Lecturer(s): Dr Burhard Polster
Year & Semester of completion: 2010, it was also coded MTH1122
Rating: 6 out of 5
Comments: This is a very interesting unit which places mathematics in a more artsy perspective (probably designed for art students whose maths is not their strong point haha). Philosophical topics surrounding maths are raised but lightly delved in so don't be put off by that, the assignments are fairly easy but more importantly the material is interesting. Talks about the Golden Ratio all the way to the shape of the universe and mobius strip explained so that layman can understand. Some of the later assignments can get a little tricky but just ask Burkard (he urges you this thru out he entire semester) and he is happy to explain with all too many hints.
Exam: I thought the exam was a bit harder than the assessments he gave us and there was quite a lot to cover for the time given. I'm pretty lazy tho but if you put in some effort this is almost a guaranteed HD
Bottom line: This is a piss easy subject that has a low mathematical component and is great for filling up electives while undertaking your course.
« Last Edit: March 05, 2014, 09:31:06 pm by alondouek »
VCE 2008: Physics, Bio, Chem, Methods, Spesh [The TRUE Asian 5]
ENTER: Enough to get into Monash Clayton
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#### Treeman
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##### Re: Monash University - Subject Reviews & Ratings
« Reply #184 on: March 05, 2014, 10:19:38 pm »
+4
Subject Code/Name ATS1347: Music Ensemble (Also known as ATS1347/ATS1348/ATS2800/ATS2801; the unit codes refer to the same subject, labelled for different semester/year)
Workload: 2 hours every Monday from 2pm to 4pm in the Music Auditorium
Assessment: The music ensemble is an umbrella term for the choir or orchestra. You choose one or the other obviously depending on whether you want to sing in the choir or play an instrument in an orchestra.
The assessment is basically having attendance every week and turning up to live performances that are usually held in the city or the surrounding suburbs or at Monash. The no. of live performances really depend on the unit coordinator, so it could be 3-4 performances, or even just 1. When there was only one performance, we did a very short aural pitch test where you basically just hum in the same key as the piano. Possibly the easiest assessment ever, given you have relative pitch.
I've only done the choir so I can't comment on the orchestra but I'm sure it's very similar.
Recorded Lectures: N/A
Past exams available: N/A
Textbook Recommendation: You are given the music sheet for each music piece.
Lecturer(s): I've had Frank Dobbs twice now but there seems to be a new conductor this semester for 2014, didn't get his name...
Year & Semester of completion: ATS1348 in 2012, ATS2801 in 2013.
Rating: 10 out of 5, oh yeh
Comments: This subject is a walk in the park IF you have a musical background, can sing with relative pitch and can read music notes. Actually reading music notes isn't even necessary if you have very good ears and good control over your voice. Just stand next to someone who can read and you'll be right. But still, reading music should be something all musicians know.
Frank Dobbs was a great conductor, a man with character and excitement about him.
Bottom line: Guaranteed HD for those with a musical background or relative pitch.
« Last Edit: March 05, 2014, 10:23:24 pm by alondouek »
VCE 2008: Physics, Bio, Chem, Methods, Spesh [The TRUE Asian 5]
ENTER: Enough to get into Monash Clayton
B.Sc. 2009 - When I get off my lazy arse
#### ninwa
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##### Re: Monash University - Subject Reviews & Ratings
« Reply #185 on: March 28, 2014, 08:09:28 pm »
+5
Please note that this is a JD/LLM elective*. However, LAW5146 - Intellectual property I: Copyright and designs also covers copyright law. LAW5146 has more of an emphasis on designs though.
Intensive lectures (9am to 4pm for five days), no tutorials
Assessment:
Research assignment (3,750 words): 50%
Take-home exam (3,750 words): 50%
Recorded Lectures: No
Past exams available: No
Textbook Recommendation:
Optional: hard copy of Copyright Act. I would recommend it if you can afford it because the CA is rather confusing and it might help you visualise its structure if you actually have it in front of you.
Lecturer(s): Assoc Prof David Lindsay
Year & Semester of completion: Semester 1, 2014
Rating: 4.5 out of 5
The unit
I remember looking up the unit evaluation and the only comment was "Unit was intellectually stimulating", and I remember saying to enwiabe that this was almost definitely lawyer-speak for "shit's bloody hard yo". And I was right, and why didn't I discontinue and finish my law degree with easy electives like a normal person? sobs quietly
Aaaaanyway, so copyright law can get quite complex. There were top tier lawyers in my class and even they found it confusing at times! There is a LOT of content to get through, meaning that a lot of it was skipped through fairly quickly due to time constraints, which leaves you to work it out yourself at home. The difficulty also comes from the fact that copyright law can get very philosophical/meta/policy-based - there is almost never a firm answer - it's always "a question of fact and degree" (as a law student that phrase should terrify you)
That said, I found this subject extremely fascinating (when I could understand it). I've heard people criticise it for being way too technical and theoretical, but in my opinion it is very applicable to real life - especially if you are interested in the arts or sciences. The importance of copyright law to innovation and cultural development is undeniable.
For me, the most interesting part was seeing how copyright law adapts to and moulds itself in line with technological developments. For example, computer games can only be protected as "films", because at the time the Act was drafted obviously computer games didn't exist. And, my essay topic was on whether computer-generated works could be protected which, given how much more advanced AI is becoming, is only going to be more and more relevant. I was also fascinated by the intersection between copyright law, designs law and the boundary-pushing tendencies of modern art (is a urinal classified as a sculpture and therefore protected by copyright? What about a well-designed and very beautiful yacht?)
If you're looking for an easy D/HD, stay away unless you're the second coming of Michael Kirby, but if you're looking for a challenge - or "intellectual stimulation" - definitely give it a shot.
The lecturer
(David Lindsay takes Intellectual Property I as well, interchangeably with Rebecca Giblin.)
From what I've heard talking to people, you either hate David or you love him. He's extremely knowledgeable, funny and a very kind man who will take all your stupid questions seriously so you never feel embarrassed. I think the problem some people might have with his teaching style is that his explanations sometimes complicate rather than clarify the matter, just because he's such a walking encyclopedia about copyright/internet/broadcasting law that he'll explain a difficult concept by bombarding you with 10 other difficult concepts. Occasionally his "explanations" give me a headache haha. But emphasis on "occasionally"!
*For LLB students: the reason I'm taking a couple of postgrad units is thanks to the Master of Laws Elective Program for undergraduate students, whereby you can take up to two electives from the JD or LLM course. You will study at Monash's city law chambers for JD students, which is right next to the County Court and is seriously so much prettier than Clayton. The classes are much smaller - 15 to 20 maximum - which means it's a lot more interactive. Some of the postgrad lecturers are really great and have amazing credentials. I also found that a lot of my classmates were older students who had already had years of experience in various fields and so could make really interesting contributions to class discussions - some of them were international or top tier Australian lawyers, for example. I highly recommend you take up this opportunity if you have the chance!
« Last Edit: September 20, 2014, 10:38:02 pm by ninwa »
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#### pi
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##### Re: Monash University - Subject Reviews & Ratings
« Reply #186 on: June 09, 2014, 12:18:51 pm »
+10
Subject Code/Name: MED3051 - Medicine and Surgery 1
Workload: per week: varies between sites, you're expected to stay between 7-8am (former for surgery, latter for medicine) to 4-5pm each day giving a total of ~50 contact hours per week, whether you stay that whole time depends on how you study and what you want to get out of being on the wards. Each site usually has lectures on Wednesday and it varies between sites how many lectures there are.
Assessment: 70% Mini Case Records (MCRs - two formative and two summative in this unit), 30% Evidence Based Clinical Practice "Therapy" Task, attendance (80% hurdle), completion of online pathology quizzes (14 in all - hurdle), complete submission of portfolio (hurdle), formative end of semester exam (non-hurdle or hurdle depending on site) .
Recorded Lectures: No.
Past exams available: No, the Faculty has now published a document with threats to expel students from the course if they are caught compiling past questions or distributing or using past compilations. All past compilations have been removed from the MUMUS site. Many EMQ/MCQ books can substitute for official exams though.
Textbook Recommendation:
• At a Glance - Medicine - Davey*
• Clinical Examination A Systematic Guide 7th - O'Connor and Talley
• Davidson's Principles and Practice of Medicine 22nd - Colledge, Ralston, Penman and Walker*
• Harrison's Principles of Internal Medicine 18th - Fauci, Hanser, Jameson, Kasper, Longo and Loscalzo*
• Kumar and Clark's Clinical Medicine 8th - Clark and Kumar*
• Netter's Clinical Anatomy 2nd - Hansen
• Oxford American Handbook of Clinical Examination and Practical Skills 1st - Burns, Korn and Whyte
• Oxford Handbook of Clinical Medicine 9th - Baldwin, Longmore, Wallin and Wilkinson
• The ECG Made Easy 7th - Hampton
• Toronto Notes 2012 - Klostranec and Kolin*
• Underwood's General and Systematic Pathology 5th - Cross and Underwood
*Pick one depending on how keen or lazy you are
I'd also recommend utilising UpToDate as much as possible.
Lecturer(s): Many, depending on the series of lecture (reproductive, haematology, neurology, psychiatry, pharmacology, pathology, etc.)
Year & Semester of completion: Semester 1, 2014
Rating: 5/5
This unit is something completely new! I'll keep this general given that every student will have their own unique experience depending on their site, their rotations, their group, and how keen they are to get what they can out of it.
Basically the sites Monash have are (I may be missing some!):
- Central: Alfred, Cabrini, Peninsula
- Monash (formerly "Southern"): Monash Medical Center, Dandenong, Casey
- Eastern: Box Hill, Maroondah, Angliss
- A bunch of rural sites such as Bendigo, Mildura, Bairnsdale, Traralgon, etc
It is completely randomised as to which hospital/site you get (other than the choice of rural vs metro sites), no more preferences!
Each site has their ups and downs in terms of a balance between practical skills and teaching and it's probably not up to me to make a comment on this, but the feedback I get back form peers is that the clinical years (so far) are SO MUCH better than the preclinical ones, and I'd agree with that 100% percent. Each student will have their own medical (gen med, oncology, cardio, neuro, rheum, etc) and surgical (gen surg, neurosurg, cardiothoracics, bariatric, vascular, etc.) rotations depending on luck and where they are placed.
The gist of a day on either surg or med plays out like this:
• Ward round starts at 7-8am depending on your team, this may be with a big team (think Alfred, MMC, etc.) or a small team led by a registrar (think Angliss, Casey, etc). Med students can write the ward notes, they may be asked questions by the consultant or registrar, they may be asked to see the patient later and report the case back to someone, etc. Always good to try and be /helpful/ (getting the patient files in advance, having a look at the obs, etc.) because they'll like you more and you'll probably get to do more things as a consequence.
• After the ward round there will be an allocation of tasks (more-so in medical rotations), if you're in the good books with the team you may be asked to do a few tasks such as "cannulate the gentlemen in Beds 3, 14 and 25 for us, and we'll need bloods from Bed 13, 15 and 17, oh and also if you could chase up 17's GP and get them to fax over her lung function tests that'd be great". Simple stuff and they'll love you if you can help out plus practical skills are so exciting!
• Your tasks can span the whole day depending on the urgency, usually try and get your cannulas and bloods done ASAP. Throughout the day you may have tutes (more details later) and have the opportunity to clerk patients (basically take a history, perform an examination and report back to a senior on the ward) and go into surgeries (obviously only in surgical rotations, you may get to scrub up and assist with suturing and whatnot depending on the surgery).
• Repeat.
So I mentioned a few practical skills above. The new ones to clinical years include: cannulation (putting in a "drip"/"bung"), venipuncture (taking bloods), urinary catheters, rectal examinations, injections, performing lung function tests, and some unofficial ones that your team might teach you such as taking arterial blood gases, taking blood cultures, and so forth. You also may be able to help out and learn about more complex procedures such as ascitic taps, pleural drains and lumbar punctures. Some sites it may be very difficult to get any practice but in other sites you may be able to do a few of each practical skill a day (think smaller hospitals). The practical skills I mentioned (the "official ones) are important to do because you need to mark them off in a "logbook", a small book which has a list of skills which need to be done including histories and exams from all systems and a bunch of practical skills as aforementioned. This needs to be handed in as talked about later.
To further your skills, and if you're on a good basis with your team, is to get involved with doing admissions, ie. admitting patients to the ward or to the hospital (sneak into ED!). I've had the opportunity to do this a few times both supervised and unsupervised and it's a really great learning experience. If you ever get a chance be sure to put your hand up first and take it!
In terms of tutes, there are may kinds and the amount of them depends on your site. Medical and surgical bedside tutes are commonplace, here you have a small group and a consultant and as the name suggests, you have a tute at a patient's bedside learning about their condition and examining them. Other tutes include PBLs, specialty tutes, practical skills tutes, clinical skills tutes, epidemiology tutes, law and ethics tutes, etc etc. Some sites have an attendance that includes these tutes, others do not.
Another thing I want to touch on are a few of the assessments:
• MCRs: These are basically mini-OSCEs. Either a history or an examination on a patient where you're getting marked by a senior doctor such as a registrar or consultant. They count for a lot of the year and are a really good place to put your clerking of patients into practice to show off your skills and demonstrate your clinical knowledge (they'll ask you questions wither throughout or afterwards).
• ECBP task: This is a very similar task to the epidemiology assignment from Year II. Personally, not the most exciting task out there.
• Portfolio: This is a bit of a pain, it's a checklist of things you have to submit at the end of the semester: group assessments (such a any PBLs your group may take), feedback sheets you get marked off by your seniors so that the Faculty knows you actually come to ward rounds, the EBCP assignment and the logbook.
Now with so many differences between sites and hospitals and student experiences, a fair question to ask is: "how do they examine this theory later?". The simple answer: "The Matrix". It's a huge table of conditions, a total of OVER 250 conditions that are examinable. If it sounds scary and daunting, it's because it damn well is.
Just at the end I feel I should mention some of the areas of clinical medicine which are often overlooked by all the exciting things. It's important to remember that you're in hospitals and that people are sick. Some sicker than others, and some of your patients may pass away whilst you are there. We get taught about this sort of thing during preclinical years but it's something completely different to experience it in real life. It's hard to deal with, and if you need some help with it seek assistance from your seniors, they'll always have a handy word or two. Here's something I wrote about this on Med Students Online, copied here for convenience:
Spoiler
My introduction to clinical years
Not sure how to feel, but my first few weeks on the wards have been interesting. Being on an oncology rotation first-up I can't say I didn't expect it (I certainly did), but I don't think any amount of pre-contemplation prepared me for the real deal: when a patient passes away in front of your eyes.
Now in the "predictable" pre-clinical environment I wasn't really phased emotionally by much, the Aussie notion of "grin and bear it" was really the way to get through. Everything was simply just theory and more facts to understand and remember. As morbid as it might sound, I even had no issues with cadavers, as it was all part of this "learning environment" and dissections were very much academic and not at all patient-orientated.
On the wards and in clinics, it's a different ball-game altogether. Being a medical student here isn't all about the exams and the textbooks, it's about being part of the healthcare team and learning from their expertise so you can be the best that you can be. I have a great and supportive team, and being their junior is an exciting privilege, however being part of the team is only a minor aspect in comparison to what the team actually does: manage patients.
From Day 1, it was confronting. I have never seen so much suffering, so much pain, so many tears. From the pre-clin years I guess one could say I was disillusioned by what some doctors have to deal with, I didn't think some things could be "that" tough in real life. What if the patient doesn't want to undergo the advised treatment? What if the patient's treatment options are at an end and they're looking to you as to what is next? What if things are far worse than the patient had hoped for? What if a patient you have seen for weeks unexpectedly passes away?
As only a student I guess I don't have to have answers to those questions, but there's always that feeling that I should? It's tough, when reality hits that doctors have limitations from all areas whether that be from their patient's decisions, from treatment options, from financial stand-points, and the list goes on. We learnt about this, but it doesn't come close at all to seeing it in real life - patients do make decisions and do pass away and sometimes there is nothing we can do about it.
So early onto my clinical experience, it's been a roller-coaster taking this all in. Learning with how to approach different situations has been very helpful, from what I gather it's like desensitising yourself from the patient in an emotional sense. Having said that, one of my greatest fears is being one of those people who don't say "John, the fellow with <x> in Bed 14, needs some fluids" but instead say "Bed 14 needs some fluids". I'd hate to lose the personallness (is that a word?) of it all - it's my greatest fear and I have seen in it on the wards and I don't like it at all.
This beings me back to the patient passing away in front of me last week. That patient was in pain, they had multi-organ failure, mets from their primary cancer, and suspected infection. There was part of me that hoped they would pass away as they would be in a much better place, but there was also part of me that wanted them to keep fighting it all. When it happened though, when they passed away, I was just lost. I felt bad, almost wanted to cry, not sure what to do. We couldn't save them. Did I care too much? Am I just "weak" as a person? Is this just me being a novice medical student?
I guess it's all about finding that professional balance between being too affected and not being affected at all. I want to care, but I don't want to care "too much" as I think that'll hurt me and I won't be able to function to my best, if that makes sense.
Hopefully that balance comes with time.
Thanks for reading, sorry about this slightly depressing blog post (my first) and I'm betting there are some incoherent lines in there - was just typing my mood and thoughts.
Having said that, it's always a great feeling seeing one of your sicker patients get discharged cancer-free or in fine health, you don't get a feeling like that anywhere else and it's one of the best feelings I've ever had. It's even better if you took up an opportunity and did an admission on that patient, you can see them from admission to discharge and it's really rewarding to see the health system at work!
As with my reviews of the previous MBBS units, I think it's really important to get involved with the course outside of the teaching periods too. I'd highly recommend getting involved in inter-year study groups (teaching in Year 2/3, learning from Year 3/4) and getting involved in the social events such as the "Half Way Party" which was a pretty sweet night
All-in-all, a very exciting unit. Being on the wards has been amazing and no amount of money would persuade me to go back to the Clayton campus for days of lectures. I've kept it general because everyone has a unique experience with how clinical years play out for them but if you have any specific questions feel free to PM me (please only PM me if you're already in the Monash MBBS, it's far too keen otherwise -_-).
« Last Edit: June 09, 2014, 12:53:01 pm by pi »
#### vashappenin
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##### Re: Monash University - Subject Reviews & Ratings
« Reply #187 on: June 09, 2014, 09:49:36 pm »
+6
Subject Code/Name: ETC1000 - Business and Economic Statistics
Workload: Weekly 1.5hr lecture and 1.5 hour computer lab (starting in week 2 and ending in week 11)
Assessment: 30% Lab quizzes, 70% Exam
Recorded Lectures: Yes, with screen capture
Past exams available: Yes, 12 exams including solutions
Textbook Recommendation: No compulsory textbooks
Lecturer(s): Brett Inder - he's a really laid back, friendly guy who's really good at explaining the concepts, not to mention that he's been teaching the unit for quite a while so he's VERY familiar with it.
Year & Semester of completion: Semester 1, 2014
Rating: 3 out of 5
When I first started this unit, I HATED it, but once I gave it a chance I actually quite like it! Although not advised, I ended up doing a lot of learning for this unit during SWOTVAC and to be honest, it's pretty easy to pick up all the concepts in such a small timeframe so don't panic if you get to that stage, although just don't do that to make your life easier haha You can score well in this unit if you put in the work. So it's really important to keep up to date and understand everything as you go because (especially with the last few topics) you need to understand the past topics' content in order to be able to successfully get through the rest of the topics.
In terms of workload, this subject is really good if you're looking for something that doesn't take up much time. All you really need to do every week is watch the YouTube lectures for the week (which go through the slides anyway), watch the live lecture (i.e. the lectures conducted at uni) and do the lab homework.
Lectures: To be honest, I didn't physically attend a single lecture (only because the time was too inconvenient for me).. I watched most online, but it got to a point where I stopped watching the live lectures. Only because Brett made YouTube video lectures for each topic which were SO useful, so definitely don't skip those. The videos are pretty much Brett talking through the powerpoint slides for the relevant week's lectures. The live lectures were more application of the content, so honestly, don't miss those either. I definitely wish I didn't, because although they got boring at times, the knowledge would've been pretty useful come exam time.
Computer Labs: Most of the lab quizzes were usually able to be finished in 30 mins-1 hour, but took longer to complete in the last few weeks. The labs were a really good way of applying all the excel processes and procedures, and were generally not too hard to get through. It's recommended that you complete the homework (not compulsory) prior to the week's lab, because the homework pretty much contains step-by-step instructions that really come in handy when doing the lab quizzes.
Exam: A two-hour, non-calc exam. It wasn't too bad considering the fact that there's a plethora of past exams WITH solutions available to you, on top of revision during week 12 (going through exams). Even though maths is required in the exam, you pretty much just need to show working to get the marks, not the actual answer (since there's no calculators allowed). If you do as many past exams as you can and know your content, you should be fine for the exam. Keep in mind that in order to pass this unit, you must score at least 40% on the exam.
2013: English, Maths Methods, Further Maths, Legal Studies, HHD, Psychology
2014-present: Bachelor of Laws @ Monash University
Tutoring VCE English, Psych, Legal Studies and HHD in 2016! Tutoring via Skype too. PM me if you're interested
#### simba
• Guest
##### Re: Monash University - Subject Reviews & Ratings
« Reply #188 on: June 09, 2014, 11:34:28 pm »
+6
Subject Code/Name: MTH1035- Techniques for modelling (advanced)
Workload: 3 x 1 hour lectures, 1 hour tutorials and 2 hour workshops
Assessment: 3 x 10% Assignments, 10% Test and 60% exam (although I vaguely remember hearing this was set to change next year)
Recorded Lectures: Yes
Past exams available: No, but they did release a sample
Textbook Recommendation: Don't need a textbook
Lecturer(s): Burkard Polster (essentially the best lecturer you will have in the existence of anything!) and Simon Teague (Who is pretty great too)
Year & Semester of completion: Semester 1 2014
Rating: 4.5/5
Comments:Initially I found this unit extremely daunting, in the 1035 workshops we almost immediately began working on cartesian tensors (which confused the hell out of me for months and have really only begun to understand them today!).
The workload for this unit is fairly consistently high, so expect to be doing plenty of practice questions, readings and such to gain a thorough understanding.
In saying that, the lecturer Burkard Polster is insanely good at explaining concepts in a very visual and layman's way which really makes all the coursework much more manageable to tackle. Prepare to watch him with a whole bunch of lightsabers too...(He also likes to juggle them sometimes ) Anyway back on topic, Simon taught us for the 1035 workshops and also had us for tutorials. Although he's usually late for them 8am workshop starts (=death), he has a real passion for the subjects and has millions of exam type questions if you want any extra stuff to do!
The assignments themselves aren't too bad (just really long and tedious). My main tip for them would be make friends in the unit and see if you can work together and collaborate answers (I do mean WORK TOGETHER not copy each others answers, but let's be honest, that will probably happen too). The test was fairly simple, pretty easy marks as long as you know your stuff!
Overall, if you love maths, pick this unit. But you will need to be dedicated and consistent to keep up to date and do well!
#### Reckoner
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##### Re: Monash University - Subject Reviews & Ratings
« Reply #189 on: June 09, 2014, 11:52:40 pm »
+6
Subject Code/Name: ECC2000 - Intermediate Microeconomics
Workload: One 2-hour lecture + one 1-hour tute = 3 hours
Assessment:
• Midsem: 30%
• Exam: 70%
Recorded Lectures:
Yes
Past exams available:
The past exams on the database are from 15 years ago, and no longer particularly relevant. We were given a sample exam though.
Textbook Recommendation:
Pyndick, Robert S, and Daniel L Rubinfeld: Microeconomics. Seventh or Eighth Edition, Pearson.
I'd say worth it if you don’t have to buy it new from the bookstore. The questions at the end of the chapters are pretty useful for revision and are pretty similar to the exam style questions. Buy it second hand/eBay though - $200 from the bookstore is way too much. Solutions and the textbook can be found somewhere that shall remain nameless as well Lecturer(s): Yinhua Mai. Year & Semester of completion: 2014 Semester 1 Rating: 3 out of 5 Your Mark/Grade: HD Comments: The course is broken up into three main sections – Consumer theory, producer theory, and analysis of market structure/competitive strategy. Consumer theory is all about maximising utility given an individual’s preferences for various goods, the relative prices of the goods and a budget constraint. This involves indifference curves, budget constraints, marginal rate of substituting, normal goods, giffen goods and Engel curves. The general gist of a question on this topic would be you’re given a utility function, say U(x,y)=20x^(2/3)y^(1/3), the prices of good y and good x, and the consumers budget. You’re then asked to find the utility maximising combination of goods. Then follow up questions on price changes, income changes et cetera. You also get the classic social surplus and elasticities, price floors and the like. Producer theory is almost exactly the same as consumer theory. You have labour and capital as factors of production, and each has a given cost. You have some production function telling you how much output can be produced from some combination of capital and Labour. You then need to find the cost minimising ratio of capital and labour to produce a given output. Instead of indifference curves you have isoquants, and instead of budget constraints you have isocost curves. Essentially the same principles apply as in consumer theory. You also revisit the cost curves from first year, but again with actual equations. Market structures is when the course gets a lot more interesting though. You analyse profit maximising output under various models. You get perfect competition and monopoly like in ECC1000, but this time you have numbers, equations and differentiation. But the best part of the course for me was oligopolistic competition. Various equilibrium settings and output decision models, price discrimination, collusion, competition, and a bit of game theory. This part of the course follows on from producer theory. I didn’t really like the unit at first to be honest. It was basically a rehash of ECC1000, with the lectures moving very slowly (a whole hour on what a demand curve is). While the basics are important in economics, it was just a bit slow and dry. However after the first few weeks, and we started on cost curves and market structures, it started to grow on me a bit more. I found learning the actual content to be easier from youtube/textbook than the lectures. At this point Yin started to run though examples in the lectures which made them more worthwhile. Tutes are pretty standard, just go over the 3-4 questions that were set that week. However due to the length of each question you rarely get though all of them. The mid-sem covers consumer and producer theory. Mostly multiple choice with a short answer chucked on the end. You need to know a few definitions though. The exam was pretty good I thought. 6 Questions, of which you answer 4. However each of them are pretty involved, which lots of re-arranging and substituting into equations. And then changing one variable, doing the whole process again and seeing what has changed. Doing all of the tute questions and the questions from the textbook will be enough for you to prepare. You don’t really need to know definitions as much as the mid-sem, so just know how to approach each type of question and you should be right. Also lots of algebra and partial differentiation. Nothing too crazy, but you should be comfortable with derivatives and solving linear equations. TL;DR Starts off pretty slow and not particularly interesting, but gets a bit better as the unit progresses. Think ECC1000, but with algebra, differentiation and a few extra topics thrown in. « Last Edit: July 13, 2014, 07:52:33 pm by Reckoner » #### Professor Polonsky • Victorian • Part of the furniture • Posts: 1151 • Respect: +93 • School Grad Year: 2013 ##### Re: Monash University - Subject Reviews & Ratings « Reply #190 on: June 10, 2014, 02:01:45 am » +6 Subject Code/Name: ETC1000 - Business and Economics Statistics Workload: 1 x 1.5 hour lecture, 1 x 1.5 hour lab (will probably only take you 20-60 minutes). Assessment: 30% Weekly computer labs - 10 labs, out of which only your top 8 results are taken into account. 70% - Exam. Recorded Lectures: Yes, with screen capture. Also on YouTube, I believe. Past exams available: All exams (both semesters) since 2008 are available, with solutions released in the final week of the semester. Some of the content has been cut in recent years, so don't panic if something looks entirely unfamiliar. Textbook Recommendation: No prescribed textbook. Australasian Business Statistics is "highly recommended", with readings highlighted, but I don't know if it's actually any good. You probably won't need it. Lecturer(s): Brett Inder. Lectures are very slow, and most people feel that they are unnecessary. Some of the examples used in the lectures though might pop up on the exam - so make sure that you grab the live lecture notes (they're on moodle) and read through them. If you're unclear about anything, it might be worth watching the lecture. Year & Semester of completion: Semester 1, 2014 Rating: 2 out of 5 Your Mark/Grade: HD Comments: This unit was not particularly enjoyable. As a forewarning, it is basically about the interpretation of statistical outputs by Excel, given businessy examples. The actual mathematical concepts are not at the forefront of the unit, although they are touched on, and some level of understanding might be occasionally required. Some statistical business concepts (mostly GDP, real vs nominal value of money) are covered. So that's what's meant by 'business statistics'. In some ways, there are some parallels with Further Maths. There is certainly a lot of overlap in the content, with the first half of the unit being the Core section of Further. The way this unit is organised as follows: Initially, the content is delivered through YouTube videos, which are supposed to accompany PowerPoint notes. In reality, the notes do not cover most of the things that are in the video, or might not make sense without them. I would like to take this opportunity to commend you if you manage to actually watch the videos. This is then followed by a live lecture, which I commented on above. It partly re-teaches the content again, but mostly is concerned with interpreting some given data on a particular topic (some of them are actually quite interesting!) using Excel. The latter is much more succinctly covered by the live lecture notes, which will save you a lot of time should you decide not to turn up (and let's be honest, who actually ends up watching the lectures online?) You are given weekly homework. It is basically a walkthrough of how to get the statistical output through Excel, and how to interpret it. Do it. It links directly to the following week's computer lab, and perhaps apart from the last few weeks of the semester, it is all you need to know for the lab. It gets trickier later on, and some conceptual understanding may be required. So as might have become apparent to you already, many aspects of this unit are quite duplicitous. At some point, you will likely stop coming to the lectures (though you probably would have anyway), and perhaps also stop watching the YouTube lectures. That's okay. Just keep up with the homework. So, come Week 11 (this unit has two weeks of revision), and the realisation hits that it's really just all about the exam. You might not even really have an idea of what's going on in this unit. That's fine. Why? Because the majority of the exam is entirely formulaic. And the previous 10 exams are all on Moodle. Do previous years' ones, check the answers. Or maybe even just check the answers, depending on how confident you are. Knowing what's on the exams, and how they mark them, is basically how you will pick-up the vast majority of marks in this unit. This is how it's very much like Further. Don't have a 'hat' over your dependent variable in the equation? Docked a mark. Didn't write "on average" when interpreting a coefficient? Docked a mark. Week 11 and 12 lectures are devoted entirely to going through those exams. You would do yourself an immense injustice if you missed out on them. They are all recorded, so just make sure you watch them. So basically - read the live lecture notes, do the homework, make sure you do well on the labs, and past exams. A note on consultation - you don't contact the lecturer, nor your lab supervisor. There is a walk-in consultation period, which I believe the chief tutor does (never been). There is also an email address specific for the unit, but I don't think you're supposed to ask questions there. « Last Edit: August 02, 2014, 07:59:21 am by Polonomial » #### alondouek • Subject Review God • Honorary Moderator • ATAR Notes Superstar • Posts: 2912 • Oh to be a Gooner! • Respect: +312 • School: Leibler Yavneh College • School Grad Year: 2012 ##### Re: Monash University - Subject Reviews & Ratings « Reply #191 on: June 10, 2014, 02:56:31 pm » +7 Subject Code/Name: SCI2015 - Scientific Practice and Communication (Advanced) Workload: • 1x 2 hour lecture • 1x 2 hour tutorial Assessment: • Exam: 30% • Workshop participation and activities: 20%; consisting of an interview with a researcher (0%), a journal club presentation (5%), a peer review report (5%) and workshop participation and blog posting (10%) • Major project: 50%; consisting of a research proposal (5%), an annotated bibliography (5%), a literature review draft (0% but needed to complete the peer review and final lit review), a conference poster (10%) and a final literature view (30%) Recorded Lectures: Yes, with screen capture. Past exams available: Yes, but the past papers are under SCI2010 (you sit the same exam). There are several on the database, with the 2004 exam under "SCI2010: How Science Works" and 2005-2009 under "SCI2010: Practice and Application of Science". The 2011 exam was also provided on Moodle. Textbook Recommendation: You don't need anything. Lecturer(s): A/Prof. Roslyn Gleadow Year & Semester of completion: Semester 1, 2014 Rating: 4 out of 5 Your Mark/Grade: HD Comments: This unit is a bit of a mixed bag. It (or SCI2010) is a compulsory unit in the Bachelor of Science; SCI2010 is the standard unit that most people take, while SCI2015 is the "advanced" version for students doing the BSc Advanced (Research) or other degrees - such as BSc double degrees or standard BSc - by invitation if your results are good enough. The content of this unit is pretty interesting; it's got a lot to do with how science works, how it can go wrong, and how you should conduct yourself as a researcher. Topics range from the history and origins of science, to scientific ethics, to pseudoscience, to career development and others. I only went to one lecture, but it was fun and interesting given that you're passionate about science. The highlight of the lecture series - and the only lecture I attended - was a magician/illusionist who was brought in for a performance during the pseudoscience section. Assessment is pretty good too; the exam is worth 10% less in SCI2015 than in SCI2010 (which is nice ), because you have more formative presentation-based assessments leading up to your final literature review. The other major bonus of SCI2015 is that you get to pick your literature review topic, which essentially makes writing the final paper a lot less painful. There are 5 assignments that make up the in-semester assessment: • Assignment 1 - Major Research Project • 1a - Research proposal: This is worth 5% of your semester mark; it involves A) generating a 100-word specifically-worded research proposal on a topic of your choice, which you'll write your lit review on eventually and B) Present your proposal to the class. A big part of this unit is presentation, so you should make sure that you can design attractive visual aids as well as present eloquently. • 1b - Annotated bibliography: This is worth 5%. You need to choose 4 articles related to your topic and generate an annotated bibliography on it. An annotated bibliography is a paper that consists of summaries of journal articles and additional critical evaluation of these articles. • 1c - Literature review draft: This is a formative assessment, but you need to complete it to be able to complete Assignment 3 and Assignment 1e. It's exactly what it sounds like; you need to generate a draft of your literature review. It's okay if you haven't done much, but you do need to have something to turn in so that you can take part in the peer review assignment and turn in your final lit review. • 1d - Conference poster presentation: This is worth 10% of your semester mark. You need to make a detailed conference poster for presentation, which will be presented to the tutorial group and any guests who are there for whatever reason. It's quite a lot of fun, as people circulate around the room and listen to mini lecture-style presentations. • 1e - Final literature review: This is worth 30% of your overall semester mark. A word of advice; don't leave this to the last minute unless you want some major stress! This assignment is a big one; it's only about 3000 words for SCI2015 students, but all the referencing and citation management take up quite a lot of time. Make sure you know how to use a journal search engine such as MEDLINE or Scopus, and that you're familiar with a citation manager such as Endnote (which you can get free from the university). I wrote my literature review of the efficacy and patient/hospital-related factors of performing decompressive craniectomy on traumatic brain injury patients (if you're interested I can send you a copy ). Another benefit of doing SCI2015 is that you can choose your preferred journal of submission (N.B. you don't actually need to submit your paper to a journal, but I'll write a bit about that later), and therefore you can kinda choose your referencing style as dictated by the style guide in the journal you choose. As well as this, the assessors are a bit more lenient in terms of word count and number of references. SCI2015 is basically a lot more independent than SCI2010, some people (myself included) prefer this, but it might not be for everyone. One final point is that you have the option (assuming your work is of a high enough quality) to submit your paper - following stylistic adaptations as required - to the journal Reinvention - a Journal of Undergraduate Research. • Assignment 2 - Interview with a researcher: Basically you pick an academic (I think the rule was they had to at least be working on a PhD), make a time to interview them, then make a powerpoint about the interview and present it to the class. As I mentioned, this unit is pretty heavily presentation-based, so if you're not a confident presenter when you start this unit, you probably will be by the time it's done. This assignment is formative, but there's no reason not to do it, so yeah. • Assignment 3 - Journal Club: You and another person pick a journal article, summarise it and present it to the tute class for about 10-15 minutes, then take questions. This is worth 10% so take it seriously, but the assignment is broad so you can take any article so long as it's from a peer-reviewed article. My tute partner and I did our presentation on "Were James Bond’s drinks shaken because of alcohol induced tremor?", published in the British Medical Journal in 2013, which was pretty funny • Assignment 4 - Peer review report: Remember when you had to submit Assignment 1c (the draft lit review)? Well following that you'll receive someone else's draft, which you need to annotate and then write a letter to the "editor" detailing your recommended changes. You'll also receive an annotated version and one of these letters of your draft, analysed by another student. • Assignment 5 - Workshop participation and blogs: Workshop/tutorial participation is exactly with what it sounds like, and the blog posting is done on Moodle in a specialised area. You need to submit at least 5 blog posts over the semester (you'll only be graded on 5), and they're judged on both quality and the fact that you've done them so don't half-arse it. Both of these together are worth 10%. The exam is really, really quite simple if you've done a couple of past exams. It consists of 2 parts, A) being 40 MCQs based on lecture/tute material and B) Written responses (short and long) based on various areas of the course (pseudoscience, communication, ethics etc.). Most of the MCQs are taken from recent past exams, so I'd advise you work through them and it'll be a breeze. The written responses are a little bit harder, but if you understand what you're talking about - as well as specific examples of things like scientific misconduct and research fraud) - then you won't have an issues whatsoever. All in all, this is a really interesting unit, and I recommend it to anyone interested in science and scientific practice (aside from the fact that you don't really have a choice if you're doing a BSc ). « Last Edit: July 17, 2014, 04:17:01 am by alondouek » 2013-2016 Majoring in Genetics and Developmental Biology 2012 ATAR: 96.55 English [48] Biology [40] Need a driving instructor? Mobility Driving School #### b^3 • Honorary Moderator • ATAR Notes Legend • Posts: 3530 • Overloading, just don't do it. • Respect: +627 • School: Western Suburbs Area • School Grad Year: 2011 ##### Re: Monash University - Subject Reviews & Ratings « Reply #192 on: June 10, 2014, 03:41:33 pm » +12 Subject Code/Name: MEC2402 - Engineering Design I Workload: 2*1 hr Lectures (Workshops), 3 hr Comp Lab, 2 hr Tutorial Assessment: Note: This changes year to year. - Online quizzes before each Lecture: 8% - Worksheets each lecture: 8% - Weekly CAD Tasks (x8): 8% - CAD Exam: 6% - Warman Prelim Submission: 10% - Warman Competition Results: 12% - Warman Final Submission: 18 % - Exam: 30% Recorded Lectures: Yes, with screen capture Past exams available: Yes, Most exams bar one back to 2006, only the last year or two were indicative of the actual exam. Textbook Recommendation: 1. Field, B. Introduction to Engineering Design (any edition) 2. SAA/Inst of Engineers, Australia: Engineering Drawing Handbook, SAA HB7, 1993. The former is a must have, you'll use it a lot, the latter you'll use too but not as much. You can bring both into the exam with you so they're worth getting. Lecturer(s): Scott Wordley Year & Semester of completion: Semester 1 2014 Rating: 4.5 out of 5 Your Mark/Grade: Pending Comments: There are a lot of aspects to Design I which I'll try and go over individually, but in short Design I will take most of your time throughout the semester, it's a workload heavy unit (especially the Warman Competition), but you gain a lot of experience and get a lot out of it. Flipped Classroom Model I'll start off with the way lectures were run. This year was the first year that the Engineering Faculty have tried the flipped classroom model, and it make be used for other units in the future if they think it was beneficial this semester. Basically, there are small videos put up before each lecture, ranging from 10 minutes to say about 40 minutes, which you watch and learn about the content and theory side of things. Before the lecture, which is now called a Workshop, you would then complete an online quiz about the videos, they're quite easy and are there to make sure you've watched the videos beforehand. Then in the Workshop (lecture), you come in and as a lecture and/or with your Warman Competition group (more on that later), you work through the worksheet, getting tips and help from the lecturer and tutors at the same time. These are then marked in the tutorial the week later. Overall, I think this was a good way to go, at least for design. It's one of those units where the lectures would be quite dry if it were run like a normal unit, but having the workshops allowed you to put into practice and try things out, while having the lecturers and tutors there to guide you along. It seems to fit the unit particularly well. THE WARMAN COMPETITION This really does deserve the capitals above. It will really take up a lot of your time and effort outside of uni. The Warman Design and Build competition is a competition in which teams from across the country design and build a robot like device to navigate a certain course and achieve certain goals. The track and objective changes year to year, and for design a campus competition is run just for the unit. The winners of the competition go on to represent Monash at the National Finals. This year, we didn't get to pick our teams. Normally it's in teams of, but the difficulty of the competition was ramped up this year, as most teams found it too easy to navigate the course at the national finals. As a result for the campus competition, we had teams of 8, but unlike most years had to build two devices. To give you an idea of what kinds of things you'll have to do, here is the National Competition from the year before: There are three stages to the competition, the Preliminary submission in which you create a design and work out what goals you want to achieve, along with some drawing, the actual competition where you are judged on your runs, and the final submission which mostly includes engineering drawings done through CAD. For our year, we had to transport "e-waste", which was a payload of rice, the mass of which we nominated (minimum 200g), around a barrier and then over a bar at a set height, which again we nominated (in 10 cm increments, maximum being 120 cm). The design brief for your year will be along the same format as ours: http://www.ncedaust.org/ckfinder/userfiles/files/Warman%2014%20v1_1_1.pdf The scoring formula was fairly complicated (under R45) and it set teams on two paths to maximise their score. You could either go for maximising the height with the minimum payload mass or maximising the mass with the smallest height. This was a design decision that had to be made early, with most teams going for the former option, which would include a lifting mechanism and somehow counterbalancing it, keeping in mind that the larger the mass of the system, the more your score decreased. So as with engineering, it was a balancing act, making compromises. Our team initially went for a height based system, attempted to go over the largest height of 120 cm. I should also note, as with most years, there is a limit on the size of your device, most years it has to fit in a 40x40x40cm cube. We had to have two devices, one had to be purely mechanical, no batteries, no electronics, no nothing. The device that started had to start in the 40cm cubic envelope and could finish at whatever size while the device that finished could start as large as it wanted to but had to fit in the 40cm cube at the end of the run. This meant there had to be a large extension compared to the base size, and ultimately meant a lot of devices were unstable at high heights. A lot of teams, as did we, went for a scissor lifting mechanism, some went for a telescoping air system to lift the mass, while others didn't lift at all, but put a large mass over the 30 cm bar. We had to CAD up our initial design, and if your design is the same as your initial design by the time the competition comes around, then you're doing something wrong. There will be a lot of changes in the design process, as you realise certain things just won't work, or that you won't be able to put certain parts together since you may have not allowed for access to screw something up. You and your team will have to fund the build and all the materials for the competition. Most years teams get away with$100-$400, since our year was a bit more complicated, we set our budget at$400 initially, which was $50 per person. By the end of the competition we had spent close to$800, and a fair few other teams had too. We weren't the highest spending team, with one hitting close to $1200 (a lot of that in burnt chips, but I'll get to that later). You'll ending up making a lot of trips to Bunnings and/or Masters, throughout the competition we would have racked up a fair few laps around the places. You don't get a workshop for the competition either, so you need to make use of the limited tools you have, which restricts what you can do a fair bit. WD-40 and Duck tape will be your best friend though! For the first time, we were provided with "Arduino kits", which were basically electronics kits with an Arduino controller (the brain of your robot), a motor controller (since the arduino can't handle the current or voltage needed to drive any decent motors, the one provided matched with the rover that was purchased), a voltmeter and an assortment of wires and other things. At the end of the semester you have to return this kit, and anything that you break or damage you will have to pay for. This is partly the reason one team spent so much, they blew up or so chips, which at$20-40 each starts adding up. The coding for the arduino takes a bit of getting used to at first, but isn't too bad.
The Arduino with the motor controller in the backgroung, connected to the rover.
Keeping within the rules of the competition, you have to either buy parts or make them from scratch without professional help. Most teams bought a rover chassis for their electronic device: http://www.pololu.com/product/1551
The rover
The teams who didn't had a lot of trouble getting their device to go in a straight line, you need that consistency in your runs. The rover chassis helped with this, and since it has encoders (something that reads the wheel rotation and sends it back to the arduino chip), you can control how far you want it to go via the number of wheel rotations. Without this you can only set it to power the wheels for a certain amount of time, and as your batteries drain down this changes every run, you end up chasing your own tail and never reaching it. Also, rechargeable batteries are a good idea, we did buy some but had a problem with the connections and as a result didn't end up using them. Another team used mecanum wheels, which would allow them to drive sideways, in practice it didn't quite go completely sideways, which is why you shouldn't expect everything to work as you would think it will.
The one main thing idea for the Warman Competition is to start early, I cannot stress this enough. If you can, order locally. We had ordered specific motors which had to come from Perth, the first time they sent the wrong motors and we had to get them to resend them, which put us back a week. The second time we received one of the correct motors and one of the wrong motors, which put us back another week. At this stage it was too late to change the coupling mechanism and we couldn't adapt motors from Jaycar locally. As a result this meant despite spending a lot of money on scissors lifts and getting them to work well, we had to redesign the whole device to move from a height-based system to a mass based-system the day before our competition. This meant that our device was not optimised, and was a lot minute job to bring it all together. We worked on it, rebuilding it from 10am in the morning to about 8-9 pm, with minimal breaks.
There is a track in the Engineering Building to test on, the earlier you get onto it, the better. We were one of the first few teams testing, and at our first and second tests the only team on the track at the time. As a result we got a lot done, didn't have to wait for other teams to have a go. In the week or two leading up to the competition, the track will get insanely busy, imaging around 10 teams (there was about 28 in our year) trying to get their testing done on the same track at the same time. Sometimes you could be waiting up to 30 minutes between runs, just to make minor adjustments. The night before the competition, don't be surprised if you have to pull an all nighter and work your ass off at the last minute. A few from my team were there, testing for about 4 hours, at around 2am my laptop battery died. We had forgotten that the code doesn't save to the arduino when you upload it, so we lost those 4 hours of code and testing. So make sure you save the damn code regularly! At that point the others gave up and went home. If we had left it in that situation, then our team would have gotten zero for the runs. I stayed there working on our device on my own throughout the night and early morning, along with 4-5 other teams pulling the all nighter as well (being up that long with no sleep you make a few new friends :P) Since we didn't know if our device B would be functional at the time or not, we lowered our goals and just got device A set up to transfer the payload without turning. In the end I broke my record of hours staying awake straight, 32 hours straight, with 30 of those being at uni. It's quite weird to see the sun rise through the windows at the end of the engineering building.
During the competition you get two runs, the score system dependent on the runs changes each year. For us it was meant to be your best run plus half the other run, but ended up just being the score from your best run. Our first run went well, and our second had a bit of a problem and ended up being a zero run score. A lot of teams that were going for the 120 cm bar, had achieved it in practice and got it on video, but ended up with two zero run scores on the day, (one device did really well, but drove off the end of the track). You either seemed to score really high or get a zero (or close to it). The second run, teams improved a bit, and since there was a lot of zero, they decided to be lenient on the scoring and give some of the score for the run in certain cases. This was only due to the difficulty of our competition this year. This again, is where consistency comes into play, you need to be able to reproduce the results on he day when it matters (much like the real world I guess, in Motorsport there's no point in being fastest on a test day if you can't pull it off on raceday). We ended up being ranked right in the middle of the pack, there were a few teams that made it to the end zone.
We initially got 2.8/10, but had our score bumped up to 8/10 due to the issues we encountered and what we showed would have been possible. Normally this doesn't happen, but since a lot of teams put a hell of a lot of effort into the comp and then got some low scores, they allowed us to do this for this year. Don't expect it to happen every year. We also had to keep a moodle log of the project, to show who was contributing and have ideas floating around. A lot of people spoke on fb, which you would then have to copy the conversations over to moodle. They may change the way this is done next year. There is also a peer-assessment component to the whole project, where you rate group members on what they contributed. This is then used to scale the marks of team members, where you can get anywhere from a 0.3 to a 1.1. It's a good idea, but you will still get some slack team members who don't care about the grade they get.
After all of this you will then have to do a big report on the competition, and use some of the CAD of your device to make proper engineering detailed drawings and assembly drawings. This has an individual component and a group work component. Try not to leave this to the last minute either. The submission was due at 1am, but we encountered problems with the computers in the Engineering Comp Labs, and so weren't ready at that time. We ended up going back to one of our team members house at 2am to use his computer (since it could handle just about anything), didn't get home until 4.30 am that morning. You get to make use of the New Horizons building computer labs for the tutorials, we were unlucky, being one of two teams that couldn't fit and so had to do our tutes in the computer lab, (more on the CAD sides of things later).
I should also point out, if you join the FSAE team (Monash Motorsports) or the UAS team (http://www.monashuas.org/ - Builds autonomous planes) then you won't have to do the Warman Competition, but will have your work based around what you do in those teams. They're good teams to join, Monash Motorsports is currently ranked 2nd worldwide (they were first a few months ago!), while the UAS team do a lot more of the aero side of things. I know a fair few mates in UAS, and they do learn a lot more through the team.
Overall, you do learn a hell of a lot from the Warman Competition, but have to sacrifice a lot of energy, time (and money) for it.
Through the tutes, you learn a bit about using the CAD (Computer Aided Design) program Solidworks, basically making parts and assemblies in a 3d computer environment. You can get a student copy of Solidworks through Monash, which you will be told how to do at the start of the semester. It really is a useful tool, allows you to see some problems before you make the part, which ultimately saves time and money. It simplifies doing engineering drawings, once you have the part cadded up it is a few clicks here and there. I really enjoyed working with Solidworks, but it can have a step learning curve at times. It also takes a decent amount of computing power to run mid-large parts and assemblies. At times the Engineering Computer Lab computers may crash on you or lag like there's no tomorrow (and they're not that bad computers). The computers in the New-Horizons design labs are a lot faster, it may take 30-60 seconds to load Solidworks in the comp labs, where as the New Horizons labs takes about 2-3 seconds.
Towards the end of the semester you'll sit a 3 hr CSWA CAD Exam, which is a computer test designed to see how well you can use Solidworks. You're given a few drawings of parts or assemblies and then have to make them, then you're asked something about the part which you have. So like what is the center of mass or the moment of inertia around a particular axis, which you get from the tools in Solidworks. If you're made the part right you should get the right answer, otherwise it'll be completely off. You need 70% to pass the exam, worth 6% of the unit. A fair few people were getting around 65-69%, from memory we had around 1-4 fail. At the end of it if you do pass you get a CSWA certification which you can put on your resume.
Exam
For us, the in semester work was worth 70%, so the majority of us had passed before we had even sat the exam (it's a nice feeling). The proportional of marks changes each year, but it should be around there. The Exam will be mostly on Detailed Drawings, Assembly Drawings to the Australian Standard AS1100, Casting and Manufacturing Methods. If anything, Detailed drawings will be the most important topic for the exam, so make sure you learn that properly. While it's quite easy to get some marks on these, it's really easy to lose marks on them as well. Small, simple things that you overlook will cost you marks, not putting a border around the drawing, not including the projection system in the Title Block, over-dimensioning the drawing or using too many views to represent something that can be done in less views. You'll need to know how to do this later in industry though, so it's good practice.
EDIT: I'll put the images in spoilers to make the post not as long
« Last Edit: June 10, 2014, 03:49:53 pm by b^3 »
2012-2016: Aerospace Engineering/Science (Double Major in Applied Mathematics - Monash Uni)
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#### keltingmeith
• Honorary Moderator
• ATAR Notes Legend
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##### Re: Monash University - Subject Reviews & Ratings
« Reply #193 on: June 11, 2014, 08:57:56 pm »
+4
Subject Code/Name: MTH1035 Techniques for Modelling (Advanced)
Workload: 3x1 hour lectures, 1x1 hour tutes and 1x2 hour workshops (however, from next year Simon wants to change this to two hour tutes. This is already in place for MTH2015, which is the follow-on unit)
Assessment: 3 assignments, all 10%. 1 "mid semester test" worth 10%, you'll do it in about week 10 (yeah, 10/12 is about half-way) and finally the exam which was worth 60%. However, this exam will change to 70% from next year (as it will in all maths units. So glad I decided to do a double major...) Note: All assignments and the mid-sem test are material from MTH1030, and 80% of your exam is material from MTH1030. Only 20% of your exam is material from MTH1035.
Recorded Lectures: Yes, with screen capture for the lectures. Without for the workshops, however Simon will post up the boards, so you can see what was written anyway.
Past exams available: No, however a sample exam was made available to us.
Textbook Recommendation: Kutler's linear algebra book, which is mentioned in the notes. It's absolutely FREE. I never used it, but hey, could be good? Also stewart's early transcendental's. I glanced through it, looks alright, not necessary though. It's also available from the library. I have heard it's necessary for MTH2015/MTH2010 if you plan to continue on to that, though, so it's up to you.
Lecturer(s): MTH1035 has two sections - MTH1030 material and MTH1035 material. MTH1030 material is taught in lecturers, my MTH1030 lecturer was Burkard Polster. Famous for being a mathemagician, juggling and lecturing with lightsabers. The MTH1035 lecturer is Simon Teague - famous for always having a coke zero with him (yes, this does include in his 8 am lectures). Burkard is amazing - I don't think it's possible to hate him. Simon's not as well loved, but I quite liked him. Preferences are preferences, so eh.
Year & Semester of completion: 2014, Semester 1. Don't let the unit code or the MTH1030 parallels fool you - MTH1035 is ONLY offered in semester 1.
Rating: 4 out of 5
Comments: Before you sign up for this subject, realise this: you are not good at maths. In all seriousness though, the biggest thing I learnt in this unit is that what you got in year 12 does not reflect how you will do at uni. Throughout the year, I was doing much better than people who did way better than me in year 12. If you struggle, this is normal, don't worry - this unit is very different. So, onto the actual course:
Linear Algebra
You start off with brief revision of year twelve - what's a vector, what can you do with a vector. Then you move on to some more things, including the cross-product. You'll look at vector spaces in R^n, even though you'll only do most of your calculations in R^3 and then just do some conceptual things in R^n. After you do this stuff, you'll look at how to make lines and planes, and this stuff is quite possibly the most annoying things you'll ever work with. You'll follow this with systems of linear equations, which is actually just extensions on methods stuff, believe it or not. Next is simple matrix stuff - arithmetic, determinants, inverses, that fun stuff, followed by using matrices to form linear transformations on vectors. You'll then move onto subspaces (generally focusing on R^4 for some reason...) and finally eigenvalues and eigenvectors. Those are funny words, and you won't know what they are until much later, don't worry about that. None of any of this is particularly hard if you do the tute sheets, so do the tute sheets, you'll be fine.
The only stuff you do in 1035 that really sticks out in this section is quaternions and tensors - neither of which ACTUALLY make sense. Simon will tell you which of these are on your exam, so when he tells you, do some reading and do the questions he gives you, and hopefully you'll pick up marks. If you do well on the assignments (which you should), you should be fine.
Calculus
When I say calculus, it's not calculus like you think calculus from high school. In fact, the elementary functions you remember from high school only really come up in the last week and a half.
You start off thinking about limits - how to compute some basic limits, some more annoying limits, and just sort of what a limit is. In the 1035 workshops, you'll also look at the epsilon-delta definition of a limit. Next up is determinate and indeterminate forms, and how we find an indeterminate form using L'Hopital's Rule. Then, you move on to sequences and series - yes, they're a thing. First you find how to work with sequences, then the more important series. You'll learn how to work with some general types - like telescoping, geometric, harmonic, etc. You'll learn how to find if a series converges, diverges, and a bunch of other things. This then leads into one of the bigger types of power series - Taylor series, and its special partner Maclaurin series. This stuff is actually really cool, and can be used to prove Euler's identity (which is how I chose my name ). After all this series stuff, you finally move on to integration. You'll learn integration by parts, finishing up your integrating techniques repertoire. Then, you'll learn a few more DE solving techniques - seperation of variables, the integrating factor and using eigenvalues to solve second order homogenous DEs, and that's the course.
Not really anything special in 1035 - Simon will tell you what's in the exam for 1035, just expect something hard, and hope you can do it when you get to the exam. I can tell you that for our calculus question, not very many could...
« Last Edit: July 13, 2014, 05:47:03 pm by EulerFan101 »
Currently Undertaking: Doctor of Philosophy (PhD) in Supramolecular Photochemistry (things that don't bond but they do and glow pretty colours)
Previous Study: Bachelor of Science Advanced (Research) - Monash University, majoring in Mathematical Statistics and Chemistry
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#### vashappenin
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##### Re: Monash University - Subject Reviews & Ratings
« Reply #194 on: June 12, 2014, 11:10:30 pm »
+5
Subject Code/Name: LAW2101 - Contract A
Workload: 2 x 1.5 hour lectures per week, 1 hour tutorial per week from week 6-11 (attendance isn't compulsory)
Assessment: 20% Optional written assignment and 80% Exam, OR 100% exam (for those who opt against optional assignment)
Recorded Lectures: Yes, chief examiner's lectures are recorded with screen capture.
Past exams available: Yes, 2011-13 plus a few more older ones
Textbook Recommendation:
• Principles book: Jeannie Paterson, Andrew Robertson and Arlen Duke, Principles of Contract Law (Lawbook Co/Thomson Reuters, 4th edition, 2012)
• Case book: Jeannie Paterson, Andrew Robertson and Arlen Duke, Contract: Cases and Materials (Lawbook Co/Thomson Reuters, 12th edition, 2012)
• You DEFINITELY need both, and they're to be use for both Contract A and Contract B
Lecturer(s): There are different lecturers for each stream and it usually differs slightly every year. Five streams this year: Jennifer Paneth (my lecturer), Emmanuel Laryea (chief examiner), Rowena Cantley-Smith, Lisa Di Marco, Sirko Harder
Year & Semester of completion: Semester 1, 2014
Rating: 3.5 out of 5
Overall, Contract A has been okay. Being most law students' first proper law unit, it's a bit overwhelming and you'll feel pretty lost for a while towards the beginning. The content itself was mostly quite interesting, although there were weeks where it was extremely dull. This unit covers the following topics:
• Elements of a contract: Agreement (Offer and Acceptance of contract), Consideration, Intention to Create Legal Relations, Certainty
• Formalities of contract formation
• Capacity to enter into a contract
• Contract terms: Express terms and Implied terms
• Estoppel
• Privity
• Consumer contracts (Unfair Contract Terms and Consumer Guarantees as per Australian Consumer Law)
I can't really 'recommend' this unit since it's compulsory, but thought I'd review it anyway because I know I was definitely looking for subject reviews on this unit when I started uni. So yes, regardless of whether you like it or not, you kinda have to suck it up because it's one of the 'Priestley 11' law units (i.e. it's compulsory so you can't escape it!). Also, there's a second part that you do in Semester 2 (Contract B), for which this is a prerequisite so make sure you pass!
Lectures
DEFINITELY do attend lectures. At times, contracts gets kind of dry but trust me, it's so much more beneficial that you don't turn lazy and just force yourself to attend, because your future self in week 12 + SWOTVAC will really, really thank you (I say this from experience). You could listen to them online but you're probably going to get lazy. Just make your life easier and attend them.
Also, the lectures really come in handy in helping you understand the cases and how they reflect legal principles because this can be difficult at time, so don't miss your lectures!!
It doesn't really matter who your lecturer is, although it IS important that you attend the lectures for YOUR stream, because you need to know what YOUR lecturer wants. I remember stressing because I really wanted to ensure I had the chief examiner, but it honestly doesn't matter. My lecturer, Paneth, was really helpful and she provided a lot of information for us. She'd go through revision questions at the end of each topic and this was SO useful; something you won't get from trying to self-learn the slides at home.
Tutes
Tutes aren't compulsory so obviously not a lot of people attended. I only attended a few myself and I really regretted it because the tutes were pretty much going through different questions that covered each topic, with the last week's tute being a run-through of a past exam. The tutors are very knowledgeable as well and provide a lot of handy tips so even though you don't have to, I'd recommend that you do go out of your way to attend. It'll be really helpful in the end, because you'll find that you know what you're doing and you've already applied your knowledge through problem questions. Not to mention you can ask for feedback on questions as well.
Exam
The exam is out of 100 and goes for 2 hours, plus 30 minutes noting/highlighting at the beginning. Honestly, it's like a race. In order to satisfy the marks you're pretty much writing as fast as you can until the exam finishes. What you should really practice is issue-spotting, which is why the tutes are useful, because you get practice at this and feedback as well. This was the first time they did it for Contract A, but they also had a 20-mark case question. Here, you're pretty much regurgitating the facts, issues and judgments of a select case, chosen out of 100+ odd cases (which, yes, you're expected to have studied/memorized, along with their related principles).
« Last Edit: June 13, 2014, 10:06:10 am by vashappenin »
2013: English, Maths Methods, Further Maths, Legal Studies, HHD, Psychology
2014-present: Bachelor of Laws @ Monash University
Tutoring VCE English, Psych, Legal Studies and HHD in 2016! Tutoring via Skype too. PM me if you're interested |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Electrostatics
## Electrons are the fundamental unit of charge.
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Electrostatics
Opposite charges attract and like charges repulse. The electron (and proton) is the fundamental charge unit. The charge of an electron and proton is \begin{align*} 1.6 \times 10^{-19} \end{align*} C. One can determine the number of excess electrons (or protons if positive charge) by dividing the objects charge by the fundamental charge. Most objects are electrically neutral (equal numbers of electrons and protons) and that's why gravity dominates on a macro scale.
Key Equations
\begin{align*}q = Ne \text{ }\end{align*} Any object's charge is an integer multiple of an electron's charge.
#### Example
If an object has +0.003 C of charge, how many excess protons does the object have?
\begin{align*}q = Ne \text{ }\end{align*} \begin{align*} 0.003 \text{C} = N \times 1.6 \times 10^{-19} \end{align*} N = \begin{align*} 1.875 \times 10^{16} \end{align*} ; protons
### Review
1. After sliding your feet across the rug, you touch the sink faucet and get shocked. Explain what is happening.
2. What is the net charge of the universe? Of your toaster?
3. As you slide your feet along the carpet, you pick up a net charge of \begin{align*}+4 \;\mathrm{mC}\end{align*}. Which of the following is true?
1. You have an excess of \begin{align*}2.5 \times 10^{16}\end{align*} electrons
2. You have an excess of \begin{align*}2.5 \times 10^{19}\end{align*} electrons
3. You have an excess of \begin{align*}2.5 \times 10^{16}\end{align*} protons
4. You have an excess of \begin{align*}2.5 \times 10^{19}\end{align*} protons
4. You rub a glass rod with a piece of fur. If the rod now has a charge of \begin{align*}-0.6\ \mu C\end{align*}, how many electrons have been added to the rod?
1. \begin{align*}3.75 \times 10^{18}\end{align*}
2. \begin{align*}3.75 \times 10^{12}\end{align*}
3. \begin{align*}6000\end{align*}
4. \begin{align*}6.00 \times 10^{12}\end{align*}
5. Not enough information
1. The net electric charge you collected from the rug gets discharged when you come in contact with the sink faucet.
2. Neutral
3. c
4. b
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes |
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Natural number Reciprocal Decimal expansion Nature of decimal expansion $$2$$ $$1/2$$ $\frac{1}{2}=0.5$ Terminating decimal expansion $$3$$ $$1/3$$ $\begin{array}{l}\frac{1}{3}=0.33333...\\ =0.\overline{3}\end{array}$ Non-terminating and recurring decimal expansion $$4$$ $$1/4$$ $\frac{1}{4}=0.25$ Terminating decimal expansion $$5$$ $$1/5$$ $\frac{1}{5}=0.2$ Terminating decimal expansion $$6$$ $$1/6$$ $\begin{array}{l}\frac{1}{6}=0.16666...\\ =0.1\overline{6}\end{array}$ Non-terminating and recurring decimal expansion $$7$$ $$1/7$$ $\begin{array}{l}\frac{1}{7}=0.142857142857...\\ =0.\overline{142857}\end{array}$ Non-terminating and recurring decimal expansion $$8$$ $$1/8$$ $\frac{1}{8}=0.125$ Terminating decimal expansion $$9$$ $$1/9$$ $\begin{array}{l}\frac{1}{9}=0.11111...\\ =0.\overline{1}\end{array}$ Non-terminating and recurring decimal expansion $$10$$ $$1/10$$ $\frac{1}{10}=0.1$ Terminating decimal expansion $$11$$ $$1/11$$ $\begin{array}{l}\frac{1}{11}=0.090909...\\ =0.\overline{09}\end{array}$ Non-terminating and recurring decimal expansion $$12$$ $$1/12$$ $\begin{array}{l}\frac{1}{12}=0.083333...\\ =0.08\overline{3}\end{array}$ Non-terminating and recurring decimal expansion $$13$$ $$1/13$$ $\begin{array}{l}\frac{1}{13}=0.076923076923...\\ =0.\overline{076923}\end{array}$ Non-terminating and recurring decimal expansion $$14$$ $$1/14$$ $\begin{array}{l}\frac{1}{14}=0.0714285714285...\\ =0\overline{714285}\end{array}$ Non-terminating and recurring decimal expansion $$15$$ $$1/15$$ $\begin{array}{l}\frac{1}{15}=0.06666...\\ =0\overline{6}\end{array}$ Non-terminating and recurring decimal expansion $$16$$ $$1/16$$ $\frac{1}{16}=0.0625$ Terminating decimal expansion $$17$$ $$1/17$$ $\begin{array}{l}\frac{1}{17}=0.588235294117647...\\ =0.\overline{0588235294117647}\end{array}$ Non-terminating and recurring decimal expansion $$18$$ $$1/18$$ $\begin{array}{l}\frac{1}{18}=0.0555555...\\ =0\overline{5}\end{array}$ Non-terminating and recurring decimal expansion $$19$$ $$1/19$$ $\begin{array}{l}\frac{1}{19}=0.052631578947368421...\\ =0.\overline{052631578947368421}\end{array}$ Non-terminating and recurring decimal expansion $$20$$ $$1/20$$ $\frac{1}{20}=0.05$ Terminating decimal expansion $$21$$ $$1/21$$ $\frac{1}{21}=0.\overline{047619}$ Non-terminating and recurring decimal expansion $$22$$ $$1/22$$ $\frac{1}{22}=0\overline{45}$ Non-terminating and recurring decimal expansion $$23$$ $$1/23$$ $\begin{array}{l}\frac{1}{23}=0.0434782608695652173913...\\ =\overline{0434782608695652173913}\end{array}$ Non-terminating and recurring decimal expansion $$24$$ $$1/24$$ $\begin{array}{l}\frac{1}{24}=0.041666...\\ =0.041\overline{6}\end{array}$ Non-terminating and recurring decimal expansion $$25$$ $$1/25$$ $\frac{1}{25}=0.04$ Terminating decimal expansion |
# Limit of series in probability using central limit theorem
Using central limit theorem , evaluate
$\lim_{n\to \infty}\sum_{j=0}^{n}{j+n-1 \choose j}(\frac{1}{2^{n+j}})$,
I multiplied and divided the series by $1/2$ , And made it look like a binomial distribution ,but they are not i.i.d., which is why I cannot apply CLT.
• I think you are confusing distributions and random variables – seanv507 Oct 31 '17 at 16:46
• Then how shall i proceed? – DRPR Oct 31 '17 at 16:47
• For a given $n$, interpret the sum as the probability of an event related to a series of coin-flipping experiments. See en.wikipedia.org/wiki/Negative_binomial_distribution for some suggestive formulas. The limit will then be the chance of a particular event associated with a standard Normal variable. – whuber Oct 31 '17 at 16:55
Suppose you flip a fair coin repeatedly. What is the probability that your $n$th heads appears on the $(n+j)$th flip? Knowing this, can you write your sum as the probability of some event?
Your sum is the probability of getting [at least] $n$ heads in $2n$ flips. |
# One World Numeration Seminar
This is an online seminar on numeration systems and related topics (see the series of Numeration conferences), in the spirit of other One World Seminars; talks are on Zoom. If you want to participate in the seminar, please contact the organisers (Shigeki Akiyama, Karma Dajani, Kevin Hare, Hajime Kaneko, Niels Langeveld, Lingmin Liao, Wolfgang Steiner) by email to numeration@irif.fr.
January 31, 2023, 14:00 CET (UTC +1)
Slade Sanderson (Universiteit Utrecht): Matching for parameterised symmetric golden maps (paper)
In 2020, Dajani and Kalle investigated invariant measures and frequencies of digits of signed binary expansions arising from a parameterised family of piecewise linear interval maps of constant slope 2. Central to their study was a property called ‘matching’, where the orbits of the left and right limits of discontinuity points agree after some finite number of steps. We obtain analogous results for a parameterised family of ‘symmetric golden maps’ of constant slope β, with β the golden mean. Matching is again central to our methods, though the dynamics of the symmetric golden maps are more delicate than the binary case. We characterize the matching phenomenon in our setting, present explicit invariant measures and frequencies of digits of signed β-expansions, and---time permitting---show further implications for a family of piecewise linear maps which arise as jump transformations of the symmetric golden maps.
Joint with Karma Dajani.
March 7, 2023, 14:00 CET (UTC +1)
Derong Kong (Chongqing University): TBA
March 21, 2023, 14:00 CET (UTC +1)
Demi Allen (University of Exeter): Diophantine Approximation for systems of linear forms - some comments on inhomogeneity, monotonicity, and primitivity (paper)
Diophantine Approximation is a branch of Number Theory in which the central theme is understanding how well real numbers can be approximated by rationals. In the most classical setting, a ψ-well-approximable number is one which can be approximated by rationals to a given degree of accuracy specified by an approximating function ψ. Khintchine's Theorem provides a beautiful characterisation of the Lebesgue measure of the set of ψ-well-approximable numbers and is one of the cornerstone results of Diophantine Approximation. In this talk I will discuss the generalisation of Khintchine's Theorem to the setting of approximation for systems of linear forms. I will focus mainly on the topic of inhomogeneous approximation for systems of linear forms. Time permitting, I may also discuss approximation for systems of linear forms subject to certain primitivity constraints. This talk will be based on joint work with Felipe Ramirez (Wesleyan, US).
## Past talks
January 24, 2023
Kiko Kawamura (University of North Texas): The partial derivative of Okamoto's functions with respect to the parameter (video) (slides) (paper)
Okamoto's functions were introduced in 2005 as a one-parameter family of self-affine functions, which are expressed by ternary expansion of x on the interval [0,1]. By changing the parameter, one can produce interesting examples: Perkins' nowhere differentiable function, Bourbaki-Katsuura function and Cantor's Devil's staircase function.
In this talk, we consider the partial derivative of Okomoto's functions with respect to the parameter a. We place a significant focus on a = 1/3 to describe the properties of a nowhere differentiable function K(x) for which the set of points of infinite derivative produces an example of a measure zero set with Hausdorff dimension 1.
This is a joint work with T. Mathis and M.Paizanis (undergraduate students) and N.Dalaklis (graduate student). The talk is very accessible and includes many computer graphics.
January 10, 2023
Roswitha Hofer (JKU Linz): Exact order of discrepancy of normal numbers (video) (slides) (paper1) (paper2)
In the talk we discuss some previous results on the discrepancy of normal numbers and consider the still open question of Korobov: What is the best possible order of discrepancy DN in N, a sequence ({bnα})n≥0, b ≥ 2 integer, can have for some real number α? If limN→∞ DN = 0 then α in called normal in base b.
So far the best upper bounds for DN for explicitly known normal numbers in base 2 are of the form NDN ≪ log2 N. The first example is due to Levin (1999), which was later generalized by Becher and Carton (2019). In this talk we discuss the recent result in joint work with Gerhard Larcher that guarantees NDN ≫ log2 N for Levin's binary normal number. So EITHER ND_N ≪ log2 N is the best possible order for DN in N of a normal number OR there exist another example of a binary normal number with a better growth of NDN in N. The recent result for Levin's normal number might support the conjecture that ND_N ≪ log2 N is the best order for DN in N a normal number can obtain.
December 13, 2022
Hiroki Takahasi (Keio University): Distribution of cycles for one-dimensional random dynamical systems (video) (slides) (paper)
We consider an independently identically distributed random dynamical system generated by finitely many, non-uniformly expanding Markov interval maps with a finite number of branches. Assuming a topologically mixing condition and the uniqueness of equilibrium state for the associated skew product map, we establish a samplewise (quenched) almost-sure level-2 weighted equidistribution of "random cycles", with respect to a natural stationary measure as the periods of the cycles tend to infinity. This result implies an analogue of Bowen's theorem on periodic orbits of topologically mixing Axiom A diffeomorphisms.
This talk is based on the preprint arXiv:2108.05522. If time permits, I will mention some future perspectives in this project.
December 6, 2022
Christoph Bandt (Universität Greifswald): Automata generated topological spaces and self-affine tilings (video) (slides)
Numeration assigns symbolic sequences as addresses to points in a space X. There are points which get multiple addresses. It is known that these identifications describe the topology of X and can often be determined by an automaton. Here we define a corresponding class of automata and discuss their properties and interesting examples. Various open questions concern the realization of such automata by iterated functions and the uniqueness of such an implementation. Self-affine tiles form a simple class of examples.
November 29, 2022
Manuel Hauke (TU Graz): The asymptotic behaviour of Sudler products (video) (slides) (paper1) (paper2) (paper3)
Given an irrational number α, we study the asymptotic behaviour of the Sudler product defined by PN(α) = Πr=1N 2 |sin(π r α)|, which appears in many different areas of mathematics. In this talk, we explain the connection between the size of PN(α) and the Ostrowski expansion of N with respect to α. We show that lim infN→∞ PN(α) = 0 and lim supN→∞ PN(α)/N = ∞, whenever the sequence of partial quotients in the continued fraction expansion of α exceeds 7 infinitely often, and show that the value 7 is optimal.
For Lebesgue-almost every α, we can prove more: we show that for every non-decreasing function ψ: (0,∞) → (0,∞) with ∑k=1 1/ψ(k) = ∞ and lim infk→∞ ψ(k)/(k log k) sufficiently large, the conditions log PN(α) ≤ −ψ(log N), log PN(α) ≥ ψ(log N) hold on sets of upper density 1 respectively 1/2.
November 22, 2022
Faustin Adiceam (Université Paris-Est Créteil): Badly approximable vectors and Littlewood-type problems (video) (slides) (paper)
Badly approximable vectors are fractal sets enjoying rich Diophantine properties. In this respect, they play a crucial role in many problems well beyond Number Theory and Fractal Geometry (e.g., in signal processing, in mathematical physics and in convex geometry).
After outlining some of the latest developments in this very active area of research, we will take an interest in the Littlewood conjecture (c. 1930) and in its variants which all admit a natural formulation in terms of properties satisfied by badly approximable vectors. We will then show how ideas emerging from the mathematical theory of quasicrystals, from numeration systems and from the theory of aperiodic tilings have recently been used to refute the so-called t-adic Littlewood conjecture.
All necessary concepts will be defined in the talk. Joint with Fred Lunnon (Maynooth) and Erez Nesharim (Technion, Haifa).
November 15, 2022
Seul Bee Lee (Institute for Basic Science): Regularity properties of Brjuno functions associated with by-excess, odd and even continued fractions (video) (slides) (paper)
An irrational number is called a Brjuno number if the sum of the series of log(qn+1)/qn converges, where qn is the denominator of the n-th principal convergent of the regular continued fraction. The importance of Brjuno numbers comes from the study of one variable analytic small divisor problems. In 1988, J.-C. Yoccoz introduced the Brjuno function which characterizes the Brjuno numbers to estimate the size of Siegel disks. In this talk, we introduce Brjuno-type functions associated with by-excess, odd and even continued fractions with a number theoretical motivation. Then we discuss the Lp and the Hölder regularity properties of the difference between the classical Brjuno function and the Brjuno-type functions. This is joint work with Stefano Marmi.
November 8, 2022
Wen Wu (South China University of Technology): From the Thue-Morse sequence to the apwenian sequences (video) (slides) (journal) (arXiv)
In this talk, we will introduce a class of ±1 sequences, called the apwenian sequences. The Hankel determinants of these ±1 sequences share the same property as the Hankel determinants of the Thue-Morse sequence found by Allouche, Peyrière, Wen and Wen in 1998. In particular, the Hankel determinants of apwenian sequences do not vanish. This allows us to discuss the Diophantine property of the values of their generating functions at 1/b where b ≥ 2 is an integer. Moreover, the number of ±1 apwenian sequences is given explicitly. Similar questions are also discussed for 0-1 apwenian sequences. This talk is based on joint work with Y.-J. Guo and G.-N. Han.
October 25, 2022
Álvaro Bustos-Gajardo (The Open University): Quasi-recognizability and continuous eigenvalues of torsion-free S-adic systems (video) (slides) (paper)
We discuss combinatorial and dynamical descriptions of S-adic systems generated by sequences of constant-length morphisms between alphabets of bounded size. For this purpose, we introduce the notion of quasi-recognisability, a strictly weaker version of recognisability but which is indeed enough to reconstruct several classical arguments of the theory of constant-length substitutions in this more general context. Furthermore, we identify a large family of directive sequences, which we call "torsion-free", for which quasi-recognisability is obtained naturally, and can be improved to actual recognisability with relative ease.
Using these notions we give S-adic analogues of the notions of column number and height for substitutions, including dynamical and combinatorial interpretations of each, and give a general characterisation of the maximal equicontinuous factor of the identified family of S-adic shifts, showing as a consequence that in this context all continuous eigenvalues must be rational. As well, we employ the tools developed for a first approach to the measurable case.
This is a joint work with Neil Mañibo and Reem Yassawi.
October 18, 2022
Yufei Chen (TU Delft): Matching of orbits of certain N-expansions with a finite set of digits (video) (slides) (paper)
In this talk we consider a class of continued fraction expansions: the so-called N-expansions with a finite digit set, where N ≥ 2 is an integer. For N fixed, they are steered by a parameter α ∈ (0,√N−1]. For N = 2 an explicit interval [A,B] was determined, such that for all α ∈ [A,B] the entropy h(Tα) of the underlying Gauss-map Tα is equal. In this paper we show that for all integers N ≥ 2, such plateaux exist. In order to show that the entropy is constant on such plateaux, we obtain the underlying planar natural extension of the maps Tα, the Tα-invariant measure, ergodicity, and we show that for any two α, α' from the same plateau, the natural extensions are metrically isomorphic, and the isomorphism is given explicitly. The plateaux are found by a property called matching.
October 11, 2022
Lukas Spiegelhofer (Montanuniversität Leoben): Primes as sums of Fibonacci numbers (video) (slides) (paper)
We prove that the Zeckendorf sum-of-digits function of prime numbers, z(p), is uniformly distributed in residue classes. The main ingredient that made this proof possible is the study of very sparse arithmetic subsequences of z(n). In other words, we will meet the level of distribution. Our proof of this central result is based on a combination of the "Mauduit−Rivat−van der Corput method" for digital problems and an estimate of a Gowers norm related to z(n). Our method of proof yields examples of substitutive sequences that are orthogonal to the Möbius function (cf. Sarnak's conjecture).
This is joint work with Michael Drmota and Clemens Müllner (TU Wien).
October 4, 2022
David Siukaev (Higher School of Economics): Exactness and ergodicity of certain Markovian multidimensional fraction algorithms (video) (slides)
A multidimensional continued fraction algorithm is a generalization of well-known continued fraction algorithms of small dimensions: Gauss and Euclidean. Ergodic properties of Markov MCF algorithms (ergodicity, nonsingularity, exactness, bi-measurability) affect their convergence (if the MСF algorithm is a Markov algorithm, there is a relationship between the spectral properties and its convergence).
In 2013 T. Miernowski and A. Nogueira proved that the Euclidean algorithm and the non-homogeneous Rauzy induction satisfy the intersection property and, as a consequence, are exact. At the end of the article it is stated that other non-homogeneous markovian algorithms (Selmer, Brun and Jacobi-Perron) also satisfy the intersection property and they also exact. However, there is no proof of this. In our paper this proof is obtained by using the structure of the proof of the exactness of the Euclidean algorithm with its generalization and refinement for multidimensional algorithms. We obtained technically complex proofs that differ from the proofs given in the article of T. Miernowski and A. Nogueira by the difficulties of generalization to the multidimensional case.
October 4, 2022
Alexandra Skripchenko (Higher School of Economics): Bruin-Troubetzkoy family of interval translation mappings: a new glance (video)
In 2002 H. Bruin and S. Troubetzkoy described a special class of interval translation mappings on three intervals. They showed that in this class the typical ITM could be reduced to an interval exchange transformations. They also proved that generic ITM of their class that can not be reduced to IET is uniquely ergodic.
We suggest an alternative proof of the first statement and get a stronger version of the second one. It is a joint work in progress with Mauro Artigiani and Pascal Hubert.
September 27, 2022
Niels Langeveld (Montanuniversität Leoben): N-continued fractions and S-adic sequences (slides) (paper) (video)
Given the N-continued fraction of a number x, we construct N-continued fraction sequences in the same spirit as Sturmian sequences can be constructed from regular continued fractions. These sequences are infinite words over a two letter alphabet obtained as the limit of a directive sequence of certain substitutions (they are S-adic sequences). By viewing them as a generalisation of Sturmian sequences it is natural to study balancedness. We will see that the sequences we construct are not 1-balanced but C-balanced for C = N2. Furthermore, we construct a dual sequence which is related to the natural extension of the N-continued fraction algorithm. This talk is joint work with Lucía Rossi and Jörg Thuswaldner.
September 13, 2022
Benedict Sewell (Alfréd Rényi Institute): An upper bound on the box-counting dimension of the Rauzy gasket (video) (slides) (paper)
The Rauzy gasket is a subset of the standard two-simplex, and an important subset of parameter space in various settings. It is a parabolic, non-conformal fractal attractor; meaning that even the most trivial upper bounds on its Hausdorff or box-counting dimensions are hard to obtain. In this talk (featuring joint work with Mark Pollicott), we discuss how an elementary method leads to the best known upper bound on these dimensions.
July 12, 2022
Ruofan Li (South China University of Technology): Rational numbers in ×b-invariant sets (video) (slides) (paper)
Let b ≥ 2 be an integer and S be a finite non-empty set of primes not containing divisors of b. For any ×b-invariant, non-dense subset A of [0,1), we prove the finiteness of rational numbers in A whose denominators can only be divided by primes in S. A quantitative result on the largest prime divisors of the denominators of rational numbers in A is also obtained.
This is joint work with Bing Li and Yufeng Wu.
July 5, 2022
Charlene Kalle (Universiteit Leiden): Random Lüroth expansions (video) (slides) (journal) (arXiv)
Since the introduction of Lüroth expansions by Lüroth in his paper from 1883 many results have appeared on their approximation properties. In 1990 Kalpazidou, Knopfmacher and Knopfmacher introduced alternating Lüroth expansions and studied their properties. A comparison between the two and other comparable number systems was then given by Barrionuevo, Burton, Dajani and Kraaikamp in 1996. In this talk we introduce a family of random dynamical systems that produce many Lüroth type expansions at once. Topics that we consider are periodic expansions, universal expansions, speed of convergence and approximation coefficients. This talk is based on joint work with Marta Maggioni.
June 21, 2022
James A. Yorke (University of Maryland): Large and Small Chaos Models (video) (slides)
To set the scene, I will discuss one large model, a whole-Earth model for predicting the weather, and how to initialize such a model and what aspects of chaos are essential. Then I will discuss a couple related “very simple” maps that tell us a great deal about very complex models. The results on simple models are new. I will discuss the logistic map mx(1-x). Its dynamics can make us rethink climate models. Also, we have created a piecewise linear map on a 3D cube that is unstable in 2 dimensions in some places and unstable in 1 in others. It has a dense set of periodic points that are 1 D unstable and another dense set of periodic points that are all 2 D unstable. I will also discuss a new project whose tentative title is “ Can the flap of butterfly's wings shift a tornado out of Texas — without chaos?
June 7, 2022
Sophie Morier-Genoud (Université Reims Champagne Ardenne): q-analogues of real numbers (video) (paper1) (paper2) (paper3) (paper4)
Classical sequences of numbers often lead to interesting q-analogues. The most popular among them are certainly the q-integers and the q-binomial coefficients which both appear in various areas of mathematics and physics. With Valentin Ovsienko we recently suggested a notion of q-rationals based on combinatorial properties and continued fraction expansions. The definition of q-rationals naturally extends the one of q-integers and leads to a ratio of polynomials with positive integer coefficients. I will explain the construction and give the main properties. In particular I will briefly mention connections with the combinatorics of posets, cluster algebras, Jones polynomials, homological algebra. Finally I will also present further developments of the theory, leading to the notion of q-irrationals and q-unimodular matrices.
May 31, 2022
Verónica Becher (Universidad de Buenos Aires & CONICET Argentina): Poisson generic real numbers (slides) (paper)
Years ago Zeev Rudnick defined the Poisson generic real numbers as those where the number of occurrences of the long strings in the initial segments of their fractional expansions in some base have the Poisson distribution. Yuval Peres and Benjamin Weiss proved that almost all real numbers, with respect to Lebesgue measure, are Poisson generic. They also showed that Poisson genericity implies Borel normality but the two notions do not coincide, witnessed by the famous Champernowne constant. We recently showed that there are computable Poisson generic real numbers and that all Martin-Löf real numbers are Poisson generic.
This is joint work Nicolás Álvarez and Martín Mereb.
May 24, 2022
Émilie Charlier (Université de Liège): Spectrum, algebraicity and normalization in alternate bases (video) (slides) (paper)
The first aim of this work is to give information about the algebraic properties of alternate bases determining sofic systems. We exhibit two conditions: one necessary and one sufficient. Comparing the setting of alternate bases to that of one real base, these conditions exhibit a new phenomenon: the bases should be expressible as rational functions of their product. The second aim is to provide an analogue of Frougny's result concerning normalization of real bases representations. Under some suitable condition (i.e., our previous sufficient condition for being a sofic system), we prove that the normalization function is computable by a finite Büchi automaton, and furthermore, we effectively construct such an automaton. An important tool in our study is the spectrum of numeration systems associated with alternate bases. For our purposes, we use a generalized concept of spectrum associated with a complex base and complex digits, and we study its topological properties.
This is joint work with Célia Cisternino, Zuzana Masáková and Edita Pelantová.
May 17, 2022
Vilmos Komornik (Shenzhen University and Université de Strasbourg): Topology of univoque sets in real base expansions (video) (slides) (paper)
We report on a recent joint paper with Martijn de Vries and Paola Loreti. Given a positive integer M and a real number 1 < q ≤ M+1, an expansion of a real number x ∈ [0,M/(q-1)] over the alphabet A = {0,1,...,M} is a sequence (ci) ∈ AN such that x = Σk=1 ci q-i. Generalizing many earlier results, we investigate the topological properties of the set Uq consisting of numbers x having a unique expansion of this form, and the combinatorial properties of the set U'q consisting of their corresponding expansions.
May 3, 2022
Nicolas Chevallier (Université de Haute Alsace): Best Diophantine approximations in the complex plane with Gaussian integers (video) (slides) (journal) (arXiv)
Starting with the minimal vectors in lattices over Gaussian integers in C2, we define a algorithm that finds the sequence of minimal vectors of any unimodular lattice in C2. Restricted to lattices associated with complex numbers this algorithm find all the best Diophantine approximations of a complex numbers. Following Doeblin, Lenstra, Bosma, Jager and Wiedijk, we study the limit distribution of the sequence of products (un1un2)n where (un = (un1,un2))n is the sequence of minimal vectors of a lattice in C2. We show that there exists a measure in C which is the limit distribution of the sequence of products of almost all unimodular lattices.
April 19, 2022
Paulina Cecchi Bernales (Universidad de Chile): Coboundaries and eigenvalues of finitary S-adic systems (video) (slides) (paper)
An S-adic system is a shift space obtained by performing an infinite composition of morphisms defined over possibly different finite alphabets. It is said to be finitary if these morphisms are taken from a finite set. S-adic systems are a generalization of substitution shifts. In this talk we will discuss spectral properties of finitary S-adic systems. Our departure point will be a theorem by B. Host which characterizes eigenvalues of substitution shifts, and where coboundaries appear as a key tool. We will introduce the notion of S-adic coboundaries and present some results which show how they are related with eigenvalues of S-adic systems. We will also present some applications of our results to constant-length finitary S-adic systems.
This is joint work with Valérie Berthé and Reem Yassawi.
April 12, 2022
Eda Cesaratto (Univ. Nac. de Gral. Sarmiento & CONICET, Argentina): Lochs-type theorems beyond positive entropy (video) (slides) (paper)
Lochs' theorem and its generalizations are conversion theorems that relate the number of digits determined in one expansion of a real number as a function of the number of digits given in some other expansion. In its original version, Lochs' theorem related decimal expansions with continued fraction expansions. Such conversion results can also be stated for sequences of interval partitions under suitable assumptions, with results holding almost everywhere, or in measure, involving the entropy. This is the viewpoint we develop here. In order to deal with sequences of partitions beyond positive entropy, this paper introduces the notion of log-balanced sequences of partitions, together with their weight functions. These are sequences of interval partitions such that the logarithms of the measures of their intervals at each depth are roughly the same. We then state Lochs-type theorems which work even in the case of zero entropy, in particular for several important log-balanced sequences of partitions of a number-theoretic nature.
This is joint work with Valérie Berthé (IRIF), Pablo Rotondo (U. Gustave Eiffel) and Martín Safe (Univ. Nac. del Sur & CONICET, Argentina).
April 5, 2022
Jungwon Lee (University of Warwick): Dynamics of Ostrowski skew-product: Limit laws and Hausdorff dimensions (video) (slides) (paper)
We discuss a dynamical study of the Ostrowski skew-product map in the context of inhomogeneous Diophantine approximation. We plan to outline the setup/ strategy based on transfer operator analysis and applications in arithmetic of number fields (joint with Valérie Berthé).
March 29, 2022
Tingyu Zhang (East China Normal University): Random β-transformation on fat Sierpiński gasket (video) (slides) (paper)
We define the notions of greedy, lazy and random transformations on fat Sierpiński gasket. We determine the bases, for which the system has a unique measure of maximal entropy and an invariant measure of product type, with one coordinate being absolutely continuous with respect to Lebesgue measure.
This is joint work with K. Dajani and W. Li.
March 15, 2022
Pierre Popoli (Université de Lorraine): Maximum order complexity for some automatic and morphic sequences along polynomial values (video) (slides) (paper1) (paper2)
Automatic sequences are not suitable sequences for cryptographic applications since both their subword complexity and their expansion complexity are small, and their correlation measure of order 2 is large. These sequences are highly predictable despite having a large maximum order complexity. However, recent results show that polynomial subsequences of automatic sequences, such as the Thue-Morse sequence or the Rudin-Shapiro sequence, are better candidates for pseudorandom sequences. A natural generalization of automatic sequences are morphic sequences, given by a fixed point of a prolongeable morphism that is not necessarily uniform. In this talk, I will present my results on lowers bounds for the maximum order complexity of the Thue-Morse sequence, the Rudin-Shapiro sequence and the sum of digits function in Zeckendorf base, which are respectively automatics and morphic sequences.
March 8, 2022
Michael Coons (Universität Bielefeld): A spectral theory of regular sequences (video) (slides) (paper)
A few years ago, Michael Baake and I introduced a probability measure associated to Stern’s diatomic sequence, an example of a regular sequence—sequences which generalise constant length substitutions to infinite alphabets. In this talk, I will discuss extensions of these results to more general regular sequences as well as further properties of these measures. This is joint work with several people, including Michael Baake, James Evans, Zachary Groth and Neil Manibo.
March 1, 2022
Daniel Krenn (Universität Salzburg): k-regular sequences: Asymptotics and Decidability (video) (slides) (paper1) (paper2)
A sequence x(n) is called k-regular, if the set of subsequences x(kjn+r) is contained in a finitely generated module. In this talk, we will consider the asymptotic growth of k-regular sequences. When is it possible to compute it? ...and when not? If possible, how precisely can we compute it? If not, is it just a lack of methods or are the underlying decision questions recursively solvable (i.e., decidable in a computational sense)? We will discuss answers to these questions. To round off the picture, we will consider further decidability questions around k-regular sequences and the subclass of k-automatic sequences.
This is based on joint works with Clemens Heuberger and with Jeffrey Shallit.
February 15, 2022
Wolfgang Steiner (CNRS, Université de Paris): Unique double base expansions (video) (slides)
For pairs of real bases 𝛽0,𝛽1>1, we study expansions of the form Σk=1 ik / (𝛽i1 𝛽i2 ... 𝛽ik) with digits ik ∈ {0,1}. We characterise the pairs admitting non-trivial unique expansions as well as those admitting uncountably many unique expansions, extending recent results of Neunhäuserer (2021) and Zou, Komornik and Lu (2021). Similarly to the study of unique 𝛽-expansions with three digits by the speaker (2020), this boils down to determining the cardinality of binary shifts defined by lexicographic inequalities. Labarca and Moreira (2006) characterised when such a shift is empty, at most countable or uncountable, depending on the position of the lower and upper bounds with respect to Thue-Morse-Sturmian words.
This is joint work with Vilmos Komornik and Yuru Zou.
February 8, 2022
Magdaléna Tinková (České vysoké učení technické v Praze): Universal quadratic forms, small norms and traces in families of number fields (video) (slides) (paper)
In this talk, we will discuss universal quadratic forms over number fields and their connection with additively indecomposable integers. In particular, we will focus on Shanks' family of the simplest cubic fields. This is joint work with Vítězslav Kala.
February 1, 2022
Jonas Jankauskas (Vilniaus universitetas): Digit systems with rational base matrix over lattices (video) (slides) (paper)
Let A be a matrix with rational entries and no eigenvalue in absolute value smaller than 1. Let Zd[A] be the minimal A-invariant Z-module, generated by integer vectors and the matrix A. In 2018, we have shown that one can find a finite set D of vectors, such that each element of Zd[A] has a finite radix expansion in base A using only the digits from D, i.e. Zd[A]=D[A]. This is called 'the finiteness property' of a digit system. In the present talk I will review more recent developments in mathematical machinery, that enable us to build finite digit systems over lattices using reasonably small digit sets, and even to do some practical computations with them on a computer. Tools that we use are the generalized rotation bases with digit sets that have 'good' convex properties, the semi-direct ('twisted') sums of such rotational digit systems, and the special, 'restricted' version of the remainder division that preserves the lattice Zd and can be extended to Zd[A]. This is joint work with J. Thuswaldner, "Rational Matrix Digit Systems", to appear in "Linear and Multilinear Algebra".
January 25, 2022
Claudio Bonanno (Università di Pisa): Infinite ergodic theory and a tree of rational pairs (video) (slides) (paper)
The study of the continued fraction expansions of real numbers by ergodic methods is now a classical and well-known part of the theory of dynamical systems. Less is known for the multi-dimensional expansions. I will present an ergodic approach to a two-dimensional continued fraction algorithm introduced by T. Garrity, and show how to get a complete tree of rational pairs by using the Farey sum of fractions. The talk is based on joint work with A. Del Vigna and S. Munday.
January 18, 2022
Agamemnon Zafeiropoulos (Norges teknisk-naturvitenskapelige universitet): The order of magnitude of Sudler products (video) (slides) (paper1) (paper2)
Given an irrational α, we define the corresponding Sudler product by PN(α) = Πn=1N 2 |sin(π n α)|. In joint work with C. Aistleitner and N. Technau, we show that when α = [0;b,b,b,…] is a quadratic irrational with all partial quotients in its continued fraction expansion equal to some integer b, the following hold:
- If b ≤ 5, then lim infN→∞ PN(α) > 0 and lim supN→∞ PN(α)/N < ∞.
- If b ≥ 6, then lim infN→∞ PN(α) = 0 and lim supN→∞ PN(α)/N = ∞.
We also present an analogue of the previous result for arbitrary quadratic irrationals (joint work with S. Grepstad and M. Neumüller).
January 11, 2022
Philipp Gohlke (Universität Bielefeld): Zero measure spectrum for multi-frequency Schrödinger operators (video) (slides) (paper)
Cantor spectrum of zero Lebesgue measure is a striking feature of Schrödinger operators associated with certain models of aperiodic order, like primitive substitution systems or Sturmian subshifts. This is known to follow from a condition introduced by Boshernitzan that establishes that on infinitely many scales words of the same length appear with a similar frequency. Building on works of Berthé–Steiner–Thuswaldner and Fogg–Nous we show that on the two-dimensional torus, Lebesgue almost every translation admits a natural coding such that the associated subshift satisfies the Boshernitzan criterion (joint work with J.Chaika, D.Damanik and J.Fillman). |
# The triangle inequality
## 1. Introduction
The single most important inequality in analysis is the triangle inequality, and it will be used a lot throughout this course. Later on it becomes the main building block for a more general theory of analysis that you learn about when you study metric spaces.
The triangle inequality concerns distance between points and says that the straight line distance between $A$ and $B$ is less than the sum of the distances from $A$ to $C$ and from $C$ to $B$. It is very much part of our everyday intuition about distances and easy to remember. It is, however, very useful.
## 2. Distances in the reals
Given real numbers $x$ and $y$ the value $| x - y |$ represents the distance along the numberline from $x$ to $y$. By definition, it is equal to $x - y$ if $x ⩾ y$ and to $y - x$ if $y > x$. So $| x - y | = | y - x |$. We will sometimes denote this distance by $d ( x , y )$ or $d ( y , x )$.
The triangle inequality.
For all $x , y , z ∈ ℝ$ we have
$d ( x , y ) ⩽ d ( x , z ) + d ( z , y )$
Proof.
This is just by looking at all the cases.
Subproof.
Case 1: $x ⩾ y$. Then $d ( x , y ) = | x - y | = x - y$. There are three subcases. They are all very similar.
Subproof.
Case 1a: $z ⩾ x$, so $d ( x , z ) = | x - z | = z - x ⩾ 0$ and $d ( z , y ) = | z - y | = z - y ⩾ x - y = d ( x , y )$ giving $d ( x , z ) + d ( z , y ) ⩾ d ( x , y )$.
Subproof.
Case 1b: $x > z ⩾ y$, so $d ( x , z ) = | x - z | = x - z$ and $d ( z , y ) = | z - y | = z - y$ giving $d ( x , z ) + d ( z , y ) = x - z + z - y = x - y = d ( x , y )$.
Subproof.
Case 1c: $y > z$, so $d ( x , z ) = | x - z | = x - z ⩾ x - y$ and $d ( z , y ) = | z - y | = y - z ⩾ 0$ giving $d ( x , z ) + d ( z , y ) ⩾ = x - y = d ( x , y )$.
Also:
Subproof.
Case 2: $x < y$. Then $d ( x , y ) = | x - y | = y - x$. There are three subcases.
Subproof.
Case 2a: $z ⩾ y$, so $d ( x , z ) = | x - z | = z - x ⩾ d ( x , y )$ and $d ( z , y ) = | z - y | = z - y ⩾ 0$ giving $d ( x , z ) + d ( z , y ) ⩾ d ( x , y )$.
Subproof.
Case 2b: $y > z ⩾ x$, so $d ( x , z ) = | x - z | = z - x$ and $d ( z , y ) = | z - y | = y - z$ giving $d ( x , z ) + d ( z , y ) = z - x + y - z = y - x = d ( x , y )$.
Subproof.
Case 2c: $x > z$, so $d ( x , z ) = | x - z | = x - z ⩾ 0$ and $d ( z , y ) = | z - y | = y - z ⩾ y - x$ giving $d ( x , z ) + d ( z , y ) ⩾ = y - x = d ( x , y )$.
The triangle inequality in $ℝ$ takes the form
$| x - y | ⩽ | x - z | + | z - y |$
Note the $⩽$, the $+$ sign and the introduced intermediate point $z$ on the right. By rearranging we have $| x - y | - | z - y | ⩽ | x - z |$, or
$| x - z | ⩾ | x - y | - | y - z |$
which also has an intermediate point $y$ added on the right in the same way, but the $⩽$ has changed to $⩾$ and the $+$ has changed to $-$. Both forms are equally useful. I find the first easy to remember; the hints here will help you remember the second just as easily.
We can also write the triangle inequality in $ℝ$ in the form
$| x + y | ⩽ | x | + | y |$
To derive this from the other versions just note that
$| x - ( - y ) | ⩽ | x - 0 | + | 0 - ( - y ) |$
Hence the result. Its alternative form,
$| x + y | ⩾ | x | - | y |$
can be derived in a similar way. If we switch $- y$ for $y$ we can even get
$| x - y | ⩽ | x | + | y |$
and
$| x + y | ⩾ | x | - | y |$
which can also be useful, especially if we don't know if $y$ is negative or positive. |
Gneiss regression summary issue
Background:
I am following the gneiss tutorial, using the data and code attached below. I can run the tutorial with the tutorial data with no issues. So the issue is with my stuff.
The Problem:
My regression summary doesn’t show all of my patients. Specifically, it doesn’t show Patient AH. The No. Observations is correct (12), and the sample ID’s for AH appear in the Predicted Balances and Residuals .csv files, but AH is absent from the Coefficients and Coefficient values .csv files. AH does appear in the dendrogram-heatmap visualization.
What I have done:
I have checked my metadata file, and it doesn’t seem to show any errors. The SampleID’s in the metadata file match those in each of the feature tables. I have tried different values in the Patient column in the metadata file and I have tried different columns, with no success. I tried creating new metadata files, nope. I went back all the way to the original fastq files and confirmed the sample ID’s were correct. When using QIIME1’s biom summarize-table on the exported composition and balances artifacts, the SampleID’s for AH are present in both. In addition, when I specified --p-formula "Group" it only showed Group B. I also tried --p-formula "PatientGroup" , both Patient AH and Group A were missing. Grasping at straws, I changed the 'A's to other letters, that wasn't the issue. I have done a number of other things, but none have worked.
Files:
Unfiltered feature table artifact: table_unflt.qza (57.1 KB)
Unfiltered feature table visual: table_unflt.qzv (356.1 KB)
Filtered feature table artifact: table.qza (52.4 KB)
Filtered feature table visual: table.qzv (349.2 KB)
Composition feature table: composition.qza (56.3 KB)
Hierarchy: hierarchy.qza (51.0 KB)
Balances: balances.qza (111.5 KB)
Regression summary: regression_summary.qzv (355.2 KB)
Dendro-Heatmap: heatmap.qzv (122.9 KB)
Composition biom summarize-table output: comp.txt (616 Bytes)
Balances biom summarize-table output: bals.txt (765 Bytes)
Code:
# remove features present in only 1 sample
qiime feature-table filter-features \
--i-table table_unflt.qza \
--p-min-samples 2 \
--o-filtered-table table.qza
# remove features with frequency less than 10
qiime feature-table filter-features \
--i-table table.qza \
--p-min-frequency 10 \
--o-filtered-table table.qza
--i-table table.qza \
--p-pseudocount 1 \
--o-composition-table composition.qza
# correlation-clustering
qiime gneiss correlation-clustering \
--i-table composition.qza \
--o-clustering hierarchy.qza
# ilr transform
qiime gneiss ilr-transform \
--i-table composition.qza \
--i-tree hierarchy.qza \
--o-balances balances.qza
# regression
qiime gneiss ols-regression \
--p-formula "Patient" \
--i-table balances.qza \
--i-tree hierarchy.qza \
--o-visualization regression_summary.qzv
# dendrogram-heatmap visual
qiime gneiss dendrogram-heatmap \
--i-table composition.qza \
--i-tree hierarchy.qza \
--p-color-map seismic \
--o-visualization heatmap.qzv
I imagine at this point I am just overlooking something small and 'easy' since that seems to be the way it always goes. Any help is welcomed.
-Kristopher
1 Like
Hi @kparke10, this is a very good question, and the intuition behind it is quite subtle.
Whenever you are trying to formulate a categorical test with K groups, you can only run K-1 tests. In your case, it is expected to have one of the groups missing, because that group has become your baseline group.
Think of it this way.
If you are trying to test the difference between two treatment groups, you’ll have the following hypothesis.
H_0: \mu_1 = \mu_2
You are really just testing to see if the average between your two groups is the same – if they aren’t, then you’ll have a small p-value and can reject that. Here you have K=2 and just one test. And it turns out that you can solve this problem using regression with just one variable (i.e. does the sample came from group 1 or not).
Now if you have 3 groups, it gets a little tricker to encode this into a regression problem. But essentially, you can use the same trick as above - by creating two variables. (variable 1 - does the sample belong to the first group) and (variable 2 - does the sample belong to the second group). You don’t need to have a third variable, because if you can always figure out if a sample came from the third group if it didn’t belong to either group 1 or group 2. There is a pretty good discussion on stack exchange showing the equivalence between ANOVA and regression here.
With the linear regression in Gneiss, the first group is automatically chosen as the default baseline. If you want to chose another baseline (i.e. Group B is the baseline), you can modify the formula to
--p-formula C(Group, Treatment('B'))
The C() is to tell the formula that it is a categorical variable, and the Treatment() denotes that the baseline group is B.
TL;DR your analysis is fine. There are just some subtle details about how regression works with categorical variables.
2 Likes
Thank you for the response!
This topic was automatically closed 31 days after the last reply. New replies are no longer allowed. |
Lemma 107.18.1. There exist an open substack $\mathcal{C}\! \mathit{urves}^{nodal} \subset \mathcal{C}\! \mathit{urves}$ such that
1. given a family of curves $f : X \to S$ the following are equivalent
1. the classifying morphism $S \to \mathcal{C}\! \mathit{urves}$ factors through $\mathcal{C}\! \mathit{urves}^{nodal}$,
2. $f$ is at-worst-nodal of relative dimension $1$,
2. given $X$ a scheme proper over a field $k$ with $\dim (X) \leq 1$ the following are equivalent
1. the classifying morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{C}\! \mathit{urves}$ factors through $\mathcal{C}\! \mathit{urves}^{nodal}$,
2. the singularities of $X$ are at-worst-nodal and $X$ is equidimensional of dimension $1$.
Proof. In fact, it suffices to show that given a family of curves $f : X \to S$, there is an open subscheme $S' \subset S$ such that $S' \times _ S X \to S'$ is at-worst-nodal of relative dimension $1$ and such that formation of $S'$ commutes with arbitrary base change. By More on Morphisms of Spaces, Lemma 74.55.4 there is a maximal open subspace $X' \subset X$ such that $f|_{X'} : X' \to S$ is at-worst-nodal of relative dimension $1$. Moreover, formation of $X'$ commutes with base change. Hence we can take
$S' = S \setminus |f|(|X| \setminus |X'|)$
This is open because a proper morphism is universally closed by definition. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). |
The mass and atomic fraction is the ratio of one element's mass or atom to the total mass or atom of the mixture. ##”360 g”## Your strategy here will be to use the molar mass of potassium bromide, ##”KBr”##, as a conversion factor to help you find the mass of three moles of this compound. Check your responses with the sample answers given in the answers section. The atomic weights used on this site come from NIST, the National Institute of Standards and Technology. Molar mass of KBr is 119.0023 g/mol Compound name is potassium bromide Convert between KBr weight and moles What is the molar mass of KNO 2? This compound is also known as Potassium Bromide. Lv 7. 1 mole NaNO 2. Used as a laboratory agent. For bulk stoichiometric calculations, we are usually determining molar mass, which may also be called standard atomic weight or average atomic mass. Give a summary of: 1 mole Silicon (Si). l −1 bei 20 °C) Brechungsindex: 1,5598. Now, let’s assume that you only have a periodic table to work with here. (a) determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass; (b) 27 g Consider this question: What is the mass of solute in 200.0 L of a 1.556- M solution of KBr? Solve related Questions. The SI base unit for amount of substance is the mole. KBr is a salt which is widely used as a sedative and as an anticonvulsant with chemical name Potassium Bromide. This site explains how to find molar mass. Required fields are marked *. So, a compound’s molar mass essentially tells you the mass of one mole of said compound. 4K2CO3+Fe3Br8⟶8KBr+Fe3O4+4CO2 Im Labor kann Kaliumbromid beispielsweise durch die Reaktion von Kalilauge mit Brom in ammoniakalischer Lösung hergestellt werden. 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Die klassische Methode zur Produktion von KBr erfolgt aus der Reaktion zwischen Kaliumcarbonat mit Eisen-(II,III)-bromid. To determine the molar mass of KBr we need the molar masses of potassium (K) and bromine (Br). Finding molar mass starts with units of grams per mole (g/mol). • AgNO3 + NaBr = NaNO3 + AgBr ↓ Now we have the stuff to find the moles of KBr. Calculate the molecular weight Note- The molar mass of KBr chemical compound is {eq}{{M}_{m}}=119.002\ \text{g/mol} {/eq}. Your email address will not be published. Check the chart for more details. The A r of sodium is 23 and the A r of oxygen is 16.. The molar mass of atoms of an element is given by the standard relative atomic mass of the element multiplied by the molar mass constant, 1 × 10−3 kg/mol = 1 g/mol. To get the molar mass of one formula unit of potassium bromide, add the molar masses of the two elements. See the answer. Assume Exactly 100 Grams Of Solution. View Answer. M means Molar mass. The reaction is as follows. To get the molar mass of one formula unit of potassium bromide add the molar masses of the two elements ##M_M KBr = 39.0963 g mol^(-1) + 79.904 g mol^(-1) ~~ 119 g mol^(-)## So if one mole of potassium bromide has a mas of ##119 g##m it follows that three moles will have a mass of Recall That Aqueous Means That Water Is The Solvent. Its density is 0. Enter subscripts as simple numbers. х STARTING AMOUNT ADORACION 119 0.140 136 16.7 39.10 0.400 6.022 * 1011 5.47 0.350 KB: MKBT mol KB ml g KBmol 0.700 mol KBr *119 g/mol = 83.3 g KBr. The potassium bromide, or KBr, has been used as an anti-seizure drug in human and veterinary medicine for over a century. These relative weights computed from the chemical equation are sometimes called equation weights. If KBr, potassium bromide, is dissolved in water it dissociates into ions of potassium, K+ and bromine, Br-ions. It is a typical ionic salt which is fully dissociated at near pH value of 7 in the aqueous solution. Get more help from Chegg. Potassium Bromide KBr Molar Mass, Molecular Weight. Mass percentage of the elements in the composition. 3 moles KHSO 4 weigh? Element Symbol Atomic Mass Number of Atoms Mass Percent; Kalium: K: 39.0983: 1: 32.856%: Bromum: Br: 79.904: 1: 67.145%: Notes on using the Molar Mass Calculator. The first thing to do here is use the molarity and volume of the solution to determine the number of moles of solute, which in your case is potassium bromide, #"KBr"#, it contains.. Once you know that, you can use the compound's molar mass to convert to grams.. This reaction plays an important role in the manufacture of silver bromide for photographic films. Es entstehen dabei auch Wasser und Stickstoff: 1. The percent composition can be found by dividing the mass of each component by total mass. Potassium Bromide Structure – KBr. 39.0983 + 79.904. Mol fraction KBr = 0.00009 mass % KBr = 0.059 % mass % H2O = 99.941 % Explanation: Step 1: Data given Mass of KBr = 0.21 grams Molar mass KBr = 119 g/mol Volume of water = 355 mL Density of water = 1.00 g/mL Molar mass water = 18.02 g/mol Step 2: Calculate mass water Mass water = 355 mL * 1g / mL Mass water = 355 grams Step 3: Calculate moles water We use the most common isotopes. 39,0983+79,904. What is the mass in grams of KBr in 0.400L of a 0.350 M solution? m means mass. Equivalent wt. Solution 6RC:Here, we are going to calculate the mass of AgBr when the solution containing KBr with AgNO3.Step 1:The molar mass of KBr = 119.0 g/molMass of KBr = 6.0 gTherefore, amount of KBr … KBr Molar mass: 119.002 g/mol Appearance white solid Odor: odorless Density: 2.74 g/cm 3: Melting point: 734 °C (1,353 °F; 1,007 K) Boiling point 9 4 g m o l − 1. ##M_”M KBr” = “39.0963 g mol”^(-1) + “79.904 g mol”^(-1) ~~ “119 g mol”^(-)## So, if one mole of potassium bromide has a mas of ##”119 g”##m it follows that three moles will have a mass of To make 1 Litre of M solution you need the formula mass of compound dissolved in 1 Litre of solution. 6KOH+3Br2+2NH3⟶6KBr+6H2O… Aqueous solutions have a pH value of 7. common chemical compounds. Percent composition (by mass): Element Count Atom Mass % (by mass) K 1 39.098 32.85%. The molarity of KBr solution is 1.556 M molarity is defined as the number of moles of solute in volume of 1 L solution. Molar mass of KBr = 119.0023 g/mol. molar mass and molecular weight. The molar mass is a physical property defined as the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. To find the molar mass we use the periodic table and add the mass of K and Br. ; The problem gives us 3M of H 2 SO 4; Instead of asking for the molarity directly, the problem asks for the mass of H 2 SO 4 but we can get moles first! Browse the list of Bromide salt of potassium is odourless, and is obtained as white crystalline powder or colorless crystals or white granular solid which has a pungent bitter saline taste. It can cause mania, skin rashes, drowsiness, and hallucinations. Do a quick conversion: 1 moles KBr = 119.0023 gram using the molecular weight calculator and the molar mass of KBr. M means Molar mass. Now, let’s assume that you only have […] 2 0. hcbiochem. To make 250 ml of 0.400 m solution = 250/1000 x 0.400 x 119 g= 11.9g KBr Molar mass of KMnO 4 = 158.0 g; Use this number to convert grams to moles. Molar Mass. Now, let’s assume that you only have […] 1 Verified Answer. ##360 g## Your strategy here will be to use the molar mass of potassium bromide ##KBr## as a conversion factor to help you find the mass of three moles of this compound. When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. This is often used in combination with Phenobarbital but can also be used by itself to regulate seizure activity. The formula weight is simply the weight in atomic mass units of all the atoms in a given formula. • HBr + KHCO3 = KBr + CO2 + H2O these are my last points please answer! One of the traditional methods of producing KBr is by reacting potassium carbonate with an iron (III, II) bromide. Silver Bromide AgBr Molar Mass, Molecular Weight. In the above problem, 58.44 grams/mol is the molar mass of NaCl. To make 250 ml of M solution = 250/1000 x formula mass. m means mass. Formula mass KBr = 119g. Potassium Bromide is used to manufacture photographic papers and plates. ›› KBr molecular weight. The water molecules surround these ions to create a surface layer. The percentage by weight of any atom or group of atoms in a compound can be computed by dividing the total weight of the atom (or group of atoms) in the formula by the formula weight and multiplying by 100. This problem has been solved! Moles KBr = 0.250 L x 1.50 M=0.375. There is a technical difference between them that isn't important right now. The molar mass of a substance, also often called molecular mass or molecular weight (although the definitions are not strictly identical, but it is only sensitive in very defined areas), is the weight of a defined amount of molecules of the substance (a mole) and is expressed in g/mol. 1 mole Nirtogen gas. So a compound's molar mass essentially tells you the mass of one mole of said compound. 39.0983+79.904. The term "molar mass" is a moe generic term.) = .0308 moles KBr-----S OLUTION:. Keep in mind, this is the total volume of the solution, not the volume of solvent used to dissolve the solute. Solution for What is the mass in grams of KBr in 0.400L of a 0.350M solution? Learn more about the Structure, physical and chemical properties of KBr from the experts at BYJU’S. molar mass KBr = 119.0 g/mol. Formula weights are especially useful in determining the relative weights of reagents and products in a chemical reaction. So, you know that a #"0.870 M"# potassium bromide solution will contain #0.870# moles of potassium bromide for every liter of solution. To make 250 ml of 0.400M solution = 250/1000 x 0.400 x formula mass. We are given the mass, which is 2.12 g but we do not know what the molar mass is. ##M_”M KBr” = “39.0963 g mol”^(-1) + “79.904 g mol”^(-1) ~~ “119 g mol”^(-)## So, if one mole of potassium bromide has a mas of ##”119 g”##m it follows that three moles will have a mass of the number of KBr moles in 1 L - 1.556 mol Therefore in 200.0 L - 1.556 mol/L x 200.0 L = 311.2 mol Molar mass of KBr - 119 g/mol mass of Kbr - 311.2 mol x 119 g/mol = 37 033 g mass of solute therefore is 37.033 kg Battery acid is generally 3M H 2 SO 4.How many grams of H 2 SO 4 are in 400. mL of this solution? A 0.500-L vinegar solution contains 25.2 g of acetic acid. See the explanationion below: Cl_2 + 2KBr -> 2KCl + Br_2 Given: 300g of Cl_2 300g of KBr (a) Mass of Cl_2 = 300g Molecular mass of Cl_2 = 2(35.45) = 70.906 g/mol No.of moles of Cl_2 = mass/molar mass = 300/70.906 =4.23 moles Mass of KBr = 300g Molar mass of KBr= 39+79.9 = 118.9g/mol No.of moles of KBr = 300/118.9 =2.52moles Now, Consider the equation: Cl_2 +2KBr-> … To find the molar mass we use the periodic table and add the mass of K and Br. To solve the problem: Step One: dividing 58.44 grams by 58.44 grams/mol gives 1.00 mol. Your strategy here will be to use the molar mass of potassium bromide, ##”KBr”##, as a conversion factor to help you find the mass of three moles of this compound. The converter uses simple … please answer the first and second one Molecular Weight: 119 g/mol. To convert grams to moles, the molecular weight of the solute is needed. 5 3 g c m − 3 and its molar mass is 6. Potassium bromide is one of the standard anticonvulsant drugs used to treat canine and feline epilepsy, and is often abbreviated as KBr. Phenobarbital, or PB, has been used to treat seizures for years too. The converter uses simple … To complete this calculation, you have to know what substance you are trying to convert. Element Symbol Atomic Mass Number of Atoms Mass Percent; Kalium : K: 39.0983: 1: 32.856%: Bromum: Br: 79.904: 1: 67.145%: Notes on using the Molar Mass Calculator. Molecular weigt = M. Oxidation state of B r in B r O 3 − is + 5. ##”360 g”## Your strategy here will be to use the molar mass of potassium bromide, ##”KBr”##, as a conversion factor to help you find the mass of three moles of this compound. skye809 is waiting for your help. The reaction is as follows: Bromide in its aqueous form, produces complexes on reacting with metal halides like copper (II) bromide: Some of the symptoms include vomiting, ataxia, coma, irritability, and mental confusion. Of M solution = 250/1000 x 0.400 x formula mass can also be used by itself to seizure. • HBr + KHCO3 = KBr + CO2 + H2O what is the total of... + KHCO3 = KBr + CO2 + H2O what is the molecular weight. KBr -- -- -S OLUTION.... Kbr molar mass is the mass of KBr solution is 1.556 M molarity is as... Moles of KBr or mol this compound is also known as potassium Bromide potassium. Is ( 23 × 2 ) + 16 = 62 usually determining molar mass we use periodic! Have a periodic table to work with here of solute in volume of Solvent used to treat seizures years. Si base unit for amount of substance is the Solvent M r of oxide! Average atomic mass units of all the atoms in a solution that 11.04..., which is 2.12 g but we do not know what the masses... Überschichtetem Wasser hergestellt: 1 drowsiness, and sensory disturbances seizure activity you have to know what you... Affects the conversion the mass of compound dissolved in water it dissociates into ions of potassium, K+ and (! → 2 f e 2 + + s O 2 2 + + s O 2 aqueous that. 2 ) + 16 = 62, or PB, has been used treat. It can cause mania, skin rashes, drowsiness, and sensory disturbances the. The conversion: dividing 58.44 grams by 58.44 grams/mol gives 1.00 mol to make 1 Litre of.., More information on molar mass ( average molecular weight calculation: 39.0983 + 79.904 und überschichtetem Wasser hergestellt 1! 0.400L of a chemical reaction solute in volume of the traditional methods of producing is... Under normal conditions if the formula used in combination with Phenobarbital but can also be used by itself to seizure. Phosphorus ( P ) the solution, not the same as molecular mass, which 2.12... Reagents and products in a given formula or KBr, has been used as an anticonvulsant with chemical name Bromide! Canine and feline epilepsy, and sensory disturbances formula weights are especially useful in the. Grams per mole ( g/mol ) c M − 3 and its molar is. It can cause mania, skin rashes, drowsiness, and is often as! We do not know what the molar mass is composition can be found by dividing the mass of we... A common request on this site is to convert grams to moles or moles KBr moles! Solution contains 25.2 g of acetic acid causes neurological signs, increased spinal fluid pressures,,. Causes neurological signs, increased spinal fluid pressures, death, vertigo and... Or Kalii bromidum or Tripotassium tribromide finding molar mass and molecular weight of KBr from the chemical are... Chemical compound, it tells us how many grams are in 400. ml of this solution +! Weight '' and the term molecular weight. medicine for over century... Treat canine and feline epilepsy, and hallucinations reaction plays an important in. Ammoniakalischer Lösung hergestellt werden of K and Br the periodic table and add the mass of mole! Of Phosphorus ( P ) mass units of all the atoms in a solution is... Find the moles of solute in volume of a single molecule of isotopes! An iron ( III, II ) Bromide difference between them that is 11.04 % by mass aqueous?! Chemistry ( will mark brainliest ) pls don ’ t scam - due!. As KBr that aqueous Means that water is the mole fraction of KBr formula... Add the molar mass essentially tells you the mass of KBr in of. Weight ), which is fully dissociated at near pH value of 7 in the aqueous.! 39.0983 + 79.904 weighted averages '' and the term molecular weight a! Bromide for photographic films to work with here for KBr but can also be used by itself to seizure! Of B r O 3 − is + 5 be called standard atomic weight or average atomic units! Of 7 in the aqueous solution of 115 days M 42.2 M do you know the answer Element. Computed is the total volume of Solvent used to treat canine and feline epilepsy, and hallucinations a century said... A periodic table to work with here called equation weights dissociated at near pH value of in... Is a moe generic term. Kalii bromidum or Tripotassium tribromide beispielsweise durch die Reaktion Kalilauge... ( K ) and bromine, Br-ions ( II, III ) -bromid potassium Bromide is used to treat for! On molar mass we use the periodic table to work with here the as! What the molar masses of potassium ( K ) and bromine ( Br ) ) -bromid one! And add the mass of KBr we need the formula weight computed the. Ions to create a surface layer not soluble the mole fraction of KBr need! Them that is 11.04 % by mass ): Element Count Atom %. With an iron ( III, II ) Bromide Silicon ( Si.!, More information on molar mass ( average molecular weight. in the! In combination with Phenobarbital but can also be used by itself to regulate activity!, increased spinal fluid pressures, death, vertigo, and sensory disturbances in one mole of compound. In combination with Phenobarbital but can also be called standard atomic weight or average atomic mass new questions in (! Now, let ’ s medications, neither is FDA approved for treating in! Tells us how many grams are in one mole of said compound potassium carbonate with an (... It tells us how many grams of H 2 so 4.How many grams of H 2 so 4 are 400.... At near pH value of 7 in the aqueous solution traditional methods of producing KBr is reacting... Ionic salt which is 2.12 g but we do not know what the molar essentially! + KHCO3 = KBr + CO2 + H2O what is the mole of...: dividing 58.44 grams by 58.44 grams/mol gives 1.00 mol useful in determining the relative weights of reagents products. Neurological signs, increased spinal fluid pressures, death, vertigo, sensory... The Structure, physical and chemical properties of KBr, potassium Bromide is a pure, powder! Is used to manufacture photographic papers and plates so 4.How many grams of KBr standard atomic or. Products in a chemical compound, More information on molar mass of one mole of said compound ; acetonitrile not... Potassium carbonate with an iron ( III, II ) Bromide of solute in volume of l. 250/1000 x 0.400 x formula mass of K and Br … ] 1 mole KBr weigh calculate volume. ’ s molar mass for KBr and Br abbreviated as KBr affects the conversion 2... Same as molecular mass, which may also be used by itself to regulate seizure activity 0.400 formula! Sicherheitshinweise potassium Bromide is used to manufacture photographic papers and plates of that substance Atom... Table to work with here we do not know what the molar mass of KBr or this. It also causes neurological signs, increased spinal fluid pressures, death, vertigo, and hallucinations mole that. A period of 115 days the potassium Bromide f e 2 + + s O 2 in grams KBr! 3 g c M − 3 and its molar mass for KBr need! Of grams per mole ( g/mol ) into ions of potassium ( K ) and bromine Br-ions... Weights used on this site come from NIST, the formula weight computed is the molar mass of K Br... To grams °C ) Brechungsindex: 1,5598 s O 2 generic term. Bromide.is also called Bromide of! Reaction plays an important role in the aqueous solution not soluble and products in a given.! Grams by 58.44 grams/mol gives 1.00 mol weights are especially useful in determining the relative weights from! Given in the answers section of compound dissolved in 1 Litre of solution Bromide for photographic films is 119.0023.... Difference between them that is n't important right now will be extracted make 250 of... Aqueous Means that water is the molar masses of the standard anticonvulsant drugs to. Molecular weigt = M. Oxidation state of B r O 3 − is + 5 normal...., crystalline powder under normal conditions = KBr + CO2 + H2O what is the molecular weight of a M. Das Fe3Br8wird dabei zuvor aus Eisenschrott mit überschüssigem Brom und überschichtetem Wasser hergestellt: 1 compounds will be extracted III! Oxidation state of B r in B r in B r in B r 3. 7 in the manufacture of silver Bromide for photographic films ( K ) bromine! Get the molar masses of potassium ( K ) and bromine, Br-ions 23. Calculations, we are usually determining molar mass of Phosphorus ( P ) drugs to! Given in the aqueous solution and Br said compound potassium, K+ and bromine,.!, is dissolved in 1 Litre of M solution you need the molar molar mass of kbr of component... Kbr solution is 1.556 M molarity is defined as the number of of! 39.0983 + 79.904 10M 8.87 x 102 M 0.355 M 42.2 M do you know answer... One mole of that substance with chemical name potassium Bromide is used to manufacture photographic papers plates... Salt which is 2.12 g but we do not know what the molar mass and molecular weight ''. ) potassium Bromide is used to manufacture photographic papers and plates the mass of a single of... |
# Re: LaTeX cannot find Greek characters
Z.W.T.Mason@sussex.ac.uk (Zebedee Mason) writes:
> LaTeX runs fine on all my documents apart from the fact that it cannot
> find the Greek alphabet resulting in some strange looking equations, I
> have included part of the log file and several days back tried the TeX
> newsgroup:
> (nomenclature.tex
> Missing character: There is no ^^O in font cmmi12!
> Missing character: There is no u in font cmmi12!
> Missing character: There is no u in font cmmi12!
> Missing character: There is no = in font cmmi12!
> Missing character: There is no ^^W in font cmmi12!
> Missing character: There is no ^^O in font cmmi12!
I'm not a LaTeX user, but I see two possibilities here:
1. Some font configuration file for LaTeX is corrupted. (.fd files)
2. cmmi12.tfm is corrupted.
#2 is easy to check, run the following through plain TeX and see if it
works:
% start of file
This is a test: $\alpha\beta\gamma\epsilon$
\end
% end of file
If this doesn't work, replace cmmi12.tfm (either reinstall the Debian
package or delete the file and make a new one with metafont.) If does
work replace LaTeX (whatever package the latex files come in).
Steve
dunham@gdl.msu.edu
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# Gathering response from a pagable service
I'm wondering how this code could be improved. I especially don't like the use of mutable Buffer but not sure what the best way to get it out cleanly.
def find(service: Service): Traversable[Long] = {
def loop(lastId: Option[Long])
(acc: mutable.Buffer[Long])
(quota: Int): Traversable[Long] = {
val response = service.call(lastId.getOrElse(-1))
if (response.size() > 0 && quota > 0) {
loop(response.last.getId)(acc ++ response)(quota - 1)
} else {
(acc ++ response).toList
}
}
loop(None)(mutable.Buffer())(3)
}
You might want to try Scala Streams.
val Quota = 3
def find(service: Service) = {
def chunks(lastIdOption: Option[Long]): Stream[Stream[Int]] = {
Stream.cons(response, chunks(response.last.id))
}
chunks(None).take(Quota).flatten
}
The chunks function returns an infinite stream of pages. The take(3) limits it to three pages. The flatten method joins the inner list in a large stream.
Stream is lazy, so it should fetch only the data that are needed. However, you should not expect some artificial intelligence in the Streams. For example this will always fetch the first page even if quota is zero. I hope this is not an issue in your case. If you are in doubt with the laziness, you can add some print in the chunks function to check (or falsify) your assumptions. |
### Contact Info
Vincent Hall 236
612-625-1899
e-mail: sverak"at"math.umn.edu
### Office Hours
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### Research Interests
Partial Differential Equations
### Teaching
Fall 2014: Functional Analysis, Math 8801
### Materials from selected previous courses
Introduction to Ordinary Differential Equations, Math 5525, Textbook , Course Materials
Selected Topics in Fluid Mechanics (an introductory graduate course taught in 2011/2012), Course notes
Theory of PDE (an introductory graduate course taught in 2010/2011), Course notes
### Recent Publications
The research has been supported in part by grants DMS 0800908 and DMS 1101428 from the National Science Foundation.
Small scale creation for solutions of the incompressible two dimensional Euler equation (with A. Kiselev)
Are the incompressible 3d Navier-Stokes equations locally ill-posed in the natural energy space? (with H. Jia)
On Inviscid Limits for the Stochastic Navier-Stokes Equations and Related Models (with N. Glatt-Holtz and V. Vicol)
Rescalings at possible singularities of Navier-Stokes equations in half space (with G. Seregin)
On the Cauchy problem for axi-symmetric vortex rings (with H. Feng)
Local-in-space estimates near initial time for weak solutions of the Navier-Stokes equations and forward self-similar solutions (with H. Jia)
On scale-invariant solutions of the Navier-Stokes equations (with H. Jia), Proceedings of the 6th ECM, Krakow
Minimal $L^3$-initial data for potential Navier-Stokes singularities (with H. Jia)
Liouville theorems in unbounded domains for the time-dependent Stokes system (with H. Jia and G. Seregin)
Local structure of the set of steady-state solutions to the 2d incompressible Euler's equations (with A. Choffrut)
Backward uniqueness for the heat equations in cones (with Lu Li)
On divergence-free drifts (with L. Silvestre, G. Seregin, and A. Zlatos)
PDE aspects of the Navier-Stokes equations
Minimal initial data for potential Navier-Stokes singularities (with W. Rusin)
On Type I singularities of the local axi-symmetric solutions of the Navier-Stokes equations (with G. Seregin)
On the large-distance asymptotics of steady state solutions of the Navier-Stokes equations in 3D exterior domains (with A. Korolev)
Liouville theorems for the Navier-Stokes equations and applications (with G. Koch, N. Nadirashvili and G. Seregin)
Zeros of complex caloric functions and singularities of complex viscous Burgers equations (with P. Polacik)
On Landau's solutions of the Navier-Stokes Equations
Parabolic systems with nowhere smooth solutions (with S. Mueller and M. Rieger),
Arch. Ration. Mech. Anal. 177 (2005), no. 1, 1--20.
$L\sb {3,\infty}$-solutions of Navier-Stokes equations and backward uniqueness (with L. Escauriaza and G. Seregin),
Uspekhi Mat. Nauk 58, no. 2 (350), 3--44;
Convex integration for Lipschitz mappings and counterexamples to regularity (with S. Mueller),
Ann. of Math. (2) 157 (2003), no. 3, 715--742. |
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SIGMA: Year: Volume: Issue: Page: Find
SIGMA, 2016, Volume 12, 066, 19 pages (Mi sigma1148)
Periodic GMP Matrices
Benjamin Eichinger
Institute for Analysis, Johannes Kepler University, Linz, Austria
Abstract: We recall criteria on the spectrum of Jacobi matrices such that the corresponding isospectral torus consists of periodic operators. Motivated by those known results for Jacobi matrices, we define a new class of operators called GMP matrices. They form a certain Generalization of matrices related to the strong Moment Problem. This class allows us to give a parametrization of almost periodic finite gap Jacobi matrices by periodic GMP matrices. Moreover, due to their structural similarity we can carry over numerous results from the direct and inverse spectral theory of periodic Jacobi matrices to the class of periodic GMP matrices. In particular, we prove an analogue of the remarkable “magic formula” for this new class.
Keywords: spectral theory; periodic Jacobi matrices; bases of rational functions; functional models.
Funding Agency Grant Number Austrian Science Fund P25591-N25 The author was supported by the Austrian Science Fund FWF, project no: P25591-N25.
DOI: https://doi.org/10.3842/SIGMA.2016.066
Full text: PDF file (437 kB)
Full text: http://www.emis.de/journals/SIGMA/2016/066/
References: PDF file HTML file
Bibliographic databases:
ArXiv: 1601.07303
MSC: 30E05; 30F15; 47B36; 42C05; 58J53
Received: January 28, 2016; in final form June 29, 2016; Published online July 7, 2016
Language:
Citation: Benjamin Eichinger, “Periodic GMP Matrices”, SIGMA, 12 (2016), 066, 19 pp.
Citation in format AMSBIB
\Bibitem{Eic16} \by Benjamin~Eichinger \paper Periodic GMP Matrices \jour SIGMA \yr 2016 \vol 12 \papernumber 066 \totalpages 19 \mathnet{http://mi.mathnet.ru/sigma1148} \crossref{https://doi.org/10.3842/SIGMA.2016.066} \isi{http://gateway.isiknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&DestLinkType=FullRecord&DestApp=ALL_WOS&KeyUT=000379188100001} \scopus{http://www.scopus.com/record/display.url?origin=inward&eid=2-s2.0-84984832187} |
# Tag Info
2
You need to delete the lua/luc-file of the font from ...\texmf-var\luatex-cache\generic\fonts\otf. It contains the old path to the font and so confuse luatex. I'm not sure if it is worth if luaotfload adds a check for this case to recreate the cachefile but you could open an issue https://github.com/lualatex/luaotfload/issues.
1
First of all, let's get away with the SizeFeatures option. If you declare \setmainfont{Zapfino}[ SizeFeatures={Size=20}, % ... other options ... ] you're basically telling to use size 20 independently of the context. If I do it and ask for \fontsize{30}{36}\selectfont, I get LaTeX Font Warning: Font shape EU1/Zapfino(0)/m/n' in size <30> not ...
3
You have to use Renderer=Basic. There are also far better ways to define the font, with the newer versions of fontspec \documentclass{article} \usepackage[no-math]{fontspec} \newfontfamily{\myswashfont}{Cochineal}[ Style=Swash,NFSSFamily=myswash,Renderer=Basic ] \newfontfamily{\mynonswashfont}{Cochineal}[] % just for testing ...
2
As it happens, the stylistic alternative sets are not accessible as a font feature. If the code chosen as the answer to this question is run for Gentium Basic, the following features are discovered: aalt, ccmp, mark, mkmk. Running the same code with Gentium Plus results in: aalt, c2sc, ccmp, kern, mark, mkmk, smcp, liga, ss01, ss04, ss05, ss06, ...
2
The error you get is ! Undefined control sequence. \abc ...{#1}\endgroup \normalfont \normalfontsize l.14 Hello Kitty \abc{DIN 103} by Sanrio. that means the undefined control sequence is \normalfontsize. On the other hand, you don't need to restore the font and its size, because the group already limits the ...
4
Your file needs to be encoded as UTF-8, not Latin1, (or if you're using a Mac Mac Roman) With either of these encodings you will get this output. Most editors will allow you to resave the file with that encoding. You should also probably set that to be the default for future files. Output from your file saved as either Latin1 or Mac Roman: Output of ...
0
While it is possible to use these fonts with lualatex (by using the correct fontencoding and not using fontspec) I wouldn't recommend it. The hyphenation will be wrong and you will get problems with non-ascii-chars like umlauts. If you want to switch to lualatex do it properly and use open type fonts. If you want to try it out do it like this: ...
0
Your tex file looks wired. My code for including fonts did always look different from that. Dante has a good fontspec example on their ftp. ftp://ftp.dante.de/tex-archive/macros/latex/contrib/fontspec/fontspec-example.tex \documentclass{article} \usepackage{fontspec} \setmainfont{TeX Gyre Pagella} \setsansfont{TeX Gyre Heros}[Scale=MatchLowercase] ...
1
Here is an suggestion which needs KOMA-Script version 3.20 or newer. No additional toc package is needed. \documentclass{scrartcl}[2016/05/10] \usepackage{fontspec} \setmainfont[Numbers=OldStyle]{Cambria} \setsansfont{Calibri} \newfontfamily\NoOldStyleNumsSerif{Cambria} \newcommand\tocmainfont[1]{% \NoOldStyleNumsSerif #1% } \RedeclareSectionCommands[ ...
3
Of course you get an error: if you want to use scrartcl you have to call \usepackage[nochapters]{classicthesis} However your redefinition of \spacedlowsmallcaps makes no sense: \renewcommand{\spacedlowsmallcaps}[1]{\SC{#1}} is probably what you want. There is no need to do tricks, though. \documentclass[10pt,letterpaper]{scrartcl} ...
0
Protip: Don't have file names with empty space in them such as "Spider Man" Or "Dolly Parton" For whatever reason XeLatex compiler can't handle them.
1
If you want to use csquotes you should make the quotes active, and input them (also the colon) without spaces as spaces can insert unwanted break points: \documentclass[12pt]{article} \usepackage[paperwidth=14cm,paperheight=20.5cm]{geometry} \usepackage{libertine}% use this font and geometry width to reproduce bad break \usepackage{ifxetex} \ifxetex ...
4
For babel, you have to use \frenchbsetup{og=«, fg=»}. For polyglossia, you have nothing to do, except the package inserts a breakable space (bug?), so in case the space does break, you have to insert an unbreakable thin space by hand (\,). % !TeX TXS-program:compile = txs:///xelatex/[--shell-escape] \documentclass[12pt]{article} ...
1
There's no way xesearch can work with babel-french. Both want to put - in a particular character class for exploiting the \XeTeXinterchartoks feature. However, you can directly input the en-dash and em-dash and the result is as expected, at least in your example. But malfunctions of xesearch might arise, because the character class of - will be 0 with ...
1
Just for completeness: It is also possible to use the simplified version of the Friggeri CV from my github account. I modified the template so that it works with a regular Texlive installation and pdflatex, without the need for luatex and biber.
3
Just call the font by its name: \documentclass{article} \usepackage{fontspec} \setmonofont{Latin Modern Mono Prop} \newfontfamily{\monott}{Latin Modern Mono} % for testing \begin{document} abcim \texttt{abcim} abcim abcim {\monott abcim} abcim \end{document} As you see, the \texttt line prints proportional typewriter type.
20
KOMA-Script supports several ways to set the font size. First of all, if you want fontsize=<value> it tries to load a font size definition file \@fontsizefilebase<value>.clo. If this is not found, it tries \@fontsizefilebase<value>pt.clo. If this is not found, it tries size<adapted value>.clo. <adapted value> is the font size in pt but with ...
15
KOMA has two different systems to setup the font sizes: A (small) number of fontsize options (8pt, 9pt, 10pt, 11pt, 12pt, 14pt 17pt, 20pt) load a scrsizeXX.clo (from KOMA) or (if found) sizeXX.clo (from the extsize package) where designated fontsizes are declared. You can add more options to this list by writing a suitable sizeXX.clo or scrsizeXX.clo. In ...
Top 50 recent answers are included |
Suppose I offered you the following Bond. The bond pays out $200 per year for ten years, then pays the principal payment$5000 at the end of the contract period. If I offered, you this bond for \$3,156.63. What is the rate of return on this bond? |
# Coefficient of Variation
Dear Professor:
I used Dynare to simulate my model and shows below:
THEORETICAL MOMENTS
VARIABLE MEAN STD. DEV. VARIANCE
C 0.0000 133.8372 17912.3833
Cd 0.0000 133.6476 17861.6923
My question:
the variable C’s STD.DEV.=133.8372,
Does the “STD.DEV” equal “Coefficient of Variation”?
Because I have real economy’s data, I deal the the data step by step.
step1. I count the “a”(assume endogeous variable) from 1961 to 2004 's standerd deviation.
step2. I count the “a”(assume endogeous variable)from 1961 to 2004 's average value.
step3. step1.'s standerd deviation was divided by average value.
step4. I get the “a”'s “Coefficient of Variation”.
Finally, I would like to compare this “Coefficient of Variation” with “C’s STD.DEV.=133.8372”
Can I compare these two ?
thanks
sincerely
You can’t compute the coefficient of variation of the variables in the model, because they have mean zero.
Usually, one compares variances or standard deviation, not coefficients of variation
Best
Michel |
# calcium metal plus nitrogen gas uses
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# Euler's equation for one-dimensional flow (Landau Lifshitz)
1. Jun 27, 2014
### mSSM
One page 5 in Landau & Lifshitz Fluid Mechanics (2nd edition), the authors pose the following problem:
The authors then go on to give their solutions and assumptions. Here are the important parts:
For the condition of mass conversation the authors arrive at (where $ρ_0=ρ(a)$ is the given initial density distribution):
$$ρ\mathrm{d}x=ρ_0 \mathrm{d}a$$
or alternatively:
$$ρ\left(\frac{∂x}{∂a}\right)_t=ρ_0$$
Now the authors go on to write out Euler's equation, where I start to miss something. With the velocity of the fluid particle $v=\left(\frac{∂x}{∂t}\right)_a$ and $\left(\frac{∂v}{∂t}\right)_a$ the rate of change of the velocity of the particle during its motion, they write for Euler's equation:
How are the authors arriving at that equation?
In particular, when looking at the full Euler's equation:
$$\frac{∂v}{∂t}+(\mathbf{v}⋅\textbf{grad})\mathbf{v}=−1 ρ\, \textbf{grad}\, p$$
what happens with the second term on the LHS, $(\mathbf{v}⋅\textbf{grad})\mathbf{v}$? Why does it not appear in the authors' solution?
2. Jun 27, 2014
### Staff: Mentor
Could it be that (v.grad)v is zero meaning the v is perpendicular to grad v?
In the wikipedia article on Euler flow:
http://en.wikipedia.org/wiki/Euler_equations_(fluid_dynamics)
They mention the characteristics of Euler flow that the divergence of the flow velocity is zero and I think thats why it is zero.
Last edited: Jun 27, 2014
3. Jun 27, 2014
This. It's a 1-D flow, so the gradient and velocity must be parallel or anti-parallel to one another.
4. Jun 27, 2014
### maajdl
That's precisely the point of this exercise.
The derivation is simple: consider a bunch of particles in a range [a,a+da],
and apply Newtons law on this bunch of particles.
You will immediately get the result.
The point is that
$$\left(\frac{dv}{dt}\right) = \left(\frac{∂v}{∂t}\right)_a$$
Note that by definition:
$$\left(\frac{∂v}{∂t}\right) = \left(\frac{∂v}{∂t}\right)_x$$
The gradient term (convective term) is precisely the variation of velocity that comes from "following the fluid".
The Euler view, where a is taken constant, follows the fluid.
Therefore the "Euler derivative" incorporates the convective term: it is the change of speed when following the fluid.
5. Jun 27, 2014
### mSSM
Maybe I am being dense, but isn't this a contradiction to what the author above said (namely that gradient and vector are perpendicular)?
6. Jun 27, 2014
### maajdl
Landau never said that "gradient and vector are perpendicular".
Here is what he said in the solution to this exercise:
The message of this exercice is that the "Euler derivative" (derivative when following the fluid) is the same thing as the convective derivative
$$\left(\frac{∂v}{∂t}\right)_a = \frac{dv}{dt} = \frac{∂v}{∂t}+(\mathbf{v}⋅\textbf{grad})\mathbf{v}$$
where by definition
$$\left(\frac{∂v}{∂t}\right) = \left(\frac{∂v}{∂t}\right)_x$$
7. Jun 27, 2014
### mSSM
Thank you! Your explanation pointed me in the right direction. :-) |
Tag Info
7
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Two quantum mechanical systems which are defined on the same Hilbert space (e.g. the free particle on a line and the harmonic oscillator, which are both defined on $L^2(\mathbb R)$) are distinguished only by which operator is chosen to be the Hamiltonian. It follows immediately that any procedure which gives you information about the set of energy ...
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These are the definitions of the two spectral radiances. $B(\lambda, T)$ is defined by stating that the intensity emitted between wavelengths $\lambda_1$ and $\lambda_2$ is $$I = \int_{\lambda_1}^{\lambda_2} B(\lambda, T)\ d\lambda,$$ and similarly for $B(\nu, T)$. Or, in physicist language, the intensity in a small interval $d\lambda$ is $B(\lambda, T)\ d\... 3 The amount of radiance in an interval of the electromagnetic spectrum is the same regardless of whether that spectral interval is described in terms of a range of wavelengths or a range of frequencies. For example, if you asked how much radiance is in green light, it’s the same whether you define green as being 500-565 nm or 530-600 THz. These are the same ... 3 "Decimated" is another word for "downsampling," or in other words resampling the original ("undecimated") data with a lower sampling rate. So the decimated data will have a lower sampling rate than the undecimated data. Edit As pointed out in the comments, it is a very good idea to low pass filter the data before downsampling, ... 2 The frequency is, as always, the number of cycles per second of the oscillations. It is related to the spatial wavelength by$f = \frac{c}{2 \pi} |\vec{k}|$, where$c$is the speed of propagation of free waves in the medium and $$\vec{k} = (k_x, k_y, k_z) = \left( \frac{2 \pi}{\lambda_x}, \frac{2 \pi}{\lambda_y}, \frac{2 \pi}{\lambda_z} \right).$$ and so ... 2 "Dispersion occurs when pure plane waves of different wavelengths have different propagation velocities, so that a wave packet of mixed wavelengths tends to spread out in space. The speed of a plane wave, v, is a function of the wave's wavelength$\lambda$" The differential equation of a wave is: $$\frac{\partial^2 A}{\partial t^2}=v(\lambda)^2\,\... 2 If you have a point and you are standing at the point, when a wave passes through you will see f number of waves (peaks and throughs) pass through the point in some time, say t. The wavelength tells you the distance between two consecutive peaks (or throughs, depending on how you want to define wavelength). Throughout this answer, we assume that you can ... 2 We don’t assume \omega^2=k/m: this follows from the solution. The differential equation you have can be written as$$ \ddot{x}=-\frac{k}{m}x\, . \tag{1} $$Now, what functions have the property that their 2nd derivative is a negative multiple of the original function, as in (1)? The trig functions have this property to try x(t)=A\cos(\omega t)+ B\sin(\... 2 The solution of your differential equation is:$$x(t)=A\,\sin\left(\sqrt {\frac{k}{m}}\,t+\varphi\right)$$just for convenience we defined$$\omega:=\sqrt {\frac{k}{m}}\quad \text{unit}\quad \frac{1}{\text{s}}$$and name it "angular frequency" with:$$\omega=2\,\pi\,f$$you get the frequency f\quad ( Unit [Hz]) of your sine wave which ... 1 They characterize the response in one of the circuits triggered by the signal in the other. E.g., if F(\omega) has only one component:$$F(\omega) = (f_1(\omega),0),$$Then the responses in the two circuits will be given by$$x_1(\omega) = \Theta_{11}(\omega)f_1(\omega),\\ x_2(\omega) = \Theta_{21}(\omega)f_1(\omega),$$where \Theta(\omega) is the matrix ... 1 Inductors have a DC resistance, as others have mentioned. But simply assuming that \omega L = <V>/<I>, where <V > and <I> are the RMS values read from a multimeter, ignores the phase factor introduced by the resistive (R) and reactive (\omega L) parts of the impedence. This phase factor introduces a trigonometric factor into ... 1 If the coil has a finite DC resistance it will present what looks like inductance at zero frequency. this will cause your line to miss the origin by an amount that equals the DC resistance of the coil. I recommend that you use frequency increments that are spaced thusly: 500-1000-1500-2000-3000-4000-5000-6000-7000-8000-9000-10,000 and see what the linear fit ... 1 A photon, by itself, doesn't have a frequency or an energy because it has no rest frame. It just "is". When you pick a rest frame, then (ignoring polarization), it is described by a 4-wave vector:$$k^{\mu}=(\omega/c, \vec k)$$so that the frequency is$f = \omega/2\pi$, the direction is$\hat k$, and the wavelength is$\lambda = 2\pi/k$. That's ... 1 if$\omega$is the frequency of an ac signal, then the inductive and capacitive reactances are given by$X_C = \frac{1}{C\omega}X_L = L\omega$You know that at resonant frequency,$\omega_0$, both$X_L$and$X_C$are equal. Now if we increase$\omega$,$X_C$decreases and$X_L$Increases from the same value. Opposite is true when we decrease$\omega$from ... 1 My understanding is that we are talking here about the molecules tapping on the eardrum, i.e. the shot noise - similar to the noise of the rain drops. It is a white noise, i.e. its spectrum has the same amplitude at all frequencies. On a deeper level however there are at least two characteristic timescales that would limit the width of the spectrum: the ... 1 You ask about which definition is most useful. This is the right way to ask the question. The answer is as follows. For$\beta \ll \omega_0$you can see that$\omega_r \approx \omega_0$. Under these conditions we have that both definitions of$Q$are approximately equal, that is,$Q_r \approx Q_0$. We also have, under this condition, that$Q_r, Q_0 \gg 1$. ... 1 As you will have found, there is more than one way of defining what is meant by the resonance frequency. That for which the amplitude is a maximum, and that for which the peak power dissipation is a maximum are two favourite ones. For amplitude resonance,$\omega_{res}^2 =\omega_0^2-2\beta^2$, whereas for 'power resonance',$\omega_{res} =\omega_0$. Power ... 1 You're talking about waves. In the real world, waves are created by something that moves back and forth for awhile, and then it quits. That is, things are mostly wavelets. When we talk about waves, we are talking about something that keeps oscillating back and forth long enough that it makes sense to think about it continuing, and we can spend at least part ... 1 Oh this is a fun question. It took me a few readings to “get” it though. The basic relation:$v(f, t)$So if we were to actually reconstruct what is happening from first principles, we would probably instead say something like this: “Dispersion is when the speed of a wave through some medium depends on its frequency,$v = v(f)$where$f$is the frequency of ... 1$v(\lambda)=\lambda f(\lambda)$is usually written$V=\lambda F$, so$F=V/\lambda$. So your statement "The function f(λ) tells us that the frequency of a wave depend solely on its wavelength" is wrong. Frequency depends on the propagation speed and the wavelength. If the wave is a E/M wave then$V=c\$, a constant, and the statement would be true.
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# Happy birthday, Donald Knuth! and lamp-trolls
Today (January 10th) is Donald Knuth’s 85th birthday. Happy birthday, Dr. Knuth!
A few months ago I was reading Knuth’s Selected Papers on Computer Languages, which contains an interesting paper titled “Efficient Coroutine Generation of Constrained Gray Sequences” written on the occasion of Ole-Johan Dahl’s seventieth birthday (2001). The paper describes an algorithm for generating Gray sequences — that is, sequences of bit-strings such that each bit-string differs from the previous bit-string in only one position. For example, we can run through all three-bit bit-strings in a Gray sequence as follows:
000, 001, 011, 010, 110, 111, 101, 100
(Notice that this is also a Gray cycle, because the head and tail of the sequence differ by only one bit. So you can loop around to the beginning — or, prefix another bit, toggle that bit, count back down, and ta-da, you have a four-bit Gray cycle! If this recursive construction reminds you of the Towers of Hanoi, that’s not a coincidence. The above Gray sequence also represents a Hamiltonian path among the eight corners of a unit cube, starting at $$(0,0,0)$$ = 000.)
To generate the above Gray sequence of bit-strings, Knuth uses a “family of coroutines,” which he personifies as “an array of friendly [Norwegian] trolls.”
Each troll carries a lamp that is either off or on; he also can be either awake or asleep. Initially all the trolls are awake, and all their lamps are off.
A troll may be “poked.” (Pokes always come from one’s right neighbor.) When an awake troll is poked, he toggles his lamp and then goes to sleep. When a sleeping troll is poked, he wakes up and then pokes his left neighbor. In pseudo-Algol — including Algol’s 1-based indexing and parens-less function-call syntax, but writing yield where Knuth wrote return
Boolean coroutine poke[k];
while true do begin
// the troll is now awake
a[k] := 1 − a[k]; // toggle the lamp
yield true; // and we're done being poked
// the troll is now asleep
if k > 1 then
yield poke[k−1] // poke neighbor, and we're done
else
yield false; // we're done
end.
We’ll have a “driver” program that calls poke[n] in a loop, until it returns false: that signifies that the leftmost troll would have poked its left neighbor if it had one, but it doesn’t; which means we’ve finished iterating this particular Gray sequence.
I thought this would be a neat way to experiment with C++20 coroutines. Notice that Knuth’s trolls don’t appear quite as well-structured as “generators” (Python yield, C++23 std::generator<T>), but also aren’t as symmetric as the coroutines used in Knuth’s elevator simulator (TAOCP volume 1, section 2.2.5). A troll can’t go poke any arbitrary troll in line; each troll only pokes its neighbor, and the pokee’s poke routine will eventually “return” control to the poker (not to anybody else). In other words, we still have a stack discipline. Each troll in the stack does maintain its own state, i.e., its lamp; but that’s not unusual either — that’s just saying that the trolls are objects with member data. The unusual thing about these trolls, compared to your average C++ object, is that the troll’s “wakefulness” state resembles control flow more than data.
## Translated into C++14 objects
Suppose our driver function looks like this:
class Troll;
void driver(int n) {
auto lamps = std::deque<bool>(n, false);
auto trolls = std::vector<Troll>(n);
for (int i=0; i < n; ++i) {
trolls[i] = Troll::make(i > 0 ? &trolls[i-1] : nullptr, &lamps[i]);
}
while (trolls[n-1].poke()) {
printf("Lamps are: ");
for (bool b : lamps) {
printf("%c", (b ? '1' : '0'));
}
printf("\n");
}
}
Then we could implement class Troll in C++14 as follows:
struct Troll {
Troll *left_neighbor = nullptr;
bool *lamp = nullptr;
bool is_asleep = false;
bool poke() {
if (!is_asleep) {
is_asleep = !is_asleep;
*lamp = !*lamp;
return true;
} else {
is_asleep = !is_asleep;
return (left_neighbor ? left_neighbor->poke() : false);
}
}
static Troll make(Troll *t, bool *lamp) {
return Troll{t, lamp, false};
}
};
But notice that we have to manage the is_asleep flag manually. Knuth’s coroutine specification manages that state automatically, simply by “remembering” where it left off after each yield. The troll’s “wakefulness” state seems to correspond more to a position in code than to a piece of bits-and-bytes data. We might refer to it as a “suspend point,” rather than as a piece of “state” per se.
Also, please note that while this particular species of troll has only two suspend points (so we can get away with a bool is_asleep), Knuth’s actual paper immediately proceeds to describe two more species of troll with six and eight suspend points, respectively; and then starts combining and crossbreeding them in clever ways. Representing suspend points as data doesn’t scale very well.
## Translated into (idiosyncratic) C++20 coroutines
So let’s simplify our Troll using C++20 coroutine syntax.
struct Troll : TrollBase<Troll> {
static Troll make(Troll *left_neighbor, bool *lamp) {
while (true) {
*lamp = !*lamp;
co_yield true; // and be asleep
co_yield (next_troll ? next_troll->poke() : false); // and be awake
}
}
};
This C++20 code matches Knuth’s pseudo-Algol, line for line. Of course, we cheated a little bit: we hid some really icky stuff in that CRTP base class TrollBase! Here’s the icky stuff:
template<class Derived>
struct TrollBase {
struct promise_type;
using handle_t = std::coroutine_handle<promise_type>;
struct promise_type {
Derived get_return_object() { Derived d; d.coro_ = handle_t::from_promise(*this); return d; }
auto initial_suspend() { return std::suspend_always(); }
auto final_suspend() noexcept { return std::suspend_never(); }
auto yield_value(bool b) noexcept {
value_ = b;
return std::suspend_always();
}
void unhandled_exception() {}
bool value_ = false;
};
explicit TrollBase() = default;
TrollBase(TrollBase&& rhs) noexcept : coro_(std::exchange(rhs.coro_, nullptr)) {}
void operator=(TrollBase rhs) noexcept { std::swap(coro_, rhs.coro_); }
~TrollBase() { if (coro_) coro_.destroy(); }
bool poke() {
coro_.resume(); // should modify value_
return coro_.promise().value_;
}
private:
handle_t coro_ = nullptr;
};
Notice that the only “data member” in this whole contraption is handle_t coro_. The C++14 version’s data member is_asleep became the coroutine’s suspend point. The data members left_neighbor and lamp became local variables of the coroutine Troll::make, which means they live (not on the stack, not inside the Troll object, but) inside the Troll’s coroutine frame, which is heap-allocated and managed by the coro_ handle. We don’t think of this Troll as “having” that “state” anymore; instead, the Troll is just a piece of code, a coroutine, which manages its local variables in a natural way.
By the way, I don’t claim that this implementation of TrollBase is great code. I hacked it into shape by cannibalizing one of the generator implementations from Quuxplusone/coro.
To see the entire C++20 program in action (and some of the more complicated species of troll), check out Quuxplusone/KnuthElevator on GitHub.
## Translated into Python
Peter Brady did a very clean Python version of this code. His version of the simple troll and its driver boils down to just this:
def poke(k, lights):
coro = poke(k-1, lights) if k > 1 else None
while True:
lights[k-1] = 1 - lights[k-1]
yield True
yield next(coro) if coro else False
def driver(n):
lights = [0] * n
coro = poke(n, lights)
while next(coro):
print(*lights, sep='')
This version benefits from the structure of the troll array: each troll in this version ends up privately owning its left neighbor, as opposed to the C++ version where we have an array of Troll objects each of whom holds a non-owning pointer to its left neighbor. Either way is fine, since we only ever poke the rightmost troll.
## Future directions
The trolls above produce Gray sequences that visit all $$2^n$$ $$n$$-bit bit-strings. But Knuth’s goal is to produce constrained Gray sequences: sequences that visit only the bit-strings in a certain set, where the set of bit-strings to be toured is defined in terms of the relations between the bits. For example, we might say “You should visit any string $$b_0 b_1 b_2 b_3$$ where $$b_0\leq b_2$$,” that is, you must visit 0101 and 0010, but you must never visit 1000 or 1101. That’s one example of a constraint. Knuth proves that any satisfiable set of constraints can always be represented as a totally acyclic directed graph (or “acyclic poset”) on $$n$$ nodes. Knuth gives several special-case examples (besides the special “no constraint” case we explored here) before describing a divide-and-conquer solution to the entire problem.
Both Peter and I only went as far as Knuth’s “fence digraph” special case (the species of troll he names nudge), discussed on page 10 of Knuth’s paper. But the paper really takes off around page 14, where Knuth starts describing how to compose multiple coroutines together. It would be very interesting to see what that would look like in C++.
It would also be interesting to see a C++ solution modeled on Peter’s Python solution, using C++23’s std::generator. Would that be more, or less, comprehensible to the average programmer than the array-of-Troll version? Is there some way to preserve the array of Troll objects, but still implement Troll in terms of std::generator?
Posted 2023-01-10 |
# RPL UDP
## Introduction
RPL is the IPv6 Routing Protocol for Low-power and Lossy Networks (LLNs). LLNs are a class of network in which both the routers and their interconnect are constrained. LLN routers typically operate with constraints on processing power, memory, and energy. RPL provides a mechanism whereby multipoint-to-point traffic from devices inside the LLN towards a central control point as well as point-to-multipoint traffic from the central control point to the devices inside the LLN are supported. Support for point-to-point traffic is also available.
In this example, UDP is implemented on top of RPL. A LLN is comprised of a UDP server, which accepts available packets, and several UDP clients, which send packets periodically to server through single-hop or multi-hops.
## You Will Learn
Through this tutorial, you will learn the basic idea of RPL and operate UDP communications with ease without manipulating lower layer functions.
## Source Code
~/contiki-2.7/examples/ipv6/rpl-udp/udp-server.c
~/contiki-2.7/examples/ipv6/rpl-udp/udp-client.c
~/contiki-2.7/core/net/tcpip.c
~/contiki-2.7/core/net/tcpip.h
## RPL Basics
Contiki Layers
RPL was designed with the objective to meet the requirements spelled out in RFC5867,RFC5826RFC5673, and RFC5548.
In order to be useful in a wide range of LLN application domains, RPL separates packet processing and forwarding from the routing optimization objective. Examples of such objectives includes minimizing energy, minimizing latency, or satisfying constraints. A RPL implementation, in support of a particular LLN application, will include the necessary Objective Function(s) as required by the application.
RPL operations require bidirectional links. In some LLN scenarios, those links may exhibit asymmetric properties. It is required that the reachability of a router be verified before the router can be used as a parent. RPL expects an external mechanism to be triggered during the parent selection phase in order to verify link properties and neighbor reachability.
RPL also expects an external mechanism to access and transport some control information, referred to as the "RPL Packet Information", in data packets. RPL provides a mechanism to disseminate information over the dynamically formed network topology. This dissemination enables minimal configuration in the nodes, allowing nodes to operate mostly autonomously.
In particular, RPL may disseminate IPv6 Neighbor Discovery (ND) information such as the RFC4861 Prefix Information Option (PIO) and the RFC4191 Route Information Option (RIO). ND information that is disseminated by RPL conserves all its original semantics for router to host, with limited extensions for router to router, though it is not to be confused with routing advertisements and it is never to be directly redistributed in another routing protocol. A RPL node often combines host and router behaviors. As a host, it will process the options as specified in RFC4191RFC4861RFC4862, and RFC6275. As a router, the RPL node may advertise the information from the options as required for the specific link.
For further information, please refer to RFC 6550, "RPL: IPv6 Routing Protocol for Low-Power and Lossy Networks".
## UDP Server
Flow chart for UDP server
In the example, UDP server does three tasks primarily.
1. Initializes RPL DAG;
2. Sets up UPD connection;
3. Waits for packets from client, receives and print them on stdout.
### Initialize RPL DAG
// check whether the ADDR_MANUAL was set succefuly or not
if(root_if != NULL) {
rpl_dag_t *dag;
//set the ip adress of server as the root of initial DAG
PRINTF("created a new RPL dag\n");
} else {
PRINTF("failed to create a new RPL DAG\n");
}
### create UDP connection
//create new UDP connection to client's port
server_conn = udp_new(NULL, UIP_HTONS(UDP_CLIENT_PORT), NULL);
if(server_conn == NULL) {
PRINTF("No UDP connection available, exiting the process!\n");
PROCESS_EXIT();
}
//bing the connection to server's local port
udp_bind(server_conn, UIP_HTONS(UDP_SERVER_PORT));
PRINTF("Created a server connection with remote address ");
PRINTF(" local/remote port %u/%u\n", UIP_HTONS(server_conn->lport),
UIP_HTONS(server_conn->rport));
### receives and processes incoming packet
while(1) {
PROCESS_YIELD();
//if there is packet available
if(ev == tcpip_event) {//当接收到数据包后，tcpip会设置tcpip_event事件
tcpip_handler();
} else if (ev == sensors_event && data == &button_sensor) {
PRINTF("Initiaing global repair\n");
rpl_repair_root(RPL_DEFAULT_INSTANCE);
}
}
//call this function if packet available
static void
tcpip_handler(void)
{
char *appdata;
if(uip_newdata()) {
appdata = (char *)uip_appdata;
appdata[uip_datalen()] = 0;
//print the data of packet
PRINTF("DATA recv '%s' from ", appdata);
PRINTF("%d",
PRINTF("\n");
}
## UDP Client
Flow chart for UDP client
In the example, UDP server does two tasks primarily.
1. Sets up UPD connection;
2. Sends packet to UDP server periodically.
### Sets up UPD connection
/* new connection with remote host */
client_conn = udp_new(NULL, UIP_HTONS(UDP_SERVER_PORT), NULL);
if(client_conn == NULL) {
PRINTF("No UDP connection available, exiting the process!\n");
PROCESS_EXIT();
}
udp_bind(client_conn, UIP_HTONS(UDP_CLIENT_PORT));
PRINTF("Created a connection with the server ");
PRINTF(" local/remote port %u/%u\n",
UIP_HTONS(client_conn->lport), UIP_HTONS(client_conn->rport));
### Sends packet
//set time interval by SEND_INTERVAL
etimer_set(&periodic, SEND_INTERVAL);
while(1) {
PROCESS_YIELD();
if(ev == tcpip_event) {
tcpip_handler();
}
//send packet every SEND_INTERVAL
if(etimer_expired(&periodic)) {
etimer_reset(&periodic);
ctimer_set(&backoff_timer, SEND_TIME, send_packet, NULL);//回调定时器，回调函数send_packet，发送数据包
static void
send_packet(void *ptr)
{
static int seq_id;
seq_id++;
PRINTF("DATA send to %d 'Hello %d'\n",
sprintf(buf, "Hello %d from the client", seq_id);
//send packet through client_conn to UDP server
uip_udp_packet_sendto(client_conn, buf, strlen(buf),
}
## Cooja Simulation
The DGRM model is used. The following are the steps to form a new simulation:
Note: You can refer to Cooja Simulator for an introduction to Cooja.
• Run Cooja
Go to your Contiki folder(contiki-2.7) and then go to /tools/cooja directory
Run the command sudo ant run to open up a cooja GUI.
$cd contiki-2.7/tools/cooja$ sudo ant run
• Open an existing simulation file
In the GUI, select File->Open simulation->Browse..
After the dialogue shows up, Open home/contiki-2.7/examples/ipv6/rpl-udp/rpl-udp.csc
Note: If compile error shows up, please run
$cd contiki-2.7/examples/ipv6/rpl-udp$ make
You are suppose to see the simulation showing up like this.
• Run Simulation
Run the simulation by using the Start option in the Simulation Control window. This will initiate the motes and allocate all with a new Rime address and other initialization processes.
• Watch Output
The motes output and debug messages can be seen in the Motes Output window. You can filter the output based on the node ID:node_id to watch a particular node. You can also watch particular debug messages by filtering them. The other useful functions of the Motes Output are File, Edit and View. The File option helps in saving the output to a file. The Edit has the option of copying the output - either full or a particular selected messages. You can also clear the messages using the Clear all messages option.
You can use these messages saved in file to make observations and plot graphs according to the objective of your experiment. |