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[ "# 424 Longest Repeating Character Replacement\n\nYou are given a string `s` and an integer `k`. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most `k` times.\n\nReturn the length of the longest substring containing the same letter you can get after performing the above operations.\n\nExample 1:\n\n``````Input: s = \"ABAB\", k = 2\nOutput: 4\nExplanation: Replace the two 'A's with two 'B's or vice versa.\n``````\n\nExample 2:\n\n``````Input: s = \"AABABBA\", k = 1\nOutput: 4\nExplanation: Replace the one 'A' in the middle with 'B' and form \"AABBBBA\".\nThe substring \"BBBB\" has the longest repeating letters, which is 4.\n``````\n\nConstraints:\n\n• `1 <= s.length <= 105`\n• `s` consists of only uppercase English letters.\n• `0 <= k <= s.length`\n `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 `````` ``````class Solution: def characterReplacement(self, s: str, k: int) -> int: win_start = 0 max_len = 0 count_map = defaultdict(int) # char count within the current window max_count = 0 for win_end, c in enumerate(s): count_map[c] += 1 max_count = max(max_count, count_map[c]) if win_end - win_start + 1 - max_count > k: count_map[s[win_start]] -= 1 win_start += 1 max_len = max(max_len, win_end - win_start + 1) return max_len``````" ]

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[ "Solutions by everydaycalculation.com\n\n1st number: 3 0/12, 2nd number: 7/90\n\n36/12 + 7/90 is 277/90.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 12 and 90 is 180\n2. For the 1st fraction, since 12 × 15 = 180,\n36/12 = 36 × 15/12 × 15 = 540/180\n3. Likewise, for the 2nd fraction, since 90 × 2 = 180,\n7/90 = 7 × 2/90 × 2 = 14/180\n540/180 + 14/180 = 540 + 14/180 = 554/180\n5. 554/180 simplified gives 277/90\n6. So, 36/12 + 7/90 = 277/90\nIn mixed form: 37/90\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]

[ "# Solving equations with brackets\n\nTo expand a bracket, multiply everything inside the bracket by the number outside.\n\n## Example\n\nExpand the bracket:", null, "Multiply everything inside the bracket by the number outside to give:", null, "Simplify this to:", null, "## Example\n\nExpand:", null, "Multiply everything inside the bracket by the number outside to give:", null, "Simplify this to:", null, "Question\n\nSolve the equation:", null, "First expand the brackets:", null, "Then subtract", null, "from both sides:", null, "Add", null, "to both sides:", null, "And divide by", null, "to give:", null, "" ]

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[ "# Voltage regulator for a AC-DC PSU\n\nStatus\nNot open for further replies.\n\n#### dennis@\n\n##### New Member\nHi\n\nWe have to build a AC to DC PSU on friday, using the following schematics\n\nThe AC input is 15 V 50Hz, and the output must be: 9V 100mA\n\nI need help with these problems:\n\n1) Dimension of the Capacitor.\n2) What voltage should the zenierdiode be on\n3 How big should R1 be\n4) Effekt lost in Q1\n\n1) Now for the capacitor i found the 10% ripple is fine (Power Supplies)\nBut if i want 5% is this the right way to do it:\nVc=Vs*(1-e^(-t/RC))\n\n15*5% = 15*(1-e^(-0.01sec/x) that gives x=0,00244 F The 0.01 sec is from the 50Hz AC to 100Hz DC.\n\n2) Q1 needs 0.7 volts to work, so the zenier should be able to handle output + VQ1\n\nThats 9v + 0.7V = 9.7V\n\nIs this right?\n\n3) R1 = (15V - 9.7)/(Izenier + IQ1) Or is it 15V - (9.7 + 0.7V)/(Izenier + IQ1) 0.7V for the zenier?\n\n4) Im lost on this one :S\n\n#### crutschow\n\n##### Well-Known Member\nThe filter capacitor discharge is not an RC exponential. Because of the regulator, the load current (and thus the filter capacitor current) is a constant i = Vcc/Rload. Thus the capacitor ripple voltage is approximately V = (i*t)/C where t = 10ms.\n\nThe zener voltage should be the desired output voltage plus the transistor base emitter voltage of 0.7V or 9.7V as you stated.\n\nR1 must supply the minimum desired zener current, plus the transistor base current required for the maximum output current with the minimum transistor gain (beta or Hfe) .\n\n#### dennis@\n\n##### New Member\nHi crutschow Thank you for you reply.\n\nIknew i messed up with the capacitor", null, "To calculate R1 do i have to consider the 0.7 Volts the diode takes or is it just 15-9.7?\n\n#### crutschow\n\n##### Well-Known Member\nTo calculate R1 do i have to consider the 0.7 Volts the diode takes or is it just 15-9.7?\nThe DC voltage on the capacitor is more than 15V, which is the RMS AC voltage. The peak AC voltage at the capacitor is about 1.4*15 minus two bridge diode's forward drop. For worst-case you would want to use a voltage slightly below the minimum voltage at the bottom of the ripple as determined by the load current and the capacitor size.\n\nEdit: If you simulate the circuit with simulator such as LTspice (free download) you can see how your circuit works before you build it.\n\nLast edited:\nStatus\nNot open for further replies.\n\nReplies\n1\nViews\n1K\nReplies\n6\nViews\n7K\nReplies\n3\nViews\n1K\nReplies\n15\nViews\n2K\nReplies\n1\nViews\n1K", null, "" ]

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[ "# Linear algebra question\n\nsweetiepi\n\n## Homework Statement\n\nLet A be a real n x n matrix. Prove that we can find a subspace V in R^N such that 1 <= dim V < = 2 and A(V) is a subset of V.\n\n## Homework Equations\n\nNone I don't think.\n\n## The Attempt at a Solution\n\nI know that the eigenspace of a matrix satisfies the condition that A(E) is a subset of E since for any vectors v in the eigenspace, Av = λv. Since we know v is in the eigenspace, any multiple of v by λ is also in the eigenspace. Now how am I supposed to incorporate the dimension into the argument?\n\n## Answers and Replies\n\nMentor\nAnother possibility is the nullspace of A. If null(A) is a line through the origin, then dim(null(A)) = 1. If null(A) is a plane containing the origin, then dim(null(A)) = 2.\n\nsweetiepi\nYes I knew that, but how does it relate to A - λI?\n\nMentor\nIt doesn't. Your problem statement doesn't mention anything about eigenvalues or eigenspaces, so why is that a concern?\n\nsweetiepi\nI guess I was just trying to stick with the eigenspace argument. So Av = λv and the set of all solutions v that satisfies this is in the eigenspace. And (A-λI)v = 0. So how do I prove that the dimension of the eigenspace is between or equal to 1 and 2?\n\nMentor\nThe dimension in your proof has to be either 1 or 2; it can't be some number in between.\n\nTo get to eigenvalues, eigenvectors, etc. you first have to get the characteristic polynomial, and then you have to factor it. The Fund. Thm. of Algebra states that every polynomial of degree n with complex coefficients has at least one complex root.\n\nsweetiepi\nOk, I'm still confused how to do this problem...\n\nSo you have a matrix A. I solve det(A-λI) = 0 and find the eigenvalues. Suppose I take an eigenvalue and plug it into A-λI and solve the equation (A-λI)x = 0. The the set of all solutions to this equation is the eigenspace. I know that the eigenspace satisfies A(V) subset of V because Ax = λx and if x is in the eigenspace, then any scalar multiple of x is also in the eigenspace by the axioms of a vector space. If the dimension is 1, then I have a line, and if the dimension is 2, I have a plane. Is that what they're asking for? That doesn't seem right or complete. What is meant by a space where the dimension is 1<= dimV <= 2?\n\nMentor\nFor your last question, 1 <= dim(V) <= 2 means that dim(V) = 1 or dim(V) = 2.\n\nWhat this problem boils down to is that you need to show that for any nxn matrix A that there is an eigenspace of dimension 1 or there is an eigenspace of dimension 2. Think about what you need to do to get the eigenvalues.\n\nsweetiepi\nOk, so I know that if A has distinct eigenvalues λ1, ..., λp, then for 1 <= k <= p the dimension of the eigenspace for λk is less than or equal to the multiplicity of the eigenvalue λi. So for this problem, that would imply that there has to be an eigenvalue with multiplicity of 1 or an eigenvalue with multiplicity of 2. But how am I supposed to show that this is the case for any real n x n matrix?\n\ncalorimetry\nWhat if A is real, but it does not have any real eigenvalue (i.e. only complex eigenvalues) like A = [0 -1],[1, 0] a 2x2 matrix?\n\nI don't think that your line of argument would apply here since there is no eigenspace in R^n with complex eigenvalues. :(\n\nsweetiepi\nGah! I still don't get it. I don't even think I have any sort of argument going... I'm just that confused. I just don't see how we can say that the dimension of V is either 1 or 2 when we don't have much information about A.\n\ncalorimetry\nRemember V can be almost anything that you want as long as you define it properly and it satisfies the given conditions. What I did was separating the problem into two cases, the first one is pretty easy, the second one, not so much.\n\nCase 1: A has some real eigenvalue\nThen there is some eigenvector v corresponding to that particular eigenvalue.\nYou can DEFINE V = span (v) and you can check that V satisfy all the given conditions.\n\nCase 2: Uber hard for me :(, but I got it, and now you going to have to think a bit\nA has ONLY complex eigenvalue a + bi\nHow would you define V? And then check it? Remember that there still exist some eigenvector (only that its complex)\n\nNow I got to do 3 and 4, want to give me some hints on those? :d\n\nMentor\nOk, so I know that if A has distinct eigenvalues λ1, ..., λp, then for 1 <= k <= p the dimension of the eigenspace for λk is less than or equal to the multiplicity of the eigenvalue λi. So for this problem, that would imply that there has to be an eigenvalue with multiplicity of 1 or an eigenvalue with multiplicity of 2. But how am I supposed to show that this is the case for any real n x n matrix?\n\nBut where do the eigenvalues come from? Look at what I said in post #6.\n\nsweetiepi\nOk, so for your case 1, we have an eigenvalue that gave us some eigenvector v. I define V to be the span of v. Because it is just the one vector, it is linearly independent (which we know is true since an eigenvector is not the zero vector). Thus, the dimension is equal to 1. And if v is in the eigenspace, then every multiple of v is also in the eigenspace by some axiom somewhere. So the A(V) is a subset of V because this property holds for an eigenspace.\n\nFor case 2, we have a complex eigenvector. If this was of the form [a + bi], couldn't we say that the eigenvector is given by v = [a] + i? And then do the same thing as before, where we define the span of v to be V? Except that would still give me a dimension of one... hmm.\n\nNot sure when you posted your response, but here's my hint for 3b: Try to prove first that Bx is also an eigenvector of A.\n\ncalorimetry\nYou can't define V to be span of v for case 2 because V is a subspace in R^n and v is a complex eigenvector (i.e. the span of a complex vector will create complex vectors outside of R^n). Try separate v into Re(v) and Im(v) and make V to be a span of those vectors. (This is where dim = 2 come from).\n\nAlso you will need A Re(x) = Re (Ax) and likewise for the other component to verify the other property of V. You can find this fact on page 340.\n\nAnd I'm already done with 3, working on 4 now.\n\nsweetiepi\nAh! Makes so much more sense! Thank you SO much for your help :)\n\nI just finished 4a. Find a basis for nul((C-lambda I)^2) and relate this to nul(C - lambda I).. if you haven't already.\n\ncalorimetry\nEh not sure what you just said, but I just make P to be the identity matrix, then C is the same as the matrix between P and P^1. After all, it asks for some matrix P.\n\ncalorimetry\nI'm stuck on 4b, you got any plan on how to do it?\n\nsweetiepi\nHmm, not entirely sure. If we consider the equation XA = BX, we must be able to show that it has a complex solution, which would account for the A = PBP^-1 and that it also has a real solution, which would account for the A = QBQ^-1. Maybe if we start with one and then work to the other?\n\ncalorimetry\nHow do you know that the equation xA = Bx has a complex solution? And what does that have to do with PBP^-1? I don't see the reasoning, care to elaborate?\n\nsweetiepi\nIn the problem it says that A and B are real matrices such that A = PBP^-1 for some invertible 2 x 2 complex matrix P. Since P is invertible, so is P^-1, and this can be rewritten as AP = PB. We can do the same thing with the real matrix Q to get AQ = PQ. So in the equation AX = XB (I think I accidentally typed XA = BX last time) we should get a complex solution (that's P) and a real solution (that's Q). We are given that there is a complex solution from the get-go, so then go from there." ]

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[ "Class 10 RD Sharma Solutions – Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.4 | Set 2\n\n• Last Updated : 29 Nov, 2021\n\nQuestion 15. 2ax + 3by = a + 2b and 3ax + 2by = 2a + b\n\nSolution:\n\nGiven that,\n\n2ax + 3by = a + 2b\n\n3ax + 2by = 2a + b\n\nOn comparing both the equation with the general form we get\n\na1 = 2a, b1 = 3b, c1 = -(a + 2b), a2 = 3a, b2 = 2b, c2 = -(2a + b)\n\nNow by using cross multiplication we get\n\nx/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)\n\nx/(-3b(2a + b) + 2b(a + 2b)) = y/(-3a(a + 2b) + 2a(2a + b)) = 1/(4ab – 9ab)\n\nx/(b2 – 4ab) = y/(a2 – 4ab) = 1/-5ab\n\nx/(-4ab + b2) = 1/-5ab\n\nx/b(b – 4a) = 1/-5ab\n\nx = (4a – b)/5a\n\nand,\n\n= -y/(-a2 + 4ab) = 1/-5ab\n\n= -y/a(-a + 4b) = 1/-5ab\n\ny = (4b – a)/5b\n\nHence, x = (4a – b)/5a and y = (4b – a)/5b\n\nQuestion 16. 5ax + 6by = 28 and 3ax + 4by = 18\n\nSolution:-\n\nGiven that,\n\n5ax + 6by = 28\n\n3ax + 4by = 18\n\nOr, 5ax + 6by – 28 = 0\n\n3ax + 4by – 18 = 0\n\nHere, a1 = 5a, b1 = 6b, c1=-28\n\na₂= 3a, b₂ = 4b, c₂ =-18\n\nOn comparing both the equation with the general form we get\n\na1 = 5a, b1 = 6b, c1 = -28, a2 = 3a, b2 = 4b, c2 = -18\n\nNow by using cross multiplication we get\n\nx/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)\n\nx/4b = -y/-6a = 1/2ab\n\nx/4b = 1/2b\n\nx = 2/a\n\nand,\n\n-y/-6a = 1/2ab\n\ny = 3/b\n\nHence, x = 2/a and y = 3/b\n\nQuestion 17. (a + 2b)x + (2a – b)y = 2 and (a – 2b)x + (2a + b)y = 3\n\nSolution:\n\nGiven that,\n\n(a + 2b)x + (2a – b)y = 2\n\n(a – 2b)x + (2a + b)y = 3\n\nOr\n\n(a + 2b) x + (2a – b)y – 2 = 0\n\n(a – 2b) x + (2a + b)y – 3 = 0\n\nOn comparing both the equation with the general form we get\n\na1 = a + 2b, b1 = 2a – b, c1 = -2, a2 = a – 2b, b2 = 2a + b, c2 = -3\n\nNow by using cross multiplication we get\n\nx/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)\n\nx/(-3(2a – b)) + (2(2a + b)) = y/(-2(a – 2b)) + (3(a + 2b)) = 1/((a + 2b)(2a + b) – (a – 2b)(2a – b))\n\nx/(5b – 2a) = y/(a + 10b) = 1/(2a2 + 5ab + 2b2 – 2a2 + 5ab – 2b2)\n\nx/(5b – 2a) = y/(a + 10b) = 1/10ab\n\nSo,\n\nx/(5b – 2a) = 1/10ab\n\nx= (5b – 2a)/10ab\n\nand,\n\ny/(a + 10b) = 1/10ab\n\ny = (a + 10b)/10ab\n\nHence, x = (5b – 2a)/10ab and y= (a + 10b)/10ab\n\nQuestion 18. x((a – b) + (ab/(a – b))) = y((a + b) – (ab/(a + b))) and x + y = 2a2\n\nSolution:\n\nGiven that,\n\nx((a – b) + (ab/(a – b))) = y((a + b) – (ab/(a + b)))\n\nOr on solving we get\n\nx((a2 + b2 – 2ab + ab)/(a – b)) = y((a2 + b2 + 2ab – ab)/(a + b))\n\n= x((a2 + b2 – 2ab + ab)/(a – b)) – y((a2 + b2 + 2ab – ab)/(a + b)) = 0\n\nand,\n\nx + y = 2a2\n\nOn comparing both the equation with the general form we get\n\na1 = (a2 + b2 – 2ab + ab)/(a – b), b1 = (a2 + b2 + 2ab – ab)/(a + b), c1 = 0,\n\na2 = 1, b2 = 1, c2 = 2a2\n\nNow by using cross multiplication we get\n\nx/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)\n\n⇒ x/(2a2((a2 + b2 – 2ab + ab)/(a – b)) – 0) = y/(0 + 2a2((a2 + b2 + 2ab – ab)/(a + b)))\n\n= 1/(((a2 + b²)/(a – b) + (a2 + b2 + ab)/(a + b)))\n\n⇒ x/(2a2((a2 + b2 – 2ab + ab)/(a – b))) = y/(-2a2)((a2 + b2 + 2ab – ab)/(a + b)) = 1/(2a3/(a2 – b2))\n\nNow,\n\nx/(2a2((a2 + b2 – 2ab + ab)/(a – b)) = 1/(2a3/(a2 – b2))\n\nx = (2a2(a2 + ab + b2)(a2 – b2)) / 2a3(a + b)\n\nx = (a3 – b3)/a\n\nand,\n\ny/(-2a2)((a2 + b2 + 2ab – ab)/(a + b)) = 1/(2a3/(a2 – b2))\n\ny = (2a2(a2 – ab + b2)(a2 – b2))/2a3(a – b)\n\ny = a3 + b3/a\n\nHence, x = (a3 – b3)/a and y = a3+ b3/a\n\nQuestion 19. bx + cy = a + b and ax[(1/(a – b)) – (1/(a + b))] + cy[(1/(b – a)) – (1/(b + a))] = 2a/(a + b)\n\nSolution:\n\nGiven that,\n\nbx + cy = a + b\n\nax[(1/(a – b)) – (1/(a + b))] + cy[(1/(b – a)) – (1/(b + a))] = 2a/(a + b)\n\nOr\n\nbx + cy -(a + b) = 0\n\nax((1/(a – b)) – (1/(a + b))) + cy(((1/(b – a)) – (1/(b + a))) – 2a/(a + b) = 0\n\nax(2b/(a2 – b2))) + cy(2a/(b2 – a2)) – 2a/(a + b) = 0\n\nOn comparing both the equation with the general form we get\n\na1 = b, b1 = c, c1 = -(a + b),\n\na2 = 2b/(a2 – b2), b2 = 2a/(b2 – a2), c2 = 2a/(a + b)\n\nNow by using cross multiplication we get\n\nx/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)\n\n⇒ x / ((-2ac/(a + b)) + ((2ac(a + b)/(b2 – a2)))) = y/((-(a + b)2ab)/(a2 – b2)) + (2ab/(a + b))\n\n= 1/((2abc/(b2– a2)) – (2abc/(a2 – b2)))\n\n⇒ x / ((-2ac/(a + b)) + ((2ac(a + b)/(b2 – a2)))) = y/((-(a + b)2ab)/(a2 – b2)) + (2ab/(a + b))\n\n= 1/(-4abc/(a2 – b2))\n\n⇒ x/(-2ac((1/(a + b)) + (1/(a – b))) = y/(2ab((-1/(a – b)) + (1/(a + b))) = 1/(-4abc/(a2 – b2))\n\n⇒ x/(-4a2c/(a2 – b2)) = y/(4ab2/(a2 – b2)) = 1/1/(-4abc/(a2 – b2))\n\nSo,\n\nx/(-4a2c/(a2 – b2)) = 1/(-4abc/(a2 – b2))\n\nx = a/b\n\nand,\n\ny/(4ab2/(a2 – b2)) = 1/(-4abc/(a2 – b2))\n\ny = b/c\n\nHence, x = a/b, y = b/c\n\nQuestion 20. (a – b) x + (a + b) y = 2a2 – 2b2 and (a + b) (x + y) = 4ab\n\nSolution:\n\nGiven that,\n\n(a – b) x + (a + b) y = 2a2 – 2b2\n\n(a + b) (x + y) = 4ab\n\n(a – b) x + (a + b) y – 2(a2 – b2) = 0\n\n(a + b)x +  (a + b)y – 4ab = 0\n\nOn comparing both the equation with the general form we get\n\na1 = a – b, b1 = a + b, c1 = -2,\n\na2 = a + b, b2 = a + b, c2 = -4ab\n\nNow by using cross multiplication we get\n\nx/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)\n\n⇒ x/(-(a + b)4ab + 2(a + b) (a2 – b2)) = y/(− 2(a2 − b2)(a + b) + 4ab(a – b))\n\n= 1/((a − b)(a + b) − (a + b)(a + b))\n\n⇒ x/(2(a + b)(a2 – b2 + 2ab)) = 1/-2b(a + b)\n\nx = (2ab – a2 + b2)/b\n\nand,\n\n= -y/(2(a – b) (a2 + b2) -2b (a + b)) = 1/ -2b(a + b)\n\ny = (a – b)(a2 + b2)/ b(a + b)\n\nHence, x = (2ab – a2 + b2)/b and y = (a – b)(a2 + b2)/ b(a + b)\n\nQuestion 21. a2x + b2y = c2 and b2x + a2y = d2\n\nSolution:\n\nGiven that,\n\na2x + b2y = c2\n\nb2x + a2y = d2\n\nOr\n\na2x + b2y  – c2 = 0\n\nb2x + a2y – d2 = 0\n\nOn comparing both the equation with the general form we get\n\na1 = a2, b1 = b2, c1 = -c2,\n\na2 = b2, b2 = a2, c2 = -d2\n\nNow by using cross multiplication we get\n\nx/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)\n\nx/(-b2d2 + a2c2) = y/(-c2b2 + a2d2) = 1/(a4-b4)\n\nx/(a2c2 – b2d2) = y/(a2d2 – c2b2) = 1/(a4-b4)\n\nTherefore,\n\n= x/(a2c2 – b2d2) = 1/(a4 – b4)\n\nx = (a2c2 – b2d2)/(a4 – b4)\n\nand,\n\n= y/(a2d2 – c2b2) = 1/(a4-b4)\n\ny = (a2c2 – b2d2) / (a4-b4)\n\nHence, x = (a2c2 – b2d2)/(a4 – b4),  y = (a2c2 – b2d2) / (a4 – b4)\n\nQuestion 22. ax + by = (a + b)/2 and 3x + 5y = 4\n\nSolution:\n\nGiven that,\n\nax + by = (a+b)/2\n\n3x + 5y = 4\n\nOr\n\nax + by – (a + b)/2 = 0\n\n3x + 5y – 4 = 0\n\nOn comparing both the equation with the general form we get\n\na1 = a, b1 = b, c1 = -(a + b)/2,\n\na2 = 3, b2 = 5, c2 = -4\n\nNow by using cross multiplication we get\n\nx/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)\n\n= x/(-4b + 5((a + b)/2)) = y/(-3((a + b)/2) + 4a) = 1/(5a – 3b)\n\n= x/((5a – 3b)/2) = y/((5a – 3b)/2) = 1/(5a – 3b)\n\nNow,\n\nx/((5a – 3b)/2) = 1/(5a – 3b)\n\nx = (5a – 3b)/(2(5a – 3b))\n\nx = 1/2\n\nand,\n\ny/((5a – 3b)/2) = 1/(5a – 3b)\n\ny = (5a – 3b)/(2(5a – 3b))\n\ny = 1/2\n\nHence, x = 1/2, y = 1/2\n\nQuestion 23. 2 (ax – by) + a + 4b = 0 and 2 (bx + ay) + b – 4a = 0\n\nSolution:\n\nGiven that,\n\n2 (ax – by) + (a + 4b) = 0\n\n2 (bx + ay) + (b – 4a) = 0\n\nOn comparing both the equation with the general form we get\n\na1 = 2a, b1 = -2b, c1 = a + 4b,\n\na2 = 2b, b2 = 2a, c2 = b – 4a\n\nNow by using cross multiplication we get\n\nx/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)\n\n= x/((-2b(a + 4b)) – (2a(b – 4a ))) = y/((2b(a + 4b)) – (2a(b – 4a))) = 1/(4a2 + 4b2)\n\n= x/(-2b2 + 8ab – 2ab + 8a2) = y/(2ab + 8b2 – 2ab + 8a2) = 1/4(a2 + b2)\n\n= x/-2(a2 + b2) = y/8(a2 + b2) = 1/4(a2 + b2)\n\nSo,\n\n= x/-2(a2 + b2) = 1/4(a2 + b2)\n\nx = -1/2\n\nand,\n\n= y/8(a2 + b2) = 1/4(a2 + b2)\n\ny = 2\n\nHence, x = -1/2 and y = 2\n\nQuestion 24. 6 (ax + by) = 3a + 2b and 6 (bx – ay) = 3b – 2a\n\nSolution:\n\ngiven that,\n\n6 (ax + by) = 3a + 2b\n\n6 (bx – ay) = 3b – 2a\n\n6 (ax + by) -(3a + 2b)=0….       (1)\n\n6 (bx – ay) -(3b – 2a) =0…..       (2)\n\nOn comparing both the equation with the general form we get\n\na1 = 6a, b1 = 6b, c1 = -(3a – 2b),\n\na2 = 6b, b2 = 66a, c2 = -(3b – 2a)\n\nNow by using cross multiplication we get\n\nx/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)\n\n= x/(-6b(3b – 2a) – 6a(3a – 2b)) = y/(-6b(3a – 2b) + 6a(3b – 2a)) = 1/(-36a2 – 36b2)\n\n= x/(-18(a2 + b2)) = y/(-12(a2 + b2)) = 1/(-36(a2 + b2))\n\nTherefore,\n\nx/(-18(a2 + b2)) = 1/(-36(a2 + b2))\n\nx = 1/2\n\nand,\n\ny/(-12(a2 + b2)) = 1/(-36(a2 + b2))\n\ny = 1/3\n\nHence, x = 1/2 and y = 1/3\n\nQuestion 25. (a2/x) − (b2/y) = 0 and (a2b/x) − (b2a/y) = a + b, x, y ≠ 0\n\nSolution:\n\nGiven that,\n\n(a2/x) − (b2/y) = 0\n\n(a2b/x) − (b2a/y) = a + b\n\nOr\n\n(a2b/x) − (b2a/y) – (a + b) = 0\n\nOn comparing both the equation with the general form we get\n\na1 = a2, b1 = -b2, c1 = 0,\n\na2 = a2b, b2 = b2a, c2 = -(a + b)\n\nNow by using cross multiplication we get\n\nx/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)\n\n= (1/x)/(b2(a + b) – 0) = (1/y)/(0 + (a2(a + b))) = 1/(a3b2 – a2b3)\n\n= (1/x)/(b2(a + b)) = (1/y)/(a2(a + b)) = 1/a2b2(a + b)\n\nSo,\n\n= (1/x)/(b2(a + b)) = 1/a2b2(a + b)\n\nx = a2\n\nand,\n\n= (1/y)/(a2(a + b)) = 1/a2b2(a + b)\n\ny = b2\n\nHence, x = a2 and y = b2\n\nQuestion 26. mx – ny = m2 + n2 and x + y = 2m\n\nSolution:\n\nGiven that,\n\nmx – ny = m2 + n2\n\nx + y = 2m\n\nOr\n\nmx – ny -(m2 + n2) = 0\n\nx + y – 2m = 0\n\nOn comparing both the equation with the general form we get\n\na1 = m, b1 = -n, c1 = -(m2 + n2),\n\na2 = 1, b2 = 1, c2 = -2m\n\nNow by using cross multiplication we get\n\nx/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)\n\n= x/(2mn + (m2 + n2)) = y/(-(m2 + n2) + 2m2) = 1/(m + n)\n\n= x/(m + n)2 = y/(m2 – n2) = 1/(m + n)\n\nTherefore,\n\nx/(m + n)2 = 1/(m + n)\n\nx = m + n\n\nand,\n\ny/(m2 – n2) = 1/(m + n)\n\ny = m – n\n\nHence, x = m + n, y = m – n\n\nQuestion 27. (ax/b) – (by/a) = a + b and ax – by = 2ab\n\nSolution:\n\nGiven that,\n\n(ax/b) – (by/a) = a + b\n\nax – by = 2ab\n\nOr\n\n(ax/b) – (by/a) – (a + b) = 0\n\nax – by – 2ab = 0\n\nOn comparing both the equation with the general form we get\n\na1 = a/b, b1 = -b/a, c1 = -(a + b),\n\na2 = a, b2 = b, c2 = -2ab\n\nNow by using cross multiplication we get\n\nx/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)\n\n= x/b(b – a) = -y/a(-a + b) = 1/(b – a)\n\nSo,\n\nx/b(b – a) = 1/(b – a)\n\nx = b\n\nand,\n\n-y/a(-a + b) = 1/(b – a)\n\ny = -a\n\nHence, x = b, y = -a\n\nQuestion 28. (b/a)x + (a/b)y = a2 + b2 and x + y = 2ab\n\nSolution:\n\nGiven that,\n\n(b/a)x + (a/b)y = a2 + b2\n\nx + y = 2ab\n\nOr\n\n(b/a)x + (a/b)y – (a2 + b2) = 0\n\nx + y – 2ab = 0\n\nOn comparing both the equation with the general form we get\n\na1 = b/a, b1 = a/b, c1 = -(a2 + b2),\n\na2 = 1, b2 = 1, c2 = -2ab\n\nNow by using cross multiplication we get\n\nx/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)\n\n= x/(b2 – a2) = y/(-b2 + a2) = 1/((b2 – a2)/ab)\n\nTherefore,\n\nx/(b2 – a2) = 1/((b2 – a2)/ab)\n\nx = ab\n\ny/(-b2 + a2) = 1/((b2 – a2)/ab)\n\ny = ab\n\nHence, x = ab, y = ab\n\nMy Personal Notes arrow_drop_up" ]

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[ "", null, "", null, "", null, "Chapter 14, Problem 14.12E\n\nChapter\nSection\nTextbook Problem\n\nDetermining missing items in return on investment computationsOne item is omitted from each of the following computations of the return on investment", null, "Determine the missing items, identifying each by the appropriate letter.\n\nTo determine\n\nConcept Introduction:\n\nProfit Margins: It is a measure which is generally used by the organization to find out the best department in the organization in terms of profit. To calculate the profit margins divide the profits with sales.\n\nReturn on investment: In this we divide the total return or profit with the investment into the assets of organization. It is the best way of calculating the % of return in the business.\n\nTo Calculate:\n\nFind out the missing figure?\n\nExplanation\n\na. = ROI/ Profit margin\n\n= 12%/8%\n\n= 1.5\n\nb. = Profit margin* Investment Turnover\n\n= 16%* 1.25\n\n= 20%\n\nc. = ROI/ Investment Turnover\n\n= 24%/ 1...\n\nStill sussing out bartleby?\n\nCheck out a sample textbook solution.\n\nSee a sample solution\n\nThe Solution to Your Study Problems\n\nBartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!\n\nGet Started\n\nAFTER-TAX COST OF DEBT The Heuser Companys currently outstanding bonds have a 10% coupon and a 12% yield to mat...\n\nFundamentals of Financial Management, Concise Edition (with Thomson ONE - Business School Edition, 1 term (6 months) Printed Access Card) (MindTap Course List)\n\nShould an economic model describe reality exactly?\n\nEssentials of Economics (MindTap Course List)\n\nWhat is a composite primary key?\n\nPkg Acc Infor Systems MS VISIO CD", null, "" ]

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[ "", null, "", null, "# GaussianStatisticsAddLanguageSpecificTextSet(\"LST9CE693A4_0?cpp=::|nu=.\");InverseComplementaryErrorFunction Method\n\nCalculates the inverse of the complementary error function.\n\nNamespace:  AGI.Foundation\nAssembly:  AGI.Foundation.Core (in AGI.Foundation.Core.dll) Version: 21.1.408.0 (21.1.408.0)", null, "Syntax\n```public static double InverseComplementaryErrorFunction(\ndouble p\n)```\n\n#### Parameters\n\np\nType: SystemDouble\nThe value to use to calculate the inverse of the complementary error function, from 0.0 to 2.0.\n\n#### Return Value\n\nType: Double\nThe inverse of the complementary error function.", null, "Remarks\nThe ComplementaryErrorFunction(Double) approaches its asymptote very quickly, at double precision its solutions are rounded to 0.0 or 2.0 at an input of +-5.93. If this method returns negative or positive infinity and a finite value must be used instead, anything outside of that range would make an acceptable substitute.", null, "See Also" ]

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[ "If you're seeing this message, it means we're having trouble loading external resources on our website.\n\nIf you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.\n\n## Algebra 1\n\n### Course: Algebra 1>Unit 13\n\nLesson 7: Factoring quadratics with difference of squares\n\n# Difference of squares intro\n\nYou might need:", null, "Calculator\n\n## Problem\n\nThe rectangle below has an area of ${x}^{2}-16$ square meters and a width of $x+4$ meters.\nWhat expression represents the length of the rectangle?\n$\\text{Length}=$\nmeters\nStuck?\nStuck?" ]

[ null, "https://cdn.kastatic.org/images/exercise-solution-tools/calculator-icon.svg", null ]

[ "Showing 25 of 102 results\n\n#### Direct virtual photon production in Au+Au collisions at $\\sqrt{s_{NN}}$ = 200 GeV\n\nThe collaboration Adamczyk, L. ; Adkins, J.K. ; Agakishiev, G. ; et al.\nPhys.Lett. B770 (2017) 451-458, 2017.\nInspire Record 1474129\n\nWe report the direct virtual photon invariant yields in the transverse momentum ranges $1\\!<\\!p_{T}\\!<\\!3$ GeV/$c$ and $5\\!<\\!p_T\\!<\\!10$ GeV/$c$ at mid-rapidity derived from the dielectron invariant mass continuum region $0.10<M_{ee}<0.28$ GeV/$c^{2}$ for 0-80\\% minimum-bias Au+Au collisions at $\\sqrt{s_{NN}}=200$ GeV. A clear excess in the invariant yield compared to the number-of-binary-collisions ($N_{bin}$) scaled $p+p$ reference is observed in the $p_T$ range $1\\!<\\!p_{T}\\!<\\!3$ GeV/$c$. For $p_T\\!>6$ GeV/$c$ the production follows $N_{bin}$ scaling. Model calculations with contributions from thermal radiation and initial hard parton scattering are consistent within uncertainties with the direct virtual photon invariant yield.\n\n22 data tables\n\nDielectron invariant mass spectra in 1.0-1.5 GeV/c.\n\nDielectron invariant mass spectra in 1.5-2.0 GeV/c.\n\nDielectron invariant mass spectra in 2.0-2.5 GeV/c.\n\nMore…\n\n#### Jet-like Correlations with Direct-Photon and Neutral-Pion Triggers at $\\sqrt{s_{_{NN}}} = 200$ GeV\n\nThe collaboration Adamczyk, L. ; Adkins, J.K. ; Agakishiev, G. ; et al.\nPhys.Lett. B760 (2016) 689-696, 2016.\nInspire Record 1442357\n\nAzimuthal correlations of charged hadrons with direct-photon ($\\gamma_{dir}$) and neutral-pion ($\\pi^{0}$) trigger particles are analyzed in central Au+Au and minimum-bias $p+p$ collisions at $\\sqrt{s_{_{NN}}} = 200$ GeV in the STAR experiment. The charged-hadron per-trigger yields at mid-rapidity from central Au+Au collisions are compared with $p+p$ collisions to quantify the suppression in Au+Au collisions. The suppression of the away-side associated-particle yields per $\\gamma_{dir}$ trigger is independent of the transverse momentum of the trigger particle ($p_{T}^{\\mathrm{trig}}$), whereas the suppression is smaller at low transverse momentum of the associated charged hadrons ($p_{T}^{\\mathrm{assoc}}$). Within uncertainty, similar levels of suppression are observed for $\\gamma_{dir}$ and $\\pi^{0}$ triggers as a function of $z_{T}$ ($\\equiv p_T^{\\mathrm{assoc}}/p_T^{\\mathrm{trig}}$). The results are compared with energy-loss-inspired theoretical model predictions. Our studies support previous conclusions that the lost energy reappears predominantly at low transverse momentum, regardless of the trigger energy.\n\n21 data tables\n\nThe Azimuthal correlation functions of charged hadrons per trigger\n\nThe Azimuthal correlation functions of charged hadrons per trigger\n\nThe Azimuthal correlation functions of charged hadrons per trigger\n\nMore…\n\n#### Exclusive $\\rho ^0$ meson photoproduction with a leading neutron at HERA\n\nThe collaboration Andreev, V. ; Baghdasaryan, A. ; Begzsuren, K. ; et al.\nEur.Phys.J. C76 (2016) 41, 2016.\nInspire Record 1387751\n\nA first measurement is presented of exclusive photoproduction of $\\rho ^0$ mesons associated with leading neutrons at HERA. The data were taken with the H1 detector in the years 2006 and 2007 at a centre-of-mass energy of $\\sqrt{s}=319$  GeV and correspond to an integrated luminosity of 1.16 pb$^{-1}$ . The $\\rho ^0$ mesons with transverse momenta $p_T<1$  GeV are reconstructed from their decays to charged pions, while leading neutrons carrying a large fraction of the incoming proton momentum, $x_L>0.35$ , are detected in the Forward Neutron Calorimeter. The phase space of the measurement is defined by the photon virtuality $Q^2 < 2$  GeV$^2$ , the total energy of the photon–proton system $20 < W_{\\gamma p}< 100$  GeV and the polar angle of the leading neutron $\\theta _n < 0.75$ mrad. The cross section of the reaction $\\gamma p \\rightarrow \\rho ^0 n \\pi ^+$ is measured as a function of several variables. The data are interpreted in terms of a double peripheral process, involving pion exchange at the proton vertex followed by elastic photoproduction of a $\\rho ^0$ meson on the virtual pion. In the framework of one-pion-exchange dominance the elastic cross section of photon-pion scattering, $\\sigma ^\\mathrm{el}(\\gamma \\pi ^+ \\rightarrow \\rho ^0\\pi ^+)$ , is extracted. The value of this cross section indicates significant absorptive corrections for the exclusive reaction $\\gamma p \\rightarrow \\rho ^0 n \\pi ^+$ .\n\n11 data tables\n\nThe $\\gamma p$ cross section integrated in the domain $0.35 < x_L < 0.95$ and $-t^\\prime < 1$~GeV$^2$ and averaged over the energy range $20 < W_{\\gamma p} < 100$ GeV for two intervals of leading neutron transverse momentum.\n\nDifferential photoproduction cross sections ${\\rm d}\\sigma_{\\gamma p}/{\\rm d}x_L$ for the exclusive process $\\gamma p \\to \\rho^0 n \\pi^+$ in two regions of neutron transverse momentum and $20 < W_{\\gamma p} < 100$ GeV. The statistical, uncorrelated and correlated systematic uncertainties, $\\delta_{stat}$, $\\delta_{sys}^{unc}$ and $\\delta_{sys}^{cor}$ respectively, are given, which does not include the global normalisation error of $4.4\\%$.\n\nDouble differential photoproduction cross sections ${\\rm d^2}\\sigma_{\\gamma p}/{\\rm d}x_L{\\rm d}p_{T,n}^2$ in the range $20 < W_{\\gamma p} < 100$ GeV. The statistical, uncorrelated and correlated systematic uncertainties, $\\delta_{stat}$, $\\delta_{sys}^{unc}$ and $\\delta_{sys}^{cor}$ respectively, are given, which does not include the global normalisation error of $4.4\\%$.\n\nMore…\n\n#### Intranuclear cascading at ultrahigh-energy in heavy ion interactions\n\nJain, P.L. ; Singh, G. ; Sengupta, K. ;\nZ.Phys. C52 (1991) 465-470, 1991.\nInspire Record 316804\n2 data tables\n\nNo description provided.\n\nNo description provided.\n\n#### Higher Moments of Net-proton Multiplicity Distributions at RHIC\n\nThe collaboration Aggarwal, M.M. ; Ahammed, Z. ; Alakhverdyants, A.V. ; et al.\nPhys.Rev.Lett. 105 (2010) 022302, 2010.\nInspire Record 853304\n\nWe report the first measurements of the kurtosis (\\kappa), skewness (S) and variance (\\sigma^2) of net-proton multiplicity (N_p - N_pbar) distributions at midrapidity for Au+Au collisions at \\sqrt(s_NN) = 19.6, 62.4, and 200 GeV corresponding to baryon chemical potentials (\\mu_B) between 200 - 20 MeV. Our measurements of the products \\kappa \\sigma^2 and S \\sigma, which can be related to theoretical calculations sensitive to baryon number susceptibilities and long range correlations, are constant as functions of collision centrality. We compare these products with results from lattice QCD and various models without a critical point and study the \\sqrt(s_NN) dependence of \\kappa \\sigma^2. From the measurements at the three beam energies, we find no evidence for a critical point in the QCD phase diagram for \\mu_B below 200 MeV.\n\n40 data tables\n\n$\\Delta N_p$ multiplicity distribution in Au+Au collisions at $\\sqrt{s_{NN}}$ = 200 GeV for 0-5 percent central collisions at midrapidity (| y |< 0.5).\n\n$\\Delta N_p$ multiplicity distribution in Au+Au collisions at $\\sqrt{s_{NN}}$ = 200 GeV for 30-40 percent central collisions at midrapidity (| y |< 0.5).\n\n$\\Delta N_p$ multiplicity distribution in Au+Au collisions at $\\sqrt{s_{NN}}$ = 200 GeV for 70-80 percent central collisions at midrapidity (| y |< 0.5).\n\nMore…\n\n#### Energy Dependence of Moments of Net-proton Multiplicity Distributions at RHIC\n\nThe collaboration Adamczyk, L. ; Adkins, J.K. ; Agakishiev, G. ; et al.\nPhys.Rev.Lett. 112 (2014) 032302, 2014.\nInspire Record 1255072\n\nWe report the beam energy (\\sqrt s_{NN} = 7.7 - 200 GeV) and collision centrality dependence of the mean (M), standard deviation (\\sigma), skewness (S), and kurtosis (\\kappa) of the net-proton multiplicity distributions in Au+Au collisions. The measurements are carried out by the STAR experiment at midrapidity (|y| < 0.5) and within the transverse momentum range 0.4 < pT < 0.8 GeV/c in the first phase of the Beam Energy Scan program at the Relativistic Heavy Ion Collider. These measurements are important for understanding the Quantum Chromodynamic (QCD) phase diagram. The products of the moments, S\\sigma and \\kappa\\sigma^{2}, are sensitive to the correlation length of the hot and dense medium created in the collisions and are related to the ratios of baryon number susceptibilities of corresponding orders. The products of moments are found to have values significantly below the Skellam expectation and close to expectations based on independent proton and anti-proton production. The measurements are compared to a transport model calculation to understand the effect of acceptance and baryon number conservation, and also to a hadron resonance gas model.\n\n46 data tables\n\n$\\Delta N_p$ multiplicity distributions in Au+Au collisions at $\\sqrt{S_{NN}}=7.7$ GeV for 0-5 percent, 30-40 percent and 70-80 percent collision centralities at midrapidity.\n\n$\\Delta N_p$ multiplicity distributions in Au+Au collisions at $\\sqrt{S_{NN}}=11.5$ GeV for 0-5 percent, 30-40 percent and 70-80 percent collision centralities at midrapidity.\n\n$\\Delta N_p$ multiplicity distributions in Au+Au collisions at $\\sqrt{S_{NN}}=19.6$ GeV for 0-5 percent, 30-40 percent and 70-80 percent collision centralities at midrapidity.\n\nMore…\n\n#### USE OF PI P ---> PI PI N REACTIONS TO STUDY PI PI SCATTERING IN THE ELASTIC INTERACTION REGION\n\nAlekseeva, E.a. ; Kartamyshev, A.a. ; Makarin, V.k. ; et al.\nSov.Phys.JETP 55 (1982) 591-600, 1982.\nInspire Record 185170\n10 data tables\n\nNo description provided.\n\nNo description provided.\n\nNo description provided.\n\nMore…\n\n#### Long-range pseudorapidity dihadron correlations in $d$+Au collisions at $\\sqrt{s_{\\rm NN}}=200$ GeV\n\nThe collaboration Adamczyk, L. ; Adkins, J.K. ; Agakishiev, G. ; et al.\nPhys.Lett. B747 (2015) 265-271, 2015.\nInspire Record 1346551\n\nDihadron angular correlations in $d$+Au collisions at $\\sqrt{s_{\\rm NN}}=200$ GeV are reported as a function of the measured zero-degree calorimeter neutral energy and the forward charged hadron multiplicity in the Au-beam direction. A finite correlated yield is observed at large relative pseudorapidity ($\\Delta\\eta$) on the near side (i.e. relative azimuth $\\Delta\\phi\\sim0$). This correlated yield as a function of $\\Delta\\eta$ appears to scale with the dominant, primarily jet-related, away-side ($\\Delta\\phi\\sim\\pi$) yield. The Fourier coefficients of the $\\Delta\\phi$ correlation, $V_{n}=\\langle\\cos n\\Delta\\phi\\rangle$, have a strong $\\Delta\\eta$ dependence. In addition, it is found that $V_{1}$ is approximately inversely proportional to the mid-rapidity event multiplicity, while $V_{2}$ is independent of it with similar magnitude in the forward ($d$-going) and backward (Au-going) directions.\n\n23 data tables\n\nCorrelated dihadron yield, per radian per unit of pseudorapidity, as a function of $\\Delta\\phi$ for 1.2 < $|\\Delta\\eta|$ < 1.8 in d+Au collisions, for low ZDC-Au activity data. Both the trigger and associated particles have 1 < $p_T$ < 3 GeV/c.\n\nCorrelated dihadron yield, per radian per unit of pseudorapidity, as a function of $\\Delta\\phi$ for 1.2 < $|\\Delta\\eta|$ < 1.8 in d+Au collisions, for high ZDC-Au activity data. Both the trigger and associated particles have 1 < $p_T$ < 3 GeV/c.\n\nCorrelated dihadron yield, per radian per unit of pseudorapidity, as a function of $\\Delta\\phi$ for -4.5 < $\\Delta\\eta$ < -2 in d+Au collisions, for low ZDC-Au activity data. Both the trigger and associated particles have 1 < $p_T$ < 3 GeV/c.\n\nMore…\n\n#### Observation of charge asymmetry dependence of pion elliptic flow and the possible chiral magnetic wave in heavy-ion collisions\n\nThe collaboration Adamczyk, L. ; Adkins, J.K. ; Agakishiev, G. ; et al.\nPhys.Rev.Lett. 114 (2015) 252302, 2015.\nInspire Record 1358666\n\nWe present measurements of π- and π+ elliptic flow, v2, at midrapidity in Au+Au collisions at sNN=200, 62.4, 39, 27, 19.6, 11.5, and 7.7 GeV, as a function of event-by-event charge asymmetry, Ach, based on data from the STAR experiment at RHIC. We find that π- (π+) elliptic flow linearly increases (decreases) with charge asymmetry for most centrality bins at sNN=27  GeV and higher. At sNN=200  GeV, the slope of the difference of v2 between π- and π+ as a function of Ach exhibits a centrality dependence, which is qualitatively similar to calculations that incorporate a chiral magnetic wave effect. Similar centrality dependence is also observed at lower energies.\n\n10 data tables\n\nThe distribution of observed charge asymmetry from STAR data.\n\nPion $v_2${2} as a function of observed charge asymmetry.\n\n$v_2$ difference between $\\pi^-$ and $\\pi^+$ as a function of charge asymmetry with the tracking efficiency correction, for 30-40% central Au+Au collisions at 200 GeV. The errors are statistical only.\n\nMore…\n\n#### Azimuthal anisotropy in U$+$U and Au$+$Au collisions at RHIC\n\nThe collaboration Adamczyk, L. ; Adkins, J.K. ; Agakishiev, G. ; et al.\nPhys.Rev.Lett. 115 (2015) 222301, 2015.\nInspire Record 1373553\n\nCollisions between prolate uranium nuclei are used to study how particle production and azimuthal anisotropies depend on initial geometry in heavy-ion collisions. We report the two- and four-particle cumulants, v2{2} and v2{4}, for charged hadrons from U+U collisions at sNN=193  GeV and Au+Au collisions at sNN=200  GeV. Nearly fully overlapping collisions are selected based on the energy deposited by spectators in zero degree calorimeters (ZDCs). Within this sample, the observed dependence of v2{2} on multiplicity demonstrates that ZDC information combined with multiplicity can preferentially select different overlap configurations in U+U collisions. We also show that v2 vs multiplicity can be better described by models, such as gluon saturation or quark participant models, that eliminate the dependence of the multiplicity on the number of binary nucleon-nucleon collisions.\n\n20 data tables\n\nNo description provided.\n\nNo description provided.\n\nNo description provided.\n\nMore…\n\n#### Measurement of Interaction between Antiprotons\n\nThe collaboration Adamczyk, L. ; Adkins, J.K. ; Agakishiev, G. ; et al.\nNature 527 (2015) 345-348, 2015.\nInspire Record 1385105\n\nOne of the primary goals of nuclear physics is to understand the force between nucleons, which is a necessary step for understanding the structure of nuclei and how nuclei interact with each other. Rutherford discovered the atomic nucleus in 1911, and the large body of knowledge about the nuclear force since acquired was derived from studies made on nucleons or nuclei. Although antinuclei up to antihelium-4 have been discovered and their masses measured, we have no direct knowledge of the nuclear force between antinucleons. Here, we study antiproton pair correlations among data taken by the STAR experiment at the Relativistic Heavy Ion Collider and show that the force between two antiprotons is attractive. In addition, we report two key parameters that characterize the corresponding strong interaction: namely, the scattering length (f0) and effective range (d0). As direct information on the interaction between two antiprotons, one of the simplest systems of antinucleons, our result provides a fundamental ingredient for understanding the structure of more complex antinuclei and their properties.\n\n2 data tables\n\nCorrelation function for proton-proton pairs (top), antiproton-antiproton pairs (middle), and the ratio of the former to the latter (bottom).\n\nMeasurements of the singlet s-wave scattering length (f0) and the effective range (d0) from this and other experiments.\n\n#### Centrality and transverse momentum dependence of elliptic flow of multistrange hadrons and $\\phi$ meson in Au+Au collisions at $\\sqrt{s_{NN}}$ = 200 GeV\n\nThe collaboration Adamczyk, L. ; Adkins, J.K. ; Agakishiev, G. ; et al.\nPhys.Rev.Lett. 116 (2016) 062301, 2016.\nInspire Record 1383879\n\nWe present high precision measurements of elliptic flow near midrapidity ($|y|<1.0$) for multi-strange hadrons and $\\phi$ meson as a function of centrality and transverse momentum in Au+Au collisions at center of mass energy $\\sqrt{s_{NN}}=$ 200 GeV. We observe that the transverse momentum dependence of $\\phi$ and $\\Omega$ $v_{2}$ is similar to that of $\\pi$ and $p$, respectively, which may indicate that the heavier strange quark flows as strongly as the lighter up and down quarks. This observation constitutes a clear piece of evidence for the development of partonic collectivity in heavy-ion collisions at the top RHIC energy. Number of constituent quark scaling is found to hold within statistical uncertainty for both 0-30$\\%$ and 30-80$\\%$ collision centrality. There is an indication of the breakdown of previously observed mass ordering between $\\phi$ and proton $v_{2}$ at low transverse momentum in the 0-30$\\%$ centrality range, possibly indicating late hadronic interactions affecting the proton $v_{2}$.\n\n23 data tables\n\nNo description provided.\n\nNo description provided.\n\nNo description provided.\n\nMore…\n\n#### Multiparticle azimuthal correlations in p -Pb and Pb-Pb collisions at the CERN Large Hadron Collider\n\nThe collaboration Abelev, Betty Bezverkhny ; Adam, Jaroslav ; Adamova, Dagmar ; et al.\nPhys.Rev. C90 (2014) 054901, 2014.\nInspire Record 1300038\n\n<p>Measurements of multiparticle azimuthal correlations (cumulants) for charged particles in <inline-formula><mml:math><mml:mi>p</mml:mi></mml:math></inline-formula>-Pb at <inline-formula><mml:math><mml:mrow><mml:msqrt><mml:msub><mml:mi>s</mml:mi><mml:mi>NN</mml:mi></mml:msub></mml:msqrt><mml:mo>=</mml:mo><mml:mn>5.02</mml:mn></mml:mrow></mml:math></inline-formula> TeV and Pb-Pb at <inline-formula><mml:math><mml:mrow><mml:msqrt><mml:msub><mml:mi>s</mml:mi><mml:mi>NN</mml:mi></mml:msub></mml:msqrt><mml:mo>=</mml:mo><mml:mn>2.76</mml:mn></mml:mrow></mml:math></inline-formula> TeV collisions are presented. They help address the question of whether there is evidence for global, flowlike, azimuthal correlations in the <inline-formula><mml:math><mml:mi>p</mml:mi></mml:math></inline-formula>-Pb system. Comparisons are made to measurements from the larger Pb-Pb system, where such evidence is established. In particular, the second harmonic two-particle cumulants are found to decrease with multiplicity, characteristic of a dominance of few-particle correlations in <inline-formula><mml:math><mml:mi>p</mml:mi></mml:math></inline-formula>-Pb collisions. However, when a <inline-formula><mml:math><mml:mrow><mml:mo>|</mml:mo><mml:mi>Δ</mml:mi><mml:mi>η</mml:mi><mml:mo>|</mml:mo></mml:mrow></mml:math></inline-formula> gap is placed to suppress such correlations, the two-particle cumulants begin to rise at high multiplicity, indicating the presence of global azimuthal correlations. The Pb-Pb values are higher than the <inline-formula><mml:math><mml:mi>p</mml:mi></mml:math></inline-formula>-Pb values at similar multiplicities. In both systems, the second harmonic four-particle cumulants exhibit a transition from positive to negative values when the multiplicity increases. The negative values allow for a measurement of <inline-formula><mml:math><mml:mrow><mml:msub><mml:mi>v</mml:mi><mml:mn>2</mml:mn></mml:msub><mml:mrow><mml:mo>{</mml:mo><mml:mn>4</mml:mn><mml:mo>}</mml:mo></mml:mrow></mml:mrow></mml:math></inline-formula> to be made, which is found to be higher in Pb-Pb collisions at similar multiplicities. The second harmonic six-particle cumulants are also found to be higher in Pb-Pb collisions. In Pb-Pb collisions, we generally find <inline-formula><mml:math><mml:mrow><mml:msub><mml:mi>v</mml:mi><mml:mn>2</mml:mn></mml:msub><mml:mrow><mml:mo>{</mml:mo><mml:mn>4</mml:mn><mml:mo>}</mml:mo></mml:mrow><mml:mo>≃</mml:mo><mml:msub><mml:mi>v</mml:mi><mml:mn>2</mml:mn></mml:msub><mml:mrow><mml:mo>{</mml:mo><mml:mn>6</mml:mn><mml:mo>}</mml:mo></mml:mrow><mml:mo>≠</mml:mo><mml:mn>0</mml:mn></mml:mrow></mml:math></inline-formula> which is indicative of a Bessel-Gaussian function for the <inline-formula><mml:math><mml:msub><mml:mi>v</mml:mi><mml:mn>2</mml:mn></mml:msub></mml:math></inline-formula> distribution. For very high-multiplicity Pb-Pb collisions, we observe that the four- and six-particle cumulants become consistent with 0. Finally, third harmonic two-particle cumulants in <inline-formula><mml:math><mml:mi>p</mml:mi></mml:math></inline-formula>-Pb and Pb-Pb are measured. These are found to be similar for overlapping multiplicities, when a <inline-formula><mml:math><mml:mrow><mml:mo>|</mml:mo><mml:mi>Δ</mml:mi><mml:mi>η</mml:mi><mml:mo>|</mml:mo><mml:mo>&gt;</mml:mo><mml:mn>1.4</mml:mn></mml:mrow></mml:math></inline-formula> gap is placed.</p>\n\n42 data tables\n\nNo description provided.\n\nNo description provided.\n\nNo description provided.\n\nMore…\n\n#### Scattering of 1-GeV Protons on Nuclei\n\nAlkhazov, G.D. ; Belostotsky, S.L. ; Vorobev, A.A ;\nPhys.Rept. 42 (1978) 89-144, 1978.\nInspire Record 135538\n\nThe 1 GeV high resolution proton nucleus scattering is reviewed. The effects from nuclear correlations are considered in detail. The sensitivity of differential cross sections to the one particle density and NN amplitude parameters are demonstrated. An analysis of the elastic proton scattering from the zero-spin nuclei and the obtained information on the neutron and matter distributions are presented. The scattering from a few nonspherical light nuclei is analysed. The first results on proton-nucleus polarization are discussed.\n\n4 data tables\n\nX ERROR D(THETA) = 0.0300 DEG.\n\nX ERROR D(THETA) = 0.0300 DEG.\n\nX ERROR D(THETA) = 0.0300 DEG.\n\nMore…\n\n#### Elastic Scattering of 1-GeV Protons by the Nuclei O-16, Ca-40, Ca-42, Ca-44, Ca-48, Ti-48 and Pb-208 in Selfconsistent Theory of Excited Nuclei\n\nAlkhazov, G.D. ; Birbrair, B.L. ; Glezer, S.I. ; et al.\nInspire Record 134442\n\nCross sections for elastic scattering of 1 GeV protons from40Ca nuclei have been calculated using the flucton model. The influence of the collective flucton nuclaon correlations on the calculated cross sections is examined. The calculated cross sections are in significant disagreement with the experimental data. This may be considered as an argument against the flucton model.\n\n5 data tables\n\nNo description provided.\n\nNo description provided.\n\nNo description provided.\n\nMore…\n\n#### $A_y$ in n-d elastic scattering: a test for three-nucleon calculations\n\nBrogli-Gysin, C. ; Campbell, J. ; Haffter, P. ; et al.\nPhys.Lett. B250 (1990) 11-14, 1990.\nInspire Record 1389638\n\nWe have measured the analyzing power A y in n-d elastic scattering at 67.0 MeV. The experiment was based on the detection of recoil deuterons, allowing for a precise measurement of the backward angular range. The results are in good agreement with recent three-nucleon calculations which are based on the Paris and Bonn NN potentials.\n\n1 data table\n\nNo description provided.\n\n#### The transverse and longitudinal cross sections for electroproduction of pions near the Δ(1236)-isobar\n\nBartel, W. ; Dudelzak, B. ; Krehbiel, H. ; et al.\nPhys.Lett. B27 (1968) 660-662, 1968.\nInspire Record 1389642\n\nThe reaction e + p → e ′+ N ∗ was studied for four momentum transfers up to 2.34 (GeV/ c ) 2 in the region of the 1236 MeV isobar. An analysis of the data in terms of the cross sections σ T and σ L for the absorption of transverse and longitudinal photons is given for invariant masses of the final pion nucleon system W =1.220 GeV and W =1.350 GeV.\n\n3 data tables\n\nTotal errors are presented.\n\nTotal errors are presented.\n\nTotal errors are presented.\n\n#### Search for solutions of the phase-shift analysis of pp interactions at 970 MeV\n\nVovchenko, V.G. ; Grebenyuk, O.G. ; Fedorov, O.Ya. ;\nInspire Record 239695\n\nA random search for solutions of the phase-shift analysis of pp scattering at 970 MeV is carried out. Solutions were selected according to the correct position of the zero of trajectory I of the Barrelet amplitude f1 in addition to the statistical criteria. Two pairs of solutions with similar phase shifts are found as a result. Two of these solutions have been found before\n\n4 data tables\n\nNo description provided.\n\nNo description provided.\n\nNo description provided.\n\nMore…\n\n#### Anisotropic flow of charged hadrons, pions and (anti-)protons measured at high transverse momentum in Pb-Pb collisions at $\\sqrt{s_{NN}}$=2.76 TeV\n\nThe collaboration Abelev, Betty ; Adam, Jaroslav ; Adamova, Dagmar ; et al.\nPhys.Lett. B719 (2013) 18-28, 2013.\nInspire Record 1116150\n16 data tables\n\nElliptic flow (v2) estimated with Event Plane method (with eta gap of 2.0) measured for unidentified charged particles as a function of transverse momentum for various centrality classes.\n\nElliptic flow (v2) estimated with four-particle cumulants measured for unidentified charged particles as a function of transverse momentum for various centrality classes.\n\nTriangular flow (v3) estimated with Event Plane method (with eta gap of 2.0) measured for unidentified charged particles as a function of transverse momentum for various centrality classes.\n\nMore…\n\n#### Combination of measurements of inclusive deep inelastic ${e^{\\pm }p}$ scattering cross sections and QCD analysis of HERA data\n\nThe & collaborations Abramowicz, H. ; Abt, I. ; Adamczyk, L. ; et al.\nEur.Phys.J. C75 (2015) 580, 2015.\nInspire Record 1377206\n9 data tables\n\nHERA combined reduced cross sections $\\sigma_{r,\\rm NC}^{+}$ for NC $e^{+}p$ scattering at $\\sqrt{s} = 318$ GeV; $\\delta_{\\rm stat}$, $\\delta_{\\rm uncor}$ and $\\delta_{\\rm cor}$ represent the statistical, uncorrelated systematic and correlated systematic uncertainties, respectively; $\\delta_{\\rm rel}$, $\\delta_{\\gamma p}$, $\\delta_{\\rm had}$ and $\\delta_{1}$ to $\\delta_{4}$ are the correlated sources of uncertainties arising from the combination procedure. The uncertainties are quoted in percent relative to $\\sigma_{r,\\rm NC}^{+}$.\n\nHERA combined reduced cross sections $\\sigma_{r,\\rm NC}^{+}$ for NC $e^{+}p$ scattering at $\\sqrt{s} = 300$ GeV; $\\delta_{\\rm stat}$, $\\delta_{\\rm uncor}$ and $\\delta_{\\rm cor}$ represent the statistical, uncorrelated systematic and correlated systematic uncertainties, respectively; $\\delta_{\\rm rel}$, $\\delta_{\\gamma p}$, $\\delta_{\\rm had}$ and $\\delta_{1}$ to $\\delta_{4}$ are the correlated sources of uncertainties arising from the combination procedure. The uncertainties are quoted in percent relative to $\\sigma_{r,\\rm NC}^{+}$.\n\nHERA combined reduced cross sections $\\sigma_{r,\\rm NC}^{+}$ for NC $e^{+}p$ scattering at $\\sqrt{s} = 251$ GeV; $\\delta_{\\rm stat}$, $\\delta_{\\rm uncor}$ and $\\delta_{\\rm cor}$ represent the statistical, uncorrelated systematic and correlated systematic uncertainties, respectively; $\\delta_{\\rm rel}$, $\\delta_{\\gamma p}$, $\\delta_{\\rm had}$ and $\\delta_{1}$ to $\\delta_{4}$ are the correlated sources of uncertainties arising from the combination procedure. The uncertainties are quoted in percent relative to $\\sigma_{r,\\rm NC}^{+}$.\n\nMore…\n\n#### High-precision measurements of $\\pi p$ elastic differential cross sections in the second resonance region\n\nThe collaboration Alekseev, I.G. ; Andreev, V.A. ; Bordyuzhin, I.G. ; et al.\nPhys.Rev. C91 (2015) 025205, 2015.\nInspire Record 1323450\n\n<p>Cross sections for <inline-formula><mml:math><mml:mrow><mml:msup><mml:mi>π</mml:mi><mml:mo>±</mml:mo></mml:msup><mml:mi>p</mml:mi></mml:mrow></mml:math></inline-formula> elastic scattering have been measured to high precision by the EPECUR Collaboration for beam momenta between 800 and 1240 MeV/<inline-formula><mml:math><mml:mi>c</mml:mi></mml:math></inline-formula> using the ITEP proton synchrotron. The data precision allows comparisons of the existing partial-wave analyses on a level not possible previously. These comparisons imply that over the covered energy range, the Carnegie-Mellon-Berkeley analysis is significantly more predictive when compared to versions of the Karlsruhe-Helsinki analyses.</p>\n\n249 data tables\n\nDifferential cross section of elastic $\\pi^+$p-scattering at P= 800.25 MeV/c. Errors shown are statistical only.\n\nDifferential cross section of elastic $\\pi^+$p-scattering at P= 803.75 MeV/c. Errors shown are statistical only.\n\nDifferential cross section of elastic $\\pi^+$p-scattering at P= 807.25 MeV/c. Errors shown are statistical only.\n\nMore…\n\n#### Detection of $B$-Mode Polarization at Degree Angular Scales by BICEP2\n\nThe collaboration Ade, P.A.R. ; Aikin, R.W. ; Barkats, D. ; et al.\nPhys.Rev.Lett. 112 (2014) 241101, 2014.\nInspire Record 1286113\n2 data tables\n\nBICEP2 TT, TE, EE, BB, TB, and EB bandpowers, ell*(ell+1)*C(ell)/(2*PI), and uncertainties, corresponding to Figure 2. Uncertainties are statistical only, the standard deviation of the constrained lensed-LambdaCDM+noise simulations, and are calculated as the square root of diagonal elements of the bandpower covariance matrix. The nature of the simulations constrains T to match the observed sky, thus TT, TE, and TB uncertainties do not include appropriate sample variance, and sample variance for a tensor BB signal is not included either. The calibration procedure uses TB and EB to constrain the polarization angle, thus TB and EB cannot be used to measure astrophysical polarization rotation.\n\nLikelihood for the tensor-to-scalar ratio, r, derived from the BICEP2 BB spectrum, corresponding to the black curve from the middle panel of Figure 10, and calculated via the \"direct likelihood\" method described in Section 11.1.\n\n#### Net-Charge Fluctuations in Pb-Pb collisions at $\\sqrt{s}_{NN} = 2.76$ TeV\n\nThe collaboration Abelev, Betty ; Adam, Jaroslav ; Adamova, Dagmar ; et al.\nPhys.Rev.Lett. 110 (2013) 152301, 2013.\nInspire Record 1123802\n7 data tables\n\nThe measured NU(+-DYN) as a function of the centrality of the collisions, expressed as the number of participating nucleons, for two values of midrapidity range.\n\nNU(+-DYN), corrected for charge conservation and finite acceptance effects, as a function of the centrality of the collisions, expressed as the number of participating nucleons, for two values of midrapidity range.\n\nThe measured and corrected NU(+-DYN) in P P collisions for two values of midrapidity range.\n\nMore…\n\n#### Charge correlations using the balance function in Pb-Pb collisions at $\\sqrt{s_{NN}}$ = 2.76 TeV\n\nThe collaboration Abelev, Betty ; Adam, Jaroslav ; Adamova, Dagmar ; et al.\nPhys.Lett. B723 (2013) 267-279, 2013.\nInspire Record 1211186\n8 data tables\n\nThe Balance Function as a function of the relative pseudorapidity of two charged particles for the centrality class 0-5%. Also shown in the second column is the result from the mixed data set.\n\nThe Balance Function as a function of the relative pseudorapidity of two charged particles for the centrality class 30-40%.\n\nThe Balance Function as a function of the relative pseudorapidity of two charged particles for the centrality class 70-80%.\n\nMore…\n\n#### Z DEPENDENCE OF DIFFERENTIAL AND TOTAL CROSS-SECTIONS IN QUASIBINARY EXCHANGE REACTIONS ON NUCLEI pi- A(z) ---> (pi0, eta, omega, f2) + A*(z-1) AT 39.1-GeV/c\n\nApokin, V.D. ; Arestov, Yu.I. ; Belikov, N.I. ; et al.\nSov.J.Nucl.Phys. 46 (1987) 877, 1987.\nInspire Record 239550\n5 data tables\nMore…" ]

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[ "Feet To Meters\n\n# 94700 ft to m94700 Foot to Meters\n\nft\n=\nm\n\n## How to convert 94700 foot to meters?\n\n 94700 ft * 0.3048 m = 28864.56 m 1 ft\nA common question is How many foot in 94700 meter? And the answer is 310695.538057 ft in 94700 m. Likewise the question how many meter in 94700 foot has the answer of 28864.56 m in 94700 ft.\n\n## How much are 94700 feet in meters?\n\n94700 feet equal 28864.56 meters (94700ft = 28864.56m). Converting 94700 ft to m is easy. Simply use our calculator above, or apply the formula to change the length 94700 ft to m.\n\n## Convert 94700 ft to common lengths\n\nUnitLength\nNanometer2.886456e+13 nm\nMicrometer28864560000.0 µm\nMillimeter28864560.0 mm\nCentimeter2886456.0 cm\nInch1136400.0 in\nFoot94700.0 ft\nYard31566.6666666 yd\nMeter28864.56 m\nKilometer28.86456 km\nMile17.9356060606 mi\nNautical mile15.5856155507 nmi\n\n## What is 94700 feet in m?\n\nTo convert 94700 ft to m multiply the length in feet by 0.3048. The 94700 ft in m formula is [m] = 94700 * 0.3048. Thus, for 94700 feet in meter we get 28864.56 m.\n\n## 94700 Foot Conversion Table", null, "## Alternative spelling\n\n94700 ft in Meter, 94700 ft to m, 94700 Foot to Meters, 94700 Foot to Meter, 94700 Foot in Meter, 94700 Feet to Meters, 94700 Feet in Meters," ]

[ null, "https://feet-to-meters.appspot.com/image/94700.png", null ]

[ "## Description of HTTP Protocol for FarCards\n\nRequirements for processing server:The server can respond using http or httpsxml encoding: only UTF-8It is mandatory to use getcardinfoex or transactionsex. Other functions are optional\nSoftware Requirements: Farcards version 6.04 or above\nDll consists of two modules: 1)the dll for Farcards, ExtDllHTTP.dll2)the licensing utility, Http_LicGen.exe\n\n### Settings\n\n#### ExtDllHttp.ini Description\n\nThe file is used by both modules.\n[Server]\n;Processing server entry point\n;address of the GetCardInfoEx function handler, i. e. what is run is http://192.168.101.141:80/getcardinfoex.php\nGetCardInfoEx=getcardinfoex.php\nTransactionsEx=transactionsex.php\nFindEmail=findemail.php\nGetCardImageEx=getcardimageex.php\n;Proxy is used for queries to the UCS licensing server\n[Proxy]\n;Use proxy (1 means yes; 0 means no)\nUseProxy=0\nBasicAuthentication=0\nServer=127.0.0.1\nPort=9944\n\n### Available Methods\n\n#### Receiving Card Information\n\nThe method is called using the address defined by the getcardinfoex parameter\nMethod: POST\nRequest body:\n<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n<ROOT>\n<QRY Card=\"8002\" Restaurant=\"9999\" UnitNo=\"2\">\n<INPBUF>\n<CHECK stationcode=\"2\" restaurantcode=\"199999999\" cashservername=\"FOCUS_MIDSERV2\" generateddatetime=\"2013-08-2116:2316:23:56\">\n<EXTINFO>\n<INTERFACES>\n<INTERFACE type=\"PDS\" id=\"1\" mode=\"0\">\n<HOLDERS>\n<ITEM cardcode=\"8002\"/>\n</HOLDERS>\n</INTERFACE>\n</INTERFACES>\n</EXTINFO>\n</CHECK>\n</INPBUF>\n</QRY>\n</ROOT>\nWhere:card is the card numberrestaurant is the restaurant ID unitno is the number of the calling POSINPBUF is a buffer containing an XML that has extended information on a card and receipt, for example. More details at XML description of a cash document for provision to external systems\nServer response:\nHTTP/1.1 200 OK\nDate: Wed, 21 Aug 2013 08:22:20 GMT\nContent-Length: 346\nKeep-Alive: timeout=5, max=100\nConnection: Keep-Alive\nContent-Type: text/xml\n\n<?xml version=\"1.0\" encoding=\"utf-8\"?>\n<Root>\n<GetCardInfoEx CardCode=\"8002\" Account=\"8002\" Deleted=\"0\" Locked=\"0\" Seize=\"0\" Discount=\"25\" Bonus=\"4\"\nSumma=\"15000\" DiscLimit=\"10000\" Holder=\"Test farcards person 8002\" unpay=\"4\"\nSum2=\"2000\" Sum3=\"3000\" Sum4=\"4000\" Sum5=\"0\"\nScrMessage=\"** Message for SCREEN **\"\nPrnMessage=\"* Message for PRINT *\" Result=\"0\" >\n\n<OutBuf OutKind=\"4\">\n<ident_info>\n<overide_info code=\"123456\"/>\n<parent_ident code=\"4547059\" printname=\"Vasily Petrovich\"/>\n<item_content hint=\"Your coupon allows to select two items from the first group and any item from the second one.\">\n<group name=\"group name\" printname=\"Group Name\" order=\"1\" maxquant=\"2\">\n<item code=\"123\" kind=\"summ\" val=\"10.00\" order=\"3\" max=\"3\" default=\"1\" disccode=\"13\"/>\n<item code=\"321\" kind=\"percent\" val=\"10.00\"order=\"2\" max=\"2\" disccode=\"123\"/>\n<item code=\"213\" kind=\"price\" val=\"10\" order=\"10\" max=\"1\" />\n</group>\n<group name=\"group name2\" printname=\"Group name2\" order=\"2\">\n<item code=\"423\" kind=\"summ\" val=\"130\" order=\"1\" line_id=\"433242\" />\n<item code=\"621\" kind=\"percent\" val=\"1230\"order=\"3\" />\n<item code=\"713\" kind=\"price\" val=\"110\" order=\"2\" />\n</group>\n</item_content>\n</ident_info>\n</OutBuf>\n</GetCardInfoEx>\n</Root>\nWhat is important here is the Result\n0 means there are no errors, the Info structure is filled in\n1 means the card does not exist\nHTTP/1.1 200 OK\nDate: Wed, 21 Aug 2013 08:23:20 GMT\nContent-Length: 346\nKeep-Alive: timeout=5, max=100\nConnection: Keep-Alive\nContent-Type: text/xml\n\n<?xml version=\"1.0\" encoding=\"utf-8\"?>\n<Root>\n<GetCardInfoEx Result=\"1\" />\n</Root>\n\n#### Receiving Card Image\n\nThe method is called using the address defined by the getcardimageex parameter\nMethod: POST\nRequest body:\n<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n<ROOT>\n<QRY CardCode=\"8002\"/>\n</ROOT>\nWhere Code is the card number\nDepending on whether an image is found or not, the response will differ.If an image is found, the server response will be the following:\nHTTP/1.1 200 OK\nDate: Wed, 21 Aug 2013 08:22:20 GMT\nContent-Length: 5862\nKeep-Alive: timeout=5, max=100\nConnection: Keep-Alive\nContent-Type: image/jpg\nThe image will be retrieved using the DLL.\nIf an image or a card is not found, the server response will be the following:\nHTTP/1.1 200 OK\nDate: Wed, 21 Aug 2013 08:22:20 GMT\nContent-Length: 125\nKeep-Alive: timeout=5, max=100\nConnection: Keep-Alive\nContent-Type: text/xml\n\n<?xml version=\"1.0\"?>\n<Root>\n</Root>\n\n#### Finding Account by E-Mail\n\nThe method is called using the address defined by the findemail parameter\nMethod: POST\nRequest body:\n<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n<ROOT>\n<QRY Email=\"director@ucs.ru\"/>\n</ROOT>\nWhere Email is the email in question\nA successful server response will be the following:\nHTTP/1.1 200 OK\nDate: Wed, 21 Aug 2013 12:40:51 GMT\nContent-Length: 123\nKeep-Alive: timeout=5, max=100\nConnection: Keep-Alive\nContent-Type: text/xml\n\n<?xml version=\"1.0\"?>\n<Root>\n<FindEmail Account=\"8002\" CardCode=\"8002\" Name=\"UCS director\" Result=\"0\"/>\n</Root>\nHTTP/1.1 200 OK\nDate: Wed, 21 Aug 2013 12:40:51 GMT\nContent-Length: 123\nKeep-Alive: timeout=5, max=100\nConnection: Keep-Alive\nContent-Type: text/xml\n\n<?xml version=\"1.0\"?>\n<Root>\n<FindEmail Result=\"1\"/>\n</Root>\n\n#### Transactions for POS Receipt\n\nThe method is called using the address defined by the transactionsex parameter\nMethod: POST\nRequest body:\n<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n<Root>\n<Transactions>\n<TransactionsEx Card=\"32\" PersonID=\"202\" Account=\"145\" Kind=\"160\" Summa=\"15\" Restaurant=\"73\" RKDate=\"2014-08-14T00-00-00\" RKUnit=\"201\" RKCheck=\"139\" VatSumA=\"86\" VatPrcA=\"143\" VatSumB=\"125\" VatPrcB=\"146\" VatSumC=\"93\" VatPrcC=\"119\" VatSumD=\"28\" VatPrcD=\"189\" VatSumE=\"161\" VatPrcE=\"72\" VatSumF=\"178\" VatPrcF=\"80\" VatSumG=\"82\" VatPrcG=\"42\" VatSumH=\"200\" VatPrcH=\"127\"/>\n<TransactionsEx Card=\"3\" PersonID=\"143\" Account=\"127\" Kind=\"157\" Summa=\"202\" Restaurant=\"73\" RKDate=\"2014-08-14T00-00-00\" RKUnit=\"201\" RKCheck=\"139\" VatSumA=\"77\" VatPrcA=\"59\" VatSumB=\"186\" VatPrcB=\"121\" VatSumC=\"7\" VatPrcC=\"125\" VatSumD=\"254\" VatPrcD=\"161\" VatSumE=\"253\" VatPrcE=\"36\" VatSumF=\"87\" VatPrcF=\"45\" VatSumG=\"178\" VatPrcG=\"81\" VatSumH=\"47\" VatPrcH=\"185\"/>\n<TransactionsEx Card=\"152\" PersonID=\"86\" Account=\"181\" Kind=\"229\" Summa=\"206\" Restaurant=\"73\" RKDate=\"2014-08-14T00-00-00\" RKUnit=\"201\" RKCheck=\"139\" VatSumA=\"160\" VatPrcA=\"210\" VatSumB=\"93\" VatPrcB=\"120\" VatSumC=\"22\" VatPrcC=\"117\" VatSumD=\"67\" VatPrcD=\"95\" VatSumE=\"199\" VatPrcE=\"95\" VatSumF=\"100\" VatPrcF=\"228\" VatSumG=\"51\" VatPrcG=\"162\" VatSumH=\"56\" VatPrcH=\"190\"/>\n</Transactions>\n\n<INPBUF>\n<CHECK stationcode=\"6\" restaurantcode=\"199999999\" cashservername=\"SHOWRK7MIDSRV1\" generateddatetime=\"2015-05-29T19:09:20\" chmode=\"0\">\n<EXTINFO reservation=\"0\">\n<INTERFACES current=\"1007682\">\n<INTERFACE type=\"PDS\" id=\"1007682\" mode=\"0\">\n<HOLDERS>\n<ITEM cardcode=\"777777\"/>\n</HOLDERS>\n</INTERFACE>\n</INTERFACES>\n</EXTINFO>\n</CHECK>\n\n</INPBUF>\n</Root>\nWhere:\ncard is a cardPersonID is a cardholder's identifierAccount is an account numberKind is a transaction typeINPBUF is a buffer containing an XML that has extended information on a card and receipt, for example. More details at XML description of a cash document for provision to external systems\n0 means a payment (withdrawing money from an account)1 means a discount2 means a bonus (crediting an account)3 means a guest's expenses (how much money a guest paid)\nSumma means the amount in kopeks,\nfor type 0 (payment):\n\n• payment by withdrawing money from a card results in a negative amount,\n• payment cancellation results in a positive amount;\n\nfor type 1 (discount):\n\n• a discount to a customer results in a negative amount,\n• discount cancellation results in a positive amount;\n\nfor type 2 (bonus):\n\n• if a customer receives a bonus, that results in a positive amount,\n• bonus cancellation results in a negative amount;\n\nfor type 3 (expenses):\n\n• if a customer has paid that results in a positive amount,\n• receipt cancellation results in a negative amount;\n\nrestaurant means a restaurant codeRKDate means a POS date ( 0 → 30/12/1899 )RKUni means a POS numberRKCheck means a receipt numberthen there is information on taxes in a receipt (8 items)VatSumA means an amount with the tax AVatPrcA means the amount of tax A as a percentage * 100 (1500 → 15.00%)VatSumB means an amount with the tax BVatPrcB means the amount of tax B as a percentage * 100VatSumC means an amount with the tax CVatPrcC means the amount of tax C as a percentage * 100VatSumD means an amount with the tax DVatPrcD means the amount of tax D as a percentage * 100VatSumE means an amount with the tax EVatPrcE means the amount of tax E as a percentage * 100VatSumF means an amount with the tax FVatPrcF means the amount of tax F as a percentage * 100VatSumG means an amount with the tax GVatPrcG means the amount of tax G as a percentage * 100VatSumH means an amount with the tax HVatPrcH means the amount of tax H as a percentage * 100\nA successful server response will be the following:\nHTTP/1.1 200 OK\nDate: Wed, 21 Aug 2013 12:56:25 GMT\nContent-Length: 65\nKeep-Alive: timeout=5, max=100\nConnection: Keep-Alive\nContent-Type: text/xml\n\n<?xml version=\"1.0\"?>\n<Root>\n<TransactionsEx Result=\"0\"/>\n<OutBuf OutKind=\"1\">\n<TRRESPONSE error_code=\"0\" err_text=\"\">\n<TRANSACTION ext_id=\"1111111\" num=\"222222\" cardcode=\"777777\" slip=\"Text for printing\" value=\"%d\" />\n</TRRESPONSE>\n</OutBuf>\n</Root>\nTransactions executed with an error:\nHTTP/1.1 200 OK\nDate: Wed, 21 Aug 2013 12:56:25 GMT\nContent-Length: 65\nKeep-Alive: timeout=5, max=100\nConnection: Keep-Alive\nContent-Type: text/xml\n\n<?xml version=\"1.0\"?>\n<Root>\n<TransactionsEx Result=\"1\"/>\n<OutBuf OutKind=\"1\">\n<TRRESPONSE error_code=\"100500\" err_text=\"Bank connection error\"/>\n</OutBuf>\n</Root>" ]

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[ "# Easy decision problem, hard search problem\n\nDeciding whether a Nash equilibrium exists is easy (it always does); however, actually finding one is believed to be difficult (it is PPAD-Complete).\n\nWhat are some other examples of problems where the decision version is easy but the search version is relatively difficult (compared the the decision version)?\n\nI would be particularly interested in problems where the decision version is non-trival (unlike the case with Nash equilibrium).\n\n• Should probably be community wiki: meta.cstheory.stackexchange.com/questions/225/… – Dave Clarke Oct 1 '10 at 14:04\n• @supercooldave: I wouldn't hurry with CW in this case. It may turn out that there are very few natural problems with a non-trivial but easy decision version and hard search version. This is not necessarily a \"big-list\". – Jukka Suomela Oct 1 '10 at 14:09\n• I went with the heuristic that big list = community wiki. – Dave Clarke Oct 1 '10 at 14:16\n• So this raises the question \"what is the natural decision problem to be associated with a search problem?\". I think existence of NE is not the natural decision problem associate with NE. – Kaveh Oct 1 '10 at 14:56\n• @Kaveh: You can define that decision problem for Nash (if you specify an encoding of a solution to Nash), but the problem is whether it is the same complexity as Nash or not, or formally, whether that decision problem is reducible to Nash. I doubt it because finding a Nash equilibrium satisfying some additional constraint is often NP-hard. – Tsuyoshi Ito Oct 1 '10 at 15:56\n\nGiven an integer, does it have a non-trivial factor? -> Non-trivially in P.\n\nGiven an integer, find a non-trivial factor, if there is one -> Not known to be in FP.\n\n• Or you could ask, does it have a prime factor? Then you don't need the PRIMES is in P paper – Bjørn Kjos-Hanssen Sep 6 '18 at 1:37\n\nHere is another example: Given a cubic graph G and a hamiltonian cycle H in G, find a different hamiltonian cycle in G. Such a cycle exists (by Smith's theorem) but, as far as I know, it is open whether it can be computed in polynomial time.\n\nIf you give the following the same \"leeway\" that you do for Nash equilibria, then:\n\n• Integer factorization, where the decision problem is \"Is there a factorized representation of this integer?\" (trivially, yes), and the search problem is to output it\n\nA number of lattice problems could conceivably fit here with the same type of generous allowance for defining the decision problem:\n\n• Shortest Vector Problem (SVP) -- decide if there's a shortest vector vs finding it\n• Closest Vector Problem (CVP) -- decide if there's a closest vector vs finding it\n\nOf course, these are all cases where the decision version I've mentioned isn't very interesting (because it's trivially the case). One problem that's not quite as trivial:\n\n• Planar graph $k$-colorability for $k \\ge 4$\n\nThe decision problem of planar graph 4-colorability is in P. But obtaining the lexicographically first such solution is NP-hard (Khuller/Vazirani).\n\nNote that the property you're really interested in is self-reducibility (or rather, non-self-reducibility). In the planar graph coloring problem, the essential issue is that the method of self-reducing the general case of $k$-colorability will destroy planarity in a graph.\n\nLet $G=G(n,1/2)$, the random graph on $1,\\ldots,n$, in which each edge is independently present with probability $1/2$. Choose $n^{1/3}$ vertices of $G$ uniformly at random and add all edges between them; call the resulting graph $H$. Then $H$ has a clique of size $n^{1/3}$.\n\nSearch problem: find a clique of size at least $10\\log n$.\n\n• Very neat! Is there a relevant paper about this? – András Salamon Oct 2 '10 at 23:05\n• @András: To give a bit more background, this is called the \"hidden clique problem\". If the hidden clique that's planted is on Omega(sqrt(n log n)) vertices, one can easily see that the vertices of the clique are those with the highest degree, almost surely. [Alon-Krivelevic-Sudakov] (tau.ac.il/~nogaa/PDFS/clique3.pdf) improve this to Omega(sqrt(n)) using spectral techniques. For hidden cliques of smaller size, such as O(log n), nothing non-trivial is known. – arnab Oct 2 '10 at 23:55\n• Another related intriguing problem, posed by Karp, is to find a clique of size (1+c)log(n) in G(n,1/2), for any constant 0<c<1. It's known that there exists a clique of size 2log(n) almost surely in G(n,1/2). The only polynomial time algorithms known (such as the greedy one) find cliques of size (1+o(1))log(n). – arnab Oct 2 '10 at 23:59\n• @arnab: Feige and Ron recently simplified the AKS result (see reference at my question cstheory.stackexchange.com/questions/1406/… ). My question to @Louigi was really about the $10\\log n$ question: what motivates the particular constant, and has this question been asked in a paper one can cite? – András Salamon Oct 3 '10 at 8:00\n\nOne more example; The Subset-sums equality: Given $a_1,a_2,a_3,...,,a_n$ natural numbers with $\\sum_1^n{a_i} \\lt 2^n -1$. The pigeon-hole principle guarantees the existence of two subsets $I, J$ in ${1,2,..., n}$ such that $\\sum_{i\\in I} a_i=\\sum_{j \\in J} a_j$ (since the are more subsets than possible sums). The existence of polynomial time algorithm for finding sets $I$ and $J$ is a famous open problem.\n\nAnother number theory example, similar to the ones above. It's known by Bertrand's postulate that for every positive integer $n$, there's a prime between $n$ and $2n$. But we have no polynomial time algorithm currently to find such a prime, given $n$. (The desired algorithm must run in polylog($n$) time.) One can easily come up with polynomial time randomized algorithms because of the prime number theorem, and one can derandomize them by assuming some standard number theoretic conjectures (such as Cramer's conjecture), but no unconditionally polynomial time deterministic algorithm is known. Related work was recently done in the Polymath4 project; Tao's blog post on the project is a good summary of it.\n\n• Even without Bertrand's postulate, you have a deterministic algorithm with expected polynomial runtime due to the Prime Number Theorem and the AKS primality test. – Joe Fitzsimons Oct 4 '10 at 14:52\n• @JoeFitzsimons, I am not sure what you mean by \"deterministic algorithm with expected polynomial runtime\". – Chandra Chekuri Jun 27 '13 at 0:14\n• @ChandraChekuri, \"deterministic\" probably is meant to say that it always gets the correct answer. – usul Jun 27 '13 at 19:36\n• @ChandraChekuri: Sorry, my choice of wording was poor. I meant that you can find a prime number with absolute certainty in expected polynomial time, rather than simply with bounded error. At least, I think that's what I meant. It was 3 years ago. – Joe Fitzsimons Jun 30 '13 at 9:33\n\nAt the risk of being slightly off-topic, let me give a simple and natural example of a theory C answer: Eulerian cycles and distributed algorithms.\n\nThe decision problem is not completely trivial, in the sense that there are are both Eulerian and non-Eulerian graphs.\n\nThere is, however, a fast and simple distributed algorithm that solves the decision problem (in the sense that for yes-instances all nodes output \"1\" and for no-instances at least one node outputs \"0\"): each node just checks the parity of its own degree and outputs 0 or 1 accordingly.\n\nBut if you would like to find a Eulerian cycle (in the sense that each node outputs the structure of the cycle in its own neighbourhood), then we need essentially global information on the graph. It shouldn't be hard to come up with a pair of examples that shows that the problem requires $\\Omega(n)$ communication rounds; on the other hand, $O(n)$ rounds is enough to solve any problem (assuming unique IDs).\n\nIn summary: $O(1)$-time decision problem, $\\Theta(n)$-time search problem, and this is the worst possible gap.\n\nEdit: This implicitly assumes that the graph is connected (or, equivalently, that we want to find an Eulerian cycle in each connected component).\n\n• This might be a stupid question (because I know almost nothing about distributed computing), but is there a promise that the graph is connected, or is the connectedness easy to check efficiently in a distributed way? – Tsuyoshi Ito Oct 2 '10 at 22:19\n• Thanks, not a stupid question at all. I clarified my answer, I had forgotten to add the assumption that we deal with connected graphs here. (Usually there is little point in studying disconnected graphs from the perspective of distributed algorithms, as by definition there is no way to transmit information between the connected components, but of course this should be made explicit.) – Jukka Suomela Oct 2 '10 at 22:43\n• Thanks! After reading your answer, I think that it should have been obvious that the graph (= network topology) was assumed to be connected. :) – Tsuyoshi Ito Oct 4 '10 at 13:51\n\nFinding Tverberg partitions is of unknown complexity:\n\nTheorem: Let $x_1,x_2,\\dots, x_m$ be points in $R^d$, $m \\ge (r-1)(d+1)+1$. Then there is a partition $S_1,S_2,\\dots, S_r$ of ${1,2,\\dots,m}$ such that $\\cap _{j=1}^r \\text{conv} (x_i: i \\in S_j) \\ne \\emptyset$.\n\nLike with Nash equilibria, the partition is guaranteed by the theorem, but it's not known if a polytime algorithm exists to find one.\n\nGil Kalai wrote a wonderful series of posts on this topic: One, Two and Three.\n\n• Actually, any problem that falls into TFNP would be a good candidate I think. When an answer is guaranteed to exist by a theorem -- then, define some apparently-harder-than-P search problem over the possible solutions to accompany it. – Daniel Apon Oct 1 '10 at 18:40\n\nIn all the examples above the decision problem is in P and the search problem is not known to be in P but not known to be NP-hard either. I want to point out that it is possible to have an NP-hard search problem whose decision version is easy.\n\nConsider the generalized satisfiability problem for given relations $R_1,\\ldots,R_k$ over Boolean domain $\\{0,1\\}$. An instance is an expression of the form $$R_{i_1}(t_{11},\\ldots,t_{1r_1}) \\wedge \\cdots \\wedge R_{i_m}(t_{m1},\\ldots,t_{mr_m})$$ where the $t_{ij}$'s are either variables or constants in ${0,1}$, and $r_1,\\ldots,r_m$ are the arities of $R_1,\\ldots,R_k$ (this is the same framework as in Schaeffer's dichotomy theorem with constants, in case you know what it is). The search problem is: given such an expression, find a lexicographically minimal solution, if there is one.\n\nIt was shown by Reith and Vollmer here that there exists a choice of relations $R_1,\\ldots,R_k$ that make this problem NP-hard (actually OptP-complete) but keep the satisfiability problem easy (quite trivial actually). An example given in the paper is $R = \\{(1,0,0),(0,1,0),(1,1,1)\\}$ (here $k = 1$). Once the satisfiability problem is solvable in polynomial-time, the question whether there exists a lexicographically minimal satisfying assignment is trivial.\n\nSee Corollary 13 and the example following it in the paper above (at least in this on-line version).\n\n• Decision version is (highly) non-trival in P: $k$-colourability ($k$ fixed) on graphs without induced path with five vertices; due to this paper.\n• Search version is NP-hard: Finding the chromatic number of graphs without induced path with five vertices; due to this paper.\n• You perhaps meant to say that for fixed $k$, the decision version is in P. – András Salamon Jun 26 '13 at 19:50\n\nTake a \"pairing-friendly\" elliptic curve. That is, a curve that has a one bilinear map $e$ associated with it - with $e (a + b, c + d) = e (a c) e (a d) e (b c) e (b d)$ such that $e$ is difficult to invert).\n\nSuch pairings are used widely in cryptography, partially since given $e$, it is trivial to solve Decisional Diffie-Hellman (given $(g, h, g^a, h^b)$, decide if $a = b$: just verify whether $e (g, h^b) = e (h, g^a)$). However, it is still conjectured that the search/computational Diffie-Hellman problem is difficult.\n\nSuch groups are also generalized to \"gap groups\".\n\nI guess Planar Perfect Matching got missed out from this list.\n\n• The decision version is in NC (even the counting version is in $\\mathsf{NC}$) by a parallel version (see Mahajan-Subramanya-Vinay) of Kastelyn's algorithm\n• The search version remains unparallelised to date i.e there is no known deterministic $\\mathsf{NC}$ algorithm for this problem (though if we drop either of the parallel or deterministic restrictions there are well known algorithms - Edmonds and Mulmuley-Vazirani-Vazirani/Karp-Upfal-Wigderson, respectively.\n\nLet's notch up the complexity a bit.\n\nMany decision problems about vector addition systems (VAS) are EXPSPACE-complete, but may require much larger witnesses. For instance, deciding whether the language of a VAS is regular is EXPSPACE-complete (e.g. Blockelet & Schmitz, 2011), but the smallest equivalent finite-state automaton might be of Ackermannian size (Valk & Vidal-Naquet, 1981). The explanation behind this huge gap is that there exist much smaller witnesses of non-regularity." ]

[ null ]

[ "", null, "Questions & Answers", null, "", null, "", null, "", null, "Question", null, "Answers\n\n# A conical vessel, with base radius 5 cm and height 24 cm, is full of water. This water is emptied into a cylindrical vessel of base radius 10 cm. Find the height to which the water will rise in the cylindrical vessel. (Use $\\pi =\\dfrac{22}{7}$)", null, "", null, "Answer Verified\nHint: Find the total volume of water using the formula for volume of a cone, given by $V=\\dfrac{1}{3}\\pi {{r}^{2}}h$, where base radius, r is 5 cm and height, h is 24 cm. Use the fact that the volume of water remains the same when it is emptied into a cylindrical vessel. For the volume of a cylindrical vessel use the formula $V=\\pi {{r}^{2}}h$. Equate both the volumes to find the value of height of the cylinder.\n\nComplete step-by-step answer:\nWe know that the volume of water in a vessel is the same as the volume of the vessel it is kept in. Thus, the total volume of water in the conical vessel can be calculated as the volume of the cone, given by $V=\\dfrac{1}{3}\\pi {{r}^{2}}h$. Using $r=5cm$ and $h=24cm$ in this formula, we get\n\n\\begin{align} & V=\\dfrac{1}{3}\\pi {{r}^{2}}h \\\\ & \\Rightarrow V=\\dfrac{1}{3}\\pi {{\\left( 5cm \\right)}^{2}}\\left( 24cm \\right) \\\\ & \\Rightarrow V=\\pi \\left( 25c{{m}^{2}} \\right)\\left( 8cm \\right) \\\\ & \\Rightarrow V=200\\pi c{{m}^{3}} \\\\ \\end{align}\n\nNow, since this entire volume is transferred to a cylindrical vessel, the volume of water would be the same as the volume of the cylinder, which can be given by $V=\\pi {{r}^{2}}h$. This volume would be equal to the volume of the cube and the base radius is given as 10 cm. Equating the two volumes thus gives us\n\n$\\pi {{r}^{2}}h=200\\pi c{{m}^{3}}$\n\nSubstituting the value of $r=10cm$ in this equation, we get\n\n\\begin{align} & \\pi {{\\left( 10cm \\right)}^{2}}h=200\\pi c{{m}^{3}} \\\\ & \\Rightarrow \\pi \\left( 100c{{m}^{2}} \\right)h=200\\pi c{{m}^{3}} \\\\ & \\Rightarrow 100h=200cm \\\\ & \\Rightarrow h=2cm \\\\ \\end{align}\n\nThus the height upto which water is filled in the cylindrical vessel is 2 cm.\n\nNote: To make calculations easier, the value of $\\pi$ has not been substituted, even though it is given in the question, because $\\pi$ occurs in the expression for both these volumes and hence, cancels out when the volumes are equated, thus reducing the calculations to a great extent.\nBookmark added to your notes.\nView Notes\nRadius of a Circle", null, "", null, "Right Circular Cone", null, "", null, "Relation Between Inch and cm", null, "", null, "Difference Between Acid and Base", null, "", null, "Lewis Acid and Base", null, "", null, "Volume of Cone", null, "", null, "Difference Between Alkali and Base", null, "", null, "Determine Radius of Curvature of a Given Spherical Surface by a Spherometer", null, "", null, "Blood Vessel", null, "", null, "Log base 2", null, "", null, "" ]

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[ "472,951 Members | 1,816 Online\n\n# Fixed precision floating point and locale facets\n\nHello,\n\nI'm writing a fixed-precision floating point class, based on the ideas\nin the example fixed_pt class in the \"Practical C++ Programming\" book\nby Steve Oualline (O' Reilly). This uses a long int to store the\nvalue, and the precision (number of decimal points) is variable (it's\na templated class):\n\ntemplate <size_t _decimal_places = 4>\nclass FixedFloat {\nprivate:\n/// The integer value.\nlong int m_value;\n[...]\n};\n\nThe value is set from a string (FixedFloat<2> ff = \"12.43\";), or via\noverloaded istream/ostream extraction and insertion operators (<< and\n) which are templated friend functions (implicit conversion to/from\n\ndouble is not allowed due to precision loss.). The conversion between\nlong int and string forms is done in the latter functions.\n\nTo output a number, I was manually splitting up the number into whole\nand fractional parts and processing them separately, using '.' as the\ndecimal point symbol. However, I've just discovered the existence of\nstd::locale::numeric and std::locale::monetary locale facets, and the\nnum_put() and num_get() methods. Ideally, I'd like to use these\nfunctions for the the conversions (FixedFloat -> std::string).\nHowever, their support for the standard numeric types (int, long,\nfloat, double) is hard-coded into the class. I can't risk conversion\nto a supported type such as double, due to loss of precision (0.60\nwould becomes 0.59 on my i686-pc-linux-gnu arch), and they will be\nused to process financial data!\n\nIs it possible to extend these to support my FixedFloat class?\n\nIs it reasonable to derive from std::num_put in this case? I had a\nlook into the GNU libstdc++ headers to see if it was possible, but\nthey were hideously complex, and I'm not keen on using the internals\nsuch as num_put<>::do_put and __convert_from_v(), since these are\npresumably non-portable.\n\nLastly, are there any standard classes that do this sort of thing?\nMany thanks!\nRoger\n\n--\nRoger Leigh\n\nPrinting on GNU/Linux? http://gimp-print.sourceforge.net/\nJul 19 '05 #1\n4", null, "7786", null, "\"Roger Leigh\" <\\${******@invalid.whinlatter.uklinux.net.invalid > wrote in message\nnews:87************@wrynose.whinlatter.uklinux.net ...\nTo output a number, I was manually splitting up the number into whole\nand fractional parts and processing them separately, using '.' as the\ndecimal point symbol. However, I've just discovered the existence of\nstd::locale::numeric and std::locale::monetary locale facets, and the\nnum_put() and num_get() methods. Ideally, I'd like to use these\nfunctions for the the conversions (FixedFloat -> std::string).\nHowever, their support for the standard numeric types (int, long,\nfloat, double) is hard-coded into the class. I can't risk conversion\nto a supported type such as double, due to loss of precision (0.60\nwould becomes 0.59 on my i686-pc-linux-gnu arch), and they will be\nused to process financial data!\n\nIs it possible to extend these to support my FixedFloat class?\nIt's possible, but probably not a rewarding exercise.\nIs it reasonable to derive from std::num_put in this case? I had a\nlook into the GNU libstdc++ headers to see if it was possible, but\nthey were hideously complex, and I'm not keen on using the internals\nsuch as num_put<>::do_put and __convert_from_v(), since these are\npresumably non-portable.\nThe do_put part is portable, the other isn't. But you're right to\nobserve that they're hideously complex.\nLastly, are there any standard classes that do this sort of thing?\n\nYou can pervert money_put and moneypunct to output a digit sequence\nstored in a string, with a specified number of decimal places, commas\nbetween thousands groups, etc.\n\nP.J. Plauger\nDinkumware, Ltd.\nhttp://www.dinkumware.com\nJul 19 '05 #2\n\"P.J. Plauger\" <pj*@dinkumware.com> writes:\n\"Roger Leigh\" <\\${******@invalid.whinlatter.uklinux.net.invalid > wrote in message\nnews:87************@wrynose.whinlatter.uklinux.net ...\nTo output a number, I was manually splitting up the number into whole\nand fractional parts and processing them separately, using '.' as the\ndecimal point symbol. However, I've just discovered the existence of\nstd::locale::numeric and std::locale::monetary locale facets, and the\nnum_put() and num_get() methods. Ideally, I'd like to use these\nfunctions for the the conversions (FixedFloat -> std::string).\nHowever, their support for the standard numeric types (int, long,\nfloat, double) is hard-coded into the class. I can't risk conversion\nto a supported type such as double, due to loss of precision (0.60\nwould becomes 0.59 on my i686-pc-linux-gnu arch), and they will be\nused to process financial data!\n\nIs it possible to extend these to support my FixedFloat class?\nIt's possible, but probably not a rewarding exercise.\n\nIf I'm correct here, I would have to add my own custom locale facet(s)\nto allow this, but I'd need to do this manually for each locale I want\nto use, which would be a pain.\n\n[do_put() and __convert_from_v()] The do_put part is portable, the other isn't. But you're right to\nobserve that they're hideously complex.\n\nOK.\nLastly, are there any standard classes that do this sort of thing?\n\nYou can pervert money_put and moneypunct to output a digit sequence\nstored in a string, with a specified number of decimal places, commas\nbetween thousands groups, etc.\n\nThis looks like what I'll do. I'll derive a \"Money\" class from\nFixedFloat and do that in there, overriding the standard ostream<< and\nistream>> operators.\n\nI've attached a copy of the working class, and a small driver program\nto show it in action (sorry it's so long). I have a few questions\n\n1. Is the header file OK style-wise? Is there anything wrong that I\nshould not be doing?\n\n2. I've noticed that the modulus (operator%) member and friend\nfunctions can be out by a small factor e.g. 0.0001 in a 4\nd.p. precision class. With fixed-point arithmetic, should I be\ndoing anything to correct this? Is it actually incorrect? For\nsome reason, I couldn't get the \"real\" % operator to work, so had\nto resort to the hack that actually gets used (subtracting the\nresult of division and subsequent multiplication from the original\nvalue).\n\n3. Looking at the compiled binary, the FixedFloat symbols have weak,\nrather than vague linkage. I thought that /all/ templated class\nmethods and functions would be vague. This is with GCC 3.3.2, GNU\nld 2.14.90.0.6 and binutils 2.14.90.0.6-5 on i686-pc-linux-gnu\nusing ELF binary format (this is probably OT).\n\n4. In the stream output and extraction friend classes, is the use of\nlocales correct? I've not used locales (in C++) before, and I've\ndone this using the Josuttis Standard Library book.\nRoger\n----begin main.cc----\n#include <iostream>\n\n#include \"fixedfloat.h\"\n\nint main()\n{\nstd::locale::global(std::locale(\"\"));\nstd::cout.imbue(std::locale());\n\nFixedFloat<4> f1(\"4.1246\");\nFixedFloat<4> f2(\"2.3443\");\n\nstd::cout << f1 << std::endl;\nstd::cout << f2 << std::endl;\n\nFixedFloat<4> n(f1);\nFixedFloat<4> o;\no = f2;\n\nstd::cout << n << std::endl;\nstd::cout << o << std::endl;\n\nstd::cout << \"Signedness\\n\";\nstd::cout << +f1 << std::endl;\nstd::cout << -f1 << std::endl;\n\nstd::cout << \"Binary arithmetic\\n\";\nstd::cout << f1 << \"-\" << f2 << \"=\" << f1-f2 << \"\\n\";\nstd::cout << f1 << \"+\" << f2 << \"=\" << f1+f2 << \"\\n\";\nstd::cout << f1 << \"*\" << f2 << \"=\" << f1*f2 << \"\\n\";\nstd::cout << f1 << \"/\" << f2 << \"=\" << f1/f2 << \"\\n\";\nstd::cout << f1 << \"%\" << f2 << \"=\" << f1%f2 << \"\\n\";\nstd::cout << -f1 << \"/\" << f2 << \"=\" << (-f1)/f2 << \"\\n\";\nstd::cout << -f1 << \"%\" << f2 << \"=\" << (-f1)%f2 << \"\\n\";\n\nstd::cout << \"Logic\\n\";\nstd::cout << f1 << \"==\" << f1 << \"=\" << (f1==f1) << \"\\n\";\nstd::cout << f1 << \"==\" << f2 << \"=\" << (f1==f2) << \"\\n\";\nstd::cout << f1 << \"!=\" << f1 << \"=\" << (f1!=f1) << \"\\n\";\nstd::cout << f1 << \"!=\" << f2 << \"=\" << (f1!=f2) << \"\\n\";\n\nstd::cout << \"Unary arithmetic\\n\";\nFixedFloat<4> f3 = f1;\nf3 += FixedFloat<4>(\"2.3430\");\nstd::cout << f1 << \"+=2.3430\" << \"=\" << f3 << \"\\n\";\n\nf3 = f1;\nf3 -= FixedFloat<4>(\"2.3430\");\nstd::cout << f1 << \"-=2.3430\" << \"=\" << f3 << \"\\n\";\n\nf3 = f1;\nf3 *= FixedFloat<4>(\"2.3430\");\nstd::cout << f1 << \"*=2.3430\" << \"=\" << f3 << \"\\n\";\n\nf3 = f1;\nf3 /= FixedFloat<4>(\"2.3430\");\nstd::cout << f1 << \"/=2.3430\" << \"=\" << f3 << \"\\n\";\n\nf3 = f1;\nf3 %= FixedFloat<4>(\"2.3430\");\nstd::cout << f1 << \"%=2.3430\" << \"=\" << f3 << \"\\n\";\n\nf3 = f1;\n++f3;\nstd::cout << \"++\" << f1 << \"=\" << f3 << \"\\n\";\n\nf3 = f1;\nstd::cout << f1 << \"++\" << \"=\" << f3++ << \" (before)\\n\";\nstd::cout << f1 << \"++\" << \"=\" << f3 << \" (after)\\n\";\n\nf3 = f1;\n--f3;\nstd::cout << \"--\" << f1 << \"=\" << f3 << \"\\n\";\n\nf3 = f1;\nstd::cout << f1 << \"--\" << \"=\" << f3-- << \" (before)\\n\";\nstd::cout << f1 << \"--\" << \"=\" << f3 << \" (after)\\n\";\n\nreturn 1;\n}\n----end main.cc----\n\n----begin fixedfloat.h----\n// fixed floating point class -*- C++ -*-\n// \\$Id: template.cc,v 1.1 2003/09/14 21:56:55 roger Exp \\$\n//\n// Copyright (C) 2003 Roger Leigh.\n//\n// Authors: Roger Leigh <ro***@whinlatter.uklinux.net>\n\n#include <iomanip>\n#include <istream>\n#include <locale>\n#include <ostream>\n#include <sstream>\n\n/**\n* A class to represent fixed floating point numbers with high\n* accuracy.\n* The float and double data types to not offer enough accuracy when\n* dealing with some types of data, for example currency values, since\n* they cannot garuantee that a particular value is representable in\n* their floating-point binary format. This class will garuantee\n* accuracy, with the restriction that there is a fixed number of\n* decimal places after the decimal point. Internally, the value is\n* held as a long integer.\n*\n* Conversion to and from the double data type is not implicit--this\n* must be done using the methods provided. However, conversion to\n* and from std::string is possible.\n*/\ntemplate <size_t _decimal_places = 2>\nclass FixedFloat {\npublic:\n/// The type used internally to hold fixed floating point values.\ntypedef long int value_type;\n\nprivate:\n/// The integer value.\nvalue_type m_value;\n/// The correction factor.\nvalue_type m_correction;\n\n/**\n* Compute the correction factor.\n* The correction value is used to correct multiplication and\n* division of fixed point numbers.\n*/\nvoid compute_correction()\n{\nm_correction = 1;\nfor (int i = 0; i < _decimal_places; ++i)\nm_correction *= 10;\n}\n\n/**\n* The constructor.\n* The initial value is set to the value provided.\n* @param value the initial value.\n*/\nFixedFloat(value_type value):\nm_value(value)\n{\ncompute_correction();\n}\n\npublic:\n/**\n* The constructor.\n* The initial value is set to 0.\n*/\nFixedFloat():\nm_value(0)\n{\ncompute_correction();\n}\n\n/**\n* The constructor.\n* The initial value is set to the value provided. If there are too\n* many numbers after the decimal place, they will be rounded to the\n* nearest representable value (0 to 4 are rounded down, 5 to 9 are\n* rounded up.\n* @param value the initial value.\n*/\nFixedFloat(const std::string& value)\n{\ncompute_correction();\n\nstd::istringstream input(value);\ninput >> *this;\n}\n\n/**\n* The copy constructor.\n*/\nFixedFloat(const FixedFloat& original):\nm_value(original.m_value),\nm_correction(original.m_correction)\n{}\n\n/// The destructor.\n~FixedFloat()\n{}\nFixedFloat& operator = (const FixedFloat& rhs)\n{\nm_value = rhs.m_value;\nreturn *this;\n}\n\nFixedFloat& operator += (const FixedFloat& rhs)\n{\nm_value += rhs.m_value;\nreturn *this;\n}\n\nFixedFloat& operator -= (const FixedFloat& rhs)\n{\nm_value -= rhs.m_value;\nreturn *this;\n}\n\nFixedFloat& operator *= (const FixedFloat& rhs)\n{\nm_value *= rhs.m_value;\nm_value /= m_correction;\nreturn *this;\n}\n\nFixedFloat& operator /= (const FixedFloat& rhs)\n{\nm_value *= m_correction;\nm_value /= rhs.m_value;\nreturn *this;\n}\n\nFixedFloat& operator %= (const FixedFloat& rhs)\n{\n*this = (*this - ((*this / rhs) * rhs));\nreturn *this;\n}\n\nFixedFloat& operator ++ ()\n{\nm_value += m_correction;\nreturn *this;\n}\n\nFixedFloat operator ++ (int)\n{\nFixedFloat ret(*this);\nm_value += m_correction;\nreturn ret;\n}\n\nFixedFloat& operator -- ()\n{\nm_value -= m_correction;\nreturn *this;\n}\n\nFixedFloat operator -- (int)\n{\nFixedFloat ret(*this);\nm_value -= m_correction;\nreturn ret;\n}\n\nfriend FixedFloat<_decimal_places> operator +<> (const FixedFloat<_decimal_places>& lhs,\nconst FixedFloat<_decimal_places>& rhs);\n\nfriend FixedFloat<_decimal_places> operator -<> (const FixedFloat<_decimal_places>& lhs,\nconst FixedFloat<_decimal_places>& rhs);\n\nfriend FixedFloat<_decimal_places> operator *<> (const FixedFloat<_decimal_places>& lhs,\nconst FixedFloat<_decimal_places>& rhs);\n\nfriend FixedFloat<_decimal_places> operator /<> (const FixedFloat<_decimal_places>& lhs,\nconst FixedFloat<_decimal_places>& rhs);\n\nfriend FixedFloat<_decimal_places> operator %<> (const FixedFloat<_decimal_places>& lhs,\nconst FixedFloat<_decimal_places>& rhs);\n\nfriend bool operator ==<> (const FixedFloat<_decimal_places>& lhs,\nconst FixedFloat<_decimal_places>& rhs);\n\nfriend bool operator !=<> (const FixedFloat<_decimal_places>& lhs,\nconst FixedFloat<_decimal_places>& rhs);\n\nfriend FixedFloat<_decimal_places> operator -<> (const FixedFloat<_decimal_places>& rhs);\n\nfriend FixedFloat<_decimal_places> operator +<> (const FixedFloat<_decimal_places>& rhs);\n\nfriend std::ostream& operator <<<> (std::ostream& output_stream,\nconst FixedFloat<_decimal_places>& rhs);\n\nfriend std::istream& operator >><> (std::istream& input_stream,\nFixedFloat<_decimal_places>& rhs);\n\n}; // class FixedFloat<>\n// Friend functions.\n\ntemplate <size_t _decimal_places>\ninline FixedFloat<_decimal_places> operator + (const FixedFloat<_decimal_places>& lhs,\nconst FixedFloat<_decimal_places>& rhs)\n{\nreturn FixedFloat<_decimal_places>(lhs.m_value + rhs.m_value);\n}\n\ntemplate <size_t _decimal_places>\ninline FixedFloat<_decimal_places> operator - (const FixedFloat<_decimal_places>& lhs,\nconst FixedFloat<_decimal_places>& rhs)\n{\nreturn FixedFloat<_decimal_places>(lhs.m_value - rhs.m_value);\n}\n\ntemplate <size_t _decimal_places>\ninline FixedFloat<_decimal_places> operator * (const FixedFloat<_decimal_places>& lhs,\nconst FixedFloat<_decimal_places>& rhs)\n{\nreturn FixedFloat<_decimal_places>((lhs.m_value * rhs.m_value) / lhs.m_correction);\n}\n\ntemplate <size_t _decimal_places>\ninline FixedFloat<_decimal_places> operator / (const FixedFloat<_decimal_places>& lhs,\nconst FixedFloat<_decimal_places>& rhs)\n{\nreturn FixedFloat<_decimal_places>((lhs.m_value * lhs.m_correction) / rhs.m_value);\n}\n\ntemplate <size_t _decimal_places>\ninline FixedFloat<_decimal_places> operator % (const FixedFloat<_decimal_places>& lhs,\nconst FixedFloat<_decimal_places>& rhs)\n{\nreturn FixedFloat<_decimal_places>(lhs - ((lhs / rhs) * rhs));\n}\n\ntemplate <size_t _decimal_places>\ninline bool operator == (const FixedFloat<_decimal_places>& lhs,\nconst FixedFloat<_decimal_places>& rhs)\n{\nreturn lhs.m_value == rhs.m_value;\n}\n\ntemplate <size_t _decimal_places>\ninline bool operator != (const FixedFloat<_decimal_places>& lhs,\nconst FixedFloat<_decimal_places>& rhs)\n{\nreturn lhs.m_value != rhs.m_value;\n}\n\ntemplate <size_t _decimal_places>\ninline FixedFloat<_decimal_places> operator - (const FixedFloat<_decimal_places>& rhs)\n{\nreturn FixedFloat<_decimal_places>(-rhs.m_value);\n}\n\ntemplate <size_t _decimal_places>\ninline FixedFloat<_decimal_places> operator + (const FixedFloat<_decimal_places>& rhs)\n{\nreturn FixedFloat<_decimal_places>(rhs.m_value);\n}\n\ntemplate <size_t _decimal_places>\ninline std::ostream& operator << (std::ostream& output_stream,\nconst FixedFloat<_decimal_places>& rhs)\n{\nbool negative = false;\nif (rhs.m_value < 0)\nnegative = true;\n\ntypename FixedFloat<_decimal_places>::value_type whole_part\n= rhs.m_value / rhs.m_correction;\ntypename FixedFloat<_decimal_places>::value_type fractional_part\n= rhs.m_value - (whole_part * rhs.m_correction);\n\nif (whole_part < 0) // turn into a positive number\nwhole_part = -whole_part;\nif (fractional_part < 0) // turn into a positive number\nfractional_part = -fractional_part;\n\nstd::ostringstream s;\ns.copyfmt(output_stream);\ns.width(0);\n\nif (negative == true)\ns << '-'; // output sign, if needed\ns << whole_part; // output the whole part\nif (_decimal_places > 0) // output the fractional part, including decimal point\n{\nstd::ostringstream fractional_string;\nfractional_string.imbue(std::locale::classic()); // \"plain\" numbers\nfractional_string << fractional_part;\n\ns << std::use_facet<std::numpunct<char> >(s.getloc()).decimal_point()\n<< std::setw(_decimal_places) << std::setfill('0')\n<< fractional_string.str();\n}\noutput_stream << s.str();\n\nreturn output_stream;\n}\n\ntemplate <size_t _decimal_places>\ninline std::istream& operator >> (std::istream& input_stream,\nFixedFloat<_decimal_places>& rhs)\n{\nbool negative = false;\ntypename FixedFloat<_decimal_places>::value_type whole_part = 0;\nchar decimal_point;\ntypename FixedFloat<_decimal_places>::value_type fractional_part = 0;\n\nstd::istream::sentry stream_sentry(input_stream, true);\n\nif (stream_sentry)\n{\n// Get the whole part of the number\nreturn input_stream;\n\n// Check signedness (would be lost if value is < 1, since -0 == 0)\nif (input_stream.peek() == '+')\nnegative = false;\nelse if (input_stream.peek() == '-')\nnegative = true;\n\ninput_stream >> whole_part;\nwhole_part *= rhs.m_correction;\nif (whole_part < 0) // turn into a positive number\nwhole_part = -whole_part;\nif (_decimal_places > 0)\n{\n// Get the decimal point.\nreturn input_stream;\n\ninput_stream >> decimal_point;\n// Check that the decimal point was the correct type for this locale\nif (decimal_point != std::use_facet<std::numpunct<char> >(input_stream.getloc()).decimal_point())\n{\nrhs.m_value = 0;\ninput_stream.setstate(std::ios::failbit);\n}\n\n// Get the fractional part of the number\nfractional_part = 0;\nfor (size_t i = _decimal_places; i > 0; --i)\n{\nreturn input_stream;\n\nchar decimal_char = '0';\ninput_stream >> decimal_char;\nif (decimal_char < '0' || decimal_char > '9')\n{\nrhs.m_value = 0;\ninput_stream.setstate(std::ios::failbit);\nreturn input_stream;\n}\nsize_t decimal_number = decimal_char - '0';\n\nsize_t multiply_factor = 1;\nfor (int j = 1; j < i; ++j)\nmultiply_factor *= 10;\n\nfractional_part += (decimal_number * multiply_factor);\n}\n}\nif (negative == false)\nrhs.m_value = whole_part + fractional_part;\nelse\nrhs.m_value = - (whole_part + fractional_part);\n}\nelse\ninput_stream.setstate(std::ios::failbit);\n\nreturn input_stream;\n}\n----end fixedfloat.h----\n\n--\nRoger Leigh\n\nPrinting on GNU/Linux? http://gimp-print.sourceforge.net/\nJul 19 '05 #3\n\"Roger Leigh\" <\\${******@invalid.whinlatter.uklinux.net.invalid > wrote in\nmessage news:87************@wrynose.whinlatter.uklinux.net ...\nYou can pervert money_put and moneypunct to output a digit sequence\nstored in a string, with a specified number of decimal places, commas\nbetween thousands groups, etc.\nThis looks like what I'll do. I'll derive a \"Money\" class from\nFixedFloat and do that in there, overriding the standard ostream<< and\nistream>> operators.\n\nI've attached a copy of the working class, and a small driver program\nto show it in action (sorry it's so long). I have a few questions\n\n1. Is the header file OK style-wise? Is there anything wrong that I\nshould not be doing?\n\nSorry, I don't have time to critique what you've done in detail.\nInstead I supply below a sample use of money_put and moneypunct\nI published in The C/C++ Users Journal (April 1998), for comparative\nanatomy studies.\n2. I've noticed that the modulus (operator%) member and friend\nfunctions can be out by a small factor e.g. 0.0001 in a 4\nd.p. precision class. With fixed-point arithmetic, should I be\ndoing anything to correct this? Is it actually incorrect? For\nsome reason, I couldn't get the \"real\" % operator to work, so had\nto resort to the hack that actually gets used (subtracting the\nresult of division and subsequent multiplication from the original\nvalue).\nNot such a hack, since it's built on the basic definition. It's very\nhard to avoid 1 or even 2 ulp errors with this sort of math. That's\nwhy people are reconsidering decimal floating point these days.\n3. Looking at the compiled binary, the FixedFloat symbols have weak,\nrather than vague linkage. I thought that /all/ templated class\nmethods and functions would be vague. This is with GCC 3.3.2, GNU\nld 2.14.90.0.6 and binutils 2.14.90.0.6-5 on i686-pc-linux-gnu\nusing ELF binary format (this is probably OT).\nI don't know anything about these forms of linkage. But I think\n\"vague linkage\" is a wonderfully surreal term, FWIW.\n4. In the stream output and extraction friend classes, is the use of\nlocales correct? I've not used locales (in C++) before, and I've\ndone this using the Josuttis Standard Library book.\n\nIf it works... The example below uses our magic locale macros to\ndeal with VC++ V6.0 compiler limitations.\n\nP.J. Plauger\nDinkumware, Ltd.\nhttp://www.dinkumware.com\n--------------\n\n#include <iomanip>\n#include <iostream>\n#include <locale>\nusing namespace std;\n\n// MONETARY TYPES\ntypedef long double MoneyVal;\n\nclass Money {\npublic:\nMoney(MoneyVal v)\n: value(v) {}\noperator MoneyVal() const\n{return (value); }\nprivate:\nMoneyVal value;\n};\n\n// Money INSERTER\ntemplate<class _E, class _Tr> inline\nbasic_ostream<_E, _Tr>& operator<<(\nbasic_ostream<_E, _Tr>& _O, Money _Y)\n{typedef ostreambuf_iterator<_E, _Tr> _Iter;\ntypedef money_put<_E, _Iter> _Mput;\n\nios_base::iostate _St = ios_base::goodbit;\nconst typename basic_ostream<_E, _Tr>::sentry _Ok(_O);\nif (_Ok)\n{try\n{const _Mput& _Fac =\n_USEFAC(_O.getloc(), _Mput);\nif (_Fac.put(_Iter(_O.rdbuf()),\n(_O.flags() & ios_base::showpos) != 0,\n_O, _O.fill(), _Y).failed())\ncatch (...)\n_O.setstate(_St);\nreturn (_O); }\n\n// moneypunct FOR USA LOCALE\nmoney_base::pattern mon_fmt = {\nmoney_base::symbol, money_base::space,\nmoney_base::sign, money_base::value};\n\nclass Mymoneypunct\n: public moneypunct<char, false> {\nprotected:\nvirtual char do_decimal_point() const\n{return ('.'); }\nvirtual char do_thousands_sep() const\n{return (','); }\nvirtual string do_grouping() const\n{return (string(\"\\3\")); }\nvirtual string do_curr_symbol() const\n{return (string(\"\\$\")); }\nvirtual string do_positive_sign() const\n{return (string(\"\")); }\nvirtual string do_negative_sign() const\n{return (string(\"-\")); }\nvirtual int do_frac_digits() const\n{return (2); }\nvirtual pattern do_pos_format() const\n{return (mon_fmt); }\nvirtual pattern do_neg_format() const\n{return (mon_fmt); }\n};\n\nint main()\n{locale loc = _ADDFAC(locale::classic(), new Mymoneypunct);\ncout.imbue(loc);\n\ncout << showbase << setw(20) << internal << setfill('*')\n<< Money(123456789.0) << endl;\nreturn (0); }\nJul 19 '05 #4\n\"P.J. Plauger\" <pj*@dinkumware.com> writes:\n\"Roger Leigh\" <\\${******@invalid.whinlatter.uklinux.net.invalid > wrote in\nmessage news:87************@wrynose.whinlatter.uklinux.net ...\n2. I've noticed that the modulus (operator%) member and friend\nfunctions can be out by a small factor e.g. 0.0001 in a 4\nd.p. precision class. With fixed-point arithmetic, should I be\ndoing anything to correct this? Is it actually incorrect? For\nsome reason, I couldn't get the \"real\" % operator to work, so had\nto resort to the hack that actually gets used (subtracting the\nresult of division and subsequent multiplication from the original\nvalue).\n\nNot such a hack, since it's built on the basic definition. It's very\nhard to avoid 1 or even 2 ulp errors with this sort of math. That's\nwhy people are reconsidering decimal floating point these days.\n\nThis looks quite exciting, and I look forward to using this in the\nfuture, when it's implemented. Are there any C++ classes implementing\nthis yet?\n\nI've got hold of the specs from the IBM Hursley site, and also some\ndocs about BCD arithmetic. If it's not too hairy, I might be able to\nknock out a C++ class for this myself, but I'm not a mathematician and\nworry about the subtleties I might get wrong when dealing with\nfinancial stuff.\n\nFor the time being, I've been looking for other classes and libraries\nout there. GNU MP (libgmp) looks like a decent choice, since it can\ndo arbitrary-precision computation, and it has a C++ binding. I\nwouldn't have to worry about overflow or underflow if I used this.\nHowever, the values still have to be stored in a fixed-precision\nbackend database (PostgreSQL numeric type), so having a\nfixed-precision class to directly represent the database types would\n3. Looking at the compiled binary, the FixedFloat symbols have weak,\nrather than vague linkage. I thought that /all/ templated class\nmethods and functions would be vague. This is with GCC 3.3.2, GNU\nld 2.14.90.0.6 and binutils 2.14.90.0.6-5 on i686-pc-linux-gnu\nusing ELF binary format (this is probably OT).\n\nI don't know anything about these forms of linkage. But I think\n\"vague linkage\" is a wonderfully surreal term, FWIW.\n\n:-)\n\nWith weak linkage, if the same symbol is present in multiple\ntranslation units, only one will be resolved by the runtime linker.\nemitted for every translation unit (I believe this is called COMDAT on\nsome platforms). I expected the latter behaviour, but didn't get it\nwith my own templates (although I see it for all the Standard Library\nones, such as basic_ofstream<> et. al.). I'll have to investigate\nthis one further.\n4. In the stream output and extraction friend classes, is the use of\nlocales correct? I've not used locales (in C++) before, and I've\ndone this using the Josuttis Standard Library book.\n\nIf it works... The example below uses our magic locale macros to\ndeal with VC++ V6.0 compiler limitations.\n\nThanks, that was quite informative. I'll be using GCC and GNU\nlibstdc++5 on all our target platforms, including Windows, so I won't\nhave to deal with VC++, at least initially. It would merit further\ninvestigation if it provided a POSIX/SUSv3 layer like Cygwin or MinGW.\nI require the Gtkmm and PostgreSQL client libraries for my current\nproject, though.\n--\nRoger Leigh\n\nPrinting on GNU/Linux? http://gimp-print.sourceforge.net/" ]

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[ "", null, "## Abstract\n\nGrid approximation of the Cauchy problem on the interval  D = {0 ≤ x ≤ d} is first studied for a linear singularly perturbed ordinary differential equation of the first order with a perturbation parameter ε multiplying the derivative in the equation where the parameter ε takes arbitrary values in the half-open interval (0, 1]. In the Cauchy problem under consideration, for small values of the parameter ε, a boundary layer of width O(ε) appears on which the solution varies by a finite value. It is shown that, for such a Cauchy problem, the solution of the standard difference scheme on a uniform grid does not converge ε-uniformly in the maximum norm; convergence occurs only under the condition h ε, where h = d N −1 , N is the number of grid intervals, h is the grid step-size. Taking into account the behavior of the singular component in the solution, a special piecewise-uniform grid is constructed that condenses in a neighborhood of the boundary layer. It is established that the standard difference scheme on such a special grid converges ε-uniformly in the maximum norm at the rate O(N −1 lnN). Such a scheme is called a robust one.\n\nFor a model Cauchy problem for a singularly perturbed ordinary differential equation, standard difference schemes on a uniform grid (a classical difference scheme) and on a piecewise-uniform grid (a special difference scheme) are constructed and investigated. The results of numerical experiments are given, which are consistent with theoretical results.\n\nHow to Cite\nShishkina, L. P., & Shishkin, G. I. (2018). Robust difference scheme for the Cauchy problem for a singularly perturbed ordinary differential equation. Mathematical Modelling and Analysis, 23(4), 527-537. https://doi.org/10.3846/mma.2018.031\nPublished in Issue\nOct 9, 2018\nAbstract Views\n196" ]

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[ "Solutions by everydaycalculation.com\n\n## Reduce 125/168 to lowest terms\n\n125/168 is already in the simplest form. It can be written as 0.744048 in decimal form (rounded to 6 decimal places).\n\n#### Steps to simplifying fractions\n\n1. Find the GCD (or HCF) of numerator and denominator\nGCD of 125 and 168 is 1\n2. Divide both the numerator and denominator by the GCD\n125 ÷ 1/168 ÷ 1\n3. Reduced fraction: 125/168\nTherefore, 125/168 simplified to lowest terms is 125/168.\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]

[ "## The Quantitative Calculation of pH in Monoacid/Monobasic Solution in Teaching of Inorganic Chemistry\n\n 基金资助: 郑州大学化学学科拔尖计划2.0项目2020郑州大学本科教学改革研究与实践项目2020郑州大学大学生创新创业训练计划项目", null, "Abstract\n\nThe quantitative calculation of pH of monoacid/monobasic solution is a basic problem for inorganic chemistry teaching. The qualitative description and quantitative calculation of pH are connected by approximation conditions. It is very important for the students to understand these approximation conditions, which will help them solve problems about ionic equilibrium in aqueous solution. However, how to use these approximation conditions correctly in solution of low concentration and weak acids is often difficult. Even in some textbooks, the description of concepts is not accurate and not consistent with the associated courses. Thus, to solve the problems mentioned above, the derivations of the approximation conditions are introduced and examples are analyzed in detail in this paper. The problems of quantitative calculation of pH of monoacid/monobasic solution which should be paid attention to in the course teaching of inorganic chemistry are discussed.\n\nKeywords： Inorganic chemistry ; Course teaching ; pH ; Quantitative calculation\n\nWang Ting. The Quantitative Calculation of pH in Monoacid/Monobasic Solution in Teaching of Inorganic Chemistry. University Chemistry[J], 2020, 35(8): 129-134 doi:10.3866/PKU.DXHX202003085\n\n## 1 一元酸碱溶液pH定量计算的近似条件的推导\n\n${\\rm H^{+} + A^{−}} \\rightleftharpoons {\\rm HA} \\\\ {K_{\\rm{a}}} = \\frac{{[{{\\rm{H}}^ + }][{{\\rm{A}}^ - }]}}{{[{\\rm{HA}}]}}$\n\n${\\rm H_{2}O} \\rightleftharpoons {\\rm H^{+} + OH^{−}} \\\\ {K_{\\rm{w}}} = [{{\\rm{H}}^ + }][{\\rm{O}}{{\\rm{H}}^ - }]$\n\n$[{\\operatorname{H} ^ + }] = [{{\\rm{A}}^ - }] + [{\\rm{O}}{{\\rm{H}}^ - }]$\n\n$[{{\\rm{H}}^ + }] = \\frac{{{K_{\\rm{a}}}[{\\rm{HA}}]}}{{[{{\\rm{H}}^ + }]}} + \\frac{{{K_{\\rm{w}}}}}{{[{{\\rm{H}}^ + }]}}$\n\n$[{{\\rm{H}}^ + }] = \\sqrt {{K_{\\rm{a}}}[{\\rm{HA}}] + {K_{\\rm{w}}}}$\n\n$c = [{\\rm{HA}}] + [{{\\rm{A}}^ - }] = [{\\rm{HA}}] + \\frac{{{K_{\\rm{a}}}[{\\rm{HA}}]}}{{[{{\\rm{H}}^ + }]}}$\n\n$[{\\rm{HA}}] = \\frac{{c[{{\\rm{H}}^ + }]}}{{[{{\\rm{H}}^ + }] + {K_{\\rm{a}}}}}$\n\n$[{{\\rm{H}}^ + }] = \\sqrt {\\frac{{{K_{\\rm{a}}}c[{{\\rm{H}}^ + }]}}{{[{{\\rm{H}}^ + }] + {K_{\\rm{a}}}}} + {K_{\\rm{w}}}}$\n\n${[{{\\rm{H}}^ + }]^3} + {K_{\\rm{a}}}{[{{\\rm{H}}^ + }]^2} - ({K_{\\rm{a}}}c + {K_{\\rm{w}}})[{{\\rm{H}}^ + }] - {K_{\\rm{a}}}{K_{\\rm{w}}} = 0$\n\n${E_1} = \\frac{{\\sqrt {{K_{\\rm{a}}}[{\\rm{HA}}]} - \\sqrt {{K_{\\rm{a}}}[{\\rm{HA}}] + {K_{\\rm{w}}}} }}{{\\sqrt {{K_{\\rm{a}}}[{\\rm{HA}}] + {K_{\\rm{w}}}} }}$\n\n 相对误差< 1% 2% 2.5% 4% 5% Ka[HA] > 49.3Kw 24.3Kw 19.3Kw 11.8Kw 9.26Kw\n\n$[{{\\rm{H}}^ + }] = \\sqrt {{K_{\\rm{a}}}[{\\rm{HA}}]}$\n\n$[{\\rm{HA}}] = c - [{{\\rm{H}}^ + }] + [{\\rm{O}}{{\\rm{H}}^ - }]$\n\n$[{\\rm{HA}}] = c - [{{\\rm{H}}^ + }]$\n\n$[{{\\rm{H}}^ + }] = \\sqrt {{K_{\\rm{a}}}(c - [{{\\rm{H}}^ + }])}$\n\n${[{{\\rm{H}}^ + }]^2} + {K_{\\rm{a}}}[{{\\rm{H}}^ + }] - {K_{\\rm{a}}}c = 0$\n\n$[{{\\rm{H}}^ + }] = \\frac{{ - {K_{\\rm{a}}} + \\sqrt {K_{\\rm{a}}^2 + 4{K_{\\rm{a}}}c} }}{2}$\n\n$[{{\\rm{H}}^ + }] = \\sqrt {{K_{\\rm{a}}}c}$\n\n${E_2} = \\frac{{\\sqrt {{K_{\\rm{a}}}c} - \\frac{{ - {K_{\\rm{a}}} + \\sqrt {{K_{\\rm{a}}}^2 + 4{K_{\\rm{a}}}c} }}{2}}}{{\\frac{{ - {K_{\\rm{a}}} + \\sqrt {{K_{\\rm{a}}}^2 + 4{K_{\\rm{a}}}c} }}{2}}}$\n\n 相对误差< 1% 2% 2.5% 4% 5% c/Ka > 2130 637 417 162 105\n\n$[{{\\rm{H}}^ + }] = \\sqrt {{K_{\\rm{a}}}c + {K_{\\rm{w}}}}$\n\n Kac c/Ka 计算方法 < 20Kw < 400 ${[{{\\rm{H}}^ + }]^3} + {K_{\\rm{a}}}{[{{\\rm{H}}^ + }]^2} - ({K_{\\rm{a}}}c + {K_{\\rm{w}}})[{{\\rm{H}}^ + }] - {K_{\\rm{a}}}{K_{\\rm{w}}} = 0$ < 20Kw > 400 $[{{\\rm{H}}^ + }] = \\sqrt {{K_{\\rm{a}}}c + {K_{\\rm{w}}}}$ > 20Kw < 400 $[{{\\rm{H}}^ + }] = \\frac{{ - {K_{\\rm{a}}} + \\sqrt {{K_{\\rm{a}}}^2 + 4{K_{\\rm{a}}}c} }}{2}$ > 20Kw > 400 $[{{\\rm{H}}^ + }] = \\sqrt {{K_{\\rm{a}}}c}$\n\n### 2.1 一元强酸pH计算实例\n\n$1.0 \\times {10^{ - 3}}{\\rm{mol}} \\cdot {\\rm{d}}{{\\rm{m}}^{ - {\\rm{3}}}} \\times \\frac{{1.0 \\times {{10}^{ - 3}}{\\rm{d}}{{\\rm{m}}^3}}}{{10.0{\\rm{ d}}{{\\rm{m}}^3} + 1.0 \\times {{10}^{ - 3}}{\\rm{ d}}{{\\rm{m}}^3}}} = 1.0 \\times {10^{ - 7}}{\\rm{mol}} \\cdot {\\rm{d}}{{\\rm{m}}^{ - {\\rm{3}}}}$\n\n\\begin{align} \\;\\;\\;\\;\\;\\; {{\\text{H}}_{\\text{2}}}\\text{O} = {{\\text{H}}^{+}} + \\text{O}{{\\text{H}}^{-}} \\\\ \\begin{array}{*{35}{l}} { {{c}_{{起始}}}(\\text{mol}\\cdot \\text{d}{{\\text{m}}^{-\\text{3}}}) } & { 1.0\\times {{10}^{-7}} }& {0} \\\\ {{{c}_{{平衡}}}(\\text{mol}\\cdot \\text{d}{{\\text{m}}^{-\\text{3}}}) } & {1.0\\times {{10}^{-7}}\\text{ + }x } & { x} \\\\ \\end{array} \\end{align}\n\n${K_{\\rm{w}}} = [{{\\rm{H}}^ + }][{\\rm{O}}{{\\rm{H}}^ - }] = (1.0 \\times {10^{ - 7}} + x)x = 1.0 \\times {10^{ - 14}}$\n\n$x = 0.6 \\times {10^{ - 7}}$\n\n$[{{\\rm{H}}^ + }] = \\left( {1.0 \\times {{10}^{ - 7}} + 0.6 \\times {{10}^{ - 7}}} \\right){\\rm{mol}} \\cdot {\\rm{d}}{{\\rm{m}}^{ - {\\rm{3}}}} = 1.6 \\times {10^{ - 7}}{\\rm{mol}} \\cdot {\\rm{d}}{{\\rm{m}}^{ - {\\rm{3}}}}$\n\n$1.0 \\times {10^{ - 3}}{\\rm{ mol}} \\cdot {\\rm{d}}{{\\rm{m}}^{ - {\\rm{3}}}} \\times \\frac{{1.0 \\times {{10}^{ - 3}}{\\rm{d}}{{\\rm{m}}^3}}}{{10.0{\\rm{ d}}{{\\rm{m}}^3} + 1.0 \\times {{10}^{ - 3}}{\\rm{ d}}{{\\rm{m}}^3}}} = 1.0 \\times {10^{ - 7}}{\\rm{mol}} \\cdot {\\rm{d}}{{\\rm{m}}^{ - {\\rm{3}}}}$\n\n$[{{\\rm{H}}^ + }] = \\frac{{ - {K_{\\rm{a}}} + \\sqrt {{K_{\\rm{a}}}^2 + 4{K_{\\rm{a}}}c} }}{2} = 1.0 \\times {10^{ - 7}}{\\rm{mol}} \\cdot {\\rm{d}}{{\\rm{m}}^{ - {\\rm{3}}}}$\n\n### 2.2 一元弱酸pH计算实例\n\n$[{{\\rm{H}}^ + }] = \\sqrt {{K_{\\rm{a}}}c + {K_{\\rm{w}}}} {\\rm{ = }}2.7 \\times {10^{ - 7}}{\\rm{mol}} \\cdot {\\rm{d}}{{\\rm{m}}^{ - {\\rm{3}}}}$\n\n$[{{\\rm{H}}^ + }] = \\frac{{ - {K_{\\rm{a}}} + \\sqrt {{K_{\\rm{a}}}^2 + 4{K_{\\rm{a}}}c} }}{2} = 2.5 \\times {10^{ - 7}}{\\rm{mol}} \\cdot {\\rm{d}}{{\\rm{m}}^{ - {\\rm{3}}}}，误差7.8\\%。$\n\n## 3 教学过程中的一些建议\n\n1) 《无机化学》教材中应适当加入一元酸碱pH计算的近似条件的推导过程，帮助学生清晰理解这些近似条件的来源。这部分内容也可作为阅读材料让学生自学，加深其对溶液中离子平衡计算问题的认识。\n\n2)在课堂讲解过程中，如果课时允许，应加入适当的推导演算环节，避免学生对公式死记硬背。同时，着重强调这些近似计算条件的适用前提。\n\n3)在课程中补充介绍一些应用计算机软件计算酸碱pH的方法，如CurTiPot等程序或软件的使用等。\n\n## 参考文献 原文顺序 文献年度倒序 文中引用次数倒序 被引期刊影响因子\n\n/\n\n 〈", null, "〉", null, "" ]

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[ "# Big numbers\n\n(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)\n\nBig numbers, known colloquially as bignums (adjectival form bignum, as in bignum arithmetic) are integers whose range exceeds those of machine registers. For example, most modern processors possess 64-bit registers which can be used to store integers up to 264-1. It is usually possible to add, subtract, multiply, or divide such integers in a single machine instruction. However, such machines possess no native implementation of arithmetic on numbers larger than this, nor any native means of representing them. In some applications, it might be necessary to work with numbers with hundreds or even thousands of digits.\n\nOf course, humans have no problems with working with numbers greater than 264-1, other than the fact that it becomes increasingly tedious as the numbers grow larger; we just write them out and use the same algorithms we use on smaller numbers: add column-by-column and carry, and so on. This turns out to be the key to working with bignums in computer programs too. When we write out a number in decimal representation, we are essentially expressing it as a string or array of digits and the algorithms we use to perform arithmetic on them, which entail examining the digits in a particular order, may be considered as loops.\n\n## Nonnegative integers\n\n### Digital representation\n\nMost of the time, and almost all of the time in contests, the only bignums the programmer will be required to implement are positive integers. As alluded to above, bignums in computer programs are represented using a radix system. This means that a bignum is stored in a base", null, "$n$ representation, where the choice of", null, "$n$ is based on the application. Precisely, the nonnegative integer", null, "$N = a_0 + a_1 n + a_2 n^2 + ... + a_k n^k$ is stored by giving the values", null, "$a_0, a_1, a_2, ..., a_k$, which we probably want to store as an array. Here are a few common possibilities, demonstrated for the example of 30! = 265252859812191058636308480000000:\n\n• If we let the radix be 10:\n• ASCII representation: the number is represented by a string which literally contains the number's digits as characters: 265252859812191058636308480000000 would be represented ['2','6','5',...,'0']\n• BCD representation: an array of digits (not their ASCII values), hence [2,6,5,...,0]\n• If we let the radix be 109, we can store nine digits in each array entry. For example, 30! could be stored [265252,859812191,058636308,480000000]. Note that we must take care to group digits starting from the ones' place and moving left, instead of starting at the most significant digit and moving right, to avoid complicating the code and eroding performance. This is because adding and subtracting bignums requires that they be aligned at the decimal point (not at the most significant digit), so it is convenient when we guarantee that this is never found within an entry, otherwise shifting would be necessary.\n• If we let the radix be 232, we use a sequence of 32-bit unsigned integer typed variables to store the bignum. We know 26525285981219105863630848000000010 = D13F6370F96865DF5DD5400000016. Again, we align at the ones' place, giving [00000D1316,F6370F9616,865DF5DD16,5400000016]. (Of course, you should not actually explicitly store the hex representations, because that would be cumbersome and slow. It's just an array of 32-bit values.)\n\n### Little-endian vs. big-endian\n\nThe byte is the fundamental addressable unit of memory for a given processor. This is distinct from the word, which is the natural unit of data for a given processor. For example, the Intel 386 processor had 8 bits to a byte but 32 bits to a word. Each byte in memory may be addressed individually by a pointer, but one cannot address the individual bits in them. That being said, when a 32-bit machine word is written to memory, there are two ways it could be done. Suppose the number CAFEBABE16 is stored at memory location DEADBEEF16. This will occupy four bytes of memory, and they must be contiguous so that the processor can read and write them as units. The important question is whether the most significant byte (in this case CA16) comes first (big-endian) or last (little-endian). The following table shows where each byte ends up in each scheme.\n\nBig-endian CA16 FE16 BA16 BE16\nLittle-endian BE16 BA16 FE16 CA16\n\nOne faces a similar choice when storing bignums: does the most significant part get stored in the first or the last position of the array? Almost every processor is either consistently little-endian or consistently big-endian, but this does not affect the programmer's ability to choose either little-endian or big-endian representations for bignums as the application requires. The importance of this is discussed in the next section.\n\n### Fixed-width versus dynamic-width bignums\n\nThere are, in principle, two kinds of bignum implementation. Suppose we know in advance the maximum size of the integers we might be working with. For example, in [email protected], we are asked to report the number of permutations which satisfy a certain property. There are only up to 30 elements, so we know that the answer will not exceed 30! = 265252859812191058636308480000000, which has 33 digits. It is not terribly difficult to implement the solution in such a way that no intermediate variable is ever larger than this. So we could, for example, use a string of length 33 to store all integers used in the computation of the answer (where numbers with fewer than 33 digits are padded with zeroes on the left), and treat all numbers as though they had 33 digits. The only time when we would care how many digits the number actually has is when outputting it. Addition can now be implemented as a loop over 33 columns, and is fairly simple. This implementation is called fixed-width.\n\nOn the other hand, sometimes it is not so easy to determine in advance the size of the numbers we might be working with, or a problem might have bundled test cases and a strict time limit, forcing the programmer to make the small cases run more quickly than the large ones. When this occurs it is a better idea to use dynamic bignums, which can expand or shrink according to their length. Dynamic bignums are trickier to code than fixed-width ones: when we add them, for example, we have to take into account that they might not be of the same length; we might then treat all the missing digits as zeroes, but in any case it requires extra code. When using dynamic bignums the difference between the little-endian and big-endian representations becomes significant. If we store the bignums little-endian, and add them, alignment is free: just look at the first entry in each of them; they are in the ones' places of their respective numbers. The code presented in this article will assume the little-endian representation.\n\n### Operations\n\nThis section describes how to actually manipulate bignums. We assume a dynamic zero-based little endian array representation but leave lots of details to the programmer.\n\nThe schoolbook addition algorithm first adds the ones places, then adds the tens (with a carry, if necessary), then the hundreds, and so on. We will likewise start by adding the ones places and proceed to more significant digits.\n\n```input bignums x,y\nn ← length of x\nm ← length of y\np ← max(m,n)\ncarry ← 0\nfor i ∈ [0..p)\ncarry ← 1\nelse\ncarry ← 0\nif carry\nz[p] ← 1\np ← p+1\n```\n\nAfter this code has completed, `z` will hold the sum of `x` and `y`, and `p` will be the length of `z` (the number of nonzero places). There are two caveats, though:\n\n• When adding bignums of equal length, the loop counter `i` will grow beyond the length of one of the two. What does `x[i]` mean when `i``n`? To make this code work, it should be treated as zero. In a working implementation, we must take care to avoid out-of-bounds array access.\n• If the radix used is the same size as a machine word, then we cannot actually check whether `x[i]+y[i]+carry ≥ radix` as shown. Instead, check whether the result is greater that or equal to both `x` and `y`. If it is not, an overflow has occurred (and the carry bit should be set.)\n\nNevertheless, this shows the basic idea behind the addition of bignums. Here is a sample of C++ code as it might actually appear, using radix 10:\n\n```// x, y, and z are vectors of digits\n{\nint n = x.length();\nint m = y.length();\nint p = max(n,m);\nz.resize(p);\nint carry = 0;\nfor (int i=0; i<p; i++)\n{\nint t=carry;\nif (i<n) t+=x[i];\nif (i<m) t+=y[i];\nz[i]=t%10;\ncarry=t/10;\n}\nif (carry)\nz.push_back(1);\n}\n```\n\n### Error conditions\n\nWhat happens if the program tries to divide a bignum by zero? In almost all popular programming languages (but not C), one can take advantage of built-in-support for exceptions. If the bignum division function raises an exception, this can be caught by the calling function and handled somehow. What we do not want is an unconditional crash of the program. There are also some other conditions that can generate unpredictable behavior, such as integer overflow. It is important to check for these in the code and take appropriate action (such as raising an exception).\n\n• Check that you don't divide by zero.\n• Check that a larger number is not subtracted from a smaller number, otherwise unpredictable behavior may result.\n• Fixed-width bignums may overflow when adding or multiplying. You will be able to detect this because a carry is generated in the left-most column when adding, or something like that. Naively failing to check this may result in out-of-bounds array access.\n• Check that the input does not overflow a fixed-width bignum. Naively failing to check this will probably result in an out-of-bounds array access.\n• If you're writing a bignum library, and intend to reuse the code, you might also want to check, with dynamic bignums, that your attempts to allocate more memory actually succeed. (You might be sure that your computer and the judge computer have enough memory for the purposes of this particular problem, but maybe you'll do another problem later with a lower memory limit, or someone else with less RAM will use your code.) If not, and you just keep going on anyway, a segfault is almost certain to occur. If you do check, you can throw an exception." ]

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[ "Becoming fluent in math tables is necessary if you want to do complex mathematical calculations fast. Memorizing the tables becomes very helpful during exams as well as in our day-to-day lives.\n\nHere’s the table of 522 which consists of the multiplication of 522 with the whole numbers:\n\n## Multiplication table of 522\n\nHere’s the multiplication table of 522 up to x20:\n\n522 times Equals\n522 x 1 522\n522 x 2 1044\n522 x 3 1566\n522 x 4 2088\n522 x 5 2610\n522 x 6 3132\n522 x 7 3654\n522 x 8 4176\n522 x 9 4698\n522 x 10 5220\n522 x 11 5742\n522 x 12 6264\n522 x 13 6786\n522 x 14 7308\n522 x 15 7830\n522 x 16 8352\n522 x 17 8874\n522 x 18 9396\n522 x 19 9918\n522 x 20 10440\n\n### 522 times table in words\n\n• 522 times 1 is equal to 522\n• 522 times 2 is equal to 1044\n• 522 times 3 is equal to 1566\n• 522 times 4 is equal to 2088\n• 522 times 5 is equal to 2610\n• 522 times 6 is equal to 3132\n• 522 times 7 is equal to 3654\n• 522 times 8 is equal to 4176\n• 522 times 9 is equal to 4698\n• 522 times 10 is equal to 5220\n• 522 times 11 is equal to 5742\n• 522 times 12 is equal to 6264\n• 522 times 13 is equal to 6786\n• 522 times 14 is equal to 7308\n• 522 times 15 is equal to 7830\n• 522 times 16 is equal to 8352\n• 522 times 17 is equal to 8874\n• 522 times 18 is equal to 9396\n• 522 times 19 is equal to 9918\n• 522 times 20 is equal to 10440\n\nAlso see:\n\n## Multiplication table of 522 in the form of addition\n\nMultiplication of natural numbers is repeated addition.\n\nIf you have to multiply 522 to 3 then it’s just the addition of 522 three times, i.e., 522 + 522 + 522 which ultimately gives you 1566.\n\nLet’s take a look at the full table of 522 below:\n\n522 x 1 = 522 522\n522 x 2 = 1044 522 + 522\n522 x 3 = 1566 522 + 522 + 522\n522 x 4 = 2088 522 + 522 + 522 + 522\n522 x 5 = 2610 522 + 522 + 522 + 522 + 522\n522 x 6 = 3132 522 + 522 + 522 + 522 + 522 + 522\n522 x 7 = 3654 522 + 522 + 522 + 522 + 522 + 522 + 522\n522 x 8 = 4176 522 + 522 + 522 + 522 + 522 + 522 + 522 + 522\n522 x 9 = 4698 522 + 522 + 522 + 522 + 522 + 522 + 522 + 522 + 522\n522 x 10 = 5220 522 + 522 + 522 + 522 + 522 + 522 + 522 + 522 + 522 + 522\n\n## FAQs\n\n### 1. How to count or read the table of 522?\n\nYou can read the table of 522 by following the below lines:\n\n• 522 times one is 522\n• 522 times two is 1044\n• 522 times three is 1566\n• 522 times four is 2088\n• 522 times five is 2610\n• 522 times six is 3132\n• 522 times seven is 3654\n• 522 times eight is 4176\n• 522 times nine is 4698\n• 522 times ten is 5220\n\n### 2. Why memorizing tables is important?\n\nMemorizing tables makes mental calculations effortless and efficient, it makes people think. The multiplication tables come in very handy while doing geometry, algebra, trigonometry, calculus, number theory, etc.\n\n## Explore\n\nExplore other interesting stuff related to math and numbers." ]

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[ "Packing and Unpacking I/O Bytes", null, "Click HERE to download a printable copy of Chapter 8\n\nPacking and Unpacking I/O Bytes\n\nBefore we proceed with more involved application examples, it is important that we explore one area of software where a little further explanation is required before the procedures become clear. This is the packing and unpacking of I/O bytes. Fundamentally, with the C/MRI we always “talk” to the railroad through the I/O cards. These cards can be the separate cards plugged into an I/O motherboard or the “I/O cards” built into the SMINI.\n\nEach I/O card type, input or output, supports either 24- or 32-I/O lines that are grouped together in sets of eight lines each. We call each group of 8-lines a port. The ports are labeled A, B and C for 24-line cards and A, B, C and D for 32-line cards. Each line within a port has two states typically referred to as being either high or low, on or off, 1 or 0, +5Vdc or 0 and so forth.\n\nIn computer-ese we often speak of an I/O line as a “binary bit” or simply “bit” for short. Eight bits, when grouped together as they are within each port, are referred to as a “byte.” Therefore, each port is equivalent to 1 byte (of data), made up of 8 individual bits.\n\nI/O HANDLING\n\nUsing an output card, we write to, or set, the values on each output line. This way the software controls devices on the railroad. With an input card we read the value that the railroad has set on each line. This way the software monitors the status of devices on the railroad.\n\nThe software needs to be able to look at each of the wired-lines on each port as a bit. Like the wire itself, each software bit can assume two states typically referred to as being either high or low, on or off, 1 or 0, etc. Each byte has 8 bits and they exhibit a one-to-one correspondence to the 8 lines for each port. Software, when it reads and writes to an I/O card’s port, must treat the grouping of the 8-bits for each port as a single entity.\n\nSoftware must read from each input card a byte at a time. Software must write to each output card a byte at a time. Neither QuickBASIC nor Visual Basic, nor most other programming languages, read or write at the individual line, or bit level. However, most software variables within our programs correspond to only a few hardware lines. For example, if SE(12) represents a 3-LED signal at the east end of Block 12, it corresponds to 3 output lines, i.e. one line for each of the 3 LEDs.\n\nAdditionally, let’s assume that SW(9) is a two-head 5-LED signal containing green, yellow and red LEDs on the upper head and yellow and red LEDs on the lower head. Thus SW(9) corresponds to 5 output lines. A typical application for this configured signal would be at the facing point end of a turnout leading into a CTC controlled passing siding.\n\nTo provide separate or independent control, each LED is wired to a separate output pin. Per the above, SE(12) needs 3 lines and SW(9) needs 5 lines. What is needed, is a simple procedure whereby we can position the SE(12) bits to use the lower 3 bits of a port while the SW(9) bits are aligned to use the upper 5 bits of a port. This procedure is called Packing. That is, we pack the bits representing different software output variables into a whole byte prior to transmitting the byte out to the railroad.\n\nWhen we read an input byte from the railroad we need a similar type of procedure to separate the various bits representing the different input devices. This procedure is called Unpacking. For example, each signal lever on a dispatcher’s CTC panel requires 2 input lines corresponding to 2 input bits. Thus we could group 4 sets of signal lever inputs on a given input port. The unpacking operation would be used to separate the different signal levers into separate software variables such as SL54, SL56, SL60 and SL62 denoting the 4 signal levers.\n\nFig. 8-1 is a flow chart, showing our real-time loop expanded to include the unpacking of input bytes and the packing of output bytes. It is possible to write subroutines to do the packing and unpacking for us and I did that with my original Heathkit computer using assembly language. However, invoking subroutines takes extra set up effort and slows down the real-time loop. Also, after extensive study, I find it is just as easy to accomplish the packing and unpacking operations using direct in-line code as it is to define a number of required variables before invoking the subroutine. I find this to be especially true when it takes but one line of code to pack, or unpack, each railroad device.", null, "Fig. 8-1. Real-time loop including unpacking and packing of I/O bytes\n\nIn actuality, using direct in-line code is especially attractive because it only takes one line of code to pack or to unpack each software variable. Also, once you become familiar with it, packing and unpacking are very easy steps to implement. The procedure is always the same. Do it a few times on a couple of different I/O cards and you will become an expert to help others with their implementation.\n\nBIT POSITION AND WIDTH CONSTANTS\n\nEvery one of our C/MRI programs requiring packing and unpacking will use two sets of constants defined by the two programming lines listed in Fig. 8-2.\n\n B0 = 1: B1 = 2: B2 = 4: B3 = 8: B4 = 16: B5 = 32: B6 = 64: B7 = 128 W1 = 1: W2 = 3: W3 = 7: W4 = 15: W5 = 31: W6 = 63: W7 = 127\n\nFig. 8-2. Packing and unpacking constants\n\nIt is important to note that: the statements in Fig. 8-2 stay exactly the same and are included in every program where we need to pack and unpack I/O bytes.\n\nI refer to the first line of constants as the “B” or Bit position constants. They are used to position, or shift, each variable right or left so that each variable lines up with the desired bit positions within the input and output bytes. The constants in the second line are referred to as the “W” or Width constants and are used to define the width of each variable, i.e. how many bits wide need to be unpacked from the input byte.\n\nFor readers desiring an understanding of why we pick these specific constants and how they are used, I will explain the mathematics of packing and unpacking. However if you would rather just go ahead and dive right into unpacking and packing operations, please feel free to skip ahead to the section Unpacking Procedure and then skip ahead to Packing Procedure.\n\nUNDERSTANDING UNPACKING\n\nIn almost every C/MRI application program we make use of integer mathematics – that is mathematics dealing with whole numbers only including 0, 1, 2, 3, 4, … and so forth. BASIC, as well as other languages, can easily handle fractional numbers such as 3¾ or 3.75 but in our railroading applications there is little need to use this capability.\n\nUsing only integer operations helps keep our C/MRI applications simple. To force QuickBASIC to treat everything as integers we can invoke the define integer statement DEFINT A-Z at the beginning of our programs. This tells the compiler to treat all program variables beginning with A through Z as integer variables. Because all program variables must begin with an alphabetical character, this statement defines all variables as integers.\n\nInteger mathematics truncates all fractions. Thus 7 divided by 3 becomes 2 and not a 2.5. A 19 divided by 6 becomes 3 and not 3.16666. An 8 divided by 3 becomes 2 not 2.6666. From the latter example it is important to stress the point that: integer mathematics does not round up or down to the nearest whole number – integer mathematics simply truncates all fractions!\n\nHowever, it turns out that integer division is exactly what we need for most of our C/MRI application programming. This is because we use division to shift a number, in our case an input byte, to the right so that we can easily sort out various bit groupings. Let’s see how this works.\n\nIn the decimal number system, dividing a given number by powers-of-ten shifts the decimal point to the left. For example taking the decimal number 17925316 and dividing it by different power-of-ten we have the relationships shown in Table 8-1.\n\nTable 8-1. Dividing a decimal number (e.g. 17925316) by powers-of-ten\n\n Power-of-ten  n Value of 10n Regular divide = 17925316 / 10n Integer divide = 17925316 \\ 10n 0 1 17925316. 17925316 1 10 1792531.6 1792531 2 100 179253.16 179253 3 1000 17925.316 17925 4 10000 1792.5316 1792 5 100000 179.25316 179 6 1000000 17.925316 17 7 10000000 1.7925316 1\n\nIn Table 8-1 the regular forward slash (/) indicates a regular, or real number, divide – i.e. where you deal with fractional numbers. The backward slash (\\) indicates an integer divide – i.e. where the fraction is truncated. These two types of divide are standard with QuickBASIC and Visual Basic and in many other languages.\n\nLooking at Table 8-1 we see that using the regular divide by powers-of-ten simply shifts the decimal point to the left by whatever power-of-ten we use. For example using n = 2, the decimal point is shifted 2 places to the left. Using n = 3, the decimal point is shifted 3 places to the left and using n = 6 yields 6 places to the left. Moving the decimal point to the left is equivalent to shifting the number to the right as is shown in the integer divide column where the fraction is truncated.\n\nThe same thing happens when we use binary mathematics to separate input bytes into different bit combinations. The only difference is that when working with binary numbers, the 1s and 0s in our input bytes, we need to use powers-of-two rather than powers-of-ten. Table 8-2 illustrates the situation for integer dividing an example input byte of 10110010 by different powers-of-two.\n\nTable 8-2. Dividing an input byte (e.g. 10110010) by powers-of-two\n\n Power-of-two  n Integer divide = 10110010 \\ 2n Decimal equivalent for power-of-two = 2n Symbol used for bit position constant 0 10110010 1 B0 1 1011001 2 B1 2 101100 4 B2 3 10110 8 B3 4 1011 16 B4 5 101 32 B5 6 10 64 B6 7 1 128 B7\n\nLooking at Table 8-2 we see that integer division by powers-of-two shifts the binary bits within our input byte to the right by whatever power-of-two we use. For example, using n = 2, the input byte is shifted 2 places to the right. Using n = 3, the input byte is shifted 3 places to the right and using n = 6 yields 6 places to the right. In digital logic, such an operation is referred to as a right-shift and the bits that get shifted out to the right, which is beyond the binary point (equivalent to the decimal point), are truncated i.e. they fall into the “bit-bucket” and are lost.\n\nIn our programming, we like to work in decimal rather than in binary, so we simply use the decimal equivalent for the powers-of-two as indicated in Table 8-2. For each of these powers-of-two, I have assigned a constant symbol denoted as B0, B1, B2, … up to B7. These bit position constants are exactly those we established in the first statement line back in Fig. 8-2; i.e. B0 = 1, B1 = 2, B2 = 4, B3 = 8 and so forth up to B7 = 128.\n\nThus, by dividing each input byte by an appropriate bit position constant, we can shift any input byte to the right however many times we need to align any railroad device to the rightmost bit positions. Once this is accomplished we simply need to retain the appropriate number of the rightmost bits, equivalent to the number of bits wide required by the railroad device, while zeroing out the leftmost bits. Mathematically this is accomplished by logic ANDing the shifted byte with the appropriate width constant defined in Table 8-3.\n\nTable 8-3. Width constant definition\n\n Symbols used for width constants Decimal value of constant Binary value of constant W1 1 00000001 W2 3 00000011 W3 7 00000111 W4 15 00001111 W5 31 00011111 W6 63 00111111 W7 127 01111111\n\nLogic ANDing two bytes is accomplished by separately ANDing each bit position within the two bytes using the relationship illustrated in Fig. 8-3.", null, "Fig. 8-3. Logic AND operation\n\nThe output of a logic AND operation for each bit position is always 0 unless both input bits are 1. Thus, if a given railroad device, such as a 5-light signal, is ANDed with W5 (which is a 00011111) the 5 signal bits that are ANDed with the 11111 are preserved. The remaining 3 bits that are ANDed with the 000 are masked away, becoming zeros in the result. In a like manner W1 is used when we need to strip off 1 bit, W2 is used to strip off 2 bits, W3 to strip off 3 bits and so forth up to using W7 to strip off 7 bits.\n\nFundamentally the AND operation, combined with an appropriate width constant directly converts all the bits that are not related to the railroad device to zeros and leaves the bits that are directly associated with the device undisturbed.\n\nThus to unpack any railroad device such as an occupancy detector, pushbutton, signal lever or whatever else is desired from its corresponding input byte we perform two steps:\n\n1. Use the bit position constant with an integer divide to shift the corresponding input byte so that the railroad device of interest is in the rightmost bit positions.\n2. Use the width constant to mask off all but the desired bits representing the railroad device of interest.\n\nIt is really more difficult to explain in general terms than it is to actually perform the operations. Therefore let’s simply dive into formulating the unpacking procedure.\n\nUNPACKING PROCEDURE\n\nTypically the first operation within the real-time loop is to read the railroad inputs and then unpack the various railroad devices from their respective input bytes. Unpacking is required because when the C/MRI software reads a port it brings in all eight bits, i.e. all 8-lines, as one entity, namely the input byte, IB(n) where the ‘n’ corresponds to the port number.\n\nAfter reading each byte, we need to separate the various bit groupings to form the separately needed software variables like BK(1), BK(2), … for block occupancy detectors and TF(18), TF(19), … for turnout position feedback, and so forth. This separating of the various railroad devices is accomplished by the unpacking procedure. Implementing the unpacking procedure requires dividing by the appropriate bit position constant and then ANDing with an appropriate width constant.\n\nUnpacking inputs requires one statement per railroad device and the format is always identical:\n\nDevice = IB(n) \\ By AND Wz\n\nTo convert the above general format to a specific statement for a given railroad device simply substitute for Device the selected variable name for the railroad device. Then substitute for “n” the port (or byte) number to which the device is connected, for “y” the starting bit position within the byte and for “z” how many input lines wide is the device.\n\nFor example if an occupancy detector for Block 15 is wired to bit Position 6 on the third port of the first input card, then the corresponding unpacking line would read:\n\nBK(15) = IB(3) \\ B6 AND W1\n\nThe railroad device is BK(15). The IB(3) denotes the port number to which the occupancy detector for Block 15 is attached. The B6 constant denotes that the occupancy detector BK(15) is connected to bit Position 6. The width constant W1 is used because the occupancy detector connection uses only 1 wire, i.e. its connection is only 1 bit wide.\n\nAs another example, let’s assume we have a CTC panel signal lever, denoted as SLEV62 and connected to Bits 2 and 3 of the same port. Two bits, or line wires, are used because, as we will see when we cover CTC panels, a signal lever uses 2 wires as inputs. The corresponding unpacking line would read:\n\nSLEV62 = IB(3) \\ B2 AND W2\n\nWe keep the same input byte, IB(3), because the signal lever is connected to the same input port. The bit position constant is changed to B2 the “starting bit” location for the wiring connections to Bits 2 and 3. The width constant W2 is used because the signal lever input is 2 bits wide.\n\n ****Important point**** Every unpacking statement must use the back-slash (\\) integer divide. If you accidentally substitute a forward-slash, the program’s input variables will become jumbled. Thus, if you run into the situation where your input variables appear to be all messed up, make sure that you are using the correct back-slash in all your unpacking instructions.\n\nUNDERSTANDING PACKING\n\nOnce unpacking is mastered, it is relatively straightforward to understanding packing. For those interested, I will explain the theory behind packing operations. If you would rather skip the theory and dive right into packing, then simply jump ahead to the next section titled Packing Procedure.\n\nPacking output bytes is much like the inverse of unpacking input bytes. We need to shift the bits in each railroad output variable so that they line up with their required positions corresponding to the lines where each device is connected. Then we fold its bits into the output byte being formulated without disturbing the bit positions used by other devices sharing the same output byte. Let’s dissect this paragraph a little to see what it really means.\n\nTo explain what is required for packing, it is easiest to simply work through an example. Assume we have two different color-light signals (i.e. signals using separate LEDs for each color) connected to Card 0 Port B as illustrated in Fig. 8-4.", null, "Fig. 8-4. Example signal connections to demonstrate packing requirements\n\nOne is a single-head 3-LED signal, which I have given the variable name SIG(9), and the other is a double-head 5-LED signal denoted as SIG(14). Also, assume that software has set SIG(9) to yellow, decimal 2, and SIG(14) to green over red, decimal 17, as indicated by their corresponding bit pattern diagrams shown in the lower right portion of Fig. 8-4.\n\nAll we need to do to create the output byte for the example situation is:\n\n1. Copy the bit pattern for SIG(9) directly into the output byte.\n2. Shift the bit pattern for SIG(14) three places to the left and then insert it into the output byte without disturbing the bit pattern already established for SIG(9).\n\nImplementing step 1 is easy, we simply use the statement:\n\nOB(2) = SIG(9)\n\nNow how do we accomplish step 2? We saw earlier, in Table 8-2, that dividing by powers-of-two shifted a bit pattern to the right. Similarly, multiplying by powers-of-two shifts a bit pattern to the left. Therefore, we can perform the required left shift – to line up our SIG(14) bits to their desired output byte positions – by simply multiplying by the appropriate bit position constant selected from the same set of constants we used for unpacking.\n\nIn this example, variable SIG(14) needs to be shifted so that its least significant bit (LSB) starts at, or aligns with, bit Position 3 of the output byte.  To accomplish this we simply need to multiply by bit position constant B3. The result of this multiply shifts SIG(14) three bits to the left, as illustrated in the lower right portion of Fig. 8-4. This lines up the SIG(14) bits so that they are ready to be inserted into the output byte. Note that the multiply operation zeroes the bit positions vacated by the shift operation.\n\nThe trick now is to insert the shifted SIG(14) bits into the output byte without disturbing the least significant bit positions already set up for the other devices sharing the same output byte – in this case the bits for SIG(9). This is accomplished by logic ORing the shifted SIG(14) byte with the output byte being formulated, i.e. already containing SIG(9). The combination multiplying and ORing operation, i.e. step 2, is written as the statement:\n\nOB(2) = SIG(14) * B3 OR OB(2)\n\nLogic ORing does the job for us because logic ORing two bytes is accomplished by separately ORing each bit position within the two bytes using the relationship illustrated in Fig. 8-4. The output of a logic OR operation for each bit position equals 0 only when both inputs are 0 otherwise the output is a 1. That is, if input A is 1 OR if input B is 1 then the output C is 1.\n\nLogic ORing a 1 or a 0 with a constant Logic 0 does not change the state of the 1 or 0. Because the shifted byte always has zeros in the bit positions used by previously packed variables their corresponding bits remain unaltered during the OR operation.\n\nSimilarly, because the bit positions not yet filled in the output byte are always zero, the ORing operation effectively copies the shifted byte into the output byte being formulated without disturbing the already filled bits. Fig. 8-4 illustrates the operations for the SIG(9) and SIG(14) example. The corresponding two statements to make all this happen are simply:\n\nOB(2) = SIG(9)\n\nOB(2) = SIG(14) * B3 OR OB(2)\n\nTo generalize the procedure for packing all railroad devices such as signals, switch motor controls, panel LEDs or whatever else you want to pack into their corresponding output bytes we need to:\n\n1. Directly copy the first device into the output byte. By first device I mean the railroad device wired to the least significant bits within a port.\n2. For each subsequent device connected to the port, multiply the device name by its appropriate bit position constant to shift the device into position and then OR it into the output byte.\n3. Repeat step 2 for each additional device attached to the port, i.e. until all devices are loaded into the output byte.\n\nThis three step process is repeated for each output byte. Once all the output bytes are packed, the complete output array OB(1), OB(2), OB(3), … up through OB(NO) are ready to be transmitted to the interface. Note that variable NO is the number of output bytes within a given node.\n\nAs with unpacking, it is really more difficult to explain in general terms than it is to actually perform the operations. Therefore let’s simply dive into formulating the packing procedure.\n\nPACKING PROCEDURE\n\nTypically the last operations within the real-time loop are to pack the various railroad devices into their corresponding output bytes and then send them out to the railroad. As with unpacking, packing is required because when the C/MRI software writes to a port it sends all eight bits, i.e. all 8-lines, as one entity, namely the output byte, OB(n) where “n” corresponds to the port number.\n\nTherefore, before we can write out the output bytes we need to combine the various bit groupings representing the railroad outputs, for example SE(1), SE521, SW(14), SW231, … for trackside signals and SM(7), SM(42), SM23, …for switch motor control and so forth. This combining of the various railroad devices to form the output bytes is accomplished by the packing procedure.\n\nPacking the first device into each output byte is directly accomplished using the relationship:\n\nOB(n) = Device\n\nThen, each additional device within a port is packed using the relationship:\n\nOB(n) = Device * By OR OB(n)\n\nTo convert the above general formats to specific statements for a given railroad device simply substitute for Device the selected variable name for the railroad device. Then substitute for “n” the port (or byte) number to which the device is connected and for “y” the device’s starting bit position within the byte. Note that the width constants are not used for packing.\n\nPacking, however, does recursively use the output byte. By this I mean we pack one device into the output byte and then we use that intermediate value of the output byte to pack in the next device. Then we use that intermediate value of the output byte to pack in the next device. For each output byte we repeatedly keep that process going forward until we have packed in all the desired devices.\n\nFor example, if switch motor SM(2) is wired to bit positions 0 and 1 on Port 5, 3-LED trackside signal SE(7) is wired to bit positions 2, 3 and 4 on the same port, and 3-LED signal SW(11) is wired to bit positions 5, 6 and 7 of the same port, then the corresponding packing statements are:\n\nOB(5) = SM(2)                'Place switch motor 2 in output byte 5\n\nOB(5) = SE(7) * B2 OR OB(5)  'Fold Signal East 7 into output byte 5\n\nOB(5) = SW(11) * B5 OR OB(5) 'Complete output byte 5 by inserting ...                                               '. . . Signal West 11\n\nThe Subscript 5 is used in the output byte OB( ) because the three railroad devices are connected to Port 5 which for an SMINI corresponds to its “Card 1” Port B location. The 3 railroad devices connected to this port are SM(2), SE(7) and SW(11). Bit position constant B2 is used when folding in SE(7) because the wiring of SE(7) starts at the port’s Bit 2 position. Constant B5 is used when folding in SW(11) because the wiring of SW(11) starts at the port’s Bit 5 position.\n\nA quick way of checking that you have the correct bit position constants is to “walk through” each of the connected devices. Switch Motor 2 being a 2-wire device uses up B0 and B1. Signal East 7, being a 3-wire device, must then start at B2 to use up B2, B3 and B4. Signal West 11, being a 3-wire device, must then start at B5 to use up B5, B6 and B7. The port is completely filled and SM(2) starts at B0, SE(7) at B2 and SW(11) at B5.\n\nThat completes our discussion covering the packing and unpacking procedures I now recommend. Once you set up the required instructions for a couple of your actual I/O ports, I am confident that you will find that writing unpacking and packing statements becomes a very straightforward operation you will perform essentially by rote.\n\nHowever, before we continue forward generating actual railroad application examples, I want to spend just a moment explaining the difference between the packing procedures I now recommend and what you may have read from my previous publications.\n\nSHORTCUT FROM PREVIOUSLY RECOMMENDED PACKING PROCEDURE\n\nMost if not all of my publications prior to the Version 3.0 of the C/MRI User’s Manual recommended a slightly different packing procedure. This previous procedure recommended that an output byte be first initialized to zero and then using the same statement for packing the first device as well as the subsequent devices. Following this procedure results in the packing statement formats:\n\nOB(x) = 0\n\nOB(x) = Device * By OR OB(x)\n\nIn this situation, the format on the second line is used for each device within the port, i.e. also including the first device. Using the examples from above, the generated statements are:\n\nOB(5) = 0                    'Initializes port to zero\n\nOB(5) = SM(2) * B0 OR OB(5)  'Places switch motor 2 in output byte 5\n\nOB(5) = SE(7) * B2 OR OB(5)  'Folds signal east 7 into output byte 5\n\nOB(5) = SW(11) * B5 OR OB(5) 'Completes output byte 5 by folding in ...\n\n'. . .signal west 11\n\nAlthough the above code is not as abbreviated as the now recommended set of four statements, the end results are identical. In fact, with some thought, it is easy to prove they are identical. First note from Fig. 8-2 that bit position constant B0 equals one and multiplying anything by 1 does not change anything. Thus it is easy to drop the * B0 term. Secondly considering that ORing any byte with a byte of zeros does not change anything. Thus it is easy to drop the OR OB(5) portion of the code in Statement 2 above. This leaves the second statement as OB(5) = SM(2) which is exactly what we wrote for the packing procedure now recommended. Doing this the initializing of OB(5) = 0 is irrelevant because the next statement simply overwrites the initialization to zero by setting OB(5) = SM(2).\n\nBecause the newly recommended approach simply defines OB(5) = SM(2), automatically placing zeros into all the bit positions not used by SM(2), there is no need to initialize OB(5) = 0 when using the newly recommended procedures.\n\nHowever, when using the previously recommended procedure it was essential to initialize each output byte to zero. Otherwise the contents of the being formulated output byte would be contaminated by the content of the previous value of the output byte.\n\nUSING BACK-SLASH VERSUS FORWARD-SLASH\n\nAs a final summary point in this chapter, always use the back-slash (\\) in your unpacking operations along with the bit position constant to shift the input byte to the right. The back-slash signifies to QuickBASIC and to Visual Basic, as well as to many other languages, that you want an integer divide.\n\nIf you should ever need a regular divide use the forward-slash (/). This way 5/2 comes out to be a 2.5. With the integer divide 5\\2 comes out to be a 2 rather than 2.5. The result of an integer divide is the same as with regular division except that the decimal fraction is dropped.\n\nI have used the integer divide, performed using a backslash, and the logic AND and OR operations for both simplicity and speed of execution. The procedures should work fine for 99+ percent of user applications. In fact, I have tested the procedures for every version of Basic that I could get a hold of including BasicA, GW Basic, QuickBASIC and Visual Basic on many different computers.\n\nThe only combination I found that would not work is on the very old Apple II series of machines when used with either Applesoft BASIC or Apple Integer BASIC. I doubt if any C/MRI user is still using such outdated machines and software. However, special packing and unpacking software procedures for this case are defined on pages 301 and 302 of the book Build Your Own Universal Computer Interface – Second Edition. It is also covered on page 216 of Build Your Own Universal Computer Interface – First Edition. This effectively outdated software is not included in this User’s Manual.\n\nPlease do not be concerned at this point if you still feel a little confused about unpacking and packing operations. As I said before, they are harder to explain in general than they are to actually implement.\n\nBelieve me, with the first couple of I/O ports under your belt, the unpacking and packing portion of I/O programming becomes an operation performed by rote and a breeze to implement. With this in mind, let’s move forward with detailed railroad examples covering signaling and turnout control with our SMINI." ]

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[ "# Two springs having spring constants 4 N/m and 6 N/m are attached in series. What is the energy stored in the system when it is compressed by .5 m", null, "The two springs have a spring constant of 4 N/m and 6 N/m respectively. They are attached in series. To calculate the potential energy of the system when the springs are compressed, first we need to determine the equivalent spring constant.\n\nThe equivalent spring constant Keq for two springs with...\n\nStart your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.\n\nThe two springs have a spring constant of 4 N/m and 6 N/m respectively. They are attached in series. To calculate the potential energy of the system when the springs are compressed, first we need to determine the equivalent spring constant.\n\nThe equivalent spring constant Keq for two springs with spring constants k1 and k2  is given by 1/Keq = 1/k1 + 1/k2.\n\nHere, (1/Keq) = 1/4 + 1/6\n\n=> 1/Keq = 10/24\n\n=> Keq = 2.4 N/m\n\nWhen the length of a spring with a spring constant k is changed by x, the potential energy in the system is given by (1/2)*k*x^2\n\nKeq has been determined to be 2.4 N/m and the length is decreased by 0.5 m.\n\nThe potential energy in the system is (1/2)*(2.4)*(0.5)^2 = 0.3 J.\n\nApproved by eNotes Editorial Team" ]

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[ "created 05/24/03; edited 11/09/2012, 06/10/2018, 01/21/2019\n\n# Chapter 24 Programming Exercises\n\nMany of your programs from previous chapters can modified by changing `while` loops into the equivalent `for` loops. Some of these exercises are repeats of previous ones.\n\n## Exercise 1 — Sheep Herd Size\n\nA breeding group of 20 bighorn sheep is released in a protected area in Colorado. It is expected that with careful management the number of sheep, `N`, after `t` years will be given by the formula:\n\n```N = 220/(1 + 10(0.83)t )\n```\n\nand that the sheep population will be able to maintain itself without further supervision once the population reaches a size of 80.\n\nWrite a program (using a `for` loop) that writes out the value of `N` for `t` starting at zero and going up to 25. How many years must the sheep heard be supervised?\n\nHint: don't calculate `(0.83)t`   \"from scratch\" each time the formula is used. Use a variable `power` that is multiplied by 0.83 in each iteration of the loop. What value should it be initialized to?\n\n(Problem from Howard Anton, Calculus, 6th ed., p. 105. )\n\n## Exercise 2 — Sum of Odd integers 1 to N\n\nWrite a program that asks the user for an integer N and then calculates two things: (a) the sum of all odd integers from 1 to N, and (b) N2. Use a `for` loop with appropriate increment.\n\n## Exercise 3 — Sum of Divisors of N\n\nWrite a program that asks the user for an integer N and then calculates the sum of all integers from 1 to N that divide N with zero left over. Use the `%` operator.\n\nFor example, if N is 10, then the sum of divisors is 1+2+5 = 8. Notice that 1 is considered a divisor and than no integer greater than N/2 is a divisor of N.\n\nAnother example: if N is 6, then the sum of divisors is 1+2+3 = 6. In this case, 6 is considered a perfect number because the sum of divisors is the number itself.\n\nInclude an `if` statement that determines if N is a perfect number.\n\n## Exercise 4 — Is an Integer the Sum of Two Squares?\n\nWrite a program that asks the user for an integer N and then determines if N is the sum of two the squares of two integers.\n\nIntegers that are a sum of two squares:\n\n• `4 = 02 + 12`\n• `5 = 22 + 12`\n• `10 = 32 + 12`\n• `25 = 32 + 42 = 02 + 52`\n\nIntegers that are not a sum of two squares:\n\n• `3`\n• `6`\n• `39`\n• `48`\n\nOne way to do this is with a doubly-nested loop that generates trial integers `a` and `b`, squares each, and tests if the sum is equal to `N`. Of course, when `a2` exceeds `N` the outer loop is done. Decide on a termination condition for the inner loop.\n\n## Exercise 5 — Is an Integer the Sum of Two Squares in two different Ways?\n\nThe integer 125 is the sum of two positive squares in two different ways. Changing the order of the squares is not regarded as a different way.\n\n```125 = 102 + 52 = 100 + 25\n125 = 112 + 22 = 121 + 4\n```\n\nWrite a program that asks the user for an integer N and then determines if it is the sum of two positive squares in two different ways.\n\nAnother way to check if an integer is the sum of two squares is with a loop that generates trial integers from one up to the square root of N. For each trial, check if (N-trial*trial) is a perfect square. Do this by calculating its square root and checking that it is an integer. `Math.sqrt()` and a type cast with `(int)` will be useful (see chapter 13). Increment a counter each time two squares sum to N.\n\n```PS>java SumTwoSquares\nN --> 50\n50 is the sum of 2 squares\n50 == 1*1 + 7*7\n50 == 5*5 + 5*5\n\nPS>\n```\n\nYou might think that using `Math.sqrt()` is more efficient than the doubly nested loop used in Exercise 4. However, `Math.sqrt()` is an expensive operation that is, in fact, implemented with a loop. You just don't see it.\n\n## Exercise 6 — Find all integers that are the Sum of Two Squares in two different Ways\n\nAsk the user for an upper limit and then write out those integers up to the limit that are the sum of two positive squares.\n\nDo this by wrapping a loop around the previous program (and making a few other changes.)\n\n```PS>java SumTwoSquares\nenter upper limit --> 100\n50 is the sum of 2 squares\n50 == 1*1 + 7*7\n50 == 5*5 + 5*5\n\n65 is the sum of 2 squares\n65 == 1*1 + 8*8\n65 == 4*4 + 7*7\n\n85 is the sum of 2 squares\n85 == 2*2 + 9*9\n85 == 6*6 + 7*7\n\nPS>\n```\n\nCan you find any integers that are the sum of two squares in three different ways?\n\n## Exercise 7 — Find cubes that are the Sum of Two Squares in two different Ways\n\nAsk the user for an upper limit, LIMIT and then writes out those integers up to the limit that are cubes which are the sum of two positive squares.\n\nDo this by modifying the previous program so that it tests if N is a perfect cube. If it is, then it tests if it is the sum of two squares as before.\n\n```PS>java CubicSumTwoSquares\n\nenter upper limit --> 1000\n125 is a cube: 5^3\n125 is the sum of 2 squares\n125 == 2*2 + 11*11\n125 == 5*5 + 10*10\n\n1000 is a cube: 10^3\n1000 is the sum of 2 squares\n1000 == 10*10 + 30*30\n1000 == 18*18 + 26*26\n\nPS>\n```\n\nWrite a program that reads 5 integers from a file, computes their sum and their maximum and prints these values to the monitor. Do this by modifying the summing program from the chapter. Use an `int` variable called `max` which is initialized to the first value in the file. This calls for an extra set of input statements before the loop starts. To compute the maximum use an `if` statement nested inside the loop." ]

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[ "# October Temperatures\n\n#### SWC75\n\n##### Bored Historian\nHere are the actual high temperatures in Syracuse for each day in 2014-2020 compared to the average high, (based on stats that go back to 1901) and the difference that day, + or – as well as a running deficit or surfeit. Here is the website I used for these numbers:\nhttps://www.accuweather.com/en/us/syracuse/13202/october-weather/329675?year=2020\n(I have now had to add this website as Accuweather has dropped the average highs from their chart:\nhttps://www.weather.gov/bgm/climateSYRDailyNormals )\nThe numbers are the date, the actual high, the historical average high, the net difference, the cumulative net since January 1st, the numbers of days we exceeded the historical average, the number of days we didn't reach it and the number of days we tied it.\n\nOctober 2020\n10/1 67 – 66 = +1 = +829 = 175-89-12\n10/2 63 – 65 = -2 = +827 = 175-90-12\n10/3 59 – 65 = -6 = +821 = 175-91-12\n10/4 63 – 65 = -2 = +819 = 175-92-12\n10/5 61 – 64 = -3 = +816 = 175-93-12\n10/6 67 – 64 = +3 = +819 = 176-93-12\n10/7 65 – 63 = +2 = +821 = 177-93-12\n10/8 57 – 63 = -6 = +815 = 177-94-12\n10/9 66 – 63 = +3 = +818 = 178-94-12\n10/10 82 – 62 = +20 = +838 = 179-94-12\n10/11 61 – 62 = -1 = +837 = 179-95-12\n10/12 67 – 61 = +6 = +843 = 180-95-12\n10/13 60 – 61 = -1 = +842 = 180-96-12\n10/14 66 – 61 = +5 = +847 = 181-96-12\n10/15 79 – 60 = +19 = +866 = 182-96-12\n10/16 52 – 60 = -8 = +858 = 182-97-12\n10/17 59 – 60 = -1 = +857 = 182-98-12\n10/18 62 – 59 = +3 = +860 = 183-98-12\n10/19 60 – 59 = +1 = +861 = 184-98-12\n10/20 61 – 58 = +3 = +864 = 185-98-12\n10/21 73 – 58 = +15 = +879 = 186-98-12\n10/22 60 – 58 = +2 = +881 = 187-98-12\n10/23 81 – 57 = +24 = +905 = 188-98-12\n10/24 69 – 57 = +12 = +917 = 189-98-12\n10/25 50 – 57 = -7 = +910 = 189-99-12\n10/26 49 – 56 = -7 = +903 = 189-100-12\n10/27 47 – 56 = -9 = +894 = 189-101-12\n10/28 53 – 56 = -3 = +891 = 189-102-12\n10/29 47 – 55 = -8 = +883 = 189-103-12\n10/30 42 – 55 = -13 = +870 = 189-104-12\n10/31 45 – 55 = -10 = +860 = 189-105-12\n\nComments: it was an up-and-down month, as October tends to be. We exceeded the average high 15 times but didn’t make it that high 16 times. But we finished +32 in degrees. November should be interesting with the incredible streak of +70 degree days but that started after the month changed so it isn’t showing up here.\n\nOctobers\n2014: +103 18-11-2\n2015: -37 12-19-0\n2016: +4 14-17-0\n2017: +224 24-7-0\n2018: -108 11-19-0\n2019 +91 20-9-2\n2020 +32 15-16-0\n\nMonth by Month in 2020:\nJanuary +220 25-6-0\nFebruary +44 44-14-2\nMarch +200 20-10-1\nApril -121 11-18-1\nMay +10 15-15-1\nJune +107 21-10-0\nJuly +189 28-2-1\nAugust +100 20-8-3\nSeptember +80 15-12-1\nOctober +32 15-16-0\n\nReplies\n0\nViews\n303\nReplies\n0\nViews\n246\nReplies\n0\nViews\n141\nReplies\n0\nViews\n214\nReplies\n0\nViews\n305\n\nMembers online\n208\nGuests online\n983\nTotal visitors\n1,191" ]

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[ "How to draw a graph to disprove this statement?\n\nThe Problem: Indicate whether the following statements are true or false:\n\n• a. If e is a minimum-weight edge in a connected weighted graph, it must be among edges of at least one minimum spanning tree of the graph.\n• b. If e is a minimum-weight edge in a connected weighted graph, it must be among edges of each minimum spanning tree of the graph\n• c. If edge weights of a connected weighted graph are all distinct, the graph must have exactly one minimum spanning tree\n• d. If edge weights of a connected weighted graph are not all distinct, the graph must have more than one minimum spanning tree\n\nI read from Min Edge that every MST must contain the minimum weighted edge and from Partial Solutions that two of the assertions are true while the rest of the two are false.\n\nFrom those two statements, I concluded that the first two statements(a, b) are true while the last two statements(c,d) are false.\n\nHere is the diagram I used to illustrate that the last option, d, is false.", null, "The edge weights are not distinct(two twos) and there is only one minimum spanning tree.\n\nCan anyone give a counterexample graph to choice option c? I tried some examples(too many to include in here) but each time with distinct weights, using Prim's algorithm, I only found one MST.\n\n• It seems to me you can just remove the parts a, b, and d from your question since you already have an answer for them. Including them makes your question unnecessarily long. Also, you already claim c is false as well. Why is that? Why do you doubt your answer? What is your question?\n– Juho\nJun 6 '15 at 20:48\n• @Juho . C is false because the solutions said that two of the following are true. The two of the following are a and b. I need parts a and b to add context to this question. I am trying to show a counterexample to c. My question is how would you find a counterexample to c? The reader would need to read the question(and thus options a,b) to understand why I am trying to show c is false Jun 6 '15 at 20:55\n\nYou have made a mistake interpreting the wording of the question compared to the Stackoverflow question (link in case of later editing).\n\nNotice that in options (a) and (b), the phrasing is a minimum weight edge, whereas in the Stackoverflow question the phrasing is the minimum weight edge. In your question, there's no guarantee that the edge is the only edge with that weight, in the Stackoverflow question, there is exactly one edge with the lowest weight.\n\nThus the mistake is assuming that (b) is true, when in fact it is false. The two true options are (a) and (c), which is why you can't find a counter example to (c).\n\nIn both (a) and (c) however, the proof is very similar to that in the Stackoverflow question. In fact (c) is given by applying that proof inductively. (a) is given by basically the same argument, but you show that the tree made by adding the given edge and removing some other edge from the cycle can't have higher weight (because the new edge has minimum weight).\n\nTo see that (b) is false, just take the graph $K_{n}$ where all edge weights are the same. Any spanning tree is a minimum spanning tree.\n\nFor (d), consider the following graph; take a tree on $n$ vertices, give all these edges weight $1$. Then add all non-edges, give these new edges weight $2$. Clearly there's only one MST, but there's $\\binom{n}{2} - n + 1$ edges with the same weight.\n\n• Is my counterexample proof for d sufficient as well? Jun 7 '15 at 3:59\n• @committedandroider, yes (I got overexcited writing the answer and forgot that you'd already solved that one). Jun 7 '15 at 4:02" ]

[ null, "https://i.stack.imgur.com/g7Ufy.png", null ]

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[ "# 0.008 Millimeters to Meters (0.008 mm to m)\n\nConvert 0.008 Millimeters to Meters (mm to m) with our conversion calculator and conversion tables. To convert 0.008 mm to m use direct conversion formula below.\n0.008 mm = 8.0E-6 m.\nYou also can convert 0.008 Millimeters to other Length (popular) units.\n\n0.008 MILLIMETERS\n\n=\n\n8.0E-6 METERS\n\nDirect conversion formula: 1 Millimeters * 1000 = 1 Meters\n\nOpposite conversion: 0.008 Meters to Millimeters\n\n## Conversion calculator\n\nAmount:\nFrom:\nTo:\n\nCheck out conversion of 0.008 mm to most popular length units:\n\n0.008 mm to Kilometers\n0.008 mm to Miles\n0.008 mm to Centimeter\n0.008 mm to Inches\n0.008 mm to Foots\n\n## Conversion table: Millimeters to Meters\n\nMILLIMETERS   METERS\n1 = 0.001\n2 = 0.002\n3 = 0.003\n4 = 0.004\n5 = 0.005\n7 = 0.007\n8 = 0.008\n9 = 0.009\n10 = 0.01\nMETERS   MILLIMETERS\n1 = 1000\n2 = 2000\n3 = 3000\n4 = 4000\n5 = 5000\n7 = 7000\n8 = 8000\n9 = 9000\n10 = 10000\n\n## Nearest numbers for 0.008 Millimeters\n\nMILLIMETERS   METERS\n0.00803 mm = 8.03E-6 m\n0.01 mm = 1.0E-5 m\n0.011 mm = 1.1E-5 m\n0.012 mm = 1.2E-5 m\n0.0123 mm = 1.23E-5 m\n0.0128 mm = 1.28E-5 m\n0.013 mm = 1.3E-5 m\n0.014 mm = 1.4E-5 m\n0.0146 mm = 1.46E-5 m\n0.017 mm = 1.7E-5 m\n0.0187 mm = 1.87E-5 m\n0.0204 mm = 2.04E-5 m\n0.021 mm = 2.1E-5 m\n0.0225 mm = 2.25E-5 m\n0.023 mm = 2.3E-5 m\n0.0295 mm = 2.95E-5 m\n0.03 mm = 3.0E-5 m\n0.031 mm = 3.1E-5 m\n0.035 mm = 3.5E-5 m\n0.0425 mm = 4.25E-5 m" ]

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[ "# C# EF common constraint writing method\n\nTake a note:\nThe article is transferred from blog Park, original address: EF common constraint writing\nInvasion and deletion!\n\n# 1, Configure constraints through Attribute\n\n## 1. Primary key constraint\n\nConfigure the primary key constraint through KeyAttribute. The code is as follows:\n\n```[Key]\npublic int PrimaryKey{ get; set; }\n```\n\n## 2. Foreign key constraint\n\nConfigure foreign key constraints through ForeignKeyAttribute. The code is as follows:\n\n```[Key]\npublic int PrimaryKey{ get; set; }\n[ForeignKey(\"ForeignKey\")]\npublic int PrimaryKey{ get; set; }\n```\n\nNote that the specified column name exists (foreign key must exist). For example, ForeignKey above, there must be an attribute named ForeignKey in the class.\n\n## 3. Length constraint\n\n(1) . normal length constraint. Configure the normal length constraint through StringLengthAttribute. The code is as follows:\n\n```[StringLength(30)]\npublic string Name { get; set; }\n```\n\n(2) . maximum length constraint, through MaxLengthAttribute, the code is as follows:\n\n```[MaxLength(30)]\npublic string Name { get; set; }\n```\n\n(3) . minimum length constraint, through MinLengthAttribute, the code is as follows:\n\n```[MinLength(30)]\npublic string Name { get; set; }\n\n```\n\n## 4. Non NULL constraint\n\nThe non NULL constraint is relatively simple. Through the RequiredAttribute, the code is as follows:\n\n```[Required]\npublic string Name{ get; set; }\n```\n\n## 5. Data type constraints\n\nConfigure data type constraints by initializing the TypeName attribute of ColumnAttribute class. The code is as follows:\n\n```[Column(TypeName=\"byte\")]\npublic string Photo{get;set;}\n\n```\n\n## 6. Field name constraint\n\nBy initializing the constructor setting with string parameter of ColumnAttribute class, the code is as follows:\n\n```[Column(\"CTime\")]\npublic DateTime CreateTime { get; set; }\n\n```\n\n## 7. Table name constraint\n\nSet through the constructor with string parameter of TableAttribute class. The code is as follows:\n\n```[Table(\"Class\")]\npublic class ClassInfo\n{}\n```\n\n## 8. GUID column value\n\nWhen the primary key value needs to be self GUID, the DatabaseGenerated feature needs to be added on the basis of setting the primary key constraint on the primary key field. The code is as follows:\n\n```[Key,DatabaseGenerated(DatabaseGeneratedOption.Identity)]\npublic GUID Id{ get; set; }\n```\n\nIf the column value GUID is not set, the database will be filled with 0, and an error will be reported in the second row, because the Id is set as the primary key.\n\n## 9. Column value + databasegeneratedoption Computed\n\n```[Key,DatabaseGenerated(DatabaseGeneratedOption.Computed)]\npublic GUID Id{ get; set; }\n```\n\nIf the attribute is marked as Computed, EF will think that the column is calculated by other columns and will not persist it to the database.\n\n## 10. Column value + databasegeneratedoption None\n\n```[Key,DatabaseGenerated(DatabaseGeneratedOption.None)]\npublic int Id{ get; set; }\n```\n\nThis is equivalent to the self increasing effect of Id primary key\n\n## 11. Ignore mapped columns\n\nWhen some fields are defined in the entity class, and these fields are obtained through some calculation or combination, we do not need to synchronize them to the database. We can configure to prevent them from being generated into the database. EF is set through the NotMappedAttribute attribute. The code is as follows:\n\n```[NotMapped]\npublic string NotNeeded { get; set; }\n```\n\n## 12. Ignore table mapping\n\nIgnoring table mapping is the same as ignoring column mapping The code is as follows:\n\n```[NotMapped]\npublic class ClassInfo\n{}\n\n```\n\n## 13. Complex type constraints\n\nThe following table is generated according to the specified constraints\n\n```[Table(\"Class\")]\npublic class ClassInfo\n{\n[Key, DatabaseGenerated(DatabaseGeneratedOption.Identity)]\npublic Guid Id { get; set; }\n\n[Required, StringLength(32)]\npublic string Name { get; set; }\n\n[Required, Column(\"CTime\")]\npublic DateTime CreateTime { get; set; }\n\n[Column(TypeName = \"ntext\"), MaxLength(20), MinLength(10)]\npublic string Remark { get; set; }\n\n[NotMapped]\npublic string NotNeed { get; set; }\n}\n```\n\n# 2, Override the OnModelCreating method of DbContext and set the constraints of the corresponding table or field\n\nCopy code\n\n```public class ClassInfo\n{\npublic Guid Id { get; set; }\n\npublic string Name { get; set; }\n\npublic DateTime CreateTime { get; set; }\n\npublic string Remark { get; set; }\n\npublic string NotNeed { get; set; }\n}\n```\n```public class EFCodeFirstDbContext : DbContext\n{\npublic EFCodeFirstDbContext()\n: base(\"name=connStr\")\n{\n\n}\n/// <summary>\n///If the entity is mapped to the database, EF will create the table name as the plural form of the entity name. Here is to force the table name to be created as the entity name\n/// </summary>\n/// <param name=\"modelBuilder\"></param>\nprotected override void OnModelCreating(DbModelBuilder modelBuilder)\n{\nmodelBuilder.Conventions.Remove<PluralizingTableNameConvention>();\nmodelBuilder.Entity<ClassInfo>().ToTable(\"Class\");//Set the table name corresponding to ClassInfo to Class\nmodelBuilder.Entity<ClassInfo>().Property(p => p.Id).HasDatabaseGeneratedOption(DatabaseGeneratedOption.Identity);//Set the Id of ClassInfo to self growth\nmodelBuilder.Entity<ClassInfo>().HasKey(p => p.Id);//Set the Id property of ClassInfo as the primary key\nmodelBuilder.Entity<ClassInfo>().Property(p => p.Name).IsRequired();//Set the Name property of ClassInfo to be non empty\nmodelBuilder.Entity<ClassInfo>().Property(p => p.Name).HasMaxLength(32);//Set the maximum length of the Name property value of ClassInfo to 32\nmodelBuilder.Entity<ClassInfo>().Property(p => p.CreateTime).IsRequired();//Set the CreateTime property of ClassInfo to be non empty\nmodelBuilder.Entity<ClassInfo>().Property(p => p.CreateTime).HasColumnName(\"CTime\");//Set the CreateTime property of ClassInfo to CTime\nmodelBuilder.Entity<ClassInfo>().Property(p => p.Remark).HasColumnType(\"ntext\");//Set the Remark property type of ClassInfo to ntext\nmodelBuilder.Entity<ClassInfo>().Property(p => p.Remark).HasMaxLength(20);//Set the maximum length of the Remark property value of ClassInfo to 32\nmodelBuilder.Entity<ClassInfo>().Ignore(p => p.NotNeed);//Ignore NotNeed field\n}\n\npublic DbSet<ClassInfo> ClassInfo { get; set; }\n}\n```\n\nKeywords: C# Database\n\nAdded by TimTimTimma on Tue, 22 Feb 2022 03:10:43 +0200" ]

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[ "# Class of Service\n\nPortworx provides the ability to specify a class of service for IOPS, throughput and more at the container granularity. Through class of service (also known as a `CoS`), a single volume’s class of service can be controlled and mapped to specific underlying storage infrastructure capabilities.\n\n### Explanation of Class of Service\n\nApplications have different storage performance requirements; some require higher IOPS/throughput performance characteristics than others. Portworx provides the ability to specify a class of service level at the container granularity. Containers operating at different classes of service can co-exist in the same node/cluster. Using class of service you can tune your volume for higher throughput and/or IOPS. The _High_CoS is optimized for IOPS, Medium is optimized for throughput.\n\n### Usage\n\nTo create a volume with a specific class of service level, use the `--io_prioirity` parameter in the volume create options. As with other parameters, this CoS parameter can also be passed in as a label via Docker or any scheduler.\n\n``pxctl volume create --io_priority high volume-name``\n\nHere is an example output from fio when measuring the CoS feature on an Intel server with NVMe and SATA drives.\n\nRandom Low CoS IOPS High CoS IOPS\n4K 768 65024\n8K 642 46848\n64K 496 9824\n\nThe graph below shows the sequential and random read and write performance on high and low CoS volume types:", null, "", null, "### Try it out on Amazon\n\n#### Create EBS volumes AWS\n\nHere, we create volumes of 3 different volume types in AWS. Refer to AWS EBS volume types for more information on the EBS volume capabilities. Portworx automatically detect the volume type and classify it into the correct service category.\n\n• Create one 500GB HDD volume\n• Create one 100GB standard volume\n• Create one 1000GB IO optimized volume", null, "Here is what you should see when you list your block devices:\n\n``lsblk``\n``````NAME MAJ:MIN RM SIZE RO TYPE MOUNTPOINT\nxvda 202:0 0 64G 0 disk\n└─xvda1 202:1 0 64G 0 part /\nxvdj 202:144 0 128G 0 disk\nxvdl 202:176 0 500G 0 disk\nxvdn 202:208 0 999G 0 disk``````\n\nCreate a `config.json` with the following drives in it… we will add the fourth standard ebs volume later\n\n``cat /etc/pwx/config.json``\n``````{\n\"clusterid\": \"cos-demo-cluster\",\n\"dataiface\": \"\",\n\"kvdb\": [\n\"etcd://localhost:4001\"\n],\n\"mgtiface\": \"\",\n\"storage\": {\n\"devices\": [\n\"/dev/xvdl\",\n\"/dev/xvdn\",\n\"/dev/xvdj\"\n]\n}\n}``````\n``pxctl status``\n``````Status: PX is operational\nNode ID: 5f794df0-b337-42d7-afc0-440c19fc4b0e\nIP: 172.31.2.134\nLocal Storage Pool: 3 pools\nPool Cos Size Used Status Zone Region\n0 COS_TYPE_LOW 500 GiB 1.1 GiB Online a us-west-1\n1 COS_TYPE_HIGH 999 GiB 1.1 GiB Online a us-west-1\n2 COS_TYPE_MEDIUM 128 GiB 1.1 GiB Online a us-west-1\nLocal Storage Devices: 3 devices\nDevice Path Media Type Size Last-Scan\n0:1 /dev/xvdl STORAGE_MEDIUM_SSD 500 GiB 14 Nov 16 02:01 UTC\n1:1 /dev/xvdn STORAGE_MEDIUM_SSD 991 GiB 14 Nov 16 02:01 UTC\n2:1 /dev/xvdj STORAGE_MEDIUM_SSD 128 GiB 14 Nov 16 02:01 UTC\ntotal - 1.6 TiB\nCluster Summary\nCluster ID: ohyeah0014\nNode IP: 172.31.2.134 - Capacity: 3.2 GiB/1.6 TiB Online (This node)\nGlobal Storage Pool\nTotal Used : 3.2 GiB\nTotal Capacity : 1.6 TiB``````\n\nThe `status` command on any node shows the pools with different classes of services listed. The format `x:y`in the Device column indicates the `pool:device` participating in that pool.\n\n#### Inspect different pools\n\n``pxctl service drive show``\n``````PX drive configuration:\nPool ID: 0\nCos: COS_TYPE_LOW\nSize: 500 GiB\nStatus: Online\nHas meta data: No\nDrives:\n1: /dev/xvdl, 4.1 GiB allocated of 500 GiB, Online\nPool ID: 1\nCos: COS_TYPE_HIGH\nSize: 991 GiB\nStatus: Online\nHas meta data: No\nDrives:\n1: /dev/xvdn, 2.1 GiB allocated of 991 GiB, Online\nPool ID: 2\nCos: COS_TYPE_MEDIUM\nSize: 128 GiB\nStatus: Online\nHas meta data: Yes\nDrives:\n1: /dev/xvdj, 2.1 GiB allocated of 128 GiB, Online``````\n\n#### Measure Performance\n\nLet’s first create three volumes with a high, medium and low class of service:\n\n``````pxctl volume create --io_priority high test-high --size 8\ntest-high\n\npxctl volume create --io_priority med test-med --size 8\ntest-med\n\npxctl volume create --io_priority low test-low --size 8\ntest-low``````\n\nNow we use fio to measure Portworx volume performance on each of these volumes. Note that backend disk performance while performance tests are running can be visualized with iostat\n\n``iostat -xm 1``\n``````Device: rrqm/s wrqm/s r/s w/s rMB/s wMB/s avgrq-sz avgqu-sz await r_await w_await svctm %util\nxvdj 30.00 114.00 660.00 380.00 10.61 43.66 106.87 48.63 93.53 1.30 253.71 0.67 70.00\nxvdl 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00\nxvdn 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00``````\n\nTest a high CoS volume on EBS\n\n``````docker run --rm --volume-driver=pxd -v test-high:/test \\\ngourao/fio /usr/bin/fio --blocksize=16k -directory=/test \\\n--size=1G --name=test --verify=meta --do_verify=1 \\\n--iodepth=128 --randrepeat=1 --end_fsync=1``````\n``````test: (g=0): rw=randread, bs=4K-4K/4K-4K/4K-4K, ioengine=libaio, iodepth=128\nfio-2.1.11\nStarting 1 process\n\ntest: (groupid=0, jobs=1): err= 0: pid=6: Tue Nov 15 07:49:40 2016\nread : io=4096.0MB, bw=41384KB/s, iops=10346, runt=101350msec\ncpu : usr=4.06%, sys=15.86%, ctx=462899, majf=0, minf=134\nIO depths : 1=0.1%, 2=0.1%, 4=0.1%, 8=0.1%, 16=0.1%, 32=0.1%, >=64=100.0%\nsubmit : 0=0.0%, 4=100.0%, 8=0.0%, 16=0.0%, 32=0.0%, 64=0.0%, >=64=0.0%\ncomplete : 0=0.0%, 4=100.0%, 8=0.0%, 16=0.0%, 32=0.0%, 64=0.0%, >=64=0.1%\nissued : total=r=1048576/w=0/d=0, short=r=0/w=0/d=0\nlatency : target=0, window=0, percentile=100.00%, depth=128\n\nRun status group 0 (all jobs):\nREAD: io=4096.0MB, aggrb=41384KB/s, minb=41384KB/s, maxb=41384KB/s, mint=101350msec, maxt=101350msec\n\npxd!pxd110428589532375940: ios=1043909/5, merge=0/2, ticks=12308307/116, in_queue=12310012, util=100.00%``````\n\nTest a medium CoS volume on EBS\n\n``````docker run --rm --volume-driver=pxd -v test-med:/test \\\ngourao/fio /usr/bin/fio --blocksize=16k -directory=/test \\\n--size=4G --name=test --direct=1 --gtod_reduce=1 \\\n--iodepth=128 --randrepeat=1 --end_fsync=1``````\n``````test: (g=0): rw=randread, bs=4K-4K/4K-4K/4K-4K, ioengine=libaio, iodepth=128\nfio-2.1.11\nStarting 1 process\n\ntest: (groupid=0, jobs=1): err= 0: pid=7: Tue Nov 15 08:00:47 2016\nread : io=4096.0MB, bw=23135KB/s, iops=5783, runt=181295msec\ncpu : usr=2.74%, sys=10.46%, ctx=591148, majf=0, minf=134\nIO depths : 1=0.1%, 2=0.1%, 4=0.1%, 8=0.1%, 16=0.1%, 32=0.1%, >=64=100.0%\nsubmit : 0=0.0%, 4=100.0%, 8=0.0%, 16=0.0%, 32=0.0%, 64=0.0%, >=64=0.0%\ncomplete : 0=0.0%, 4=100.0%, 8=0.0%, 16=0.0%, 32=0.0%, 64=0.0%, >=64=0.1%\nissued : total=r=1048576/w=0/d=0, short=r=0/w=0/d=0\nlatency : target=0, window=0, percentile=100.00%, depth=128\n\nRun status group 0 (all jobs):\nREAD: io=4096.0MB, aggrb=23135KB/s, minb=23135KB/s, maxb=23135KB/s, mint=181295msec, maxt=181295msec\n\npxd!pxd230469319006318075: ios=1045946/4, merge=0/1, ticks=22656364/162, in_queue=22658103, util=100.00%``````\n\nTest a low CoS volume on EBS\n\n``````docker run --rm --volume-driver=pxd -v test-low:/test \\\ngourao/fio /usr/bin/fio --blocksize=4k -directory=/test \\\n--size=1G --name=test --direct=1 --gtod_reduce=1 \\\n--iodepth=128 --randrepeat=1 --end_fsync=1``````\n``````test: (g=0): rw=randread, bs=4K-4K/4K-4K/4K-4K, ioengine=libaio, iodepth=128\nfio-2.1.11\nStarting 1 process\n\ntest: (groupid=0, jobs=1): err= 0: pid=6: Tue Nov 15 09:20:17 2016\nread : io=1024.0MB, bw=365111B/s, iops=89, runt=2940858msec\ncpu : usr=0.05%, sys=0.19%, ctx=160992, majf=0, minf=134\nIO depths : 1=0.1%, 2=0.1%, 4=0.1%, 8=0.1%, 16=0.1%, 32=0.1%, >=64=100.0%\nsubmit : 0=0.0%, 4=100.0%, 8=0.0%, 16=0.0%, 32=0.0%, 64=0.0%, >=64=0.0%\ncomplete : 0=0.0%, 4=100.0%, 8=0.0%, 16=0.0%, 32=0.0%, 64=0.0%, >=64=0.1%\nissued : total=r=262144/w=0/d=0, short=r=0/w=0/d=0\nlatency : target=0, window=0, percentile=100.00%, depth=128\n\nRun status group 0 (all jobs):\nREAD: io=1024.0MB, aggrb=356KB/s, minb=356KB/s, maxb=356KB/s, mint=2940858msec, maxt=2940858msec" ]

[ null, "https://2.1.docs.portworx.com/img/cos-random.png", null, "https://2.1.docs.portworx.com/img/cos-seq.png", null, "https://2.1.docs.portworx.com/img/cos.png", null ]

[ "## 7.3 – Factoring Trinomials\n\n### Learning Objectives\n\n• (7.3.1) – Factoring out the greatest common factor\n• Identify the greatest common factor of a polynomial\n• Factor the greatest common factor out of a polynomial\n• Factoring out a greatest common factor that is a binomial\n• (7.3.2) – Factor a Trinomial with Leading Coefficient = 1\n• Use a method to factor a trinomial with a leading coefficient of 1\n• Recognize when a trinomial cannot be factored (prime)\n• (7.3.3) – Factor a Trinomial with Leading Coefficient $\\neq 1$\n• Factor by trial and error\n• Factor by Grouping\n\nFactors are the building blocks of multiplication. They are the numbers that you can multiply together to produce another number: 2 and 10 are factors of 20, as are 4, 5, 1, 20. To factor a number is to rewrite it as a product. $\\displaystyle 20=4\\cdot5� . In algebra, we use the word factor as both a noun – something being multiplied – and as a verb – the action of rewriting a sum or difference as a product. Factoring is very helpful in simplifying expressions and solving equations involving polynomials. # (7.3.1) – Factoring out the Greatest common factor The greatest common factor (GCF) of two numbers is the largest number that divides evenly into both numbers. For instance, [latex]4$ is the GCF of $16$ and $20$ because it is the largest number that divides evenly into both $16$ and $20$ The GCF of polynomials works the same way: $4x$ is the GCF of $16x$ and $20{x}^{2}$ because it is the largest polynomial that divides evenly into both $16x$ and $20{x}^{2}$.\n\nWhen factoring a polynomial expression, our first step should be to check for a GCF. Look for the GCF of the coefficients, and then look for the GCF of the variables.\n\n### Greatest Common Factor\n\nThe greatest common factor (GCF) of two or more expressions is the largest expression that is a factor of all the expressions (both the coefficient and the variable with the highest degree.)\n\n### Example\n\nFind the greatest common factor of $25b^{3}$ and $10b^{2}$.\n\nThe video that follows gives an example of finding the greatest common factor of two monomials with only one variable.\n\nSometimes you may encounter a polynomial with more than one variable, so it is important to check whether both variables are part of the GCF. In the next example we find the GCF of two terms which both contain two variables.\n\n### Example\n\nFind the greatest common factor of $81c^{3}d$ and $45c^{2}d^{2}$.\n\nThe video that follows shows another example of finding the greatest common factor of two monomials with more than one variable.\n\nNow that you have practiced identifying the GCF of a term with one and two variables, we can apply this idea to factoring the GCF out of a polynomial. Notice that the instructions are now “Factor” instead of “Find the greatest common factor”.\n\nTo factor a polynomial, first identify the greatest common factor of the terms. You can then use the distributive property to rewrite the polynomial in a factored form. Recall that the distributive property of multiplication over addition states that a product of a number and a sum is the same as the sum of the products.\n\n#### Distributive Property Forward and Backward\n\nForward: Product of a number and a sum: $a\\left(b+c\\right)=a\\cdot{b}+a\\cdot{c}$. You can say that “$a$ is being distributed over $b+c$.”\n\nBackward: Sum of the products: $a\\cdot{b}+a\\cdot{c}=a\\left(b+c\\right)$. Here you can say that “$a$ is being factored out.”\n\nWe first learned that we could distribute a factor over a sum or difference, now we are learning that we can “undo” the distributive property with factoring.\n\n### Example\n\nFactor $25b^{3}+10b^{2}$.\n\nThe factored form of the polynomial $25b^{3}+10b^{2}$ is $(5b^{2})\\left(5b+2\\right)$. You can check this by doing the multiplication. $(5b^{2})\\left(5b+2\\right)=25b^{3}+10b^{2}$.\n\nNote that if you do not factor the greatest common factor at first, you can continue factoring, rather than start all over.\n\nFor example:\n\n$\\begin{array}{l}25b^{3}+10b^{2}=5\\left(5b^{3}+2b^{2}\\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Factor out }5.\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=(5b^{2})\\left(5b+2\\right) \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Factor out }b^{2}.\\end{array}$\n\nNotice that you arrive at the same simplified form whether you factor out the GCF immediately or if you pull out factors individually.\n\nIn the following video we show two more examples of how to find and factor the GCF from binomials.\n\nWe will show one last example of finding the GCF of a polynomial with several terms and two variables. No matter how large the polynomial, you can use the same technique described below to factor out it’s GCF.\n\n### How To: Given a polynomial expression, factor out the greatest common factor.\n\n1. Identify the GCF of the coefficients.\n2. Identify the GCF of the variables.\n3. Combine to find the GCF of the expression.\n4. Determine what the GCF needs to be multiplied by to obtain each term in the expression.\n5. Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by.\n\n### Example\n\nFactor $6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy$.\n\nIn the following video you will see two more example of how to find and factor out the greatest common factor of a polynomial.\n\n### Factoring out a GCF that is a binomial\n\nNext we present two examples where we can factor out a binomial term from both expressions.\n\n### ExAMPLE\n\nFactor $3y(y+7)-4(y+7)$.\n\n### ExAMPLE\n\nFactor $y(x+3)+2(x+3)$.\n\n# (7.3.2) – Factor a Trinomial with Leading Coefficient = 1\n\nTrinomials are polynomials with three terms. We are going to show you a method for factoring a trinomial whose leading coefficient is 1.  Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that trinomials can be factored. The trinomial ${x}^{2}+5x+6$ has a GCF of 1, but it can be written as the product of the factors $\\left(x+2\\right)$ and $\\left(x+3\\right)$.\n\nRecall how to use the distributive property to multiply two binomials:\n\n$\\left(x+2\\right)\\left(x+3\\right) = x^2+3x+2x+6=x^2+5x+6$\n\nWe can reverse the distributive property and return $x^2+5x+6\\text{ to }\\left(x+2\\right)\\left(x+3\\right)$ by finding two numbers with a product of $6$ and a sum of $5$.\n\n### Factoring a Trinomial with Leading Coefficient 1\n\nIn general, for a trinomial of the form ${x}^{2}+bx+c$ you can factor a trinomial with leading coefficient 1 by finding two numbers,$p$ and $q$ whose product is $c$, and whose sum is $b$. The trinomial will then factor into $(x+p)(x+q)$, with $pq=c$ and $p+q=b$.\n\nLet’s put this idea to practice with the following example.\n\n### Example\n\nFactor ${x}^{2}+2x - 15$.\n\nIn the following video we present two more examples of factoring a trinomial with a leading coefficient of 1.\n\nTo summarize our process consider these steps:\n\n### How To: Given a trinomial in the form ${x}^{2}+bx+c$, factor it.\n\n1. List factors of $c$.\n2. Find $p$ and $q$, a pair of factors of $c$ with a sum of $b$.\n3. Write the factored expression $\\left(x+p\\right)\\left(x+q\\right)$.\n\nWe will now show an example where the trinomial has a negative c term. Pay attention to the signs of the numbers that are considered for p and q.\n\nIn our next example, we show that when c is negative, either p or q will be negative.\n\n### Example\n\nFactor $x^{2}+x-12$.\n\nWhich property of multiplication can be used to describe why $\\left(x+4\\right)\\left(x-3\\right) =\\left(x-3\\right)\\left(x+4\\right)$. Use the textbox below to write down your ideas before you look at the answer.\n\nIn our next example we will show how to factor a trinomial whose $b$ term is negative.\n\n### Example\n\nFactor ${x}^{2}-7x+6$.\n\nIn the next example, we factor a trinomial with two variable, but the leading coefficient is still 1.\n\n### ExAMPLE\n\nFactor: $x^2-8xy-9y^2$\n\n### Recognize when a trinomial cannot be factored (prime)\n\nCan every trinomial be factored as a product of binomials?\n\nMathematicians often use a counter example to prove or disprove a question. A counter example means you provide an example where a proposed rule or definition is not true. Can you create a trinomial with leading coefficient 1 that cannot be factored as a product of binomials?\n\nUse the textbox below to write your ideas.\n\n### Definition\n\nA polynomial is prime if it cannot be factored using integer coefficients. A polynomial is factored completely when it is written as the product of prime polynomials.\n\nFor example, the polynomial $x^2+x+1$ is a prime.\n\n# (7.3.3) – Factor a Trinomial with Leading Coefficient $\\neq 1$\n\n### Factor by Trial and Error\n\nWhat happens when the leading coefficient is not 1 and there is no GCF? There are several methods that can be used to factor these trinomials. First we will use the Trial and Error method.\n\nLet’s factor the trinomial using Trial and Error: $3y^2+22y+7$.", null, "", null, "", null, "", null, "", null, "", null, "Try another example, this time with a negative leading coefficient: (Hint: try factoring out $-1$ out first.)\n\n### Factor by Grouping\n\nTrinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can factor by grouping by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial $2{x}^{2}+5x+3$ can be rewritten as $\\left(2x+3\\right)\\left(x+1\\right)$ using this process. We begin by rewriting the original expression as $2{x}^{2}+2x+3x+3$ and then factor each portion of the expression to obtain $2x\\left(x+1\\right)+3\\left(x+1\\right)$. We then pull out the GCF of $\\left(x+1\\right)$ to find the factored expression.\n\nThe first step in this process is to figure out what two numbers to use to re-write the x term as the sum of two new terms. Making a table to keep track of your work is helpful. We are looking for two numbers with a product of $2\\cdot3=6$ and a sum of $5$\n\nFactors of $2\\cdot3=6$ Sum of Factors\n$1,6$ $7$\n$-1,-6$ $-7$\n$2,3$ $5$\n$-2,-3$ $-5$\n\nThe pair $p=2,\\text{ and }q=3$ will give the correct x term, so we will rewrite it using the new factors:\n\n$2{x}^{2}+5x+3=2x^2+2x+3x+3$\n\nNow we can group the polynomial into two binomials.\n\n$2x^2+2x+3x+3=(2x^2+2)+(3x+3)$\n\nIdentify the GCF of each binomial.\n\n2x is the GCF of $(2x^2+2)$ and 3 is the GCF of $(3x+3)$, use this to rewrite the polynomial:\n\n$(2x^2+2x)+(3x+3)=2x(x+1)+3(x+1)$\n\nNote how we leave the signs in the binomials and the addition that joins them, be careful with signs when you factor out the GCF. The GCF of our new polynomial is $(x+1)$, we factor this out as well:\n\n$2x(x+1)+3(x+1)=(x+1)(2x+3)$.\n\nSometimes it helps visually to write the polynomial this way $(x+1)2x+(x+1)3$ before you factor out the GCF. This is purely a matter of preference, multiplication is commutative, so order doesn’t matter.\n\n### A General Note: Factor by Grouping\n\nTo factor a trinomial in the form $a{x}^{2}+bx+c$ by grouping, we find two numbers with a product of $ac$ and a sum of $b$. We use these numbers to divide the $x$ term into the sum of two terms and factor each portion of the expression separately, then factor out the GCF of the entire expression.\n\n### Example\n\nFactor $5{x}^{2}+7x - 6$ by grouping.\n\nWe can summarize our process in the following way:\n\n### Given a trinomial in the form $a{x}^{2}+bx+c$, factor by grouping.\n\n1. List factors of $ac$.\n2. Find $p$ and $q$, a pair of factors of $ac$ with a sum of $b$.\n3. Rewrite the original expression as $a{x}^{2}+px+qx+c$.\n4. Pull out the GCF of $a{x}^{2}+px$.\n5. Pull out the GCF of $qx+c$.\n6. Factor out the GCF of the expression.\n\nIn the following video we present one more example of factoring a trinomial whose leading coefficient is not 1 using the grouping method.\n\nFactoring trinomials whose leading coefficient is not 1 becomes quick and kind of fun once you get the idea.  Give the next example a try on your own before you look at the solution.\n\nWe will show a few more examples so you can become acquainted with the variety of possible outcomes for factoring this type of trinomial.\n\n### Example\n\nFactor $2{x}^{2}+9x+9$.\n\nHere is an example where the $x$ term is positive and $c$ is negative.\n\n### Example\n\nFactor $6{x}^{2}+x - 1$.\n\nIn the following video example, we will factor a trinomial whose leading term is negative.\n\nFor our last example, you will see that sometimes, you will encounter polynomials that, despite your best efforts, cannot be factored into the product of two binomials.\n\n### Example\n\nFactor $7x^{2}-16x-5$." ]

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[ "", null, "The cuboctahedron is an Archimedean solid. It was drawn by Leonardo da Vinci for Luca Pacioli's book 'De divina proportione'.", null, "Leonardo da Vinci made several drawings of polyhedra for Luca Pacioli's book 'De divina proportione'. Here we can see an adaptation of the cuboctahedron.\n\nWe can see this polyhedron as a truncated cube or as a truncated octahedron and we can easily calculate its volume.", null, "A cuboctahedron is an Archimedean solid. It can be seen as made by cutting off the corners of a cube.", null, "A cuboctahedron is an Archimedean solid. It can be seen as made by cutting off the corners of an octahedron.\n\nThe first stellation of the cuboctahedron is the compound of a cube and its dual octahedron, with the vertices of the cuboctahedron located at the midpoints of the edges. It is the same to say that the solid common to both the cube and the octahedron in a cube-octahedron compound is a cuboctahedron.", null, "", null, "", null, "You can build the stellated cuboctahedron using Zome:", null, "Can you calculate the relative sizes of these polyhedra?", null, "The vertices of a stellated cuboctahedron are the vertices of a rhombic dodecahedron (that is the dual polyhedra of a cuboctahedron).", null, "REFERENCES\n\nHugo Steinhaus - Mathematical Snapshots - Oxford University Press - Third Edition (p. 197)\nMagnus Wenninger - 'Polyhedron Models', Cambridge University Press.\nPeter R. Cromwell - 'Polyhedra', Cambridge University Press, 1999.\nH.Martin Cundy and A.P. Rollet, 'Mathematical Models', Oxford University Press, Second Edition, 1961 (p. 87).\nW.W. Rouse Ball and H.S.M. Coxeter - 'Matematical Recreations & Essays', The MacMillan Company, 1947.\nLuca Pacioli - De divina proportione - (La divina proporción) Ediciones Akal, 4ª edición, 2004. Spanish translation by Juan Calatrava.", null, "A cuboctahedron is an Archimedean solid. It can be seen as made by cutting off the corners of a cube.", null, "The volume of an octahedron is four times the volume of a tetrahedron. It is easy to calculate and then we can get the volume of a tetrahedron.", null, "The volume of a tetrahedron is one third of the prism that contains it.", null, "The first drawing of a plane net of a regular tetrahedron was published by Dürer in his book 'Underweysung der Messung' ('Four Books of Measurement'), published in 1525 .", null, "We study different prisms and we can see how they develop into a plane net. Then we explain how to calculate the lateral surface area.", null, "Plane nets of prisms with a regular base with different side number cut by an oblique plane.", null, "We study different cylinders and we can see how they develop into a plane. Then we explain how to calculate the lateral surface area.", null, "We study different cylinders cut by an oblique plane. The section that we get is an ellipse.", null, "Plane net of pyramids and pyramidal frustrum. How to calculate the lateral surface area.", null, "Plane developments of cones and conical frustum. How to calculate the lateral surface area.", null, "Plane developments of cones cut by an oblique plane. The section is an ellipse.", null, "The first drawing of a plane net of a regular dodecahedron was published by Dürer in his book 'Underweysung der Messung' ('Four Books of Measurement'), published in 1525 .", null, "The twelve vertices of an icosahedron lie in three golden rectangles. Then we can calculate the volume of an icosahedron", null, "Some properties of this platonic solid and how it is related to the golden ratio. Constructing dodecahedra using different techniques.", null, "The truncated octahedron is an Archimedean solid. It has 8 regular hexagonal faces and 6 square faces. Its volume can be calculated knowing the volume of an octahedron.", null, "We can cut in half a cube by a plane and get a section that is a regular hexagon. Using eight of this pieces we can made a truncated octahedron.", null, "Using eight half cubes we can make a truncated octahedron. The cube tesselate the space an so do the truncated octahedron. We can calculate the volume of a truncated octahedron.", null, "These polyhedra pack together to fill space, forming a 3 dimensional space tessellation or tilling.", null, "The truncated octahedron is an Archimedean solid. It has 8 regular hexagonal faces and 6 square faces. Its volume can be calculated knowing the volume of an octahedron.", null, "You can chamfer a cube and then you get a polyhedron similar (but not equal) to a truncated octahedron. You can get also a rhombic dodecahedron.", null, "Leonardo da Vinci made several drawings of polyhedra for Luca Pacioli's book 'De divina proportione'. Here we can see an adaptation of the truncated octahedron.", null, "Leonardo da Vinci made several drawings of polyhedra for Luca Pacioli's book 'De divina proportione'. Here we can see an adaptation of the cuboctahedron.", null, "Leonardo da Vinci made several drawings of polyhedra for Luca Pacioli's book 'De divina proportione'. Here we can see an adaptation of the dodecahedron.", null, "Leonardo da Vinci made several drawings of polyhedra for Luca Pacioli's book 'De divina proportione'. Here we can see an adaptation of the stellated octahedron (stella octangula).", null, "The truncated tetrahedron is an Archimedean solid made by 4 triangles and 4 hexagons.", null, "When you truncate a cube you get a truncated cube and a cuboctahedron. If you truncate an octahedron you get a truncated octahedron and a cuboctahedron." ]

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[ "# [R] Using evaluate-deparse-substitute\n\nUwe Ligges ligges at statistik.uni-dortmund.de\nThu Feb 20 11:08:03 CET 2003\n\n```Adaikalavan Ramasamy wrote:\n> Being the lazy soul I am, I wish to write a function to replace saying\n> ls(pattern=...) everytime. Here is what I have:\n>\n> lsp <- function(x){\n> y <- eval(deparse(substitute(x)))\n> print(y) # CHECK\n>\n> print( ls(pattern = eval(y)) ) # TRY 1\n> print( ls(pattern = eval(deparse(substitute(x)))) ) # TRY 2\n> }\n>\n> Suppose I have\n> rubbish.in = rubbish.out = grub <- 1\n>\n> I get the following when I try\n>\n>\n>>lsp(rub)\n>\n> \"rub\"\n> character(0)\n> character(0)\n>\n> Can someone explain/help with this? Thank you very much.\n>\n> ______________________________________________\n> R-help at stat.math.ethz.ch mailing list\n> http://www.stat.math.ethz.ch/mailman/listinfo/r-help\n\nYou have to think about environments and scoping rules.\nYou define another function and ls() is invoked in that functions, thus\nit is called in another environment than you obviously expected.\nSee ?ls, particularly its argument \"name\", why the following prints the\nthings you expect from ls().\n\nlsp <- function(x){\ny <- deparse(substitute(x))\nprint(y)\nprint(ls(sys.frame(sys.parent()), pattern = y))\n}" ]

[ null ]

[ "", null, "Computational Geometry\nDepartment of Computer Science\nStatistical Data Depth at Tufts\n\nStatisticians have recently developed the notion of data depth for non-parametric multivariate data analysis (see for example , ). This new concept provides center-outward orderings of points in Euclidean space of any dimension and leads to a new non-parametric multivariate statistical analysis in which no distributional assumptions are needed.\n\nA data depth measures how deep (or central) a given point x in Rd is relative to F, a probability distribution in Rd (assuming {X1,.., Xn } is a random sample from F) or relative to a given data cloud.\n\nSome examples of data depth are Halfspace Depth , Simplicial Depth , the Convex Hull Peeling Depth and Regression depth (which is the depth of a hyperplane relative to a set of points). All of these depths are affine invariant: each depth value remains the same after the data are transformed by any affine transformation. Different notions of data depth capture different statistical characteristics of the underlying distribution.\n\nDepth contours, constructed by enclosing all points of depth d or higher, are especially powerful for visualizing and quantifying data. Simple (in many cases 2D) graphs can be used to visualize these parameters for the data set. The potential is enormous for analysis of massive data sets in such areas as quality control and aviation safety analysis, clinical data mining, biological imaging analysis, and statistical process control.\n\n Data Depth Sub Pages (This will be enhanced)\nFigure", null, "Halfspace depth contours for a data set that consists of 50 points, drawn from a bivariate normal distribution with mean (0,0) and covariance 4 times identity. (rousseeuw.eps)\n\nPapers and Presentations\n\n\"Multivariate analysis by data depth: descriptive statistics, graphics and inference \", Liu, R. The Annals of Statistics (27) 783-858,1999\n\n\"On a notion of data depth based on random simplices\", Liu, R. The Annals of Statistics (18) 405-414,1990\n\n\" Efficient Computation of Location Depth Contours by Methods of Combinatorial Geometry\", K. Miller, S. Ramaswami, P. Rousseeuw, T. Sellares, D. Souvaine, I. Streinu, A. Struyf. Statistics and Computing, 2003, Postscript version.\n\n \"Fast implementation of depth contours using topological sweep,'' K. Miller, S. Ramaswami, P. Rousseeuw, T. Sellares, D. Souvaine, I. Streinu, A. Struyf. Proceedings of the Twelfth ACM-SIAM Symposium on Discrete Algorithms, Washington, DC, January, 2001. Postscript version\n\n\"Regression depth\", Rousseeuw, P. J. and M. Hubert,\nJ. Amer. Statist. Assoc. (94),388-433, 1999\n\n \"Mathematics and the picturing of data\", Tukey, John W., Proceedings of the International Congress of Mathematicians, Vancouver, B. C., 1974, Vol. 2, 523--531\n\n\"Convex hull peeling\", Eddy, W in \"COMPSTAT\", 42-47, 1982\n\n\" Computational Geometry and Statistical Depth Measures\" , E. Rafalin, D. Souvaine, Theory and Applications of Recent Robust Methods, edited by M. Hubert, G. Pison, A. Struyf and S. Van Aelst, 2004 in Series: Statistics for Industry and Technology, Birkhauser, Basel. Postscript version Pdf version\n\n\" Computational Geometry, Data Depth and Robust Statistics\" , E. Rafalin, D. Souvaine, Interface 2004, Baltimore, MD. Postscript version Pdf version\n\n[*] A powerpoint presentation describing the code developed in the department for analysis based on the notion of data depth (presented in a DIMACS workshop on Data Depth, May 2003)\n\n[*] A powerpoint presentation surveying the connection between computational geometry and depth based statistics (presented in The International Conference on Robust Statistics, ICORS 03, Antwerp, Belgium, July 2003)" ]

[ null, "http://www.cs.tufts.edu/r/geometry/images/Tufts_univ_blue.gif", null, "http://www.cs.tufts.edu/r/geometry/depth/images/contours.gif", null ]

[ "# How to tell Solve[] to satisfy as many constraints as it can?\n\nIs there any way to tell Solve (or something similar) to return values for variables that solved as many constraints as it happened to satisfy?\n\n### Edit:\n\nI had trouble finding a small example that wasn't trivial, but I finally did. Here's one:\n\nSolve[{{-1 + Abs[q2 (-(7/4) + t1)], q1, -1 + Abs[q1 (-(7/4) + t2)], 3 + q2} == 0,\n{t1, t2} >= 7/4},\n{t1, t2, q1, q2}]\n\n\nThe kind of output that I want (I don't care if it's not unique; I can deal with that):\n\n{{t1 -> 25/12, t2 -> Indeterminate, q1 -> 0, q2 -> -3}}\n\n\nThe output that I get, but don't want:\n\n{}\n\n• Please paste in a minimal example. – Jack LaVigne Mar 10 '16 at 14:38\n• @JackLaVigne: I'm afraid if I post an example then people are going to suggest a method that solves that particular example and not the actual problem... – user541686 Mar 10 '16 at 22:06\n• @JackLaVigne: But here's a trivial example: Solve[{x^2 + 1 == 0, y^2 - 1 == 0}, {x, y}, Reals] gives no solution, while Minimize[{Abs[x^2 + 1] + Abs[y^2 - 1]}, {x, y}, Reals] finds a solution for y which is still useful to me. (If you read the above, please don't tell me to separate the equations and solve them independently. I know I can do that for this particular example, but that also completely misses the point of my question, which I think is clear.) – user541686 Mar 10 '16 at 22:14\n• If your system is indeed overdetermined, the ArgMin[] route would be more expedient than forcing Solve[] to do your bidding. FWIW, your last snippet gives the same output as With[{p = 1}, ArgMin[Norm[Subtract @@@ {x^2 + 1 == 0, y^2 - 1 == 0}, p], {x, y}, Reals]]. Do you really need to use the $1$-norm? That is a more difficult problem than minimizing with respect to the usual $2$-norm. – J. M.'s technical difficulties Mar 13 '16 at 3:05\n• Okay, I think I understand what you're after now. I'll have to think about that. – Mr.Wizard Mar 13 '16 at 7:00\n\nIf your equations and constraints are linear (or can be expressed as linear), and if a machine precision solution is enough you can use functions like LinearProgramming or NMinimize to get a solution:\n\nM = 50;\nNMinimize[{\nz1 + z2 + z3,\n{\n-1 + Abs[q2 (-(7/4) + t1)] == 0,\nq1 == 0,\n-1 + Abs[q1 (-(7/4) + t2)] >= 0 - M*z3,\n-1 + Abs[q1 (-(7/4) + t2)] <= 0 + M*z3,\n3 + q2 == 0,\nt1 >= 7/4 - z1*M,\nt2 >= 7/4 - z2*M,\nz1 \\[Element] Integers, 0 <= z1 <= 1,\nz2 \\[Element] Integers, 0 <= z2 <= 1,\nz3 \\[Element] Integers, 0 <= z3 <= 1\n}\n},\n{t1, t2, q1, q2, z1, z2, z3}\n]\n\n\n{2., {t1 -> 2.08333, t2 -> 549.224, q1 -> 0, q2 -> -3., z1 -> 1, z2 -> 1, z3 -> 0}}\n\nBasically, you introduce some binary ($0-1$) decision variables $z_i$ and one or more costant $M$ big enough so that you can rewrite the desired $i$ constraint in a way that when $z_i$ is $1$ constraint is always satisfied. You then minimze the sum of $z_i$.\n\nMore references to LinearProgramming in my answer here and\n\nand another example of \"conditional constraints\", here:\n\nWith NMinimize you can also handle some non-linear constraint. Unfortunately Minimize is not available because:\n\nMinimize::mixdom: Exact optimization with mixed real and integer variables is not yet implemented.\n\nYou can of course try to implement by yourself this strategy (a search on a suitable tree) to get an exact answer. If there are only few \"conditional constraints\" you can also try to Solve for all possible $(z_1z_2\\ldots)$ until you find a solution. In this case, with at most $2^3=8$ cases is not too difficult.\n\nAs an idea of how we can do a search on a tree, start building a suitable tree. This tree is such that the binary digits of the node are used to identify a subset of constraints. There is also an interesting ordering in DepthFirstScan visit order; see the picture.\n\nvt[n_] := Module[{l, p},\nl = Range[0, 2^n - 1];\np = FromDigits[#, 2] & /@\nReplace[IntegerDigits[l,\n2], {a : 0 ..., 1, b___} :> {a, 0, b}, {1}];\nTreeGraph[l, p, VertexShapeFunction -> \"Name\",\nVertexLabels -> Thread[l -> IntegerString[l, 2, n]]]\n]\nvt", null, "A simple (?), but far from optimal, search function is:\n\nssolve[eqns_, cons_, vars_, dom : _ : Reals] :=\nModule[{n, t, m, c, s, l, w, r},\nn = Length@cons;\nt = vt[n];\nm = -\\[Infinity]; c = Indeterminate;\nr = {};\nDepthFirstScan[t, 0,\n\"DiscoverVertex\" -> Function[{u, v, d},\nWhich[\ns[v], s[u] = True,\nd > m,\nWith[{sol =\nQuiet@Solve[\nJoin[eqns, Pick[cons, IntegerDigits[u, 2, n], 1]], vars,\ndom]},\nIf[sol == {}, s[u] = True,\nc = u; r = sol; m = d\n]]\n]]];\n{Pick[cons, IntegerDigits[c, 2, n], 1], r}\n]\n\n\nA sample (all your equations and inequalities are considered optional):\n\neqns = Thread /@ {{-1 + Abs[q2 (-(7/4) + t1)],\nq1, -1 + Abs[q1 (-(7/4) + t2)], 3 + q2} == 0, {t1, t2} >= 7/4} //\nFlatten\nvars = {t1, t2, q1, q2};\nssolve[{}, eqns, vars, Reals]\n\n\nThe return value is a list with the set of constraints fulfilled and the return value of Solve.\n\n{{-1 + Abs[q2 (-(7/4) + t1)] == 0, -1 + Abs[q1 (-(7/4) + t2)] == 0,\n3 + q2 == 0, t1 >= 7/4,\nt2 >= 7/4}, {{t1 -> ConditionalExpression[25/12, q1 > 0],\nt2 -> ConditionalExpression[(4 + 7 q1)/(4 q1), q1 > 0],\nq2 -> ConditionalExpression[-3, q1 > 0]}, {t1 ->\nConditionalExpression[25/12, q1 < 0],\nt2 -> ConditionalExpression[(-4 + 7 q1)/(4 q1), q1 < 0],\nq2 -> ConditionalExpression[-3, q1 < 0]}}}\n\n\nI didn't fully tested the code but the basic idea should work, and should be more efficient than testing all the possible subsets of constraints. The use of built-in function DepthFirstScan is easy, but unfortunately at present doesn't allow to really skip the visit of a subtree.\n\nEdit. In the way I used DepthFirstScan, nodes are not processed in DFS order. To fix this problem, I think a more involved code is required. At this point, I don't really see any reason to build a TreeGraph and use DepthFirstScan. I think it's better to use another strategy. I'll try to post an update when I have time.\n\n• Thanks but they're definitely nonlinear... even the example I gave isn't linear. – user541686 Mar 13 '16 at 9:24\n• @Mehrdad As you can see NMinimize handle some nonlinear constraint. And the example given can be expressed as linear. You can combine this approach with Solve and a search on binary trees or, more simply, with a full search to get what you want. – unlikely Mar 13 '16 at 9:35" ]

[ null, "https://i.stack.imgur.com/2PwJS.png", null ]

[ "# question\n\nAccepted\n48 3 10 9\n\n## LEVEL 2 EQUITY DATA - WHERE IS TRADE TYPE?\n\nHi Im subscribing to LEVEL 2 MBO equity pricing from Oslo exchange, (example RIC AKA.NO). How can I find the trade/order type in this price stream?\nDo I need to change my domain to get trade type or is it the case that trade data is delivered in the L1 MARKET_PRICE domain?", null, "Up to 2 attachments (including images) can be used with a maximum of 512.0 KiB each and 1.0 MiB total.\n\nCan you please confirm the RIC code you are using? I cannot find AKA.NO - unless you have a direct feed and not sourcing from Elektron.\n\nAlso, I am not a content expert, but my understanding of MBO data is that it represents 'open' orders i.e. with Quote date and Quote Time and therefore would not be expected to contain Trade-related data.\n\nIf I have misunderstood the question please correct my understanding.\n\nAccepted\n20.2k 53 14 20\n\n@duncan_kerr,\n\nThe message I posted previously was the MBO as that is what you were referring to in your question. The Market By Price message is different and looks like this:\n\n```MapEntry action=\"Add\" key dataType=\"Buffer\" value=\"372E ****\" dataType=\"FieldList\"\nFieldList FieldListNum=\"32767\" DictionaryId=\"1\"\nFieldEntry fid=\"3427\" name=\"ORDER_PRC\" dataType=\"Real\" value=\"7.98\"\nFieldEntry fid=\"3428\" name=\"ORDER_SIDE\" dataType=\"Enum\" value=\"2\"\nFieldEntry fid=\"3430\" name=\"NO_ORD\" dataType=\"UInt\" value=\"1\"\nFieldEntry fid=\"3886\" name=\"ORDER_TONE\" dataType=\"Rmtes\" value=\"0\"\nFieldEntry fid=\"4356\" name=\"ACC_SIZE\" dataType=\"Real\" value=\"2000.0\"\nFieldEntry fid=\"6527\" name=\"LV_TIM_MS\" dataType=\"UInt\" value=\"54302722\"\nFieldEntry fid=\"6529\" name=\"LV_DATE\" dataType=\"Date\" value=\"11 NOV 2020\"\nFieldEntry fid=\"14268\" name=\"LV_TIM_NS\" dataType=\"Time\" value=\"15:05:02:722:025:000\"\nFieldListEnd\nMapEntryEnd\n\n```", null, "Up to 2 attachments (including images) can be used with a maximum of 512.0 KiB each and 1.0 MiB total.\n\n20.2k 53 14 20\n\nThe Market by Order (order book) data only contains the orders. Any matches on the BID and ASK side are removed from the book and show up as trades. Trade information is not available on the Level 2 stream, and you will need to subscribe to Market Price (L1) domain for that.\n\nHere is what an OMM message for an order for AKAS.OL on Oslo exchange looks like:\n\n```MapEntry action=\"Add\" key dataType=\"Buffer\" value=\"3531***\" dataType=\"FieldList\"\nFieldList FieldListNum=\"32767\" DictionaryId=\"1\"\nFieldEntry fid=\"3426\" name=\"ORDER_ID\" dataType=\"Rmtes\" value=\"51486573296191456\"\nFieldEntry fid=\"3427\" name=\"ORDER_PRC\" dataType=\"Real\" value=\"7.79\"\nFieldEntry fid=\"3428\" name=\"ORDER_SIDE\" dataType=\"Enum\" value=\"2\"\nFieldEntry fid=\"3429\" name=\"ORDER_SIZE\" dataType=\"Real\" value=\"10000.0\"\nFieldEntry fid=\"3886\" name=\"ORDER_TONE\" dataType=\"Rmtes\" value=\"0\"\nFieldEntry fid=\"6520\" name=\"PR_TIM_MS\" dataType=\"UInt\" value=\"47103936\"\nFieldEntry fid=\"6521\" name=\"PR_TIM_MSP\" dataType=\"UInt\" value=\"936880\"\nFieldEntry fid=\"6522\" name=\"PR_DATE\" dataType=\"Date\" value=\"10 NOV 2020\"\nFieldEntry fid=\"6524\" name=\"OR_TIM_MS\" dataType=\"UInt\" value=\"55200295\"\nFieldEntry fid=\"6525\" name=\"OR_TIM_MSP\" dataType=\"UInt\" value=\"295304\"\nFieldEntry fid=\"6526\" name=\"OR_DATE\" dataType=\"Date\" value=\"10 NOV 2020\"\nFieldEntry fid=\"14268\" name=\"LV_TIM_NS\" dataType=\"Time\" value=\"15:20:00:295:304:000\"\nFieldEntry fid=\"14275\" name=\"PR_TIM_NS\" dataType=\"Time\" value=\"13:05:03:936:880:000\"\nFieldListEnd\nMapEntryEnd```\n\nAn order does not have a type, the whole book does. The market trading status etc, is available in the summary message of the order book.", null, "Up to 2 attachments (including images) can be used with a maximum of 512.0 KiB each and 1.0 MiB total.\n\n48 3 10 9\n\nHi Gurpreet - Im accessing the MBP tickstream from Bloomberg over TREP, and I see an ADD message for oslo exchange like this:\n\nMapEntry action=\"Add\" key dataType=\"Buffer\" value=\"31 1\"\n\ndataType=\"FieldList\"\n\nFieldList\n\nFieldEntry fid=\"3428\" name=\"ORDER_SIDE\" dataType=\"Enum\" value=\"2\"\n\nFieldEntry fid=\"3427\" name=\"ORDER_PRC\" dataType=\"Real\" value=\"153.55\"\n\nFieldEntry fid=\"4356\" name=\"ACC_SIZE\" dataType=\"Real\" value=\"1346.0\"\n\nFieldEntry fid=\"3430\" name=\"NO_ORD\" dataType=\"UInt\" value=\"5\"\n\nFieldEntry fid=\"6527\" name=\"LV_TIM_MS\" dataType=\"UInt\" value=\"52840822\"\n\nFieldListEnd\n\nMapEntryEnd\n\nIs see a UPDATE like this:\n\nMapEntry action=\"Update\" key dataType=\"Buffer\" value=\"6566 ef\"\n\ndataType=\"FieldList\"\n\nFieldList\n\nFieldEntry fid=\"3428\" name=\"ORDER_SIDE\" dataType=\"Enum\" value=\"2\"\n\nFieldEntry fid=\"3427\" name=\"ORDER_PRC\" dataType=\"Real\" value=\"153.65\"\n\nFieldEntry fid=\"4356\" name=\"ACC_SIZE\" dataType=\"Real\" value=\"630.0\"\n\nFieldEntry fid=\"3430\" name=\"NO_ORD\" dataType=\"UInt\" value=\"2\"\n\nFieldEntry fid=\"6527\" name=\"LV_TIM_MS\" dataType=\"UInt\" value=\"53327612\"\n\nFieldListEnd\n\nMapEntryEnd\n\nI suspect I need to talk to bloomberg about the missing FIDs.", null, "Up to 2 attachments (including images) can be used with a maximum of 512.0 KiB each and 1.0 MiB total.", null, "" ]

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[ "$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n\n# 6.1E: Exercises\n\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n\n## Exercise $$\\PageIndex{1}$$ functional value\n\nFor the following exercises, evaluate each function at the indicated values.\n\n1) $$\\displaystyle W(x,y)=4x^2+y^2.$$ Find $$\\displaystyle W(2,−1), W(−3,6).$$\n\nSolution:$$\\displaystyle 17,72$$\n\n2) $$\\displaystyle W(x,y)=4x^2+y^2$$. Find $$\\displaystyle W(2+h,3+h).$$\n\n3) The volume of a right circular cylinder is calculated by a function of two variables, $$\\displaystyle V(x,y)=πx^2y,$$ where $$\\displaystyle x$$ is the radius of the right circular cylinder and $$\\displaystyle y$$ represents the height of the cylinder. Evaluate $$\\displaystyle V(2,5)$$ and explain what this means.\n\nSolution:$$\\displaystyle 20π.$$ This is the volume when the radius is $$\\displaystyle 2$$ and the height is $$\\displaystyle 5$$.\n\n4) An oxygen tank is constructed of a right cylinder of height $$\\displaystyle y$$ and radius $$\\displaystyle x$$ with two hemispheres of radius $$\\displaystyle x$$ mounted on the top and bottom of the cylinder. Express the volume of the cylinder as a function of two variables, $$\\displaystyle x$$ and $$\\displaystyle y$$, find $$\\displaystyle V(10,2)$$, and explain what this means.\n\n## Exercise $$\\PageIndex{2}$$ Domain\n\nFor the following exercises, find the domain of the function.\n\n1) $$\\displaystyle V(x,y)=4x^2+y^2$$\n\nSolution:All points in the $$\\displaystyle xy-plane$$\n\n2) $$\\displaystyle f(x,y)=\\sqrt{x^2+y^2−4}$$\n\n3) $$\\displaystyle f(x,y)=4ln(y^2−x)$$\n\nSolution:$$\\displaystyle x<y^2$$\n\n4) $$\\displaystyle g(x,y)=\\sqrt{16−4x^2−y^2}$$\n\n5) $$\\displaystyle z(x,y)=y^2−x^2$$\n\nSolution:All real ordered pairs in the $$\\displaystyle xy-plane$$ of the form $$\\displaystyle (a,b)$$\n\n6) $$\\displaystyle f(x,y)=\\frac{y+2}{x^2}$$\n\n## Exercise $$\\PageIndex{3}$$ range\n\nFind the range of the functions.\n\n1) $$\\displaystyle g(x,y)=\\sqrt{16−4x^2−y^2}$$\n\nSolution:$$\\displaystyle \\{z|0≤z≤4 \\}$$\n\n2) $$\\displaystyle V(x,y)=4x^2+y^2$$\n\n3) $$\\displaystyle z=y^2−x^2$$\n\nSolution:The set $$\\displaystyle R$$.\n\n## Exercise $$\\PageIndex{4}$$ Level Curves\n\nFor the following exercises, find the level curves of each function at the indicated value of $$\\displaystyle c$$ to visualize the given function.\n\n1) $$\\displaystyle z(x,y)=y^2−x^2, c=1$$\n\n2) $$\\displaystyle z(x,y)=y^2−x^2, c=4$$\n\nSolution:$$\\displaystyle y^2−x^2=4,$$ a hyperbola\n\n3) $$\\displaystyle g(x,y)=x^2+y^2;c=4,c=9$$\n\n4) $$\\displaystyle g(x,y)=4−x−y;c=0,4$$\n\nSolution:$$\\displaystyle 4=x+y,$$ a line; $$\\displaystyle x+y=0,$$ line through the origin\n\n5) $$\\displaystyle h(x,y)=2x−y;c=0,−2,2$$\n\nSolution:$$\\displaystyle 2x−y=0,2x−y=−2,2x−y=2;$$ three lines\n\n6) $$\\displaystyle f(x,y)=x^2−y;c=1,2$$\n\n7) $$\\displaystyle g(x,y)=\\frac{x}{x+y};c=−1,0,2$$\n\nSolution:$$\\displaystyle \\frac{x}{x+y}=−1,\\frac{x}{x+y}=0,\\frac{x}{x+y}=2$$\n\n8) $$\\displaystyle g(x,y)=x^3−y;c=−1,0,2$$\n\n9) $$\\displaystyle g(x,y)=e^{xy};c=\\frac{1}{2},3$$\n\nSolution:$$\\displaystyle e^{xy}=\\frac{1}{2},e^{xy}=3$$\n\n10) $$\\displaystyle f(x,y)=x^2;c=4,9$$\n\n12) $$\\displaystyle f(x,y)=xy−x;c=−2,0,2$$\n\nSolution:$$\\displaystyle xy−x=−2,xy−x=0,xy−x=2$$\n\n13) $$\\displaystyle h(x,y)=ln(x^2+y^2);c=−1,0,1$$\n\n14) $$\\displaystyle g(x,y)=ln(\\frac{y}{x^2});c=−2,0,2$$\n\nSolution:$$\\displaystyle e^{−2}x^2=y,y=x^2,y=e^2x^2$$\n\n15) $$\\displaystyle z=f(x,y)=\\sqrt{x^2+y^2}, c=3$$\n\n16) $$\\displaystyle f(x,y)=\\frac{y+2}{x^2}, c=$$any constant\n\nSolution:The level curves are parabolas of the form $$\\displaystyle y=cx^2−2.$$\n\n## Exercise $$\\PageIndex{5}$$ Vertical Traces\n\nFor the following exercises, find the vertical traces of the functions at the indicated values of $$\\displaystyle x$$ and $$\\displaystyle y$$, and plot the traces.\n\n1) $$\\displaystyle z=4−x−y;x=2$$\n\n2) $$\\displaystyle f(x,y)=3x+y^3,x=1$$\n\nSolution:$$\\displaystyle z=3+y^3,$$ a curve in the zy-plane with rulings parallel to the $$\\displaystyle x-axis$$", null, "3) $$\\displaystyle z=cos\\sqrt{x^2+y^2} x=1$$\n\n## Exercise $$\\PageIndex{6}$$ Domain\n\nFind the domain of the following functions.\n\n1) $$\\displaystyle z=\\sqrt{100−4x^2−25y^2}$$\n\nSolution:$$\\displaystyle \\frac{x^2}{25}+\\frac{y^2}{4}≤1$$\n\n2) $$\\displaystyle z=ln(x−y^2)$$\n\n3) $$\\displaystyle f(x,y,z)=\\frac{1}{\\sqrt{36−4x^2−9y^2−z^2}}$$\n\nSolution:$$\\displaystyle \\frac{x^2}{9}+\\frac{y^2}{4}+\\frac{z^2}{36}<1$$\n\n4) $$\\displaystyle f(x,y,z)=\\sqrt{49−x^2−y^2−z^2}$$\n\n5) $$\\displaystyle f(x,y,z)=\\sqrt{16−x^2−y^2−z^2}$$\n\nSolution:All points in $$\\displaystyle xyz-space$$\n\n6) $$\\displaystyle f(x,y)=cos\\sqrt{x^2+y^2}$$\n\n## Exercise $$\\PageIndex{7}$$ Graph\n\nFor the following exercises, plot a graph of the function.\n\n1) $$\\displaystyle z=f(x,y)=\\sqrt{x^2+y^2}$$\n\nSolution:", null, "2) $$\\displaystyle z=x^2+y^2$$\n\n3) Use technology to graph $$\\displaystyle z=x^2y.$$\n\nSolution:", null, "## Exercise $$\\PageIndex{8}$$ Level curves\n\nSketch the following by finding the level curves. Verify the graph using technology.\n\n1) $$\\displaystyle f(x,y)=\\sqrt{4−x^2−y^2}$$\n\n2) $$\\displaystyle f(x,y)=2−\\sqrt{x^2+y^2}$$", null, "3) $$\\displaystyle z=1+e^{−x^2−y^2}$$\n\n4) $$\\displaystyle z=cos\\sqrt{x^2+y^2}$$\n\nSolution:", null, "5) $$\\displaystyle z=y^2−x^2$$\n\n## Exercise $$\\PageIndex{9}$$ Contour lines\n\n1) Describe the contour lines for several values of $$\\displaystyle c$$ for $$\\displaystyle z=x^2+y^2−2x−2y.$$\n\nSolution:The contour lines are circles.\n\n## Exercise $$\\PageIndex{10}$$ level surface\n\nFind the level surface for the functions of three variables and describe it.\n\n1) $$\\displaystyle w(x,y,z)=x−2y+z,c=4$$\n\n2) $$\\displaystyle w(x,y,z)=x^2+y^2+z^2,c=9$$\n\nSolution:$$\\displaystyle x^2+y^2+z^2=9$$, a sphere of radius $$\\displaystyle 3$$\n\n3) $$\\displaystyle w(x,y,z)=x^2+y^2−z^2,c=−4$$\n\n4) $$\\displaystyle w(x,y,z)=x^2+y^2−z^2,c=4$$\n\nSolution:$$\\displaystyle x^2+y^2−z^2=4,$$ a hyperboloid of one sheet\n\n5) $$\\displaystyle w(x,y,z)=9x^2−4y^2+36z^2,c=0$$\n\n## Exercise $$\\PageIndex{11}$$ level curve at a given point\n\nFor the following exercises, find an equation of the level curve of $$\\displaystyle f$$ that contains the point $$\\displaystyle P$$.\n\n1) $$\\displaystyle f(x,y)=1−4x^2−y^2,P(0,1)$$\n\nSolution:$$\\displaystyle 4x^2+y^2=1,$$\n\n2) $$\\displaystyle g(x,y)=y^2arctanx,P(1,2)$$\n\n3) $$\\displaystyle g(x,y)=e^{xy}(x^2+y^2),P(1,0)$$\n\nSolution:$$\\displaystyle 1=e^{xy}(x^2+y^2)$$\n\n## Exercise $$\\PageIndex{12}$$ Applications\n\n1) The strength $$\\displaystyle E$$ of an electric field at point $$\\displaystyle (x,y,z)$$ resulting from an infinitely long charged wire lying along the $$\\displaystyle y-axis$$ is given by $$\\displaystyle E(x,y,z)=k/\\sqrt{x^2+y^2}$$, where $$\\displaystyle k$$ is a positive constant. For simplicity, let $$\\displaystyle k=1$$ and find the equations of the level surfaces for $$\\displaystyle E=10$$ and $$\\displaystyle E=100.$$\n\n2) A thin plate made of iron is located in the $$\\displaystyle xy-plane.$$ The temperature $$\\displaystyle T$$ in degrees Celsius at a point $$\\displaystyle P(x,y)$$ is inversely proportional to the square of its distance from the origin. Express $$\\displaystyle T$$ as a function of $$\\displaystyle x$$ and $$\\displaystyle y$$.\n\nSolution:$$\\displaystyle T(x,y)=\\frac{k}{x^2+y^2}$$\n3) Refer to the preceding problem. Using the temperature function found there, determine the proportionality constant if the temperature at point $$\\displaystyle P(1,2)$$ is $$\\displaystyle 50°C.$$ Use this constant to determine the temperature at point $$\\displaystyle Q(3,4).$$\n4) Refer to the preceding problem. Find the level curves for $$\\displaystyle T=40°C$$ and $$\\displaystyle T=100°C,$$ and describe what the level curves represent.\nSolution:$$\\displaystyle x^2+y^2=\\frac{k}{40}, x^2+y^2=\\frac{k}{100}$$. The level curves represent circles of radii $$\\displaystyle \\sqrt{10k}/20$$ and $$\\displaystyle \\sqrt{k}/10$$" ]

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[ "# Questions Asked onMay 25, 2015\n\n1. ## Algebra\n\nA baker bakes a batch of muffins and splits the batch evenly onto six different trays. She then adds five croissants to each tray. If each tray now contains at least twenty baked goods, what is the least possible number of muffins in the baker's original\n\n2. ## Precalculus\n\nA 60-foot tree casts a shadow 85 feet long. The sine of the angle of elevation of the top of the tree to the sun is approximately ___. A. .82 B. .75 C. .58 D. .33\n\n3. ## physics\n\nAt what height above the ground must a body of mass 10kg be situated in order to have potential energy in value to the kinetic energy possessed by another body of mass 10kg moving with a velocity of 10m-s\n\n4. ## Math\n\nA model rocket is launched from a roof into a large field. The path the rocket can be modeled by the equation y=-0.04x^2+8.3x+4.3 where x is the horizontal distance in meters, from the starting point on the roof and y is the height in meters, of the rocket\n\n5. ## Chemistry\n\nUsing the balanced equation below, how many grams of cesium fluoride would be required to make 73.1 g of cesium xenon heptafluoride? CsF+XeF6 ---> CsXeF7\n\n6. ## algbra 1\n\nA farm has only sheep and geese. It has a total of 58 animals and a total of 150 legs on the animals. How many sheep and geese are on the farm?\n\n7. ## chemistry\n\n39.3 grams of c6h6 are allowed to react with 108.3 grams of o2. How much co2 will be produced by this reaction?\n\n8. ## English\n\nI am stuck on these two questions(can you check #1 one over for me?)! For the poem Annabel Lee; 1. What romanticized explanation does the speaker give for the death of his love, Annabel Lee? He tells us that angels from heaven were jealous of their love\n\n9. ## Life orientation\n\nEvaluate the extent to which the service delivery protest you have described in 2 above met or did not meet with the processes that you have researched in 3 above.\n\n10. ## Algebra\n\nMr. Johnston needs a shelf to hold a set of textbooks, each 1 3/4 in. wide. How many books will fit on a shelf that is 35 in long?\n\n11. ## Algebra 1\n\nSince opening night, attendance at Play A has increased steadily, while Play B first rose then fell. Equations modeling the daily attendance y at each play are shown below, where x is the number of days since opening night. On what day(s) was the\n\n12. ## Algebra 1\n\nWhat are the solutions to the system? y=x^2+5x-9 y+2x+1\n\n13. ## Chemistry\n\nDetermine the pH of a buffer made from 250 ml of 0.50M HF and 150 ml of 0.75 M NaF.\n\n14. ## criminal Justice\n\nThe ACLU has publicly: A. supported the USA PATRIOT Act. B. filed suit to stop the USA PATRIOT Act. C. is neutral about the USA PATRIOT Act. D. has altered the USA PATRIOT Act. is it A 2. Terrorism is frequently committed: A. impulsively. B. with extensive\n\n15. ## Math\n\nMrs. Murphy separates her class into groups of 4 students each, 1 student is left over. If she separates her class into groups of 5 students each, 2 students are left over. What is the least number of students the class could have?\n\n16. ## Trigonometry\n\nUse De Moivre's Theorem to write the complex number in trigonometric form (cos(2pi)/7)+ i sin((2pi)/7))^5\n\n17. ## math\n\nmargaret Hillman invested \\$4,000 at 1.2% compounded quarterly for one year. find the futrue value and the interest earned for the year\n\n18. ## chemistry\n\nwhy does water,oil,acetone and isopropanol have the same concave meniscus shape?\n\n19. ## Algebra\n\nA microscope can magnify the specimen 10^4 times. How many is that? My answer 10 x 10 x 10 x 10 = 10,000 times Am I right?\n\n20. ## Algebra\n\nWrite 806,000,000 in scientific notation\n\n21. ## English\n\nAnother question regarding the poem Annabel Lee. Can you check my response for me and help me elaborate on the rhythm explanation please; 1. What is the overall tone of the poem? How does the rhythm of the poem affect the tone? my ans: The tone of the poem\n\n22. ## Dynamics\n\nThe flat circular disc rotates about a vertical axis through O with a constant angular velocity of 240rpm. Prior to rotation, each of the 0.5kg sliding blocks has the position x=25mm with no force in its attached spring. Each spring has a stiffness of\n\n23. ## math\n\na circular maize field has a radius of 200m .What is the length of fence around this field\n\n24. ## Chemistry\n\nMethane gas burns in air to give carbon dioxide and water. a) give a chemical equation b) How can the mixture of these gases be separated b) Which gases does it mean?? Can anyone tell me.. Pleease.\n\n25. ## Chemistry ( I NEED HELP REAL QUICK )\n\nHow much Al is needed to react with 675 mL of 0.655M HCl solution according to the reaction => 2Al + 6HCl→2AlCl3+ 3H2 Help your friend!\n\n26. ## math\n\non monday 464 students went on a trip to zoo .all 9 buses were filled and 5 students had to travel in cars. how many students were in each bus?\n\n27. ## algebra\n\nassume x > 0. what percent of 8x is 5x? i know that the equation is 5x/8x = p/100 and i have to multiply both sides by 100, but i'm not sure how to multiply 100 * 5x / 8x? what is 65% of 80x? again, i know that the equation is a/80x = 65/100 and i have to\n\n28. ## Algebra\n\nA salesperson receives a 3% commission on sales. The salesperson receives \\$180 in commission. What is the amount of sales? The percent equation is: a/w = p/100 3% substitutes for p, but I'm not sure what 180 would substitute for, a or w?\n\n29. ## biology\n\nIf A represents atomic mass and Z represents atomic number, which of the following describes an atom after a proton is emitted from its nucleus? A. A+ 1, Z + 1 B. A-1,Z-1 C. A+ 1, Z-1 D. A-1, Z + 1 My answer is B.\n\n30. ## Chemistry check my answers\n\nPlease check my answers For each of the 7 reactions in this experiment, write a balanced equation and classify it as a synthesis, decomposition, single replacemnt, neutralization, or double replacement reaction. Reaction 1 1)Adjust a burner flae to high\n\n31. ## English help\n\nWhat is the resolution in the taming of the shrew?\n\n32. ## Algebra\n\nFind the GCF of 14abc and 28a^2 b^2 c^3 How do I answer this??\n\n33. ## Criminal Justice\n\nWhich of the following changed the laws on investigating terrorism? A. The USA PATRIOT Act B. The Sarbanes-Oxley Act C. The Homeland Security Act D. The Anti-Terrorism and Effective Death Penalty Act is it A\n\n34. ## l.o\n\ndo you agree with the statement below ? coverage of sport ,sport personalities and recreational activities vary in media .the media tend to focus on the male dominated sport and sports personalities as well as certain sports codes and neglect others ,\n\n35. ## American Government\n\nThe fourth branch of government refers to\n\n36. ## physics\n\nA body A of mass 2.0kg makes an elastic collision with another body B at rest. After collision,A continues to move in the original direction but with one-fourth of its original speed. Determine the value of mass B\n\n37. ## Algebra\n\nFind the product Simplify if possible: 12y/13 * 11/24\n\n38. ## Algebra\n\nA set of encyclopedia has 27 volumes. each volume is 1 3/8 inches thick. If the volumes are placed side by side, how long will the set be?\n\n39. ## electronic medical records\n\nif a physician is charging for a mole removal procedure based on what he and other physicans gerally say for that procedures,hes probably using\n\n40. ## math\n\nI need to calculate the effective interest rate of a simple discount note \\$21,750 at an ordinary bank discount rate of 7.91% for 120 days find the effective rate\n\n41. ## algebra\n\n14.2 is 35.5% of what number? 3/5 is 60% of what number? what number is 25% of 7/8?\n\n42. ## Foreign language, French\n\nI need help with speaking portion of the VHL Central Lesson 2B Enregitrez as I can't really hear the speaker quite properly. If someone is currently doing the course ( the course is French 2) that will be ideal but any and all help I will find acceptable.\n\n43. ## Foreign language, French\n\nI also need help understanding what the speaker for Lesson 4A Decrivez is saying because she kind of mumbles and I can't catch it. If someone is currently doing the course ( the course is French 2) that will be ideal but any and all help I will find\n\n44. ## math packet\n\nhow do I solve the following: 2x(x-3)+3=4\n\n45. ## Physics\n\nA car drives over a hump in the road. The radius of the (circular) hump is 45m. How fast (m/s) would the car need to travel, so that the wheels just barely lose contact with the road?\n\n46. ## algebra\n\n-2(x-3) 6x Is that the right answer?\n\n47. ## First Nations Studies 12\n\nActivity Sheet 5-10. Analysis: Louis Riel's summation to the Jury speech. Please someone help me with these questions if you can please!! I REALLY NEED THE HELP!!!??? What information is left out of his speech? What other sources of information could you\n\n48. ## chemistry\n\n15g of calcium hydroxide is dissolved in 500cm^3 of water 300cm^3 of the solution is used to neutralise hydrocloric acid of/5mol.dm^3 how do i calculate the volume of the acid\n\n49. ## Chemistry\n\nWhat is the molarity of a bleach solution containing 6.6 g of NaOCl per liter of bleach? answer please\n\n50. ## Math\n\nWhich number is NOT divisible by 9?\n\n51. ## Algebra\n\nWhich fraction is not equivalent to 6/10? 3/5 EQUIVELANT 12/20 EQUIVELANT -6/-10 - 6/10 I know the first two are equivalent to 6/10 but I'm not sure about the last two.\n\n52. ## Algebra\n\nYou have two types of bread, three types of meat, and four types of cheese. How many different sandwiches can you make if each sandwich has one type of bread, one type of meat, and one type of cheese? 9 sandwiches *My answer* 16 sandwiches 24 sandwiches 27\n\n53. ## science\n\nIf A represents atomic mass and Z represents atomic number, which of the following describes an atom after a proton is emitted from its nucleus? A. A+ 1, Z + 1 B. A-1,Z-1 C. A+ 1, Z-1 D. A-1, Z + 1 My answer is B\n\n54. ## English help\n\nWhat is the resolution and exposition in 10 things I hate about you? I need help!!\n\n55. ## Algebra\n\nWhat is the prime factorization of 180? My answer 2 x 2 x 3 x 3 x 5 Am I right?\n\n56. ## chemistry\n\nhow many grams of solid magnesium nitrate should be added to 5.0 grams of solid sodium nitrate in a 500 mL volumetric flask in order to produce an aqueous 0.25M solution of NO3- ions (once the flask is filled to the 500mL volume line)? No idea where to\n\n57. ## math\n\nI am having trouble understanding how to solve this problem. I found an answer, about 74.2%, but I am not sure if that is correct. Any help would be greatly appreciated! A chess board has 64 squares, 32 white and 32 black, and is played with 16 black and\n\n58. ## Algebra\n\nEvaluate. Write in simplest form. 6-b/3a a=10 a=-9 My answer is 2 or 5...I'm not sure\n\n59. ## Algebra\n\nx^7 * y^3 * x^8 * y^2 can you help me find the answer?\n\n60. ## Algebra\n\nWrite 0.000000474 in scientific notation.\n\n61. ## Algebra\n\nWrite 9.79 x 10^5 in standard notation.\n\n62. ## Algebra\n\nOrder 3.21 x 10, 42 x 10^4, and 0.11 x 10^10 from least to greatest\n\n63. ## Algebra (VERY IMPORTANT)\n\n2/3(4x+2)^2 +1= 9\n\n64. ## Math\n\nA cylinder is 12 cm tall and has a diameter of 3 cm. What is the total surface area? Pi x radius x radius=3.14 x 1.5 x 2=9.42 cm Pi x diameter= 3.14 x 3= 9.42cm length x width= 9.42 x 12= 113.04cm Total Surface Area= 9.42+9.42+113.04=131.8cmsquared. The\n\n65. ## science\n\nA ping pong ball and a penny are placed in water. The ping pong ball floats while the penny sinks? What property of water does this demonstrate? I think it has to do with the density. The reason the ping pong ball floats is because it has a lower density\n\n66. ## Math\n\ntriangle abc has sides that are 20 centimeters 38 centimeters and 50 centimeters in length . it is related to triangle cde by a scale factor of 0.5 what are the lengths of the sides of triangle CDE. I need help to figure this out. Thanks\n\n67. ## Math 7\n\nA photo studio offers different sized prints of the same subject. The original print measures 8 inches by 10 inches. A printer enlarges the orginal the orginal by a scale factor of 1.5, and then enlarges the second image by a scale factor of 3. What are\n\n68. ## science\n\nwhat happens to insulin when kidney failure happens?\n\n69. ## maths\n\ncalculet the volume and total surface area of aright circular cylinder of height 1m and radius 70cm.\n\n70. ## Algebra\n\nPaul needs 3 1/4 yards of fabric to make one tablecloth. How many tablecloths can he make from 20 1/2 yards of fabric. 67 7 My answer 6 66\n\n71. ## Algebra\n\nMary thinks that a quarter-pound hamburger is heavier than a 5 oz hamburger. Explain why she is incorrect.\n\n72. ## Algebra\n\nFind the least common multiple: 2b^2 and 12c^3\n\n73. ## Algebra\n\nAnita owns a beauty supply store. Every two weeks she receives a supply of shampoo. Every four weeks she receives a carton of nail polishes. Every eight weeks she receives a box of combs, and every twelve weeks she receives a box of hair dyes. If Anita\n\n74. ## maths\n\ntwo chords,AB and CD ,of acircle intersect at right angles at a point inside the circle. If m(angleBAC):35,FIND m(angleABD).\n\n75. ## computer\n\nThe merge process involves which two types of files? A. Main document and data source B. Text and merge fields C. Mail merge template and mailings D. Primary and secondary\n\n76. ## henry ford\n\nA rocket is fired at a speed of 74.0 m/s from ground level, at an angle of 68.0 ° above the horizontal. The rocket is fired toward an 49.8-m high wall, which is located 32.0 m away. The rocket attains its launch speed in a negligibly short period of time,\n\n77. ## math\n\ni have 5/6 yards of fabric to make shirts I need 1/12 yards of fabric for each chirt. How many shirts can I make? 5/6 x 1/12\n\n78. ## algebra (check my work)\n\n60% of what number is 24? a/w = p/100 24/w = 60/100 24 * 100 = 60* w 2400 / 60 = 60w / 60 w = 40 am i correct?\n\n79. ## Algebra\n\nJane and Jeff are selling cheesecakes for a school fundraiser. Customers can buy New York style cheesecakes and strawberry cheesecakes. Jane sold 14 New York Style Cheesecakes and 4 Strawberry cheesecakes for a total of \\$250. Jeff sold 7 New York Style\n\n80. ## Chemistry\n\nHow many milliliters of a 5.0 M H2SO4 stock solution would you need to prepare 111.0 mL of 0.36 M H2SO4?\n\n81. ## Statistics\n\nThe performance of students is given by the following result: 52,74,54,12,47,36,48,84,65,42,15,45,74,36, 47,53,57,32,41 and 58. If the pass mark is 50. What is the probability that a student should score 84?\n\n82. ## Algebra\n\nSimplify 3/7 - 4/m\n\n83. ## Probability\n\nLet θ be an unknown constant. Let W1,…,Wn be independent exponential random variables each with parameter 1. Let Xi=θ+Wi. What is the maximum likelihood estimate of θ based on a single observation X1=x1? Enter your answer in terms of x1 (enter as x_1)\n\n84. ## math\n\nrosa needs \\$4,000in three years to make a down payment on a car how much must she invest today is she reeceives 0.5% interest annually compounded annually.\n\n85. ## math\n\nrosa needs \\$3,000in three years to make a down payment on a car how much must she invest today is she reeceives 0.5% interest annually compounded annually.\n\n86. ## Physics/Dynamics\n\nA truck has wheels of diameter 0.9m, a weight of 8000kg, a wheel-base of 4m, and a tray height of approximately 1.2m. The truck’s weight distribution is 50% on the rear wheels and 50% on the front wheels. The truck’s front springs have a combined\n\n87. ## American Government\n\nThe voting rate among 18 - 24 year olds in 2004:\n\n88. ## mat/116\n\na student scores 74 out of 100 on a test. if the maximum on the next test is also 100, what does the student to maintain an average of 78\n\n89. ## math\n\nHello Sir, I have a question . A bird collector wants to buy 100 birds and to spend exactly \\$100. 1 Blue Bird= 7 dollars 1 Green Bird= 5 dollars 20 Yellow Birds= 1 dollar How many blue, green, and yellow birds can he buy? Please sir i'm waiting for your\n\n90. ## math\n\ncompute the compound amount and the interest on a loan of \\$19,400 compounded annually for 6 years at 12%\n\n91. ## Algebra\n\nSolve for h: h + 3/8 = - 1/4\n\n92. ## Algebra\n\nSolve for t: 15 3/4 = t + 4 5/8\n\n93. ## Algebra\n\nSolve for h: 3/7 h = 9\n\n94. ## economics\n\nthe south african government is not providing enough welfare or subsidies to the poor and underpriviledged\n\n95. ## math\n\nhow to cut a 10 ft board into 9 smaller boards\n\n96. ## earth science\n\n_____ currents assist in the transfer of energy from warmer climates to cooler climates\n\n97. ## Math\n\nExpress the given equation in rectangular coordinate equation and sketch the curve. r^2=1+sin(theta)\n\n98. ## chemistry\n\nWhy does acetone evapourate faster than water in terms of intermolecular forces\n\n99. ## Algebra\n\nOrder from least to greatest: 2.1, 3/5, -0.7, 7/10\n\n100. ## Social Studies\n\nSoutherners justified secession with the theory of Constitutional rights. Federal rights. the Union's errors State's rights. I think the answer is state's rights. Please correct me if i am wrong.\n\n101. ## algebra\n\n3 shirts and 2 hats cost \\$52. 2 shirts and 1 hat cost \\$32. Find the cost of each.\n\n102. ## Math\n\nA catapult launches a boulder with an upward velocity of 92 m/s. The height of the boulder, h, in meters after 1 seconds is given in the function h= -5t^2+92t+16. How many seconds does it take to reach maximum height? What is the boulders maximum height?\n\n103. ## Social Studies\n\nIs there a website that talks about rebuilding the twin towers?\n\n104. ## Physics\n\nA person struggles to read by holding a book at arms length, a distance of 50.0 cm. What power of reading glasses should be prescribed for him, assuming they will be placed 2.0 cm from the eye and he wanted to read at normal near point of 25.0 cm?\n\n105. ## MATH Help (NOT MULTIPLE CHOICE) PLZZ\n\nThe O’Toole’s leave their house in Canada to vacation in Mexico for a week over winter break. In their absence, their house receives an average of 8 inches of snow per day each day during the seven-day period. However, the average air temperature rises\n\n106. ## Math Thanks\n\n2 Eight children from the Schickel family are combining their savings to purchase a gaming system to connect to and play on their television. The gaming system is on sale at their favorite discount electronics store for 10 percent off of its regular price\n\n107. ## Ethics and Social respo.\n\nWhat is one of the most fundamental questions in the issue of whether or not one can ethically use stem cells from embryos for research? (Points : 1) How many embryos are there? What is the dividing time for cells in an embryo? From whom did the embryo\n\n108. ## Algebra\n\nSuppose that a virus doubles every 3 hours. If a person has 30 virus cells, how many will he have after 9 hours? I got 810,000. 30*30=900 900*900=810,000 or would it be 240 since 30*2=60 60*2=120 120*2=240\n\n109. ## criminal justice\n\nwhat are all the procedlural and substantive ways that a prosecutor and a defense attorney can resolve a case before it goes to a grand jury?\n\n110. ## Math\n\nI brought a book and paid \\$1.80 in sales tax. If the tax was 8%, what was the cost of the book. How do I solve this problem?\n\n111. ## Biology\n\nHow a radioactive element will decay over time?\n\nMarket environment assay\n\n113. ## chemistry\n\nHow would this problem be worked using dimensional analysis? Answer is 1.32g and I can work it (.900/13.6) * 20 = 1.32 But i need it in dimensional analysis and the hint was to first find volume of mercury. How many grams of alcohol with a density of 0.900\n\n114. ## math\n\nwhat is a probability of rolling a total of 5 when a pair of number cubes is rolled once\n\n115. ## algebra\n\n6 slices of pizza and 4 drinks cost \\$12.60 while 3 slices of pizza and 4 drinks cost \\$8.10. What is the cost of each?\n\n116. ## Math\n\nWhat is the equation of the parabola with a focus of (4, 5) and a vertex of (4, 2) opening upwards? y - 2 = 1/12(x - 4)^2?\n\n117. ## ap chemistry\n\na cell is based on the following reaction Fe(s)+2Fe+3(aq) -----> 3Fe+2 (aq) E cell= 1.18 V. Calculate the concentration of iron (II), in M, if the cell emf is 1.28 V. when [Fe^(3+)]= 0.5 M the answer is 0.047 M\n\n118. ## maths\n\nsolve for x if 2cos(x-30)=1, £(-90;270)\n\n119. ## Math\n\nIdentify the sampling,method. You want to determine the number of text messages students at your school send in a month. You go to the cafeteria and ask every 4th student that walks in. a- random b- systemic c- stratified d-m none of these My answer is\n\n120. ## Algebra\n\nHow can you write the expression with a rationalized denominator? ^3 sqrt 3 / ^3 sqrt 4" ]

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[ "# diff()\n\nHere’s how to use diff() function.\n\n## A .diff()\n\nDescription:\n\nGenerate a new sequence by removing from the first sub-sequence of a sequence the members that exist in the other sub-sequences of the sequence.\n\nSyntax:\n\nA .diff()\n\nNote:\n\nGenerally sequence A contains multiple sub-sequences. The function  calculates the difference between its sequence-type members to ensure that the new sequence doesn’t include any member of the other sub-sequences.\n\nThe algorithm is to compute the difference between the first sub-sequence and the second one, then compute the difference between the result and the third sub-sequence, and so on and so forth.\n\nParameters:\n\n A A sequence whose members are sequences\n\nReturn value:\n\nA new sequence created by performing difference operations on A’s first sub-sequence and the other sub-sequences.\n\nExample:\n\n A 1 =[[1,2,3,4,5],[3,7,8]].diff() [1,2,4,5] 2 =[[1,2,3],[3,2],1].diff() [] 3 =[[1,2,2,3],2].diff() [1,2,3] remove only one of the duplicate members 4 =demo.query(\"select top 2 * from EMPLOYEE\")", null, "5 =demo.query(\"select top 1 * from EMPLOYEE\")", null, "6 =[A4,A5].diff()", null, "Since A4 and A5 come from different table sequences and have different store addresses, so the same records are regarded as different members\n\nNote:\n\nIf the sub-sequences are record sequences, we can judge if they are duplicate by their store addresses.\n\nRelated functions:\n\nA.xunion()\n\n## A .diff( x )\n\nDescription:\n\nCalculate difference between the sequence-type members in a specified sequence to generate a new sequence where the first subsequence won’t contain members that exist in the other sub-sequences.\n\nSyntax:\n\nA.diff(x)\n\nNote:\n\nGenerally sequence A contains multiple sub-sequences. The function loops through each sub-sequence of A to compute expression x and get the difference between these sub-sequences, ensuring that the difference result doesn’t include any member of the other sub-sequences.\n\nThe algorithm is to compute the difference between the first sub-sequence and the second one, then compute the difference between the result and the third sub-sequence, and so on and so forth.\n\nParameters:\n\n A A sequence whose members are sequences x An expression that returns a sequence\n\nReturn value:\n\nA sequence\n\nExample:\n\n A 1 =demo.query(\"select * from EMPLOYEE where GENDER = 'M' order by NAME\") 2 =demo.query(\"select * from EMPLOYEE where GENDER = 'F' order by NAME\") 3 =[A1,A2].diff(~.(NAME)) Remove only one of the duplicate members\n\nRelated functions:" ]

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[ "# How Do You Write Numbers in Expanded Form?", null, "Kristin Lee/Getty Images\n\nTo write a number in expanded form, break apart a number, and write it as the sum of each place value. For example, 8324 = 8000 + 300 + 20 + 4. Decimals may also be expanded in a similar format: 0.239 = 0.2 + 0.03 + 0.009.\n\nExpansion may be made simpler for teaching purposes or for beginners by setting up a chart or by first expressing the number using words.\n\nThe chart method uses a chart with the various place value names, and the number is written correctly into the chart. Multiply each digit by the appropriate power of 10 for the place value, before listing all values as a sum. For instance, 4827.65 expands as such:\n\nThousands: 4 x 1000 = 4000.\n\nHundreds: 8 x 100 = 800.\n\nTens: 2 x 10 = 20.\n\nUnits: 7 x 1 = 7.\n\nTenths: 6 x 1/10 = 0.6.\n\nHundredths: 5 x 1/100 = 0.05.\n\n4000 + 800 + 20 + 7 + .6 + 0.05.\n\nFor the word method, write the number first using only words, and express each part individually. The sum is the expanded form of the number. For example, 4827.65 is written as \"four thousand, eight hundred, twenty seven, six tenths, and five-hundredths,\" resulting in 4000 + 800 + 20 + 7 + 0.6 + 0.05." ]

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[ "", null, "MORE IN Object Oriented and Multicore Programming\nSPPU Computer Engineering (Semester 4)\nObject Oriented and Multicore Programming\nDecember 2014\nTotal marks: --\nTotal time: --\nINSTRUCTIONS\n(1) Assume appropriate data and state your reasons\n(2) Marks are given to the right of every question\n(3) Draw neat diagrams wherever necessary\n\nAnswer any one question from Q1 and Q2\n1 (a) What is a friend function? How do you declare one? When is friend function compulsory? Give an example.\n6 M\n1 (b) What is virtual function? Why do we need virtual functIon When do we make a virtual function 'pure' ? What are the implications of making a function a pure virtual function?\n6 M\n\n2 (a) Write short notes on:\ni) 'this' pointer\nii) Copy constructor.\n6 M\n2 (b) What is operator overloading? Name the operators that cannot be overloaded in C++? How do you declare an overloaded stream insertion and extraction operator?\n6 M\n\nAnswer any one question from Q3 and Q4\n3 (a) Distinguish between overloaded functions and function templates. Write a function template for finding the minimum value contained in an array.\n6 M\n3 (b) Draw the state diagram for the process. Explain each process state briefly.\n6 M\n\n4 (a) What is an exception ? How is an exception handled in C++ ? What are the advantages of using exception handling mechanism in program?\n6 M\n4 (b) How can we create a child process from parent process ?Describe parent-child relation.\n6 M\n\nAnswer any one question from Q5 and Q6\n5 (a) What do you mean by threads? Write a threaded program in C++ and explain Pthread_join( ) function.\n7 M\n6 M\n\n6 (a) How can we pass command line arguments to the thread function? Determine the number of threads using command line argument.\n7 M\n6 (b) What are the types of thread? Describe in brief.\n6 M\n\nAnswer any one question from Q7 and Q8\n7 (a) What is meant by critical section? How can we manage critical section using mutex semaphor?\n6 M\n7 (b) Enlist concurrency models. Write a short note on any two.\n7 M\n\n8 (a) Can all threads in the process share global data, variables, and data structure? If yes, explain with an example.\n6 M\n8 (b) Describe the basic functions of POSIX message queue.\n7 M\n\nMore question papers from Object Oriented and Multicore Programming" ]

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[ "", null, "Download", null, "Download Presentation", null, "Equilibrium\n\n# Equilibrium\n\nTélécharger la présentation", null, "## Equilibrium\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -\n##### Presentation Transcript\n\n1. Equilibrium\n\n2. Equilibrium Some reactions (theoretically all) are reversible reactions, in which the products take part in a separate reaction to reform the reactants.\n\n3. Equilibrium We show equilibrium in a reaction using an arrow pointing both ways (  ). For example: 2NO2 (g)  N2O4 (g)\n\n4. Equilibrium • Chemical equilibrium is when the rate of the forward reaction is equal to the rate of the reverse reaction.\n\n5. Equilibrium • At equilibrium, the concentrations of reactants and products remain constant over time.\n\n6. Equilibrium The equilibrium constant expresses the relative concentrations of reactants and products at equilibrium. It is Keq.\n\n7. Equilibrium For the reaction aA + bB  cC + dD Keq =[C]c[D]d [A]a[B]b\n\n8. Equilibrium Keq =[C]c[D]d [A]a[B]b • [A],[B],[C] and [D] are molar concentrations • a, b, c and d are the coefficients in the balanced equation\n\n9. Equilibrium Keq =[C]c[D]d [A]a[B]b • When writing equilibrium constant expressions, it is customary to omit units\n\n10. Equilibrium For example: Write the equilibrium expression for this reaction: 2CO(g) + O2 (g)  2CO2(g) Keq =[CO2]2 [CO]2[O2]\n\n11. Equilibrium By plugging in values for concentration, we can determine whether a system is at equilibrium and how it might change as it heads toward equilibrium.\n\n12. Equilibrium Consider this reaction: COCl2 (g)  CO(g) + Cl2(g) Keq = 170 If the concentrations of CO and Cl2 are each 0.15 M and the concentration of COCl2 is 1.1 x 10-3 M, is the reaction at equilibrium? If not, in which direction will it proceed?\n\n13. Equilibrium Set up the equilibrium expression and plug in values: Keq =[CO][Cl2]=(0.15) (0.15) [COCl2] (1.1 x 10-3)\n\n14. Equilibrium Q = 20, which is not the same as 170, so the system is not at equilibrium. To be at equilibrium, we would like to have it get bigger.\n\n15. Equilibrium If Q is to get bigger, we would need more CO and Cl2, so the reaction needs to shift to the right." ]

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[ "# Pandas和Cassandra：numpy数组格式不兼容\n\nNumpyProtocolHander：将结果直接反序列化为NumPy数组 . 这有助于与分析工具包（如Pandas）的高效集成 .\n\n``````<class 'cassandra.cluster.ResultSet'>\n``````\n\n``````{u'reversals_rejected': array([0, 0]), u'revenue': array([ 0, 10]), u'reversals_revenue': array([0, 0]), u'rejected': array([3, 1]), u'impressions_positive': array([3, 3]), u'site_user_id': array([226226, 354608], dtype=int32), u'error': array([0, 0]), u'impressions_negative': array([0, 0]), u'accepted': array([0, 2])}\n``````\n\n（我限制了查询结果，我正在使用更大量的数据 - 因此想要使用numpy和pandas） .\n\n``````rslt = cassandraSession.execute(\"SELECT accepted FROM table\")\n\n``````\n\n``````Traceback (most recent call last):\nFile \"/UserStats.py\", line 27, in <module>\nFile \"cassandra/cluster.py\", line 3380, in cassandra.cluster.ResultSet.__getitem__ (cassandra/cluster.c:63998)\nTypeError: list indices must be integers, not list\n``````\n\n### 回答(1)\n\n``````df = pd.DataFrame(rslt)\n``````\n\nrslt 为Python dict提供数据，可以轻松转换为Pandas数据帧 .\n\n``````import pandas as pd\nfrom cassandra.cluster import Cluster\nfrom cassandra.protocol import NumpyProtocolHandler\nfrom cassandra.query import tuple_factory\n\ncluster = Cluster(\ncontact_points=['your_ip'],\n)\nsession = cluster.connect('your_keyspace')\nsession.row_factory = tuple_factory\nsession.client_protocol_handler = NumpyProtocolHandler\n\nprepared_stmt = session.prepare ( \"SELECT * FROM ... WHERE ...;\")\nbound_stmt = prepared_stmt.bind([...])\nrslt = session.execute(bound_stmt)\ndf = pd.DataFrame(rslt)\n``````\n\nNote: 如果查询很大，上述解决方案只会为您提供部分数据 . 所以你应该这样做：\n\n``````df = pd.DataFrame()\nfor r in rslt:\ndf = df.append(r)\n``````" ]

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[ "# Excel CSV輔助工具強化版－支援換行符號\n\n1. CSV內含日文，使用Shift-JIS編碼（ANSI）而非UTF8，當初將所有ANSI檔案視為BIG5，形成亂碼\n2. 部分欄位內容夾帶換行符號（如黃底所示），擾亂原本以\"\\r\\n\"分隔資料列的解析邏輯\n3. 程式未考慮CSV部分欄位自帶雙引號的情況，造成雙引號重複（=\"\"…\"\"）。", null, "", null, "", null, "` using (StreamReader sr = new StreamReader(fn, encoding, true))` ` {` ` StringBuilder sb = new StringBuilder();` ` bool quotMarkMode = false;` ` string newLineReplacement = \"\\x07\";` ` string commaReplacement = \"\\x08\";` ` //支援CSV雙引號內含換行符號規則，採逐字讀入解析` ` //雙引號內如需表示\", 使用\"\"代替` ` while (sr.Peek() >= 0)` ` {` ` var ch = (char)sr.Read();` ` if (quotMarkMode)` ` {` ` //雙引號包含區段內遇到雙引號有兩種情境` ` if (ch == '\"')` ` {` ` //連續兩個雙引號，為欄位內雙引號字元` ` if (sr.Peek() == '\"')` ` sb.Append((char)sr.Read());` ` //遇到結尾雙引號，雙引號包夾模式結束` ` else` ` quotMarkMode = false;` ` sb.Append(ch);` ` }` ` //雙引號內遇到換行符號，先置換成特殊字元，稍後換回` ` else if (ch == '\\r' && sr.Peek() == '\\n')` ` {` ` sr.Read();` ` sb.Append(newLineReplacement);` ` }` ` //雙引號內遇到逗號，先置換成特殊字元，稍後換回` ` else if (ch == ',')` ` sb.Append(commaReplacement);` ` //否則，正常插入字元` ` else ` ` sb.Append(ch);` ` }` ` else` ` {` ` sb.Append(ch);` ` if (ch == '\"') quotMarkMode = true;` ` }` ` }` ` var fixedCsv = sb.ToString();` ` sb.Length = 0;` ` string line;` ` using (var lr = new StringReader(fixedCsv))` ` {` ` while ((line = lr.ReadLine()) != null)` ` {` ` string[] p = line.Split(',');` ` sb.AppendLine(string.Join(\",\",` ` //若欄位以0起首，重新組裝成=\"....\"格式` ` p.Select(o => ` ` o.StartsWith(\"0\") ? ` ` string.Format(\"=\\\"{0}\\\"\", o) : ` ` //還原換行符號及逗號` ` o.StartsWith(\"\\\"\") ? ` ` o.Replace(newLineReplacement, \"\\r\\n\")` ` .Replace(commaReplacement, \",\") : o` ` ).ToArray()));` ` }` ` }` ` ` ` //調整結果另存為同目錄下*.fixed.csv檔` ` string fixedFile = Path.Combine(` ` Path.GetDirectoryName(fn), ` ` Path.GetFileNameWithoutExtension(fn) + \".fixed.csv\");` ` //一律存為UTF8` ` File.WriteAllText(fixedFile, sb.ToString(), Encoding.UTF8);` ` //開啟CSV` ` Process proc = new Process();` ` proc.StartInfo = new ProcessStartInfo(fixedFile);` ` proc.Start();` ` }`", null, "", null, "", null, "### # 2015-08-09 11:07 PM by Jeffrey\n\nto Fish, 恭喜尋獲「錯字彩蛋」！（感謝指正）" ]

[ null, "https://blog.darkthread.net/img/loading.svg", null, "https://blog.darkthread.net/img/loading.svg", null, "https://blog.darkthread.net/img/loading.svg", null, "https://blog.darkthread.net/img/loading.svg", null, "https://blog.darkthread.net/img/loading.svg", null, "https://blog.darkthread.net/img/loading.svg", null ]

[ "Get instant live expert help with Excel or Google Sheets", null, "“My Excelchat expert helped me in less than 20 minutes, saving me what would have been 5 hours of work!”\n\n#### Post your problem and you'll get expert help in seconds\n\nYour message must be at least 40 characters\nOur professional experts are available now. Your privacy is guaranteed.\n\n# Calculate years between dates\n\nWe often compare two dates and calculate the difference between them in terms of days, weeks, months and years. In Excel, we can calculate years between dates using specialized functions like YEARFRAC and DATEDIF functions. This step by step tutorial will help all levels of Excel users calculate the number of years between two dates.\n\nFigure 1. Calculating Years Between Dates\n\n## Using YEARFRAC Function to Calculate Years Between Dates\n\nThe YEARFRAC function calculates years between two dates, including the fractional value. The decimal number represents the fraction of the year between the dates. The formula to get the difference between dates in terms of years, including fraction, is;\n\n`=YEARFRAC(B2, C2)`\n\nApply this formula in cell D2 and drag the fill handle to get the other calculations.\n\nFigure 2. Applying the YEARFRAC Function\n\nTo round the decimal number or fractional value to nearest whole number we can wrap the above formula in the ROUND function, such as;\n\n`=ROUND(YEARFRAC(B2,C2),0)`\n\nFigure 3. Rounding the Fractional Value to Nearest Whole Number\n\nIf we want to return only the integer part of the year’s value without fractional value, then we need to wrap the YEARFRAC function inside the INT function, such as;\n\n`=INT(YEARFRAC(B2,C2))`\n\nFigure 4. Getting the Year Value Without Fractional Value\n\n## Using DATEDIF Function to Calculate Years Between Dates\n\nThe DATEDIF function calculates the number of years between two dates when the interval argument is specified as “Y” as per its syntax, such as;\n\n`=DATEDIF( start_date, end_date, “Y” )`\n\nWhen we apply this function with “Y” interval then it calculates years between two dates a complete calendar year without fractional value. As per our example, we need to apply the following DATEDIF formula in cell D2, and copy or drag down the fill handle to other cells;\n\n`=DATEDIF(B2,C2,\"Y\")`\n\nFigure 5. Calculating Years Between Dates with DATEDIF Function\n\n## Instant Connection to an Expert through our Excelchat Service\n\nMost of the time, the problem you will need to solve will be more complex than a simple application of a formula or function. If you want to save hours of research and frustration, try our live Excelchat service! Our Excel Experts are available 24/7 to answer any Excel question you may have. We guarantee a connection within 30 seconds and a customized solution within 20 minutes.\n\n### Did this post not answer your question? Get a solution from connecting with the expert.", null, "" ]

[ null, "https://www.got-it.ai/solutions/excel-chat/wp-content/themes/seocms/assets/images/seo-avatars/user-52.jpg", null, "https://secure.gravatar.com/avatar/", null ]

[ "## Section13.4Atoms of a Boolean Algebra\n\nIn this section we will look more closely at something we've hinted at, which is that every finite Boolean algebra is isomorphic to an algebra of sets. We will show that every finite Boolean algebra has $2^n$ elements for some $n$ with precisely $n$ generators, called atoms.\n\nConsider the Boolean algebra $[B; \\lor , \\land, \\bar{}{\\hspace{3 pt}}]\\text{,}$ whose ordering diagram is depicted in Figure 13.4.1\n\nWe note that $1 = a_1 \\lor a_2 \\lor a_3\\text{,}$ $b_1 = a_1 \\lor a_2\\text{,}$ $b_2 = a_1 \\lor a_3\\text{,}$ and $b_3 = a_2 \\lor a_3\\text{;}$ that is, each of the elements above level one can be described completely and uniquely in terms of the elements on level one. The $a_i$'s have uniquely generated the non-least elements of $B$ much like a basis in linear algebra generates the elements in a vector space. We also note that the $a_i$'s are the immediate successors of the minimum element, 0. In any Boolean algebra, the immediate successors of the minimum element are called atoms. For example, let $A$ be any nonempty set. In the Boolean algebra $[\\mathcal{P}(A); \\cup , \\cap, \\hspace{1 mm}^c]$ (over $\\subseteq$), the singleton sets are the generators, or atoms, of the algebraic structure since each element $\\mathcal{P}(A)$ can be described completely and uniquely as the join, or union, of singleton sets.\n\n###### Definition13.4.2.Atom.\n\nA non-least element $a$ in a Boolean algebra $[B; \\lor , \\land, \\bar{}\\hspace{3 pt}]$ is called an atom if for every $x \\in B\\text{,}$ $x \\land a = a$ or $x \\land a = 0\\text{.}$\n\nThe condition that $x \\land a = a$ tells us that $x$ is a successor of $a\\text{;}$ that is, $a \\preceq x\\text{,}$ as depicted in Figure 13.4.3(a)\n\nThe condition $x \\land a = 0$ is true only when $x$ and $a$ are “not connected.” This occurs when $x$ is another atom or if $x$ is a successor of atoms different from $a\\text{,}$ as depicted in Figure 13.4.3(b).\n\nAn alternate definition of an atom is based on the concept of “covering.”\n\n###### Definition13.4.4.The Covering Relation.\n\nGiven a Boolean algebra $[B; \\lor , \\land, \\bar{}\\hspace{3 pt}]\\text{,}$ let $x, z \\in B\\text{.}$ We say that $z$ covers $x$ iff $x \\prec z$ and there does not exist $y \\in B$ with $x \\prec y \\prec z\\text{.}$\n\nIt can be proven that the atoms of Boolean algebra are precisely those elements that cover the zero element.\n\nThe set of atoms of the Boolean algebra $\\left[D_{30}; \\lor , \\land, \\bar{\\hspace{5 mm}}\\right]$ is $M = \\{2, 3, 5\\}\\text{.}$ To see that $a = 2$ is an atom, let $x$ be any non-least element of $D_{30}$ and note that one of the two conditions $x \\land 2 = 2$ or $x \\land 2 = 1$ holds. Of course, to apply the definition to this Boolean algebra, we must remind ourselves that in this case the 0-element is 1, the operation $\\land$ is greatest common divisor, and the poset relation is “divides.” So if $x = 10\\text{,}$ we have $10 \\land 2 = 2$ (or $2 \\mid 10$), so Condition 1 holds. If $x = 15\\text{,}$ the first condition is not true. (Why?) However, Condition 2, $15 \\land 2 = 1\\text{,}$ is true. The reader is encouraged to show that 3 and 5 also satisfy the definition of an atom. Next, if we should compute the join (the least common multiple in this case) of all possible combinations of the atoms 2, 3, and 5 to generate all nonzero (non-1 in this case) elements of $D_{30}\\text{.}$ For example, $2 \\lor 3 \\lor 5 = 30$ and $2 \\lor 5 = 10\\text{.}$ We state this concept formally in the following theorem, which we give without proof.\n\nThe least element in relation to this theorem bears noting. If we consider the empty set of atoms, we would consider the join of elements in the empty set to be the least element. This makes the statement of the theorem above a bit more tidy since we don't need to qualify what elements can be generated from atoms.\n\nWe now ask ourselves if we can be more definitive about the structure of different Boolean algebras of a given order. Certainly, the Boolean algebras $\\left[D_{30}; \\lor , \\land, \\land\\bar{\\hspace{5 mm}} \\right]$ and $[\\mathcal{P}(A); \\cup , \\cap, \\hspace{1 mm}^c]$ have the same graph (that of Figure 13.4.1), the same number of atoms, and, in all respects, look the same except for the names of the elements and the operations. In fact, when we apply corresponding operations to corresponding elements, we obtain corresponding results. We know from Chapter 11 that this means that the two structures are isomorphic as Boolean algebras. Furthermore, the graphs of these examples are exactly the same as that of Figure 13.4.1, which is an arbitrary Boolean algebra of order $8 = 2^3\\text{.}$\n\nIn these examples of a Boolean algebra of order 8, we note that each had 3 atoms and $2^3 = 8$ number of elements, and all were isomorphic to $[\\mathcal{P}(A ); \\cup , \\cap, \\hspace{1 mm}^c]\\text{,}$ where $A = \\{a, b, c\\}\\text{.}$ This leads us to the following questions:\n\n• Are there any different (nonisomorphic) Boolean algebras of order 8?\n\n• What is the relationship, if any, between finite Boolean algebras and their atoms?\n\n• How many different (nonisomorphic) Boolean algebras are there of order 2? Order 3? Order 4? etc.\n\nThe answers to these questions are given in the following theorem and corollaries.\n\nAn isomorphism that serves to prove this theorem is $T:\\mathcal{P}(A) \\to B$ defined by $T(S)= \\bigvee_{a \\in S}{a}\\text{,}$ where $T(\\emptyset)$ is interpreted as the zero of $\\mathcal{B}\\text{.}$ We leave it to the reader to prove that this is indeed an isomorphism.\n\nLet $A$ be the set of all atoms of $\\mathcal{B}$ and let $\\left| A\\right| = n\\text{.}$ Then there are exactly $2^n$ elements (subsets) in $\\mathcal{P}(A)\\text{,}$and by Theorem 13.4.6, $[B; \\lor, \\land, \\bar{\\hspace{5 mm}} ]$ is isomorphic to $[\\mathcal{P}(A); \\cup , \\cap \\hspace{1 mm}^c]$ and must also have $2^n$ elements.\n\nEvery Boolean algebra of order $2^n$ is isomorphic to $[\\mathcal{P}(A); \\cup , \\cap, \\hspace{1 mm}^c ]$ when $\\lvert A \\rvert= n\\text{.}$ Hence, if $\\mathcal{B}_1$ and $\\mathcal{B}_2$ each have $2^n$ elements, they each have $n$ atoms. Suppose their sets of atoms are $A_1$ and $A_2\\text{,}$ respectively. We know there are isomorphisms $T_1$ and $T_2\\text{,}$ where $T_i:\\mathcal{B}_i \\to \\mathcal{P}(A_i)\\text{,}$ $i=1,2\\text{.}$ In addition we have an isomorphism, $N$ from $\\mathcal{P}(A_1)$ into $\\mathcal{P}(A_2)\\text{,}$ which we ask you to prove in Exercise 13.4.1.9. We can combine these isomorphisms to produce the isomorphism $T_{2}^{-1}\\circ N \\circ T_1:\\mathcal{B}_1 \\to \\mathcal{B}_2\\text{,}$ which proves the corollary.\n\nThe above theorem and corollaries tell us that we can only have finite Boolean algebras of orders $2^1, 2^2, 2^3,. . , 2^n\\text{,}$ and that all finite Boolean algebras of any given order are isomorphic. These are powerful tools in determining the structure of finite Boolean algebras. In the next section, we will discuss one of the easiest ways of describing a Boolean algebra of any given order.\n\n### Exercises13.4.1Exercises\n\n###### 1.\n1. Show that $a = 2$ is an atom of the Boolean algebra $\\left[D_{30}; \\lor , \\land, - \\right]\\text{.}$\n\n2. Repeat part a for the elements 3 and 5 of $D_{30}\\text{.}$\n\n3. Verify Theorem 13.4.5 for the Boolean algebra $\\left[D_{30}; \\lor , \\land, - \\right]\\text{.}$\n\n1. For $a = 3$ we must show that for each $x \\in D_{30}$ one of the following is true: $x\\land 3=3$ or $x\\land 3=1\\text{.}$ We do this through the following table:\n\n\\begin{equation*} \\begin{array}{cc} x & \\textrm{ verification} \\\\ \\hline \\begin{array}{c} 1 \\\\ 2 \\\\ 3 \\\\ 5 \\\\ 6 \\\\ 10 \\\\ 15 \\\\ 30 \\\\ \\end{array} & \\begin{array}{c} 1\\land 3=1 \\\\ 2\\land 3=1 \\\\ 3\\land 3=3 \\\\ 5\\land 3=1 \\\\ 6\\land 3=3 \\\\ 20\\land 3=1 \\\\ 15\\land 3=3 \\\\ 30\\land 3=3 \\\\ \\end{array} \\\\ \\end{array} \\end{equation*}\n\nFor $a=5\\text{,}$ a similar verification can be performed.\n\n2. $6 = 2 \\lor 3\\text{,}$ $10 = 2 \\lor 5\\text{,}$ $15 = 3 \\lor 5\\text{,}$ and $30 = 2 \\lor 3 \\lor 5\\text{.}$\n\n###### 2.\n\nLet $A = \\{a, b, c\\}\\text{.}$\n\n1. Rewrite the definition of atom for $[\\mathcal{P}(A); \\cup , \\cap, c ]\\text{.}$ What does $a \\leq x$ mean in this example?\n\n2. Find all atoms of $[\\mathcal{P}(A); \\cup , \\cap, c ]\\text{.}$\n\n3. Verify Theorem 13.4.5 for $[\\mathcal{P}(A); c, \\cup , \\cap ]\\text{.}$\n\n###### 3.\n\nVerify Theorem 13.4.6 and its corollaries for the Boolean algebras in Exercises 1 and 2 of this section.\n\nIf $B = D_{30}\\textrm{ }$ 30 then $A = \\{2, 3, 5\\}$ and $D_{30}$ is isomorphic to $\\mathcal{P}(A)\\text{,}$ where $\\begin{array}{cc} 1\\leftrightarrow \\emptyset \\textrm{ } & 5\\leftrightarrow \\{5\\} \\\\ 2\\leftrightarrow \\{2\\}\\textrm{ } & 10\\leftrightarrow \\{2,5\\} \\\\ 3\\leftrightarrow \\{3\\}\\textrm{ } & 15\\leftrightarrow \\{3,5\\} \\\\ 6\\leftrightarrow \\{2,3\\}\\textrm{ } & 30\\leftrightarrow \\{2,3,5\\} \\\\ \\end{array}$ and $\\begin{array}{c} \\textrm{ Join} \\leftrightarrow \\textrm{ Union} \\\\ \\textrm{ Meet}\\leftrightarrow \\textrm{ Intersection} \\\\ \\textrm{ Complement}\\leftrightarrow \\textrm{ Set} \\textrm{ Complement} \\\\ \\end{array}$\n\n###### 4.\n\nGive an example of a Boolean algebra of order 16 whose elements are certain subsets of the set $\\{1, 2, 3, 4, 5, 6, 7\\}$\n\n###### 5.\n\nCorollary 13.4.7 implies that there do not exist Boolean algebras of orders 3, 5, 6, 7, 9, etc. (orders different from $2^n$). Without this corollary, directly show that we cannot have a Boolean algebra of order 3.\n\nHint\n\nAssume that $[B; \\lor , \\land, - ]$ is a Boolean algebra of order 3 where $B = \\{0, x, 1\\}$ and show that this cannot happen by investigating the possibilities for its operation tables.\n\nAssume that $x \\neq 0 \\textrm{ or } 1$ is the third element of a Boolean algebra. Then there is only one possible set of tables for join and meet, all following from required properties of the Boolean algebra.\n\n\\begin{equation*} \\begin{array}{lr} \\begin{array}{c|c} \\lor & \\begin{array}{ccc} 0 & x & 1 \\\\ \\end{array} \\\\ \\hline \\begin{array}{c} 0 \\\\ x \\\\ 1 \\\\ \\end{array} & \\begin{array}{ccc} 0 & x & 1 \\\\ x & x & 1 \\\\ 1 & 1 & 1 \\\\ \\end{array} \\\\ \\end{array} & \\begin{array}{c|c} \\land & \\begin{array}{ccc} 0 & x & 1 \\\\ \\end{array} \\\\ \\hline \\begin{array}{c} 0 \\\\ x \\\\ 1 \\\\ \\end{array} & \\begin{array}{ccc} 0 & 0 & 0 \\\\ 0 & x & x \\\\ 0 & x & 1 \\\\ \\end{array} \\\\ \\end{array} \\end{array} \\end{equation*}\n\nNext, to find the complement of $x$ we want $y$ such that $x \\land y = 0$ and $x \\lor y = 1\\text{.}$ No element satisfies both conditions; hence the lattice is not complemented and cannot be a Boolean algebra. The lack of a complement can also be seen from the ordering diagram from which $\\land$ and $\\lor$ must be derived.\n\n###### 6.\n1. There are many different, yet isomorphic, Boolean algebras with two elements. Describe one such Boolean algebra that is derived from a power set, $\\mathcal{P}(A)\\text{,}$ under $\\subseteq\\text{.}$ Describe a second that is described from $D_n\\text{,}$ for some $n \\in P\\text{,}$ under “divides.”\n\n2. Since the elements of a two-element Boolean algebra must be the greatest and least elements, 1 and 0, the tables for the operations on $\\{0, 1\\}$ are determined by the Boolean algebra laws. Write out the operation tables for $[\\{0, 1\\}; \\lor , \\land, -]\\text{.}$\n\n###### 7.\n\nFind a Boolean algebra with a countably infinite number of elements.\n\nLet $X$ be any countably infinite set, such as the integers. A subset of $X$ is cofinite if it is finite or its complement is finite. The set of all cofinite subsets of $X$ is:\n\n1. Countably infinite - this might not be obvious, but here is a hint. Assume $X=\\left\\{x_0,x_1,x_2,\\ldots \\right\\}\\text{.}$ For each finite subset $A$ of $X\\text{,}$ map that set to the integer $\\sum _{i=0}^{\\infty } \\chi _A \\left(x_i\\right)2^i$ You can do a similar thing to sets that have a finite complement, but map them to negative integers. Only one minor adjustment needs to be made to accommodate both the empty set and $X\\text{.}$\n\n2. Closed under union\n\n3. Closed under intersection, and\n\n4. Closed under complementation.\n\nTherefore, if $B =\\{A \\subseteq X : A \\textrm{ is cofinite}\\}\\text{,}$ then $B$ is a countable Boolean algebra under the usual set operations.\n\n###### 8.\n\nProve that the direct product of two Boolean algebras is a Boolean algebra.\n\nHint\n\n“Copy” the corresponding proof for groups in Section 11.6.\n\n###### 9.\n\nProve if two finite sets $A_1$ and $A_2$ both have $n$ elements then $[\\mathcal{P}(A_1); \\cup , \\cap, \\hspace{1 mm}^c]$ is isomorphic to $[\\mathcal{P}(A_2); \\cup , \\cap, \\hspace{1 mm}^c]$\n\n###### 10.\n\nProve an element of a Boolean algebra is an atom if and only if it covers the zero element." ]

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[ "# Temperature Coefficient of Resistance : Formula and Measuring Method\n\nIn electrical or electronic engineering, when the flow of current supplies through a wire then it gets heat because of the wire’s resistance. In perfect condition, resistance must be ‘0’ however that doesn’t take place. When the wire gets heat up, then the wire resistance changes according to the temperature. Even though it is preferred that resistance must stay stable & it must be independent for the temperature. So, the resistance change for every degree change within temperature is termed as the temperature coefficient of resistance (TCR). Generally, it is denoted with a symbol alpha (α). The TCR of the pure metal is positive because when the temperature increases then resistance will be increased. Therefore, to make highly accurate resistances wherever resistance does not modify alloys is necessary.\n\n## What is the Temperature Coefficient of Resistance (TCR)?\n\nWe know that there are many materials and they have some resistance. The resistance of material changes based on the variation of temperature. The main relation between the modify in temperature & modification in resistance can be given by the parameter called TCR (temperature coefficient of resistance). It is signified with the symbol α (alpha).\n\nBased on the material obtainable, TCR is separated into two types such as a positive temperature coefficient of resistance (PTCR) and a negative temperature coefficient of resistance (NTCR).\n\nIn PTCR, when the temperature is increased, then the material resistance will be increased. For instance, in conductors when the temperature increases then the resistance also increases. For the alloys like constantan & manganin, the resistance is pretty low over a particular temperature range. For semiconductors such as insulators (rubber, wood), silicon & germanium & electrolytes. the resistance reduces then the temperature will be increased thus they have negative TCR.\n\nIn metallic conductors, when the temperature increases then the resistance will be increased due to the following factors which include the following.\n\n• Straightly on the early resistance\n• Rise of temperature.\n• Based on the life of the material.\n\n### The Formula for Temperature Coefficient of Resistance\n\nThe conductor resistance can be calculated at any specified temperature from the temperature data, it’s TCR, its resistance at the typical temperature & the operation of temperature. In general term, the temperature coefficient of the resistance formula can be expressed as\n\nR = Rref (1+α (T−Tref))\n\nWhere\n\n‘R’ is the resistance at ‘T’ temperature\n\n‘Rref’ is the resistance at ‘Tref ‘temperature\n\n‘α’ is the TCR of the material\n\n‘T’ is the temperature of the material in ° Celsius\n\n‘Tref’ is the reference temperature used for which the coefficient of temperature is stated.\n\nThe SI unit of the temperature coefficient of resistivity is per degree celsius or ( /°C)\n\nThe unit of the temperature coefficient of resistance is ° Celsius\n\nNormally, the TCR (temperature coefficient of resistance) is consistent with a 20°C temperature. So normally this temperature is taken as normal room temperature. Thus the temperature coefficient of resistance derivation normally takes this into the description:\n\nR = R20 (1+α20 (T−20) )\n\nWhere\n\n‘R20’ is the resistance at 20°C\n\n‘α20’ is the TCR at 20°C\n\nThe TCR of resistors is positive, negative otherwise constant over a fixed range of temperature. Selecting the correct resistor could stop the need for temperature compensation. A large TCR is required to measure temperature in some applications. Resistors intended for these applications are known as thermistors, which have a PTC (positive temperature coefficient of resistance) or NTC (negative temperature coefficient of resistance).\n\n#### Positive Temperature Coefficient of Resistance\n\nA PTC refers to some materials which experience once their temperature raised then electrical resistance also increased. The materials which have a higher coefficient then show a quick rise with temperature. A PTC material is designed to achieve the utmost temperature used for a given i/p voltage because at a particular point when the temperature increases then electrical resistance will be increased. The positive temperature coefficient of resistance materials is self-limiting naturally not like NTC materials or linear resistance heating. Some of the materials like PTC rubber also have exponentially rising temperature coefficient\n\n#### Negative Temperature Coefficient of Resistance\n\nAn NTC refers to some materials which experience once their temperature raised then electrical resistance will be decreased. The materials which have a lower coefficient then they show a quick decrease with temperature. NTC materials are mainly used for making current limiters, thermistors and temperature sensors.\n\n#### Measuring Method of TCR\n\nThe TCR of a resistor can be decided through calculating the resistance values over a suitable range of temperatures. The TCR can be measured when the normal slope of the resistance value is above this interval. For linear relations, this is precise as the temperature coefficient of the resistance is stable at each temperature. But, there are several materials that have a coefficient like non-linear. For example, a Nichrome is a popular alloy used for resistors, and the main relation among the TCR and temperature is not linear.\n\nAs the TCR is measured like normal slope thus, it is very significant to identify the interval of TCR & the temperature. The TCR can be calculated using a standardized method like MIL-STD-202 technique for the range of temperature from -55°C to 25°C and 25°C to 125°C. Because the maximum calculated value is identified as TCR. This technique frequently effects above indicating a resistor intended for low demanding applications.\n\n### Temperature Coefficient of Resistance for Some Materials\n\nThe TCR of some materials at 20°C temperature is listed below.\n\n• For Silver (Ag) material, the TCR is 0.0038°C\n• For Copper (Cu) material, the TCR is 0.00386°C\n• For Gold (Au) material, the TCR is 0.0034°C\n• For Aluminum (Al) material, the TCR is 0.00429°C\n• For Tungsten (W) material, the TCR is 0.0045°C\n• For Iron (Fe) material, the TCR is 0.00651°C\n• For Platinum (Pt) material, the TCR is 0.003927°C\n• For Manganin (Cu = 84% + Mn = 12% + Ni = 4%) material, the TCR is 0.000002°C\n• For Mercury (Hg) material, the TCR is 0.0009°C\n• For Nichrome (Ni = 60% + Cr = 15% + Fe = 25%) material, the TCR is 0.0004°C\n• For Constantan (Cu = 55% + Ni = 45%) material, the TCR is 0.00003°C\n• For Carbon (C) material, the TCR is – 0.0005°C\n• For Germanium (Ge) material, the TCR is – 0.05°C\n• For Silicon (Si) material, the TCR is – 0.07°C\n• For Brass (Cu = 50 – 65% + Zn = 50 – 35%) material, the TCR is 0.0015°C\n• For Nickel (Ni) material, the TCR is 0.00641°C\n• For Tin (Sn) material, the TCR is 0.0042°C\n• For Zinc (Zn) material, the TCR is 0.0037°C\n• For Manganese (Mn) material, the TCR is 0.00001°C\n• For Tantalum (Ta) material, the TCR is 0.0033°C\n\n### TCR Experiment\n\nThe temperature coefficient of the resistance experiment is explained below.\n\nObjective\n\nThe main objective of this experiment is to discover the TCR of a given coil.\n\nApparatus\n\nThe apparatus of this experiment mainly include connecting wires, Carey foster bridge, resistance box, lead accumulator, one-way key, unknown low resistor, jockey, galvanometer, etc.\n\nDescription\n\nA Carey foster bridge is mainly similar to a meter bridge because this bridge can be designed with 4 resistances like P, Q, R & X and these are connected to each other.\n\nIn the above Whetstone’s bridge, the galvanometer (G), a lead accumulator (E) & the keys of the galvanometer and accumulator are K1&K respectively.\n\nIf the resistance values are changed then there is no flow current through the ‘G’ and the unknown resistance can be determined by any of three known resistances like P, Q, R & X. The following relationship is used to determine the unknown resistance.\n\nP/Q =R/X\n\nThe Carey foster bridge can be used to calculate the disparity among two almost equal resistances & knowing the one value, the other value can be calculated. In this kind of bridge, the last resistances are removed in computation. It is a benefit and thus it can easily use to calculate a known resistance.\n\nThe equal resistances like P & Q are connected in the internal gaps 2 & 3, the typical resistance ‘R’ can be connected within gap1 & the ‘X’ (unknown resistance) is connected within the gap4. The ED is the balancing length which can be calculated from the ‘E’ end. According to the Whetstone Bridge principle\n\nP/Q = R + a + l1ρ/X + b + (100- l1)ρ\n\nIn the above equation, a & b are the end modifications at the E & F end & is the resistance for the length of every unit in bridge wire. If this testing is continual by changing X & R, the balancing length ‘l2’ is calculated from the end E.\n\nP/Q = X + a + 12 ρ/ R + b +(100-12) ρ\n\nFrom the above two equations,\n\nX = R + ρ (11 -12)\n\nLet l1 & l2 are the balancing lengths once the above testing is done through a typical resistance ‘r’ instead of ‘R’ & instead of X, a broad copper strip of ‘0’resistance.\n\n0 = r + ρ (11’ -12’) or ρ = r/11’ -12’\n\nIf the coil resistances are X1 & X2 at the temperatures like t1oc & t2oc, then the TCR is\n\nΑ = X2 – X1/(X1t2 – X2t1)\n\nAnd also if the coil resistances are X0 & X100 at the temperatures like 0oc & 100oc, then the TCR is\n\nΑ = X100 – X0/(X0 x 100)\n\nThus, this is all about the temperature coefficient of resistance. From the above information finally, we can conclude that this is the calculation of modify in any substance of electrical resistance for every level of temperature change. Here is a question for you, what is the unit of the temperature coefficient of resistance?" ]

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[ "", null, "### X and Y are partners in a business. X contributed $\\Large\\frac{1}{3}$ of the capital for 9 months and Y received $\\Large\\frac{2}{5}$ of the profit. For how long was Y’s money used in the business ?\n\nA. 2 months B. 3 months C. 4 months D. 5 months Answer: Option B\n\n### Solution(By Apex Team)\n\n\\begin{aligned}&\\text{Let the total profit be Rs. x}\\\\ &\\text{Then,}\\\\ &\\text{Y’s share = Rs.}\\frac{2z}{5}\\\\ &\\text{X’s share = Rs.}\\left(z-\\frac{2z}{5}\\right)\\\\ &=\\text{Rs.}\\ \\frac{3z}{5}\\\\ &\\therefore\\mathrm{X}:\\mathrm{Y}&\\\\ &=\\frac{3z}{5}:\\frac{2z}{5}&\\\\ &=3:2&\\end{aligned} Let the total capital be Rs. x and Suppose Y’s money was used for y months Then, $\\begin{array}{l} \\Rightarrow \\Large\\frac{\\frac{1}{3} x \\times 9}{\\frac{2}{3} x \\times y}=\\frac{3}{2} \\\\ \\Rightarrow 18 x=6 x y \\\\ \\Rightarrow y=3 \\end{array}$ Hence, Y’s money was used for 3 months.\n\nA. 5 : 7 : 8\nB. 20 : 49 : 64\nC. 38 : 28 : 21\nD. None of these\n\nA. Rs. 4000\nB. Rs. 5000\nC. Rs. 6000\nD. Rs. 7000\n\nA. Rs. 2380\nB. Rs. 2300\nC. Rs. 2280\nD. Rs. 2260" ]

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[ " Greg Maxwell：CSW再次伪造中本聪签名 | 北竹林\n\nGreg Maxwell：CSW再次伪造中本聪签名", null, "“如果有人想要发布一堆神秘的ECDSA签名，使得公众认为这些签名是来自比特币的创造者，以此破坏比特币市场，从人们那里榨取钱财，或以其他方式说服人们听取他的意见，他们要怎样才能实现呢？”\n\nMaxwell自答道：\n\n“不幸的是，鉴于公众对密码学的理解有限，这显然是一件容易造成欺诈的事情。", null, "[同样的漏洞也被带到了BCH中的原始OP_DSV操作码中，它最初没有对输入数据进行哈希处理，而是将其留给了用户。但我上报了它，并且在部署之前，开发者们似乎已修复了它。]\n\nF = FiniteField (0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F)\nC = EllipticCurve ([F (0), F (7)])\nG = C.lift_x(0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798)\nN = FiniteField (C.order())\nP = P=-C.lift_x(0x11db93e1dcdb8a016b49840f8c53bc1eb68a382e97b1482ecad7b148a6909a5c) # block 9 coinbase payout key.\ndef forge(c, a=-1): # Create a forged 'ECDSA' (hashless) signature\n# set a to something other than -1 to be less obvious\na = N(a)\nR = c*G + int(a)*P\ns = N(int(R.xy()))/a\nm = N(c)*N(int(R.xy()))/a\nprint 'hash1 = %d'%m\nprint 'r1 = %d'%(int(R.xy()))\nprint 's1 = %d'%s\nfor c in range(1,10):\nforge(c)\n\nhash1 = 25292222169426362969760742810503101183086560848420849767135309758511048414376\nr1 = 61518691557461623794232686914770715342344584505217074682876722883231084339701\ns1 = 54273397679854571629338298093917192510492979773857829699728440258287077154636" ]

[ null, "https://www.beizhulin.com/wp-content/uploads/2019/05/005BGvbFly1fqc7c5b04xg30dc08c3yb.gif", null, "https://www.beizhulin.com/wp-content/uploads/2019/05/005BGvbFly1fqc7c5b04xg30dc08c3yb.gif", null ]

[ "# Point process model diagnostic: Nearest-Neighbor Distance Distribution or Pair Correlation Function?\n\nI have a point pattern which is clearly inhomogeneous. Furthermore, the inhomogeneity has two components: a large scale effect and a local scale effect. I have constructed a Markov point process model to capture these two first order intensity variations. However, it is clear that the empirical pair correlation function should not be used to determine the second order behavior of the point pattern. So I simply modelled the second order behavior as a Poisson process. Now I fitted the model, obtained simulations from fitted model and try to compare the descrepancies in second order behavior between the simulated data and the real data.\n\nI don't want to use intensity reweighted second order summary statistics as it will very likely to give highly biased results due to potential misfit of the first order intensity from the model.\n\nMy question is, in this scenario, does it make sense to compare second order behavior between the simulated data and the real data using any kind of unweighted second order summary statistics? If so, which one would be better? NND distribution or PCF?" ]

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[ "Solutions by everydaycalculation.com\n\n## Divide 6/4 with 9/3\n\n1st number: 1 2/4, 2nd number: 3 0/3\n\n6/4 ÷ 9/3 is 1/2.\n\n#### Steps for dividing fractions\n\n1. Find the reciprocal of the divisor\nReciprocal of 9/3: 3/9\n2. Now, multiply it with the dividend\nSo, 6/4 ÷ 9/3 = 6/4 × 3/9\n3. = 6 × 3/4 × 9 = 18/36\n4. After reducing the fraction, the answer is 1/2\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]

[ "# Proof of link between the OLS slope estimate and two sample t test statistic (categorical Xvar)\n\nRegarding a univariate OLS regression with a single categorical predictor (coded 0,1).\n\nI am wrestling with the proof that $$t =\\frac{b_1}{s(b_1)}$$\n\nstarting from the basic OLS estimator for the slope which is $$b_1= \\frac{\\sum (X_i - \\bar{X})Y_i}{\\sum(X_i - \\bar{X})^2}.$$\n\nI know that the first step is to show that the denominator $\\sum(X_i - \\bar{X})^2$ is equal to\n\n$$\\frac{n_1n_0}{n}$$\n\nwhere $n_0$ and $n_1$ are the groups A and B where $X_i = 0$ and $1$ respectively.\n\nI just can't get there, and know my partial summation algebra is lacking.\n\nI am comfortable that $\\sum(X_i - \\bar{x})^2 = \\sum(X_A - \\bar{x})^2 + \\sum(X_B - \\bar{x})^2$ and then expanding each of these to the form $\\sum X_i^2 - n\\bar{x} ^2$ but can't get further than this:\n\n$$\\sum_{i=1}^{n_0} X_i^2 - n_0\\bar{x} ^2 + \\sum_{i=1}^{n_1} X_i^2 - n_1\\bar{x} ^2.$$\n\nI think the next step hinges on the fact that for the zero group $\\sum X_i^2 = 0$ and for the 1 group $\\sum X_i^2 = 1$.\n\nAny advice on what I am missing to move forward here? Or if anyone can point me to a complete proof I'd be really grateful.\n\n• ok, I see, thanks, i thought it should be one sample t test. Aug 9, 2015 at 13:14\n• You are right about the sums by group. You would also need to figure out what $\\bar x$ is in terms of $n_0$, $n_1$ and $n$. Aug 9, 2015 at 14:15\n• I don't think that it directly helps, but still: In the last formula, the result is not 1, but rather $n_1$. And furthermore, in your formula for $b_1$, it should be $Y_i-\\bar Y_i$ (or maybe you are assuming its average is 0). Aug 10, 2015 at 10:21\n• His formula for $b_1$ is correct when $X_i$ is binary, at the begginning, I also think it is wrong. Also I don't know what $X_A$ and $X_B$ stand for in his formula. Aug 10, 2015 at 11:57\n\nThis becomes easy when you reparameterize the problem.\n\nInstead of using a slope and intercept, notice that when there are just two distinct values of the $x_i$ you can describe the fit by giving its value $\\eta_0$ for $x=0$ and its value $\\eta_1$ for $x=1$.", null, "This example shows the data as red dots, the OLS fit as a dashed line, and summarizes the two groups with boxplots. Group $A$ is at the left and group $B$ at the right. The slope of the line is precisely the amount needed to go from the mean of group $A$, with $\\eta_0$ near $10$, to the mean of group $B$, with $\\eta_1$ near $13$.\n\nLeast squares requires you to choose values of these parameters that minimize the sum of squares of residuals. Since the value of $\\eta_0$ affects the residuals only for group $A$ (where $x_i=0$) and $\\eta_1$ affects the residuals only for group $B$ (where $x_i=1$), each will be estimated as the mean of its associated group. Because these means also happen to be the Maximum Likelihood estimates (as well as the OLS estimates), the ML estimate of the slope (which is also its OLS estimate) must be\n\n$$b_1 = \\frac{\\hat\\eta_1 - \\hat\\eta_0}{1-0} = \\hat\\eta_1 -\\hat\\eta_0,$$\n\nwhich is just the difference in the group means. The OLS estimate of its variance (which does differ from the ML estimate, so we cannot exploit ML at this point) is the sum of squared residuals divided by the degrees of freedom, which is $n-2$. It should be equally obvious that this is precisely the pooled variance for the two-sample t-test. Consequently, $b_1/se(b_1)$ is exactly the same--and computed in exactly the same way--as the Student t statistic.\n\n• Thanks so much for this insight. It is one thing to solve algebraically but another to demonstrate in such an intuitive way... Aug 15, 2015 at 1:09\n\nThanks everyone for helping me out. Of course the first thing I needed to do was express $\\bar{x}$ in terms of the ns. My final solution is below:\n\n$$\\sum_{i=1}^{n}(X_i - \\bar{X})^2 = \\sum_{i=1}^{(n_0 + n_1)} (X_i - \\bar{X}) ^2 = \\sum_{i=1}^{(n_0)} (X_i - \\bar{X}) ^2 + \\sum_{i=(n_0 + 1)}^{n} (X_i - \\bar{X}) ^2$$\n\n$$= \\sum_{i=1}^{(n_0)} (0 - \\frac{n_1}{n_0 + n_1}) ^2 + \\sum_{i=(n_0 + 1)}^{n} (1 - \\frac{n_1}{n_0 + n_1}) ^2$$\n\n$$= \\sum_{i=1}^{(n_0)} ( - \\frac{n_1}{n_0 + n_1}) ^2 + n_1 (1 - \\frac{n_1}{n_0 + n_1}) ^2$$\n\n$$= n_0( - \\frac{n_1}{n_0 + n_1}) ^2 + n_1 (1 - \\frac{n_1}{n_0 + n_1}) ^2$$\n\n$$= ( - \\frac{n_0n_1}{n_0 + n_1}) ^2 + n_1 (1^2 - \\frac{2n_1}{n_0 + n_1} + \\frac{n_1^2}{(n_0 + n_1)^2} )$$\n\n$$= \\frac{n_0n_1^2}{(n_0 + n_1)^2} + \\frac{n_1(n_0+n_1)^2}{(n_0 + n_1)^2} - \\frac{2n_1^2 (n_0 + n_1)}{(n_0 + n_1)^2} + \\frac{n_1^3}{(n_0 + n_1)^2}$$\n\n$$= \\frac{n_0n_1^2 + n_1(n_0^2 + 2n_0n_1 + n_1^2) - 2n_1^2n_0 - 2n_1^3 + n_1^3}{n^2}$$\n\n$$= \\frac{n_0n_1^2 + n_0^2n_1}{n^2}$$\n\n$$= \\frac{n_0n_1(n_1 + n_0)}{n^2}$$\n\n$$= \\frac{n_0n_1n}{n^2}$$\n\n$$= \\frac{n_0n_1}{n}$$\n\nIf you just want to show that $\\frac{b_1}{s(b_1)}$ has a $t$ distribution\n\nSuppose your linear regression model is $Y=Xb+\\epsilon$ and your $\\epsilon$ has a $N(0,\\sigma^2)$ distribution. Therefore, $Y$ also has a $N(0,\\sigma^2)$ distribution.\n\nAnd solve the linear equation by matrix notation:\n\nWe can show that $b=(X'X)^{-1}X'Y$\n\nAnd just note that $(X'X)^{-1}X'$ is a scalar vector, so $b$ has a normal distribution with $E(b)=0$ and variance is $(X'X)^{-1}X'\\sigma^2((X'X)^{-1}X')'$\n\n(If you just have one predictor and do not consider intercept, $(X'X)^{-1}X'$ is just a number.)\n\nYou will estimate the variance of $b$ by sample variance $s^2$\n\nThen by Student theorem:\n\n$\\frac{b_1}{s(b1)}$ has a t distribution. Then you can use one sample t test to test whether $b_1=0$ i.e $T=\\frac{b_1-0}{s(b_1)}$.\n\nI think the key is to show $b$ is normal distributed through $Y$.\n\nBut I don't know how to prove by your method." ]

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[ "## Reciprocal rank fusionedit\n\nThis functionality is in technical preview and may be changed or removed in a future release. The syntax will likely change before GA. Elastic will apply best effort to fix any issues, but features in technical preview are not subject to the support SLA of official GA features.\n\nReciprocal rank fusion (RRF) is a method for combining multiple result sets with different relevance indicators into a single result set. RRF requires no tuning, and the different relevance indicators do not have to be related to each other to achieve high-quality results.\n\nRRF uses the following formula to determine the score for ranking each document:\n\n```score = 0.0\nfor q in queries:\nif d in result(q):\nscore += 1.0 / ( k + rank( result(q), d ) )\nreturn score\n\n# where\n# k is a ranking constant\n# q is a query in the set of queries\n# d is a document in the result set of q\n# result(q) is the result set of q\n# rank( result(q), d ) is d's rank within the result(q) starting from 1```\n\n### Reciprocal rank fusion APIedit\n\nYou can use RRF as part of a search to combine and rank documents using multiple result sets from\n\n• 1 query and 1 or more kNN searches\n• 2 or more kNN searches\n\nThe `rrf` parameter is an optional object defined as part of a search request’s rank parameter. The `rrf` object contains the following parameters:\n\n`rank_constant`\n(Optional, integer) This value determines how much influence documents in individual result sets per query have over the final ranked result set. A higher value indicates that lower ranked documents have more influence. This value must be greater than or equal to `1`. Defaults to `60`.\n`window_size`\n(Optional, integer) This value determines the size of the individual result sets per query. A higher value will improve result relevance at the cost of performance. The final ranked result set is pruned down to the search request’s <<search-size-param, size>. `window_size` must be greater than or equal to `size` and greater than or equal to `1`. Defaults to `100`.\n\nAn example request using RRF:\n\n```GET example-index/_search\n{\n\"query\": {\n\"term\": {\n\"text\": \"shoes\"\n}\n},\n\"knn\": {\n\"field\": \"vector\",\n\"query_vector\": [1.25, 2, 3.5],\n\"k\": 50,\n\"num_candidates\": 100\n},\n\"rank\": {\n\"rrf\": {\n\"window_size\": 50,\n\"rank_constant\": 20\n}\n}\n}```\n\nIn the above example, we first execute the kNN search to get its global top 50 results. Then we execute the query to get its global top 50 results. Afterwards, on a coordinating node, we combine the knn search results with the query results and rank them based on the RRF method to get the final top 10 results.\n\nNote that if `k` from a knn search is larger than `window_size`, the results are truncated to `window_size`. If `k` is smaller than `window_size`, the results are `k` size.\n\n### Reciprocal rank fusion supported featuresedit\n\nRRF does support:\n\nRRF does not currently support:\n\nUsing unsupported features as part of a search using RRF will result in an exception.\n\n### Reciprocal rank fusion full exampleedit\n\nWe begin by creating a mapping for an index with a text field, a vector field, and an integer field along with indexing several documents. For this example we are going to use a vector with only a single dimension to make the ranking easier to explain.\n\n```PUT example-index\n{\n\"mappings\": {\n\"properties\": {\n\"text\" : {\n\"type\" : \"text\"\n},\n\"vector\": {\n\"type\": \"dense_vector\",\n\"dims\": 1,\n\"index\": true,\n\"similarity\": \"l2_norm\"\n},\n\"integer\" : {\n\"type\" : \"integer\"\n}\n}\n}\n}\n\nPUT example-index/_doc/1\n{\n\"text\" : \"rrf\",\n\"vector\" : ,\n\"integer\": 1\n}\n\nPUT example-index/_doc/2\n{\n\"text\" : \"rrf rrf\",\n\"vector\" : ,\n\"integer\": 2\n}\n\nPUT example-index/_doc/3\n{\n\"text\" : \"rrf rrf rrf\",\n\"vector\" : ,\n\"integer\": 1\n}\n\nPUT example-index/_doc/4\n{\n\"text\" : \"rrf rrf rrf rrf\",\n\"integer\": 2\n}\n\nPUT example-index/_doc/5\n{\n\"vector\" : ,\n\"integer\": 1\n}\n\nPOST example-index/_refresh```\n\nWe now execute a search using RRF with a query, a kNN search, and a terms aggregation.\n\n```GET example-index/_search\n{\n\"query\": {\n\"term\": {\n\"text\": \"rrf\"\n}\n},\n\"knn\": {\n\"field\": \"vector\",\n\"query_vector\": ,\n\"k\": 5,\n\"num_candidates\": 5\n},\n\"rank\": {\n\"rrf\": {\n\"window_size\": 5,\n\"rank_constant\": 1\n}\n},\n\"size\": 3,\n\"aggs\": {\n\"int_count\": {\n\"terms\": {\n\"field\": \"integer\"\n}\n}\n}\n}```\n\nAnd we receive the response with ranked `hits` and the terms aggregation result. Note that `_score` is `null`, and we instead use `_rank` to show our top-ranked documents.\n\n```{\n\"took\": ...,\n\"timed_out\" : false,\n\"_shards\" : {\n\"total\" : 1,\n\"successful\" : 1,\n\"skipped\" : 0,\n\"failed\" : 0\n},\n\"hits\" : {\n\"total\" : {\n\"value\" : 5,\n\"relation\" : \"eq\"\n},\n\"max_score\" : null,\n\"hits\" : [\n{\n\"_index\" : \"example-index\",\n\"_id\" : \"3\",\n\"_score\" : null,\n\"_rank\" : 1,\n\"_source\" : {\n\"integer\" : 1,\n\"vector\" : [\n3\n],\n\"text\" : \"rrf rrf rrf\"\n}\n},\n{\n\"_index\" : \"example-index\",\n\"_id\" : \"2\",\n\"_score\" : null,\n\"_rank\" : 2,\n\"_source\" : {\n\"integer\" : 2,\n\"vector\" : [\n4\n],\n\"text\" : \"rrf rrf\"\n}\n},\n{\n\"_index\" : \"example-index\",\n\"_id\" : \"4\",\n\"_score\" : null,\n\"_rank\" : 3,\n\"_source\" : {\n\"integer\" : 2,\n\"text\" : \"rrf rrf rrf rrf\"\n}\n}\n]\n},\n\"aggregations\" : {\n\"int_count\" : {\n\"doc_count_error_upper_bound\" : 0,\n\"sum_other_doc_count\" : 0,\n\"buckets\" : [\n{\n\"key\" : 1,\n\"doc_count\" : 3\n},\n{\n\"key\" : 2,\n\"doc_count\" : 2\n}\n]\n}\n}\n}```\n\nLet’s break down how these hits were ranked. We start by running the query and the kNN search separately to collect what their individual hits are.\n\nFirst, we look at the hits for the query.\n\n```\"hits\" : [\n{\n\"_index\" : \"example-index\",\n\"_id\" : \"4\",\n\"_score\" : 0.16152832,\n\"_source\" : {\n\"integer\" : 2,\n\"text\" : \"rrf rrf rrf rrf\"\n}\n},\n{\n\"_index\" : \"example-index\",\n\"_id\" : \"3\",\n\"_score\" : 0.15876243,\n\"_source\" : {\n\"integer\" : 1,\n\"vector\" : ,\n\"text\" : \"rrf rrf rrf\"\n}\n},\n{\n\"_index\" : \"example-index\",\n\"_id\" : \"2\",\n\"_score\" : 0.15350538,\n\"_source\" : {\n\"integer\" : 2,\n\"vector\" : ,\n\"text\" : \"rrf rrf\"\n}\n},\n{\n\"_index\" : \"example-index\",\n\"_id\" : \"1\",\n\"_score\" : 0.13963442,\n\"_source\" : {\n\"integer\" : 1,\n\"vector\" : ,\n\"text\" : \"rrf\"\n}\n}\n]```\n rank 1, `_id` 4 rank 2, `_id` 3 rank 3, `_id` 2 rank 4, `_id` 1\n\nNote that our first hit doesn’t have a value for the `vector` field. Now, we look at the results for the kNN search.\n\n```\"hits\" : [\n{\n\"_index\" : \"example-index\",\n\"_id\" : \"3\",\n\"_score\" : 1.0,\n\"_source\" : {\n\"integer\" : 1,\n\"vector\" : ,\n\"text\" : \"rrf rrf rrf\"\n}\n},\n{\n\"_index\" : \"example-index\",\n\"_id\" : \"2\",\n\"_score\" : 0.5,\n\"_source\" : {\n\"integer\" : 2,\n\"vector\" : ,\n\"text\" : \"rrf rrf\"\n}\n},\n{\n\"_index\" : \"example-index\",\n\"_id\" : \"1\",\n\"_score\" : 0.2,\n\"_source\" : {\n\"integer\" : 1,\n\"vector\" : ,\n\"text\" : \"rrf\"\n}\n},\n{\n\"_index\" : \"example-index\",\n\"_id\" : \"5\",\n\"_score\" : 0.1,\n\"_source\" : {\n\"integer\" : 1,\n\"vector\" : \n}\n}\n]```\n rank 1, `_id` 3 rank 2, `_id` 2 rank 3, `_id` 1 rank 4, `_id` 5\n\nWe can now take the two individually ranked result sets and apply the RRF formula to them to get our final ranking.\n\n```# doc | query | knn | score\n_id: 1 = 1.0/(1+4) + 1.0/(1+3) = 0.4500\n_id: 2 = 1.0/(1+3) + 1.0/(1+2) = 0.5833\n_id: 3 = 1.0/(1+2) + 1.0/(1+1) = 0.8333\n_id: 4 = 1.0/(1+1) = 0.5000\n_id: 5 = 1.0/(1+4) = 0.2000```\n\nWe rank the documents based on the RRF formula with a `window_size` of `5` truncating the bottom `2` docs in our RRF result set with a `size` of `3`. We end with `_id: 3` as `_rank: 1`, `_id: 2` as `_rank: 2`, and `_id: 4` as `_rank: 3`. This ranking matches the result set from the original RRF search as expected." ]

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[ "# Introduction to Semiconductor Materials and Devices\n\n## Introduction to Semiconductor Materials and Devices:\n\nAn electronic device controls the movement of electrons. The study of electronic devices requires a basic understanding of the relationship between electrons and the other components of an atom. The movement of electrons within a solid, and the bonding forces between atoms can then be investigated. This leads to a knowledge of the differences between conductors, insulators, and semiconductors, and to an understanding of p-type and n-type in the introduction to semiconductor material.\n\nJunctions of p-type and n-type material (pn-junctions) are basic to all but a very few semiconductor devices. Forces act upon electrons that are adjacent to a pn-junction, and these forces are altered by the presence of an external bias voltage.\n\n#### Atomic Theory of Semiconductors:\n\nThe atom can be thought of as consisting of a central nucleus surrounded by orbiting electrons (see Fig. 1-1). Thus, it may be compared to a planet with orbiting satellites. Just as satellites are held in orbit by the attractive force of gravity due to the mass of the planet, so each electron is held in orbit by an electrostatic force of attraction between it and the nucleus.", null, "Each electron has a negative electrical charge of 1.602 x 10-19 coulombs (C), and some particles within the nucleus (protons) have a positive charge of the same magnitude. Since opposite charges attract, a force of attraction exists between the oppositely charged electron and nucleus. Compared to the mass of the nucleus, electrons are relatively tiny particles of almost negligible mass. In fact, they can be considered to be little particles of negative electricity having no mass at all.", null, "The nucleus of an atom (Fig. 1-2) is largely a cluster of two types of particles, protons and neutrons. Protons have a positive electrical charge, equal in magnitude (but opposite in polarity) to the negative charge on an electron. A neutron has no charge at all. Protons and neutrons each have masses about 1800 times the mass of an electron. For a given atom, the number of protons in the nucleus normally equals the number of orbiting electrons.\n\nSince the protons and orbital electrons are equal in number and equal and opposite in charge, they neutralise each other electrically. For this reason, all atoms are normally electrically neutral. If an atom loses an electron, it has lost some negative charge. Thus, it becomes positively charged and is referred to as a positive ion, [see Fig 1-3(a)]. Similarly, if an atom gains an additional electron, it becomes negatively charged and is termed a negative ion, [Fig. 1-3(b)].", null, "The differences among atoms consist largely of dissimilar numbers and arrangements of the three basic types of particles. However, all electrons are identical, as are all protons and all neutrons. An electron from one atom could replace an electron in any other atom. Different materials are made up of different types of atoms, or differing combinations of several types of atoms.\n\nThe number of protons in an atom is referred to as the atomic number of the atom. The atomic weight is approximately equal to the total number of protons and neutrons in the nucleus of the atom. The atom of the Introduction to Semiconductor Materials silicon has 14 protons and 14 neutrons in its nucleus, as well as 14 orbital electrons. Therefore, the atomic number for silicon is 14, and its atomic weight is approximately 28.\n\n#### Electron Orbits and Energy Levels:\n\nAtoms may be conveniently represented by the two-dimensional diagrams shown in Figs. 1-4. It has been found that electrons can occupy only certain orbital rings or shells at fixed distances from the nucleus, and that each shell can contain only a particular number of electrons. The electrons in the outer shell determine the electrical (and chemical) characteristics of each particular type of atom. These electrons are usually referred to as valence electrons. An atom may have its outer shell, or valence shell, completely filled or only partially filled.", null, "The atoms represented in Fig. 1-4 are those of two important Introduction to Semiconductor materials, silicon (Si) and germanium (Ge). It is seen that each of these atoms has four electrons in a valence shell that can contain a maximum of eight. Thus, the valence shells have four electrons and four holes. A hole is defined as an absence of an electron in a shell where one could exist. Even though the valence shells of silicon and germanium have four holes, both types of atoms are electrically neutral because the total number of orbiting (negatively charged) electrons equals the total number of (positively charged) protons in the nucleus.\n\nThe closer an electron is to the nucleus, the stronger are the forces that bind it. Each shell has an energy level associated with it that represents the amount of energy required to extract an electron from the atom. Since the electrons in the valence shell are farthest from the nucleus, they require the least amount of energy to extract them, [see Fig 1-5(a)]. Conversely, those electrons closest to the nucleus require the greatest energy application to extract them from the atom.\n\nThe energy levels of the orbiting electrons are measured in electron volts (eV). An electron volt is defined as the amount of energy required to move one electron through a voltage difference of one volt.\n\n#### Energy Band Diagram of Semiconductor:", null, "So far, the discussion has concerned a system of electrons in one isolated atom. The electrons of an isolated atom are acted upon only by the forces within that atom. However, when atoms are brought closer together, as In a solid, the electrons come under the influence of forces from other atoms. Under these circumstances, the energy levels that may be occupied by electrons merge into bands of energy levels. Within any given material there are two distinct energy bands in which electrons may exist; the valence band and the conduction band. Separating these two bands is an energy gap, termed the forbidden gap, in which no electrons can normally exist. The valence band, conduction band, and forbidden gap are shown diagrammatically in Fig. 1-5(b).\n\nElectrons within the conduction band have become disconnected from atoms and are drifting around within the material. Conduction band electrons may be easily moved by the application of relatively small amounts of energy. Much larger amounts of energy must be applied to move an electron in the valence band. Electrons in the valence band are usually in orbit around a nucleus. For any given type of Introduction to Semiconductor Materials, the forbidden gap may be large, small, or nonexistent. The distinction between conductors, insulators, and semiconductors is largely concerned with the relative widths of the forbidden gap.\n\nIt is important to note that the energy band diagram is simply a graphic representation of the energy levels associated with electrons. To repeat; those electrons in the valence band are actually in orbit around the nucleus of an atom; those in the conduction band are drifting in the spaces between atoms.\n\nUpdated: April 19, 2020 — 11:57 pm" ]

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[ "TL;DR\n\nWhere we remove all programming joy from this nice puzzle.\n\nDo you see any pattern?\n\n$for N in$(seq 10 20) ; do ./run.sh 04-luckier-sum \"$N\" ; done solution => [6,2,1,0,0,0,1,0,0,0] solution => [7,2,1,0,0,0,0,1,0,0,0] solution => [8,2,1,0,0,0,0,0,1,0,0,0] solution => [9,2,1,0,0,0,0,0,0,1,0,0,0] solution => [10,2,1,0,0,0,0,0,0,0,1,0,0,0] solution => [11,2,1,0,0,0,0,0,0,0,0,1,0,0,0] solution => [12,2,1,0,0,0,0,0,0,0,0,0,1,0,0,0] solution => [13,2,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0] solution => [14,2,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0] solution => [15,2,1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0] solution => [16,2,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0] It seems that this would always be a solution, at least for$N$sufficiently large: • 0 contains value$N - 4$• 1 contains value$2$• 2 contains value$1$• N-4 contains value$1$• everything else is$0$. When$N > 6$, then$N - 4 > 2$which is the condition in which slot N-4 does not overlap with any of the other three slots that have non-zero values. Is this always a solution for$N > 6$a.k.a.$N - 4 > 2$? Yes it is: • 0, 1, 2, and N-1 are 4 distinct slots, because$N-4>2$; • these are the only slots holding a value different from$0$; • all the other slots (i.e.$N - 4$of all slots) hold value$0$, which is consistent with the value at slot 0; • value$1$appears exactly 2 times (in slot 2 and N-4), and slot 1 contains value$2$; • value$2$appears exactly once (in slot 1), and slot 2 contains value$1$; • value$N-4$appears exactly once (in slot 0), and slot N-4 contains value$1$. So there’s no need for complicated searches for$N > 6$: just provide the solution according to the pattern above. sub autobiographical_numbers ($n) {\nmy @solution;\nif ($n == 4) { @solution = (1, 2, 1, 0); # also good: (2, 0, 2, 0) } elsif ($n > 6) {\n@solution = (0) x $n; @solution[0, 1, 2,$n - 4] = ($n - 4, 2, 1, 1); } return {solution => [map {+{$_ => 1}} @solution]};\n}\n\n\nFind all of this at stage 5.\n\nHow boring. And yet… are these the only solutions?!? E.g. $N = 4$ allows two different solutions… is it possible elsewhere?!?\n\n# The end of it\n\nCurious about the whole series? Here it is:" ]

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[ "• A\n• A\n• A\n• АБB\n• АБB\n• АБB\n• А\n• А\n• А\n• А\n• А\nОбычная версия сайта\nБакалавриат 2018/2019\n\n## Компьютерные вычисления\n\nСтатус: Курс обязательный (Математика)\nНаправление: 01.03.01. Математика\nКогда читается: 2-й курс, 4 модуль\nФормат изучения: Full time\nЯзык: английский\nКредиты: 3\n\n### Course Syllabus\n\n#### Abstract\n\nIn this course, we learn how to use Mathematica and similar computer algebra systems (such as SageMath) in various mathematical problems. Mathematica is great for visualizing mathematical objects (such as functions, sets, polytopes etc.), for collecting empirical data and for testing conjectures. In particular, we focus on 3D graphics tools provided by Mathematica. These tools are ideal for drawing high precision pictures for mathematical papers. This is compulsory course. Pre-requisites: First year courses in Algebra, Analysis, Geometry, Discrete Mathematics and Topology", null, "#### Learning Objectives\n\n• To study main principles of computer algebra systems\n• To apply Mathematica tools for solving problems from Algebra, Analysis, Combinatorics, Geometry, Number Theory and Topology", null, "#### Expected Learning Outcomes\n\n• Can collect empirical data and test conjectures\n• Can draw high precision pictures for mathematical papers\n• Can visualize mathematical objects", null, "#### Course Contents\n\n• Integer, rational, real and complex numbers and their presentation in Mathematica\n• Geometry in plane and 3-space and Mathematica tools for 2D and 3D graphics\n• Tables and lists in Mathematica; commands for working with lists; applications to matrices and polynomials\n• Mathematica tools for Analysis and Differential Equations; real and complex valued functions; continued fractions; animations for solutions of differential equations\n• Tools for Number Theory; breaking the RSA code\n• Combinatorics; tools for drawing graphs and computing their invariants\n• Tools for working with groups and permutations", null, "#### Assessment Elements\n\n• Midterm Test\n• Project", null, "#### Interim Assessment\n\n• Interim assessment (4 module)\n0.7 * Midterm Test + 0.3 * Project", null, "" ]

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[ "# Week 4: Rational Functions – Answers\n\nWeek 4: Rational Functions – Day 5\n\nHere is an organized way to do this.\n\nA. Find the zeros and undefined x-values of the function (places where numerator or denominator are zero).\nB. Arrange these values in increasing order.\nC. Divide the x-axis into intervals around these values. For example, if the values are -1 and 0, the intervals are:\n-1 < x, -1 < x < 0, and 0 < x. If there are N values, there will be N+1 intervals.\nD. Test the sign of the function on each interval. You don’t have to compute exact values of the function, just check to see if it is positive or negative.\n\nHere is a set of problems that guide through the process:\n\n1. f(x) = (x2 + x) / (x2 + 6x + 5)\nFactor the numerator and denominator of f(x).\n\nf(x) = x(x + 1) / (x + 1)(x + 5)\nAs long as x does not equal -1, f(x) = x/(x + 5)\n\n2. Where does f(x) have zero value(s)?\n\nf(x) has only one zero value, at x = 0. (It might seem like f(x) is zero at x = -1, but f(x) is undefined there.)\n\n3. Where does f(x) have vertical asymptote(s)?\n\nf(x) has only one vertical asymptote, at x = -5. (It might seem like f(x) has a vertical asymptote at x = -1, but f(x) does not approach infinity or minus infinity near x = -1.)\n\n4. Where does f(x) have a hole?\n\nf(x) has a hole at its undefined point, x = -1. Because f(x) has defined, bounded (not unbounded) values near x = -1, that point is a hole in the graph.\n\n5. Using the information from problems 1-4, test the sign of f(x) over the appropriate intervals.\n\nThe values we have to pay attention to are: -5, -1, and 0. The intervals are:\nx < -5\n-5 < x < -1\n-1 < x < 0\n0 < x\n\nIf x < -5, f(x) is positive.\nIf – 5 < x < -1, f(x) is negative.\nIf – 1 < x < 0, f(x) is negative.\nIf 0 < x, f(x) is positive.\n\nWeek 4: Rational Functions – Day 4\n\nWhat is the end behavior of each of these functions? If there is a horizontal or slant asymptote, what is its equation?\n\n1. f(x) = (2x + 1) / (x – 4)\n\nThe degree of the numerator and denominator is the same. There is a horizontal asymptote. To find its equation, we divide numerator and denominator by the highest power of x (in effect, multiply f(x) by the x-1 / x-1)\n\nf(x) = (2 + 1/x) / (1 – 4/x)\n\nfor large values of x, the 1/x and 4/x have very little effect and f(x) is very close to 2/1 = 2.\n\nTherefore the horizontal asymptote is y = 2.\n\n2. f(x) = (x2 + 3) / (x – 1)\n\nThe degree of the numerator is 2, the degree of the denominator is 1. Since 2-1 = 1, there is a slant asymptote.\nTo find the equation of the slant asymptote, we use long division:", null, "3. f(x) = (3x4 – x3 + 5) / (x2 + 7)\n\nDegree of numerator is 4, degree of denominator is 2. Since 4-2 = 2 > 1, there is no linear asymptote. The end behavior of this function is similar to 3x4 / x2 = 3x2.\n\nIn other words, the end behavior is that of the quadratic 3x2 ; it goes to infinity as x goes to infinity or negative infinity.\n\n4. f(x) = 17 / (x2 + 4x – 5)\n\nThe degree of the numerator is zero, and the degree of the denominator is 2. This function tends to zero for large positive or negative values of x. The horizontal asymptote is y=0.\n\n5. f(x) = (6x2 + 1) / (3x2 – 1)\n\nThe degree of the numerator is 2, and degree of denominator is 2. Therefore there is a horizontal asymptote.\nThis function is equivalent to:\n\nf(x) = ( 6 + 1/x2) / (3 – 1/x2)\n\nFor large values of x, f(x) is very close to 6/3 = 2. The horizontal asymptote is y = 2.\n\nWeek 4: Rational Functions – Day 3\n\nWhich functions below have vertical asymptotes? If there are any vertical asymptotes, what are their equations?\n\n1. f(x) = 3 / (x – 2)\n\nOne vertical asymptote, x = 2.\n\n2. f(x) = (x3 + x)/x\n\nThis function is equivalent to x2 + 1 for x not equal to zero. There is no vertical asymptote, even though it is undefined at x = 0.\n\n3. f(x) = x / (x2 + 5)\n\nThis function is defined for all x. It does not approach infinity for finite values of x (it has a defined, finite value of y for every finite value of x). There is no vertical asymptote.\n\n4. f(x) = (2x3 – 32)/(x2 – 6x + 5)\n\nThis function has a denominator equivalent to (x – 5)(x – 1). So it is undefined at x = 5 and x = 1. The numerator is nonzero at x = 5 and x = 1, and therefore there are two asymptotes, x = 5 and x = 1.\n\n5. f(x) = 1 – 1/x + 2/(x – 1)\n\nWeek 4: Rational Functions – Day 2\n\nThe range is the set of all values taken by the function. State the range of the following rational functions, in interval notation.\n\n1. f(x) = 1/(x – 1)\n\nThis function never takes the value zero because the numerator is 1, and never zero (a quotient is zero only if the numerator is zero). This function reaches negative values if x < 1 and positive values if x > 1. The function approaches negative infinity as x approaches 1 form the left, and positive infinity as x approaches 1 from the right (check this by evaluating f(x) for numbers such as x = 0.9999 and x = 1.0001 ; you will see that f approaches negative and positive infinity).\n\nThe range is the set (-∞, 0) U (0, ∞)\n\n2. f(x) = 3\n\nThe range is the single value 3.\n\n3. f(x) = 3/(x2 + 1)\n\nThis function never takes the value zero because the numerator is 3, which is never zero. This function is always positive. It has a maximum value of 3, achieved when x = 0. It reaches very small values when x is large (plug x = 1000 into f(x) for example).\n\nThe range is (0, 3]\n\n4. f(x) = 1/x2\n\nThis function never takes the value zero. It is always positive. It approaches infinity for x close to zero (plug x = 0.0001 or x = -0.00001 into f(x) for example). For large values of x, it approaches zero.\n\nThe range is (0, ∞)\n\n5. f(x) = (x3 + 1)/x\n\nTo see what the range is, it helps to divide through by x.\n\nf(x) = x2 + 1/x\n\nThis function has multiple effects happening. The x2 part is positive for all x, but the 1/x part is negative for x\n\nThe range is (-∞, ∞).\n\nWeek 4: Rational Functions – Day 1\n\nDefinition, Domain\n\nThe definition of a rational function is: it can be written as a quotient of two polynomials.\n\n1. Suppose f(x) = 1/x and g(x) = 1/(x + 1). Are f(x) and g(x) both rational functions?\n\nYes. f(x) is the quotient of the polynomials 1 and x. g(x) is the quotient of the polynomials 1 and (x + 1).\n\n2. Is h(x) = f(x) + g(x) from problem 1 above a rational function? Explain.\n\nYes. Finding a common denominator, h(x) = (2x + 1) / (x2 + x) and is therefore the quotient of two polynomials.\n\n3. What is the domain of f(x) = 1/x?\n\nf(x) is only undefined for x = 0. The domain of f(x) is (-∞, 0) U (0, ∞) in interval notation. Put another way the domain of f is the set of x such that x does not equal zero.\n\n4. What is the domain of the rational function f(x) = (x – 1) / (x2 + 1) ?\n\nThis function is defined for all x. The domain is (-∞, ∞).\n\n5. What is the domain of h(x) = 1/x + 1/(x + 1)?\n\nh(x) is not defined for x = 0 and x = -1. The domain of h is the set of all x not equal to 0 or -1. In interval notation: (-∞, -1) U (-1, 0) U (0, ∞)" ]

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[ "# A general defender‐attacker risk model for networks\n\nA general defender‐attacker risk model for networks Purpose – The purpose of this paper is to present a competitive defender‐attacker risk model that assumes a dual exponential relationship between defender ( C i ) and attacker ( A i ) resource allocation: v i ( A i , C i )=e − α i c i −e − α i C i − γ i A i . Design/methodology/approach – Network risk is defined in terms of degree sequence, g , node/link damage, d , and probability of failure, v : R =∑ g i v i d i . The paper finds the optimal allocation of resources ( A i , C i ) that minimizes R from the defender's point of view, and maximizes R from the attacker's point of view. Findings – The effectiveness of the optimal min‐max strategy is compared with three allocation strategies: random, non‐network, and network. It is shown that total network risk is minimized by the non‐network strategy, because this strategy considers damage values and ignores network topology in the definition of risk. Originality/value – The method is illustrated by applying it to critical infrastructure – a hypothetical water‐and‐power network. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png The Journal of Risk Finance Emerald Publishing\n\n# A general defender‐attacker risk model for networks\n\n, Volume 9 (3): 18 – May 23, 2008\n18 pages", null, "", null, "", null, "", null, "", null, "", null, "/lp/emerald-publishing/a-general-defender-attacker-risk-model-for-networks-zPDz7S5G5l\nPublisher\nEmerald Publishing\nISSN\n1526-5943\nDOI\n10.1108/15265940810875577\nPublisher site\nSee Article on Publisher Site\n\n### Abstract\n\nPurpose – The purpose of this paper is to present a competitive defender‐attacker risk model that assumes a dual exponential relationship between defender ( C i ) and attacker ( A i ) resource allocation: v i ( A i , C i )=e − α i c i −e − α i C i − γ i A i . Design/methodology/approach – Network risk is defined in terms of degree sequence, g , node/link damage, d , and probability of failure, v : R =∑ g i v i d i . The paper finds the optimal allocation of resources ( A i , C i ) that minimizes R from the defender's point of view, and maximizes R from the attacker's point of view. Findings – The effectiveness of the optimal min‐max strategy is compared with three allocation strategies: random, non‐network, and network. It is shown that total network risk is minimized by the non‐network strategy, because this strategy considers damage values and ignores network topology in the definition of risk. Originality/value – The method is illustrated by applying it to critical infrastructure – a hypothetical water‐and‐power network.\n\n### Journal\n\nThe Journal of Risk FinanceEmerald Publishing\n\nPublished: May 23, 2008\n\nKeywords: Risk analysis; Risk management; Resource allocation" ]

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[ "# Project Euler 38\n\n• 問題\n\nProblem 38：Pandigital multiples\nTake the number 192 and multiply it by each of 1, 2, and 3:\n\n192 × 1 = 192\n192 × 2 = 384\n192 × 3 = 576\nBy concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)\n\nThe same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).\n\nWhat is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n > 1?\n\n• 解答例\n```max = 0\nfor i in range(1, 10000):\ntempStr = []\nj = 1\nwhile True:\ntempStr.extend(list(str(i * j)))\nif len(tempStr) >= 9:\nif \"\".join(sorted(tempStr)) == '123456789':\nif max < int(\"\".join(tempStr)):\nmax = int(\"\".join(tempStr))\nprint(i)\nbreak\nj += 1\n\nprint(max)\n```" ]

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[ "# Gravity from self-interaction redux\n\nS. Deser Physics Department, Brandeis University, Waltham, MA 02454 and\nLauritsen Laboratory, California Institute of Technology, Pasadena, CA 91125\nNovember 28, 2020\n###### Abstract\n\nI correct some recent misunderstandings about, and amplify some details of, an old explicit non-geometrical derivation of GR.\n\ngraviton self-coupling, stress-tensors, spin 2 sources\n###### pacs:\n04.20.-q, 04.20.Cv, 04.20.Fy, 04.60.-m\npreprint: BRX-TH-613preprint: CALT 68-2752\n\nLong ago , I presented a compact derivation of GR from an initial free flat space long-range symmetric spin two field: Since special relativity replaces the matter Newtonian scalar mass density by its stress-tensor, a tensor must likewise replace the scalar ”potential”. Consistency then forces this field to couple to its its own stress tensor if it is to allow any matter coupling: it either stays free– and dull– or its stress- tensor must be added to that of matter as the field’s source. This bootstrap was then explicitly performed in GR by exploiting its first derivative, cubic, , rather than its more familiar second-order non-polynomial , form. The process was also illustrated in the simpler, but precisely analogous, context of deriving (nonlinear)YM from a multiplet of free Maxwell fields, which must likewise self-couple to accept non-abelian sources. Subsequently, two extensions of were found: First, it was generalized to allow starting from any constant curvature background, where spin 2 is consistently defined . The cosmological term could then also be included in the bootstrap. Second, a tree-level quantum derivation (later generalized to include SUGRA ) provided an alternate framework, where the irrelevance of inherent field redefinition ambiguities and freedoms is particularly clear.\n\nRecently, however, there have appeared lengthy, (if not mutually consistent) critiques [5,6] of . This note addresses and resolves their concerns, both conceptual and technical, by expanding on the, perhaps too concise, original. For orientation, we start with the list of main worries and the short answers.\n\n1. The self-coupling idea, while appealing, does not work out concretely; also,the gravitational stress-tensor is ill-defined.\n\nThese worries stem from too narrow a view of self-coupling and a too broad one of non-uniqueness. Self-coupling means that the right-hand side of the original free-field equations, in one of its possible incarnations, acquires as a source the field’s own total stress tensor. This will be (re-)derived below, using the equivalent but more convenient Ricci, rather than Einstein, form of the equations. A related complaint was that the coupling did not appear in the naive, , form in the action. True, but irrelevant: to repeat, the only physical requirement is that, in the field equations, the full become the source of the originally free field; the action’s sole job is to yield these, and it does — see (13) below. Non-uniqueness of the stress tensor: it is indeed always undetermined up to identically conserved super-potentials. Further, while the one place where this non-uniqueness is relevant, namely when the stress tensors become local sources, is here, it is also precisely here that all such ambiguities can be absorbed, as we shall see, by harmless field redefinitions. Another non-uniqueness pseudo-problem is that free gauge fields of spin cannot possess (abelian) gauge-invariant stress tensors; this truism actually turns out to be a plus: only full GR recaptures the initial invariance, but now in non-abelian form, at the (satisfactory!) price of forfeiting any physical significance for its own stress-tensors, a fact also known as the equivalence principle. The only restriction on the initial stress-tensor(s) is that they be symmetric so they can drive the graviton’s symmetric field equations; further, only they can define angular momentum.\n\n2. The GR action’s non-analytic dependence on the Einstein constant cannot be obtained perturbatively starting from the, , free field.\n\nThis worry will be easily dispatched in its place; simply, the final dependence arises from a constant field rescaling of the (analytic) result to connect the field theoretical and geometrical variables’ dimensions.\n\n3. The theory’s second derivative order was an assumption.\n\nThis is as true here as it was for Einstein and Newton! Formally, GR is but one of an infinite set of geometrical models, with as as many derivatives as desired (e.g., )…Observation determines the initial kinematics, excluding (to leading order at least) scalar-tensor mixtures and higher derivative terms. Most relevant for us, second derivative order together with infinite range (any finite range makes qualitatively wrong weak-field predictions ) means that a gauge invariant (i.e., ghost-free) massless tensor field is the initial, special relativistic, mediator of matter-matter forces (their attractive sign then being a built-in bonus ).\n\n4. Total divergences and surface terms are important.\n\nYes, but not to obtain Euler-Lagrange equations from an action. Surface terms are indeed physically useful in GR, but not because of their presence in its action, contrary to myth.\n\n5. As (correctly) noted in , there have many other attempts at deriving GR from self-coupling, none of which succeeded: their approach being purely metric, the infinite summations needed to reach non-polynomial metric GR have never been performed. Instead, they were replaced by such statements as ”what else could it sum to?” and ”the sum must be general covariant, ergo GR”.\n\nAgreed. In particular the covariance of the final result, in the strong sense of being achieved without involving an external metric, does emerge here without being postulated; likewise, ”summation” is trivial.\n\nFor maximum clarity, we focus on the logic, with a minimum of formalism and indices; that can be found in . The flat space, first order, Fierz-Pauli massless spin 2 Lagrangian is\n\n L2=hμν(∂αΓαμν−∂μΓααν)+ημν(ΓαμνΓββα−ΓαβμΓβαν) (1)\n\nThe two independent variables are the Minkowski tensors (, ), with dimension (, ) as befits their ”” nature; is the Minkowski metric. The resulting first order field equations\n\n ∂αΓαμν−12(∂μΓααν+∂νΓααμ) = 0 (2) ∂αhμν−∂μhνα−12ημν∂αhββ = 2Γαμν−ηαμΓββν−ηανΓββμ (3)\n\nare equivalent to\n\n 2RLμν(h)≡∂β∂β(hμν−12ημνhαα)−∂ν∂αhμα−∂μ∂αhνα (4)\n\nin terms of the linearized Ricci (rather than Einstein) tensor.111For comparison, the first order vector theory equivalents are the initial, and as final, forms; they are spelled out in . [Our is related to the usual covariant metric deviation by .] Note however that our is not the start of an expansion, but is the total deviation, from its Minkowski value, of the full contravariant metric density.\n\nThe full GR, Palatini, Lagrangian we want to derive is\n\n LEH(G,Γ)=κ−2GμνRμν(Γ)=κ−2Gμν(∂αΓαμν−∂μΓααν+ΓαμνΓββα−ΓαβμΓβαν); (5)\n\nis the contravariant metric density, the (independent) affinity. The chief differences between (1) and (5) are that there is no background space dependence in (5), and that it is cubic (rather than quadratic) in the fields. This latter property is its compelling attraction for us, in contrast to the second order metric formulation’s non-polynomial dependence on both the metric and its inverse through the affinity’s metric dependence. The GR equations, from varying and independently, are\n\n Rμν(Γ)≡∂αΓαμν−12∂μΓααν−12∂νΓααμ+(ΓαμνΓββα−ΓαβμΓβαν) = 0, (6) −∂αGμν+GμνΓλλα−GμρΓναρ−GνρΓμαρ = 0, (7)\n\nand reduce to upon inserting into (6). Note that the geometrical variables’ dimensions are (, ). We will see that the non-analyticity of (5) is purely apparent, being removable by constant rescalings. It is useful for the sequel to express this desired answer in flat space notation by expanding (5) in terms of ( restores ’s original dimension ) and to restore its old dimension to , by defining ; we now drop all indices to concentrate on the form and logic:\n\n LEH(h,¯¯¯¯Γ)=κ−1η∂¯¯¯¯Γ+(h∂¯¯¯¯Γ+η¯¯¯¯Γ ¯¯¯¯Γ)+κh¯¯¯¯Γ ¯¯¯¯Γ. (8)\n\nThe first term being an irrelevant total divergence, now appears quite tamely in the rest of (8), disposing nicely of that worry. The middle terms are precisely the quadratic free field Lagrangian (1). The cubic term, is of course supposed to supply the heralded self-coupling of to its stress tensor in the field equations (as we will check it does), the very reason S is not itself the stress tensor. Given this flat space form of GR, it remains to show that the cubic term in (8) is the right choice: does it provide just the right (whatever that is) stress tensor source of the free field–middle terms’–field equation? The justification has three parts: first obtaining the stress tensor(s) of the middle terms’ action , then showing why its non-uniqueness (including abelian gauge-variance) is harmless, and finally verifying that the chosen cubic term (the one that agrees with ) indeed produces this stress tensor.\n\nFirst, the stress tensor: We use the Belinfante prescription: write the flat space action covariantly with respect to a fictitious auxiliary metric (for us a contravariant density) , vary the resulting action with respect to it, then set it back to in the resulting variation. The result is a symmetric on-shell, trace-shifted, stress tensor. In (1), there are two places to covariantize: the obvious and , where is the covariant tensor derivative involving the auxiliary Christoffel symbols to first order. Manifestly,\n\n ¯¯¯¯Tμν≡Tμν−12ημνtrT≡(δI/δγ)|γ=η∼∂(hΓ)+ΓΓ. (9)\n\nNext (non-)uniqueness: to the Belinfante tensor (of any system) may be added any identically conserved super-potential\n\n Δμν=∂α∂βH[μα][νβ]=Δνμ,   ∂μΔμν≡0, (10)\n\nwhere is any 4-index function with the symmetries of the Riemann tensor, to keep symmetric. [These contributions may also be thought of as the result of adding non-minimal couplings to the original action (before varying ).] But identical conservation of means precisely that it can be absorbed by field redefinition: the usual linearized Einstein equation is of the form\n\n GLμν(h)=Oμναβhαβ,   ∂μOμναβ≡0. (11)\n\nHence any identically conserved source can simply be removed by a corresponding shift in . [The initial Belinfante part, not being a super-potential, cannot be shifted away.] Finally, we must show that the cubic term in (8) indeed yields the desired field equation, with the stress tensor (9) as source of the free field. That is, we want to verify that the full field equation reads . The Einstein equations (6,7) are, dropping the overbars and expanding ,\n\n ∂Γ+κΓΓ=0,   Γ=∂h+κhΓ. (12)\n\nDifferentiating the second and inserting it into the first equation gives precisely the promised second order form\n\n ∂2h=κ[∂(hΓ)+ΓΓ]≡κ¯T. (13)\n\nMore explicitly, the left side is , while the right is just the of (9) if (and only if) we use the cubic term of the GR action (5). Equally important, the bootstrap stops here because this cubic term in the action does not generate any further (cubic) stress-tensor correction, being both -and derivative-independent. This completes our exegesis.\n\nSources: it is rather obvious that any matter action must couple to the final GR through its variables or alone, and do so covariantly in order to respect the GR equation’s Bianchi identities by having an (on-shell) covariantly conserved metric variation. But this is just Noether’s theorem: any system’s stress-tensor, namely the variation of its action with respect to the metric that makes it invariant, is covariantly conserved by virtue of its own field equations, irrespective of the equations (if any), satisfied by the metric.\n\nIn summary, I have annotated the steps involved in the non-geometric derivation of GR from special relativistic field theory as the unique consistent self-interacting system, (13) extending the initial free massless spin 2. The main ingredients were: computing the field’s standard Belinfante stress tensor, invoking field-redefinition freedom to neutralize its non-uniqueness, performing a constant field rescaling to relate geometric and field theoretic variables, and (most important) employing the cubic, Palatini, first order forms to permit explicit, trivial, summation. It goes without saying that this non-geometrical interpretation of GR, far from replacing Einstein’s original geometrical vision, is a tribute to its scope.\n\nThis work was supported by NSF grant PHY 07-57190 and DOE grant DE-FG02-92ER40701\n\nReferences:\n\n1. S.Deser, Gen Rel Grav 1 9(1970), reprinted as gr-qc/0411023.\n\n2. S.Deser, Class Quantum Grav 4 L99(1987).\n\n3. D.Boulware and S.Deser, Ann Phys 89 193(1975).\n\n4. D.Boulware, S.Deser and J.Kay, Physica 96A 141(1979).\n\n5. T.Padmanabhan, Int J Mod Phys D17 367(2008), gr-qc/0409089.\n\n6. L.Butcher, M.Hobson and A.Lasenby, Phys. Rev D80 084014(2009), gr-qc/0906.0926.\n\n7. H.van Dam and M.Veltman, Nucl. Phys. B22 397(1970); V.Zakharov, JETP Lett. 12, 312(1970); L.Faddeev and A.Slavnov, Theor.Math.Phys. 3 18(1970); S.Wong, Phys. Rev. D3, 945(1971); I.Kogan, S.Mouslopoulos and A.Papazoglou, Phys.Lett.B503 173(2001), hep-th/0011138; M.Porrati, Phys. Lett. B498 92(2001), hep-th/0011152.\n\n8. S Deser, Am. J Phys. 73 6(2005), gr-qc/0411026." ]

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[ "# 餘割\n\n 餘割", null, "性質 奇偶性 奇 定義域 {x|x≠kπ，k∈Z} 到達域 |csc x|≥1 周期 2π 特定值 當x=0 ∞ 當x=+∞ N/A 當x=-∞ N/A 最大值 +∞ 最小值 -∞ 其他性質 渐近线 N/A 根 無實根 臨界點 kπ-π/2 拐點 kπ k是一個整數。\n\n## 定义\n\n### 直角三角形中\n\n$\\csc \\theta ={\\frac {\\mathrm {c} }{\\mathrm {a} }}$\n\n### 直角坐标系中\n\n$\\alpha$ 是平面直角坐标系xOy中的一个象限角$P\\left({x,y}\\right)$ 是角的终边上一点，$r={\\sqrt {x^{2}+y^{2}}}>0$ 是P到原点O的距离，则$\\alpha$ 的余割定义为：\n\n$\\csc \\alpha ={\\frac {r}{y}}$\n\n### 单位圆定义\n\n$\\csc \\theta =\\csc \\left(\\theta +2\\pi k\\right)$\n\n### 與其他函數定義\n\n$\\csc x={\\frac {1}{\\sin x}}$\n\n### 級數定義\n\n$\\csc x={\\frac {1}{x}}+{\\frac {x}{6}}+{\\frac {7x^{3}}{360}}+{\\frac {31x^{5}}{15120}}+{\\frac {127x^{7}}{604800}}+{\\frac {73x^{9}}{3421440}}+.....$\n\n### 微分方程定义\n\n$\\csc 'x=-\\csc x\\cot x$\n$\\csc x=\\left(\\ln \\left|\\csc x-\\cot x\\right|\\right)'$\n\n### 指数定义\n\n$\\csc \\theta ={\\frac {2\\mathrm {i} }{e^{{\\mathrm {i} }\\theta }-e^{-{\\mathrm {i} }\\theta }}}\\,$\n\n## 恆等式\n\n### 和差角公式\n\n$\\csc(\\theta \\pm \\psi )={\\frac {\\csc \\theta \\csc \\psi }{\\cot \\psi \\pm \\cot \\theta }}$" ]

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[ " Olve the following simultaneous equations for x and y.(ii) Solve the tom(x + y) + n(x - y) - (m2 + mn + n2) = 0;fr + y) + m(x - y) - (m? - mn", null, ", 26.02.2020 09:10, siya1472006\n\n# Olve the following simultaneous equations for x and y.(ii) Solve the tom(x + y) + n(x - y) - (m2 + mn + n2) = 0;fr + y) + m(x - y) - (m? - mn + nạ) = 0.(1) Let x + y = a, x - y = b.(2) Transfer the terms not containing x and y to RHS.(3) Add equations, substitute the values of a and b and solve​", null, "", null, "", null, "### Other questions on the subject: Math", null, "Math, 18.08.2019 16:00, veer1277gmailcom\nFind the point which bisects the line joining of a(-2,-5) and (3,-8)", null, "Math, 18.08.2019 18:00, Firdaus17f\n19. in how many ways can a class elect a president, vice president, secretary and treasures from a class of 100 students? ​", null, "Math, 19.08.2019 10:00, asthasen18\nPlz solve this immediate plz experts​", null, "Math, 19.08.2019 11:00, success124\nThe prime factorization of 24 is 2 x 2 x 2 x 3 (which yes i am aware of the exponent way) but in the process, how do you get the 6 x 4? why can't i use 8 x 3, 12 x 2 etc.?\nDo you know the correct answer?\nOlve the following simultaneous equations for x and y.\n(ii) Solve the to\nm(x + y) + n(x -...\n\n### Questions in other subjects:", null, "", null, "History, 16.01.2021 07:05", null, "", null, "Science, 16.01.2021 07:05", null, "", null, "", null, "", null, "", null, "Math, 16.01.2021 07:05", null, "Total solved problems on the site: 29784358" ]

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[ "# what does += mean in java\n\nIn this post, we will see what does += mean in java.\n\n## Java += operator\n\n`+=` is compound addition assignment operator which adds value of right operand to variable and assign the result to variable. Types of two operands determine the behavior of `+=` in java.\n\nIn the case of number, `+=` is used for addition and concatenation is done in case of String.\n\n`a+=b` is similar to `a=a+b` with one difference which we will discuss later in the article.\n\nLet’s see with the help of example:\n\nOutput: // Prints 6\n\n2\n\n## Using += in loops\n\nYou can use `+=` in for loop when you want to increment value of variable by more than 1. In general, you might have used `i++`, but if you want to increment it by 2, then you can use `i+=2`.\n\nLet’s understand with the help of example:\nIf you want to print even number from 0 to 10, then you could use `+=` operator as below:\n\nOutput:\n\n0 2 4 6 8 10\n\n## Difference between `a+=b` and `a=a+b`\n\nIf `a` and `b` are of different types, the behavior of `a+=b` and `a=a+b` will differ due to rule of java language.\n\nLet’s understand with the help of example:\n\nHere, `+=` does implicit cast, where as `+` operator requires explicit cast for second operand, otherwise it won’t compile.\n\nAs per oracle docs for compound assignment expression:\n\nA compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T) ((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once.\n\nSo,\n\nis equivalent to:\n\n## Using += for String concatenation\n\nYou can use `+=` operator for String concatenation as well.\n\nOutput:\n\njava2blog\n\nThat’s all about what does += mean in java.\n\n## Related Posts\n\n•", null, "20 May\n\n### Question mark operator in java\n\nLearn about Question mark operator in java. It is also know as ternary operator.\n\n•", null, "29 March\n\n### Modulo operator in java\n\nTable of ContentsModulo operator SyntaxModulo operator usagesEven odd programPrime number programModulo operator behaviour with negative integer In this post, we will see about modulo or modulus operator in java. Modulo operator(%) is used to find the remainder when one integer is divided by another integer. Modulo operator Syntax [crayon-632f726ec038d681203260/] We can’t use modulo with any […]\n\n•", null, "22 March\n\n### XOR operator in java\n\nIn this tutorial, we will see about XOR operator in java. XOR operator or exclusive OR takes two boolean operands and returns true if two boolean operands are different. XOR operator can be used when both the boolean conditions can’t be true simultaneously. Here is truth table for XOR operator. Let’s say you have Person […]\n\n## Subscribe to our newletter\n\nGet quality tutorials to your inbox. Subscribe now." ]

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[ "#### `dom` method\n\nThis method translates a dominance literal `dom(X R Y)` into the corresponding problem-specific constraint (as described in Section 4.3.2). Remember that since", null, "is the inverse of", null, ", we only represent one of them: we represent", null, "when", null, "are encoded by `I,J` and `I>J`.\n\n`meth dom(X R Y)   {self var2node(X _)}   {self var2node(Y _)}   I = @var2int.X   J = @var2int.Yin    if I==J then 1::{Encode R}   elseif I>J then       @choices.(I*1000+J)::{Encode R}   else       @choices.(J*1000+I)::{Encode {Inverse R}}   end end`\n\nHere is how to inverse and encode the symbolic representation of a dominance specification.\n\n<<\n`fun {Encode R}   case R   of eq    then 1   [] above then 2   [] below then 3   [] side  then 4   [] _|_   then {Map R Encode}   end end fun {Inverse R}   case R   of eq    then eq   [] above then below   [] below then above   [] side  then side   [] _|_   then {Map R Inverse}   end end`\n\nDenys Duchier\nVersion 1.2.0 (20010221)" ]

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[ "# What is the compound interest on $39040 at 13% over 24 years? If you want to invest$39,040 over 24 years, and you expect it will earn 13.00% in annual interest, your investment will have grown to become $733,487.05. If you're on this page, you probably already know what compound interest is and how a sum of money can grow at a faster rate each year, as the interest is added to the original principal amount and recalculated for each period. The actual rate that$39,040 compounds at is dependent on the frequency of the compounding periods. In this article, to keep things simple, we are using an annual compounding period of 24 years, but it could be monthly, weekly, daily, or even continuously compounding.\n\nThe formula for calculating compound interest is:\n\n$$A = P(1 + \\dfrac{r}{n})^{nt}$$\n\n• A is the amount of money after the compounding periods\n• P is the principal amount\n• r is the annual interest rate\n• n is the number of compounding periods per year\n• t is the number of years\n\nWe can now input the variables for the formula to confirm that it does work as expected and calculates the correct amount of compound interest.\n\nFor this formula, we need to convert the rate, 13.00% into a decimal, which would be 0.13.\n\n$$A = 39040(1 + \\dfrac{ 0.13 }{1})^{ 24}$$\n\nAs you can see, we are ignoring the n when calculating this to the power of 24 because our example is for annual compounding, or one period per year, so 24 × 1 = 24.\n\n## How the compound interest on $39,040 grows over time The interest from previous periods is added to the principal amount, and this grows the sum a rate that always accelerating. The table below shows how the amount increases over the 24 years it is compounding: Start Balance Interest End Balance 1$39,040.00 $5,075.20$44,115.20\n2 $44,115.20$5,734.98 $49,850.18 3$49,850.18 $6,480.52$56,330.70\n4 $56,330.70$7,322.99 $63,653.69 5$63,653.69 $8,274.98$71,928.67\n6 $71,928.67$9,350.73 $81,279.40 7$81,279.40 $10,566.32$91,845.72\n8 $91,845.72$11,939.94 $103,785.66 9$103,785.66 $13,492.14$117,277.80\n10 $117,277.80$15,246.11 $132,523.91 11$132,523.91 $17,228.11$149,752.02\n12 $149,752.02$19,467.76 $169,219.78 13$169,219.78 $21,998.57$191,218.35\n14 $191,218.35$24,858.39 $216,076.74 15$216,076.74 $28,089.98$244,166.72\n16 $244,166.72$31,741.67 $275,908.39 17$275,908.39 $35,868.09$311,776.48\n18 $311,776.48$40,530.94 $352,307.42 19$352,307.42 $45,799.96$398,107.39\n20 $398,107.39$51,753.96 $449,861.35 21$449,861.35 $58,481.98$508,343.32\n22 $508,343.32$66,084.63 $574,427.95 23$574,427.95 $74,675.63$649,103.59\n24 $649,103.59$84,383.47 $733,487.05 We can also display this data on a chart to show you how the compounding increases with each compounding period. As you can see if you view the compounding chart for$39,040 at 13.00% over a long enough period of time, the rate at which it grows increases over time as the interest is added to the balance and new interest calculated from that figure.\n\n## How long would it take to double $39,040 at 13% interest? Another commonly asked question about compounding interest would be to calculate how long it would take to double your investment of$39,040 assuming an interest rate of 13.00%.\n\nWe can calculate this very approximately using the Rule of 72.\n\nThe formula for this is very simple:\n\n$$Years = \\dfrac{72}{Interest\\: Rate}$$\n\nBy dividing 72 by the interest rate given, we can calculate the rough number of years it would take to double the money. Let's add our rate to the formula and calculate this:\n\n$$Years = \\dfrac{72}{ 13 } = 5.54$$\n\nUsing this, we know that any amount we invest at 13.00% would double itself in approximately 5.54 years. So $39,040 would be worth$78,080 in ~5.54 years.\n\nWe can also calculate the exact length of time it will take to double an amount at 13.00% using a slightly more complex formula:\n\n$$Years = \\dfrac{log(2)}{log(1 + 0.13)} = 5.67\\; years$$\n\nHere, we use the decimal format of the interest rate, and use the logarithm math function to calculate the exact value.\n\nAs you can see, the exact calculation is very close to the Rule of 72 calculation, which is much easier to remember.\n\nHopefully, this article has helped you to understand the compound interest you might achieve from investing \\$39,040 at 13.00% over a 24 year investment period." ]

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[ "Change 30.9 to a fraction\n\nWelcome! Here is the answer to the question: Change 30.9 to a fraction or maybe \"What is 30.9 percent as a fraction?\". Use the percent to fraction calculator below to write any percentage in fraction form.\n\nPercent to Fraction Calculator\n\n Enter a percent value:  Ex.: 62.5, 7.5, 87.5, etc. Equivalent fraction: Result here Equivalent fraction Explained: Equivalent fraction explained here" ]

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[ "## You searched in “Electrical appliances”\n\nShow me:\nand\n\n• Explore\n• Explore\n• 2 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 1 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 1 items Explore\n• Explore\n• Explore\n• 1 items Explore\n• 1 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 5 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 2 items Explore\n• Explore\n• 1 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 1 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 383 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 1 items Explore\n• 1 items Explore\n• Explore\n• Explore\n• 3 items Explore\n• Explore\n• Explore\n• 1 items\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 5 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 3 items Explore\n• 2 items Explore\n• Explore\n• 2 items Explore\n• Explore\n• 5 items Explore\n• Explore\n• 2 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 2 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 1 items Explore\n• Explore\n• Explore\n• Explore\n• 2 items Explore\n• Explore\n• Explore\n• 1 items Explore\n• Explore\n• Explore\n• 1 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 10 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 3 items Explore\n• 3 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 3 items Explore" ]

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[ "#### Vol. 12, No. 3, 2019\n\n Download this article", null, "For screen For printing", null, "", null, "Recent Issues", null, "", null, "The Journal About the Journal Editorial Board Editors’ Interests Subscriptions Submission Guidelines Submission Form Policies for Authors Ethics Statement ISSN: 1948-206X (e-only) ISSN: 2157-5045 (print) Author Index To Appear Other MSP Journals", null, "", null, "The BMO-Dirichlet problem for elliptic systems in the upper half-space and quantitative characterizations of VMO\n\n### José María Martell, Dorina Mitrea, Irina Mitrea and Marius Mitrea\n\nVol. 12 (2019), No. 3, 605–720\n##### Abstract\n\nWe prove that for any homogeneous, second-order, constant complex coefficient elliptic system $L$ in ${ℝ}^{n}$, the Dirichlet problem in ${ℝ}_{+}^{n}$ with boundary data in $BMO\\left({ℝ}^{n-1}\\right)$ is well-posed in the class of functions $u$ for which the Littlewood–Paley measure associated with $u$, namely\n\n$d{\\mu }_{u}\\left({x}^{\\prime },t\\right):=|\\nabla u\\left({x}^{\\prime },t\\right){|}^{2}\\phantom{\\rule{0.3em}{0ex}}t\\phantom{\\rule{0.3em}{0ex}}d{x}^{\\prime }\\phantom{\\rule{0.3em}{0ex}}dt,$\n\nis a Carleson measure in ${ℝ}_{+}^{n}$.\n\nIn the process we establish a Fatou-type theorem guaranteeing the existence of the pointwise nontangential boundary trace for smooth null-solutions $u$ of such systems satisfying the said Carleson measure condition. In concert, these results imply that the space $BMO\\left({ℝ}^{n-1}\\right)$ can be characterized as the collection of nontangential pointwise traces of smooth null-solutions $u$ to the elliptic system $L$ with the property that ${\\mu }_{u}$ is a Carleson measure in ${ℝ}_{+}^{n}$.\n\nWe also establish a regularity result for the BMO-Dirichlet problem in the upper half-space, to the effect that the nontangential pointwise trace on the boundary of ${ℝ}_{+}^{n}$ of any given smooth null-solutions $u$ of $L$ in ${ℝ}_{+}^{n}$ satisfying the above Carleson measure condition actually belongs to Sarason’s space $VMO\\left({ℝ}^{n-1}\\right)$ if and only if ${\\mu }_{u}\\left(T\\left(Q\\right)\\right)∕|Q|\\to 0$ as $|Q|\\to 0$, uniformly with respect to the location of the cube $Q\\subset {ℝ}^{n-1}$ (where $T\\left(Q\\right)$ is the Carleson box associated with $Q$, and $|Q|$ denotes the Euclidean volume of $Q$).\n\nMoreover, we are able to establish the well-posedness of the Dirichlet problem in ${ℝ}_{+}^{n}$ for a system $L$ as above in the case when the boundary data are prescribed in Morrey–Campanato spaces in ${ℝ}^{n-1}$. In such a scenario, the solution $u$ is required to satisfy a vanishing Carleson measure condition of fractional order.\n\nBy relying on these well-posedness and regularity results we succeed in producing characterizations of the space $VMO$ as the closure in BMO of classes of smooth functions contained in BMO within which uniform continuity may be suitably quantified (such as the class of smooth functions satisfying a Hölder or Lipschitz condition). This improves on Sarason’s classical result describing $VMO$ as the closure in BMO of the space of uniformly continuous functions with bounded mean oscillations. In turn, this allows us to show that any Calderón–Zygmund operator $T$ satisfying $T\\left(1\\right)=0$ extends as a linear and bounded mapping from $VMO$ (modulo constants) into itself. In turn, this is used to describe algebras of singular integral operators on $VMO$, and to characterize the membership to $VMO$ via the action of various classes of singular integral operators.\n\nHowever, your active subscription may be available on Project Euclid at\nhttps://projecteuclid.org/apde\n\nWe have not been able to recognize your IP address 3.236.13.53 as that of a subscriber to this journal.\nOnline access to the content of recent issues is by subscription, or purchase of single articles.\n\nor by using our contact form.", null, "##### Keywords\nBMO Dirichlet problem, VMO Dirichlet problem, Carleson measure, vanishing Carleson measure, second-order elliptic system, Poisson kernel, Lamé system, nontangential pointwise trace, Fatou-type theorem, Hardy space, Holder space, Morrey–Campanato space, square function, quantitative characterization of VMO, dense subspaces of VMO, boundedness of Calderón–Zygmund operators on VMO\n##### Mathematical Subject Classification 2010\nPrimary: 35B65, 35C15, 35J47, 35J57, 35J67, 42B37\nSecondary: 35E99, 42B20, 42B30, 42B35" ]

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[ "## Conversion formula\n\nThe conversion factor from hours to minutes is 60, which means that 1 hour is equal to 60 minutes:\n\n1 hr = 60 min\n\nTo convert 128 hours into minutes we have to multiply 128 by the conversion factor in order to get the time amount from hours to minutes. We can also form a simple proportion to calculate the result:\n\n1 hr → 60 min\n\n128 hr → T(min)\n\nSolve the above proportion to obtain the time T in minutes:\n\nT(min) = 128 hr × 60 min\n\nT(min) = 7680 min\n\nThe final result is:\n\n128 hr → 7680 min\n\nWe conclude that 128 hours is equivalent to 7680 minutes:\n\n128 hours = 7680 minutes\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 minute is equal to 0.00013020833333333 × 128 hours.\n\nAnother way is saying that 128 hours is equal to 1 ÷ 0.00013020833333333 minutes.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that one hundred twenty-eight hours is approximately seven thousand six hundred eighty minutes:\n\n128 hr ≅ 7680 min\n\nAn alternative is also that one minute is approximately zero times one hundred twenty-eight hours.\n\n## Conversion table\n\n### hours to minutes chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from hours to minutes\n\nhours (hr) minutes (min)\n129 hours 7740 minutes\n130 hours 7800 minutes\n131 hours 7860 minutes\n132 hours 7920 minutes\n133 hours 7980 minutes\n134 hours 8040 minutes\n135 hours 8100 minutes\n136 hours 8160 minutes\n137 hours 8220 minutes\n138 hours 8280 minutes" ]

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[ "## 27.11.14\n\n### Excercise-18 : Recursive Function Progamming\n\nWrite a program using recursive function in C++ to find factorial of a given integer number.\n\n1.", null, "#include\nusing namespace std;\nclass recursive\n\n{\npublic:\n\nint rec(int x)\n{\nif(x<=1)\nreturn 1;\n\nelse\nreturn(x*rec(x-1));\n}\n};\n\nint main()\n{\nrecursive rc;\nint a;\ncin>>a;\ncout<<\"the factorial is: \"<<rc.rec(a);\n}\n\n2.", null, "Noted. Found correct. It is quite appreciable that you did it using recursive with class.\n\n3.", null, "#include\nusing namespace std;\nint fact(int n);\nint main()\n{\nint n,result=0;\ncin>>n;\ncout<<\"the factorial is:\"<<fact(n);\n}\nint fact(int inp)\n{\nif(inp<=1)\nreturn 1;\nelse\nreturn(inp*fact(inp-1));\n}\n\n1.", null, "4.", null, "http://itlearn24.blogspot.com/2014/11/recursive-function-progamming.html\n\n1.", null, "5.", null, "#include\nusing namespace std;\nint fact(int r);\nint main()\n{\nint r;\ncin>>r;\ncout<< \"The factorial is:\"<<fact(r);\nreturn 0;\n}\nint fact(int r)\n{\nif(r!=1)\nreturn r*fact(r-1);\n}\n\nComment Here" ]

[ null, "http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35", null, "http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35", null, "http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35", null, "http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35", null, "http://3.bp.blogspot.com/-5iuPIlAGicY/VFDurEYE4aI/AAAAAAAAAGg/Bz-RS17qn60/s35/1408550010193.jpg", null, "http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35", null, "http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35", null ]

[ "# Adding outlier to a matrix\n\n18 views (last 30 days)\nNiki on 22 May 2014\nCommented: Star Strider on 22 May 2014\nHi , I have a matrix of intensities (n*p). I want to add an outlier to it. Do you have any idea about how to simulate an outlier? I was thinking to take the mean of matrix and then multiply it to some very high value?\nDo you have any idea ?\n\nStar Strider on 22 May 2014\nI suggest:\nM = rand(4,5); % Create data\nMsts = [mean(M(:)) std(M(:))];\nOutlier = Msts(1)+5*Msts(2);\nDefine the outlier by a multiple of the standard deviation from the mean. An value of 5*std is quite far out.\n\nStar Strider on 22 May 2014\nSee if this does what you want:\nOutsXR = mean(X,2) + 5*std(X,[],2);\nfigure(1)\nplot([X OutsXR])\nNiki on 22 May 2014\nNot that one but this one helps\nOutsXR = mean(X,1) + 5*std(X,[],1);\nStar Strider on 22 May 2014\nMy pleasure, and thanks!\n\nRoger Wohlwend on 22 May 2014\nActually you answered your question yourself. Instead of the mean I would use the maximum value of the matrix. Multiply that number by a certain value. That's all. It is quite easy.\n\n#### 1 Comment\n\nNiki on 22 May 2014\nIn fact, I want to know whether it is scientifically correct or not? It is not a number. As I explained it is a matrix of n*p so I wont have a value as outlier but a row which corresponds as an outlier\nOn the other hand, do you know any way to show it as a outlier ?" ]

[ null ]

[ "Cody\n\n# Problem 23. Finding Perfect Squares\n\nSolution 490598\n\nSubmitted on 26 Aug 2014 by kunal\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\n%% a = [2 3 4]; assert(isequal(isItSquared(a),true))\n\n2   Pass\n%% a = [20:30]; assert(isequal(isItSquared(a),false))\n\n3   Pass\n%% a = ; assert(isequal(isItSquared(a),true))\n\n4   Pass\n%% a = [6 10 12 14 36 101]; assert(isequal(isItSquared(a),true))\n\n5   Pass\n%% a = [6 10 12 14 101]; assert(isequal(isItSquared(a),false))" ]

[ null ]

[ "### Polynomial Interpolation Matlab Help\n\nPolynomial interpolation is a concept, which is defined as the reconstruction of the functions, coefficient or graphs of a polynomial according to the different data test points.\n\nAccording to our matlab assignment experts, engineering and\n\nthe polynomial interpolation using matlab are highly interconnected with each other. Nowadays, polynomial interpolation using matlab has become more popular in different areas, which include electronics engineering, thermodynamics, chemical engineering, metallurgical engineering and industrial engineering.\n\nOur online polynomial interpolation tutors and our experts of polynomial interpolation in matlab have years of experience. Our experts of polynomial interpolation in matlab provide the guidance to the students, which include polynomial interpolation assignment help, polynomial interpolation homework help, polynomial interpolation in matlab quizzes preparation help, polynomial interpolation project paper help, polynomial interpolation dissertation help, etc.\n\nAt our matlab assignment experts, our experts are highly professional and qualified, in which some of them are polynomial interpolation helpers and others are expert in polynomial interpolation solvers. Our services are available 24×7, which help the students of universities and colleges in order to make their matlab polynomial interpolation assignments. We are also providing matlab polynomial interpolation tutoring at our matlab assignment experts, which is of a high quality and it can be provided to the students of colleges, universities, or PhDs.\n\nFurthermore, our experts at the matlab assignment experts are well educated and they also have degrees of bachelors, masters, or PhDs. All of them are well aware from the referencing styles such as AML, APA, Harvard, or many others. The topics which have been covered by experts in our matlab polynomial interpolation assignment are listed below:\n\n•  Simpson’s Rule\n•  Evaluative interpolation using divided coefficients\n•  Constructive Interpolation using divided coefficients" ]

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[ "# Solved – Noise in the Fourier transform\n\nI have a smooth but rather complex curve, sampled with a good frequency. I apply the discrete Fourier transform to it using the fast Fourier transform (FFT) algorithm and get its Fourier image. I need to find peaks on the resulting Fourier transform curve, but the image I get contains a substantial amount of noise, which is a real problem because peaks can't be clearly seen. So, my question is as follows: What are the sources of noise in the Fourier transform, and how it can be reduced?\n\nTypical plots look like this:\n\nInitial function", null, "Initial function, narrow range", null, "Fourier transform", null, "Fourier transform, narrow range", null, "Before transformation a constant was subtracted from the the function so that it goes to zero.\n\nContents" ]

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[ "# Number 2761\n\nNumber 2,761 spell 🔊, write in words: two thousand, seven hundred and sixty-one . Ordinal number 2761th is said 🔊 and write: two thousand, seven hundred and sixty-first. The meaning of number 2761 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 2761. What is 2761 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 2761.\n\n## What is 2,761 in other units\n\nThe decimal (Arabic) number 2761 converted to a Roman number is MMDCCLXI. Roman and decimal number conversions.\n The number 2761 converted to a Mayan number is", null, "Decimal and Mayan number conversions.\n\n#### Weight conversion\n\n2761 kilograms (kg) = 6086.9 pounds (lbs)\n2761 pounds (lbs) = 1252.4 kilograms (kg)\n\n#### Length conversion\n\n2761 kilometers (km) equals to 1716 miles (mi).\n2761 miles (mi) equals to 4444 kilometers (km).\n2761 meters (m) equals to 9059 feet (ft).\n2761 feet (ft) equals 841.563 meters (m).\n2761 centimeters (cm) equals to 1087.0 inches (in).\n2761 inches (in) equals to 7012.9 centimeters (cm).\n\n#### Temperature conversion\n\n2761° Fahrenheit (°F) equals to 1516.1° Celsius (°C)\n2761° Celsius (°C) equals to 5001.8° Fahrenheit (°F)\n\n#### Power conversion\n\n2761 Horsepower (hp) equals to 2030.44 kilowatts (kW)\n2761 kilowatts (kW) equals to 3754.43 horsepower (hp)\n\n#### Time conversion\n\n(hours, minutes, seconds, days, weeks)\n2761 seconds equals to 46 minutes, 1 second\n2761 minutes equals to 1 day, 22 hours, 1 minute\n\n### Zip codes 2761\n\n• Zip code 2761 ESTANCIA SANTA CATALINA, BUENOS AIRES, Argentina a map\n• Zip code 2761 SANTA COLOMA, BUENOS AIRES, Argentina a map\n\n### Codes and images of the number 2761\n\nNumber 2761 morse code: ..--- --... -.... .----\nSign language for number 2761:", null, "", null, "", null, "", null, "Number 2761 in braille:", null, "Images of the number\nImage (1) of the numberImage (2) of the number", null, "", null, "More images, other sizes, codes and colors ...\n\n#### Number 2761 infographic", null, "### Gregorian, Hebrew, Islamic, Persian and Buddhist year (calendar)\n\nGregorian year 2761 is Buddhist year 3304.\nBuddhist year 2761 is Gregorian year 2218 .\nGregorian year 2761 is Islamic year 2205 or 2206.\nIslamic year 2761 is Gregorian year 3300 or 3301.\nGregorian year 2761 is Persian year 2139 or 2140.\nPersian year 2761 is Gregorian 3382 or 3383.\nGregorian year 2761 is Hebrew year 6521 or 6522.\nHebrew year 2761 is Gregorian year 999 a. C.\nThe Buddhist calendar is used in Sri Lanka, Cambodia, Laos, Thailand, and Burma. The Persian calendar is official in Iran and Afghanistan.\n\n## Share in social networks", null, "## Mathematics of no. 2761\n\n### Multiplications\n\n#### Multiplication table of 2761\n\n2761 multiplied by two equals 5522 (2761 x 2 = 5522).\n2761 multiplied by three equals 8283 (2761 x 3 = 8283).\n2761 multiplied by four equals 11044 (2761 x 4 = 11044).\n2761 multiplied by five equals 13805 (2761 x 5 = 13805).\n2761 multiplied by six equals 16566 (2761 x 6 = 16566).\n2761 multiplied by seven equals 19327 (2761 x 7 = 19327).\n2761 multiplied by eight equals 22088 (2761 x 8 = 22088).\n2761 multiplied by nine equals 24849 (2761 x 9 = 24849).\nshow multiplications by 6, 7, 8, 9 ...\n\n### Fractions: decimal fraction and common fraction\n\n#### Fraction table of 2761\n\nHalf of 2761 is 1380,5 (2761 / 2 = 1380,5 = 1380 1/2).\nOne third of 2761 is 920,3333 (2761 / 3 = 920,3333 = 920 1/3).\nOne quarter of 2761 is 690,25 (2761 / 4 = 690,25 = 690 1/4).\nOne fifth of 2761 is 552,2 (2761 / 5 = 552,2 = 552 1/5).\nOne sixth of 2761 is 460,1667 (2761 / 6 = 460,1667 = 460 1/6).\nOne seventh of 2761 is 394,4286 (2761 / 7 = 394,4286 = 394 3/7).\nOne eighth of 2761 is 345,125 (2761 / 8 = 345,125 = 345 1/8).\nOne ninth of 2761 is 306,7778 (2761 / 9 = 306,7778 = 306 7/9).\nshow fractions by 6, 7, 8, 9 ...\n\n### Calculator\n\n 2761\n\n#### Is Prime?\n\nThe number 2761 is not a prime number. The closest prime numbers are 2753, 2767.\n2761th prime number in order is 24979.\n\n#### Factorization and factors (dividers)\n\nThe prime factors of 2761 are 11 * 251\nThe factors of 2761 are 1 , 11 , 251 , 2761\nTotal factors 4.\nSum of factors 3024 (263).\n\n#### Powers\n\nThe second power of 27612 is 7.623.121.\nThe third power of 27613 is 21.047.437.081.\n\n#### Roots\n\nThe square root √2761 is 52,545219.\nThe cube root of 32761 is 14,028852.\n\n#### Logarithms\n\nThe natural logarithm of No. ln 2761 = loge 2761 = 7,923348.\nThe logarithm to base 10 of No. log10 2761 = 3,441066.\nThe Napierian logarithm of No. log1/e 2761 = -7,923348.\n\n### Trigonometric functions\n\nThe cosine of 2761 is -0,896078.\nThe sine of 2761 is 0,443897.\nThe tangent of 2761 is -0,495377.\n\n## Number 2761 in Computer Science\n\nCode typeCode value\nPIN 2761 It's recommendable to use 2761 as a password or PIN.\n2761 Number of bytes2.7KB\nUnix timeUnix time 2761 is equal to Thursday Jan. 1, 1970, 12:46:01 a.m. GMT\nIPv4, IPv6Number 2761 internet address in dotted format v4 0.0.10.201, v6 ::ac9\n2761 Decimal = 101011001001 Binary\n2761 Decimal = 10210021 Ternary\n2761 Decimal = 5311 Octal\n2761 Decimal = AC9 Hexadecimal (0xac9 hex)\n2761 BASE64Mjc2MQ==\n2761 MD5566a9968b43628588e76be5a85a0f9e8\n2761 SHA112dc672132ef15bbfdba913435b4ffe1581d6978\n2761 SHA2249debef36ba4d671c3c6fdd95208c19c032e10c98c120f5ccb06a4821\n2761 SHA256d2cc2cd378a0cbf24293e9059fff3a2435ec802d01893fc38edf444fb9f4decc\n2761 SHA384d558f3dd8efd3b280cffc1cf0d9449b786a0e5a2b39c64e1fd5488d17653c81a885d1fde4d13181860fbd02614f89161\nMore SHA codes related to the number 2761 ...\n\nIf you know something interesting about the 2761 number that you did not find on this page, do not hesitate to write us here.\n\n## Numerology 2761\n\n### The meaning of the number 7 (seven), numerology 7\n\nCharacter frequency 7: 1\n\nThe number 7 (seven) is the sign of the intellect, thought, psychic analysis, idealism and wisdom. This number first needs to gain self-confidence and to open his/her life and heart to experience trust and openness in the world. And then you can develop or balance the aspects of reflection, meditation, seeking knowledge and knowing.\nMore about the meaning of the number 7 (seven), numerology 7 ...\n\n### The meaning of the number 6 (six), numerology 6\n\nCharacter frequency 6: 1\n\nThe number 6 (six) denotes emotional responsibility, love, understanding and harmonic balance. The person with the personal number 6 must incorporate vision and acceptance in the world. Beauty, tenderness, stable, responsible and understanding exchange, the sense of protection and availability also define the meaning of the number 6 (six).\nMore about the meaning of the number 6 (six), numerology 6 ...\n\n### The meaning of the number 2 (two), numerology 2\n\nCharacter frequency 2: 1\n\nThe number two (2) needs above all to feel and to be. It represents the couple, duality, family, private and social life. He/she really enjoys home life and family gatherings. The number 2 denotes a sociable, hospitable, friendly, caring and affectionate person. It is the sign of empathy, cooperation, adaptability, consideration for others, super-sensitivity towards the needs of others.\n\nThe number 2 (two) is also the symbol of balance, togetherness and receptivity. He/she is a good partner, colleague or companion; he/she also plays a wonderful role as a referee or mediator. Number 2 person is modest, sincere, spiritually influenced and a good diplomat. It represents intuition and vulnerability.\nMore about the meaning of the number 2 (two), numerology 2 ...\n\n### The meaning of the number 1 (one), numerology 1\n\nCharacter frequency 1: 1\n\nNumber one (1) came to develop or balance creativity, independence, originality, self-reliance and confidence in the world. It reflects power, creative strength, quick mind, drive and ambition. It is the sign of individualistic and aggressive nature.\nMore about the meaning of the number 1 (one), numerology 1 ...\n\n## Interesting facts about the number 2761\n\n### Asteroids\n\n• (2761) Eddington is asteroid number 2761. It was discovered by E. L. G. Bowell from Anderson Mesa on 1/1/1981.\n\n### Distances between cities\n\n• There is a 2,761 miles (4,443 km) direct distance between Addis Ababa (Ethiopia) and Rawalpindi (Pakistan).\n• There is a 2,761 miles (4,443 km) direct distance between Addis Ababa (Ethiopia) and Volgograd (Russia).\n• There is a 2,761 miles (4,443 km) direct distance between Baghdad (Iraq) and Kolkata (India).\n• There is a 2,761 miles (4,442 km) direct distance between Bandarlampung (Indonesia) and Kānpur (India).\n• More distances between cities ...\n• There is a 1,716 miles (2,761 km) direct distance between Bangkok (Thailand) and Shangyu (China).\n• There is a 1,716 miles (2,761 km) direct distance between Bucharest (Romania) and Chelyabinsk (Russia).\n• There is a 1,716 miles (2,761 km) direct distance between Bursa (Turkey) and Perm (Russia).\n• There is a 2,761 miles (4,443 km) direct distance between Datong (China) and Rājkot (India).\n• There is a 2,761 miles (4,443 km) direct distance between Gwangju (South Korea) and Johor Bahru (Malaysia).\n• There is a 2,761 miles (4,442 km) direct distance between Hegang (China) and Kathmandu (Nepal).\n• There is a 1,716 miles (2,761 km) direct distance between Hegang (China) and Xiamen (China).\n• There is a 2,761 miles (4,443 km) direct distance between Hohhot (China) and Johor Bahru (Malaysia).\n• There is a 2,761 miles (4,442 km) direct distance between Hong Kong (Hong Kong) and Rājkot (India).\n• There is a 2,761 miles (4,442 km) direct distance between Changzhou (China) and Indore (India).\n• There is a 2,761 miles (4,442 km) direct distance between Chennai (India) and Medina (Saudi Arabia).\n• There is a 2,761 miles (4,443 km) direct distance between Chennai (India) and Nanjing (China).\n• There is a 2,761 miles (4,442 km) direct distance between Jeddah (Saudi Arabia) and Paris (France).\n• There is a 1,716 miles (2,761 km) direct distance between Jieyang (China) and Kuala Lumpur (Malaysia).\n• There is a 2,761 miles (4,443 km) direct distance between Kaduna (Nigeria) and Prague (Czech Republic).\n• There is a 2,761 miles (4,442 km) direct distance between Medellín (Colombia) and Santiago (Chile).\n• There is a 2,761 miles (4,442 km) direct distance between Meerut (India) and Qiqihar (China).\n• There is a 1,716 miles (2,761 km) direct distance between Milano (Italy) and Saratov (Russia).\n• There is a 2,761 miles (4,443 km) direct distance between Muzaffarābād (Pakistan) and Shangyu (China).\n• There is a 1,716 miles (2,761 km) direct distance between Nagoya-shi (Japan) and Yunfu (China).\n• There is a 1,716 miles (2,761 km) direct distance between Phnom Penh (Cambodia) and Varanasi (India).\n• There is a 2,761 miles (4,443 km) direct distance between Pune (India) and Rostov-na-Donu (Russia).\n• There is a 2,761 miles (4,443 km) direct distance between Quetta (Pakistan) and Xinyang (China).\n• There is a 2,761 miles (4,442 km) direct distance between Quetta (Pakistan) and Zhongshan (China).\n• There is a 2,761 miles (4,442 km) direct distance between Rostov-na-Donu (Russia) and Shivaji Nagar (India).\n• There is a 1,716 miles (2,761 km) direct distance between Sapporo (Japan) and Wuhan (China).\n• There is a 1,716 miles (2,761 km) direct distance between Ürümqi (China) and Wuhan (China).\n\n## Number 2,761 in other languages\n\nHow to say or write the number two thousand, seven hundred and sixty-one in Spanish, German, French and other languages. The character used as the thousands separator.\n Spanish: 🔊 (número 2.761) dos mil setecientos sesenta y uno German: 🔊 (Anzahl 2.761) zweitausendsiebenhunderteinundsechzig French: 🔊 (nombre 2 761) deux mille sept cent soixante et un Portuguese: 🔊 (número 2 761) dois mil, setecentos e sessenta e um Chinese: 🔊 (数 2 761) 二千七百六十一 Arabian: 🔊 (عدد 2,761) ألفان و سبعمائةواحد و ستون Czech: 🔊 (číslo 2 761) dva tisíce sedmset šedesát jedna Korean: 🔊 (번호 2,761) 이천칠백육십일 Danish: 🔊 (nummer 2 761) totusinde og syvhundrede og enogtreds Hebrew: (מספר 2,761) שנים אלף שבע מאות ששים ואחד Dutch: 🔊 (nummer 2 761) tweeduizendzevenhonderdeenenzestig Japanese: 🔊 (数 2,761) 二千七百六十一 Indonesian: 🔊 (jumlah 2.761) dua ribu tujuh ratus enam puluh satu Italian: 🔊 (numero 2 761) duemilasettecentosessantuno Norwegian: 🔊 (nummer 2 761) to tusen, syv hundre og seksti-en Polish: 🔊 (liczba 2 761) dwa tysiące siedemset sześćdziesiąt jeden Russian: 🔊 (номер 2 761) две тысячи семьсот шестьдесят один Turkish: 🔊 (numara 2,761) ikibinyediyüzaltmışbir Thai: 🔊 (จำนวน 2 761) สองพันเจ็ดร้อยหกสิบเอ็ด Ukrainian: 🔊 (номер 2 761) двi тисячi сiмсот шiстдесят одна Vietnamese: 🔊 (con số 2.761) hai nghìn bảy trăm sáu mươi mốt Other languages ...\n\n## News to email\n\nPrivacy Policy.\n\n## Comment\n\nIf you know something interesting about the number 2761 or any natural number (positive integer) please write us here or on facebook." ]

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[ "# Find the domain of the functions y = ln (x-7)\n\nWe need to find the domain of the function y = ln (x – 7). Let’s remember the definition. The domain of definition or the domain of the function y = f (x) is the set of values x for which there are values y = f (x). The domain of the function is denoted – D (f) or D (y). If the function has the form y = ln (f (x)), then the domain of definition will be the set of solutions to the inequality f (x)> 0. The domain of this function will be the solution to the inequality x – 7> 0. x> 7; Answer: D (y): x Є (7; + infinity).", null, "One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities." ]

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[ "# #0013e5 Color Information\n\nIn a RGB color space, hex #0013e5 is composed of 0% red, 7.5% green and 89.8% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 91.7% magenta, 0% yellow and 10.2% black. It has a hue angle of 235 degrees, a saturation of 100% and a lightness of 44.9%. #0013e5 color hex could be obtained by blending #0026ff with #0000cb. Closest websafe color is: #0000cc.\n\n• R 0\n• G 7\n• B 90\nRGB color chart\n• C 100\n• M 92\n• Y 0\n• K 10\nCMYK color chart\n\n#0013e5 color description : Pure (or mostly pure) blue.\n\n# #0013e5 Color Conversion\n\nThe hexadecimal color #0013e5 has RGB values of R:0, G:19, B:229 and CMYK values of C:1, M:0.92, Y:0, K:0.1. Its decimal value is 5093.\n\nHex triplet RGB Decimal 0013e5 `#0013e5` 0, 19, 229 `rgb(0,19,229)` 0, 7.5, 89.8 `rgb(0%,7.5%,89.8%)` 100, 92, 0, 10 235°, 100, 44.9 `hsl(235,100%,44.9%)` 235°, 100, 89.8 0000cc `#0000cc`\nCIE-LAB 29.718, 69.324, -97.451 14.373, 6.122, 74.548 0.151, 0.064, 6.122 29.718, 119.593, 305.427 29.718, -9.094, -116.401 24.742, 60.393, -161.322 00000000, 00010011, 11100101\n\n# Color Schemes with #0013e5\n\n• #0013e5\n``#0013e5` `rgb(0,19,229)``\n• #e5d200\n``#e5d200` `rgb(229,210,0)``\nComplementary Color\n• #0086e5\n``#0086e5` `rgb(0,134,229)``\n• #0013e5\n``#0013e5` `rgb(0,19,229)``\n• #6000e5\n``#6000e5` `rgb(96,0,229)``\nAnalogous Color\n• #86e500\n``#86e500` `rgb(134,229,0)``\n• #0013e5\n``#0013e5` `rgb(0,19,229)``\n• #e56000\n``#e56000` `rgb(229,96,0)``\nSplit Complementary Color\n• #13e500\n``#13e500` `rgb(19,229,0)``\n• #0013e5\n``#0013e5` `rgb(0,19,229)``\n• #e50013\n``#e50013` `rgb(229,0,19)``\n• #00e5d2\n``#00e5d2` `rgb(0,229,210)``\n• #0013e5\n``#0013e5` `rgb(0,19,229)``\n• #e50013\n``#e50013` `rgb(229,0,19)``\n• #e5d200\n``#e5d200` `rgb(229,210,0)``\n• #000d99\n``#000d99` `rgb(0,13,153)``\n• #000fb2\n``#000fb2` `rgb(0,15,178)``\n• #0011cc\n``#0011cc` `rgb(0,17,204)``\n• #0013e5\n``#0013e5` `rgb(0,19,229)``\n• #0015ff\n``#0015ff` `rgb(0,21,255)``\n• #192cff\n``#192cff` `rgb(25,44,255)``\n• #3343ff\n``#3343ff` `rgb(51,67,255)``\nMonochromatic Color\n\n# Alternatives to #0013e5\n\nBelow, you can see some colors close to #0013e5. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #004ce5\n``#004ce5` `rgb(0,76,229)``\n• #0039e5\n``#0039e5` `rgb(0,57,229)``\n• #0026e5\n``#0026e5` `rgb(0,38,229)``\n• #0013e5\n``#0013e5` `rgb(0,19,229)``\n• #0000e5\n``#0000e5` `rgb(0,0,229)``\n• #1300e5\n``#1300e5` `rgb(19,0,229)``\n• #2600e5\n``#2600e5` `rgb(38,0,229)``\nSimilar Colors\n\n# #0013e5 Preview\n\nText with hexadecimal color #0013e5\n\nThis text has a font color of #0013e5.\n\n``<span style=\"color:#0013e5;\">Text here</span>``\n#0013e5 background color\n\nThis paragraph has a background color of #0013e5.\n\n``<p style=\"background-color:#0013e5;\">Content here</p>``\n#0013e5 border color\n\nThis element has a border color of #0013e5.\n\n``<div style=\"border:1px solid #0013e5;\">Content here</div>``\nCSS codes\n``.text {color:#0013e5;}``\n``.background {background-color:#0013e5;}``\n``.border {border:1px solid #0013e5;}``\n\n# Shades and Tints of #0013e5\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #00010d is the darkest color, while #f9f9ff is the lightest one.\n\n• #00010d\n``#00010d` `rgb(0,1,13)``\n• #000321\n``#000321` `rgb(0,3,33)``\n• #000434\n``#000434` `rgb(0,4,52)``\n• #000648\n``#000648` `rgb(0,6,72)``\n• #00085c\n``#00085c` `rgb(0,8,92)``\n• #00096f\n``#00096f` `rgb(0,9,111)``\n• #000b83\n``#000b83` `rgb(0,11,131)``\n• #000c97\n``#000c97` `rgb(0,12,151)``\n• #000eaa\n``#000eaa` `rgb(0,14,170)``\n• #0010be\n``#0010be` `rgb(0,16,190)``\n• #0011d1\n``#0011d1` `rgb(0,17,209)``\n• #0013e5\n``#0013e5` `rgb(0,19,229)``\n• #0015f9\n``#0015f9` `rgb(0,21,249)``\n• #0d21ff\n``#0d21ff` `rgb(13,33,255)``\n• #2133ff\n``#2133ff` `rgb(33,51,255)``\n• #3445ff\n``#3445ff` `rgb(52,69,255)``\n• #4857ff\n``#4857ff` `rgb(72,87,255)``\n• #5c69ff\n``#5c69ff` `rgb(92,105,255)``\n• #6f7bff\n``#6f7bff` `rgb(111,123,255)``\n• #838dff\n``#838dff` `rgb(131,141,255)``\n• #979fff\n``#979fff` `rgb(151,159,255)``\n• #aab1ff\n``#aab1ff` `rgb(170,177,255)``\n• #bec3ff\n``#bec3ff` `rgb(190,195,255)``\n• #d1d5ff\n``#d1d5ff` `rgb(209,213,255)``\n• #e5e7ff\n``#e5e7ff` `rgb(229,231,255)``\n• #f9f9ff\n``#f9f9ff` `rgb(249,249,255)``\nTint Color Variation\n\n# Tones of #0013e5\n\nA tone is produced by adding gray to any pure hue. In this case, #6a6b7b is the less saturated color, while #0013e5 is the most saturated one.\n\n• #6a6b7b\n``#6a6b7b` `rgb(106,107,123)``\n• #616484\n``#616484` `rgb(97,100,132)``\n• #585c8d\n``#585c8d` `rgb(88,92,141)``\n• #4f5596\n``#4f5596` `rgb(79,85,150)``\n• #464e9f\n``#464e9f` `rgb(70,78,159)``\n• #3e46a7\n``#3e46a7` `rgb(62,70,167)``\n• #353fb0\n``#353fb0` `rgb(53,63,176)``\n• #2c38b9\n``#2c38b9` `rgb(44,56,185)``\n• #2330c2\n``#2330c2` `rgb(35,48,194)``\n• #1a29cb\n``#1a29cb` `rgb(26,41,203)``\n• #1222d3\n``#1222d3` `rgb(18,34,211)``\n``#091adc` `rgb(9,26,220)``\n• #0013e5\n``#0013e5` `rgb(0,19,229)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0013e5 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "Introduction\n\nThe Adafruit Ultimate GPS Breakout board is an excellent way to get started with GPS and Arduino. Adafruit does an excellent job providing tutorials and code for the user. I would suggest checking out their provided tutorials and code before looking elsewhere.\n\nWhen I was working with the GPS, I made a few changes to the code that Adafruit provided based on how the Arduino handles floats (or doesn't handle them). The changes that I made are not particularly necessary depending on what it is being used for, but does increase the accuracy of the module on the code level. However, I suggest that you become familiar with how the unit works first before attempting to alter the code.\n\nSchematic\n\nBelow is a schematic for connecting the GPS module to the Arduino.", null, "There are several pins on the breakout board that are unused in this schematic. These other pins are not necessary for communication to the GPS module, but may be useful in other respects.\n\nHow to Read Latitude and Longitude from Output\n\nAdafruit has provided a number of different codes for a variety of applications. Specifically, I am going to go over the parsing code they provide and how to read the data for latititude and longitude. Below is the portion of the code that deals with the latitude and longitude data.\n\n// Test code for Adafruit GPS modules using MTK3329/MTK3339 driver\n//\n// This code shows how to listen to the GPS module in an interrupt\n// which allows the program to have more 'freedom' - just parse\n// when a new NMEA sentence is available! Then access data when\n// desired.\n//\n// Tested and works great with the Adafruit Ultimate GPS module\n// using MTK33x9 chipset\n// Pick one up today at the Adafruit electronics shop\n// and help support open source hardware & software! -ada\n\nSerial.print(\"Location: \");\nSerial.print(GPS.latitude, 4); Serial.print(GPS.lat);\nSerial.print(\", \");\nSerial.print(GPS.longitude, 4); Serial.println(GPS.lon);\n\nThis code uses the libraries that Adafruit has made available where they parse the data for the user. Here, you can see that GPS.latitude will call the float holding the latitude for the user (GPS being an instance of the Adafruit_GPS class and GPS.latitude being a property of the class). While GPS.lat will get you either 'N' or 'S' depending on the hemisphere.\n\nTranslating GPS Data\n\nHere we need to analyze the actual output more closely. The sample output that we will be working with is a latitude reading of 1234.5678 and a longitude reading of 12345.6789.\n\nIf latitude reading = 1234.5678, then the actual latitude is read as 12° 34.5678'.\n\nIf longitude reading = 12345.6789, then the actual longitude is read as 123º 45.6789'.\n\nThis means that the floats can require some processing depending on what data is needed. Personally, I prefer turning all the data into minutes of degrees.\n\nThus, 12° 34.5678' = 12*60 + 34.5678 = 754.5678' and 123º 45.6789' = 123*60 + 45.6789 = 7425.6789'.\n\nThis can then be converted into meters knowing that 1' ≈ 1852 meters (going off of the fact that each degree is approximately 111km).\n\nWith this knowledge, we can take two latitude and two longitude readings and determine the difference is meters between the two coordinates. I show how to do this in a tutorial on Making a Map (pending).\n\nProblem with Floats\n\nIf you look in the library provided by Adafruit, you will see that the data is parsed as follows (this is only a small portion of the entire code):\n\n/***********************************\nThis is our GPS library\n\nAdafruit invests time and resources providing this open source code,\n\nAll text above must be included in any redistribution\n****************************************/\n\n// parse out latitude\np = strchr(p, ',')+1;\nlatitude = atof(p);\n\np = strchr(p, ',')+1;\nif (p == 'N') lat = 'N';\nelse if (p == 'S') lat = 'S';\nelse if (p == ',') lat = 0;\nelse return false;\n\n// parse out longitude\np = strchr(p, ',')+1;\nlongitude = atof(p);\n\np = strchr(p, ',')+1;\nif (p == 'W') lon = 'W';\nelse if (p == 'E') lon = 'E';\nelse if (p == ',') lon = 0;\nelse return false;\n\nAll the data starts off as a string and is parsed according to what the data actually is. Here, you can see that the string containing the latitude and longitude are turned into floats with the atof(p) command.\n\nWhat I have found is that Arduino was not made to handle floats well (and doubles are not different than floats to the Arduino). Arduino floats are really only accurate up to seven digits while the GPS floats have up to eight or nine digits. So, what the Arduino does is given seven accurate digits and then guesses on the other two. Below is a picture of an experiment where the string from the GPS is printed followed by the atof(p) float that the original code is producing.", null, "Looking at this data, you can see that the difference between the true string value and the float estimate is as large as 0.0005, which may not seem like much, but this can translate to approximately 0.926 meters (using the math from above), which could have an effect on the accuracy of the GPS. Although, admittedly, this may not be a large enough difference to matter and the approximation of meters to degrees will probably not be accurate enough to warrant the 0.926m difference. My suggestion would be to skip the rest of the tutorial if this does not matter. However, if this does matter, there is a work around to get the string values.\n\nAltering the Library\n\nTo fix the problem, I introduce new variables to hold the latitude and longitude into the header file. This variable is a character array. Below is the line added to the header file:\n\nchar latit, longit; //added to get strings for latitude and longitude\n\nThe code the was displayed before is then altered to look like the following:\n\n/***********************************\nThis is our GPS library\n\nAdafruit invests time and resources providing this open source code,\n\nAll text above must be included in any redistribution\n\nAlterations by Jennifer Case\n****************************************/\n\n// parse out latitude\np = strchr(p, ',')+1;\nfor(int i=0;i<9;i++) {\nlatit[i] = p[i];\n}\nlatit = '\\0';\nlatitude = atof(p);\n\np = strchr(p, ',')+1;\nif (p == 'N') lat = 'N';\nelse if (p == 'S') lat = 'S';\nelse if (p == ',') lat = 0;\nelse return false;\n\n// parse out longitude\np = strchr(p, ',')+1;\nfor(int i=0;i<10;i++) {\nlongit[i] = p[i];\n}\nlongit = '\\0';\nlongitude = atof(p);\n\nThe new variables, latit and longit, can then be called similar to how the example parsing code called latitude and longitude previously. The only difference being that latit and longit are treated as strings rather than floats.\n\nBecause these coordinates are now strings, that can complicate the processing of them, so I have made a Coordinate class to handle the strings and distribute them in a more controlled manner.\n\nCoordinate Library\n\nThe coordinate library has been developed to work with the strings that will get parsed out with these modifications to the Adafruit library. I have attached the library to this tutorial.\n\nThere are uses in this library that will not be necessary to use (such as a few constuctors). Some of them have been made to function with the construction of a map.\n\nThe main constructor that will be of interest is the following:\n\n//constructs latitude and longitude from GPS string\n//degreeLen is determined by whether you are constructing\n//a latitude or a longitude. Use a degreeLen of 2 when\n//constructing a latitude and a degreeLen of 3 when\n//constructing a longitude.\nCoordinate::Coordinate(char* gpsString, int degreeLen) {\nint i=0;\nchar degree;\nchar minute;\nchar minflt;\n\n//Process string\nint j;\nfor(j=0;j<degreeLen;j++) {\ndegree[j] = gpsString[i];\ni++;\n}\ndegree[j] = '\\0';\n\nfor(j=0;j<2;j++) {\nminute[j] = gpsString[i];\ni++;\n}\nminute[j] = '\\0';\n\nminflt='0';\nfor(j=1;j<6;j++) {\nminflt[j]=gpsString[i];\ni++;\n}\nminflt[j] = '\\0';\n\n//calculate for and aft\nfore = 60*atol(degree) + atol(minute);\naft = atof(minflt); //atof is acceptable because of reduced digits\n}\n\nThis method takes two arguments, a string and an integer. The string is latit or longit from the previous code. The integer lets the class know whether to process the string as a latitude or longitude. As stated previously, longitudes have one extra digit in the degree portion. So while the degree portion of a latitude is two digits long (degreeLen = 2), the degree portion of a longitude is three digits long (degreeLen = 3).\n\nThis converts everything into minutes for the user and manages it in two sections a fore part and an aft. The fore portion is the whole minutes while the aft part is the decimal parts of the minutes.\n\nFor communication purposes, there is also a method that allows printing a coordinate through the Arduino's serial:\n\n//convert coordinate to a string and print through serial\nvoid Coordinate::print() {\nchar strFore, strAft, tempAft;\n\ndtostrf(aft,7,4,strAft);\n\nint i = 0;\nfor(i=0;i<6;i++) {\ntempAft[i] = strAft[i+2];\n}\ntempAft[i] = '\\0';\n\nsprintf(strFore, \"%02d\", fore);\nSerial.print(strFore); //this serial can be changed if using a mega or soft serial\nSerial.print(tempAft); //this serial can be changed if using a mega or soft serial\n}\n\nThere are more methods that can be used as well or added to the library to fit individual needs.\n\nJenn Case Tue, 11/26/2013 - 13:44\nAttachments\nCoordinate.h819 bytes" ]

[ null, "https://robotic-controls.com/sites/default/files/styles/large/public/images/Arduino-GPS.png", null, "https://robotic-controls.com/sites/default/files/styles/large/public/images/GPSFloatString.png", null ]

[ "Amelia (version 1.8.0)\n\namelia: AMELIA: Multiple Imputation of Incomplete Multivariate Data\n\nDescription\n\nRuns the bootstrap EM algorithm on incomplete data and creates imputed datasets.\n\nUsage\n\namelia(x, ...)\n\n# S3 method for amelia amelia(x, m = 5, p2s = 1, frontend = FALSE, ...)\n\n# S3 method for molist amelia(x, ...)\n\n# S3 method for default amelia(x, m = 5, p2s = 1, frontend = FALSE, idvars = NULL, ts = NULL, cs = NULL, polytime = NULL, splinetime = NULL, intercs = FALSE, lags = NULL, leads = NULL, startvals = 0, tolerance = 1e-04, logs = NULL, sqrts = NULL, lgstc = NULL, noms = NULL, ords = NULL, incheck = TRUE, collect = FALSE, arglist = NULL, empri = NULL, priors = NULL, autopri = 0.05, emburn = c(0, 0), bounds = NULL, max.resample = 100, overimp = NULL, boot.type = \"ordinary\", parallel = c(\"no\", \"multicore\", \"snow\"), ncpus = getOption(\"amelia.ncpus\", 1L), cl = NULL, ...)\n\nArguments\n\nx\n\neither a matrix, data.frame, a object of class \"amelia\", or an object of class \"molist\". The first two will call the default S3 method. The third a convenient way to perform more imputations with the same parameters. The fourth will impute based on the settings from moPrep and any additional arguments.\n\n...\n\nfurther arguments to be passed.\n\nm\n\nthe number of imputed datasets to create.\n\np2s\n\nan integer value taking either 0 for no screen output, 1 for normal screen printing of iteration numbers, and 2 for detailed screen output. See \"Details\" for specifics on output when p2s=2.\n\nfrontend\n\na logical value used internally for the GUI.\n\nidvars\n\na vector of column numbers or column names that indicates identification variables. These will be dropped from the analysis but copied into the imputed datasets.\n\nts\n\ncolumn number or variable name indicating the variable identifying time in time series data.\n\ncs\n\ncolumn number or variable name indicating the cross section variable.\n\npolytime\n\ninteger between 0 and 3 indicating what power of polynomial should be included in the imputation model to account for the effects of time. A setting of 0 would indicate constant levels, 1 would indicate linear time effects, 2 would indicate squared effects, and 3 would indicate cubic time effects.\n\nsplinetime\n\ninterger value of 0 or greater to control cubic smoothing splines of time. Values between 0 and 3 create a simple polynomial of time (identical to the polytime argument). Values k greater than 3 create a spline with an additional k-3 knotpoints.\n\nintercs\n\na logical variable indicating if the time effects of polytime should vary across the cross-section.\n\nlags\n\na vector of numbers or names indicating columns in the data that should have their lags included in the imputation model.\n\na vector of numbers or names indicating columns in the data that should have their leads (future values) included in the imputation model.\n\nstartvals\n\nstarting values, 0 for the parameter matrix from listwise deletion, 1 for an identity matrix.\n\ntolerance\n\nthe convergence threshold for the EM algorithm.\n\nlogs\n\na vector of column numbers or column names that refer to variables that require log-linear transformation.\n\nsqrts\n\na vector of numbers or names indicating columns in the data that should be transformed by a sqaure root function. Data in this column cannot be less than zero.\n\nlgstc\n\na vector of numbers or names indicating columns in the data that should be transformed by a logistic function for proportional data. Data in this column must be between 0 and 1.\n\nnoms\n\na vector of numbers or names indicating columns in the data that are nominal variables.\n\nords\n\na vector of numbers or names indicating columns in the data that should be treated as ordinal variables.\n\nincheck\n\na logical indicating whether or not the inputs to the function should be checked before running amelia. This should only be set to FALSE if you are extremely confident that your settings are non-problematic and you are trying to save computational time.\n\ncollect\n\na logical value indicating whether or not the garbage collection frequency should be increased during the imputation model. Only set this to TRUE if you are experiencing memory issues as it can significantly slow down the imputation process\n\narglist\n\nan object of class \"ameliaArgs\" from a previous run of Amelia. Including this object will use the arguments from that run.\n\nempri\n\nnumber indicating level of the empirical (or ridge) prior. This prior shrinks the covariances of the data, but keeps the means and variances the same for problems of high missingness, small N's or large correlations among the variables. Should be kept small, perhaps 0.5 to 1 percent of the rows of the data; a reasonable upper bound is around 10 percent of the rows of the data.\n\npriors\n\na four or five column matrix containing the priors for either individual missing observations or variable-wide missing values. See \"Details\" for more information.\n\nautopri\n\nallows the EM chain to increase the empirical prior if the path strays into an nonpositive definite covariance matrix, up to a maximum empirical prior of the value of this argument times n, the number of observations. Must be between 0 and 1, and at zero this turns off this feature.\n\nemburn\n\na numeric vector of length 2, where emburn is a the minimum EM chain length and emburn is the maximum EM chain length. These are ignored if they are less than 1.\n\nbounds\n\na three column matrix to hold logical bounds on the imputations. Each row of the matrix should be of the form c(column.number, lower.bound,upper.bound) See Details below.\n\nmax.resample\n\nan integer that specifies how many times Amelia should redraw the imputed values when trying to meet the logical constraints of bounds. After this value, imputed values are set to the bounds.\n\noverimp\n\na two-column matrix describing which cells are to be overimputed. Each row of the matrix should be a c(row,column) pair. Each of these cells will be treated as missing and replaced with draws from the imputation model.\n\nboot.type\n\nchoice of bootstrap, currently restricted to either \"ordinary\" for the usual non-parametric bootstrap and \"none\" for no bootstrap.\n\nparallel\n\nthe type of parallel operation to be used (if any). If missing, the default is taken from the option \"amelia.parallel\" (and if that is not set, \"no\").\n\nncpus\n\ninteger: the number of processes to be used in parallel operation: typically one would choose the number of available CPUs.\n\ncl\n\nan optional parallel or snow cluster for use if parallel = \"snow\". If not supplied, a cluster on the local machine is created for the duration of the amelia call.\n\nValue\n\nAn instance of S3 class \"amelia\" with the following objects:\n\nimputations\n\na list of length m with an imputed dataset in each entry. The class (matrix or data.frame) of these entries will match x.\n\nm\n\nan integer indicating the number of imputations run.\n\nmissMatrix\n\na matrix identical in size to the original dataset with 1 indicating a missing observation and a 0 indicating an observed observation.\n\ntheta\n\nAn array with dimensions \\((p+1)\\) by \\((p+1)\\) by \\(m\\) (where \\(p\\) is the number of variables in the imputations model) holding the converged parameters for each of the m EM chains.\n\nmu\n\nA \\(p\\) by \\(m\\) matrix of of the posterior modes for the complete-data means in each of the EM chains.\n\ncovMatrices\n\nAn array with dimensions \\((p)\\) by \\((p)\\) by \\(m\\) where the first two dimensions hold the posterior modes of the covariance matrix of the complete data for each of the EM chains.\n\ncode\n\na integer indicating the exit code of the Amelia run.\n\nmessage\n\nan exit message for the Amelia run\n\niterHist\n\na list of iteration histories for each EM chain. See documentation for details.\n\narguments\n\na instance of the class \"ameliaArgs\" which holds the arguments used in the Amelia run.\n\novervalues\n\na vector of values removed for overimputation. Used to reformulate the original data from the imputations.\n\nNote that the theta, mu and covMatrcies objects refers to the data as seen by the EM algorithm and is thusly centered, scaled, stacked, tranformed and rearranged. See the manual for details and how to access this information.\n\nMethods (by class)\n\n• amelia: Run additional imputations for Amelia output\n\n• molist: Perform multiple overimputation from moPrep\n\n• default: Run core Amelia algorithm\n\nDetails\n\nMultiple imputation is a method for analyzing incomplete multivariate data. This function will take an incomplete dataset in either data frame or matrix form and return m imputed datatsets with no missing values. The algorithm first creates a bootstrapped version of the original data, estimates the sufficient statistics (with priors if specified) by EM on this bootstrapped sample, and then imputes the missing values of the original data using the estimated sufficient statistics. It repeats this process m times to produce the m complete datasets where the observed values are the same and the unobserved values are drawn from their posterior distributions.\n\nThe function will start a \"fresh\" run of the algorithm if x is either a incomplete matrix or data.frame. In this method, all of the options will be user-defined or set to their default. If x is the output of a previous Amelia run (that is, an object of class \"amelia\"), then Amelia will run with the options used in that previous run. This is a convenient way to run more imputations of the same model.\n\nYou can provide Amelia with informational priors about the missing observations in your data. To specify priors, pass a four or five column matrix to the priors argument with each row specifying a different priors as such:\n\none.prior <- c(row, column, mean,standard deviation)\n\nor,\n\none.prior <- c(row, column, minimum, maximum, confidence).\n\nSo, in the first and second column of the priors matrix should be the row and column number of the prior being set. In the other columns should either be the mean and standard deviation of the prior, or a minimum, maximum and confidence level for the prior. You must specify your priors all as distributions or all as confidence ranges. Note that ranges are converted to distributions, so setting a confidence of 1 will generate an error.\n\nSetting a priors for the missing values of an entire variable is done in the same manner as above, but inputing a 0 for the row instead of the row number. If priors are set for both the entire variable and an individual observation, the individual prior takes precedence.\n\nIn addition to priors, Amelia allows for logical bounds on variables. The bounds argument should be a matrix with 3 columns, with each row referring to a logical bound on a variable. The first column should be the column number of the variable to be bounded, the second column should be the lower bounds for that variable, and the third column should be the upper bound for that variable. As Amelia enacts these bounds by resampling, particularly poor bounds will end up resampling forever. Amelia will stop resampling after max.resample attempts and simply set the imputation to the relevant bound.\n\nIf each imputation is taking a long time to converge, you can increase the empirical prior, empri. This value has the effect of smoothing out the likelihood surface so that the EM algorithm can more easily find the maximum. It should be kept as low as possible and only used if needed.\n\nAmelia assumes the data is distributed multivariate normal. There are a number of variables that can break this assumption. Usually, though, a transformation can make any variable roughly continuous and unbounded. We have included a number of commonly needed transformations for data. Note that the data will not be transformed in the output datasets and the transformation is simply useful for climbing the likelihood.\n\nAmelia can run its imputations in parallel using the methods of the parallel package. The parallel argument names the parallel backend that Amelia should use. Users on Windows systems must use the \"snow\" option and users on Unix-like systems should use \"multicore\". The multicore backend sets itself up automatically, but the snow backend requires more setup. You can pass a predefined cluster from the parallel::makePSOCKcluster function to the cl argument. Without this cluster, Amelia will attempt to create a reasonable default cluster and stop it once computation is complete. When using the parallel backend, users can set the number of CPUs to use with the ncpus argument. The defaults for these two arguments can be set with the options \"amelia.parallel\" and \"amelia.ncpus\".\n\nReferences\n\nHonaker, J., King, G., Blackwell, M. (2011). Amelia II: A Program for Missing Data. Journal of Statistical Software, 45(7), 1--47. URL http://www.jstatsoft.org/v45/i07/." ]

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[ "# Is the answer 6, 9, 300 or 11809.8?", null, "It was the analysis of question such as the one above that led to one of my early insights into mathematical misunderstandings. I discovered then that a common answer to this question was 11809.8. This is caused by students finding 310 and then dividing the whole thing by 5. So either students misunderstand the rules of precedence or they just can’t use their calculator. As you’ll see from above, I have since added targeted feedback in response to errors of this type.\n\nA more recent analysis has shown that the most common error is now to calculate 3 x 10/5 (which gives an answer of 6).  These students don’t really  understand what powers are at all – it is like saying that 23 is the same as 2 x 3.\n\nAnd there’s another common error, namely to find  3 x (1010)1/5 i.e. 3 x 102(which equals 300), demonstrating a confusion between powers and powers of 10 – or just using the wrong calculator button.\n\nThe correct answer, of course, is 9 (i.e. 32).\n\nIn addition to leading to additional insight into specific misunderstandings, this question demonstrates several principles of the whole approach. Firstly, although I have given the correct and incorrect answers as a series of options in the title of this blog post, it is a free text question. So students work out their own response rather than selecting one. I believe this gives more powerful evidence of the fact that these responses are the result of genuine misunderstandings (or careless calculator use) rather than being guesses. Of course, were I devising a mulitple-choice question now, I might use these as distractors – but I would not have thought of those particular options when the original question was written 10 years ago.\n\nThe fact that similar errors occur in different variants of the question also adds to the strength of the findings. Having said that, sometimes the fact that different variants behave differently can add insight too. In this case, the first error I have described (giving an answer of 11809.8) is less common in variants such as this one, than when the error results in a nice round number. It appears that students realise that an answer like 11809.8 is unlikely to be a correct solution – and so they double check.\n\nFinally, note that what was once the most common incorrect response does not have that honour any more. I’d like to think that this is because of the targeted feedback that has been added. In addition, I am adding a screencast to Maths for Science in an attempt to resolve the other misunderstandings. When you begin to appreciate the common misunderstandings of your students, you can attempt to do something about them.\n\nThis entry was posted in mathematical misunderstandings and tagged , . Bookmark the permalink." ]

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[ "Cody\n\n# Problem 660. Find a subset that divides the vector into equal halves\n\nSolution 98535\n\nSubmitted on 14 Jun 2012 by @bmtran (Bryant Tran)\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\n%% x = [1 2 3 4 5 6 7]; xi = split_it(x); assert(isequal(sum(x(xi)),sum(x)/2));\n\n2   Pass\n%% x = [2 2 2 2 2 2]; xi = split_it(x); assert(isequal(sum(x(xi)),sum(x)/2));\n\n3   Pass\n%% x = [2 5 4 5 4]; xi = split_it(x); assert(isequal(sum(x(xi)),sum(x)/2));\n\n4   Pass\n%% x = [1 3 1 1 9 7]; xi = split_it(x); assert(isequal(sum(x(xi)),sum(x)/2));\n\n5   Pass\n%% x = primes(100); xi = split_it(x); assert(isequal(sum(x(xi)),sum(x)/2));" ]

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[ "# 理解K近邻¶\n\n### 理论¶\n\nkNN是可用于监督学习的最简单的分类算法之一。这个想法是在特征空间中搜索测试数据的最近邻。我们将用下面的图片来研究它。", null, "### OpenCV中的kNN¶\n\nimport cv2 as cv\nimport numpy as np\nimport matplotlib.pyplot as plt\n# 包含(x,y)值的25个已知/训练数据的特征集\ntrainData = np.random.randint(0,100,(25,2)).astype(np.float32)\n# 用数字0和1分别标记红色或蓝色\nresponses = np.random.randint(0,2,(25,1)).astype(np.float32)\n# 取红色族并绘图\nred = trainData[responses.ravel()==0]\nplt.scatter(red[:,0],red[:,1],80,'r','^')\n# 取蓝色族并绘图\nblue = trainData[responses.ravel()==1]\nplt.scatter(blue[:,0],blue[:,1],80,'b','s')\nplt.show()\n\n\n1. 给新样本的标签取决于我们之前看到的kNN理论。如果要使用“最近邻居”算法，只需指定k=1即可，其中k是邻居数。\n2. k最近邻的标签。\n3. 衡量新加入到每个最近邻的相应距离。\n\nnewcomer = np.random.randint(0,100,(1,2)).astype(np.float32)\nplt.scatter(newcomer[:,0],newcomer[:,1],80,'g','o')\nknn = cv.ml.KNearest_create()\nknn.train(trainData, cv.ml.ROW_SAMPLE, responses)\nret, results, neighbours ,dist = knn.findNearest(newcomer, 3)\nprint( \"result: {}\\n\".format(results) )\nprint( \"neighbours: {}\\n\".format(neighbours) )\nprint( \"distance: {}\\n\".format(dist) )\nplt.show()\n\n\nresult: [[ 1.]]\nneighbours: [[ 1. 1. 1.]]\ndistance: [[ 53. 58. 61.]]", null, "# 10个新加入样本\nnewcomers = np.random.randint(0,100,(10,2)).astype(np.float32)\nret, results,neighbours,dist = knn.findNearest(newcomer, 3)\n# 结果包含10个标签\n\n\n### 附加资源¶\n\n1. NPTEL关于模式识别的注释，第11章" ]

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[ "# Java String lines() Method Tutorial\n\nIn this section, we will learn what the String lines() method is and how to use it in Java.\n\n## What is Java String lines() Method?\n\nThe Java String lines() method is used to create a stream of lines from a string value.\n\nBasically, this method looks for newline characters in a string value and splits the entire string into multiple lines using these new line characters in order to create a stream.\n\n## Java lines() Method Syntax:\n\n`public Stream lines()`\n\n## lines() Method Parameters:\n\nThis method does not take an argument.\n\n## lines() Method Return Value:\n\nThe return value of this method is a stream object that contains the lines of the target string value.\n\n## Example: using String lines() method\n\n```import java.util.stream.Stream;\n\npublic class Main {\n\npublic static void main(String[] args) throws Exception{\n\nStream stream = \"This is the line one \\n This is the line two \\n This is the line three \\n This is the line four\".lines();\nstream.forEach(System.out::println);\n}\n}\n```\n\nOutput:\n\n```This is the line one\n\nThis is the line two\n\nThis is the line three\n\nThis is the line four```" ]

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[ "# C# - Type Conversion\n\nType conversion is converting one type of data to another type. It is also known as Type Casting. In C#, type casting has two forms −\n\n• Implicit type conversion − These conversions are performed by C# in a type-safe manner. For example, are conversions from smaller to larger integral types and conversions from derived classes to base classes.\n\n• Explicit type conversion − These conversions are done explicitly by users using the pre-defined functions. Explicit conversions require a cast operator.\n\nThe following example shows an explicit type conversion −\n\n```using System;\n\nnamespace TypeConversionApplication {\nclass ExplicitConversion {\nstatic void Main(string[] args) {\ndouble d = 5673.74;\nint i;\n\n// cast double to int.\ni = (int)d;\nConsole.WriteLine(i);\n}\n}\n}\n```\n\nWhen the above code is compiled and executed, it produces the following result −\n\n```5673\n```\n\n## C# Type Conversion Methods\n\nC# provides the following built-in type conversion methods −\n\nSr.No. Methods & Description\n1\n\nToBoolean\n\nConverts a type to a Boolean value, where possible.\n\n2\n\nToByte\n\nConverts a type to a byte.\n\n3\n\nToChar\n\nConverts a type to a single Unicode character, where possible.\n\n4\n\nToDateTime\n\nConverts a type (integer or string type) to date-time structures.\n\n5\n\nToDecimal\n\nConverts a floating point or integer type to a decimal type.\n\n6\n\nToDouble\n\nConverts a type to a double type.\n\n7\n\nToInt16\n\nConverts a type to a 16-bit integer.\n\n8\n\nToInt32\n\nConverts a type to a 32-bit integer.\n\n9\n\nToInt64\n\nConverts a type to a 64-bit integer.\n\n10\n\nToSbyte\n\nConverts a type to a signed byte type.\n\n11\n\nToSingle\n\nConverts a type to a small floating point number.\n\n12\n\nToString\n\nConverts a type to a string.\n\n13\n\nToType\n\nConverts a type to a specified type.\n\n14\n\nToUInt16\n\nConverts a type to an unsigned int type.\n\n15\n\nToUInt32\n\nConverts a type to an unsigned long type.\n\n16\n\nToUInt64\n\nConverts a type to an unsigned big integer.\n\nThe following example converts various value types to string type −\n\n```using System;\n\nnamespace TypeConversionApplication {\nclass StringConversion {\nstatic void Main(string[] args) {\nint i = 75;\nfloat f = 53.005f;\ndouble d = 2345.7652;\nbool b = true;\n\nConsole.WriteLine(i.ToString());\nConsole.WriteLine(f.ToString());\nConsole.WriteLine(d.ToString());\nConsole.WriteLine(b.ToString());\n\n}\n}\n}\n```\n\nWhen the above code is compiled and executed, it produces the following result −\n\n```75\n53.005\n2345.7652\nTrue\n```" ]

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[ "## Ordinals in googology\n\nA forum where anything goes. Introduce yourselves to other members of the forums, discuss how your name evolves when written out in the Game of Life, or just tell us how you found it. This is the forum for \"non-academic\" content.\ngameoflifemaniac\nPosts: 1191\nJoined: January 22nd, 2017, 11:17 am\nLocation: There too\n\n### Re: Ordinals in googology\n\nMoosey wrote:The Veblen function is actually pretty simple--\nphi_0(n) = w^n,\nphi_n+1(0) = the 0th fixed point of phi_n(a),\nphi_n(0) for limit ordinal n = the limit of {phi_(n)(0), phi_(n), (0) phi_(n)(0),...},\nphi_n(m+1) = the next fixed point like phi_n(m),\nphi_n(m) for limit ordinal m = the limit of {phi_n(m),phi_n(m),phi_n(m),...}.\n\nThe multivariable Veblen function is basically the same, but it's organized like so: phi(mth entry, m-1th entry,...3rd entry,2nd entry, 1st entry)\nThe (n+1)th entry indicates fixed points in the nth entry, so gamma_0 = phi(1,0,0) because it's a|->phi_a(0)\nCan you give some examples? I still don't really get it.\nI was so socially awkward in the past and it will haunt me for the rest of my life.\n\nCode: Select all\n\nx = 21, y = 5, rule = B3/S23\nb2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!\n\n\nMoosey\nPosts: 3555\nJoined: January 27th, 2019, 5:54 pm\nLocation: A house, or perhaps the OCA board. Or [click to not expand]\nContact:\n\n### Re: Ordinals in googology\n\ngameoflifemaniac wrote:\nNovember 2nd, 2019, 10:15 am\nMoosey wrote:The Veblen function is actually pretty simple--\nphi_0(n) = w^n,\nphi_n+1(0) = the 0th fixed point of phi_n(a),\nphi_n(0) for limit ordinal n = the limit of {phi_(n)(0), phi_(n), (0) phi_(n)(0),...},\nphi_n(m+1) = the next fixed point like phi_n(m),\nphi_n(m) for limit ordinal m = the limit of {phi_n(m),phi_n(m),phi_n(m),...}.\n\nThe multivariable Veblen function is basically the same, but it's organized like so: phi(mth entry, m-1th entry,...3rd entry,2nd entry, 1st entry)\nThe (n+1)th entry indicates fixed points in the nth entry, so gamma_0 = phi(1,0,0) because it's a|->phi_a(0)\nCan you give some examples? I still don't really get it.\nphi_1(0) = e_0\nphi_2(5) = z_5\nphi_4(0) = a|->h_a\nphi_w(0) = sup{e_0,z_0,h_0,...}\nphi_w+1(0) = a|->phi_w(a)\nI am a prolific creator of many rather pathetic googological functions\n\nMy CA rules can be found here\n\nAlso, the tree game\nBill Watterson once wrote: \"How do soldiers killing each other solve the world's problems?\"\n\nNanho walåt derwo esaato?\n\ngameoflifemaniac\nPosts: 1191\nJoined: January 22nd, 2017, 11:17 am\nLocation: There too\n\n### Re: Ordinals in googology\n\nMoosey wrote:\nNovember 2nd, 2019, 10:34 am\ngameoflifemaniac wrote:\nNovember 2nd, 2019, 10:15 am\nMoosey wrote:The Veblen function is actually pretty simple--\nphi_0(n) = w^n,\nphi_n+1(0) = the 0th fixed point of phi_n(a),\nphi_n(0) for limit ordinal n = the limit of {phi_(n)(0), phi_(n), (0) phi_(n)(0),...},\nphi_n(m+1) = the next fixed point like phi_n(m),\nphi_n(m) for limit ordinal m = the limit of {phi_n(m),phi_n(m),phi_n(m),...}.\n\nThe multivariable Veblen function is basically the same, but it's organized like so: phi(mth entry, m-1th entry,...3rd entry,2nd entry, 1st entry)\nThe (n+1)th entry indicates fixed points in the nth entry, so gamma_0 = phi(1,0,0) because it's a|->phi_a(0)\nCan you give some examples? I still don't really get it.\nphi_1(0) = e_0\nphi_2(5) = z_5\nphi_4(0) = a|->h_a\nphi_w(0) = sup{e_0,z_0,h_0,...}\nphi_w+1(0) = a|->phi_w(a)\nSo\nphi_0(0) = w\nphi_0(1) = w^w\nphi_0(2) = w^w^w\nphi_1(0) = e0\nphi_1(1) = e1\nphi_1(2) = e2\nphi_2(0) = z0\nphi_2(1) = z1\nphi_2(2) = z2\n.\n.\n.\nand\nphi_w(0) = Gamma(0), kind of a limit of new kinds of fixed points?\n\nAlso, is\nz1 = z0^z0^z0^z0^z0^...\nz2 = z1^z1^z1^z1^z1^...\n.\n.\n.\n?\n\nAnd how do you write\nphi_3(n) = ?\nphi_4(n) = ?\n.\n.\n.\nI was so socially awkward in the past and it will haunt me for the rest of my life.\n\nCode: Select all\n\nx = 21, y = 5, rule = B3/S23\nb2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!\n\n\nMoosey\nPosts: 3555\nJoined: January 27th, 2019, 5:54 pm\nLocation: A house, or perhaps the OCA board. Or [click to not expand]\nContact:\n\n### Re: Ordinals in googology\n\ngameoflifemaniac wrote:\nNovember 2nd, 2019, 11:19 am\nMoosey wrote:\nNovember 2nd, 2019, 10:34 am\ngameoflifemaniac wrote:\nNovember 2nd, 2019, 10:15 am\n\nCan you give some examples? I still don't really get it.\nphi_1(0) = e_0\nphi_2(5) = z_5\nphi_4(0) = a|->h_a\nphi_w(0) = sup{e_0,z_0,h_0,...}\nphi_w+1(0) = a|->phi_w(a)\nSo\nphi_0(0) = w\nphi_0(1) = w^w\nphi_0(2) = w^w^w\nNo, phi_0(0) = 1, phi_0(1) = w, phi_0(2) = w^2\ngameoflifemaniac wrote:\nNovember 2nd, 2019, 10:15 am\nphi_1(0) = e0\nphi_1(1) = e1\nphi_1(2) = e2\nphi_2(0) = z0\nphi_2(1) = z1\nphi_2(2) = z2\nYes\ngameoflifemaniac wrote:\nNovember 2nd, 2019, 10:15 am\n.\n.\n.\nand\nphi_w(0) = Gamma(0), kind of a limit of new kinds of fixed points?\nNo. gamma_0 = phi_gamma_0(0). phi_w(0) << gamma_0. Actually I don't know why phi_w(0), which is a rather important ordinal being the supremum of {1,e_0, z_0, ...} does not have a name besides phi_w(0), but that's a real problem because everyone (myself included) that knows about ordinals has (or is) mistakenly thought that gamma_0 was the supremum of {1,e_0,z_0}...\ngameoflifemaniac wrote:\nNovember 2nd, 2019, 10:15 am\nAlso, is\nz1 = z0^z0^z0^z0^z0^...\nz2 = z1^z1^z1^z1^z1^...\n.\n.\n.\n?\nI think those are e_(z_0 +1) and e_(z_1 +1), not the much larger z_1 and z_2.\nI think that for limit ordinal n > some point between e_0 and z_0, e_(n+1) = the 0th a|->n^a, or n^^w, which is why e_(W+1) = the 0th a|->W^a\ngameoflifemaniac wrote:\nNovember 2nd, 2019, 10:15 am\nAnd\nphi_3(n) = ?\nphi_4(n) = ?\n.\n.\n.\nthe former is η_n, also known as the nth a|->z_a. η_0 can be though of as z_z_z_z_z_z_z...\nThe latter is the nth a|->η_a.\nGenerally we call things beyond eta-naught by their names in the Veblen function anyways, so there isn't a much better (or any other canonical) name for phi_4(n) than phi_4(n).\nLast edited by Moosey on November 2nd, 2019, 11:39 am, edited 1 time in total.\nI am a prolific creator of many rather pathetic googological functions\n\nMy CA rules can be found here\n\nAlso, the tree game\nBill Watterson once wrote: \"How do soldiers killing each other solve the world's problems?\"\n\nNanho walåt derwo esaato?\n\ngameoflifemaniac\nPosts: 1191\nJoined: January 22nd, 2017, 11:17 am\nLocation: There too\n\n### Re: Ordinals in googology\n\nOk, now I understand everything, but this\nhow do you construct e1, e2, e3... from e0; z1, z2, z3... from z0; and η1, η2, η3... from η0?\nAnd how do you define Gamma(1), Gamma(2), Gamma(3)...?\nWhat's the ϑ function, θ function, ψ function and Ω? And can you explain how does the multivariable phi function work?\nI was so socially awkward in the past and it will haunt me for the rest of my life.\n\nCode: Select all\n\nx = 21, y = 5, rule = B3/S23\nb2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!\n\n\ntestitemqlstudop\nPosts: 1367\nJoined: July 21st, 2016, 11:45 am\nLocation: in catagolue\nContact:\n\n### Re: Ordinals in googology\n\nIs the well ordering of w_1 provable?\n\nMoosey\nPosts: 3555\nJoined: January 27th, 2019, 5:54 pm\nLocation: A house, or perhaps the OCA board. Or [click to not expand]\nContact:\n\n### Re: Ordinals in googology\n\ngameoflifemaniac wrote:\nNovember 2nd, 2019, 11:38 am\nOk, now I understand everything, but this\nhow do you construct e1, e2, e3... from e0; z1, z2, z3... from z0; and η1, η2, η3... from η0?\ne_1 = sup{e_0 +1,w^(e_0 +1), w^w^(e_0 +1),...}\ne_2 = sup{e_1 +1,w^(e_1 +1), w^w^(e_1 +1),...}\n...\nz_1 = sup{z_0 +1,e_(z_0 +1),e_e_(z_0 +1)...}\n...\nh_1 = sup{h_0 +1,z_(h_0 +1),z_z_(h_0 +1)...}\nFor the epsilons, you could alternatively say that, very loosely,\ne_(n+1) = (e_n)^^w\ngameoflifemaniac wrote:\nNovember 2nd, 2019, 11:38 am\nAnd how do you define Gamma(1), Gamma(2), Gamma(3)...?\nThe next fixed points of phi_a(0) after gamma_0\ngameoflifemaniac wrote:\nNovember 2nd, 2019, 11:38 am\nWhat's the ϑ function, θ function, ψ function and Ω?\nThose are OCFs, which have gotten a bit of discussion here. Ω is w_1 = the first uncountable ordinal\ngameoflifemaniac wrote:\nNovember 2nd, 2019, 11:38 am\nAnd can you explain how does the multivariable phi function work?\nMoosey wrote:\nNovember 2nd, 2019, 7:16 am\nThe multivariable Veblen function is basically the same, but it's organized like so: phi(mth entry, m-1th entry,...3rd entry,2nd entry, 1st entry)\nThe (n+1)th entry indicates fixed points in the nth entry, so gamma_0 = phi(1,0,0) because it's a|->phi_a(0)\nSo the 3rd entry does to the second entry what the second entry does to the first--series of fixed points for the previous entry\nI am a prolific creator of many rather pathetic googological functions\n\nMy CA rules can be found here\n\nAlso, the tree game\nBill Watterson once wrote: \"How do soldiers killing each other solve the world's problems?\"\n\nNanho walåt derwo esaato?\n\ngameoflifemaniac\nPosts: 1191\nJoined: January 22nd, 2017, 11:17 am\nLocation: There too\n\n### Re: Ordinals in googology\n\nThanks now I understand everything. But what about w-1? Does it exist?\nI was so socially awkward in the past and it will haunt me for the rest of my life.\n\nCode: Select all\n\nx = 21, y = 5, rule = B3/S23\nb2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!\n\n\ntoroidalet\nPosts: 1138\nJoined: August 7th, 2016, 1:48 pm\nLocation: My computer\nContact:\n\n### Re: Ordinals in googology\n\ntestitemqlstudop wrote:\nNovember 2nd, 2019, 11:47 pm\nIs the well ordering of w_1 provable?\nIsn't well-orderedness part of the definition of an ordinal?\n\nIf I'm missing something, then see this proof (set X to be the set of natural numbers to get the conclusion that ω_1 exists and is well-ordered)\ngameoflifemaniac wrote:\nNovember 3rd, 2019, 12:33 pm\nThanks now I understand everything. But what about w-1? Does it exist?\nIf such an ordinal existed, it would be an infinite ordinal less than ω, which contradicts its definition as the smallest infinite ordinal. Limit ordinals are not successors and so subtraction is undefined in general.\nHowever, surreal numbers include all ordinals and real numbers and things like ω-1, 1/ω, and √ω.\n\"Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life.\"\n\n-Terry Pratchett\n\ngameoflifemaniac\nPosts: 1191\nJoined: January 22nd, 2017, 11:17 am\nLocation: There too\n\n### Re: Ordinals in googology\n\nInteresting. If Gamma(0) is phi_phi_phi...(0)(0)(0), is Gamma(1) phi_phi_phi...(1)(1)(1), Gamma(2) phi_phi_phi...(2)(2)(2) etc.?\nAlso, I heard somewhere that w+2 and 2+w aren't the same. Is that true?\nI was so socially awkward in the past and it will haunt me for the rest of my life.\n\nCode: Select all\n\nx = 21, y = 5, rule = B3/S23\nb2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!\n\n\nMoosey\nPosts: 3555\nJoined: January 27th, 2019, 5:54 pm\nLocation: A house, or perhaps the OCA board. Or [click to not expand]\nContact:\n\n### Re: Ordinals in googology\n\ngameoflifemaniac wrote:\nNovember 4th, 2019, 9:11 am\nInteresting. If Gamma(0) is phi_phi_phi...(0)(0)(0), is Gamma(1) phi_phi_phi...(1)(1)(1), Gamma(2) phi_phi_phi...(2)(2)(2) etc.?\nno. Probably gamma_1 = sup{gamma_0+1,phi_gamma_0+1(0),phi_phi_gamma_0+1(0)(0),etc.}\ngameoflifemaniac wrote:\nNovember 4th, 2019, 9:11 am\nAlso, I heard somewhere that w+2 and 2+w aren't the same. Is that true?\nHOW DID YOU MAKE IT THIS FAR WITHOUT KNOWING THAT WAS TRUE?\nYes, w+2 = {0,1,2,3...w,w+1}, which is > 2+w = {0,0,0,1,2,3,...}\nThe same goes for w+1 and 1+w, or 2w and w2 (the former is = w, and the latter is larger)\n\nOn an unrelated note, I should update the definition of // from ah:\n\nCode: Select all\n\nah_n{$_0//(b+1),$_1} = ah_n((($_0)(@^_n)ah_n{$_0//b})#{//b,$_1}) ah_n{$_0//0} = ah_n{$_0} ah_n{$_0//b,$_1} = ah_n{$_0//(b[n]),$_1} for lim ord b Else: apply ah's rules. (e.g. ah_{$_0//,0,0,w+1} = ah_{$_0//,w+1,w+1,w}) Where ($_1)(@^_n)a =\n($_1)@(ah_n(($_1)(@^_n)(a-1))), a > 1,\nah_n{$_1}, a=1 This makes expressions such as ah_5{w+5//7,w+100223} valid. Additionally, it allows for a better @@: Code: Select all ah_n{$_0@@0} = ah_n{$_0} ah_n{$_0@@a,$_1} = ah_n{$_0@@a[n],$_1},a a lim ord ah_n{$_0@@(a+1),$_1} = ah_n({$_0//}#{$_0@@(a),$_1}\nApply basic ah rules otherwise\n\nThis, of course, can be extended indefinitely:\n\nCode: Select all\n\nah_n{$_0(@@b)0} = ah_n{$_0}\nah_n{$_0((@@b)a,$_1} = ah_n{$_0(@@b)a[n],$_1},a a lim ord\nah_n{$_0(@@0)(a+1),$_1} = ah_n({$_0//}#{$_0@@(a),$_1} ah_n{$_0(@@b+1)(a+1),$_1} = ah_n({$_0(@@b)}#{$_0@@(a),$_1}\nah_n{$_0(@@b)$_2} = ah_n{$_0(@@b[n])$_2}, if b is a lim ord\nApply basic ah rules otherwise\nWhich of course leads to my new fastest-growing-function (by Moosey):\nah_n{gamma_0(@@gamma_0)gamma_0}\nLast edited by Moosey on November 5th, 2019, 7:26 am, edited 1 time in total.\nI am a prolific creator of many rather pathetic googological functions\n\nMy CA rules can be found here\n\nAlso, the tree game\nBill Watterson once wrote: \"How do soldiers killing each other solve the world's problems?\"\n\nNanho walåt derwo esaato?\n\ngameoflifemaniac\nPosts: 1191\nJoined: January 22nd, 2017, 11:17 am\nLocation: There too\n\n### Re: Ordinals in googology\n\nMoosey wrote:\nNovember 4th, 2019, 9:54 am\nHOW DID YOU MAKE IT THIS FAR WITHOUT KNOWING THAT WAS TRUE?\nUmm what do you mean? I knew that, but I wanted to be 100% sure.\nMoosey wrote:\nNovember 4th, 2019, 9:54 am\nno. Probably gamma_1 = sup{gamma_0+1,phi_gamma_0+1(0),phi_phi_gamma_0+1(0)(0),etc.}\nProbably?\nI was so socially awkward in the past and it will haunt me for the rest of my life.\n\nCode: Select all\n\nx = 21, y = 5, rule = B3/S23\nb2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!\n\n\nPosts: 1971\nJoined: November 8th, 2014, 8:48 pm\nLocation: Getting a snacker from R-Bee's\n\n### Re: Ordinals in googology\n\ngameoflifemaniac wrote:\nNovember 4th, 2019, 10:19 am\nMoosey wrote:\nNovember 4th, 2019, 9:54 am\nno. Probably gamma_1 = sup{gamma_0+1,phi_gamma_0+1(0),phi_phi_gamma_0+1(0)(0),etc.}\nProbably?\nYes. Remember, phi(1,0,a) enumerates fixed points of phi(0,a,0).\nThen phi(1,1,a) enumerates fixed points of phi(1,0,a), phi(1,2,a) the fixed points of phi(1,1,a), and so on until\nphi(2,0,a) enumerates the fixed points of (1,a,0), phi(3,0,a) enumerates the fixed points of phi(2,a,0), and finally\nphi(1,0,0,a) enumerates the fixed points of phi(a,0,0).\nThis continues until you reach the Small Veblen Ordinal, psi(W^W^w).\nLifeWiki: Like Wikipedia but with more spaceships. [citation needed]", null, "testitemqlstudop\nPosts: 1367\nJoined: July 21st, 2016, 11:45 am\nLocation: in catagolue\nContact:\n\n### Re: Ordinals in googology\n\ntoroidalet wrote:\nNovember 3rd, 2019, 3:43 pm\ntestitemqlstudop wrote:\nNovember 2nd, 2019, 11:47 pm\nIs the well ordering of w_1 provable?\nIsn't well-orderedness part of the definition of an ordinal?\n\nIf I'm missing something, then see this proof (set X to be the set of natural numbers to get the conclusion that ω_1 exists and is well-ordered)\nSeveral things:\n\nThe well ordering of w_1 implies the well ordering of all countable ordinals.\n\nAm I dumb or does that immediately prove the infinite-time computability of w_1^CK?\n\nAll proof systems have PTO < w_1, so wouldn't a proof be part of the most powerful proof system?\n\ngameoflifemaniac\nPosts: 1191\nJoined: January 22nd, 2017, 11:17 am\nLocation: There too\n\n### Re: Ordinals in googology\n\nCan you give me some examples of values of Weiermann's ϑ? Maybe I'll understand it this way\nI was so socially awkward in the past and it will haunt me for the rest of my life.\n\nCode: Select all\n\nx = 21, y = 5, rule = B3/S23\nb2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!\n\n\nMoosey\nPosts: 3555\nJoined: January 27th, 2019, 5:54 pm\nLocation: A house, or perhaps the OCA board. Or [click to not expand]\nContact:\n\n### Re: Ordinals in googology\n\ngameoflifemaniac wrote:\nNovember 5th, 2019, 1:56 pm\nCan you give me some examples of values of Weiermann's ϑ? Maybe I'll understand it this way\nWhy that one? I've never even heard of that one! Maybe I can explain some simple OCF like Madore's ψ or some simpler one.\n\nHere's a simple example:\n\nCode: Select all\n\nC_0(α)={0,1,Ω}\nC_n+1(α)={γ+δ,ψ(η)|γ,δ,η∈C_n(α);η<α}\nC(α)=⋃(n<ω)C_n(α)\nψ(α)=min{β∈Ω|β∉C(α)}\nψ(0) = w (the minimum value you can't get by adding finitely many finite numbers together)\nψ(1) = w^2 (The minimum value you cannot get by adding finitely many ws together)\nψ(n) = w^(n+1) for finite n\nψ(w) = w^w\netc.\nψ(e_0) = e_0\nψ(e_0 + 1) = e_0\n... epsilon_0 for a long time\nψ(W) = e_0\nψ(W+1) = (e_0)w (I think)\nThe limit of ψ = e_w\n\nSuppose we introduce multiplication:\n\nCode: Select all\n\nC_0(α)={0,1,Ω}\nC_n+1(α)={γ+δ,γδ,ψ(η)|γ,δ,η∈C_n(α);η<α}\nC(α)=⋃(n<ω)C_n(α)\nψ(α)=min{β∈Ω|β∉C(α)}\nNow, it starts off at ψ(0):\nStill w\nψ(1) = the smallest ordinal you can't get by multiplying 0,1, or w finitely many times:\nw^w\nψ(n) for finite n = w^^(n+1)\nψ(w) = e_0\nψ(w2) = I think e_1\nψ(wn) for finite n = I think e_(n-1)\nψ(w^2) = e_w in that case\n...\n\nCode: Select all\n\nThat\nC_0(α)={0,1,Ω}\nC_n+1(α)={γ+δ,γδ,γ^δ,ψ(η)|γ,δ,η∈C_n(α);η<α}\nC(α)=⋃(n<ω)C_n(α)\nψ(α)=min{β∈Ω|β∉C(α)}\n\nCode: Select all\n\nMadore's psi\nC_0(α)={0,1,w,Ω}\nC_n+1(α)={γ+δ,γδ,γ^δ,ψ(η)|γ,δ,η∈C_n(α);η<α}\nC(α)=⋃(n<ω)C_n(α)\nψ(α)=min{β∈Ω|β∉C(α)}\nIt would start off more slowly but would be the same for all practical purposes.\n\nIf you want to see someone analyse it go to that link on the words Madore's psi.\n\nUnfortunately you can't just extend it thru the hyper operators:\n\nCode: Select all\n\nC_0(α)={0,1,w,Ω}\nC_n+1(α)={γ+δ,γδ,γ^δ,γ{ξ}δ,ψ(η)|γ,δ,ξ,η∈C_n(α);η<α}\nC(α)=⋃(n<ω)C_n(α)\nψ(α)=min{β∈Ω|β∉C(α)}\nUnless you define a system for ordinal hyperoperators. If you do, awesome. (Technically since BEAF arrays of {a,b,c} = a{c}b you could use ordinal BEAF, mentioned here, as a system for this. Don't ask me how ordinal beaf works because, I hate to tell you this, but I hardly understand normal BEAF.)\n\nBut you can define this extremely simple one:\n\nCode: Select all\n\nC_0(α)={0,Ω}\nC_n+1(α)={γ+δ,γδ,γ^δ,φ_γ(δ),ψ(η)|γ,δ,η∈C_n(α);η<α}\nC(α)=⋃(n<ω)C_n(α)\nψ(α)=min{β∈Ω|β∉C(α)}\nWhich would get you to gamma_0 at ψ(0). Yes, ψ of zero. Without so much as a 1 in the starter set.\n\nWith the finitely-many-entry Veblen function instead of the Veblen function, you could get to the SVO at 0, and with the any-amount-of-entries-in-the-function-including-infinitely-many you could get to the LVO.\n\nProbably another good system would be\n\nCode: Select all\n\nC_0(α)={0,Ω}\nC_n+1(α)={γ+δ,γδ,γ^δ,φ_γ(δ),ψ_0(η),w_γ|γ,δ,η∈C_n(α);η<α}\nC(α)=⋃(n<ω)C_n(α)\nψ_0(α)=min{β∈Ω|β∉C(α)}\nψ_1(α)=sup(C(α))\nIn this case you could have two functions in one (and also the first one, a modification of the one in the code box above the one I'm talking about right now, would be more powerful than the other one in the other code box), and the uncountable-generating one would reach a|->w_a which I think is about inaccessible.\nIn LaTeX:\n\nCode: Select all\n\n\\begin{eqnarray*}\nC_0(\\alpha) &=& \\{0, \\Omega\\}\\\\\nC_{n+1}(\\alpha) &=& \\{\\gamma + \\delta, \\gamma\\delta,\\gamma^\\delta,\\phi_{\\gamma}(\\delta),\\omega_\\gamma,\\psi_0(\\eta) | \\gamma, \\delta, \\eta \\in C_n (\\alpha); \\eta < \\alpha\\} \\\\\nC(\\alpha) &=& \\bigcup_{n < \\omega} C_n (\\alpha) \\\\\n\\psi_0(\\alpha) &=& \\min\\{\\beta \\in \\Omega|\\beta \\notin C(\\alpha)\\} \\\\\n\\psi_1(\\alpha) &=& \\sup(C(\\alpha)) \\\\\n\\end{eqnarray*}\nI am a prolific creator of many rather pathetic googological functions\n\nMy CA rules can be found here\n\nAlso, the tree game\nBill Watterson once wrote: \"How do soldiers killing each other solve the world's problems?\"\n\nNanho walåt derwo esaato?\n\ngameoflifemaniac\nPosts: 1191\nJoined: January 22nd, 2017, 11:17 am\nLocation: There too\n\n### Re: Ordinals in googology\n\nMoosey wrote:\nNovember 5th, 2019, 5:38 pm\nWhy that one? I've never even heard of that one!\nBecause it's used to write any countale ordinal larger than Gamma(0) and smaller than w_1CK in the googology wiki.\nAckermann's ordinal is Weiermann's ϑ(Ω^3), for example, but what does that mean?\nI was so socially awkward in the past and it will haunt me for the rest of my life.\n\nCode: Select all\n\nx = 21, y = 5, rule = B3/S23\nb2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!\n\n\nMoosey\nPosts: 3555\nJoined: January 27th, 2019, 5:54 pm\nLocation: A house, or perhaps the OCA board. Or [click to not expand]\nContact:\n\n### Re: Ordinals in googology\n\ngameoflifemaniac wrote:\nNovember 6th, 2019, 2:57 am\nMoosey wrote:\nNovember 5th, 2019, 5:38 pm\nWhy that one? I've never even heard of that one!\nBecause it's used to write any countale ordinal larger than Gamma(0) and smaller than w_1CK in the googology wiki.\nAckermann's ordinal is Weiermann's ϑ(Ω^3), for example, but what does that mean?\nIf you need that one specifically...\n... Then they have a nice explanation on the wiki:\nEssentially, it wrote: C(a,b) is the set of all ordinals constructible using only the following:\n0, all ordinals less than b, and Ω.\nFinite applications of c+d, κ↦ωκ [or w^c in other terms], and ϑ(n) for n < a, assuming n, c, and d are produceable using those methods or are 0 or W.\n\nThen, ϑ(a) is the smallest ordinal b such that α∈C(a,b), and b is greater than all the countable ordinals in C(a,b)\nI am a prolific creator of many rather pathetic googological functions\n\nMy CA rules can be found here\n\nAlso, the tree game\nBill Watterson once wrote: \"How do soldiers killing each other solve the world's problems?\"\n\nNanho walåt derwo esaato?\n\ngameoflifemaniac\nPosts: 1191\nJoined: January 22nd, 2017, 11:17 am\nLocation: There too\n\n### Re: Ordinals in googology\n\nMoosey wrote:\nNovember 6th, 2019, 7:39 am\ngameoflifemaniac wrote:\nNovember 6th, 2019, 2:57 am\nMoosey wrote:\nNovember 5th, 2019, 5:38 pm\nWhy that one? I've never even heard of that one!\nBecause it's used to write any countale ordinal larger than Gamma(0) and smaller than w_1CK in the googology wiki.\nAckermann's ordinal is Weiermann's ϑ(Ω^3), for example, but what does that mean?\nIf you need that one specifically...\n... Then they have a nice explanation on the wiki:\nEssentially, it wrote: C(a,b) is the set of all ordinals constructible using only the following:\n0, all ordinals less than b, and Ω.\nFinite applications of c+d, κ↦ωκ [or w^c in other terms], and ϑ(n) for n < a, assuming n, c, and d are produceable using those methods or are 0 or W.\n\nThen, ϑ(a) is the smallest ordinal b such that α∈C(a,b), and b is greater than all the countable ordinals in C(a,b)\nI don't really understand the definition. I would get it if you'd gave me some examples.\n\nAlso, am I bothering you in this topic? You all are asking much more complicated things than me and I don't want to disturb you.\nI was so socially awkward in the past and it will haunt me for the rest of my life.\n\nCode: Select all\n\nx = 21, y = 5, rule = B3/S23\nb2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!\n\n\nMoosey\nPosts: 3555\nJoined: January 27th, 2019, 5:54 pm\nLocation: A house, or perhaps the OCA board. Or [click to not expand]\nContact:\n\n### Re: Ordinals in googology\n\ngameoflifemaniac wrote:\nNovember 6th, 2019, 1:16 pm\nI don't really understand the definition. I would get it if you'd gave me some examples.\nI'm not that good with more complicated OCFs, so I can't answer that question.\ngameoflifemaniac wrote:\nNovember 6th, 2019, 1:16 pm\nAlso, am I bothering you in this topic? You all are asking much more complicated things than me and I don't want to disturb you.\nNo. Personally I like the activity. It keeps this thread up and I desperately want a response to my ah stuff (like this):\nMoosey wrote:\nNovember 4th, 2019, 9:54 am\nOn an unrelated note, I should update the definition of // from ah:\n\nCode: Select all\n\nah_n{$_0//(b+1),$_1} = ah_n((($_0)(@^_n)ah_n{$_0//b})#{//b,$_1}) ah_n{$_0//0} = ah_n{$_0} ah_n{$_0//b,$_1} = ah_n{$_0//(b[n]),$_1} for lim ord b Else: apply ah's rules. (e.g. ah_{$_0//,0,0,w+1} = ah_{$_0//,w+1,w+1,w}) Where ($_1)(@^_n)a =\n($_1)@(ah_n(($_1)(@^_n)(a-1))), a > 1,\nah_n{$_1}, a=1 This makes expressions such as ah_5{w+5//7,w+100223} valid. Additionally, it allows for a better @@: Code: Select all ah_n{$_0@@0} = ah_n{$_0} ah_n{$_0@@a,$_1} = ah_n{$_0@@a[n],$_1},a a lim ord ah_n{$_0@@(a+1),$_1} = ah_n({$_0//}#{$_0@@(a),$_1}\nApply basic ah rules otherwise\n\nThis, of course, can be extended indefinitely:\n\nCode: Select all\n\nah_n{$_0(@@b)0} = ah_n{$_0}\nah_n{$_0((@@b)a,$_1} = ah_n{$_0(@@b)a[n],$_1},a a lim ord\nah_n{$_0(@@0)(a+1),$_1} = ah_n({$_0//}#{$_0@@(a),$_1} ah_n{$_0(@@b+1)(a+1),$_1} = ah_n({$_0(@@b)}#{$_0@@(a),$_1}\nah_n{$_0(@@b)$_2} = ah_n{$_0(@@b[n])$_2}, if b is a lim ord\nApply basic ah rules otherwise\nWhich of course leads to my new fastest-growing-function (by Moosey):\nah_n{gamma_0(@@gamma_0)gamma_0}\nI am a prolific creator of many rather pathetic googological functions\n\nMy CA rules can be found here\n\nAlso, the tree game\nBill Watterson once wrote: \"How do soldiers killing each other solve the world's problems?\"\n\nNanho walåt derwo esaato?\n\ngameoflifemaniac\nPosts: 1191\nJoined: January 22nd, 2017, 11:17 am\nLocation: There too\n\n### Re: Ordinals in googology\n\nWhat is the large Veblen ordinal? I know it's defined by α↦φ_Ω^α(0), but can you even talk about phi of an uncountable cardinal?\nWhat is phi(1,1,0), phi(2,0,0)?\nAnd is the Ackermann ordinal equal to Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ...?\nLast edited by gameoflifemaniac on November 8th, 2019, 4:23 pm, edited 1 time in total.\nI was so socially awkward in the past and it will haunt me for the rest of my life.\n\nCode: Select all\n\nx = 21, y = 5, rule = B3/S23\nb2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!\n\n\nMoosey\nPosts: 3555\nJoined: January 27th, 2019, 5:54 pm\nLocation: A house, or perhaps the OCA board. Or [click to not expand]\nContact:\n\n### Re: Ordinals in googology\n\ngameoflifemaniac wrote:\nNovember 8th, 2019, 11:42 am\nWhat is the large Veblen ordinal? I know it's defined by α↦φ_Ω^α(0), but can you even talk about phi of an uncountable cardinal?\nNo, itś defined like this:\n\nCode: Select all\n\n1. w\n2. phi(1,0,0,0,0,0,0,0,...) w/ w 0´s (SVO)\n3. phi(1,0,0,0,0,0,...) w/ that many 0s\n4. phi(1,0,0,0,...) w/ that many 0s\n...\n\nThe LVO is the supremum of all of these\nI am a prolific creator of many rather pathetic googological functions\n\nMy CA rules can be found here\n\nAlso, the tree game\nBill Watterson once wrote: \"How do soldiers killing each other solve the world's problems?\"\n\nNanho walåt derwo esaato?\n\ngameoflifemaniac\nPosts: 1191\nJoined: January 22nd, 2017, 11:17 am\nLocation: There too\n\n### Re: Ordinals in googology\n\ngameoflifemaniac wrote:\nNovember 8th, 2019, 11:42 am\nWhat is phi(1,1,0), phi(2,0,0)?\nAnd is the Ackermann ordinal equal to Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ...?\nThis\nAlso, I can't appreciate your ah function cuz small brain\nI was so socially awkward in the past and it will haunt me for the rest of my life.\n\nCode: Select all\n\nx = 21, y = 5, rule = B3/S23\nb2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!\n\n\nMoosey\nPosts: 3555\nJoined: January 27th, 2019, 5:54 pm\nLocation: A house, or perhaps the OCA board. Or [click to not expand]\nContact:\n\n### Re: Ordinals in googology\n\ngameoflifemaniac wrote:\nNovember 8th, 2019, 11:42 am\nWhat is phi(1,1,0), phi(2,0,0)?\nAnd is the Ackermann ordinal equal to Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ...?\nAnd is the Ackermann ordinal equal to Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ...?\nNo.\nI think min a|->gamma_a = phi(1,1,0) but I could be wrong.\nphi(2,0,0) = min a|->phi(1,a,0) much like how gamma_0 = phi(1,0,0) = min a|->phi(a,0)\ngameoflifemaniac wrote:\nNovember 8th, 2019, 3:57 pm\nAlso, I can't appreciate your ah function cuz small brain\nSimply put, ah is my extension of something by pkmnQ\n\nSince Ę_0,a(n) is hardly more powerful than f_a(n), I extended it much farther-- for instance, ah_n(a@w) is at least f_a*n-ish, and ah_n(a(@@a)a) is probably much more powerful for large enough a. Using the extension of ah rules I will write now.\n(Can someone check how powerful it is? Is it e_(a+1)-ish?)\n\nah rules:\n\nCode: Select all\n\nPart one:\n\ndefine $_n as any entries (including no entries) in an array. It’s my symbol for we-don’t-care entries. In any one use of any one rule, if n is the same,$_n is the same\n\na#b = concatenation of a and b\n\nn@m =\nn, m = 1\n{n}#(n@(m-1)), m > 1, m not a lim ord\nn@(m[n]) if m is a lim ord\n\ng(a,n,B) =\nah_g(a-1,n,B) {B}, a > 1;\nn, a = 0\n\nah definition:\nRule 1.\nah^a_n {$_0} = g(a,n,$_0)\nRule 2.\nah_n ({$_1}#{z}) = ah_n {$_1}, z = 0\nRule 3.\nah_n{} = n+1\nRule 4.\nah_n{a+1,$_2} = ah^n_n{a,$_2}\nRule 5.\nah_n{a,$_3} = ah_n{a[n],$_3}, a a lim ord\nRule 6.\nah_n ((0@b)#{a+1,$_4}) = ah_(n+1) (((a+1)@b)#{a,$_4}), b > 0\nRule 7.\nah_n ((0@b)#{a,$_5}) = ah_n ((a@b)#{a[n],$_5}), a a lim ord & b > 0\n\nThe rest of the array rules:\nah_n{$_0//(b+1),$_1} = ah_n((($_0)(@^_n)ah_n{$_0//b})#{//b,$_1}) ah_n{$_0//0} = ah_n{$_0} ah_n{$_0//b,$_1} = ah_n{$_0//(b[n]),$_1} for lim ord b Else: apply ah's 7 rules, starting after legion bar. (e.g. ah_{$_0//,0,0,w+1} = ah_{$_0//,w+1,w+1,w}) ($_1)(@^_n)a =\n($_1)@(ah_n(($_1)(@^_n)(a-1))), a > 1,\nah_n{$_1}, a=1 ah_n{$_0@@0} = ah_n{$_0} ah_n{$_0@@a,$_1} = ah_n{$_0@@a[n],$_1},a a lim ord ah_n{$_0@@(a+1),$_1} = ah_n({$_0//}#{$_0@@(a),$_1}\nApply basic ah rules otherwise to everything after the @@\n\nah_n{$_0(@@b,$_2)0} = ah_n{$_0} ah_n{$_0((@@b,$_2)a,$_1} = ah_n{$_0(@@b)a[n],$_1},a a lim ord\nah_n{$_0(@@0)(a+1),$_1} = ah_n({$_0//}#{$_0(@@0)(a),$_1} ah_n{$_0(@@b+1,$_2)(a+1),$_1} = ah_n({$_0(@@b,$_2)}#{$_0@@(a),$_1}\nah_n{$_0(@@b,$_2)$_3} = ah_n{$_0(@@b[n],$_2)$_3}, if b is a lim ord\nApply basic ah rules otherwise to everything after the (@@$) or everything inside the (@@$) depending on what is necessary\n\nThese array rules can be applied without the ah_n of the array, but any n refers to the n in the ah subscript.\n\nNote that we can now nest things as powerful as (@@$) to make extremely powerful functions. What is the power of ah_n{gamma_0(@@gamma_0)gamma_0} ? EDIT: Redefine legion bars and the like in a way that formalizes the array rules without the ah: Code: Select all if unspecified, n is the subscript in the first ah that is in front of the array. Now we can redefine the rest of the array rules: The rest of the array rules: {$_0//(b+1),$_1} = ((($_0)(@^_n)ah_n{$_0//b})#{//b,$_1})\n{$_0//0} = {$_0}\n{$_0//b,$_1} = {$_0//(b[n]),$_1} for lim ord b\nElse: apply ah's 7 rules, starting after legion bar. This includes nesting and incrementing the subscript (e.g. ah_n{$_0//,0,0,w+1} = ah_(n+1){$_0//,w+1,w+1,w})\n\n($_1)(@^_n)a = ($_1)@(ah_n(($_1)(@^_n)(a-1))), a &gt; 1, ah_n{$_1}, a=1\n\n{$_0@@0} = {$_0}\n{$_0@@a,$_1} = {$_0@@a[n],$_1},a a lim ord\n{$_0@@(a+1),$_1} = ({$_0//}#{$_0@@(a),$_1} Apply basic ah rules otherwise to everything after the @@. See legion bar notes for more details {$_0(@@b,$_2)0} = {$_0}\n{$_0((@@b,$_2)a,$_1} = {$_0(@@b)a[n],$_1},a a lim ord {$_0(@@0)(a+1),$_1} = ({$_0//}#{$_0(@@0)(a),$_1}\n{$_0(@@b+1,$_2)(a+1),$_1} = ({$_0(@@b,$_2)}#{$_0@@(a),$_1} {$_0(@@b,$_2)$_3} = {$_0(@@b[n],$_2)$_3}, if b is a lim ord Apply basic ah rules otherwise to everything after the (@@$) or everything inside the (@@$) depending on what is necessary. See legion bar notes for more details. (Now things such as ah_n{a(@@(a(@@a)a))a} are well defined.) {$_0@@@(a+1),$_1} = {$_0(@@($_0@@@a,$_1))$_0,$_1}\n{$_0@@@0} = {$_0}\n{$_0@@@a,$_1} = {$_0@@@a[n],$_1}, a a lim ord.\nLast edited by Moosey on November 16th, 2019, 8:28 am, edited 2 times in total.\nI am a prolific creator of many rather pathetic googological functions\n\nMy CA rules can be found here\n\nAlso, the tree game\nBill Watterson once wrote: \"How do soldiers killing each other solve the world's problems?\"\n\nNanho walåt derwo esaato?\n\ngameoflifemaniac\nPosts: 1191\nJoined: January 22nd, 2017, 11:17 am\nLocation: There too\n\n### Re: Ordinals in googology\n\nWow.\n\nWhat is the Bachmann-Howard ordinal?\nI was so socially awkward in the past and it will haunt me for the rest of my life.\n\nCode: Select all\n\nx = 21, y = 5, rule = B3/S23\nb2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!" ]

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[ "# 0.6 Discrete structures recursion  (Page 3/8)\n\n Page 3 / 8\n\nThe function f is the function that satisfies the following two clauses:\n\nBasis Clause: f(0) = 0! = 1\n\nInductive Clause: For all natural number n,  f(n+1) = (n+1) f(n).\n\nNote that here Extremal Clause is not necessary, because the set of natural numbers can be defined recursively and that has the extremal clause in it. So there is no chance of other elements to come into the function being defined.\n\nUsing this definition, 3! can be found as follows:\n\nSince 0 ! = 1,   1 ! = 1 * 0 ! = 1 * 1 = 1 ,\n\nHence 2 ! = 2 * 1 ! = 2 * 1 = 2 .\n\nHence 3 ! = 3 * 2 ! = 3 * 2 * 1 = 6 .\n\nExample 6: The function f(n) = 2n + 1 for natural numbers n can be defined recursively as follows:\n\nThe function f is the function that satisfies the following two clauses:\n\nBasis Clause: f(0) = 1\n\nInductive Clause: For all natural number n,  f(n+1) = f(n) + 2 .\n\nSee above for the extremal clause.\n\nExample 7: The function f(n) = 2n for natural numbers n can be defined recursively as follows:\n\nThe function f is the function that satisfies the following two clauses:\n\nBasis Clause: f(0) = 1\n\nInductive Clause: For all natural number n,  f(n+1) = 2 f(n) .\n\nSee Example 5 for the extremal clause.\n\nExample 8: The function L from the set S of strings over {a, b} to the set of natural numbers that gives the length of a string can be defined recursively as follows:\n\nThe function L is the function that satisfies the following two clauses:\n\nBasis Clause: For symbols a and b of the alphabet,   L(a) = 1 and L(b) = 1.\n\nInductive Clause: For any string x and y of S,  L(xy) = L(x) + L(y) ,  where xy is the concatenation of strings x and y.\n\nSee Example 5 for the extremal clause.\n\nThis function L gives the number of a's and b's.\n\n## Recursive algorithm\n\nA recursive algorithm is an algorithm which calls itself with \"smaller (or simpler)\" input values, and which obtains the result for the current input by applying simple operations to the returned value for the smaller (or simpler) input. More generally if a problem can be solved utilizing solutions to smaller versions of the same problem, and the smaller versions reduce to easily solvable cases, then one can use a recursive algorithm to solve that problem. For example, the elements of a recursively defined set, or the value of a recursively defined function can be obtained by a recursive algorithm.\n\nIf a set or a function is defined recursively, then a recursive algorithm to compute its members or values mirrors the definition. Initial steps of the recursive algorithm correspond to the basis clause of the recursive definition and they identify the basis elements. They are then followed by steps corresponding to the inductive clause, which reduce the computation for an element of one generation to that of elements of the immediately preceding generation.\n\nIn general, recursive computer programs require more memory and computation compared with iterative algorithms, but they are simpler and for many cases a natural way of thinking about the problem.\n\nExample 1: Algorithm for finding the k-th even natural number Note here that this can be solved very easily by simply outputting 2*(k - 1) for a given k . The purpose here, however, is to illustrate the basic idea of recursion rather than solving the problem.\n\nwhat are the products of Nano chemistry?\nThere are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..\nlearn\nEven nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level\nlearn\nPreparation and Applications of Nanomaterial for Drug Delivery\nApplication of nanotechnology in medicine\nwhat is variations in raman spectra for nanomaterials\nI only see partial conversation and what's the question here!\nwhat about nanotechnology for water purification\nplease someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.\nDamian\nyes that's correct\nProfessor\nI think\nProfessor\nwhat is the stm\nis there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?\nRafiq\nindustrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong\nDamian\nHow we are making nano material?\nwhat is a peer\nWhat is meant by 'nano scale'?\nWhat is STMs full form?\nLITNING\nscanning tunneling microscope\nSahil\nhow nano science is used for hydrophobicity\nSantosh\nDo u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq\nRafiq\nwhat is differents between GO and RGO?\nMahi\nwhat is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq\nRafiq\nif virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION\nAnam\nanalytical skills graphene is prepared to kill any type viruses .\nAnam\nAny one who tell me about Preparation and application of Nanomaterial for drug Delivery\nHafiz\nwhat is Nano technology ?\nwrite examples of Nano molecule?\nBob\nThe nanotechnology is as new science, to scale nanometric\nbrayan\nnanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale\nDamian\nIs there any normative that regulates the use of silver nanoparticles?\nwhat king of growth are you checking .?\nRenato\nWhat fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?\nwhy we need to study biomolecules, molecular biology in nanotechnology?\n?\nKyle\nyes I'm doing my masters in nanotechnology, we are being studying all these domains as well..\nwhy?\nwhat school?\nKyle\nbiomolecules are e building blocks of every organics and inorganic materials.\nJoe\nanyone know any internet site where one can find nanotechnology papers?\nresearch.net\nkanaga\nsciencedirect big data base\nErnesto\nGot questions? Join the online conversation and get instant answers!", null, "By David Bourgeois", null, "By Samuel Madden", null, "By OpenStax", null, "By Sarah Warren", null, "By Sarah Warren", null, "By Madison Christian", null, "By John Gabrieli", null, "By Vongkol HENG", null, "By Danielrosenberger", null, "By OpenStax" ]

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[ "## Monday, May 28, 2012\n\n### Linear regression for a binary outcome: is it Kosher?\n\nRegression models are the most popular tool for modeling the relationship between an outcome and a set of inputs. Models can be used for descriptive, causal-explanatory, and predictive goals (but in very different ways! see Shmueli 2010 for more).\n\nThe family of regression models includes two especially popular members: linear regression and logistic regression (with probit regression more popular than logistic in some research areas). Common knowledge, as taught in statistics courses, is: use linear regression for a continuous outcome and logistic regression for a binary or categorical outcome. But why not use linear regression for a binary outcome? the two common answers are: (1) the linear regression can produce predictions that are not binary, and hence \"nonsense\" and (2) inference based on the linear regression coefficients will be incorrect.", null, "I admit that I bought into these \"truths\" for a long time, until I learned never to take any \"statistical truth\" at face value. First, let us realize that problem #1 relates to prediction and #2 to description and causal explanation. In other words, if issue #1 can be \"fixed\" somehow, then I might consider linear regression for prediction even if the inference is wrong (who cares about inference if I am only interested in predicting individual observations?). Similarly, if there is a fix for issue #2, then I might consider linear regression as a kosher inference mechanism even if it produces \"nonsense\" predictions.\n\nThe 2009 paper Linear versus logistic regression when the dependent variable is a dichotomy by Prof. Ottar Hellevik from Oslo University de-mystifies some of these issues. First, he gives some tricks that help avoid predictions outside the [0,1] range. The author identifies a few factors that contribute to \"nonsense predictions\" by linear regression:\n\n• interactions that are not accounted for in the regression\n• non-linear relationships between a predictor and the outcome\nThe suggested remedy for these issues is including interaction terms for categorical variables, and if numerical predictors are involved, then bucket them into bins and include those as dummies + interactions. So, if the goal is predicting a binary outcome, linear regression can be modified and used.\n\nNow to the inference issue. \"The problem with a binary dependent variable is that the homoscedasticity assumption (similar variation on the dependent variable for units with different values on the independent variable) is not satisfied... This seems to be the main basis for the widely held opinion that linear regression is inappropriate with a binary dependent variable\". Statistical theory tells us that violating the homoscedasticity assumption results in biased standard errors for the coefficients, and that the coefficients might not be the most precise in terms of variance. Yet, the coefficients themselves remain unbiased (meaning that with a sufficiently large sample they are \"on target\"). Hence, with a sufficiently large sample we need not worry! Precision is not an issue in very large samples, and hence the on-target coefficients are just what we need.\nI will add that another concern is that the normality assumption is violated: the residuals from a regression model on a binary outcome will not look very bell-shaped... Again, with a sufficiently large sample, the distribution does not make much difference, since the standard errors are so small anyway.", null, "Chart from Hellevik (2009)\nHellevik's paper pushes the envelope further in an attempt to explore \"how small can you go\" with your sample before getting into trouble. He uses simulated data and compares the results from logistic and linear regression for fairly small samples. He finds that the differences are minuscule.\n\nThe bottom line: linear regression is kosher for prediction if you take a few steps to accommodate non-linear relationships (but of course it is not guaranteed to produce better predictions than logistic regression!). For inference, for a sufficiently large sample where standard errors are tiny anyway, it is fine to trust the coefficients, which are in any case unbiased." ]

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[ "# 概述\n\nJava 语言中提供的数组是用来存储固定大小的同类型元素。\n\n# 声明数组变量\n\n``````dataType[] arrayRefVar; // 首选的方法\n\n//或\n\ndataType arrayRefVar[]; // 效果相同，但不是首选方法\n``````\n\n## 实例\n\n``````double[] myList; // 首选的方法\n\n//或\n\ndouble myList[]; // 效果相同，但不是首选方法\n``````\n\n# 创建数组\n\nJava 语言使用 new 操作符来创建数组，语法如下：\n\n``````arrayRefVar = new dataType[arraySize];\n``````\n\n• 一、使用 dataType [arraySize] 创建了一个数组。\n• 二、把新创建的数组的引用赋值给变量 arrayRefVar。\n\n``````dataType[] arrayRefVar = new dataType[arraySize];\n``````\n\n``````dataType[] arrayRefVar = {value0, value1, ..., valuek};\n``````\n\n## 实例\n\n``````public class TestArray {\npublic static void main(String[] args) {\n// 数组大小\nint size = 10;\n// 定义数组\ndouble[] myList = new double[size];\nmyList = 5.6;\nmyList = 4.5;\nmyList = 3.3;\nmyList = 13.2;\nmyList = 4.0;\nmyList = 34.33;\nmyList = 34.0;\nmyList = 45.45;\nmyList = 99.993;\nmyList = 11123;\n// 计算所有元素的总和\ndouble total = 0;\nfor (int i = 0; i < size; i++) {\ntotal += myList[i];\n}\nSystem.out.println(\"总和为： \" + total);\n}\n}\n``````\n\n``````总和为： 11367.373\n``````", null, "# 处理数组\n\n## 示例\n\n``````public class TestArray {\n\npublic static void main(String[] args) {\ndouble[] myList = {1.9, 2.9, 3.4, 3.5};\n\n// 打印所有数组元素\nfor (int i = 0; i < myList.length; i++) {\nSystem.out.println(myList[i] + \" \");\n}\n// 计算所有元素的总和\ndouble total = 0;\nfor (int i = 0; i < myList.length; i++) {\ntotal += myList[i];\n}\nSystem.out.println(\"Total is \" + total);\n// 查找最大元素\ndouble max = myList;\nfor (int i = 1; i < myList.length; i++) {\nif (myList[i] > max) max = myList[i];\n}\nSystem.out.println(\"Max is \" + max);\n}\n}\n``````\n\n``````1.9\n2.9\n3.4\n3.5\nTotal is 11.7\nMax is 3.5\n``````\n\n# foreach 循环\n\nJDK 1.5 引进了一种新的循环类型，被称为 foreach 循环或者加强型循环，它能在不使用下标的情况下遍历数组。\n\n``````for (type element: array) {\nSystem.out.println(element);\n}\n``````\n\n## 实例\n\n``````public class TestArray {\n\npublic static void main(String[] args) {\ndouble[] myList = {1.9, 2.9, 3.4, 3.5};\n\n// 打印所有数组元素\nfor (double element: myList) {\nSystem.out.println(element);\n}\n}\n}\n``````\n\n``````1.9\n2.9\n3.4\n3.5\n``````\n\n# 数组作为函数的参数\n\n``````public static void printArray(int[] array) {\nfor (int i = 0; i < array.length; i++) {\nSystem.out.print(array[i] + \" \");\n}\n}\n``````\n\n``````printArray(new int[]{3, 1, 2, 6, 4, 2});\n``````\n\n# 数组作为函数的返回值\n\n``````public static int[] reverse(int[] list) {\nint[] result = new int[list.length];\n\nfor (int i = 0, j = result.length - 1; i < list.length; i++, j--) {\nresult[j] = list[i];\n}\nreturn result;\n}\n``````\n\n# 多维数组\n\n``````String str[][] = new String;\n``````\n\n## 多维数组的动态初始化（以二维数组为例）\n\n1. 直接为每一维分配空间，格式如下：\n\n``````type[][] typeName = new type[typeLength1][typeLength2];\n``````\n\ntype 可以为基本数据类型和复合数据类型，typeLength1 和 typeLength2 必须为正整数，typeLength1 为行数，typeLength2 为列数。\n\n例如：\n\n``````int a[][] = new int;\n``````\n\n解析：\n\n二维数组 a 可以看成一个两行三列的数组。\n\n2. 从最高维开始，分别为每一维分配空间，例如：\n\n``````String s[][] = new String[];\ns = new String;\ns = new String;\ns = new String(\"Good\");\ns = new String(\"Luck\");\ns = new String(\"to\");\ns = new String(\"you\");\ns = new String(\"!\");\n``````\n\n解析：\n\ns = new Strings = new String 是为最高维分配引用空间，也就是为最高维限制其能保存数据的最长的长度，然后再为其每个数组元素单独分配空间 s0 = new String(“Good”) 等操作。\n\n## 多维数组的引用（以二维数组为例）\n\n``````num;\n``````\n\n# Arrays 类\n\njava.util.Arrays 类能方便地操作数组，它提供的所有方法都是静态的。具有以下功能：\n\n• 给数组赋值：通过 fill 方法。\n• 对数组排序：通过 sort 方法，按升序。\n• 比较数组：通过 equals 方法比较数组中元素值是否相等。\n• 查找数组元素：通过 binarySearch 方法能对排序好的数组进行二分查找法操作。\n\n1 public static int binarySearch(Object[] a, Object key) 用二分查找算法在给定数组中搜索给定值的对象 (Byte,Int,double 等)。数组在调用前必须排序好的。如果查找值包含在数组中，则返回搜索键的索引；否则返回 (-(插入点) - 1)。\n2 public static boolean equals(long[] a, long[] a2) 如果两个指定的 long 型数组彼此相等，则返回 true。如果两个数组包含相同数量的元素，并且两个数组中的所有相应元素对都是相等的，则认为这两个数组是相等的。换句话说，如果两个数组以相同顺序包含相同的元素，则两个数组是相等的。同样的方法适用于所有的其他基本数据类型（Byte，short，Int 等）。\n3 public static void fill(int[] a, int val) 将指定的 int 值分配给指定 int 型数组指定范围中的每个元素。同样的方法适用于所有的其他基本数据类型（Byte，short，Int 等）。\n4 public static void sort(Object[] a) 对指定对象数组根据其元素的自然顺序进行升序排列。同样的方法适用于所有的其他基本数据类型（Byte，short，Int 等）。" ]

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[ "# Reichish Phase\n\nThis is a tribute to the early music of Steve Reich. It was realized on the Sonic Pi, using the Bohlen-Pierce scale, equal tempered as MIDI notes.\n\nSource code:\n\n``````# Reichish Phase\n\ntritave = 19.02 # a 3/2 tritave is 1902 cents, or 19.02 in MIDI\nhalf = tritave / 13 # size of the BP half step in MIDI\nwhole = half * 2\n\na3 = 69 # MIDI 69 is A440\nb3 = a3 + whole\nc3 = b3 + half\nd3 = c3 + whole\ne3 = d3 + half\nf3 = e3 + half\ng3 = f3 + whole\nh3 = g3 + half\nj3 = h3 + whole\na4 = a3 + tritave\n\na2 = a3 - tritave\nb2 = b3 - tritave\nc2 = c3 - tritave\nd2 = d3 - tritave\ne2 = e3 - tritave\nf2 = f3 - tritave\ng2 = g3 - tritave\nh2 = h3 - tritave\nj2 = j3 - tritave\n\nnotes = (ring a2, a3, f2, c2, c3, h2, e2, e3, b2, g2, g3, d2, j2)\n\nuse_synth :pretty_bell\nfor i in 0..326\nprint i\nn = notes.tick\nnp = n / a4\nplay n, amp: 1.0 - np, pan: -np / 2\nsleep rrand(0.298,0.302)\ni +=1\nend\nend" ]

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[ "# Strange consequences of Axiom of Choice in Zermelo set theory\n\nRecently I read the following quote (for the umpteenth time):\n\nThe Axiom of Choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma? — Jerry Bona\n\nNormally, the equivalence of these three over ZF is given as a justification for Zorn's lemma, since AC is 'intuitively' true. However, I have always found it lacking because the equivalence crucially depends on transfinite induction via the axiom schema of replacement, which I find unintuitive, unlike the axioms of Zermelo set theory (which lacks both Replacement and Regularity). So my question is:\n\nWhat are 'counter-intuitive' consequences of AC in ZC (Zermelo set theory)?\n\nI understand that this is not a precise question, but I am looking for examples of the same sort as mentioned here and here. I consider these examples as 'near misses', because while many people consider them counter-intuitive, I do not since they all rely on being able to handle arbitrary real numbers or even sets of real numbers, which is not possible in the first place. Specifically, there is no issue with a conceptual choice function that picks one representative from every equivalence class even if it is not implementable, just as there is no issue with a conceptual oracle machine to solving the halting problem even though it is not implementable.\n\nAs for the Banach-Tarski paradox, I do not find it counter-intuitive because we cut up the ball not into nice pieces at all, but instead sets of point dust that can be rearranged. It is not much different from taking the countable set $S = \\{ \\exp(i{\\large \\frac{k}{2^m}}) : k,m \\in \\mathbb{N} \\land m2^m \\le k < (m+1)2^m \\}$ as points in the complex plane and rotating it by $\\exp(i)$, which is a rigid transformation of a bounded point set that has just made countably many points vanish (in fact, ordered around the unit circle, every other point has vanished), completely constructively.\n\nRelated to this question, based on the iterative conception of sets, we get roughly Replacement for countable sequences as stated by Boolos in the quote here. Thus I am also interested to know any 'counter-intuitive' consequences of AC in ZC plus Replacement for countable sequences.\n\n• Entirely too subjective, I'm afraid. Also, your intuition is probably calibrated to choice. So its consequences seem normal. Apr 17, 2017 at 14:39\n• \"while many people consider them counter-intuitive, I do not since they all rely on being able to handle arbitrary real numbers or even sets of real numbers, which is not possible in the first place.\" Why is it impossible to handle arbitrary reals/sets of reals? While it certainly is impossible to do so in the physical universe, it's quite easy to do in the mathematical universe; and this kind of constructive restriction seems especially odd in the choicey context. I do find this question interesting, but I don't really have a good sense of what you're looking for right now. Apr 17, 2017 at 14:50\n• @NoahSchweber: Well I don't disagree with anything in your comment. I might have been too imprecise, but what I meant was that people bring up those puzzle games as 'counter-intuitive' examples, but the reasoning that leads to classifying them as counter-intuitive requires one to be able to handle reals impossibly. Specifically consider the game of guessing a function's value based on all other points. It is clearly impossible if you can pick each point arbitrarily independently so the problem there is with the impossibility of uniformly randomly picking a real from $[0,1]$. Apr 17, 2017 at 15:06\n• @AsafKaragila: Aww.. I was hoping you might have some examples to share! I originally thought only DC was intuitive, but later I realized that perhaps my intuitive issue wasn't with the full AC but with Replacement. I'm still unsure that's why I would like to see the most bizarre theorems of AC in the absence of full Replacement. =) Apr 17, 2017 at 15:10\n• @user254665: Literally the consequence that the OP mentions as not-strange. Apr 17, 2017 at 20:48\n\nTo sum up some of the discussion in the comments, I think you are greatly overestimating the role of Replacement in applications of the axiom of choice. The vast majority of transfinite induction arguments do not use Replacement in any essential way. One key to many applications is the following theorem of Hartogs (which is valid in Zermelo set theory without Choice):\n\nTheorem: Let $$X$$ be a set. Then there exists a well-ordered set $$W$$ such that there is no injection $$f:W\\to X$$.\n\nProof: Define $$W$$ to be the set of isomorphism classes of well-orderings of subsets of $$X$$, ordered by length. A well-ordering of a subset of $$X$$ is a subset of $$X\\times X$$, and an isomorphism class is a set of such subsets, so $$W\\subset\\mathcal{P}(\\mathcal{P}(X\\times X))$$ and is a set by Separation. Note that the well-ordering of $$W$$ is longer than any element of $$W$$, since each element of $$W$$ is isomorphic to a proper initial segment of $$W$$ (namely, its own set of proper initial segments). If there existed an injection $$f:W\\to X$$, the ordering on $$W$$ would give a well-ordering of the image $$f(W)\\subseteq X$$ which has the same length as the well-ordering of $$W$$. But the well-ordering of $$f(W)$$ is an element of $$W$$. This is a contradiction.\n\nThis theorem can, for instance, be used to prove Zorn's lemma without Replacement. Given a poset $$X$$ which is a counterexample to Zorn's lemma, let $$W$$ be as in the theorem. Now just follow the usual transfinite recursion argument, using $$W$$ as the index set of the recursion rather than the ordinals. The argument constructs a strictly increasing function $$W\\to X$$, which is in particular an injection. This is a contradiction.\n\nBy a similar argument, the theorem can be used to prove the well-ordering principle: using a choice function on nonempty subsets of $$X$$, just define a function $$f:W\\to X$$ by transfinite recursion which is injective as long as possible (that is, $$f(w)$$ is different from $$f(v)$$ for all $$v unless such values $$f(v)$$ are already all of $$X$$). Since $$f$$ cannot be an injection, this means that it must be surjective, and this gives a well-ordering of $$X$$.\n\nYou also mentioned in a comment:\n\nI also find theorems that use the cardinality of the reals to be unintuitive, such as that there is a subset of the plane that intersects every straight line in exactly 2 points.\n\nResults like these do not use Replacement either: they merely use a well-ordering of the reals. Given that a well-ordering of the reals exists, you can take one of minimal length, and then use that well-ordering as a substitute for the cardinality of the reals everywhere.\n\n• Not exactly related, but I can't help but notice that Hartog theorem sounds a lot like the Axiom of Archimedes. Any relation? May 30, 2017 at 20:28\n• Thanks for your answer; I'll accept it. I now wonder whether I'm finding this kind of results unintuitive only because I keep thinking of them as providing a completed collection, though I actually can reinterpret them constructively in the same way I did for Banach Tarski. May 31, 2017 at 16:38\n\nJust a thought. In ZFC, let $k$ be the first uncountable strong limit cardinal. Then $(V_k,\\in)$ satisfies ZC (includung Regularity) but not Relacement. Because with $a_0=\\omega$ and $a_{n+1}=2^{a_n}$ we have $k=\\sup_{n\\in \\omega}a_n,$ and we can write a formula of the form $\\forall n\\in \\omega \\exists ! x\\; (F(x,n)),$ where $F(x,n)$ iff $x=a_n$ when $n\\in \\omega$, and such that this formula is absolute between $V$ and $(V_k,\\in).$\n\nIntuitively the formula \"should\" define a countable sequence. But $\\{a_n\\}_{n\\in \\omega}\\not \\in V_k$.... I am just wondering whether we could find some more-interesting properties of $V_k$ that make use of the occasional failure of certain countable sequences to exist in $V_k.$ So that, assuming Con(ZFC), these properties would be undecidable without Replacement.\n\n• This also happens much earlier than strong limits - $V_{\\omega+\\omega}$ is a model of ZC. Also, I don't think this really is connected with choice? May 30, 2017 at 20:17" ]

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[ "", null, "## Quicksort\n\nA sorting algorithm with O(n log n) average time complexity.\n\nOne element, x of the list to be sorted is chosen and the other elements are split into those elements less than x and those greater than or equal to x. These two lists are then sorted recursively using the same algorithm until there is only one element in each list, at which point the sublists are recursively recombined in order yielding the sorted list.\n\nThis can be written in Haskell:\n\n``` qsort :: Ord a => [a] -> [a]\nqsort [] = []\nqsort (x:xs) = qsort [ u | u<-xs, u<x ] ++\n[ x ] ++\nqsort [ u | u<-xs, u>=x ]\n\n```\n[Mark Jones, Gofer prelude.]\n\n### Nearby terms:\n\nTry this search on Wikipedia, Wiktionary, Google, OneLook." ]

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[ "# Data Science from Scratch\n\nToday’s messy glut of data holds answers to questions no one’s even thought to ask.\n\nMay 4, 2015", null, "## A Crash Course in Python\n\nPeople are still crazy about Python after twenty-five years, which I find hard to believe.\n\nMichael Palin\n\n## Learn faster. Dig deeper. See farther.\n\nJoin the O'Reilly online learning platform. Get a free trial today and find answers on the fly, or master something new and useful.\n\nAll new employees at DataSciencester are required to go through new employee orientation, the most interesting part of which is a crash course in Python.\n\nThis is not a comprehensive Python tutorial but instead is intended to highlight the parts of the language that will be most important to us (some of which are often not the focus of Python tutorials).\n\n## The Basics\n\n### Getting Python\n\nYou can download Python from python.org. But if you don’t already have Python, I recommend instead installing the Anaconda distribution, which already includes most of the libraries that you need to do data science.\n\nAs I write this, the latest version of Python is 3.4. At DataSciencester, however, we use old, reliable Python 2.7. Python 3 is not backward-compatible with Python 2, and many important libraries only work well with 2.7. The data science community is still firmly stuck on 2.7, which means we will be, too. Make sure to get that version.\n\nIf you don’t get Anaconda, make sure to install pip, which is a Python package manager that allows you to easily install third-party packages (some of which we’ll need). It’s also worth getting IPython, which is a much nicer Python shell to work with.\n\n(If you installed Anaconda then it should have come with pip and IPython.)\n\nJust run:\n\npip install ipython\n\nand then search the Internet for solutions to whatever cryptic error messages that causes.\n\n### The Zen of Python\n\nPython has a somewhat Zen description of its design principles, which you can also find inside the Python interpreter itself by typing import this.\n\nOne of the most discussed of these is:\n\nThere should be one—​and preferably only one—​obvious way to do it.\n\nCode written in accordance with this “obvious” way (which may not be obvious at all to a newcomer) is often described as “Pythonic.” Although this is not a book about Python, we will occasionally contrast Pythonic and non-Pythonic ways of accomplishing the same things, and we will generally favor Pythonic solutions to our problems.\n\n### Whitespace Formatting\n\nMany languages use curly braces to delimit blocks of code. Python uses indentation:\n\nfor i in [1, 2, 3, 4, 5]:\nprint i # first line in \"for i\" block\nfor j in [1, 2, 3, 4, 5]:\nprint j # first line in \"for j\" block\nprint i + j # last line in \"for j\" block\nprint i # last line in \"for i\" block\nprint \"done looping\"\n\nThis makes Python code very readable, but it also means that you have to be very careful with your formatting. Whitespace is ignored inside parentheses and brackets, which can be helpful for long-winded computations:\n\nlong_winded_computation = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 +\n13 + 14 + 15 + 16 + 17 + 18 + 19 + 20)\n\nand for making code easier to read:\n\nlist_of_lists = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]\n\neasier_to_read_list_of_lists = [ [1, 2, 3],\n[4, 5, 6],\n[7, 8, 9] ]\n\nYou can also use a backslash to indicate that a statement continues onto the next line, although we’ll rarely do this:\n\ntwo_plus_three = 2 + \\\n3\n\nOne consequence of whitespace formatting is that it can be hard to copy and paste code into the Python shell. For example, if you tried to paste the code:\n\nfor i in [1, 2, 3, 4, 5]:\n\n# notice the blank line\nprint i\n\ninto the ordinary Python shell, you would get a:\n\nIndentationError: expected an indented block\n\nbecause the interpreter thinks the blank line signals the end of the for loop’s block.\n\nIPython has a magic function %paste, which correctly pastes whatever is on your clipboard, whitespace and all. This alone is a good reason to use IPython.\n\n### Modules\n\nCertain features of Python are not loaded by default. These include both features included as part of the language as well as third-party features that you download yourself. In order to use these features, you’ll need to import the modules that contain them.\n\nOne approach is to simply import the module itself:\n\nimport re\nmy_regex = re.compile(\"[0-9]+\", re.I)\n\nHere re is the module containing functions and constants for working with regular expressions. After this type of import you can only access those functions by prefixing them with re..\n\nimport re as regex\nmy_regex = regex.compile(\"[0-9]+\", regex.I)\n\nYou might also do this if your module has an unwieldy name or if you’re going to be typing it a lot. For example, when visualizing data with matplotlib, a standard convention is:\n\nimport matplotlib.pyplot as plt\n\nIf you need a few specific values from a module, you can import them explicitly and use them without qualification:\n\nfrom collections import defaultdict, Counter\nlookup = defaultdict(int)\nmy_counter = Counter()\n\nIf you were a bad person, you could import the entire contents of a module into your namespace, which might inadvertently overwrite variables you’ve already defined:\n\nmatch = 10\nfrom re import * # uh oh, re has a match function\nprint match # \"<function re.match>\"\n\nHowever, since you are not a bad person, you won’t ever do this.\n\n### Arithmetic\n\nPython 2.7 uses integer division by default, so that 5 / 2 equals 2. Almost always this is not what we want, so we will always start our files with:\n\nfrom __future__ import division\n\nafter which 5 / 2 equals 2.5. Every code example in this book uses this new-style division. In the handful of cases where we need integer division, we can get it with a double slash: 5 // 2.\n\n### Functions\n\nA function is a rule for taking zero or more inputs and returning a corresponding output. In Python, we typically define functions using def:\n\ndef double(x):\n\"\"\"this is where you put an optional docstring\nthat explains what the function does.\nfor example, this function multiplies its input by 2\"\"\"\nreturn x * 2\n\nPython functions are first-class, which means that we can assign them to variables and pass them into functions just like any other arguments:\n\ndef apply_to_one(f):\n\"\"\"calls the function f with 1 as its argument\"\"\"\nreturn f(1)\n\nmy_double = double # refers to the previously defined function\nx = apply_to_one(my_double) # equals 2\n\nIt is also easy to create short anonymous functions, or lambdas:\n\ny = apply_to_one(lambda x: x + 4) # equals 5\n\nYou can assign lambdas to variables, although most people will tell you that you should just use def instead:\n\nanother_double = lambda x: 2 * x # don't do this\ndef another_double(x): return 2 * x # do this instead\n\nFunction parameters can also be given default arguments, which only need to be specified when you want a value other than the default:\n\ndef my_print(message=\"my default message\"):\nprint message\n\nmy_print(\"hello\") # prints 'hello'\nmy_print() # prints 'my default message'\n\nIt is sometimes useful to specify arguments by name:\n\ndef subtract(a=0, b=0):\nreturn a - b\n\nsubtract(10, 5) # returns 5\nsubtract(0, 5) # returns -5\nsubtract(b=5) # same as previous\n\nWe will be creating many, many functions.\n\n### Strings\n\nStrings can be delimited by single or double quotation marks (but the quotes have to match):\n\nsingle_quoted_string = 'data science'\ndouble_quoted_string = \"data science\"\n\nPython uses backslashes to encode special characters. For example:\n\ntab_string = \"\\t\" # represents the tab character\nlen(tab_string) # is 1\n\nIf you want backslashes as backslashes (which you might in Windows directory names or in regular expressions), you can create raw strings using r””:\n\nnot_tab_string = r\"\\t\" # represents the characters '\\' and 't'\nlen(not_tab_string) # is 2\n\nYou can create multiline strings using triple-[double-]-quotes:\n\nmulti_line_string = \"\"\"This is the first line.\nand this is the second line\nand this is the third line\"\"\"\n\n### Exceptions\n\nWhen something goes wrong, Python raises an exception. Unhandled, these will cause your program to crash. You can handle them using try and except:\n\ntry:\nprint 0 / 0\nexcept ZeroDivisionError:\nprint \"cannot divide by zero\"\n\nAlthough in many languages exceptions are considered bad, in Python there is no shame in using them to make your code cleaner, and we will occasionally do so.\n\n### Lists\n\nProbably the most fundamental data structure in Python is the list. A list is simply an ordered collection. (It is similar to what in other languages might be called an array, but with some added functionality.)\n\ninteger_list = [1, 2, 3]\nheterogeneous_list = [\"string\", 0.1, True]\nlist_of_lists = [ integer_list, heterogeneous_list, [] ]\n\nlist_length = len(integer_list) # equals 3\nlist_sum = sum(integer_list) # equals 6\n\nYou can get or set the nth element of a list with square brackets:\n\nx = range(10) # is the list [0, 1, ..., 9]\nzero = x # equals 0, lists are 0-indexed\none = x # equals 1\nnine = x[-1] # equals 9, 'Pythonic' for last element\neight = x[-2] # equals 8, 'Pythonic' for next-to-last element\nx = -1 # now x is [-1, 1, 2, 3, ..., 9]\n\nYou can also use square brackets to “slice” lists:\n\nfirst_three = x[:3] # [-1, 1, 2]\nthree_to_end = x[3:] # [3, 4, ..., 9]\none_to_four = x[1:5] # [1, 2, 3, 4]\nlast_three = x[-3:] # [7, 8, 9]\nwithout_first_and_last = x[1:-1] # [1, 2, ..., 8]\ncopy_of_x = x[:] # [-1, 1, 2, ..., 9]\n\nPython has an in operator to check for list membership:\n\n1 in [1, 2, 3] # True\n0 in [1, 2, 3] # False\n\nThis check involves examining the elements of the list one at a time, which means that you probably shouldn’t use it unless you know your list is pretty small (or unless you don’t care how long the check takes).\n\nIt is easy to concatenate lists together:\n\nx = [1, 2, 3]\nx.extend([4, 5, 6]) # x is now [1,2,3,4,5,6]\n\nIf you don’t want to modify x you can use list addition:\n\nx = [1, 2, 3]\ny = x + [4, 5, 6] # y is [1, 2, 3, 4, 5, 6]; x is unchanged\n\nMore frequently we will append to lists one item at a time:\n\nx = [1, 2, 3]\nx.append(0) # x is now [1, 2, 3, 0]\ny = x[-1] # equals 0\nz = len(x) # equals 4\n\nIt is often convenient to unpack lists if you know how many elements they contain:\n\nx, y = [1, 2] # now x is 1, y is 2\n\nalthough you will get a ValueError if you don’t have the same numbers of elements on both sides.\n\nIt’s common to use an underscore for a value you’re going to throw away:\n\n_, y = [1, 2] # now y == 2, didn't care about the first element\n\n### Tuples\n\nTuples are lists’ immutable cousins. Pretty much anything you can do to a list that doesn’t involve modifying it, you can do to a tuple. You specify a tuple by using parentheses (or nothing) instead of square brackets:\n\nmy_list = [1, 2]\nmy_tuple = (1, 2)\nother_tuple = 3, 4\nmy_list = 3 # my_list is now [1, 3]\n\ntry:\nmy_tuple = 3\nexcept TypeError:\nprint \"cannot modify a tuple\"\n\nTuples are a convenient way to return multiple values from functions:\n\ndef sum_and_product(x, y):\nreturn (x + y),(x * y)\n\nsp = sum_and_product(2, 3) # equals (5, 6)\ns, p = sum_and_product(5, 10) # s is 15, p is 50\n\nTuples (and lists) can also be used for multiple assignment:\n\nx, y = 1, 2 # now x is 1, y is 2\nx, y = y, x # Pythonic way to swap variables; now x is 2, y is 1\n\n### Dictionaries\n\nAnother fundamental data structure is a dictionary, which associates values with keys and allows you to quickly retrieve the value corresponding to a given key:\n\nempty_dict = {} # Pythonic\nempty_dict2 = dict() # less Pythonic\ngrades = { \"Joel\" : 80, \"Tim\" : 95 } # dictionary literal\n\nYou can look up the value for a key using square brackets:\n\njoels_grade = grades[\"Joel\"] # equals 80\n\nBut you’ll get a KeyError if you ask for a key that’s not in the dictionary:\n\ntry:\nexcept KeyError:\nprint \"no grade for Kate!\"\n\nYou can check for the existence of a key using in:\n\njoel_has_grade = \"Joel\" in grades # True\nkate_has_grade = \"Kate\" in grades # False\n\nDictionaries have a get method that returns a default value (instead of raising an exception) when you look up a key that’s not in the dictionary:\n\njoels_grade = grades.get(\"Joel\", 0) # equals 80\nno_ones_grade = grades.get(\"No One\") # default default is None\n\nYou assign key-value pairs using the same square brackets:\n\ngrades[\"Tim\"] = 99 # replaces the old value\nnum_students = len(grades) # equals 3\n\nWe will frequently use dictionaries as a simple way to represent structured data:\n\ntweet = {\n\"user\" : \"joelgrus\",\n\"text\" : \"Data Science is Awesome\",\n\"retweet_count\" : 100,\n\"hashtags\" : [\"#data\", \"#science\", \"#datascience\", \"#awesome\", \"#yolo\"]\n}\n\nBesides looking for specific keys we can look at all of them:\n\ntweet_keys = tweet.keys() # list of keys\ntweet_values = tweet.values() # list of values\ntweet_items = tweet.items() # list of (key, value) tuples\n\n\"user\" in tweet_keys # True, but uses a slow list in\n\"user\" in tweet # more Pythonic, uses faster dict in\n\"joelgrus\" in tweet_values # True\n\nDictionary keys must be immutable; in particular, you cannot use lists as keys. If you need a multipart key, you should use a tuple or figure out a way to turn the key into a string.\n\n#### defaultdict\n\nImagine that you’re trying to count the words in a document. An obvious approach is to create a dictionary in which the keys are words and the values are counts. As you check each word, you can increment its count if it’s already in the dictionary and add it to the dictionary if it’s not:\n\nword_counts = {}\nfor word in document:\nif word in word_counts:\nword_counts[word] += 1\nelse:\nword_counts[word] = 1\n\nYou could also use the “forgiveness is better than permission” approach and just handle the exception from trying to look up a missing key:\n\nword_counts = {}\nfor word in document:\ntry:\nword_counts[word] += 1\nexcept KeyError:\nword_counts[word] = 1\n\nA third approach is to use get, which behaves gracefully for missing keys:\n\nword_counts = {}\nfor word in document:\nprevious_count = word_counts.get(word, 0)\nword_counts[word] = previous_count + 1\n\nEvery one of these is slightly unwieldy, which is why defaultdict is useful. A defaultdict is like a regular dictionary, except that when you try to look up a key it doesn’t contain, it first adds a value for it using a zero-argument function you provided when you created it. In order to use defaultdicts, you have to import them from collections:\n\nfrom collections import defaultdict\n\nword_counts = defaultdict(int) # int() produces 0\nfor word in document:\nword_counts[word] += 1\n\nThey can also be useful with list or dict or even your own functions:\n\ndd_list = defaultdict(list) # list() produces an empty list\ndd_list.append(1) # now dd_list contains {2: }\n\ndd_dict = defaultdict(dict) # dict() produces an empty dict\ndd_dict[\"Joel\"][\"City\"] = \"Seattle\" # { \"Joel\" : { \"City\" : Seattle\"}}\n\ndd_pair = defaultdict(lambda: [0, 0])\ndd_pair = 1 # now dd_pair contains {2: [0,1]}\n\nThese will be useful when we’re using dictionaries to “collect” results by some key and don’t want to have to check every time to see if the key exists yet.\n\n#### Counter\n\nA Counter turns a sequence of values into a defaultdict(int)-like object mapping keys to counts. We will primarily use it to create histograms:\n\nfrom collections import Counter\nc = Counter([0, 1, 2, 0]) # c is (basically) { 0 : 2, 1 : 1, 2 : 1 }\n\nThis gives us a very simple way to solve our word_counts problem:\n\nword_counts = Counter(document)\n\nA Counter instance has a most_common method that is frequently useful:\n\n# print the 10 most common words and their counts\nfor word, count in word_counts.most_common(10):\nprint word, count\n\n### Sets\n\nAnother data structure is set, which represents a collection of distinct elements:\n\ns = set()\ns.add(1) # s is now { 1 }\ns.add(2) # s is now { 1, 2 }\ns.add(2) # s is still { 1, 2 }\nx = len(s) # equals 2\ny = 2 in s # equals True\nz = 3 in s # equals False\n\nWe’ll use sets for two main reasons. The first is that in is a very fast operation on sets. If we have a large collection of items that we want to use for a membership test, a set is more appropriate than a list:\n\nstopwords_list = [\"a\",\"an\",\"at\"] + hundreds_of_other_words + [\"yet\", \"you\"]\n\n\"zip\" in stopwords_list # False, but have to check every element\n\nstopwords_set = set(stopwords_list)\n\"zip\" in stopwords_set # very fast to check\n\nThe second reason is to find the distinct items in a collection:\n\nitem_list = [1, 2, 3, 1, 2, 3]\nnum_items = len(item_list) # 6\nitem_set = set(item_list) # {1, 2, 3}\nnum_distinct_items = len(item_set) # 3\ndistinct_item_list = list(item_set) # [1, 2, 3]\n\nWe’ll use sets much less frequently than dicts and lists.\n\n### Control Flow\n\nAs in most programming languages, you can perform an action conditionally using if:\n\nif 1 > 2:\nmessage = \"if only 1 were greater than two...\"\nelif 1 > 3:\nmessage = \"elif stands for 'else if'\"\nelse:\nmessage = \"when all else fails use else (if you want to)\"\n\nYou can also write a ternary if-then-else on one line, which we will do occasionally:\n\nparity = \"even\" if x % 2 == 0 else \"odd\"\n\nPython has a while loop:\n\nx = 0\nwhile x < 10:\nprint x, \"is less than 10\"\nx += 1\n\nalthough more often we’ll use for and in:\n\nfor x in range(10):\nprint x, \"is less than 10\"\n\nIf you need more-complex logic, you can use continue and break:\n\nfor x in range(10):\nif x == 3:\ncontinue # go immediately to the next iteration\nif x == 5:\nbreak # quit the loop entirely\nprint x\n\nThis will print 0, 1, 2, and 4.\n\n### Truthiness\n\nBooleans in Python work as in most other languages, except that they’re capitalized:\n\none_is_less_than_two = 1 < 2 # equals True\ntrue_equals_false = True == False # equals False\n\nPython uses the value None to indicate a nonexistent value. It is similar to other languages’ null:\n\nx = None\nprint x == None # prints True, but is not Pythonic\nprint x is None # prints True, and is Pythonic\n\nPython lets you use any value where it expects a Boolean. The following are all “Falsy”:\n\n• False\n• None\n• [] (an empty list)\n• {} (an empty dict)\n• “”\n• set()\n• 0\n• 0.0\n\nPretty much anything else gets treated as True. This allows you to easily use if statements to test for empty lists or empty strings or empty dictionaries or so on. It also sometimes causes tricky bugs if you’re not expecting this behavior:\n\ns = some_function_that_returns_a_string()\nif s:\nfirst_char = s\nelse:\nfirst_char = \"\"\n\nA simpler way of doing the same is:\n\nfirst_char = s and s\n\nsince and returns its second value when the first is “truthy,” the first value when it’s not. Similarly, if x is either a number or possibly None:\n\nsafe_x = x or 0\n\nis definitely a number.\n\nPython has an all function, which takes a list and returns True precisely when every element is truthy, and an any function, which returns True when at least one element is truthy:\n\nall([True, 1, { 3 }]) # True\nall([True, 1, {}]) # False, {} is falsy\nany([True, 1, {}]) # True, True is truthy\nall([]) # True, no falsy elements in the list\nany([]) # False, no truthy elements in the list\n\n## The Not-So-Basics\n\nHere we’ll look at some more-advanced Python features that we’ll find useful for working with data.\n\n### Sorting\n\nEvery Python list has a sort method that sorts it in place. If you don’t want to mess up your list, you can use the sorted function, which returns a new list:\n\nx = [4,1,2,3]\ny = sorted(x) # is [1,2,3,4], x is unchanged\nx.sort() # now x is [1,2,3,4]\n\nBy default, sort (and sorted) sort a list from smallest to largest based on naively comparing the elements to one another.\n\nIf you want elements sorted from largest to smallest, you can specify a reverse=True parameter. And instead of comparing the elements themselves, you can compare the results of a function that you specify with key:\n\n# sort the list by absolute value from largest to smallest\nx = sorted([-4,1,-2,3], key=abs, reverse=True) # is [-4,3,-2,1]\n\n# sort the words and counts from highest count to lowest\nwc = sorted(word_counts.items(),\nkey=lambda (word, count): count,\nreverse=True)\n\n### List Comprehensions\n\nFrequently, you’ll want to transform a list into another list, by choosing only certain elements, or by transforming elements, or both. The Pythonic way of doing this is list comprehensions:\n\neven_numbers = [x for x in range(5) if x % 2 == 0] # [0, 2, 4]\nsquares = [x * x for x in range(5)] # [0, 1, 4, 9, 16]\neven_squares = [x * x for x in even_numbers] # [0, 4, 16]\n\nYou can similarly turn lists into dictionaries or sets:\n\nsquare_dict = { x : x * x for x in range(5) } # { 0:0, 1:1, 2:4, 3:9, 4:16 }\nsquare_set = { x * x for x in [1, -1] } # { 1 }\n\nIf you don’t need the value from the list, it’s conventional to use an underscore as the variable:\n\nzeroes = [0 for _ in even_numbers] # has the same length as even_numbers\n\nA list comprehension can include multiple fors:\n\npairs = [(x, y)\nfor x in range(10)\nfor y in range(10)] # 100 pairs (0,0) (0,1) ... (9,8), (9,9)\n\nand later fors can use the results of earlier ones:\n\nincreasing_pairs = [(x, y) # only pairs with x < y,\nfor x in range(10) # range(lo, hi) equals\nfor y in range(x + 1, 10)] # [lo, lo + 1, ..., hi - 1]\n\nWe will use list comprehensions a lot.\n\n### Generators and Iterators\n\nA problem with lists is that they can easily grow very big. range(1000000) creates an actual list of 1 million elements. If you only need to deal with them one at a time, this can be a huge source of inefficiency (or of running out of memory). If you potentially only need the first few values, then calculating them all is a waste.\n\nA generator is something that you can iterate over (for us, usually using for) but whose values are produced only as needed (lazily).\n\nOne way to create generators is with functions and the yield operator:\n\ndef lazy_range(n):\n\"\"\"a lazy version of range\"\"\"\ni = 0\nwhile i < n:\nyield i\ni += 1\n\nThe following loop will consume the yielded values one at a time until none are left:\n\nfor i in lazy_range(10):\ndo_something_with(i)\n\n(Python actually comes with a lazy_range function called xrange, and in Python 3, range itself is lazy.) This means you could even create an infinite sequence:\n\ndef natural_numbers():\n\"\"\"returns 1, 2, 3, ...\"\"\"\nn = 1\nwhile True:\nyield n\nn += 1\n\nalthough you probably shouldn’t iterate over it without using some kind of break logic.\n\n###### Tip\n\nThe flip side of laziness is that you can only iterate through a generator once. If you need to iterate through something multiple times, you’ll need to either recreate the generator each time or use a list.\n\nA second way to create generators is by using for comprehensions wrapped in parentheses:\n\nlazy_evens_below_20 = (i for i in lazy_range(20) if i % 2 == 0)\n\nRecall also that every dict has an items() method that returns a list of its key-value pairs. More frequently we’ll use the iteritems() method, which lazily yields the key-value pairs one at a time as we iterate over it.\n\n### Randomness\n\nAs we learn data science, we will frequently need to generate random numbers, which we can do with the random module:\n\nimport random\n\nfour_uniform_randoms = [random.random() for _ in range(4)]\n\n# [0.8444218515250481, # random.random() produces numbers\n# 0.7579544029403025, # uniformly between 0 and 1\n# 0.420571580830845, # it's the random function we'll use\n# 0.25891675029296335] # most often\n\nThe random module actually produces pseudorandom (that is, deterministic) numbers based on an internal state that you can set with random.seed if you want to get reproducible results:\n\nrandom.seed(10) # set the seed to 10\nprint random.random() # 0.57140259469\nrandom.seed(10) # reset the seed to 10\nprint random.random() # 0.57140259469 again\n\nWe’ll sometimes use random.randrange, which takes either 1 or 2 arguments and returns an element chosen randomly from the corresponding range():\n\nrandom.randrange(10) # choose randomly from range(10) = [0, 1, ..., 9]\nrandom.randrange(3, 6) # choose randomly from range(3, 6) = [3, 4, 5]\n\nThere are a few more methods that we’ll sometimes find convenient. random.shuffle randomly reorders the elements of a list:\n\nup_to_ten = range(10)\nrandom.shuffle(up_to_ten)\nprint up_to_ten\n# [2, 5, 1, 9, 7, 3, 8, 6, 4, 0] (your results will probably be different)\n\nIf you need to randomly pick one element from a list you can use random.choice:\n\nmy_best_friend = random.choice([\"Alice\", \"Bob\", \"Charlie\"]) # \"Bob\" for me\n\nAnd if you need to randomly choose a sample of elements without replacement (i.e., with no duplicates), you can use random.sample:\n\nlottery_numbers = range(60)\nwinning_numbers = random.sample(lottery_numbers, 6) # [16, 36, 10, 6, 25, 9]\n\nTo choose a sample of elements with replacement (i.e., allowing duplicates), you can just make multiple calls to random.choice:\n\nfour_with_replacement = [random.choice(range(10))\nfor _ in range(4)]\n# [9, 4, 4, 2]\n\n### Regular Expressions\n\nRegular expressions provide a way of searching text. They are incredibly useful but also fairly complicated, so much so that there are entire books written about them. We will explain their details the few times we encounter them; here are a few examples of how to use them in Python:\n\nimport re\n\nprint all([ # all of these are true, because\nre.search(\"a\", \"cat\"), # * 'cat' has an 'a' in it\nnot re.search(\"c\", \"dog\"), # * 'dog' doesn't have a 'c' in it\n3 == len(re.split(\"[ab]\", \"carbs\")), # * split on a or b to ['c','r','s']\n\"R-D-\" == re.sub(\"[0-9]\", \"-\", \"R2D2\") # * replace digits with dashes\n]) # prints True\n\n### Object-Oriented Programming\n\nLike many languages, Python allows you to define classes that encapsulate data and the functions that operate on them. We’ll use them sometimes to make our code cleaner and simpler. It’s probably simplest to explain them by constructing a heavily annotated example.\n\nImagine we didn’t have the built-in Python set. Then we might want to create our own Set class.\n\nWhat behavior should our class have? Given an instance of Set, we’ll need to be able to add items to it, remove items from it, and check whether it contains a certain value. We’ll create all of these as member functions, which means we’ll access them with a dot after a Set object:\n\n# by convention, we give classes PascalCase names\nclass Set:\n\n# these are the member functions\n# every one takes a first parameter \"self\" (another convention)\n# that refers to the particular Set object being used\n\ndef __init__(self, values=None):\n\"\"\"This is the constructor.\nIt gets called when you create a new Set.\nYou would use it like\ns1 = Set() # empty set\ns2 = Set([1,2,2,3]) # initialize with values\"\"\"\n\nself.dict = {} # each instance of Set has its own dict property\n# which is what we'll use to track memberships\nif values is not None:\nfor value in values:\n\ndef __repr__(self):\n\"\"\"this is the string representation of a Set object\nif you type it at the Python prompt or pass it to str()\"\"\"\nreturn \"Set: \" + str(self.dict.keys())\n\n# we'll represent membership by being a key in self.dict with value True\nself.dict[value] = True\n\n# value is in the Set if it's a key in the dictionary\ndef contains(self, value):\nreturn value in self.dict\n\ndef remove(self, value):\ndel self.dict[value]\n\nWhich we could then use like:\n\ns = Set([1,2,3])\nprint s.contains(4) # True\ns.remove(3)\nprint s.contains(3) # False\n\n### Functional Tools\n\nWhen passing functions around, sometimes we’ll want to partially apply (or curry) functions to create new functions. As a simple example, imagine that we have a function of two variables:\n\ndef exp(base, power):\nreturn base ** power\n\nand we want to use it to create a function of one variable two_to_the whose input is a power and whose output is the result of exp(2, power).\n\nWe can, of course, do this with def, but this can sometimes get unwieldy:\n\ndef two_to_the(power):\nreturn exp(2, power)\n\nA different approach is to use functools.partial:\n\nfrom functools import partial\ntwo_to_the = partial(exp, 2) # is now a function of one variable\nprint two_to_the(3) # 8\n\nYou can also use partial to fill in later arguments if you specify their names:\n\nsquare_of = partial(exp, power=2)\nprint square_of(3) # 9\n\nIt starts to get messy if you curry arguments in the middle of the function, so we’ll try to avoid doing that.\n\nWe will also occasionally use map, reduce, and filter, which provide functional alternatives to list comprehensions:\n\ndef double(x):\nreturn 2 * x\n\nxs = [1, 2, 3, 4]\ntwice_xs = [double(x) for x in xs] # [2, 4, 6, 8]\ntwice_xs = map(double, xs) # same as above\nlist_doubler = partial(map, double) # *function* that doubles a list\ntwice_xs = list_doubler(xs) # again [2, 4, 6, 8]\n\nYou can use map with multiple-argument functions if you provide multiple lists:\n\ndef multiply(x, y): return x * y\n\nproducts = map(multiply, [1, 2], [4, 5]) # [1 * 4, 2 * 5] = [4, 10]\n\nSimilarly, filter does the work of a list-comprehension if:\n\ndef is_even(x):\n\"\"\"True if x is even, False if x is odd\"\"\"\nreturn x % 2 == 0\n\nx_evens = [x for x in xs if is_even(x)] # [2, 4]\nx_evens = filter(is_even, xs) # same as above\nlist_evener = partial(filter, is_even) # *function* that filters a list\nx_evens = list_evener(xs) # again [2, 4]\n\nAnd reduce combines the first two elements of a list, then that result with the third, that result with the fourth, and so on, producing a single result:\n\nx_product = reduce(multiply, xs) # = 1 * 2 * 3 * 4 = 24\nlist_product = partial(reduce, multiply) # *function* that reduces a list\nx_product = list_product(xs) # again = 24\n\n### enumerate\n\nNot infrequently, you’ll want to iterate over a list and use both its elements and their indexes:\n\n# not Pythonic\nfor i in range(len(documents)):\ndocument = documents[i]\ndo_something(i, document)\n\n# also not Pythonic\ni = 0\nfor document in documents:\ndo_something(i, document)\ni += 1\n\nThe Pythonic solution is enumerate, which produces tuples (index, element):\n\nfor i, document in enumerate(documents):\ndo_something(i, document)\n\nSimilarly, if we just want the indexes:\n\nfor i in range(len(documents)): do_something(i) # not Pythonic\nfor i, _ in enumerate(documents): do_something(i) # Pythonic\n\nWe’ll use this a lot.\n\n### zip and Argument Unpacking\n\nOften we will need to zip two or more lists together. zip transforms multiple lists into a single list of tuples of corresponding elements:\n\nlist1 = ['a', 'b', 'c']\nlist2 = [1, 2, 3]\nzip(list1, list2) # is [('a', 1), ('b', 2), ('c', 3)]\n\nIf the lists are different lengths, zip stops as soon as the first list ends.\n\nYou can also “unzip” a list using a strange trick:\n\npairs = [('a', 1), ('b', 2), ('c', 3)]\nletters, numbers = zip(*pairs)\n\nThe asterisk performs argument unpacking, which uses the elements of pairs as individual arguments to zip. It ends up the same as if you’d called:\n\nzip(('a', 1), ('b', 2), ('c', 3))\n\nwhich returns [('a','b','c'), ('1','2','3')].\n\nYou can use argument unpacking with any function:\n\ndef add(a, b): return a + b\n\nadd(*[1, 2]) # returns 3\n\nIt is rare that we’ll find this useful, but when we do it’s a neat trick.\n\n### args and kwargs\n\nLet’s say we want to create a higher-order function that takes as input some function f and returns a new function that for any input returns twice the value of f:\n\ndef doubler(f):\ndef g(x):\nreturn 2 * f(x)\nreturn g\n\nThis works in some cases:\n\ndef f1(x):\nreturn x + 1\n\ng = doubler(f1)\nprint g(3) # 8 (== ( 3 + 1) * 2)\nprint g(-1) # 0 (== (-1 + 1) * 2)\n\nHowever, it breaks down with functions that take more than a single argument:\n\ndef f2(x, y):\nreturn x + y\n\ng = doubler(f2)\nprint g(1, 2) # TypeError: g() takes exactly 1 argument (2 given)\n\nWhat we need is a way to specify a function that takes arbitrary arguments. We can do this with argument unpacking and a little bit of magic:\n\ndef magic(*args, **kwargs):\nprint \"unnamed args:\", args\nprint \"keyword args:\", kwargs\n\nmagic(1, 2, key=\"word\", key2=\"word2\")\n\n# prints\n# unnamed args: (1, 2)\n# keyword args: {'key2': 'word2', 'key': 'word'}\n\nThat is, when we define a function like this, args is a tuple of its unnamed arguments and kwargs is a dict of its named arguments. It works the other way too, if you want to use a list (or tuple) and dict to supply arguments to a function:\n\ndef other_way_magic(x, y, z):\nreturn x + y + z\n\nx_y_list = [1, 2]\nz_dict = { \"z\" : 3 }\nprint other_way_magic(*x_y_list, **z_dict) # 6\n\nYou could do all sorts of strange tricks with this; we will only use it to produce higher-order functions whose inputs can accept arbitrary arguments:\n\ndef doubler_correct(f):\n\"\"\"works no matter what kind of inputs f expects\"\"\"\ndef g(*args, **kwargs):\n\"\"\"whatever arguments g is supplied, pass them through to f\"\"\"\nreturn 2 * f(*args, **kwargs)\nreturn g\n\ng = doubler_correct(f2)\nprint g(1, 2) # 6\n\n### Welcome to DataSciencester!\n\nThis concludes new-employee orientation. Oh, and also, try not to embezzle anything.\n\n## For Further Exploration\n\nPost topics: Data science\nShare:" ]

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[ "## Poisson Autoregression\n\nPublikation: Working paperForskning\n\n### Dokumenter\n\n• DP 08-35\n\nForlagets udgivne version, 535 KB, PDF-dokument\n\nThis paper considers geometric ergodicity and likelihood based inference for linear and nonlinear Poisson autoregressions. In the linear case the conditional mean is linked linearly to its past values as well as the observed values of the Poisson process. This also applies to the conditional variance, implying an interpretation as an integer valued GARCH process. In a nonlinear conditional Poisson model, the conditional mean is a nonlinear function of its past values and a nonlinear function of past observations. As a particular example an exponential autoregressive Poisson model for time series is considered. Under geometric ergodicity the maximum likelihood estimators of the parameters are shown to be asymptotically Gaussian in the linear model. In addition we provide a consistent estimator of the asymptotic covariance, which is used in the simulations and the analysis of some transaction data. Our approach to verifying geometric ergodicity proceeds via Markov theory and irreducibility. Finding transparent conditions for proving ergodicity turns out to be a delicate problem in the original model formulation. This problem is circumvented by allowing a perturbation of the model. We show that as the perturbations can be chosen to be arbitrarily small, the differences between the perturbed and non-perturbed versions vanish as far as the asymptotic distribution of the parameter estimates is concerned.\nOriginalsprog Engelsk Department of Economics, University of Copenhagen 37 Udgivet - 2008\n\n### Antal downloads er baseret på statistik fra Google Scholar og www.ku.dk\n\nIngen data tilgængelig\n\nID: 9508644" ]

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[ "# Fraction Worksheets For Grade 7 Fraction Worksheets For Grade 7 Medium To Large Size Of Complex Fractions Worksheet Grade Worksheets Fraction Fractions Decimals And Percents Worksheets 7th Grade Pdf", null, "fraction worksheets for grade 7 fraction worksheets for grade 7 medium to large size of complex fractions worksheet grade worksheets fraction fractions decimals and percents worksheets 7th grade pdf.\n\nfraction decimal percent worksheet grade 7 fractions decimals worksheets comparing and percentages percents with answers,fractions and decimals class 7 cbse worksheets pdf grade year percentages,fractions and decimals worksheets year 7 converting fraction decimal percentage with answer class cbse pdf grade,percent worksheets grade 7 6 fractions decimals percents and percentages year class cbse with answers worksheet,fractions decimals percentages worksheet year 7 and worksheets grade with answers template,fractions decimals and percents worksheets 7th grade pdf 7 a site with series of free pages on converting percentages worksheet year,fractions decimals worksheets grade 7 and pdf converting to percentages worksheet year,fractions and decimals worksheets grade 7 with answers fraction decimal percent worksheet year percents,comparing fractions and decimals percentages worksheet year 7 worksheets percents 7th grade pdf,math worksheets for grade 7 fractions and decimals word problems or pdf percentages year with answers." ]

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[ "# Java String toCharArray() with example\n\n• Difficulty Level : Basic\n• Last Updated : 04 Dec, 2018\n\nThe java string toCharArray() method converts the given string into a sequence of characters. The returned array length is equal to the length of the string.\n\nSyntax :\n\n```public char[] toCharArray()\nReturn : It returns a newly allocated character array.\n```\n\n `// Java program to demonstrate``// working of toCharArray() method`` ` `class` `Gfg {``    ``public` `static` `void` `main(String args[])``    ``{``        ``String s = ``\"GeeksforGeeks\"``;``        ``char``[] gfg = s.toCharArray();``        ``for` `(``int` `i = ``0``; i < gfg.length; i++) {``            ``System.out.println(gfg[i]);``        ``}``    ``}``}`\n\nOutput:\n\n```G\ne\ne\nk\ns\nf\no\nr\nG\ne\ne\nk\ns\n```\nMy Personal Notes arrow_drop_up" ]

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[ "# Thermodynamics and gravity\n\n## Main Question or Discussion Point\n\nA gas cloud shrinks due to gravitational forces. The gas gets hotter because the velocity (hence kinetic energy) of the gas particles gets larger. At the same time the potential energy gets smaller because the gas particles are closer to each other.\n\nOnce you get past the fact that the decrease in potential energy is greater than the increase in kinetic energy (hence the gas gets hotter even though the total energy decreases), the next question is this: what happens to the lost energy?\n\nCan this question be answered without resorting to GR?\n\nRelated Classical Physics News on Phys.org\nI don´t understand your position clearly. But I think there´s thousands of ways for the ´lost energy´ to go: photons, internal energy for gas particles,etc. Otherwise, the increase of kinetic energy should be equal to the decrease of potential energy.\n\nruss_watters\nMentor\nddesai said:\nthe decrease in potential energy is greater than the increase in kinetic energy (hence the gas gets hotter even though the total energy decreases),\nWhere do you get that from?\n\nSee: http://math.ucr.edu/home/baez/entropy.html for the source of the question. Baez show that the the decrease in potential energy is greater than the increase in kinetic energy: overall the internal energy decreases.\n\nThis is an ideal* gas in which particles are pointlike. We don't have em radiation since we are assuming the particles don't really have internal state (or we are neglecting this effect) and there are no significant photon sources to interact with.\n\nThe only thing left is gravity. In GR you have energy radiated away as gravity waves.\n\nBut in newtonian physics? There is no such thing as gravity waves. So how do we explain the decrease in total energy?\n\n*Not really an ideal gas, since the particles do interact with each other via gravity.\n\nGravity and the 1st law of thermodynamics\n\nI have been reading the responses to the possibility of non-linear energy conversion and I too am skeptical of many, if not all of the claims out there. However, I have recently published a web page that you might find interesting http://www.geocities.com/joeiii63/\nMy angle is simple and straight forward. If the value of the inertial mass (read apparent mass) in an isolated system is varied during the system's cycle, then the effect is exactly the same as if the gravitational field affecting it were fluctuated between weak and strong. This can be used to cause a weight to behave as though it were lighter on its way up and heavier on its way down.\nThis invitation has been extended to researchers, university physics depts. (including MIT and Cambridge) and skeptics worldwide. As of yet, no one has indicated why this system wouldn't or can't work (invoking the 1st Law of Thermodynamics in this case is no different than saying that since humans can't fly, airplanes must be impossible because they would allow humans to fly).\nSo far the responses have been mostly polite, but similar; they all want more time to review the idea. If you understand physics, its just not that complicated.\n\nWhat is meant by reversing the system is this: The weight doesn't fall up, it is lifted by the kinetic energy stored in the flywheel and the upward momentum of the weight. At the start, the weight falls in a normal flywheel configuration and its potential energy is divided between the flywheel and the weight. At the end of this half cycle, the motion of the weight is reversed, possibly by a spring, and it re-enters the system on the other side of the flywheel axle (the upward side). Yes, some velocity (and thus kinetic energy) will be lost, but not enough to overcome the large advantage provided by the fact that as the weight is raised, its deceleration due to the pull of gravity diminishes as it gets higher. Because the weight is being lifted by the variable inertia configuration, its deceleration drops from 0.5g at the bottom to 0.2g at the top. It's exactly as though the gravitational field acting on the weight is weakening.\nRemember, when the weight fell, it was connected to the flywheel in the non-variable (normal) configuration, and it had a constant acceleration of 0.5g.\nThis means that using the variable inertia configuration, a weight can be lifted to a higher elevation than it falls from, using nothing more than the kinetic energy produced by its own fall through a non-varying inertia configuration.\n\nIf anyone should insist on knowing where the extra energy comes from when the system is reversed, they need only to figure out where the missing energy goes when the system is operated in the direction described on the web page. Therein lies the answer. It doesn't go anywhere. It just disappears. And when the system is reversed, energy simply appears. It was never anywhere.\n\nddesai said:\nSee: http://math.ucr.edu/home/baez/entropy.html for the source of the question. Baez show that the the decrease in potential energy is greater than the increase in kinetic energy: overall the internal energy decreases.\n\nThis is an ideal* gas in which particles are pointlike. We don't have em radiation since we are assuming the particles don't really have internal state (or we are neglecting this effect) and there are no significant photon sources to interact with.\nBaez is not claiming that the energy of an isolated system can change. What that page means is that if the overall energy of the gas cloud decreases (which it will do by radiation) then the temperature will increase.\n\nYes, never assumed it was an isolated system. But you are right.\n\nHe uses the Virial Theorem to show that, under certain conditions, the total energy is equal to minus the kinetic energy or equal to the one-half the potential energy. This means the rate of change of kinetic energy with respect to total energy is -1 and the rate of change of potential energy with respect to energy is 2.\n\nThis means the potential energy will decrease faster than the kinetic energy will increase as the clound looses energy (and hence volume).\n\nBaez relaxes the \"energy is constant\" constraint assuming that there is some process by which energy is taken away from the system.\n\nAt the end of ther article he asks \"where does the lost energy go?\" This was the question I was trying to answer.\n\nI think I have a tentative answer. In newtonian physics, if the gas is ideal, then the energy must be constant: there is no energy loss, because there is nothing to carry the energy away.\n\nIn quantum mechanics, where atoms have internal states, you could have energy loss via inelastic collisions and the emission of photons.\n\nBut if we ignore QM and treat this as an ideal gas as Baez does, then what does he mean by \"where does the energy go?\" The only thing I could think of was gravity waves. Where else can the energy go in this model?\n\nAnd it is surprising that you can conclude, amoung other things, that as potential energy decreases faster than the kinetic energy increases just by resorting to the Virial Theorem and not specifying how the energy is lost.\n\nAlso, the Virial theorem here assumes the existence of an inverse square law but does it still hold if there are other processes (inelastic collisions, gravity waves, or whatever) at work? Have to think about this." ]

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[ "The square root of 14 is expressed together √14 in the radical type and as (14)½ or (14)0.5 in the exponent form. The square source of 14 rounded as much as 5 decimal locations is 3.74166. The is the positive solution the the equation x2 = 14.\n\nYou are watching: What is the square root of 14\n\nSquare root of 14: 3.7416573867739413Square source of 14 in exponential form: (14)½ or (14)0.5Square root of 14 in radical form: √14\n 1 What Is the Square root of 14? 2 IsSquare source of 14Rational or Irrational? 3 How to discover the Square source of 14? 4 Thinking the end of the Box! 5 Important note on Square source of 14 6 FAQs top top Square root of 14\n\n## What Is the Square source of 14?\n\n4 deserve to be expressed together 22 = 2×2. Here 2is dubbed the square root of 4 and we recognize that 4is a perfect square. Non-square numbers additionally have a square root, the only distinction is that they room not totality numbers.Similarly the square root of 14 is expressed together √14in the radical type and together 14½in the exponent form. The square source of 14 rounded to 5 decimal areas is+3.74165, -3.74165.\n\n## Is the Square root of 14Rational orIrrational?\n\nA number the cannot be expressed together a ratio of two integers is one irrational number. The decimal type of the irrational number is non-terminating (i.e., it never ever ends) and also non-recurring (i.e., the decimal component of the number never ever repeats a pattern). Now let us look in ~ the square source of 14.\n\n√14= 3.74165738677.\n\nDo you think the decimal component stops after ~ 3.74165738677? No, that is never-ending and also you cannot notice a pattern in the decimal part.Hence, √14is an irrational number.\n\n## How to find the Square source of 14?\n\n### Square source of 14ByLong Division\n\nThe worth of the square source of 14by long division method consists of the following steps:\n\nStep 1: starting from the right, we will pair up the digits by placing a bar above them.Repeat this procedure to get the decimal areas you want.\n\nThe picture shows step by step long division method to find the square root of 14", null, "Can you shot and refer the square source of 15in a comparable way? The prize of √15should it is in 3.872.\n\nTo simplify the square source of 14, permit us an initial express 14 together a product of its prime factors. The prime factorization that 14 = 2× 7. Therefore, √14is in the lowest form and can not be streamlined further. Thus, we have expressed the square source of 14 in the radical form.\n\nExplore Square roots utilizing illustrations and also interactive examples\n\nThink Tank:\n\nCan girlfriend think of a quadratic equation which has actually a root together √14?Since (-(√14)2=14, deserve to we say the -√14is additionally a square source of 14?\n\nImportant Notes:\n\nThe square source of 14 is expressed as √14in the radical form.In the exponent form, the square root of 14is expressed together 14½.The actual roots that √14are±3.74165.\n\n## Square root of 14Solved Examples\n\nExample 1: Allan told his friends the the worth of -√14is exact same as the worth of √-14. What execute you think?\n\nSolution\n\nNegative square source cannot be real numbers. -√14is a real number. However √-14is an imagine number. Hence, they space not the same.\n\n-√14and √-14are no same.\n\nExample 2: Joel had a doubt. He knew that 6.403is the square source of 14 yet wanted to know if -6.403is also a square root of 14? can you help him?\n\nSolution:\n\nLet us take an instance of a perfect square number and extend the very same logic come clarify she doubt. We recognize that 3 is a square source of 9 since when 3 is multiply to chin it offers 9. But what around -3? As, -3×-3 = 9. Therefore, -3 is additionally a square source of 9. Walking by the exact same logic,-3.741 isthe square root of 14.\n\ngo to slidego come slidego come slide", null, "How can your child grasp math concepts?\nMath mastery comes with practice and also understanding the ‘Why’ behind the ‘What.’ endure the chrischona2015.org difference.\n\nBook a totally free Trial Class\n\n## FAQs on the Square root of 14\n\n### What is the value of the Square source of 14?\n\nThe square source of 14 is 3.74165.\n\n### Why is the Square source of 14 one Irrational Number?\n\nUpon element factorizing 14 i.e. 21 × 71, 2 is in strange power. Therefore, the square root of 14 is irrational.\n\n### What is the worth of 15 square root 14?\n\nThe square source of 14 is 3.742. Therefore, 15 √14 = 15 × 3.742 = 56.125.\n\n### What is the Square source of -14?\n\nThe square root of -14 is an imagine number. It have the right to be created as √-14 = √-1 × √14 = ns √14 = 3.741iwhere ns = √-1 and it is called the imaginary unit.\n\nSee more: 2007 Pt Cruiser Oil Pressure Sensor Replacement Costs, Chrysler Pt Cruiser Oil Pressure Switch & Sender\n\n### Evaluate 15 plus 6 square source 14\n\nThe provided expression is 15 + 6 √14. We recognize that the square root of 14 is 3.742. Therefore, 15 + 6 √14 = 15 + 6 × 3.742 = 15 + 22.450 = 37.450\n\n### If the Square root of 14 is 3.742. Find the worth of the Square root of 0.14.\n\nLet us represent √0.14 in p/q type i.e. √(14/100) = 0.14/10 = 0.374. Hence, the worth of √0.14 = 0.374\n\nExplore math program\nExplore coding program\nMake her child naturally math minded\nBook A cost-free Class" ]

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[ "", null, "", null, "", null, "BACK\n Lagrange Interpolation with time interval of day(s)       First date (optional)", null, "Produce file of the selected parameters", null, "Draw data   Dynamic graph   Show errors Spectral analysis (FFT, complex for 2D signal) Amplitude PSD min. frequency or period       max. frequency or period x linear scale x log scale y linear scale y log scale frequency spectrum (in cycle / unit of time) period spectrum (in unit of time) First derivative Produce data Plot Weighted least square fit of periodic components (periods in unit of time) Positive periods         Polynomial of degree Negative periods       Weighted least square (take negative periods only for 2 dimensional signal)         Draw residuals and input data     Draw fit and input data     Print residuals In-phase and out-of-phase terms (a, b) are estimated, as well as amplitude A and phase φ : 1-D :$$\\small X = A \\cos[2\\pi/T (t-t_0) + \\phi] = a \\cos[2\\pi/T (t-t_0)] + b \\cos[2\\pi/T (t-t_0)]$$ 2-D : $$\\small X +i Y = A e^{i [2\\pi/T (t-t_0) + \\phi]$$ with the reference epoch $$\\small t_0$$ = 1/1/2000 0hUT that is : $$\\small X = a \\cos[2\\pi/T (t-t_0)] - b \\sin[2\\pi /T (t-t_0)$$    $$\\small Y = b cos[2\\pi/T (t-t_0)] + a sin[2\\pi/T (t-t_0) ]$$ with $$\\small a = A \\cos\\phi$$      $$\\small b = A\\sin\\phi$$ Vondrak low/high pass filter Remove parabolic trend Produce data file Draw   with input data Select band above Select band below (P0) time unit     Transfer coefficient for P0:T0= % The Vondrak filter transfer function at another period P is given by T=1/(1+(P0/P)6 (1-T0)/T0) For the case \"Select band around\" the periods in [P0-0.1*P0, P0+0.1*P0] are transmitted with the rate > T0%. For the case \"Remove band around\" the periods outside [P0-0.1*P0, P0+0.1*P0] are transmitted with the rate > T0 %. Panteleev band pass filter (for 2D signal) Produce data file Draw filtered data and envelope / phase referred to 2πfct   Select band around fc= cycle/time unit     Band width f0= cycle/time unit This band pass filter was designed by Russian astronomer and gravimetrist V. L. Panteleev. Its frequency transfer function is given by $$\\small T(f) = \\frac{f_0^4}{ ( f - f_c)^4 + f_0^4 }$$. At the edges of the window $$\\small |f - f_c| = f_0$$ and $$\\small T = 0.5$$. Singular Spectral Analysis (SSA) -  Zoom between and    The extracted components are decorellated over time windows of (in the time unit) with the interpolation lag (in the time unit). Firt step consists in the determination of the eigenvalues and eigenvectors printed by decreasing weight. Then 5 singular components are reconstructed according to the following combinations of eigenvectors, to be stated from the analysis of the eigenvalues . RC1 Reconstructed Component (RC) based upon eigenvectors N1 and N2 RC2 RC3 RC4 RC5    produce time series (date, signal, RC1,RC2,RC3,RC4,RC5,residuals) draw Graphic dimension x   output graphics png   pdf   ps       Partial Interface with the C-Fortran Libraries SLAVA (C. Bizouard) & MIMOSA (S. Lambert). Thank you for bringing to our knowledge any possible mistake, mail to : christian.bizouard at obspm.fr 1 Variations produced by the solid Earth zonal tides (IERS 2000 model) 2 TT=TAI+32.184 s (for parameter UT1-TAI) 3 Reference precession-nutation model for celestial pole offsets: either IAU 2000 or IAU 1980" ]

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[ "", null, "# I'm a gameplay programmer focused on developing engaging and unique experiences!\n\nOne of the most common tasks for a vertex shader is to convert a vertex position from local space to projected space. A simple approach may be to cache the matrices LocalToWorld, WorldToCamera, and CameraToProjected in constant buffers for the vertex shader to use. Then, the shader could calculate the LocalToProjected matrix and output the correct position for the fragment shader to use. However, by doing this calculation in the vertex shader, LocalToProjected will be computed for every vertex of a mesh! This can be optimized by calculating LocalToProjected on the CPU and storing that in the constant buffer for the vertex shader to use at will.\n\nOn that note, we store LocalToProjected per draw call because LocalToWorld will be unique for every object that needs to be drawn.\n\n## Matrix Multiplication Order\n\nA further optimization on generating LocalToProjected is the order we multiply the matrices LocalToWorld, WorldToCamera, and CameraToProjected. An initial approach may be to simply multiply these matrices in order resulting in this code:\n\n``````Matrix LocalToCamera = WorldToCamera * LocalToWorld;\nMatrix LocalToProjected = CameraToProjected * LocalToCamera;``````\n\nA subtle problem arises in that this order of operations requires every object to use two Matrix multiplications for computing LocalToProjected. Let's see what happens when multiplying in the opposite order\n\n``````Matrix WorldToProjected = CameraToProjected * WorldToCamera;\nMatrix LocalToProjected = WorldToProjected * LocalToWorld;``````\n\nHmm, this still requires two matrix multiplications. However, now the matrix WorldToProjected can be reused for every other object that needs to be drawn! Thus, we cut down the number of matrix multiplications from 2N to N + 1. The WorldToProjected matrix could also be stored in a constant buffer for shaders to use because it it will be the same for any vertex.\n\n### Instruction Count\n\nOne way to see the results of this optimization is to look at the disassembly for a shader that calculates LocalToProjected rather than pulling it from a constant buffer. Below, we see that kind of shader takes 24 instructions.\n\n``````mul r1.xyzw, r0.xxxx, cb2.xyzw\nmul r2.xyzw, r0.yyyy, cb2.xyzw\nmul r2.xyzw, r0.zzzz, cb2.xyzw\nmul r0.xyzw, r0.wwww, cb2.xyzw\nadd r0.xyzw, r0.xyzw, r1.xyzw // r0.x <- vertexPosition_world.x; r0.y <- vertexPosition_world.y; r0.z <- vertexPosition_world.z; r0.w <- vertexPosition_world.w\n\nmul r1.xyzw, r0.xxxx, cb0.xyzw\nmul r2.xyzw, r0.yyyy, cb0.xyzw\nmul r2.xyzw, r0.zzzz, cb0.xyzw\nmul r0.xyzw, r0.wwww, cb0.xyzw\nadd r0.xyzw, r0.xyzw, r1.xyzw // r0.x <- vertexPosition_camera.x; r0.y <- vertexPosition_camera.y; r0.z <- vertexPosition_camera.z; r0.w <- vertexPosition_camera.w\n\nmul r1.xyzw, r0.xxxx, cb0.xyzw\nmul r2.xyzw, r0.yyyy, cb0.xyzw\nmul r2.xyzw, r0.zzzz, cb0.xyzw\nmul r0.xyzw, r0.wwww, cb0.xyzw\n\nret\n// Approximately 24 instruction slots used``````\n\nNow let's look at the same shader that has been optimized to use a cached LocalToProjected matrix.\n\n``````mov r0.xyz, v0.xyzx // r0.x <- vertexPosition_local.x; r0.y <- vertexPosition_local.y; r0.z <- vertexPosition_local.z\nmov r0.w, l(1.000000) // r0.w <- vertexPosition_local.w\n\nmul r1.xyzw, r0.xxxx, cb2.xyzw\nmul r2.xyzw, r0.yyyy, cb2.xyzw" ]

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[ "", null, "###### Jonathan Fong\n\nU.C.Berkeley\nM.Ed.,San Francisco State Univ.\n\nJonathan has been teaching since 2000 and currently teaches chemistry at a top-ranked high school in San Francisco.\n\n##### Thank you for watching the video.\n\nTo unlock all 5,300 videos, start your free trial.\n\n# Energy in Hydrogen Emission Spectrum Problem - Concept\n\nJonathan Fong", null, "###### Jonathan Fong\n\nU.C.Berkeley\nM.Ed.,San Francisco State Univ.\n\nJonathan has been teaching since 2000 and currently teaches chemistry at a top-ranked high school in San Francisco.\n\nShare\n\nHere we have a challenging problem about identifying elements on the periodic table. So here we have four different cases. Letter A says we want to identify a member of the same family as oxygen, whose most stable ion contains 36 electrons. What I always tell my students is, figure out what kind of information you have that can help you figure out how the answer to the problem.\n\nHere we actually have two pieces of information. So the same family of oxygen. If you take a look on the periodic table, oxygen is in group 6A, the same family that would be group 6A. That narrows it down to Oxygen, Sulphur, Selenium, Tellurium, or Lead. You have five choices. 20% chance you could guess, but let’s make it a little bit easier.\n\nThe most stable ion contains 36 electrons. Look at which period which would have closest to 36. So 34 you have Selenium, a neutral atom. And then Sulphur would have 16, Oxygen would have 8, Tellurium would have 42 and then Polonium would have 84. When you take a look here, the closest one would be Selenium but let’s check take a look. Group 6A all of them form ions with a -2 charge for its ion.\n\nIf you have a -2 charge, that means you’re going to add two electrons to each of those numbers that we have here. So instead of 34 electrons, for an atom of selenium, it would be 34 plus 2 which is 36. Which is what we wanted to find here. And so for each of these, remember electrons are negative, so if you have a -2 charge, for the ions, that means you’re adding 2 electrons. So Oxygen Oxide would have 10 and then Sulphur would have 18. Tellurium, the ion would have 54 and then Polonium for the ion would have 86 just like the normal gases that you would have there.\n\nSo the answer for letter A would be selenium, because it’s in group 6A which satisfies the first part, and the ion would have 34 plus 2, which is 36 electrons. That would satisfy the second part. So letters b, c and d that we have here, basically we do the same thing.\n\nAlkali metals. Let’s take a look. They are in group 1A and then they all form a +1 charge for the ion. So if the ion contains 54 electrons, that means that the real element would have 55 electrons. 55 electrons for the atom and then since it has a +1 charge, so we’re going to take away one electron because of +1 charge. So we'd have ion 54 and atom 55. So the atom would have 55. You take a look at 55 on the periodic table and that would be Cesium, because Cesium has 55 for the atom. And then its stable ion has a +1 charge, so you take away 1 from the ion charge and then you’ll get 54 electrons, total for the ion.\n\nFor the noble gas, noble gases they are in group 8. So this one’s easy. 54 protons, so you take a look at the one with the atomic number 54. And 54 is Xenon. That one was simple.\n\nHalogen, group 7A. 53 protons, so you take a look at the atomic number 53, and you actually didn’t need the information halogen here. And it is Iodine. On these last two examples, instead of looking at the group number we could have gone straight for the number of protons. Remember the protons always represent the element that you have there.\n\nRemember elements have unique number of protons, so the protons always tell you which element it is. Hopefully, this helps you out in identifying elements on the periodic table. Remember, if you have the protons, go to the protons first, because that will definitely tell you what the element is, without even spending time looking for the other information. But sometimes you need the other clues to help you in isolating the element. Have a good one." ]

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[ "Preprint Article Version 1 Preserved in Portico This version is not peer-reviewed\n\n# Mathematical Model Describing HIV Infection with Time-Delayed CD4 T-Cell Activation\n\nVersion 1 : Received: 27 April 2020 / Approved: 28 April 2020 / Online: 28 April 2020 (08:43:56 CEST)\n\nA peer-reviewed article of this Preprint also exists.\n\nToro-Zapata, H.D.; Trujillo-Salazar, C.A.; Carranza-Mayorga, E.M. Mathematical Model Describing HIV Infection with Time-Delayed CD4 T-Cell Activation. Processes 2020, 8, 782. Toro-Zapata, H.D.; Trujillo-Salazar, C.A.; Carranza-Mayorga, E.M. Mathematical Model Describing HIV Infection with Time-Delayed CD4 T-Cell Activation. Processes 2020, 8, 782.\n\nJournal reference: Processes 2020, 8, 782\nDOI: 10.3390/pr8070782\n\n## Abstract\n\nA mathematical model, composed of two non-linear differential equations that describe the population dynamics of CD4 T cells in the human immune system, as well as viral HIV particles, is proposed. The invariance region is determined, classical equilibria stability analysis is performed using the basic reproduction number, and numerical simulations are carried out, in order to illustrate stability results. Later, the model is modified with a delay term, which describes the time that cells require for immunological activation. This generates a two-dimensional integro-differential system, which is transformed into a system with three ordinary differential equations, via auxiliary variable use. For the new model, equilibrium points are determined, their local stability is examined, and results are studied by way of numerical simulation.\n\n## Keywords\n\nmathematical model; delay differential equations; HIV; immune system\n\n## Subject\n\nMATERIALS SCIENCE, Other\n\nViews 0" ]

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