docid
stringlengths 8
16
| text
stringlengths 3
29.7k
| randval
float64 0
1
| before
int64 0
99
|
|---|---|---|---|
science336 | I think that ignoring limits is problematic. If there was a limit of the function $f(x,y)=x/y$ for $x,y \to 0$ regardless of how the limit is performed, then one would define that value to be $f(0,0)$, even if everything else is strange. Since the limiting value depends on the way the limit is done, choosing a value for $f(0,0)$ is counterproductive as it gives a non-continuous function. Better to have a continuous function over a slightly smaller domain. This also forces the point that if you do have a limit process that results in the evaluation of $f(0,0)$, you realize early on that you should examine the limit carefully rather than use $\lim g(x) = g(\lim x)$ (which is only true for continuous functions, of course.) By the way, this might be too trivial, but I'll give an example of how the limiting value depends on the limit: $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x} = 1$, (both $\sin(x)$ and $x$ go to 0) $\displaystyle \lim_{x \to 0} \frac{\cos x - 1}{x} = 0$ (again, both numerator and denominator go to 0, but numerator goes "faster") $\displaystyle \lim_{x \to 0} \frac{\sqrt{x}}{x} =+\infty$. | 0.389874 | 99 |
science338 | I'll try it for 2D and then we can get 1D as a corollary [excercise!]... This is the only proof I know of, there may be a more intuitive (and less messy without tex!) proof out there, but I like this one- it uses generating functions in a really nifty way. Consider the probability of being at the origin after 2n steps (notice we cannot return in an odd number of steps): $u_0=1$ $u_{2n} = p(n,n,0,0)+ p(n-1,n-1,1,1)+...+p(n-k,n-k,k,k)+...+p(0,0,n,n)$ (when $n\neq0$) Here $p(u,d,l,r)$ is the probability of the first 2n steps being u up, d down, l left and r right in any order. Each order has probability $\frac{1}{4^{2n}}$, and there are $\frac{(2n)!}{u!d!l!r!}$ distinct orders, giving $p(n-k,n-k,k,k)=\frac{1}{4^{2n}} \frac{(2n)!}{(n-k)!(n-k)!k!k!}$ Now, since $(2n)!=n! n! \binom{2n}{n}$ we have $p(n-k,n-k,k,k)=\frac{1}{4^{2n}} \binom{2n}{n} \binom{n}{k}^2$ giving $u_{2n}= \frac{1}{4^{2n}} \Sigma_k \binom{2n}{n} \binom{n}{k}^2$ which, by one of those silly binomial results, can be contracted to $u_{2n}= \frac{1}{4^{2n}} \binom{2n}{n}^2$ Let us put that in our back pocket for now and consider instead the probability of first returning after 2n steps $f_{2n}$ --- this is rather difficult to tackle directly but we can make ourselves a cunning little formula involving it, jazz out some generating function fun and seal the deal with some. The formula in question is: $u_{2n}= f_2 u_{2n-2}+ f_4 u_{2n-4} +....+f_{2n-2}u_{2} + f_{2n}$ Which we shall not prove so much as explain: to return to the origin after 2n steps (LHS) you must either first return after 2 steps and do a 'return to the origin in 2n-2 steps walk' (first term RHS) or first return after 4 steps and do a 'return to the origin in 2n-4 steps walk' (2nd term) or... or first return after 2n-2 steps and do a 'return to the origin in 2 steps walk or first return to the origin after 2n steps. We shall now tweak our formula just a tiny bit so it has the right symmetry properties for what is to come, we do this by adding $f_0=0$ and $u_0=1$ to give $u_{2n}= f_0 u_{2n}+ f_2 u_{2n-2}+ f_4 u_{2n-4} +....+f_{2n-2}u_{2} + f_{2n}u_0$ Which is secretly a statement about generating functions (this is the sweet bit!), see look at the generating functions of $u_n$, $f_n$: $U(x)= \Sigma_m u_{2m} x^{2m}$, $F(x)= \Sigma_m f_{2m} x^{2m}$ We see: $U(x)= 1+ U(x)F(x)$ (where the '1+' is to compensate for the fact that $u_0$ does not appear in the product) Rearranging to: $F(x)= \frac{U(x)-1}{U(x)}$ Observe that the probability of return is $\Sigma f_n= F(1)$, which comes out as 1 because $U(1)$ diverges by some tedious stirlings formula bounds that I forget. Edit: Until Tex comes online, this is pretty unreadable, so here's a link to some lecture notes I found with the same proof (and, fortunately, the same notation!). Enjoy! Edit$^2$: Hooray, Tex has come online!!! Enjoy. | 0.369291 | 0 |
science339 | For an old example, Mersenne made the following conjecture in 1644: The Mersenne numbers, $M_n=2^n − 1$, are prime for n = 2, 3, 5, 7, 13, 17, 19, 31, 67, 127 and 257, and no others. Pervushin observed that the Mersenne number at $M_{61}$ is prime, so refuting the conjecture. $M_{61}$ is quite large by the standards of the day: 2 305 843 009 213 693 951. According to Wikipedia, there are 51 known Mersenne primes as of 2018 | 0.24222 | 0 |
science342 | If we use the idea of set exponentiation to define exponentiation of cardinals, we have the following natural idea: $$A^B:=\{f:B\to A\}$$ We define the exponential of cardinals as follows: $|A|^{|B|}:=|A^B|$. It's easy to check that this agrees with our intuition for exponentiation of natural numbers when $B$ is nonempty. There is only one set representing the cardinal $0$, namely the empty set. Then we may look at $0^0$ as follows: $$0^0=|\emptyset|^{|\emptyset|}=|\emptyset^\emptyset| =|\{f:\emptyset\to\emptyset\}|=1$$ | 0.348797 | 99 |
science344 | If I'm not mistaken, the integral domain of holomorphic functions on a connected open set $U \subset \mathbb{C}$ works. It is a theorem (in Chapter 15 of Rudin's Real and Complex Analysis, and essentially a corollary of the Weierstrass factorization theorem), that every finitely generated ideal in this domain is principal. This implies that if $a,b$ have no common factor, they generate the unit ideal. However, for instance, the ideal of holomorphic functions in the unit disk that vanish on all but finitely many of ${1-\frac{1}{n}}$ is nonprincipal. | 0.273842 | 99 |
science345 | See Durrett, Probability: Theory and Examples (link goes to online copy of the fourth edition; original defunct link). On p. 164 Durrett gives a proof that simple random walk is recurrent in two dimensions. First find the probability that simple random walk in one dimension is at $0$ after $2n$ steps; this is clearly $\rho_1(2n) = \binom{2n}{n}/2^{2n}$, since $\binom{2n}{n}$ is the number of paths with $n$ right steps and $n$ left steps. Next, the probability that simple random walk in two dimensions -- call this $\rho_2(2n)$ -- is at $0$ after $2n$ steps is the square of the previous probability. Consider the simple random walk which makes steps to the northeast, northwest, southeast, and southwest with equal probability. The projections of this walk onto the x- and y-axes are independent simple random walks in one dimension. Rotating and rescaling gives the "usual" SRW in two dimensions (with steps north, east, south and west) and doesn't change the probability of being at $0$. So $\rho_2(2n) = \left(\binom{2n}{n}/2^{2n}\right)^2$. This is asymptotic to $1/(\pi n)$, and the expected number of returns to the origin is the $\sum_{n \geq 1} \rho_2(2n)$, so the expected number of returns to the origin is infinite. It's not hard to show (and is in fact true, but the proof is unenlightening so I'll leave it to Durrett) that in this case the probability of eventually returning to the origin is $1$. | 0.206115 | 0 |
science346 | Markov chains, especially hidden Markov models are hugely important in computation linguistics. A hidden Markov model is one where we can't directly view the state, but we do have some information about what the state might be. For example, consider breaking down a sentence into what is called "parts of speech" such as verbs, adjectives, ect. We don't know what the parts of speech are, but we can attempt to deduce them from the word. For example, the word run might be used 80% as a verb, 18% of the time as a noun and 2% of the time as an adjective. We also have (Markov) relations between the parts of speech, so for example an adjective might be followed by a noun 70% of the time and another adjective 30% of the time. We can use the Viterbi algorithm to decide which sequence is most likely to have generated the observed sentence. This algorithm takes into account two factors: the probability of such a sequence of parts of speech occurring together in a sentence the relative chance that such a sequence of parts of speech would be responsible for us observing the given sentence. | 0.33634 | 99 |
science347 | This is close to being the flipside of the geometric mean, which is the nth root of the product of the numbers, and can be expressed as the exponential of the sum of the logarithms. Another pair of dual mean measures is the regular mean and the harmonic mean (n divided by the sum of the reciprocals). I say the soft maximum is close to being the flipside of the geometric mean, but it lacks the good property that all of the others have of taking a list of the same value to that value (for definedness, let all values be positive). Let's call the hyperbolic mean the "soft maximum" of the nth roots of the terms in the list: then this has that good property. The hyperbolic mean the emphasises large values in a roughly symmetric manner to the way that the geometric mean emphasises small values (which is always smaller than the regular mean), and is, of course, much smaller for long list of large values. So I say, consider it an amplified version of a useful addition to the family of of mean operators. | 0.3271 | 99 |
science352 | Edited to clear some things up: Simplicial and singular (co)homology were invented to detect holes in spaces. To get an intuitive idea of how this works, consider subspaces of the plane. Here the 2-chains are formal sums of things homeomorphic to the closed disk, and 1-chains are formal sums of things homeomorphic to a line segment. The operator d takes the boundary of a chain. For example, the boundary of the closed disk is a circle. If we take d of the circle we get $0$ since a circle has no boundary. And in general it happens that $d^2 = 0$, that is boundaries always have no boundaries themselves. Now suppose we remove the origin from the plane and take a circle around the origin. This circle is in the kernel of d since it has no boundary. However, it does not bound any 2-chain in the space (since the origin is removed) and so it is not in the image of the boundary operator on two-dimensions. Thus the circle represents a non-trivial element in the quotient space $\ker( d ) / \operatorname{im} (d)$. The way I have defined things makes the above a homology theory simply because the d operator decreases dimension. Cohomology is the same thing only the operator increases dimension (for example the exterior derivative on differential forms). Thus algebraically there really is no difference between cohomology and homology since we can just change the grading from $i$ to $-i$. From a homology we can get a corresponding cohomology theory by dualizing, that is by looking at maps from the group of chains to the underlying group (e.g. $\Bbb Z$ or $\Bbb R$). Then d on the cohomology theory becomes the adjoint of the previous boundary operator and thus increases degrees. | 0.422543 | 99 |
science353 | Most of us know that, being deterministic, computers cannot generate true random numbers. However, let's say you have a box which generates truly random binary numbers, but is biased: it's more likely to generate either a 1 or a 0, but you don't know the exact probabilities, or even which is more likely (both probabilities are > 0 and sum to 1, obviously) Can you use this box to create an unbiased random generator of binary numbers? | 0.245033 | 0 |
science354 | Assuming you have unlimited time and cash, is there a strategy that's guaranteed to win at roulette? | 0.117398 | 0 |
science355 | Fork in the road 2 You're once again at a fork in the road, and again, one path leads to safety, the other to doom. There are three natives at the fork. One is from a village of truth-tellers, one from a village of liars, one from a village of random answerers. Of course you don't know which is which. Moreover, the natives answer "pish" and "posh" for yes and no, but you don't know which means "yes" and which means "no." You're allowed to ask only two yes-or-no questions, each question being directed at one native. What do you ask? | 0.301053 | 0 |
science356 | I've always thought this sort of puzzle is a wonderful example of the difference between behaviour as you approach a limit, versus behaviour AT the limit. You can make the puzzle more revealing like this: Say I have an infinite number of banknotes, each with a unique serial number. I give you bill number 1. Now, you have two options, either you keep that bill and the game ends, or I give you 10 more banknotes, but you have to burn the lowest numbered bill. It seems obvious that option two is much better. Now, we repeat this game over and over. You keep getting more and more money. But at the limit, every banknote has been burned and you're worse off than you would have been had you just taken the one banknote and left. | 0.145264 | 0 |
science357 | Adding additional axioms would make more truths provable. But it wouldn't make all truths provable (unless the axiom was inconsistent with the already given ones, in which case all falsehoods would also be provable too). So adding additional axioms isn't going to help make all the truths provable. I guess you could just add, say, Goldbach's conjecture and the Riemann hypothesis as extra axioms and carry on doing mathematics-PLUS! but why would you want to? | 0.092186 | 99 |
science359 | A function can be represented as a power series if and only if it is complex differentiable in an open set. This follows from the general form of Taylor's theorem for complex functions. Being real differentiable--even infinitely many times--is not enough, as the function $e^{-1/x^2}$ on the real line (equal to 0 at 0) is $C^\infty$ yet does not equal its power series expansion since all its derivatives at zero vanish. The reason is that the complexified version of the function is not even continuous at the origin. | 0.364187 | 99 |
science361 | Gauss-Jordan elimination can be used to determine when a matrix is invertible and can be done in polynomial (in fact, cubic) time. The same method (when you apply the opposite row operation to identity matrix) works to calculate the inverse in polynomial time as wel. | 0.191336 | 99 |
science363 | Let $p$ be a prime congruent to 1 mod 4. Then to write $p = x^2 + y^2$ for $x,y$ integers is the same as writing $p = (x+iy)(x-iy) = N(x+iy)$ for $N$ the norm. It is well-known that the ring of Gaussian integers $\mathbb{Z}[i]$ is a principal ideal domain, even a euclidean domain. Now I claim that $p$ is not prime in $\mathbb{Z}[i]$. To determine how a prime $p$ of $\mathbb{Z}$ splits in $\mathbb{Z}[i]$ is equivalent to determining how the polynomial $X^2+1$ splits modulo $p$. First off, $-1$ is a quadratic residue modulo $p$ because $p \equiv 1 \mod 4$. Consequently, there is $t \in \mathbb{Z}$ with $t^2 \equiv -1 \mod p$, so $X^2+1$ splits modulo $p$, and $p$ does not remain prime in $\mathbb{Z}[i]$. (Another way of seeing this is to note that if $p$ remained prime, then we'd have $p \mid (t+i)(t-i)$, which means that $p \mid t+i$ or $t \mid t-i$.) Anyway, as a result there is a non-unit $x+iy$ of $\mathbb{Z}[i]$ that properly divides $p$. This means that the norms properly divide as well. In particular, $N(x+iy) = x^2+y^2$ properly divides $p^2$, so is $p$ or $1$. It cannot be the latter since otherwise $x+iy$ would be a unit. So $x^2+y^2 = p$. | 0.215505 | 99 |
science364 | A matrix is invertible iff its determinant is non-zero. There are algorithms which find the determinant in slightly worse than O(n2) | 0.278024 | 99 |
science367 | Grigory has already answered your particular question. However, I wanted to point out that your question "How do you prove that a group specified by a presentation is infinite?" has no good answer in general. Indeed, in general the question of whether a group presentation defines the trivial group is undecidable. | 0.334836 | 99 |
science368 | 100%, for the same reason as the 1-D walk In fact (again for the same reason), your chances are 100% of eventually reaching X-greater heads than tails (or tails than heads), where X is any non-negative integer. | 0.542989 | 99 |
science372 | The following slideshow gives an explanation of how algebraic geometry can be used in phylogenetics. See also this post of Charles Siegel on Rigorous Trivialties. This is not an area I've looked at in much detail at all, but it appears that the idea is to use a graph to model evolutionary processes, and such that the "transition function" for these processes is given by a polynomial map. In particular, it'd be of interest to look at the potential outcomes, namely the image of the transition function; that corresponds to the image of a polynomial map (which is not necessarily an algebraic variety, but it is a constructible set, so not that badly behaved either). (In practice, though, it seems that one studies the closure, which is a legitimate algebraic set.) | 0.232686 | 99 |
science375 | The trains take half an hour to collide, which, at a rate of 75kph, leads to the fly travelling 37.5km. | 0.200401 | 99 |
science377 | The representation theory of finite groups can be used to prove results about finite groups themselves that are otherwise much harder to prove by "elementary" means. For instance, the proof of Burnside's theorem (that a group of order $p^a q^b$ is solvable). A lot of the classification proof of finite simple groups relies on representation theory (or so I'm told, I haven't read the proof...). Mathematical physics. Lie algebras and Lie groups definitely come up here, but I'm not familiar enough to explain anything. In addition, the classification of complex simple Lie algebras relies on the root space decomposition, which is a significant (and nontrivial) fact about the representation theory of semisimple Lie algebras. Number theory. The nonabelian version of L-functions (Artin L-functions) rely on the representations of the Galois group (in the abelian case, these just correspond to sums of 1-dimensional characters). For instance, the proof that Artin L-functions are meromorphic in the whole plane relies on (I think) Artin Brauer's theorem (i.e., a corollary of the usual statement) that any irreducible character is an rational integer combination of induced characters from cyclic subgroups -- this is in Serre's Linear Representations of Finite Groups. Also, the Langlands program studies representations of groups $GL_n(\mathbb{A}_K)$ for $\mathbb{A}_K$ the adele ring of a global field. This is a generalization of standard "abelian" class field theory (when $n=1$ and one is determining the character group of the ideles). Combinatorics. The representation theory of the symmetric group has a lot of connections to combinatorics, because you can parametrize the irreducibles explicitly (via Young diagrams), and this leads to the problem of determining how these Young diagrams interact. For instance, what does the tensor product of two Young diagrams look like when decomposed as a sum of Young diagrams? What is the dimension of the irreducible representation associated to a Young diagram? These problems have a combinatorial flavor. I should add the disclaimer that I have not formally studied representation theory, and these are likely to be an unrepresentative sample of topics (some of which I have only vaguely heard about). | 0.464935 | 1 |
science379 | Yes; it's possible to write a system of equations that can be solved to find the correct coefficients. Here's an example for the given formula. We're trying to find coefficients A, B, and C such that $A (\mathrm{Al}) + B (\mathrm{O_2}) \rightarrow C (\mathrm{Al_2 O_3})$ In order to do this, we can write an equation for each element based on how many atoms are on each side of the equation. for Al: $A = 2C$ for O: $2B = 3C$ This is an uninteresting example, but these will always be linear equations in terms of the coefficients. Note that we have fewer equations than variables. This means that there's more than one way to correctly balance the equation (and there is, because any set of coefficients can be scaled by any factor). We just need to find one integral solution to these equations. To solve, we can arbitrarily set one of the variables to 1 and we'll get a solution with (probably fractional) coefficients. If we add $A=1$, the solution is $(A,B,C) = (1,\frac{3}{4},\frac{1}{2})$. To get the smallest solution with integer coefficients, just multiply by the least common multiple of the denominators ($4$ in this case), giving us $(4,3,2)$. If the set of equations has no solution where the coefficients are nonzero, then you know that the equation cannot be balanced. | 0.237478 | 99 |
science380 | You can use Pappus's centroid theorem as in my answer here, but it does not provide much insight. If instead of a cylinder and a cone, you consider a cube and a square-based pyramid where the "top" vertex of the pyramid (the one opposite the square base) is shifted to be directly above one vertex of the base, you can fit three such pyramids together to form the complete cube. (I've seen this as physical toy/puzzle with three pyramidal pieces and a cubic container.) This may give some insight into the 1/3 "pointy thing rule" (for pointy things with similar, linearly-related cross-sections) that Katie Banks discussed in her comment. | 0.33258 | 1 |
science385 | A visual demonstration for the case of a pyramid with a square base. As Grigory states, Cavalieri's principle can be used to get the formula for the volume of a cone. We just need the base of the square pyramid to have side length $ r\sqrt\pi$. Such a pyramid has volume $\frac13 \cdot h \cdot \pi \cdot r^2. $ Then the area of the base is clearly the same. The cross-sectional area at distance a from the peak is a simple matter of similar triangles: The radius of the cone's cross section will be $a/h \times r$. The side length of the square pyramid's cross section will be $\frac ah \cdot r\sqrt\pi.$ Once again, we see that the areas must be equal. So by Cavalieri's principle, the cone and square pyramid must have the same volume:$ \frac13\cdot h \cdot \pi \cdot r^2$ | 0.204304 | 1 |
science386 | The games aren't just about winning or losing, but also about utility. Here is a more accurate table for chicken: B swerves B straight A swerves No gain for either A loses, B wins A straight B loses, A wins Both have large loss Here is one for prisoners dilemma: B silent B testify A silent Both have small loss A large loss, B loses nothing A testify B loses nothing, A large loss, both medium lose In the prisoners dilemma, an individual prisoner will always do better by testifying (look at the table), however, by both testifying they end up in a worse position than if both were silent. In contrast, in chicken, going straight will be better if the other swerves and swerving will be better if the other goes straight. Your tables represent some strategies as being equally good for players when they are not. | 0.470689 | 99 |
science389 | A probability problem I love. Take a shuffled deck of cards. Deal off the cards one by one until you reach any Ace. Turn over the next card, and note what it is. The question: which card has a higher probability of being turned over, the Ace of Spades or the Two of Hearts? | 0.006028 | 0 |
science390 | I don't know if you consider General Relativity "outside acadamia"(and I don't care to argue the point!) but if you do, the group of symmetries with respect to the Lorentzian Metric can be written as Matrices containing hyperbolic trig functions as elements. Note Kenny's comment. | 0.087408 | 99 |
science391 | I've read a fair amount of Sets for Mathematics and found it to be a gentle introduction. http://www.amazon.com/Sets-Mathematics-F-William-Lawvere/dp/0521010608/ref=pd_sim_b_5 | 0.346795 | 0 |
science393 | If you know your relationship is going to be a polynomial, then there are some pretty (conceptually) simple ways you can do this. If you know what degree your polynomial is (line, parabola, cubic, etc.) then your job will be much easier. But if not, then you simply need to look at the number of points you have. If you are given one point, the best you can do is of degree 0 ( y = k ) If you are given two points, the best you can do is of degree 1 ( y = A x + B ) If you are given three points, the best you can do is of degree 2 ( y = A x2 + B x + C ) If you are given four points, the best you can do is of degree 3 ( y = A x3 + B x2 + C x + D ) etc. When I say "the best you can do", what I mean is -- if you have a parabola, but are only given two points, then you really can't identify the parabola. But you can say that it's a simple line. Let's assume you have three points. The "best you can do" is assume that it is degree 2. If it is actually of degree one, your answer will magically turn into a line ( your x^2 coefficient will be 0 ) The basic idea of solving relationships/equations is: If you have n unknowns, you need n equations/points. Notice how, in the form of the Parabola ( y = A x2 + B x + C ), you have three unknowns? And also three equations! (points) Let's pick three arbitrary points x 1 2 4 y 6 7 3 You would set up three equations: 6 = 12 * A + 1 * B + C 7 = 22 * A + 2 * B + C 3 = 42 * A + 4 * B + C Three equations, three unknowns. You should be able to solve this with a combination of most system-of-equation-solving rules. In our case, we find: A = -1 B = 4 C = 3 So our equation is y = -x2 + 4 x + 3 Note that, if your original three points formed a line, your $A$ would $= 0$ However, if your equation is NOT a polynomial, then you are left with little more than guess and check, plugging in various coefficients and trials (exponential? trigonometric?) The beauty of the polynomial approach is that a polynomial of high enough degree will always fit any list of points. (provided that the points form a function) | 0.49119 | 99 |
science394 | It is actually the other way round. A generating function is generally defined to have an addition operation where the components are added and a multiplication operation like that you mentioned. Once we have made these definitions, we observe that polynomials obey the same laws and so that it is convenient to represent generating functions as infinite polynomials rather than just an infinite tuple. | 0.270176 | 99 |
science395 | Casebash is correct that this is a definition and not a theorem. But the motivation from 3.48 (Defintion of product of series) of little Rudin may convince you that this is a good definition: $\sum_{n=0}^{\inf} a_n z^n \cdot \sum_{n=0}^{\inf} b_n z^n = (a_0+a_1z+a_2z^2+ \cdots)(b_0+b_1z+b_2z^2+ \cdots)$ $=a_0b_0+(a_0b_1 + a_1b_0)z + (a_0b_2+a_1b_1+a_2b_0)z^2 + \cdots$ $=c_0+c_1z+c_2z^2+ \cdots $ where $c_n=\sum_{k=0}^n a_k b_{n-k}$ | 0.360424 | 99 |
science396 | Given a list of terms of a sequence as you describe, one technique that may be of use (supplementary to Justin's answer) is finite differences. Calculate the differences between successive terms. If these first differences are constant, then a linear equation fits the terms you have. If not, compute the differences of the differences. If these second differences are constant, then a quadratic equation fits the terms you have. If not, you can continue to compute differences until you reach a constant difference (in the nth differences means an nth degree polynomial), differences that are a constant multiple of the previous differences (exponential of some sort), or you run out of terms. In any case, what you find is limited to matching the terms you know, as without some kind of general rule for the sequence, the first unknown term could be anything and completely alter the pattern (and with n known terms, a polynomial of degree n-1 will always perfectly fit). | 0.210653 | 99 |
science397 | In the general case of trying to predict some infinite sequence of integers, there is no formula. This is because there is no reason to expect a pattern to continue, since all sequences are possible. However, you can say a certain number is more likely given a set of functions. For instance if you considered all Turing machines, you could say that given n elements of a sequence look at all the Turing machines that predict the current sequence, and then find the most predicted next number. There still isn't a efficient way to compute what the most likely next number is. Ray Solomonoff called this "Universal Probabilistic Induction" This is explained in more depth here: http://en.wikipedia.org/wiki/Inductive_inference | 0.4212 | 99 |
science398 | I believe the right generalization of dual spaces is that of a dual object in a tensor category, which I will assume symmetric for convenience. Recall what makes a dual space of a vector space work: We have a map $V \times V^* \to k$ (for $k$ the ground field). The problem is, this isn't a homomorphism in the category of vector spaces; it is rather a bilinear map. So you can think of it as a map $V \otimes V^* \to k$ instead. This is why you need a tensor structure to think of duals. This isn't enough, though, because we need to know that the pairing is nondegenerate. One way to express this is that there's a map $k \to V \times V^*$ mapping 1 to the "Casimir element" (which is the sum $\sum e_i \otimes e_i^{\vee}$ where $e_i$ ranges over a basis of $V$ and $e_i^{\vee}$ the dual basis; it is independent of the choice of $e_i$ as a quick computation shows). The Casimir morphism satisfies the condition that $V \to (k) \otimes V \to (V \otimes V^*) \otimes V \to V \otimes (V^* \otimes V)$ is just the identity. Conversely, this is enough to show that the pairing is nondegenerate. So, anyway, how does this make sense in a symmetric tensor category? Basically, $V$ is the object, $V^*$ the putative dual, and $k$ replaced by the unital object. This definition is entirely arrow-theoretic, and it all goes through as usual. It is an exercise to check that the dual is unique. Some examples: This coincides with the usual dual in the category of vector spaces This coincides with the dual sheaf if one is working in the category of locally free sheaves on a scheme This corresponds to the dual (contragredient) representation in the (tensor) category of representations of any Hopf algebra (so this includes representations of finite groups and Lie algebras) Oh, and what happens if you don't have a symmetric tensor category? Then you have to worry about "left" and "right" duals, respectively. For more about all this, I recommend the notes of Pavel Etingof on tensor categories. | 0.218035 | 99 |
science400 | Your trouble with determinants is pretty common. They’re a hard thing to teach well, too, for two main reasons that I can see: the formulas you learn for computing them are messy and complicated, and there’s no “natural” way to interpret the value of the determinant, the way it’s easy to interpret the derivatives you do in calculus at first as the slope of the tangent line. It’s hard to believe things like the invertibility condition you’ve stated when it’s not even clear what the numbers mean and where they come from. Rather than show that the many usual definitions are all the same by comparing them to each other, I’m going to state some general properties of the determinant that I claim are enough to specify uniquely what number you should get when you put in a given matrix. Then it’s not too bad to check that all of the definitions for determinant that you’ve seen satisfy those properties I’ll state. The first thing to think about if you want an “abstract” definition of the determinant to unify all those others is that it’s not an array of numbers with bars on the side. What we’re really looking for is a function that takes N vectors (the N columns of the matrix) and returns a number. Let’s assume we’re working with real numbers for now. Remember how those operations you mentioned change the value of the determinant? Switching two rows or columns changes the sign. Multiplying one row by a constant multiplies the whole determinant by that constant. The general fact that number two draws from: the determinant is linear in each row. That is, if you think of it as a function $\det: \mathbb{R}^{n^2} \rightarrow \mathbb{R}$, then $$ \det(a \vec v_1 +b \vec w_1 , \vec v_2 ,\ldots,\vec v_n ) = a \det(\vec v_1,\vec v_2,\ldots,\vec v_n) + b \det(\vec w_1, \vec v_2, \ldots,\vec v_n),$$ and the corresponding condition in each other slot. The determinant of the identity matrix $I$ is $1$. I claim that these facts are enough to define a unique function that takes in N vectors (each of length N) and returns a real number, the determinant of the matrix given by those vectors. I won’t prove that, but I’ll show you how it helps with some other interpretations of the determinant. In particular, there’s a nice geometric way to think of a determinant. Consider the unit cube in N dimensional space: the set of vectors of length N with coordinates 0 or 1 in each spot. The determinant of the linear transformation (matrix) T is the signed volume of the region gotten by applying T to the unit cube. (Don’t worry too much if you don’t know what the “signed” part means, for now). How does that follow from our abstract definition? Well, if you apply the identity to the unit cube, you get back the unit cube. And the volume of the unit cube is 1. If you stretch the cube by a constant factor in one direction only, the new volume is that constant. And if you stack two blocks together aligned on the same direction, their combined volume is the sum of their volumes: this all shows that the signed volume we have is linear in each coordinate when considered as a function of the input vectors. Finally, when you switch two of the vectors that define the unit cube, you flip the orientation. (Again, this is something to come back to later if you don’t know what that means). So there are ways to think about the determinant that aren’t symbol-pushing. If you’ve studied multivariable calculus, you could think about, with this geometric definition of determinant, why determinants (the Jacobian) pop up when we change coordinates doing integration. Hint: a derivative is a linear approximation of the associated function, and consider a “differential volume element” in your starting coordinate system. It’s not too much work to check that the area of the parallelogram formed by vectors $(a,b)$ and $(c,d)$ is $\Big|{}^{a\;b}_{c\;d}\Big|$ either: you might try that to get a sense for things. | 0.456271 | 0 |
science401 | These implications are reached by considering the three, different cases for the roots $\{ r_1, r_2, r_3 \}$ of the polynomial: repeated root, all distinct real roots, or two complex roots and one real root. When one of the roots is repeated, say $r_1$ and $r_2$, then it is clear that the discriminant is $0$ because the $r_1 - r_2$ term of the product is $0$. When one root is a complex number $\rho = x+ yi$, then by the complex conjugate root theorem, $\overline{\rho} = x - yi$ is also a root. By the same theorem, the remaining third root must be real. Evaluating the product in the discriminant for this case, $$ \begin{align*} (\rho - \overline{\rho})^2 (\rho - r_3)^2 (\overline{\rho} - r_3)^2 &= (2yi)^2 (x + yi - r^3)^2 (x - yi - r^3)^2 \\ &= -4y^2 [((x - r_3) + yi) ((x - r_3) - yi) ]^2 \\ &= -4y^2 ((x - r_3)^2 + y^2)^2 \end{align*} $$ which is less than or equal to $0$. Finally, when all roots are real, the product is clearly positive. Putting it all together, $\Delta$: less than $0$ implies that one root is complex; equal to $0$ implies that one root is repeated; greater than $0$ implies that all roots are distinct and real. | 0.279802 | 99 |
science403 | Suppose we shuffle a deck and get a permutation p. For each previous shuffling there is a 1-1/52! chance that p doesn't match it. Each previous shuffling is independent, in that regardless of what p and the other permutations are, the chance of p matching the shuffling is 1-1/52! When probabilities are independent we can simply multiple them to find the chance of all the events happening. In this case, the each event is actually a match not happening, so the chance of no matches given n previous shuffles is (1-1/52!)^n. We can then complete the calculations as Michael did. | 0.314351 | 99 |
science405 | The three triangle inequalities are \begin{align} x + y &> 1-x-y \\ x + (1-x-y) &> y \\ y + (1-x-y) &> x \\ \end{align} Your problem is that in picking the smaller number first from a uniform distribution, it's going to end up being bigger than it would if you had just picked two random numbers and taken the smaller one. (You'll end up with an average value of $1/2$ for the smaller instead of $1/3$ like you actually want.) Now when you pick $y$ on $[0, 1-x]$, you're making it smaller than it should be (ending up with average value of $1/4$). To understand this unequal distribution, we can substitute $y (1-x)$ for $y$ in the original inequalities and we'll see the proper distribution. (Note that the $y$-axis of the graph doesn't really go from $0$ to $1$; instead the top represents the line $y=1-x$. I'm showing it as a square because that's how the probabilities you were calculating were being generated.) Now the probability you're measuring is the area of the strangely-shaped region on the left, which is $$\int_0^{1/2}\frac1{2-2x}-\frac{2x-1}{2x-2}\,dx=\ln 2-\frac12\approx0.19314$$ I believe that's the answer you calculated. | 0.043418 | 99 |
science407 | The answer is, it's just a fact “cone over a simplex is a simplex” rewritten in terms of the generating function: observe that because n-simplex is a cone over (n-1)-simplex $\frac{\partial}{\partial x}vol(\text{n-simplex w. edge x}) = vol(\text{(n-1)-simplex w. edge x})$; in other words $e(x):=\sum_n vol\text{(n-simplex w. edge x)}$ satisfies an equvation $e'(x)=e(x)$. So $e(x)=Ce^x$ -- and C=1 because e(0)=1. | 0.483889 | 99 |
science408 | Knuth says to look at it as generating all nested parentheses in lexicographic order. Look here for the details http://www-cs-faculty.stanford.edu/~uno/fasc4a.ps. | 0.444221 | 99 |
science409 | Basic Theory The way to solve this problem is to calculate how much each payment reduces your debt after you have been repaying your loan for $n$ years. Let $r=1+R/100$, ie. this converts the interest rate from a percentage to a value you can multiply your debt by to calculate how much you owe after adding one time period's interest. If I make a payment of $P$ at the end of the $k$th year, then we avoid paying interest on this money $n-k$ times and so we reduce our debt by $Pr^{n-k}$. We sum up the future values of all our payments: $\sum\limits_{k=1}^n Pr^{n-k}$ If we reverse this, it is equivalent to: $\sum\limits_{k=0}^{n-1} Pr^k$ This is a geometric series, which can be solved using the formula $\frac{ar^{n-1}}{r-1}$ where $a$ is the first term, $r$ is the factor and $n$ is the number of terms being summed. We then attempt to equate this with the debt owed after $n$ years, which is $Mr^n$. We now compare the two equations: $\frac{Pr^{n-1}}{r-1} = Mr^n$ Calculating $n$ We group the $r^n$ terms: $\frac{P}{r-1} = r^n\frac{M-P}{r-1}$ $r^n = \frac{P}{M(r-1)-P}$ So we just take the $n$th log of the right hand side. Calculating repayments Given the principal ($M$) and the interest rate ($r$), what will my payment-per-term ($P$) be over $n$ accruation terms? $P=\frac{Mr^n(r-1)}{r^{n-1}}$ Payments made at the start of the year In this case, the future values of our interest payment simply become: $\sum\limits_{k=1}^n Pr^k$ We proceed as we did before. Notes We could also solve this problem using present value instead of future value. | 0.036323 | 99 |
science410 | Let g(n,k) = # combinations of cakes. Notice that: g(n,1) = 1. (all the cakes are the same) g(n,2) = n+1. (e.g. for 5 cakes, the # of cakes of type 1 can be 0, 1, 2, 3, 4, 5) g(1,k) = k. g(2,k) = k*(k-1)/2 + k (the first term is two different cakes; the second term is when both cakes are the same), as long as k > 1. (otherwise g(2,1) = 1) g(3,k) = k * (k-1) * (k-2)/6 + k*(k-1)/2 * 2 + k (the first term is 3 different cakes; the second term is 2 different cakes, with a *2 since there are two choices for which one to duplicate, the third term is when all 3 cakes are the same), as long as k > 2. If we think of k as a radix rather than the # of cakes, then this problem is equivalent to expressing the # of distinct n-digit numbers in base k whose digits are in sorted order. (e.g. 1122399 is equivalent to 9921231) I think I can express it as a nonrecursive sum: g(n,k) = sum from j=1 to max(n,k) of { (k choose j) * h(n,j) } where h(n,j) is the # of ways to partition N cakes using j different types. (the term in the sum is when there are j distinct cakes actually chosen.) But that's about as far as I can get... :/ edit: looks like it's combinations with repetitions = ((k+n-1) choose n). (same as the wikipedia article with n and k swapped) | 0.040683 | 99 |
science411 | Using a method that's often called "stars and bars": We draw $n$ stars in a row to represent the cakes, and $k-1$ bars to divide them up. All of the stars to the left of the first bar are cakes of the first type; stars between the first two bars are of the second type; . . . . **|***||*| Here's an example with $n=6$ and $k=5$. We're getting 2 of the first type, 3 of the second type, 0 of the third type, 1 of the fourth type, and 0 of the fifth type. In order to solve the problem, we just need to reorder the stars and bars by choosing the $k-1$ spots for the bars out of the $n+k-1$ total spots, so our answer is: $$ \binom{n+k-1}{k-1}. $$ | 0.332754 | 99 |
science414 | Of course there are infinite equation (even if you require them to be infinitely differentiable...) that satisfy the given constraints. As Isaac and Justin already wrote, you may always find a polynomial of degree at most n-1 (where n is the number of points given) which satisfies the given data; but you cannot be sure that this is the right answer. Moreover, if data is not exact but approximate the resulting polynomial may be quite different from the correct function, since it would likely hade huge peaks and falls. In such cases, an approximate method like least squares could be more useful. | 0.368875 | 99 |
science416 | Couldn't all transformation which send each point (x,y) to another point (x',y') which can be computed from the first one by performing only the four operations and extraction of square root? | 0.206132 | 99 |
science417 | Without the unit-length segment--that is, without something to compare the first segment to--its length is entirely arbitrary, so can't be valued, so there's no value of which to take the square root. Let the given segment (with length x) be AB and let point C be on ray AB such that BC = 1. Construct the midpoint M of segment AC, construct the circle with center M passing through A, construct the line perpendicular to AB through B, and let D be one of the intersections of that line with the circle centered at M (call the other intersection E). BD = sqrt(x). AC and DE are chords of the circle intersecting at B, so by the power of a point theorem, AB * BC = DB * BE, so x * 1 = x = DB * BE. Since DE is perpendicular to AC and AC is a diameter of the circle, AC bisects DE and DB = BE, so x = DB^2 or DB = sqrt(x). edit: this is a special case of the more general geometric-mean construction. Given two lengths AB and BC (arranged as above), the above construction produces the length BD = sqrt(AB * BC), which is the geometric mean of AB and BC. | 0.165066 | 99 |
science418 | If you have a segment $AB$, place the unit length segment on the line where $AB$ lies, starting with $A$ and in the direction opposite to $B$; let $C$ be the other point of the segment. Now draw a semicircle with diameter $BC$ and the perpendicular to $A$; this line crosses the semicircle in a point $D$. Now $AD$ is the square root of $AB$. $\triangle BCD$ is a right triangle, like $\triangle ACD$ and $\triangle ABD$; all of these are similar, so you find out that $AC/AD = AD/AB$. But $AC=1$, so $AD = \sqrt{AB}$. See the drawing below: | 0.361817 | 99 |
science420 | Parabolas of the form you described (y = ...) are symmetric over a vertical line through their vertex. Let's call that line x = k. This means that if the graph crosses the x-axis (meaning that $ax^2+bx+c=0$ has real solution(s)), they must be equidistant from x = k, so (k,0) must be the midpoint of the segment with endpoints at the zeros of the quadratic or k is the average of the zeros. From the quadratic formula, the two zeros of the quadratic are $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, so their sum is $-\frac{b}{a}$ and their average is $k=-\frac{b}{2a}$. So, the x-coordinate of the vertex must be $-\frac{b}{2a}$. If the parabola does not cross the x-axis (no real solutions), there is another parabola with equation $y=ax^2+bx+c'$ for some $c'$ for which the graph is a vertical translation of the graph of the original quadratic, but crosses the x-axis. Its axis of symmetry is $x=-\frac{b}{2a}$ and because it is a vertical translation of the original, the axis of symmetry of the original is also $x=-\frac{b}{2a}$, so the vertices of both have x-coordinate $-\frac{b}{2a}$. | 0.509402 | 99 |
science421 | If you use the formula you describe for squares, your measurements are coming from one corner of the square. Imagine those measurements growing slowly. The square will grow, but only along the two sides opposite the corner from which you measured, so the derivative of the area formula is only the perimeter on those two sides. Alternately, consider measuring the size of a square by the distance d from its center to the midpoint of a side. This would make the side length 2d, the perimeter 8d and the area 4d^2. Now, the derivative of the area is the perimeter. (Also, if you imagine growing the square slowly with this measurement, it grows from the center outward, growing on all four sides.) | 0.296902 | 99 |
science424 | Complementary to Mau's answer: Call a series $a_n$ absolutely convergent if $\sum|a_n|$ converges. If $a_n$ converges but is not absolutely convergent we call $a_n$ conditionally convergent The Riemann series theorem states that any conditionally convergent series can be reordered to converge to any real number. Morally this is because both the positive and negative parts of your series diverge but the divergences cancel each other out, one or other's canceling the other can be staggered by adding on, say, the negative bits every third term in stead of every other term. This means that in the race for the two divergences to cancel each other out, we give the positive bit something of a head-start and will get a larger positive outcome. Notice how, even in this rearranged version of the series, every term will still come up exactly once. It is also worth noting, on the Wikipedia link Mau provided, that the convergence to $\ln 2$ of your series is at the edge of the radius of convergence for the series expansion of $\ln(1-x)$- this is a fairly typical occurrence: at the boundary of a domain of convergence of a Taylor series, the series is only just converging- which is why you see this conditional convergence type behavior. | 0.322974 | 0 |
science427 | Let's say you have a sequence of nonnegative numbers $a_1 \geq a_2 \geq \dots$ tending to zero. Then it is a theorem that the alternating sum $\sum (-1)^i a_i$ converges (not necessarily absolutely, of course). This in particular applies to your series. Incidentally, if you're curious why it converges to $\log(2)$ (which seems somewhat random), it's because of the Taylor series of $\log(1+x)$ while letting $x \to 1$. | 0.40866 | 0 |
science429 | I had an answer before, but I looked into it a bit more and my answer was incorrect so I removed it. This link may be of interest: Circle Packing in a Square (wikipedia) It was suggested by KennyTM that there may not be an optimal solution yet to this problem in general. Further digging into this has shown me that this is probably correct. Check out this page: Circle Packing - Best Known Packings. As you can see, solutions up to only 30 circles have been found and proven optimal. (Other higher numbers of circles have been proven optimal, but 31 hasn't) Note that although problem defined on the wikipedia page and the other link is superficially different than the question asked here, the same fundamental question is being asked, which is "what is the most efficient way to pack circles in a square/rectangle container?". ...And it seems the answer is "we don't really know" :) | 0.405653 | 99 |
science430 | A problem is in class NP if its solution may be verified in polynomial time, that is if the dimension of the problem is n you may be sure that for large enough n you need less than r·nk operations to verify the solution. A problem is in class P if its solution may be found in polynomial time, instead. A problem in P is in NP by definition, but the converse may not be the case; probably the most important open question in computer science is whether classes P and NP are the same, that is P=NP. NP-complete is a family of NP problems for which you know that if one of them had a polynomial solution then everyone of them has. (EDITED) For the time being, only known algorithms for NP-complete problems are exponential in number of operations, so they are not practically solvable for n large. | 0.257348 | 99 |
science431 | To expand on Mau's answer, you should care about NP-complete problems because there is an entire family of them that spans a large number of seemingly basic algorithms across a wide range of disciplines. These aren't obscure problems, but extremely important and highly practical questions. For examples, consider the following: Travelling salesman problem - finding the shortest path (on a graph) that allows you to visit every city exactly once. Bin packing problem - there are a number of fixed (integer) size bins and objects of varying sizes. Minimise the number of bins required to hold all of the objects Knapsack problem - given objects of various sizes and values and a knapsack with a fixed integer size, choose the objects that can fit inside with the most value Minimal Vertex cover - finding the smallest set of vertices such that every edge contains at least one chosen vertice Clique - finding that largest group of people who all know each other Subgraph isomorphism - does one graph contain a subgraph isomorphic to another? Set packing - given a number of sets, what is the maximum number of disjoint sets that can be selected? This is related to set cover, where we are trying to choose sets so that every element is within at least one set Subset sum - Given a set of integers, does some subset sum to 0? Although many of these problems may seem abstract, many more complicated problems can't be efficiently solved with current techniques, as they are equivalent to one of these. The problem of NP completeness has received a huge amount of attention. Once you've reduced a problem to NP-complete, you know to give up on an efficient fast algorithm and to start looking at approximations. | 0.082653 | 99 |
science432 | Any problem for which a solution (once found) can be quickly verified as a solution is said to be "in NP" (Here, "quickly" means in polynomial-time). Any problem for which a solution can be found quickly is said to be "in P." P is a subset of NP - that is, any problem for which a solution can be quickly found can also be quickly verified. A problem is NP-complete if it is the hardest problem in NP. Surprisingly, there are many NP-complete problems, which are all equivalent - here, equivalent means that a quick (polynomial-time) solution to any one of them would give you a quick solution to all the rest. Also somewhat surprisingly, a quick solution to any NP-complete problem would also give you a quick solution to any problem in NP. So is there a quick (polynomial-time) algorithm to solve NP-complete problems? That is the P=NP problem, one of the greatest unsolved problems of our time. However, most sane mathemeticians believe (and hope!) that P≠NP, because proving math-theorems is NP-Complete; so if P=NP, we'd be out of a job! :) | 0.26361 | 99 |
science433 | Another option would be to use eigendecomposition . It allows you to raise the eigenvalues in the diagonal of the decomposition A = VDV^-1 to a power. It changes the problem from matrix multiplication to the multiplication of the eigenvalues. Once in eigendecomposition form you could perform the same addition-chain exponentiation technique but it would be with scalars, not matrices. Much more efficient, because each matrix multiplication has n^3 multiplies, but with ed you would only have n. More is explained here: http://en.wikipedia.org/wiki/Matrix_decomposition#Eigendecomposition | 0.27148 | 99 |
science434 | Given that the reals are uncountable (which can be shown via Cantor diagonalization) and the rationals are countable, the irrationals are the reals with the rationals removed, which is uncountable. (Or, since the reals are the union of the rationals and the irrationals, if the irrationals were countable, the reals would be the union of two countable sets and would have to be countable, so the irrationals must be uncountable.) | 0.398639 | 0 |
science435 | If you were to try to measure instantaneous speed as you described, you would in fact have traveled 0 miles in 0 time and 0/0 is undefined If, however, you look at your average speed over smaller and smaller periods of time around the instant you care about--that is, (distance traveled from $t=t_0$ to $t=t_0+\epsilon$ for various small values of ε--and these average speeds "converge" (they all get closer to a single value as ε gets closer to 0), then we say that the instantaneous speed is that single value upon which the average speeds around that point converge. This is, of course, a somewhat informal explanation; to be more precise requires getting into differential calculus. | 0.184886 | 99 |
science437 | I think there is a very clear meaning in the physical world: If, at some moment, you were going 40 mph, if you were to stop de/accelerating and just hold that velocity, you would cover 40 miles in 1 hour. | 0.10288 | 99 |
science439 | A state-variable kind of approach: (represents the physical states of a system) Newton's first law of motion says that without external forces, masses will move at constant velocity. Every change in velocity (for a car, whether a change in speed due to braking, or a change in direction due to friction of the tires and the steering mechanism) is due to an external force. So the "state" of the car at any instant in time is its velocity; distance traveled can be derived from velocity: While you are going 60mi/hr = 26.8m/sec (see conversion factors) in a particular direction at constant velocity, you can multiply a change in time times velocity and get distance: 26.8m/sec = 26.8 meters in one second. Or 26.8mm in one millisecond. Or 26.8 micrometers in one microsecond. The process of going from velocity to (directed) distance is called integration, and for those two variables (velocity and directed distance) is just the summing up of distance increments (= velocity * time increments) over infinitesimally short periods of time. For smoothly varying velocities, coarse time increments are a good approximation to reality -- differential equation solvers use this or similar approaches (e.g. Simpson's Rule) to perform integration. But instantaneous velocity...you aren't measuring average speed for a given amount of time Well, instantaneous velocity is just the limit of average velocity as the time period becomes infinitely small. | 0.441697 | 99 |
science440 | BlueRaja's answer is certainly more complete than this one (and gives good references), but here's a rough overview of linear programming. Suppose that you have a linear function (in high school courses, it's typically a function of two variables) that you want to optimize on a convex "feasible" region bounded by linear equations (again, in high school, the bounds are typically described using linear inequalities in two variables). Because the function to optimize is linear, the set of points for which the function has a particular value, say c, is a line (and all such lines are parallel) and the value of the function anywhere to one side of the line is greater than c and anywhere to the other side of the line is less than c. So, you can think about moving through this set of constant-value lines to increase/decrease the value of the target function as appropriate to optimize it. Since the feasible region is bounded by linear equations, as the constant-value line moves through and out of the feasible region, it last touches the feasible region at a vertex (or possibly at all points on an edge connecting two vertices), so the optimal solution must occur at a vertex of the feasible region. Given all that, linear programming comes down to evaluating the target function at all the vertices of the feasible region to find the optimal value. | 0.423518 | 99 |
science441 | With two stamps, you can do it in linear pseudo-linear time - O(totalCost/costOfLargerStamp) - by simply enumerating every possibility (there is only one possible count of the smaller stamp for each count of the larger stamp). In general, however, solving this is equivalent to solving a general integer linear programming problem written in standard form, which is NP-complete. | 0.371992 | 99 |
science443 | The Dirac delta "function." It's not a "function," strictly speaking, but rather a very simple example of a distribution that isn't a function. | 0.280477 | 99 |
science444 | This is similar to other answers. Imagine another car beside yours. That car is covering a distance of 40 miles over the next hour at a constant velocity. At the point at which you keep pace with that car (relative velocity = 0), you are traveling at 40 mph. | 0.020576 | 99 |
science447 | Yes it can. In fact, as Jamie Banks noted, a determinant is an intuitive way of thinking about volumes. To summarise the argument, if we consider the vectors as a matrix, switching two rows, multiplying one by a constant or adding a linear combination will have the same effect on the volume as on the determinate. We can use these operations to transform any n-parallelotope to cube and note that the determinate matches the signed volume here, so it will match it everywhere as well. | 0.276902 | 99 |
science448 | It is important to understand that NP-completeness is defined in terms of input size, not in terms of either cost to be paid, C, or the number of types of stamps, S, when we are dealing with integers rather than real numbers. This algorithm can actually be solved in time CS. We take the first stamp value an mark all multiple of it as obtainable (up to the least number greater than the cost). Then, for each subsequent stamp value, we mark as obtainable all numbers that can be obtained by adding a multiple of its value with an obtainable number (up to the least number greater than the cost). We can keep track of the lowest cost postage amount that we have found so far. | 0.523488 | 99 |
science449 | The Weierstrass function is continuous everywhere and differentiable nowhere. The Dirichlet function (the indicator function for the rationals) is continuous nowhere. A modification of the Dirichlet function is continuous at all irrational values and discontinuous at rational values. The Devil's Staircase is uniformly continuous but not absolutely. It increases from 0 to 1, but the derivative is 0 almost everywhere. | 0.109088 | 99 |
science450 | I would recommend against getting a degree from any online-only university. Even if you happen to find one that's not shady, everyone else who hasn't heard of it will assume it is some kind of diploma mill without bothering to do much research. Instead, I think you'd be better off going to a nearby university you're interested in, and ask them if they would be willing to make some kind of special arrangement for you. Many universities allow reduced course loads for students that have families or work full time, and aren't very good about advertising it. | 0.093427 | 99 |
science452 | $$\begin{align*} \sum F &= ma\\ \frac{dv}{dt} &= a\\ &= \frac{\sum F}{m}\\ \sum F &= mg - kv\\ \frac{dv}{dt} &= g - \frac{k}{m} v\end{align*}$$ This is a differential equation with a solution of $$\begin{align*} v &= A + B \space exp\left(\frac{-k}{m} t\right)\\ \frac{dv}{dt} &= - B \cdot \frac{k}{m} \cdot \exp\left(\frac{-k}{m} t\right) \end{align*}$$ Match terms and initial conditions ($v = 0$ at time $t = 0$) and you get $$\begin{align*} &g - \frac{k}{m}A - \frac{k}{m} B\cdot \exp\left(\frac{-k}{m} t\right)\\ \implies&-B \frac{k}{m} \cdot \exp\left(\frac{-k}{m} t\right)\\ \implies& A = g \cdot \frac{m}{k} \space \text{and} \space B = -g \cdot \frac{m}{k}\\ \implies& v = \frac{mg}{k} \cdot \left(1 - \exp\left(\frac{-k}{m} t\right)\right)\end{align*}$$ That's a linear differential equation ($\frac{dv}{dt}$ is a linear function of $v$); the $\sum F = -kv^2$ is a nonlinear differential equation (can't remember off the top of my head how to deal with that one; it may not have a closed form solution). It's a bit difficult to summarize the techniques in general, but any good book on differential equations would cover them. | 0.410266 | 99 |
science455 | $dm$ takes density fluctuations into account, it's just $dm = \rho(\vec r) d^3 \mathbf r$. E.g. for a homogeneous cylinder with the rotational axis align parallel to the z-axis it is $\rho = \frac{m}{V}\cdot\theta(R-r)\theta(a^2-z^2)$, where $\theta(x) = \begin{cases} 0 \;\text{ for } x<0 \text{ and}\\ 1 \;\text{ for } x>0 \end{cases}$ and $r = \sqrt{x^2+y^2}$ is the radial coordinate in cylindrical coordinates. The result is that your integration over $r$ is from $0$ to $R$ and for $z$ from $-a$ to $a$. $V=\pi R^2 \cdot 2a$ is the volume of the cylinder, $m$ its weight. In cylindrical coordinates, $d^3\mathbf r = r\, dr \,d\phi\, dz$, so you got $$I = \int_{r=0}^{R} \int_{z=-a}^a \int_{\phi=0}^{2\pi} r^2 \frac{m}{V} \cdot r\, d\phi\, dz\, dr = 4a\pi \frac{m}{V} \int_{r=0}^R r^3 dr = \frac{am\pi}{V}R^4 = \frac{1}{2}mR^2.$$ | 0.245131 | 99 |
science456 | First, some background. A "line" (the object containing the path of least distance between two points) on a sphere is a great circle; a great circle is the intersection of a sphere with a plane passing through the center of the sphere. A spherical triangle is the triangle formed by the spherical line segments connecting three non-collinear points on a sphere. Below is the Spherical Law of Cosines as it appears in UCSMP Functions, Statistics, and Trigonometry, 3rd ed., copied here because the diagram is good and helps with clarity. If ABC is a spherical triangle with arcs a, b, and c (meaning the measures of the arcs, not the lengths), then cos c = cos a * cos b + sin a * sin b * cos C. Now, to the specific problem at hand. Let's use the diagram below, also from UCSMP Functions, Statistics, and Trigonometry, 3rd ed., for reference. Let B be the initial point and A be the final point of the movement (and for simplicity, I'll assume they are both in the northern hemisphere, and leave extending the solution to any points as an exercise); N and S are the north and south poles, respectively; C and D are the points on the equator that are on the same line of longitude as A and B, respectively. Consider spherical triangle ABN. a = (90° - latitude of point A); b = (90° - latitude of point B). N = positive difference in longitude between points A and B. Use the Spherical Law of Cosines (cos n = ... form) to determine n, then use the Spherical Law of Cosines (cos b = ... form) to determine B. If the points are arranged as pictured, the heading of travel from B to A is 360° - B; for other arrangements of the points, the exact method of determining the heading based on the measure of angle B may vary. (graphics from Lesson 5-10 of UCSMP Functions, Statistics, and Trigonometry, 3rd ed., © 2010 Wright Group/McGraw Hill) | 0.01316 | 99 |
science457 | This is a very interesting question. I don't have a complete solution yet, but I did get some results. Consider an arbitrary distribution of points. Let $P(i,b)=i+\frac{b}{2}-1$ where $i$ is the number of internal points and $P$ is the number of boundary points. Let $P(A)=P(i_A,b_A)$, where $A$ is a simple polygon with all its vertices on the points. Wikipedia shows that $P(C)=P(A)+P(T)$ where $T$ is a triangle that shares a single edge with $A$ and $C$ a simple polygon formed by the union of $A$ and $T$. Since all simple polygons can be triangulated, $P(C)=sum P(t)$ for all $t$ in the triangulation of $C$. The proof then has a second part that shows $P(t)$ equals the area for any triangle. So if we want to generalize it for other grids, we have to find a property equal to $P(T)$ for any triangle. | 0.024148 | 99 |
science460 | If W is randomly chosen with the PDF P(x), then the expectation value should be $E[e^{-\gamma W}]=\int_{-\infty}^\infty P(x) e^{-\gamma x} dx$ http://mathcache.appspot.com/?tex=%5cpng%5c%5bE%5Be%5E%7B-%5Cgamma%20W%7D%5D%3D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20P%28x%29%20e%5E%7B-%5Cgamma%20x%7D%20dx%5c%5d And I think that equation (E[e-γW] = e-γ(E[W] - ½γVar[W])) is correct only when W is a normal distribution. | 0.46733 | 99 |
science461 | Nonstandard proof Consider the semi-circle with endpoints A and C and center O and the inscribed angle ∠ABC (B on the semi-circle) together with the rotation image of both about O by 180°. The image of A is C and vice versa; let B' be the image of B. The image of a line under a 180° rotation is parallel to the original line, AB is parallel to CB' and BC is parallel to B'A, so ABCB' is a parallelogram. BO and its image must be parallel, but the image of O is itself, since it is the center of rotation, and if BO and B'O are parallel and contain a point in common, they must lie on the same line, so BB' passes through O. AC and BB' (the diagonals of ABCB') are both diameters of the circle, so they are congruent. A parallelogram with congruent diagonals is a rectangle. Thus, ∠ABC is a right angle (and has measure 90°). diagram http://www.imgftw.net/img/762828246.png Standard proof (or, at least, my guess at it based on the description in the question) As above, consider the semi-circle with endpoints A and C and center O and the inscribed angle ∠ABC (B on the semi-circle). Draw in radius OB. OA = OB, so △AOB is isosceles and ∠OAB≅∠OBA. OB = OC, so △BOC is isosceles and ∠OBC≅∠OCB. Let α=m∠OAB=m∠OBA and β=m∠OBC=m∠OCB. In △ABC, the measures of the angles are α, α+β, and β, so α+(α+β)+β=180° or 2(α+β)=180° or α+β=90°, so ∠ABC has measure 90° and is a right angle. diagram http://www.imgftw.net/img/319527897.png edit: Another Nonstandard proof Use the labeling as above and apply Stewart's Theorem to △ABC: $$(AB)^2(OC) + (BC)^2(AO) = (AC)((BO)^2 + (AO)(OC))$$ Substituting the length r of the radius of the semicircle as appropriate: $$(AB)^2r + (BC)^2r = 2r(r^2 + r^2)=4r^3$$ Dividing both sides by r: $$(AB)^2+(BC)^2=(2r)^2=(AC)^2$$ So, by the converse of the Pythagorean Theorem, ∠ABC is a right angle. | 0.375109 | 99 |
science462 | Look at this answer on MathOverflow: Yes, there is a way to guess a number asking 14 questions in worst case. To do it you need a linear code with length 14, dimension 10 and distance at least 3. One such code can be built based on Hamming code (see http://en.wikipedia.org/wiki/Hamming_code). Here is the strategy. Let us denote bits of first player's number as ai, i ∈ [1..10]. We start with asking values of all those bits. That is we ask the following questions: "is it true that i-th bit of your number is zero?" Let us denote answers on those questions as bi, i ∈ [1..10]. Now we ask 4 additional questions: Is it true that a1 ⊗ a2 ⊗ a4 ⊗ a5 ⊗ a7 ⊗ a9 is equal to zero? (⊗ is sumation modulo 2). Is it true that a1 ⊗ a3 ⊗ a4 ⊗ a6 ⊗ a7 ⊗ a10 is equal to zero? Is it true that a2 ⊗ a3 ⊗ a4 ⊗ a8 ⊗ a9 ⊗ a10 is equal to zero? Is it true that a5 ⊗ a6 ⊗ a7 ⊗ a8 ⊗ a9 ⊗ a10 is equal to zero? Let q1, q2, q3 and q4 be answers on those additional questions. Now second player calculates ti (i ∈ [1..4]) --- answers on those questions based on bits bj which he previously got from first player. Now there are 16 ways how bits qi can differ from ti. Let di = qi ⊗ ti (hence di = 1 iff qi ≠ ti ). Let us make table of all possible errors and corresponding values of di: position of error -> (d1, d2, d3, d4) no error -> (0, 0, 0, 0) error in b1 -> (1, 1, 0, 0) error in b2 -> (1, 0, 1, 0) error in b3 -> (0, 1, 1, 0) error in b4 -> (1, 1, 1, 0) error in b5 -> (1, 0, 0, 1) error in b6 -> (0, 1, 0, 1) error in b7 -> (1, 1, 0, 1) error in b8 -> (0, 0, 1, 1) error in b9 -> (1, 0, 1, 1) error in b10 -> (0, 1, 1, 1) error in q1 -> (1, 0, 0, 0) error in q2 -> (0, 1, 0, 0) error in q3 -> (0, 0, 1, 0) error in q4 -> (0, 0, 0, 1) All the values of (d1, d2, d3, d4) are different. Hence we can find where were an error and hence find all ai. Answered by falagar. | 0.54286 | 99 |
science465 | For the case of Bendford's Law, of course scale invariance is a necessary condition; it the law must be true either if we measure things in meters or in feet or in furlongs, thus multiplying given data for a constant, the only distribution which allows this is the logarithmic one. But its being necessary does not mean that it is the answer, of course. Scale invariance is not relevant for Zipf's Law, however, since we have an absolute rank. | 0.23298 | 99 |
science467 | There is, in fact, a general formula for solving quartic (4th degree polynomial) equations. As the cubic formula is significantly more complex than the quadratic formula, the quartic formula is significantly more complex than the cubic formula. Wikipedia's article on quartic functions has a lengthy process by which to get the solutions, but does not give an explicit formula. Beware that in the cubic and quartic formulas, depending on how the formula is expressed, the correctness of the answers likely depends on a particular choice of definition of principal roots for nonreal complex numbers and there are two different ways to define such a principal root. There cannot be explicit algebraic formulas for the general solutions to higher-degree polynomials, but proving this requires mathematics beyond precalculus (it is typically proved with Galois Theory now, though it was originally proved with other methods). This fact is known as the Abel-Ruffini theorem. Also of note, Wolfram sells a poster that discusses the solvability of polynomial equations, focusing particularly on techniques to solve a quintic (5th degree polynomial) equation. This poster gives explicit formulas for the solutions to quadratic, cubic, and quartic equations. edit: I believe that the formula given below gives the correct solutions for x to $ax^4+bx^3+cx^2+d+e=0$ for all complex a, b, c, d, and e, under the assumption that $w=\sqrt{z}$ is the complex number such that $w^2=z$ and $\arg(w)\in(-\frac{\pi}{2},\frac{\pi}{2}]$ and $w=\sqrt[3]{z}$ is the complex number such that $w^3=z$ and $\arg(w)\in(-\frac{\pi}{3},\frac{\pi}{3}]$ (these are typically how computer algebra systems and calculators define the principal roots). Some intermediate parameters $p_k$ are defined to keep the formula simple and to help in keeping the choices of roots consistent. Let: \begin{align*} p_1&=2c^3-9bcd+27ad^2+27b^2e-72ace \\\\ p_2&=p_1+\sqrt{-4(c^2-3bd+12ae)^3+p_1^2} \\\\ p_3&=\frac{c^2-3bd+12ae}{3a\sqrt[3]{\frac{p_2}{2}}}+\frac{\sqrt[3]{\frac{p_2}{2}}}{3a} \end{align*} $\quad\quad\quad\quad$ \begin{align*} p_4&=\sqrt{\frac{b^2}{4a^2}-\frac{2c}{3a}+p_3} \\\\ p_5&=\frac{b^2}{2a^2}-\frac{4c}{3a}-p_3 \\\\ p_6&=\frac{-\frac{b^3}{a^3}+\frac{4bc}{a^2}-\frac{8d}{a}}{4p_4} \end{align*} Then: $$\begin{align} x&=-\frac{b}{4a}-\frac{p_4}{2}-\frac{\sqrt{p_5-p_6}}{2} \\\\ \mathrm{or\ }x&=-\frac{b}{4a}-\frac{p_4}{2}+\frac{\sqrt{p_5-p_6}}{2} \\\\ \mathrm{or\ }x&=-\frac{b}{4a}+\frac{p_4}{2}-\frac{\sqrt{p_5+p_6}}{2} \\\\ \mathrm{or\ }x&=-\frac{b}{4a}+\frac{p_4}{2}+\frac{\sqrt{p_5+p_6}}{2} \end{align}$$ (These came from having Mathematica explicitly solve the quartic, then seeing what common bits could be pulled from the horrifically-messy formula into parameters to make it readable/useable.) | 0.134613 | 1 |
science468 | Yes, there is a quartic formula. There is no such solution by radicals for higher degrees. This is a result of Galois theory, and follows from the fact that the symmetric group $S_5$ is not solvable. It is called Abel's theorem. In fact, there are specific fifth-degree polynomials whose roots cannot be obtained by using radicals from $\mathbb{Q}$. | 0.16556 | 1 |
science470 | When you try to solve a degree $n$ equation, there are $n$ roots you have to find (in principle) and none of them is favoured over any of the others, which (in some metaphorical sense) means that you have to break an $n$-fold symmetry in order to write down the roots. Now the symmetry group of the n roots becomes more and more complicated the larger $n$ is. For $n = 2$, it is abelian (and very small!); for $n = 3$ and $4$ it is still solvable (in the technical sense of group theory), which explains the existence of an explicit formula involving radicals (this is due to Galois, and is a part of so-called Galois theory); for $n = 5$ or more this group is non-solvable (in the technical sense of group theory), and this corresponds to the fact that there is no explicit formula involving radicals. Summary: The complexity of the symmetry group of the $n$ roots leads to a corresponding complexity in explicitly solving the equation. | 0.238783 | 99 |
science472 | Differentiation and integration is precisely why it is considered natural, but not just because $$\displaystyle\int \frac{1}{x} dx=\ln x$$ $e^x$ has the two following nice properties $$ \frac{d}{dx} e^x=e^x $$ $$ \int e^x dx=e^x+c $$ If we looked at $a^x$ instead, we would get: $$\frac {d} {dx} a^x= \frac{d}{dx} e^{x\ln(a)}=\ln(a) \cdot a^x$$ $$\int a^x dx= \int e^{x\ln(a)} dx=\frac{a^x}{\ln(a)}+c$$ So $e$ is vital to the integration and differentiation of exponentials. | 0.349519 | 0 |
science473 | The Wikipedia article about e lists many properties of the constant that make it naturally occurring. I think the biggest reason it is natural when it comes to exponentiation/logarithms is that it is the only number that satisfies $$ \frac{d}{dt} e^t =e^t $$ while every other number satisfies $$ \frac{d}{dt} a^t = c \cdot a^t$$ where $c$ is some constant, different than 1. This makes it "normalized" in a sense. | 0.277424 | 0 |
science475 | The square of the discriminant is symmetric in the roots of the polynomial; if you permute the roots, it stays the same. By the fundamental theorem of symmetric functions, that means it can be expressed as a polynomial, with integer coefficients, in the coefficients of the original polynomial. (I am using both "coefficients" and "polynomial" in two senses here, so let me know if this doesn't make sense.) This is a basic observation which will become important if you ever study Galois theory. An example will probably make this clearer. Suppose I have a quadratic polynomial $x^2 + bx + c$ with two roots $r_1, r_2$. Then $x^2 + bx + c = (x - r_1)(x - r_2)$, so $b = - r_1 - r_2$ and $c = r_1 r_2$. These are the elementary symmetric functions in two variables, and the theorem above implies that every polynomial function of $r_1$ and $r_2$ which is invariant under switching the two is actually a polynomial in $b$ and $c$. For example, the discriminant is $$(r_1 - r_2)^2 = r_1^2 - 2r_1 r_2 + r_2^2 = (r_1 + r_2)^2 - 4r_1 r_2 = b^2 - 4c.$$ | 0.040616 | 99 |
science477 | The proof that uses the fact that shearing a parallelogram parallel to one of its sides preserves area is my favorite. Here's an animation from this site: | 0.0387 | 2 |
science479 | There is a combinatorial proof. If $A(x) = \sum_{n \ge 0} a_n \frac{x^n}{n!}$ is an exponential generating function for the number of ways $a_n$ to put a certain structure on a set of size $n$, where $a_0 = 0$, then $\exp A(x) = \sum_{n \ge 0} b_n \frac{x^n}{n!}$ has coefficients $b_n$ which count the number of ways to split a set of size $n$ into subsets, each of which is given an $A$-structure. For example, $A(x) = e^x - 1$ is the structure of "being a nonempty set," and $\exp A(x)$ counts the number of ways to partition a set into nonempty subsets. Now $\log \frac{1}{1 - x} = \sum_{n \ge 1} \frac{x^n}{n}$ can be thought of as the structure of "being a cycle," which is to say there are $(n-1)!$ ways to arrange $n$ objects into a cycle $s_1 \to s_2 \to ... \to s_n \to s_1$. So $\exp \log \frac{1}{1 - x}$ counts the number of ways to split $n$ objects up into disjoint cycles, which is the same as the number of permutations of $n$ by cycle decomposition. So $$\exp \log \frac{1}{1 - x} = \frac{1}{1 - x}.$$ A generalization of this argument lets you compute the cycle index polynomials of the symmetric groups. Note that all of the above manipulations take place in the ring of formal power series over $\mathbb{Z}$. | 0.23009 | 99 |
science480 | A nice application is Wedderburn's theorem: every finite skewfield is necessarily commutative. Here a skewfield is something which satisfies the same axioms as field, except that multiplication is not required to be commutative; the typical example are quaternions. To see this, let $F$ be a finite skewfield, $Z$ be its center. It is easy to see that $Z$ is a field, hence it must be $\mathbb{F}_q$ for some prime power $q = p^k$. $F$ will then be a vector space of finite dimension $n$ over $\mathbb{F}_q$, hence it will have $q^n$ elements. Now write the class equation for the multiplicative group of $F$: $$q^n - 1 = q - 1 + \sum_i \frac{q^n - 1}{q^{t_i} - 1}$$ Here $q - 1$ appears as the cardinality of the center, while the sum extends over a set of representatives of the non-trivial conjugacy classes. Note that for $q^{t_i} - 1$ to divide $q^n - 1$, $t_i$ must divide $n$. Indeed the order of $q$ modulo $q^{t_i} - 1$ is $t_i$, and $q^n = 1 \pmod{q^{t_i} - 1}$. Now let $f_n$ be the $n$-th cyclotomic polynomial. Then $f_n(q)$ divides $q^n - 1$ and it also divides each term in the sum, so $f_n(q)$ divides $q - 1$. But this is impossible unless $n = 1$ and the sum is empty, in which case $F$ is commutative. Indeed $f_n(q)$ is a product of terms of the form $q - \omega$, where $\omega$ is a root of unity, and this product will have bigger absolute value than $q - 1$ as soon as $n > 1$. | 0.089832 | 99 |
science484 | This is not really a proof, in the sense that it depends on certain physical assumptions, but it forces you to think very hard about what those physical assumptions are. I learned it from Mark Levi's The Mathematical Mechanic, although unfortunately I don't have a diagram. Consider a fish tank in the shape of a triangular prism, where the triangle on top has vertices $A, B, C$ ($C$ is the right angle) and the triangle on the bottom has opposite vertices $A', B', C'$. Let $a, b, c$ denote the lengths of the opposite sides. Run a rod through the edge $A A'$ on which the fish tank is allowed to pivot, and fill it with water. By conservation of energy, this system is in mechanical equilibrium. On the other hand, I claim (and it is not hard to see) that the total torque about the rod is proportional to $c^2 - b^2 - a^2$, each contribution coming from the pressure of the water in the tank against the corresponding side. | 0.051901 | 2 |
science485 | As Akhil mentions, the keyword is elliptic regularity. Since I don't know anything about this, let me just say some low-level things and maybe they'll make sense to you. A differentiable function $f : \mathbb{R} \to \mathbb{R}$ can be thought of as a function which behaves locally like a linear function $f(x) = ax + b$. So, very roughly, it is a collection of tiny vectors which fit together. These tiny vectors can, however, fit together in a very erratic manner. That's because since you only have to fit one vector to the two vectors that are its neighbors, there is a lot of room for bad behavior. A differentiable function $f : \mathbb{C} \to \mathbb{C}$ has to satisfy a much more stringent requirement: locally, it has to behave like a linear function $f(z) = az + b$ where $z, a, b$ are complex, which is a rotation (and scale, and translation). So, very roughly, it is a collection of tiny rotations which fit together. Now one rotation has a continuum of neighbors to worry about, and it becomes much harder for erratic behavior to persist. | 0.294307 | 2 |
science486 | Another definition of discriminant is as the resultant of the polynomial with it's first derivative (up to a scalar), and the resultant of two polynomials vanishes if and only if they have a common root. So when does a polynomial have a common root with it's derivative? That's when a zero is also a local max or min, precisely a multiple root. | 0.451088 | 99 |
science487 | Regarding the inability to solve the quintic, this is sort-of true and sort-of false. No, there is no general solution in terms of $+$, $-$, $\times$ and $\div$, along with $\sqrt[n]{}$. However, if you allow special theta values (a new operation, not among the standard ones!) then yes, you can actually write down the solutions of arbitrary polynomials this way. Also, you can do construct lengths equal to the solutions by intersecting lower degree curves (for a quintic, you can do so with a trident and a circle.) | 0.287103 | 1 |
science489 | Maybe this is helpful, a formula for the volume of Lp balls in R^n. | 0.131115 | 99 |
science493 | Answer edited, in response to the comment and a second wind for explaining mathematics: Let $F(x_0,x_1,x_2)=0$ be the equation for your curve, and take $(y_0,y_1,y_2)$ to be coordinates on $(\mathbb{P}^2)^*$. Also, assume that $F$ is irreducible and has no linear factors. Then $y_0 x_0+y_1 x_1+y_2 x_2=0$ is the equation of a general line in $\mathbb{P}^2$ (recall, here $y_0,y_1,y_2$ are fixed, and the $x_i$ are the coordinates on the plane) and we look at the open set of $(\mathbb{P}^2)^*$ where $y_2\neq 0$. On this open set, we can solve the equation of the line for $x_2$, and look at $g(x_0,x_1)=y_2^n F(x_0,x_1,-\frac{1}{y_2}(y_0x_0+y_1x_1))$, a homogeneous polynomial of degree $n$ in $x_0,x_1$ with coefficients homogeneous polynomials in the $y_i$. This polynomial has zeros the intersections of our curve $C$ with the line $L$ we're looking at. So we want to find points of multiplicity at least two. So how do we find multiple roots of a polynomial? We take the discriminant! Specifically, we do it for an affinization, and we get a homogeneous polynomial of degree $2n^2-n$ in the $y_i$. All that's left is to factor the polynomial, and kill all the linear factors, just throw them away, the reasons are explained in more computational detail on my blogpost, and there I also do explicit examples, but the method for calculating the equation of the dual curve is as above. | 0.222157 | 99 |
science494 | Benoît Mandelbrot Although he provided many valuable contributions to the field, I am most in love with his work on Fractals. I find math to be quite beautiful, and the Mandelbrot Set (magnified portion shown below) is a perfect example: | 0.000082 | 99 |
science498 | Well, class field theory states that the class number is the degree of the largest everywhere-unramified abelian extension of a number field (namely, the Hilbert class field). But class field theory really says a lot more: it says that there's an isomorphism between the Galois group and the ideal class group. And in general, for any abelian extension $L/K$, there's an isomorphism between $G(L/K)$ and a certain "generalized ideal class group" of $K$ where you quotient by ideals that are norms from $L$. This can be stated somewhat more elegantly using ideles. But, you asked about the class number. In class field theory, it's difficult (to my knowledge) to prove directly that the Artin reciprocity map defined from ideals to the Galois group is an isomorphism. You can show that it is surjective and injective, however; to do this, you have to estimate the order of these generalized ideal class groups, which you can do using either analytic (L-function) methods or algebraic (e.g., using the Herbrand quotient) methods. So this is where the size and seemingly "decategorified" properties (like the order) become more important than the groups themselves: the proofs. | 0.415504 | 99 |
science500 | On an intuition level: Complex numbers are somewhat equivalent to $\mathbb R^2$ (homeomorphic to a $2$-dimensional real space), on the other hand they are also a $1$-dimensional complex space. This gives them an extra structure with consequences like this theorem. Another view on this is that differentiable complex functions must also satisfy the Cauchy-Riemann equations and this extra hypothesis is what makes them holomorphic. | 0.212831 | 2 |
science501 | Suppose you have a large collection of books, all of the same size. Balance one of them on the edge of a table so that one end of the book is as far from the table as possible. Balance another book on top of that one, and again try to get as far from the table as possible. Take $n$ of them and try to balance them on top of each other so that the top book is as far as possible away from the edge of the table horizontally. Theorem: With enough books, you can get arbitrarily far from the table. If you are really careful. This is a consequence of the divergence of the harmonic series. I think if you haven't heard this one before it's very hard to tell whether it's true or false. | 0.392304 | 0 |
science503 | This isn't a complete answer, partly because I've had discussions with other grad students and we weren't able to work it out satisfactorily, but I've been told that you can actually put a topology on logical statements such that the compactness theorem translates to "The set of true statements is a compact subset of the set of all statements" or something similar. But as I said, a few of my friends and I weren't able to work out the topology. | 0.127612 | 0 |
science505 | The analogy for the compactness theorem for propositional calculus is as follows. Let $p_i $ be propositional variables; together, they take values in the product space $2^{\mathbb{N}}$. Suppose we have a collection of statements $S_t$ in these boolean variables such that every finite subset is satisfiable. Then I claim that we can prove that they are all simultaneously satisfiable by using a compactness argument. Let $F$ be a finite set. Then the set of all truth assignments (this is a subset of $2^{\mathbb{N}}$) which satisfy $S_t$ for $t \in F$ is a closed set $V_F$ of assignments satisfying the sentences in $F$. The intersection of any finitely many of the $V_F$ is nonempty, so by the finite intersection property, the intersection of all of them is nonempty (since the product space is compact), whence any truth in this intersection satisfies all the statements. I don't know how this works in predicate logic. | 0.496508 | 0 |
science506 | A further sense in which Higher Order Logics with standard or saturated semantics (HOL, hereafter) are less well-behaved than First Order Logic (FOL, hereafter) is a direct consequences of the failure of Completeness (and thus, as explained in other answers, of Compactness). The set of logical truths and the set of correct claims of semantic consequence for these logics are not recursively enumerable. FOL is Complete, yet not Decidable. So, we determine of an arbitrary sentences and sets of sentences of the language of FOL if those sentences are logical truths or if a set has a given sentence as a consequence. But, since FOL is Complete and proofs are finitely long, we can (in the mathematicians sense of "can") enumerate the proofs and inspect one by one, checking what sentence the proof shows as a theorem or what sentence the proof derives from what set of assumptions. This gets us a recursive enumeration of the truths and the sentence/set pairs that stand in the consequence relation. (This does not contradict the failure of decidability as we cannot conclude that since we've yet to come across a proof in out enumeration, there isn't one if only we kept looking.) Since HOL's are not Complete, this means of showing them recursively enumerable is not available. Indeed, there can be no means; were there such a means, it could be exploited to induce a Complete proof system, and there cannot be such a Complete proof system as the HOL's are not Compact. | 0.426096 | 99 |
Subsets and Splits