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How is prisoner's dilemma different from chicken?
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?
Okay, so this question was bound to come up sooner or later- the hope was to ask it well before someone asked it badly... ###We all love a good puzzle To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of sudoku for example is hidden in [latin squares][1]). Mathematicians and puzzles get on, it seems, rather well. ###But what is a good puzzle? Okay, so in order to make this question worthwhile (and not a ten-page wadeathon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify for answerhood- to do so it must - Not be widely known: If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that hilarious scene in the film 21, where kevin spacey explains the monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disemvowelled. - Be mathematical: as much as possible- true: logic is mathematics, but puzzles beginning 'There is a street where everyone has a different coloured house...' are much of a muchness and tedious as hell. Note: there is a happy medium between this and trig substitutions. - Not be too hard: any level is cool but if the answer requires more than two sublemmas, you are misreading your audience - Actually have an answer: crank questions will not be appreciated! You can post the answers/hints in [Rot-13][2] underneath as comments as on MO if you fancy. And should - Ideally include where you found it: so we can find more cool stuff like it - Have that indefinable spark that makes a puzzle awesome: a situation that seems familiar, requiring unfamiliar thought... For ease of voting- one puzzle per post is bestest. ###Some examples to set the ball rolling > Simplify $\sqrt{2+\sqrt{3}}$ From: problem solving magazine Hint: GEL N GJB GREZ FBYHGVBA > Can one make an equilateral triangle with all vertices at integer coordinates? From: Durham distance maths challenge 2010 Hint: GUVF VF RDHVINYRAG GB GUR ENGVBANY PNFR >nxn [Magic squares][3] form a vector space over $\mathbb{R}$ prove this, and by way of a linear transformation, derive the dimension of this vector space. From: Me, I made this up (you can tell, can't you!) Hint: NCCYL GUR ENAX AHYYVGL GURBERZ Happy puzzling! [1]: http://en.wikipedia.org/wiki/Latin_squares [2]: http://en.wikipedia.org/wiki/Rot_13 [3]: http://en.wikipedia.org/wiki/Magic_squares
Can someone help me with a math/logic problem my friend and I are struggling with? Dealing with poisoned wines and servants
I covered homotopy theory in a recent maths course. However I was never presented with any reasons as to why (or even if) it is useful. Is there any good examples of its use outside academia?
I just got out from my Math and Logic class with my friend. During the lecture, a well-known math/logic puzzle was presented: >The King has $1000$ wines, $1$ of which is poisoned. He needs to identify the poisoned wine as soon as possible, and with the least resources, so he hires the protagonist, a Mathematician. The king offers you his expendable servants to help you test which wine is poisoned. >The poisoned wine is very potent, so much that one molecule of the wine will cause anyone who drinks it to die. However, it is slow-acting. The nature of the slow-acting poison means that there is only time to test one "drink" per servant. (A drink may be a mixture of any number of wines) (Assume that the King needs to know within an hour, and that any poison in the drink takes an hour to show any symptoms) >What is the minimum amount of servants you would need to identify the poisoned wine? Upon hearing this problem, the astute mathematics student will see that this requires at most ten ($10$) servants (in fact, you could test 24 more wines on top of that 1000 before requiring an eleventh servant). The proof/procedure is left to the reader. My friend and I, however, was not content with resting upon this answer. My friend added the question: >What would be different if there were $2$ wines that were poisoned out of the 1000? What is the new minimum then? We eventually generalized the problem to this: > Given $N$ bottles of wine ($N \gt 2$) and, of those, $k$ poisoned wines ($0 \lt k \lt N$), what is the optimum method to identify the all of the poisoned wines, and how many servants are required ($s(N,k)$)? After some mathsing, my friend and I managed to find some (possibly unhelpful) lower and upper bounds: $ log_2 {N \choose k} \le s(N,k) \le N-1 $ This is because $log_2 {N \choose k}$ is the minimum number of servants to uniquely identify the $N \choose k$ possible configurations of $k$ poisoned wines in $N$ total wines. Can anyone help us find an optimum strategy? Besides the trivial one requiring $N-1$ servants. How about a possible approach to start? Would this problem be any different if you were only required to find a strategy that would for sure find a wine that is not poisoned, instead of identifying all poisoned wines?
Can someone help me with a math/logic problem my friend and I are struggling with? Dealing with identifying poisoned wines with minimal servants.
I've learned that multiplying two generating functions f(x) and g(x) gives you the result $\sum_{k=0}^\infty\left(\sum_{j=0}^k a_j\,b_{k-j}\right)x^k$. I've used the result, but it was presented in my class without proof and I'm having some trouble tracking one down. Weak google-foo today, I suppose. Can anyone give me a pointer to a proof? If this is a question better answered in book form, that is fine as well.
One thing I have been wondering about lately is how to write an equation that describes a pattern of numbers. What I mean is: x 0 1 2 y 1 5 9 If I have this, I can tell that an equation that describes this would be y=4x+1. In fact, I don't even need the 3rd pair of numbers. It's very easy when the equation is a straight line. But when the equation is a parabola, its not always that easy. For example: x 0 1 2 y 1 2 5 I can tell this is y=x<sup>2</sup>+1, because I recognize the pattern. But I can't always tell just by looking at the numbers what the right equation should be. Is there some way to always know the right equation? I know that if you get the x=0 term you can get the c in y=ax<sup>2</sup>+bx+c, but that's not enough to let me solve it like I can when the equation is just a line. For example, can someone show me how you would do it for x 0 1 2 y 5 4 7 It's not a homework question, I promise!
How can I write an equation that matches any sequence?
EDIT: For some reason, the jsMath plugin isn't displaying plus signs. How odd. 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 ( $y = A x^2 + B x + C$ ) - If you are given four points, the best you can do is of degree 3 ( $y = A x^3 + B x^2 + C x + D$ ) - 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 ( $y = A x^2 + B x + C$ ), you have three unknowns? And also three equations! (points) Let's pick three arbitrary points <pre> x 1 2 3 y 4 6 1 </pre> You would set up three equations: $ 4 = A1^2 + B1 + C $ $ 6 = A2^2 + B2 + C $ $ 1 = A3^2 + B3 + C $ Three equations, three unknowns. You should be able to solve this with a combination of most system-of-equation-solving rules. 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)
###Motivatation I am presently looking at a structure that I am trying to pin down- my strategy being to pull the thing up into the greatest possible generality (based on the bits I'm sure about) and narrow it down from there. The situation I have is somewhat similar to the dual space structure in vector spaces, though almost certainly less well-behaved and vector spaces alone will not cut it. ###The Construction Consider a vector space $V$ over a field $k$- its dual space $V^$ appears naturally as the set linear of maps $$w^ : V \to k$$ which, coincidentally, form a vector space in themselves. We can generalise this to arbitrary categories $A$, $B$, $C$ by setting $B=hom(A,C). Then, at least in some sense, $B=A^$. So far so standard, but I want more: a nice property of dual spaces is that an element of $V \otimes V^$ can be canonically seen as an element of $hom(V,V)$- this is because of the way that $k$ acts on $V$ by multiplication. We can mimic this by letting $C$ be a monoid acting on $A$. > In summary: A category $A$ acted on by a monoid $C$ and a dual A $A^*:=hom(A,C)$ I am particularly interested in when $A$ is also a monoid, especially so when $A$ is a space of stochastic matrices. ###Questions: So this isn't too unlikely a construction, in fact it's probably forehead-slappingly well known, so: - What is it called, if anything? - In what cases can we have $A=B=C$? Is that necessarily a permutation group for example? - Are there any useful canonical examples, besides vector spaces? - Better yet, theorems??? Papers??? As you can probably tell, I am no category theorist, so any help would be awesome.
I believe the right generalization of dual spaces is that of a [dual object][1] in a [tensor category][2], 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: 1. This coincides with the usual dual in the category of vector spaces 2. This coincides with the dual sheaf if one is working in the category of locally free sheaves on a scheme 3. 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][3] of Pavel Etingof on tensor categories. [1]: http://en.wikipedia.org/wiki/Dual_object [2]: http://en.wikipedia.org/wiki/Tensor_category [3]: http://www-math.mit.edu/~etingof/tenscat1.pdf
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 = Ax + B` ) - If you are given three points, the best you can do is of degree 2 ( `y = Ax^2 + Bx + C` ) - If you are given four points, the best you can do is of degree 3 ( `y = Ax^3 + Bx^2 + Cx + 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 x^2 + B x + C` ), you have three unknowns? And also three equations! (points) Let's pick three arbitrary points <pre> x 1 2 3 y 4 6 1 </pre> You would set up three equations: <pre> 4 = A1^2 + B1 + C 6 = A2^2 + B2 + C 1 = A3^2 + B3 + C </pre> Three equations, three unknowns. You should be able to solve this with a combination of most system-of-equation-solving rules. 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)
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 = Ax + B` ) - If you are given three points, the best you can do is of degree 2 ( `y = Ax^2 + Bx + C` ) - If you are given four points, the best you can do is of degree 3 ( `y = Ax^3 + Bx^2 + Cx + 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 x^2 + B x + C` ), you have three unknowns? And also three equations! (points) Let's pick three arbitrary points <pre> x 1 2 4 y 6 7 3 </pre> You would set up three equations: <pre> 6 = A1^2 + B1 + C 7 = A2^2 + B2 + C 3 = A4^2 + B4 + C </pre> 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: <pre> A = -1 B = 4 C = 3 </pre> So our equation is `y = -x^2 + 4x + 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)
I just got out from my Math and Logic class with my friend. During the lecture, a well-known math/logic puzzle was presented: >The King has $1000$ wines, $1$ of which is poisoned. He needs to identify the poisoned wine as soon as possible, and with the least resources, so he hires the protagonist, a Mathematician. The king offers you his expendable servants to help you test which wine is poisoned. >The poisoned wine is very potent, so much that one molecule of the wine will cause anyone who drinks it to die. However, it is slow-acting. The nature of the slow-acting poison means that there is only time to test one "drink" per servant. (A drink may be a mixture of any number of wines) (Assume that the King needs to know within an hour, and that any poison in the drink takes an hour to show any symptoms) >What is the minimum amount of servants you would need to identify the poisoned wine? Upon hearing this problem, the astute mathematics student will see that this requires at most ten ($10$) servants (in fact, you could test 24 more wines on top of that 1000 before requiring an eleventh servant). The proof/procedure is left to the reader. My friend and I, however, was not content with resting upon this answer. My friend added the question: >What would be different if there were $2$ wines that were poisoned out of the 1000? What is the new minimum then? We eventually generalized the problem to this: > Given $N$ bottles of wine ($N \gt 2$) and, of those, $k$ poisoned wines ($0 \lt k \lt N$), what is the optimum method to identify the all of the poisoned wines, and how many servants are required ($s(N,k)$)? After some mathsing, my friend and I managed to find some (possibly unhelpful) lower and upper bounds: $ log_2 {N \choose k} \le s(N,k) \le N-1 $ This is because $log_2 {N \choose k}$ is the minimum number of servants to uniquely identify the $N \choose k$ possible configurations of $k$ poisoned wines in $N$ total wines. Can anyone help us find an optimum strategy? Besides the trivial one requiring $N-1$ servants. How about a possible approach to start? Would this problem be any different if you were only required to find a strategy that would for sure find a wine that is not poisoned, instead of identifying all poisoned wines? (other than the slightly trivial solution of $k$ servants)
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 x<sup>2</sup> + B x + C ) - If you are given four points, the best you can do is of degree 3 ( y = A x<sup>3</sup> + B x<sup>2</sup> + 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 x<sup>2</sup> + B x + C ), you have three unknowns? And also three equations! (points) Let's pick three arbitrary points <pre> x 1 2 4 y 6 7 3 </pre> You would set up three equations: 6 = 1<sup>2</sup> * A + 1 * B + C 7 = 2<sup>2</sup> * A + 2 * B + C 3 = 4<sup>2</sup> * 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: <pre> A = -1 B = 4 C = 3 </pre> So our equation is y = -x<sup>2</sup> + 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)
I just got out from my Math and Logic class with my friend. During the lecture, a well-known math/logic puzzle was presented: >The King has $1000$ wines, $1$ of which is poisoned. He needs to identify the poisoned wine as soon as possible, and with the least resources, so he hires the protagonist, a Mathematician. The king offers you his expendable servants to help you test which wine is poisoned. >The poisoned wine is very potent, so much that one molecule of the wine will cause anyone who drinks it to die. However, it is slow-acting. The nature of the slow-acting poison means that there is only time to test one "drink" per servant. (A drink may be a mixture of any number of wines) (Assume that the King needs to know within an hour, and that any poison in the drink takes an hour to show any symptoms) >What is the minimum amount of servants you would need to identify the poisoned wine? Upon hearing this problem, the astute mathematics student will see that this requires at most ten ($10$) servants (in fact, you could test 24 more wines on top of that 1000 before requiring an eleventh servant). The proof/procedure is left to the reader. My friend and I, however, was not content with resting upon this answer. My friend added the question: >What would be different if there were $2$ wines that were poisoned out of the 1000? What is the new minimum then? We eventually generalized the problem to this: > Given $N$ bottles of wine ($N \gt 1$) and, of those, $k$ poisoned wines ($0 \lt k \lt N$), what is the optimum method to identify the all of the poisoned wines, and how many servants are required ($s(N,k)$)? After some mathsing, my friend and I managed to find some (possibly unhelpful) lower and upper bounds: $ log_2 {N \choose k} \le s(N,k) \le N-1 $ This is because $log_2 {N \choose k}$ is the minimum number of servants to uniquely identify the $N \choose k$ possible configurations of $k$ poisoned wines in $N$ total wines. Can anyone help us find an optimum strategy? Besides the trivial one requiring $N-1$ servants. How about a possible approach to start? Would this problem be any different if you were only required to find a strategy that would for sure find a wine that is not poisoned, instead of identifying all poisoned wines? (other than the slightly trivial solution of $k$ servants)
I just came back from my Introduction to Rotational Kinematics class, and one of the important concepts they described was Rotational Inertia, or Moment of Inertia. It's basically the equivalent of mass in Netwon's $F = m a$ in linear motion. The equivalent rotational equation is $\tau = I \alpha$, where $\tau$ is rotational force, $\alpha$ is rotational acceleration, and $I$ is rotational inertia. For a point about an axis, $I$ is $m r^2$, where $r$ is the distance from the point to the axis of rotation. For a continuous body, this is an integral -- $I = \int r^2 \,dm$. This really doesn't make any sense to me...you have two independent variables? I am only used to having one independent variable and one constant. So I would solve this, using my experience with calculus (which encompasses a read through the Sparks Notes packet) as $ I = m r^2 $ But obviously, this is wrong? $r$ is not a constant! How do I deal with it? Do I need to replace $r$ with an expression that varies with $m$? But how could $r$ possibly vary with $m$? Isn't it more likely the other way around? But how can $m$ vary with $r$? It's all rather confusing me. Could someone help me figure out what to do with all these substitutions for, example, figuring out the Moment of Inertia of a hoop with no thickness and width $w$, with the axis of rotation running through its center orthogonal to its plane?
Finding Moment of Inertia (rotional inertia?) $I$ using integration?
I have heard that one can think of stacks on a site as taking sheaves but instead of the restrictions being equal, we just loosen it to isomorphic, and treat the sheaf conditions with the "obvious" coherence relations. How seriously can one take this analogy? Note that my background is stacks is feeble at best. I hope this question isn't too vague. One may choose to respond to this question with "That analogy is stupid because of ______ gotcha". Thanks in advance!
In my linear algebra class we just talked about determinants. So far I’ve been understanding the material okay, but now I’m very confused. I get that when the determinant is zero, the matrix doesn’t have an inverse. I can find the determinant of a 2 by 2 matrix by the formula. Our teacher showed us how to compute the determinant of an N by N matrix by breaking it up into the determinants of smaller matrices, and apparently there’s a way by summing over a bunch of permutations. But the notation is really hard for me and I don’t really know what’s going on with them anymore. Can someone help me figure out what a determinant is intuitively, and how all those definitions of it are related?
What's an intuitive way to think about the determinant?
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? (1) Switching two rows or columns changes the sign. (2) Multiplying one row by a constant multiplies the whole determinant by that constant. (3) The general fact that number two draws from: the determinant is linear in each row. I claim that these facts, together with the fact that the determinant of the identity matrix is one, is 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 approximations 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 det((a,b),(c,d)), either: you might try that to get a sense for things.
In my AP chem class, I often have to balance chemical equations like the following: Al + O<sub>2</sub> ---> Al<sub>2</sub>O<sub>3</sub> The goal is to make both side of the arrow have the same amount of atoms by adding compounds in the equation to each side. An solution 4Al + 3O<sub>2</sub> ---> 2Al<sub>2</sub>O<sub>3</sub> When the subscripts become really large, or there are lot of atoms involved, trial and error is impossible unless performed by a computer. What if some chemical equation can not be balanced(does such equations exist)? I tried one for a long time only to realize the problem was wrong. My teacher said trial and error is the only way. Are there other methods?
I just came back from a class on Probability in Game Theory, and was musing over something in my head. Assuming, for the sake of the question: - Playing cards in their current state have been around for approximately eight centuries - A deck of playing cards is shuffled to a random configuration one billion times per day - Every shuffle ever is completely (theoretically) random and unaffected by biases caused by human shuffling and the games the cards are used for - By "deck of cards", I refer to a stack of unordered 52 unique cards, with a composition that is identical from deck to deck. This would, approximately, be on the order of 3 x 10<sup>14</sup> random shuffles in the history of playing cards. If I were to shuffle a new deck today, completely randomly, what are the probabilistic odds (out of 1) that you create a new unique permutation of the playing cards that has never before been achieved in the history of 3 x 10<sup>14</sup> similarly random shuffles? My first thought was to think that it was a simple matter of $\frac{1}{52!} * 3 * 10^{14}$, but then I ran into things like [Birthday Paradox](http://en.wikipedia.org/wiki/Birthday_Paradox). While it is not analogous (I would have to be asking about the odds that any two shuffled decks in the history of shuffled decks ever matched), it has caused me to question my intuitive notions of Probability. What is wrong in my initial approach, if it is wrong? What is the true probability? And, if the probability is less than 0.5, if we how many more years (centuries?) must we wait, assuming the current rate of one billion shuffles per day, until we reach a state where the probability is 0.5+? 0.9+?
When you randomly shuffle a deck of cards, what is the probability that it is a unique permutation never before configured?
I just came back from a class on Probability in Game Theory, and was musing over something in my head. Assuming, for the sake of the question: - Playing cards in their current state have been around for approximately eight centuries - A deck of playing cards is shuffled to a random configuration one billion times per day - Every shuffle ever is completely (theoretically) random and unaffected by biases caused by human shuffling and the games the cards are used for - By "deck of cards", I refer to a stack of unordered 52 unique cards, with a composition that is identical from deck to deck. This would, approximately, be on the order of 3 x 10<sup>14</sup> random shuffles in the history of playing cards. If I were to shuffle a new deck today, completely randomly, what are the probabilistic odds (out of 1) that you create a new unique permutation of the playing cards that has never before been achieved in the history of 3 x 10<sup>14</sup> similarly random shuffles? My first thought was to think that it was a simple matter of $\frac{1}{52!} * 3 * 10^{14}$, but then I ran into things like [Birthday Paradox](http://en.wikipedia.org/wiki/Birthday_Paradox). While it is not analogous (I would have to be asking about the odds that any two shuffled decks in the history of shuffled decks ever matched), it has caused me to question my intuitive notions of Probability. What is wrong in my initial approach, if it is wrong? What is the true probability? And, if the probability is less than 0.5, if we how many more years (centuries?) must we wait, assuming the current rate of one billion shuffles per day, until we reach a state where the probability is 0.5+? 0.9+? (Out of curiosity, it would be neat to know the analogous birthday paradox answer, as well)
Randomly break a stick (or a piece of dry spaghetti, etc.) in two places, forming three pieces. The probability that these three pieces can form a triangle is 1/4 (coordinatize the stick form 0 to 1, call the breaking points x and y, consider the unit square of the coordinate plane, shade the areas that satisfy the triangle inequality). The other day in class<sup></sup>, my professor was demonstrating how to do a Monte Carlo simulation of this problem on a calculator and wrote a program that, for each trial did the following: 1. Pick a random number x between 0 and 1. This is the first side length. 2. Pick a random number y between 0 and 1 - x (the remaning part of the stick). This is the second side length. 3. The third side length is 1 - x - y. 4. Test if the three side lengths satisfy the triangle inequality (in all three permutations). He ran around 1000 trials and was getting 0.19, which he said was probably just random-chance error off 0.25, but every time the program was run, no matter who's calculator we used, the result was around 0.19. What's wrong with the simulation method? What is the theoretical answer to the problem actually being simulated? (<sup></sup> the other day was more than 10 years ago)
About two years ago, I started thinking about the following problem: You're given an N and an S, positive integers. You start with an N-gon that has positive integer labels at each vertex, such that the labels sum to S. A "move" is replacing the label at some vertex with the positive difference of the labels at its two adjacent vertices. You win if you can exhibit a sequence of moves such that all of the vertex labels are zero at the end. This is a generalization of 2003 USAMO problem B3, "A positive integer is written at each vertex of a hexagon. A move is to replace a number by the (non-negative) difference between the two numbers at the adjacent vertices. If the starting numbers sum to 2003^2003, show that it is always possible to make a sequence of moves ending with zeros at every vertex." The USAMO problem itself can be answered in the affirmative by doing a modified greedy algorithm for most of the moves, showing there's always a move where you can reduce the problem to a smaller sum while changing the parities systematically, then doing casework for the "end game" to show that each configuration of 1s and 0s (or k's and zeros, the cases are of course equivalent) can be gotten down to all zeros. It's easy to check that for n=4 and S odd, there's a winning strategy. I have lots of simulation/numerical evidence and some half baked ideas to suggest that, in general, there's a winning strategy iff N and S have opposite parities. But in two years of coming back to it on and off, I haven't really made progress. It's open, as far as I know, though I knew that more surely two years ago than I do now. Ideas?
How do Lagrange multipliers work to find the lowest value of a function subject to a constraint?
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? (1) Switching two rows or columns changes the sign. (2) Multiplying one row by a constant multiplies the whole determinant by that constant. (3) 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: R^(n^2) -> R, det(av_1 +bw_1, v_2,...,v_n)=adet(v_1,...,v_n)+bdet(w_1,...,w_n), and the corresponding condition in each other slot. I claim that these facts, together with the fact that the determinant of the identity matrix is one, is 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 approximations 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 det((a,b),(c,d)), either: you might try that to get a sense for things.
Two mathematicians each come into a coffee shop at a random time between 8:00 a.m. and 9:00 a.m. each day. Each orders a cup of coffee then sits at a table, reading a newspaper for 20 minutes before leaving to go to work. On any day, what is the probability that both mathematicians are at the coffee shop at the same time (that is, their arrival times are within 20 minutes of each other)?
Randomly break a stick (or a piece of dry spaghetti, etc.) in two places, forming three pieces. The probability that these three pieces can form a triangle is 1/4 (coordinatize the stick form 0 to 1, call the breaking points x and y, consider the unit square of the coordinate plane, shade the areas that satisfy the triangle inequality edit: see comments on the question, below, for a better explanation of this). The other day in class<sup></sup>, my professor was demonstrating how to do a Monte Carlo simulation of this problem on a calculator and wrote a program that, for each trial did the following: 1. Pick a random number x between 0 and 1. This is the first side length. 2. Pick a random number y between 0 and 1 - x (the remaning part of the stick). This is the second side length. 3. The third side length is 1 - x - y. 4. Test if the three side lengths satisfy the triangle inequality (in all three permutations). He ran around 1000 trials and was getting 0.19, which he said was probably just random-chance error off 0.25, but every time the program was run, no matter who's calculator we used, the result was around 0.19. What's wrong with the simulation method? What is the theoretical answer to the problem actually being simulated? (<sup></sup> the other day was more than 10 years ago)
What property of certain regular polygons allows them to be faces of the Platonic Solids?
How can I randomly generate trees?
Knuth says to look at it as generating all nested parentheses in lexicographic order. Look here for the details http://www.google.com/url?sa=t&source=web&cd=1&ved=0CBIQFjAA&url=http%3A%2F%2Fwww-cs-faculty.stanford.edu%2F~uno%2Ffasc4a.ps.gz&ei=l_NLTOagLIHCsAPS4oHkCw&usg=AFQjCNGFPnsHAYgcsoaKZ2ieiwJ-pn6hiw&sig2=-dVevIzkUotrbwWuk_QRPg
Suppose that I am buying cakes for a party. There are k different types and I intend to buy a total of n cakes. How many different combinations of cakes could I possibly bring to the party?
Suppose we model traffic flow between two points with a directed graph. Each route has a maximum capacity of how much traffic it can hold and how long it takes to travel that route. We assume that each driver wishes to minimise their own travel time and we assume that the drivers form a Nash equilibria. Can removing a route ever decrease the average travelling time? Note that the existence of multiple Nash equilibria makes this question a bit complicated. To clarify, I am looking for a route removal that will guarantee a decrease in the average traveling time regardless of the Nash equilibria that are chosen before and after.
Complex numbers are the final step in a sequence of increasingly "unreal" extensions to the number system that humans have found it necessary to add over the centuries in order to express significant numerical concepts. The first such "unreal number" was zero, back in the mists of time. It seems obvious to us now, but it must have seemed strange at first. How can the number of sheep I have be zero, when I don't actually have any sheep? Negative numbers are the next most obvious addition to the family of numbers. But what does it mean to have -2 apples? If I have 3 apples and you have 5, it's convenient to be able to say that I have -2 more apples than you. Even so, during the middle ages many mathematicians were very uncomfortable with the idea of negative numbers and tried to arrange their equations so that they didn't occur. Rationals (fractions) seem real enough, since I'm happy to have 2/5 of a pizza. However, this is not the number of pizzas that I have, just a ratio between 0 and 1 pizzas, and so is further removed from the concept of counting. More significant philosophical problems problems arose when irrational numbers were first discovered by the classical Greeks. They were astonished when Euclid (or one of his predecessors) proved that the diagonal of a unit square could not be represented by ratio of two integers. This was such an outrageous idea to them that they called these numbers "irrational". However, they couldn't easily deny their existence since they have a direct geometric representation. Irrational numbers were first understood as the solutions to algebraic equations such as x^2 = 2 but this still doesn't cover all the possible numbers that we need. For example, the ratio of a circle's diameter to its circumference is π and so has a direct geometric representation. However, in 1882 π was proved to be transcendental, meaning that it can't be defined as the solution to a specific algebraic equation. This begins to seem a lot less "real", especially when you consider that there are many important transcendental numbers that, unlike π, don't have a geometric interpretation. There are of course many algebraic equations that don't have a solution even among the irrational numbers, and in some ways there's no reason why they should. However, when 16th century mathematicians like Gerolamo Cardano began working on solutions to cubic equations they found that the square roots of negative numbers began cropping up very naturally in their procedures, even though the solutions themselves were purely real. This eventually led people to explore the arithmetic of complex numbers and they were surprised to find that it produced a consistent and beautiful theory. However, complex numbers don't have such an intuitively obvious geometrical meaning as the numbers that came before. They are typically represented graphically as points in the 2D plane, and the rules of addition and multiplication are equivalent to certain operations on lengths and angles, but those operations aren't driven by geometrical necessity in quite the way that squares and circles are. Even so, complex numbers are a perfect representation for various physical phenomena such as the state of particles in quantum mechanics and the behaviour of varying currents in electrical circuits. They are also very useful for reducing the cost of computation in 3D computer graphics. The really special thing about complex numbers, though, is that they are the end of this journey that has been going on for millenia. There is no need to invent further number systems to provide solutions to problems expressed in terms of complex numbers because now every non-contradictory equation, algebraic or transcendental, has a solution within the complex numbers. They are a self-sufficient, consistent system. Physicists have been breaking matter down into smaller and smaller particles over the years. Molecules, atoms, nuclei, protons, quarks. When will the process stop? For several decades it was felt that quarks could be the final, indivisible particle. However, new theoretical frameworks such as string theory are suggesting that there may be more fundamental entities than quarks. Because physics ultimately relies on experimental verification, we can never be sure that there aren't going to be more steps in the sequence. In mathematics, however, we can prove things for all time. Complex numbers are the final step in the sequence, the numbers that we have been reaching for since before the beginning of recorded history. Every other number system is just a subset of the complex numbers, just a part of the true picture. Complex numbers are the real thing.
Complex numbers involve the square root of negative one, and most non-mathematicians find it hard to accept that such a number is meaningful. In contrast, they feel that real numbers have an obvious and intuitive meaning. What's the best way to explain to a non-mathematician that complex numbers are necessary and meaningful, in the same way that real numbers are? This is not a Platonic question about the reality of mathematics, or whether abstractions are as real as physical entities, but an attempt to bridge a comprehension gap that many people experience when encountering complex numbers for the first time. The wording, although provocative, is deliberately designed to match the way that many people do in fact ask this question.
Do complex numbers really exist?
I need n cakes for a party. I go to the cake shop and there are k different kinds of cake. For variety, I'd like to get at least one of every cake. How many ways can I do this?
In least-squares approximations the <a href="">normal equations</a> act to project a vector existing in N-dimensional space onto a lower dimensional space, where our problem actually lies, thus providing the "best" solution we can hope for (the orthogonal projection of the N-vector onto our solution space). The "best" solution is the one that minimizes the Euclidean distance (two-norm) between the N-dimensional vector and our lower dimensional space. There exist other norms and other spaces besides $\mathbb{R}^d$, what are the analogues of least-squares under a different norm, or in a different space?
Are there variations on least-squares approximations?
Recently I discussed an experiment with a friend. Assume we start a random experiment. At first there is an array with size 100 000, all set to 0. We calculate at each round a random number modulo 2 and select one random position in that array. If the number in the array is 1, nothing is changed and otherwise the pre-computed value is set. The question is: how many distinct hash values would we have added in 1%, 5%, 50%, 95%, 99% of all cases? Example: 4 rounds with array of size 10: Array Position random number [0,...,0] 5 0 [0,...,0] 7 1 [0,...0,1,0,0,0] 6 1 [0,..0,.1,1,0,0,0] 6 0 [0,..0,.1,1,0,0,0] 2 0 First we considered this a somehow simple problem. But after thinking about for some hours, searching the web and asking some math students we couldn't find a solution.So do you know a probability distribution for this problem? Remark: Was also posted on [Math Overflow][1]. [1]: http://mathoverflow.net/questions/33335/looking-for-a-probability-distribution
The calculus of relations is an algebra of operations over sets of pairs of individuals, where for any relation R, we can express the relation in the usual infix manner: x R y iff (x,y) ∈ R. This allows all of the properties of relations to be expressed in equational form. There are three fundamental relations we want to define: + Eq = {(x,x) \| for each individual x}. This is the diagonal or equality relation that holds only between something and itself; + All = {(x,y) \| for all individuals x,y}. The whenever relation; + Neq = {(x,y) \| for all individuals x,y for which x ≠ y}: the inequality relation that holds between anything different. + Never is the empty set. Three basic binary operations on relations, and one unary relation: + R∩S = {(x,y) \| (x,y) ∈ R and (x,y) ∈ S}; + R–S = {(x,y) \| (x,y) ∈ R and (x,y) ∉ S}; + R⋅S = {(x,z) \| there is some y such that (x,y) ∈ R and (y,z) ∈ S} (composition); + tr(R) = {(y,x) \| (x,y) ∈ R} (transpose). Observe that Never = All–All, and Neq = All–Eq. We say R -> S when S holds whenever R holds. This is the same as saying either that R is a subset of S, or that R∩S=R, or that R-S=Never. So Never -> Eq, and Eq -> All. Then we can express your five relations: 1. R is reflexive when Eq -> R; dually we have an additional property of a relation, where R is anti-reflexive when R -> Neq; 2. R is symmetric when R -> tr(R); 3. R is asymmetric when R ∩ tr(R) -> Eq, or equivalently when R ∩ tr(R) ∩ Neq = 0. 4. R is anti-symmetric when R ∩ tr(R)=Never: a relation is anti-symmetric when it is asymmetric and anti-reflexive; 5. R is transitive when R⋅R -> R. Also, how can a relation be a- and antisymmetrical at the same time? Don't they cancel each other out? — Look at Eq: it is both symmetric and asymmetric, although it is not anti-symmetric. The difference is that asymmetric and anti-symmetric differ in what they assert about the diagonal: asymmetric doesn't care about what pairs there might be along the diagonal, whilst anti-symmetric insists that there are no pairs along the diagonal, which is irreflexivity.
The calculus of relations is an algebra of operations over sets of pairs of individuals, where for any relation R, we can express the relation in the usual infix manner: x R y iff (x,y) ∈ R. This allows all of the properties of relations to be expressed in equational form. There are four fundamental relations we want to define, each of which make non-useless examples of three of your five properties of relations, plus one other useful property, irreflexivity: + Eq = {(x,x) \| for each individual x}. This is the diagonal or equality relation that holds only between something and itself. It is reflexive, symmetric and transitive, which is to say it is an equivalence relation. + All = {(x,y) \| for all individuals x,y}. The whenever relation, it is also an equivalence relation. + Neq = {(x,y) \| for all individuals x,y for which x ≠ y}: the inequality relation that holds between anything different. It is symmetric and irreflexive, but not transitive. + Never is the empty set. It is symmetric and transitive, and irreflexive. Three basic binary operations on relations, and one unary relation: + R∩S = {(x,y) \| (x,y) ∈ R and (x,y) ∈ S}; + R–S = {(x,y) \| (x,y) ∈ R and (x,y) ∉ S}; + R⋅S = {(x,z) \| there is some y such that (x,y) ∈ R and (y,z) ∈ S} (composition); + tr(R) = {(y,x) \| (x,y) ∈ R} (transpose). Observe that Never = All–All, and Neq = All–Eq. We say R -> S when S holds whenever R holds. This is the same as saying either that R is a subset of S, or that R∩S=R, or that R-S=Never. So Never -> Eq, and Eq -> All. Then we can express your five relations: 1. R is reflexive when Eq -> R; dually we have an additional property of a relation, where R is anti-reflexive when R -> Neq; 2. R is symmetric when R -> tr(R); 3. R is asymmetric when R ∩ tr(R) -> Eq, or equivalently when R ∩ tr(R) ∩ Neq = 0. 4. R is anti-symmetric when R ∩ tr(R)=Never: a relation is anti-symmetric when it is asymmetric and anti-reflexive; 5. R is transitive when R⋅R -> R. Also, how can a relation be a- and antisymmetrical at the same time? Don't they cancel each other out? — Look at Eq: it is both symmetric and asymmetric, although it is not anti-symmetric. In fact, all anti-symmetric relations are asymmetric, but not vica versa: the difference is that asymmetric and anti-symmetric differ in what they assert about the diagonal: asymmetric doesn't care about what pairs there might be along the diagonal, whilst anti-symmetric insists that there are no pairs along the diagonal, which is irreflexivity.
I've just came back from my Mathematics of Packing and Shipping lecture, and I've run into a problem I've been trying to figure out. Let's say I have a rectangle of length $l$ and width $w$. Is there a simple equation that can be used to show me how many circles of radius $r$ can be packed into the rectangle? So that no circles overlap. ($r$ is less than both $l$ and $w$) I'm rather in the dark as to what the optimum method of packing circles together in the least amount of space is, for a given shape. An equation with a non-integer output is useful to me as long as the truncated value is the true answer.
How many circles of a given can be packed into a given rectangular box?
I've just came back from my Mathematics of Packing and Shipping lecture, and I've run into a problem I've been trying to figure out. Let's say I have a rectangle of length $l$ and width $w$. Is there a simple equation that can be used to show me how many circles of radius $r$ can be packed into the rectangle? So that no circles overlap. ($r$ is less than both $l$ and $w$) I'm rather in the dark as to what the optimum method of packing circles together in the least amount of space is, for a given shape. An equation with a non-integer output is useful to me as long as the truncated value is the true answer. (I'm not that interested in how the circles would be packed, as I am going to go into business and only want to know how much I can demand from the packers I hire to pack my product)
How many circles of a given radius can be packed into a given rectangular box?
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?
I've just came back from my Mathematics of Packing and Shipping lecture, and I've run into a problem I've been trying to figure out. Let's say I have a rectangle of length $l$ and width $w$. Is there a simple equation that can be used to show me how many circles of radius $r$ can be packed into the rectangle, in the optimal way? So that no circles overlap. ($r$ is less than both $l$ and $w$) I'm rather in the dark as to what the optimum method of packing circles together in the least amount of space is, for a given shape. An equation with a non-integer output is useful to me as long as the truncated (rounded down) value is the true answer. (I'm not that interested in how the circles would be packed, as I am going to go into business and only want to know how much I can demand from the packers I hire to pack my product)
Is there symbolism (or even a name) for groups of statements in which any fixed-number of them imply all the rest? For example, in linear algebra, a basis is sometimes defined as a set of n-dimensional vectors which: * Is linearly independent * Spans the vector space * Contains exactly n elements However, it is then shown that any TWO of these statements being true implies the third is true. How would you write this symbolically? (other than $A \wedge B \Rightarrow C, A \wedge C \Rightarrow B, B \wedge C \Rightarrow A$, which does not scale well...)
Given 1. A straight line of arbitrary length 2. The ability to construct a straight line in any direction from any starting point with the "unit length", or the length whose square root of its magnitude yields its own magnitude. Is there a way to geometrically construct (using only a compass and straightedge) the a line with the length of the square root of the arbitrary-lengthed line? What is the mathematical basis? Also, why can't this be done without the unit length line?
Compass-and-straightedge construction of the square root of a given line?
Say there are three jars, $j_1, j_2, j_3$ filled with different binary sequences of length two. The distribution of the binary sequences in each of the jars is given by the $p_i^k(1-p_i)^{n-k}$, where $p_i = \frac{i}{m}$ where m is the number of jars, i is the jar index, k is number of 1's and n is the length of the string. So for three jars we have $p_1 = 0.25, p_2 = 0.5$, and $p_3 = 0.75$ for $j_1, j_2, j_3$ respectively. Here are the sequences and their probabilities for $j_1$ with $p_1 = 0.25$: P(00) = 9 / 16 P(10) = 3 / 16 P(01) = 3 / 16 P(11) = 1 / 16 If I tell you that I have selected a binary sequence and the first element is 1 what is the expected value of X over $p_i$? Well, this can be calculated by looking at each of the jars and adding up the probability of candidate sequences times the value of $p_i$. Thus E(X) = (4/16 * 1/4) + (2/4 * 1/2) + (12/16 * 3/4) = 7/8 = $\left (\frac{1}{4} \right)^{2} + \left (\frac{1}{2} \right)^{2} + \left (\frac{3}{4} \right)^{2}$ So the question is ... what is E(X) when the numbers of jars goes to infinity (or alternatively, when p can take on values between 0 and 1)? Also what happens when the size of the binary strings goes to infinity? Does it have an effect on the outcome? If it does, does the order we take the limits change the answer? And most importantly what is the general case for when I have $s$ 1's and $r$ 0's?, with a continuous p from 0 to 1 and infinite sequences?
Given 1. A straight line of arbitrary length 2. The ability to construct a straight line in any direction from any starting point with the "unit length", or the length whose square root of its magnitude yields its own magnitude. Is there a way to geometrically construct (using only a compass and straightedge) the a line with the length of the square root of the arbitrary-lengthed line? What is the mathematical basis? Also, why can't this be done without the unit line length?
I'm taking a course on Basic Conic Sections, and one of the ones we are discussing is of a parabola of the form y = a x<sup>2</sup> + b x + c My teacher gave me the formula: $x = -\frac{b}{2a}$ as the x coordinate of the vertex. I asked her why, and she told me not to tell her how to do her job. My smart friend mumbled something about it involving calculus, but I've always found him a rather odd fellow and I doubt I'd be able to understand a solution involving calculus, because I have no background in it. If you use something you know from calculus, explain it to someone who has no background in it. Because I sure don't. Is there a purely algebraic or geometrical yet elegant derivation for the x coordinate of a parabola of the above form?
I read [this question][1] the other day and it got me thinking: the area of a circle is $\pi r^2$, which differentiates to $2 \pi r$, which is just the perimeter of the circle. >Why doesn't the same thing happen for squares? If we start with the area formula for squares, $l^2$, this differentiates to $2l$ which is sort of right but only half the perimeter. I asked my calculus teacher and he couldn't tell me why. Can anyone explain??? [1]: http://math.stackexchange.com/questions/625/why-is-the-derivative-of-a-circles-area-its-perimeter-and-similarly-for-spheres
Why does area differentiate to perimeter for circles and not for squares?
The notion (rank-2) "tensor" appears in many different parts of physics, e.g. stress tensor, moment of inertia tensor, etc. I know mathematically a tensor can be represented by a 3x3 matrix. But I can't grasp its geometrical picture — unlike scalar (a number) and vector (an arrow with direction and magnitude) which I can easily see what's going on. How to visualize a tensor?
How to visualize a rank-2 tensor?
I know that the harmonic series 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/n + ... diverges, but what about the alternating harmonic series 1/1 - 1/2 + 1/3 - 1/4 + 1/5 - ... + (-1)^n/n + ...? Does it converge? If so, what is its sum?
I have two WGS-84 coordinates on the earths surface with times $t$ and $t+1$ and would like to calculate a heading vector for those. How do I accomplish that?
How to calculate a heading vector on the earths surface?
I have two WGS-84 coordinates on the earths surface with times $t$ and $t+1$ and would like to calculate a heading vector for those using the "True North" as reference. How do I accomplish that?
Is there an algorithm for working out the best way (i.e. fewest multiplications) of calculating A^n in a structure where multiplication is associative? For example, suppose A is a square matrix. Matrix multiplication is associative, and I can compute A^9 with 4 multiplications: A^2 = AA A^3 = A^2A A^6 = A^3A^3 A^9 = A^6A^3 One method which works is to compute A^(2^i) and use the binary representation of n, but this is not always optimal, e.g. with n=23, we can do it in 6 multiplications: A^2 = AA A^3 = A^2A A^5 = A^3A^2 A^10 = A^5A^5 A^20 = A^10A^10 A^23 = A^23A^3 rather than 7: A^2 = AA A^4 = A^2A^2 A^8 = A^4A^4 A^16 = A^8A^8 A^20 = A^16A^4 A^22 = A^20A^2 A^23 = A^22*A Is there an algorithm which gives the quickest way?
There seems to be no efficient algorithm for this. Quoting Wikipedia on [Addition-chain exponentiation](http://en.wikipedia.org/wiki/Addition-chain_exponentiation): > … the addition-chain method is much more complicated, since the determination of a shortest addition chain seems quite difficult: no efficient optimal methods are currently known for arbitrary exponents, and the related problem of finding a shortest addition chain for a given set of exponents has been proven NP-complete. Even given a shortest chain, addition-chain exponentiation requires more memory than the binary method, because it must potentially store many previous exponents from the chain simultaneously. In practice, therefore, shortest addition-chain exponentiation is primarily used for small fixed exponents for which a shortest chain can be precomputed and is not too large. On the same page, however, it does link to [a reference](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6WH3-45J55SP-1M&_user=10&_coverDate=04%2F30%2F1998&_rdoc=1&_fmt=high&_orig=search&_sort=d&_docanchor=&view=c&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=36cb5100aa8887c3cded79d2dd80b01f) of some methods better than binary exponentiation.
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)][1] 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][2]. 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" :) [1]: http://en.wikipedia.org/wiki/Circle_packing_in_a_square [2]: http://hydra.nat.uni-magdeburg.de/packing/csq/csq.html
I keep hearing questions about whether something is NP-complete, but they never really mention what it is. Why do people care so much about NP-complete problems?
What are NP-complete problems and why are they so important?
To expand on Mau's answer, you should care about NP-complete problems because there is an [entire family][1] 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][2] - finding the shortest path (on a graph) that allows you to visit every city exactly once. - [Bin packing problem][3] - 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][4] - 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][5] - finding the smallest set of vertices such that every edge contains at least one chosen vertice - [Clique][6] - finding that largest group of people who all know each other - [Subgraph isomorphism][7] - does one graph contain a subgraph isomorphic to another? - [Set packing][8] - 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][9] - 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 solved as they are effectively the same as one of theses. 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. [1]: http://en.wikipedia.org/wiki/List_of_NP-complete_problems [2]: http://en.wikipedia.org/wiki/Travelling_salesman_problem [3]: http://en.wikipedia.org/wiki/Bin_packing_problem [4]: http://en.wikipedia.org/wiki/Knapsack_problem [5]: http://en.wikipedia.org/wiki/Vertex_cover [6]: http://en.wikipedia.org/wiki/Clique_problem [7]: http://en.wikipedia.org/wiki/Subgraph_isomorphism_problem [8]: http://en.wikipedia.org/wiki/Set_packing [9]: http://en.wikipedia.org/wiki/Subset_sum_problem
To expand on Casebash's answer: Any problem for which a solution (once found) can be verified as a solution quickly 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 one of them 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][1], one of the [greatest unsolved problems][2] 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! :) [1]: http://en.wikipedia.org/wiki/P%3DNP [2]: http://en.wikipedia.org/wiki/Millennium_Prize_Problems
To expand on Casebash's answer: Any problem for which a solution (once found) can be verified as a solution quickly 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][1], one of the [greatest unsolved problems][2] 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! :) [1]: http://en.wikipedia.org/wiki/P%3DNP [2]: http://en.wikipedia.org/wiki/Millennium_Prize_Problems
Is there a way of taking a number known to limited precision (e.g. 1.644934) and finding out an "interesting" real number (e.g. $\pi^2/6$) that's close to it? I'm thinking of something like Sloane's Online Encyclopedia of Integer Sequences, only for real numbers. The intended use would be: write a program to calculate an approximation to $\sum_{i=0}^\infty 1/n^2$, look up the answer ("looks close to $\pi^2/6$") and then use the likely answer to help find a proof that the sum really is $\pi^2/6$. Does such a thing exist?
Can someone point me to a proof that the set of irrational numbers is uncountable. I know how to show that the set Q is countable but how would you then show that the irrationals are uncountable?
Are there any real online mathematics (applied math, statistics, ...) degree programs out there? I'm full-time employed, thus not having the flexibility of attending an on campus programs. I also already have a MSc in Computer Science. My motivation for a math degree is that I like learning, and am interested in the subject. I've studied through number of OCW courses on my own, but it would be nice if I could actually be able to have my studying count towards something. I've done my share of Googling for this, but searching for online degrees seems to bring up a lot of institutions that (at least superficially) seem a bit shady (diploma mills?).
I was coming back from my Driver's Education class, and something mathsy really stuck out to me. One of the essential properties of a car is its current speed. Or speed at a current time. For example, at a given point in time in my drive, I could be traveling 40 mph. But what does that mean? From my basic algebra classes, I've learned that speed = distance/time. So if I travel ten miles in half an hour, my average speed would be 20 mph ($\frac{10 mi}{25 h}$). But instantaneous velocity...you aren't measuring average speed for a given amount of time. You're measuring instantaneous speed over an...instantaneous amount of time. That would be something like (miles) / (time), where time = 0? Isn't that infinite? And perhaps, in a difference of time = 0, then I'd be traveling 0 miles. So would I be said to be going 0 mph at an instantaneous moment in time? I'd like to be able to tell that to any cops pull me over for "speeding"! But then if miles = 0 and time = 0, then you have 0/0? This is all rather confusing. What does it mean to be going 40 mph at a given moment in time, exactly? (I figured that my problem had numbers in it, and therefore has to do with Maths.)
What does it mean to be going 40 mph (or 64 kph, etc.) at a given moment?
Are there any real online mathematics (applied math, statistics, ...) degree programs out there? I'm full-time employed, thus not having the flexibility of attending an on campus program. I also already have a MSc in Computer Science. My motivation for a math degree is that I like learning and am interested in the subject. I've studied through number of OCW courses on my own, but it would be nice if I could actually be able to have my studying count towards something. I've done my share of Googling for this, but searching for online degrees seems to bring up a lot of institutions that (at least superficially) seem a bit shady (diploma mills?).
I was coming back from my Driver's Education class, and something mathsy really stuck out to me. One of the essential properties of a car is its current speed. Or speed at a current time. For example, at a given point in time in my drive, I could be traveling 40 mph. But what does that mean? From my basic algebra classes, I've learned that speed = distance/time. So if I travel ten miles in half an hour, my average speed would be 20 mph ($\frac{10 mi}{25 h}$). But instantaneous velocity...you aren't measuring average speed for a given amount of time. You're measuring instantaneous speed over an...instantaneous amount of time. That would be something like (miles) / (time), where time = 0? Isn't that infinite? And perhaps, in a difference of time = 0, then I'd be traveling 0 miles. So would I be said to be going 0 mph at an instantaneous moment in time? I'd like to be able to tell that to any cops pull me over for "speeding"! But then if miles = 0 and time = 0, then you have 0/0? This is all rather confusing. What does it mean to be going 40 mph at a given moment in time, exactly? I've heard this explained using this strange art called "calculus" before, and it's all gone over my head. Can anyone explain this using terms I (a High School Algebra and Geometry and Driving student) will understand? (I figured that my problem had numbers in it, and therefore has to do with Maths.)
To expand on Mau's answer, you should care about NP-complete problems because there is an [entire family][1] 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][2] - finding the shortest path (on a graph) that allows you to visit every city exactly once. - [Bin packing problem][3] - 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][4] - 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][5] - finding the smallest set of vertices such that every edge contains at least one chosen vertice - [Clique][6] - finding that largest group of people who all know each other - [Subgraph isomorphism][7] - does one graph contain a subgraph isomorphic to another? - [Set packing][8] - 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][9] - 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. [1]: http://en.wikipedia.org/wiki/List_of_NP-complete_problems [2]: http://en.wikipedia.org/wiki/Travelling_salesman_problem [3]: http://en.wikipedia.org/wiki/Bin_packing_problem [4]: http://en.wikipedia.org/wiki/Knapsack_problem [5]: http://en.wikipedia.org/wiki/Vertex_cover [6]: http://en.wikipedia.org/wiki/Clique_problem [7]: http://en.wikipedia.org/wiki/Subgraph_isomorphism_problem [8]: http://en.wikipedia.org/wiki/Set_packing [9]: http://en.wikipedia.org/wiki/Subset_sum_problem
What are some particularly well-known functions that exhibit pathological behavior at or near at least one value and are particularly useful as examples? For instance, if f'(a) = b, then f(a) exists, f is continuous at a, f is differentiable at a, but f' need not be continuous at a. A function for which this is true is f(x) = x^2 sin(1/x) at x=0.
I asked this question on Stack Overflow but it was closed as "not programming related". So I think this is probably the best place for it... --- I read over the wikipedia [article][1], but it seems to be beyond my comprehension. It says it's for optimization, but how is it different than any other method for optimizing things? An answer that introduces me to linear programming so I can begin diving into some less beginner-accessible material would be most helpful. [1]: http://en.wikipedia.org/wiki/Linear_programming
What is linear programming?
> You have sheets of 42-cent stamps and > 29-cent stamps, but you need at least > $3.20 to mail a package. What is the > least amount you can make with the 42- > and 29-cent stamps that is sufficient > to mail the package? A contest problem such as this is probably most easily solved by tabulating the possible combinations, using 0 through ceiling(total/greater value) of the greater-value stamp and computing the necessary number of the smaller stamp and the total postage involved. The particular example above would be solved with a 9-row table, showing the minimum to be $3.23, made with 7 42-cent stamps and 1 29-cent stamp. Is there a better algorithm for solving this kind of problem? What if you have more than two values of stamps?
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][1], one of the [greatest unsolved problems][2] 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! :) [1]: http://en.wikipedia.org/wiki/P%3DNP [2]: http://en.wikipedia.org/wiki/Millennium_Prize_Problems
The [standard form][1] (and [example][2]) sections pretty well describe what it is. > How is it different than any other method for optimizing things? It's, well, just another method. However, it is somewhat special in that many other optimization algorithms either [use linear programming][3] as part of their solution, or are in reality a [specialized solution to a linear programming problem][4]. In fact, [integer linear programming][5] is NP-complete, meaning that any problem in NP [can be stated as an (integer) linear programming problem][6]. [1]: http://en.wikipedia.org/wiki/Linear_programming#Standard_form [2]: http://en.wikipedia.org/wiki/Linear_programming#Example [3]: http://en.wikipedia.org/wiki/Linear_programming_relaxation [4]: http://en.wikipedia.org/wiki/Flow_network [5]: http://en.wikipedia.org/wiki/Linear_programming#Integer_unknowns [6]: http://math.stackexchange.com/questions/726/what-are-np-complete-problems-and-why-are-they-so-important/729#729
The [standard form][1] (and [example][2]) sections pretty well describe what it is. > How is it different than any other method for optimizing things? It's, well, just another method. However, it is somewhat special in that many other optimization algorithms either [use linear programming][3] as part of their solution, or are in reality a [specialized solution to a linear programming problem][4]. In fact, [integer linear programming][5] is NP-complete, meaning that any problem in NP [can be stated as an (integer) linear programming problem][6]. (this also means solving your typical integer linear programming problem is much* more difficult than if we didn't restrict ourselves to integers..)* [1]: http://en.wikipedia.org/wiki/Linear_programming#Standard_form [2]: http://en.wikipedia.org/wiki/Linear_programming#Example [3]: http://en.wikipedia.org/wiki/Linear_programming_relaxation [4]: http://en.wikipedia.org/wiki/Flow_network [5]: http://en.wikipedia.org/wiki/Linear_programming#Integer_unknowns [6]: http://math.stackexchange.com/questions/726/what-are-np-complete-problems-and-why-are-they-so-important/729#729
The absolute value of a 2 by 2 matrix determinant is the area of a corresponding parallelogram with the 2 row vectors as sides. The absolute value of a 3 by 3 matrix determinant is the volume of a corresponding parallelepiped with the 3 row vectors as sides. Can it be generalized to n-D? The absolute value of a n by n matrix determinant is the volume of a corresponding n-parallelotope?
Say there are three jars, $j_1, j_2, j_3$ filled with different binary sequences of length two. The distribution of the binary sequences in each of the jars is given by the $p_i^k(1-p_i)^{n-k}$, where $p_i = \frac{i}{m + 1}$ where m is the number of jars, i is the jar index, k is number of 1's and n is the length of the string. So for three jars we have $p_1 = 0.25, p_2 = 0.5$, and $p_3 = 0.75$ for $j_1, j_2, j_3$ respectively. Here are the sequences and their probabilities for $j_1$ with $p_1 = 0.25$: P(00) = 9 / 16 P(10) = 3 / 16 P(01) = 3 / 16 P(11) = 1 / 16 If I tell you that I have selected a binary sequence and the first element is 1 what is the expected value of X over $p_i$? Well, this can be calculated by looking at each of the jars and adding up the probability of candidate sequences times the value of $p_i$. Thus E(X) = (4/16 * 1/4) + (2/4 * 1/2) + (12/16 * 3/4) = 7/8 = $\left (\frac{1}{4} \right)^{2} + \left (\frac{1}{2} \right)^{2} + \left (\frac{3}{4} \right)^{2}$ So the question is ... what is E(X) when the numbers of jars goes to infinity (or alternatively, when p can take on values between 0 and 1)? Also what happens when the size of the binary strings goes to infinity? Does it have an effect on the outcome? If it does, does the order we take the limits change the answer? And most importantly what is the general case for when I have $s$ 1's and $r$ 0's?, with a continuous p from 0 to 1 and infinite sequences?
Here's one that popped into my mind when I was thinking about binary search. I'm thinking of an integer between 1 and n. You have to guess my number. You win as soon as you guess the correct number. If your guess is not correct, I'll give you a hint by saying "too high" or "too low". What's your best strategy? This is an easy problem if I always tell the truth: by guessing the (rounded) mean of the lower bound and the upper bound, you can find my number in roughly log<sub>2</sub> n guesses at most. But what if I'm allowed to cheat once? What is your best strategy then? To clarify: if you guess my number, you always win instantly. But I'm allowed, at most once, to tell you "too high" when your guess is actually too low, or the opposite. I can also decide not to lie. Here's a rough upper bound: you can ask each number twice to make sure I'm not cheating, and if I ever give two different answers, just ask a third time and from then on, play the regular game. In this way, you can win with about 2 log<sub>2</sub> n guesses at most. I'm pretty sure that bound can be improved. Any ideas?
I have a 3-D sphere of radius R, centered at the origin. (x<sub>1</sub>,y<sub>1</sub>,z<sub>1</sub>) and (x<sub>2</sub>,y<sub>2</sub>,z<sub>2</sub>) are two points on the sphere. The Euclidean distance is easy to calculate, but what if I were to restrict myself to traveling on the surface of the sphere? Two approaches come to mind: use [arc-length](http://en.wikipedia.org/wiki/Arc_length) in some way, or simply use trigonometry: calculate the angle between the two points and get a distance from that. Will both/either of these methods work? Which would be easier? Somewhat related to [this question](http://math.stackexchange.com/questions/720/how-to-calculate-a-heading-vector-on-the-earths-surface). Maybe it will inspire someone to go answer it!
How do I measure distance on a globe?
I've come back from a Skydiving Training lecture, and I've run into a conceptual problem that I can't seem to grasp. If I'm given a certain, non-varying acceleration, I can graph velocity over time easily. In fact, if the acceleration varies over depending on a known function (constant, polynomial, exponential, trigonometric, etc.), I can do it pretty easily, too, by integrating over time. But consider the case `w(t) = kv` or `w(t) = kv^2`, where `k` is a constant, and the sign of `w` is always so that it is the opposite direction of velocity. The total acceleration will be `g + w(t)` How can I find a (closed form?) equation that will give me `v(t)` (for both cases)? (with `v(0) = 0`) I have tried integrating $\int k v \, dt$ and $\int k v^2 \, dt$, but I get stuck. Somewhere along the substitution of variables, things always fall apart. It's because the acceleration varies depending on the velocity, which varies depending on the acceleration. It's like an infinite feedback loop? Common sense tells me that this should be some sort of exponential decay function, but aside from that, I have no experience dealing with differential equations, which I fear this is. Can anybody show me how?
Wind resistance -- upwards acceleration, typically varies either linearly or quadratically by the current velocity. There is a constant downward acceleration due to gravity. How can we model the velocity over time of a falling object, subject only to wind resistance and downwards gravity? I don't have much experience with differential equations, but I do know that this answer necessarily involves it, so could you possibly explain every step? Thank you.
RHS is a well known test for determining the [congruency of triangles][1]. It is easy enough to prove it works, simply use Pythagorus' theorem to reduce to SSS. I thought that it seems strange that this only works for an angle being 90 degrees - or does it? What if I tried changed the given angle to 89 degrees or 91 degrees, would it still be uniquely identified up to congruence? [1]: http://en.wikipedia.org/wiki/Congruence_(geometry)#Determining_congruence
RHS Congruency test - What makes 90 degrees different?

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