Greg Graviton
• Member for 11 years, 4 months
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## 36 Answers

20 answers
165 votes
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145 votes

I always liked the derivation of Taylor's formula with error term: $$\begin{array}{rl} f(x) &= f(0) + \int_0^x f'(x-t) \,dt\\ &= f(0) + xf'(0) + \int_0^x tf''(x-t)\,dt\\ &... View answer 6 answers 165 votes 29k views 58 votes Mathematicians and physicists use very different languages when they talk about tensors. Fortunately, they are talking about the same thing, but unfortunately, this is not obvious at all. Let me ... View answer 9 answers 90 votes 29k views 46 votes This is a difficult question to answer, mainly because any advice must be very personal to be useful for you. I'll try anyway. Before learning mathematics, one has to learn how to learn mathematics. ... View answer 7 answers 77 votes 23k views 34 votes Go back one step and add the defining equation for i to the ideal. In other words, consider your ring as a quotient of the ring of polynomials \mathbb Z[x]:$$ \mathbb Z[i] / (3-i) = \mathbb Z [x]...

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Here a worked out example: What is the Lie algebra of the group of rotations in 3-dimensional space, $SO(3)$? Matrices $A\in SO(3)$ are defined by the property that they are invertible and that the ...

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If you want an intuition that explains why Farka's Lemma should be true, you will have to use the geometric interpretation; there's no way around that. If you want an intuition that shows what Farka'...

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Monads in Haskell and monads in category theory are very much the same: A monad consists of a functor $T: C \to C$ and two natural transformations $\eta_X : X \to T(X)$ (return in Haskell) and $\mu_X :... View answer 3 answers 14 votes 4k views Accepted answer 12 votes To sum up the comments: when Poincaré worked on the beginnings of algebraic topology, he originally thought that a space with trivial homology groups must be contractible. (More precisely, he thought ... View answer 3 answers 10 votes 4k views Accepted answer 11 votes It turns out that there are two different but related notions of differentiation for a function$f:\mathbb R^n\to\mathbb R$: the total derivative$df$and the gradient$\nabla f$. The total ... View answer 11 answers 37 votes 15k views 11 votes The beauty of straightedge and compass constructions, as opposed to the use of, say, a protractor, is that you don't measure anything. With ruler and compass you can bisect an angle without knowing ... View answer 5 answers 58 votes 7k views 11 votes Determinants are best understood in the context of exterior algebra, which goes back to the work of Hermann Grassmann. Here a down-to-earth description of the intuition behind it. Consider an$n$-... View answer 1 answers 5 votes 3k views Accepted answer 9 votes One way is to note that$x_1y_2 - x_2y_1$is a signed area, i.e. it may positive or negative. Adding up all the signed areas of the triangles formed by the points$O$,$P_k$and$P_{k+1}$will cancel ... View answer 2 answers 17 votes 604 views 8 votes I'll answer the first question by considering a slightly different problem where it is easier to explicitly construct an irrational number not in$S_\epsilon$. Namely, consider the interval$(0,1)$. ... View answer 3 answers 13 votes 1k views Accepted answer 8 votes It was pointed out in the comments that you can interpret the matrix as a bilinear form and obtain the inequality $$x^T A x = \cos(\pi/N) \sum_{i=1}^N x_i^2 - \sum_{i=1}^N x_i x_{i+1} \cos\theta_i \... View answer 2 answers 3 votes 347 views Accepted answer 6 votes You don't need to assume that g is non-zero, and it could be strictly decreasing as well. Furthermore, the conditions on g only need to hold on the image of f (which doesn't need to the be whole ... View answer 1 answers 6 votes 208 views Accepted answer 5 votes The Weyl equidistribution theorem says that for irrational \alpha and sufficiently many k, the fractional parts \{\alpha k\} will be equidistributed in the interval [0,1]. To apply this to ... View answer 1 answers 1 votes 994 views Accepted answer 5 votes If all entries of A^{-1} are positive numbers, then A has the property you desire. (Edit: This condition is both sufficient and necessary. If one entry A^{-1}_{ij} is negative, then the choice ... View answer 1 answers 6 votes 483 views Accepted answer 4 votes Concerning the physical meaning, I take it that f_1 and f_2 represent the fractions of the two phases in the alloy (this implies f_1 + f_2 = 1). I imagine x_1 and x_2 to correspond to an ... View answer 2 answers 5 votes 2k views 4 votes There exists a bijection between the one-point compactification \mathbb{N}\cup\lbrace\infty\rbrace and \mathbb N, for instance by mapping$$ \infty \mapsto 0 \text{ and } n \mapsto n+1 .$$Use ... View answer 2 answers 4 votes 1k views Accepted answer 4 votes The identity$$ \int_V d\vec x\, \langle \nabla \phi, \nabla u \rangle = -\int_V d\vec x\ \phi\Delta u $$is a consequence of integration by parts and the divergence theorem. Namely, we have$$ \... View answer 2 answers 7 votes 2k views Accepted answer 4 votes Apparently, Mario Serna has produced pictures of$U(1)$-bundles on his webpage and in his paper "Riemannian Gauge Theory and Charge Quantization". Here an example The image represents a trivial$\...

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3 votes
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I don't know much about its utility in economics, but Fourier analysis is invaluable in physics. The main reason for that is that it transforms linear differential equations into simple algebraic ...

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3 votes

Here a fancy proof that uses operators and differentiation. Let $T_a$ be the translation operator which maps polynomials to their translates $$T_a p(x) = p(x + a) .$$ Restricting attention to ...

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Algebra is a very large field, so you probably want to be a bit more specific. In case you are wondering about Galois Theory, and want to learn its history for the purpose of understanding it, I ...

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As other answers have noted, you are about to discover the calculus of finite differences. For practical calculations, here a most useful fact: the rule $\frac{d}{dx} x^n = n x^{n-1}$ corresponds ...

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540 views
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Since it's an equation of degree 2, the curve in question is a conic section. Looking at the coefficients, it's probably an ellipse. (The criterion for that is that the quadratic form should always be ...

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Hint: removing $n$ points will also give something that only consists of 1-dimensional things. You can also use van Kampen's theorem to calculate the fundamental group directly. EDIT: Err, you can't, ...

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It's not possible if you want the isomorphism to be compatible with the product structure. Namely, choose $A=B$ and $H = \lbrace (a,a) : a\in A\rbrace$ the "diagonal subgroup". Clearly, the subgroup $... View answer 1 answers 1 votes 147 views 2 votes This is simply the triangle inequality:$\$ \lVert\frac{dC}{dt}\rVert = \lVert\frac{dC}{dt} + \beta - \beta\rVert \le \lVert\frac{dC}{dt} + \beta \rVert + \lVert-\beta\rVert < N + \lVert\beta\rVert ...

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Concerning the Hopf fibration, the film "Dimensions" does a superb job of visualizing it (chapters 7-8 in the table of contents). Concerning the visualization of the glueing, maybe it helps to ...

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