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## Hot answers tagged fourier-transform

42

From the point of view of physics, Fourier transforms are ubiquitous because they are expansions in eigenfunctions of the derivative operator - and the derivative operator is fundamental in many aspects. Just to give two examples: The derivative operator is the generator of translations (in space or time), and to learn about the natural world, it is crucial ...

27

This is basically a lightly transformed version of Euler's cotangent partial fraction expansion $$\pi \cot(\pi z) = \frac{1}{z} + \sum_{n=1}^\infty \frac{1}{z-n} + \frac{1}{z+n}$$ (the log derivative of his famous sine product formula $\frac{\sin \pi z}{\pi z} = \prod_{n=1}^\infty \big(1-\frac{z^2}{n^2}\big)$). By telescoping series one can rewrite this as ...

25

A simple argument that the Fourier transform cannot be nonnegative for any $m>1$, integer or not, is given in my 1991 paper with Odlyzko and Rush: Noam D. Elkies, Andrew M. Odlyzko, and Jason A. Rush: On the packing densities of superballs and other bodies, Invent. Math. 105 (1991), 613-639. The Fourier transform $f$ of a nonnegative even function ...

24

It suffices that $f$ be (the restriction to $[0,2\pi]$ of) a completely monotone real-valued function defined on $[0,\infty)$. Indeed, then for some finite measure $\mu$ on $[0,\infty)$ and all real $x\ge0$ we have $$f(x)=\int_0^\infty\mu(da) e^{-a x},$$ whence for natural $n$ $$\hat f(n)=\int_0^\infty\mu(da) \int_0^{2\pi}dx\,\cos(nx)e^{-a x} =\int_0^\infty\... 23$$ \hat f(\xi) = \prod_{j=1}^\infty \operatorname{sinc}(\xi/2^j)$$will work (it is rapidly decreasing, and obeys the hypotheses of the Paley-Wiener theorem; alternatively, its inverse Fourier transform can be written explicitly as an infinite convolution that is manifestly compactly supported). 21 The version number 2 is the only one that makes the Fourier transform both a unitary operator on L^2 and an algebra homomorphism from the convolution algebra in L^1 to the product algebra in L^\infty . It is not, however, of widespread use in analysis as far as I know. From the point of view of semiclassical analysis, it amounts roughly speaking to ... 21 I would like to add the point of view from a more general aspect. In the general situation, the Fourier transform can be defined for any locally compact abelian group G. Let G be a locally compact abelian group. Let \hat{G} = \mathrm{Hom}_{\mathrm{cont}}(G, \mathbb{S}^1) be the Pontryagin dual of G, where \mathbb{S}^1 = \{z\in\mathbb{C}: |z|=1\} ... 21 To add a representation theory perspective: if G is a Lie group, and f is a function (or more precisely a distribution) on G then (under certain mild conditions on f and G), the function f is uniquely determined by its unitary matrix coefficients, i.e. the coefficients of the matrix \rho(f) where \rho:G\to GL_n goes over all isomorphism ... 20 Given a square-integrable, positive semi-definite function f, with its Fourier transform \hat{f}, then the function$$F=f^2+\hat{f}\star\hat{f},$$with \star the convolution, is its own Fourier transform: \hat{F}=F. If we require that F is a probability density (absolutely integrable and positive semi-definite), then any F with \hat{F}=F is ... 20 Set A = L^1([0, \infty)), equipped with the structure of a Banach *-algebra via convolution. The spectrum of this algebra is the half plane \text{Re}(z) \geq 0, and the Gelfand transform is the Laplace transform. 18 At the risk of answering my own question, here is what I have since found: For general n, formula (1) seems to occur first on p. 177 of S. Bochner, Summation of multiple Fourier series by spherical means, Trans. AMS 40 (1936) 175-207. Bochner exposes it again on pp. 73-74 of Fourier Transforms (Princeton UP 1949). For n=3, Burkhardt (Trigonometrische ... 17 Yes, \mathscr O_M is bornological. Grothendieck calls this "not trivial" and proves it in his thesis via a tensor product representation and quite general results about tensor products. Together with Julian Larcher, I recently gave an alternative proof which is still in the refereeing process. If you are interested I can send you the preprint by email. ... 16 The exponentials used in Fourier series are eigenvalues of shifts, and thus of any operator commuting with shifts, not just Laplacian. Similarly, spherical harmonics carry irreducible representations of SO(3), and so they are eigenfunctions of any rotationally invariant operator. If the underlying space has symmetries, it's no wonder that a basis ... 16 it is positive for m=1, but not for m=2, see this Mathematica output: 16 I think the result goes back to Polya, see "Some theorems on stable processes" by Blumenthal and Getoor. Another reference is Paul Lévy "Sur une application de la dérivée d’ordre non entier au calcul des probabilités" page 1118 of CRAS 1923. It might be good to also give the gist of Levy's very simple argument: if the Fourier ... 16 please show me how to prove for |\omega|>20 With great pleasure. We shall just show that F(y)=\int_0^\infty e^{-x^a}x^a\log x\cos(yx)\,dx>0 for large enough y>0. Note that \cos(yx)=\Re e^{iyx}, so we have the real part of a contour integral from 0 to +\infty of e^{-z^a}z^a\log z e^{iyz}\,dz. The integrand oscillates like crazy on the ... 11 The op$$T_x = \frac{D_x}{e^{D_x}-1} = e^{b.D_x},$$where (b.)^n = b_n are the Bernoulli numbers, is (mod signs) often referred to as the Todd operator (maybe originally given that name by Hirzebruch, who used it to construct his Todd characteristic class). It has a discretizing (or derivational) property that can be expressed in the following useful ways ... 10 \def\Z{\mathbf{Z}}If N is prime and f(x) has the form \zeta^{g(x)}, where \zeta is an Nth root of unity and g is some function from \Z/N\Z to \Z, then g must be quadratic. The following argument is very similar to the argument given in the answer https://mathoverflow.net/a/136046/20598, so I feel at liberty to be brief on details. For ... 10 The exercise is stated in Dym and McKean with a mistake. The correct statement is If f is real, even, the finite limits f(x\pm 0) exist for all x, and \hat{f} does not change sign for |x|>A, then f is continuous at all points. The statement is in the paper of M. Kac (1938) to which Dym and McKean refer. The proof is simple: inverse Fourier ... 10 The following example is interesting$$\widehat{\sin(x)\over x \sqrt{|x|}} = \sqrt{2\pi} \ \left\{ { \matrix{ \scriptstyle \sqrt{|\xi+1|} + \sqrt{|\xi-1|} & \hbox{ if } |\xi| \leq 1, \cr {2\over \sqrt{|\xi+1|} + \sqrt{|\xi-1|}} & \hbox{ if } |\xi| \geq 1.\cr}} \right.$$because it is in ... 10 Well you seem to have worked it out but I wrote most of this before your comment happened: I claim there isn't any material difference between your "Minkowski space Fourier transform" and the usual Fourier transform on {\mathbb R}^n: in fact write$$ \hat f(\xi)\equiv \int e^{i\eta(x,\xi)} f(x) dx $$for any non-degenerate bilinear form \eta. Then there ... 9 Care should be taken because of the pole at k=0, let me first take the principal value of the integral. I note that I(\alpha,-r)=\bar{I}(\alpha,r) (complex conjugate), for convenience I will restrict myself to r>0. The principal value integral evaluates to$$I(\alpha,r)=\int_{-\infty}^\infty dk\, e^{ikr} \cfrac{\alpha^2 + \beta k^2}{k(k^2+\alpha^2)}...

9

From the point of view of engineering, sin and cos are eigenfunctions of LTI (linear time-invariant) systems, which makes the Fourier transform immanently important for system theory - and thus for control theory, signal processing and many other fields that make use of LTI systems

9

Note: I am not sure if I understand the word "converges" correctly. This is completely analogous to the similar question regarding convergence of Fourier series, which is classical. Let $$g(x,r) = \int_{-r}^r \hat f(\zeta) e^{2\pi i \zeta x} d\zeta$$ by "partial sums" of the inverse Fourier transform, and denote by $$h(x, r) = \int_0^1 g(... 9 This is true. This was proved by Hardy and Littlewood and the proof is reproduced in Zygmund's Trigonometric Series (which I don't have access to at the moment). (Contradicting my prior "answer") The answer to this question is negative. Such counterexamples are known as "Pauli partners" and are studied in, among other places, the quantum ... 8 The correct formula is, for \chi primitive of conductor q,$$ \sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) = \frac{A}{x}\sum_{n\in\mathbb{Z}}\bar\chi(n)\hat f\left(\frac{n/x}{\sqrt{q}}\right). $$Here A:=\sqrt{q}/\tau(\bar\chi) is the so-called root number, it is of modulus 1. This formula is essentially equivalent to the ... 8 I personally consider a "character" to be function on the group, whereas you speak about it living on the convex hull of the weights, which is in the dual of the Lie algebra of the torus. Anyway, the point is that you already understand Heckman's asymptotics of the weight multiplicity function on the weight lattice, as a measure on the vector space ... 8 As noted in my comment above, the Poisson summation formula would give$$ 1= f(0) = \sum_{n\in {\Bbb Z}} f(n) = \sum_{k\in {\Bbb Z}} {\hat f}(k) \ge {\hat f}(0),  since ${\hat f}(k) \ge 0$ by assumption. Since ${\hat f}(0) = \int_{-1}^{1} f(x) dx$, this proves the desired bound. If you are worried about using the Poisson summation formula on just a ...

8

The space $L^2(a,b)$ is a Hilbert space of infinite dimension. Therefore there is an element $f\neq 0$ of this space which is orthogonal to $e^{i\lambda_j x}$. Take this $f$. You can also find such $f$ is any finite dimensional subspace whose dimension is $>n$. Just choose a basis and solve a system of linear equations.

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