## Hot answers tagged fourier-analysis

41

So I guess my initial intuition was wrong; there is enough "room at infinity" to concoct such a function $f$. The key lemma is
Lemma. Let $m_1,m_2,m_3,\dots$ be an enumeration of the integers. Then there exists an increasing sequence $0 = f_0 \leq f_1 \leq f_2 \leq \dots$ of finitely supported functions $f_n: {\bf Z} \to {\bf R}^+$ such that for any ...

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).

10

Denote $z_k=e^{i\theta_k}$, where $i$ is imaginary unit. Up to some non-zero factor like power of 2, this is the integral of $\prod_j z_j^{-8}\cdot \prod_{j<k}(z_j-z_k)\cdot \prod_j (z_j+z_{j+1})^2 (z_j+z_{j+2})^2$, indices are modulo 9, integral with respect to $\prod d\theta_j$. Only constant term survives after integration, that is, your question is ...

10

I am somewhat confused that, despite saying many true and useful things, no one has said directly that $H^{-1}$ on the circle can be characterized as the set of distributions $\theta$ such that $\sum_{n\in \mathbb Z}(1+n^2)^{-1}\cdot |\theta(x\to e^{inx})|^2<\infty$. (All distributions on the circle are compactly supported, so can be applied to the ...

7

The simplest infinitely differentiable function with compact support is
$f(x)=\exp\left(-1/x(1-x)\right), \; 0<x<1$. This is the first example given in textbooks.
Its Fourier transform is
$$F(z)=\int_0^1e^{-izx}f(x)dx.$$
Is this an "explicit" formula? In not, why is the infinite product in Terence Tao's answer "explicit"?
(I think the word "explicit" ...

4

Note first of all that $H^{-1}$ will contain objects that are not functions, such as the Dirac $\delta$ concentrated at a point. (Think that $2\delta_0=\frac{d^2}{dx^2}|x|$.)
The correct definition is that $H^{-1}$ is the completion of $C^\infty(S^1)$ with respect to the norm $\newcommand{\ii}{\boldsymbol{i}}$
$$\Vert ...

2

Related questions have been considered in some depth by Dan Rockmore and collaborators. For more, check out Rockmore's web page.

2

Mathematica says: $$i \sqrt{\frac{\pi }{2}} \text{csch}\left(\frac{\pi t}{2}\right).$$

2

The function $\sigma\mapsto f(\sigma+it)$ is periodic with period $1$, for any value
of $t$. Therefore
$$f(\sigma+it)=\sum_{n\in Z} c_n(t)e^{2\pi i n\sigma}.$$
It follows that
$$c_n(t)=\int_0^1 f(x+it) e^{-2\pi i n x}\,dx.$$
The $n$-th term in the Fourier expansion is, with $s=\sigma+it$
\begin{multline*}
f_m(s)=f_m(\sigma+it)=c_n(t)e^{2\pi i ...

2

You ask for the spectral analysis of a nonstationary stochastic process. Because the autocorrelation function $C(s,t)$ now depends on the two times $s$ and $t$ separately, and not only on their difference, the power spectral density $P(\omega,\omega')$ will depend on two frequencies, and not just on a single frequency. Alternatively, one can define a power ...

1

For disjoint compact sets $A$ and $B$ in $G$ you can find an open set $U$ containing $A$ and an identity nbd $V$ such that both have compact closures and $UV^{−1}$ is disjoint from $B$. You can find $[0,1]$-valued continuous functions $\phi$ and $\psi$ supported on $U$ and $V$ correspondingly such that $\phi|_A=1$ and $\psi(e)=1$. Then it is easy to check ...

1

in the second transformation the integer $k$ runs from $0$ to $\tilde{N}-1$, not to $N-1$; so you have twice as many data points $X_k$ than you have moments $x_n$, which is perfectly OK (oversampling); to use the conventional formulas for the DCT, just extend the list of moments $x_n$ by padding it with zeros ($x_n=0$ for $N\leq n\leq \tilde{N}-1$, so that ...

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