New answers tagged

2

Yes. In fact one derivative is enough. The Fourier transform of $f'$ is $itF(t)$ (or something like that, depending on how your definition of the Fourier transform is normalized). Since $f'\in L^1$ this shows that $tF(t)$ is bounded. So $$\int_{|t|>1}|F(t)|^2\,dt\le c\int_{|t|>1}\frac{dt}{t^2}<\infty.$$ We certainly have $\int_{|t|\le1}|F(t)|^2\,dt&...


1

by using Fourier series of $f(x)=1, x\in[0,1]$ $$1=\sum_{n=1}^\infty\frac{4}{(2n-1)\pi}\sin (2n-1)\pi x$$ integrate both sides when integration limits are $x=0 \rightarrow 1$ $$\int_{0}^{1}1.dx=\int_{0}^{1} \sum_{n=1}^\infty\frac{4}{(2n-1)\pi}\sin (2n-1)\pi x dx$$ $$1=\sum_{n=1}^\infty\frac{8}{(2n-1)^2\pi^2}$$ $$\sum_{n=1}^\infty\frac{1}{(2n-1)^2}=\frac{\pi^...


0

See my answer to your other question: Riesz basis of Paley-Wiener space.. Consider the Paley-Wiener space $$ PW_{\frac{1}{2}} := \{ f \in L^2 (\mathbb{R}) \; | \; \text{supp} \;\hat{f} \subseteq [-\frac{1}{2}, \frac{1}{2}] \} $$ Then the collection $\{\text{sinc} (x - n) \}_{n \in \mathbb{Z}}$ is an orthonormal basis for $PW_{\frac{1}{2}}$. ...


1

The idea being discussed doesn't need a rigorous definition of wavelet, or even to use wavelets at all; the notion of a wavelet is, I think, mainly a clever way of generating a convenient basis in a systematic fashion starting from a basic shape. (the particular basis you listed could probably stand to be scaled so as to be orthonormal basis rather than just ...


1

The argument establishing the inequality for $\varphi\in\mathcal S$ would work just as well if it had been assumed only that $\varphi\in L^1$, but the author wanted to focus on $\mathcal S$ for a reason. The space $\mathcal S$ is actually closed under the Fourier transform. So $\widehat\varphi$, the Fourier transform of $\varphi\in\mathcal S$, is ...


1

Let $\tau_t g$ denote, for a fixed $t$, the function $x \mapsto g(x-t)$. We have: $$F_g f (\omega, t) = \int_{\Bbb R} f(x) \overline{\tau_t g(x)} e^{-i\omega x } dx = \mathcal F({f\overline{\tau_t g}})(\omega)$$ We know that $\mathcal F: L^2 \to L^2$ is an isometry up to a $2\pi$ factor, so (wrt $\omega$) $\|F_g f\|_{L^2(\Bbb R)}^2 = 2 \pi \|f \overline{\...


1

If $s_n$ converges uniformly on a set of full measure then $s_n$ converges uniformly, period. (If $|s_n-s_m|<\epsilon$ on $E$ then $|s_n-s_m|\le\epsilon$ everywhere.) So at a minimum you need $f=g$ almost everywhere, with $g$ continuous. Of course that's far from sufficient. But the question is really about uniformly convergent Fourier series, without ...


1

This is true assuming just that $f$ is bounded and has one-sided limits at the origin. The simplest proof is by a cheap trick: A change of variables shows that $$f*K_\epsilon(0)=\int f(\epsilon t)K_1(-t)\,dt.$$Now apply dominated convergence...


1

Suppose $f\in L^1(\mathbb{R})$. For $0 < R < \infty$ and $f \in L^1(\mathbb{R})$, the function $$ \hat{f_R}(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}f(t)e^{-ist}dt $$ is infinitely differentiable with $$ \hat{f_{R}}^{(n)}(s)=\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}f(t)(-it)^{n}e^{-ist}dt. $$ Furthermore, one has the uniform ...


1

If $f\in L^2(\mathbb{R})$, then $f \in L^1[-R,R]\cap L^2[-R,R]$ for any $0 < R < \infty$. Consequently $$ \hat{f_{R}}(s)=\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}f(t)e^{-ist}dt \in L^2(\mathbb{R}) $$ is a continuous function of $s$, and, by Parseval's identity, $$ \left\|\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}f(t)e^{-ist}dt\right\|_{L^2}^2=\int_{-R}^{R}...


1

Following proof rely on this integral identity : $$\int_{a}^{1}\frac{\arccos x}{\sqrt{x^2-a^2}}\mathrm{d}x=-\frac{\pi}{2}\ln a\qquad ;\,a\in(0,1]$$ We will prove it later on. Now, let's make a power series : $$\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}=\int_0^1\frac{1}{x}\sum_{n=1}^{\infty}\frac{x^n}{n}\,\mathrm{d}x=-\int_0^1\frac{\ln(1-x)}{x}\,\mathrm{d}x=...


2

Let $X$ be an independent Laplace random variable with $X\sim L(0,1) = \frac12 \exp{(-|x|)}$, then its characteristic function : $$\varphi_X(t)=\mathbb{E}[e^{itX}]=\frac{1}{1+t^2} \newcommand{\var}[1]{\mathrm{var}\left[#1\right]}$$ By symmetry $\mathbb{E}[X]=0$ we write (generally) : $$\varphi_X(t)=\mathbb{E}[e^{itX}]=\mathbb{E}[1+itX-t^2X^2+\cdots\,]=1-\...


1

If you break everything into intervals, then you do have a proper orthogonal analysis. For example, $$ f(t)=\sum_{n=-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}\int_{(n-1/2)\delta}^{(n+1/2)\delta}\hat{f}(\omega)e^{i\omega t}d\omega = \sum_{n=-\infty}^{\infty}f_n(t). $$ In this case, $$ (f_n,f_m)=\int_{-\infty}^{\infty}f_n(t)\overline{f_m(t)}dt=0,\;...


1

Your density lemma says it suffices to prove the result for a function of the form $q(x)=1_{[a,b]}(x)$ $[a,b] \subset [0,2\pi]$. Now \begin{align*} \int _{0}^{2\pi} q(x) p(nx) d x &= \int _{0}^{2n\pi} \frac{1}{n}q(x/n)p(x) d x \\ &= \sum _{k=0}^{n-1} \int _{2(k)\pi}^{2(k+1)\pi}\frac{1}{n} q(x/n)p(x) d x \\ &= \sum _{k=0}^{n-1} \int _{0}^{...


3

The theorem you stated is definitely related to the question. Let $s$ be a step function so that $\| p-s\|_{L^1} <\epsilon$, then $$\bigg|\int_0^{2\pi} p(x) q(nx) - \int_0^{2\pi}s(x) q(nx)\bigg|\le C\int_0^{2\pi}|p(x)-s(x)| <C\epsilon,$$ where $C$ is the bounded of $q$, that is $|q(y)|\le C$ for all $y\in [0,2\pi]$. Now if you have shown the theorem ...


2

If you differentiate, \begin{align} \left(-\frac{d^2}{ds^2}+a^2\right)\int_{0}^{\infty}\frac{1}{t(t^2+a^2)}\sin(st)dt &= \int_{0}^{\infty}\frac{\sin(st)}{t}dt \\ &=\int_{0}^{\infty}\frac{\sin(st)}{(st)}d(st) \\ &=\int_{0}^{\infty}\frac{\sin(u)}{u}du=\frac{\pi}{2}. \end{align} Therefore, $f(s)=\int_{0}^{\infty}\frac{1}{t(t^2+a^2)}\sin(st)...


2

When doing the change of variable $u = sx$, it is implicitly assumed that $s \neq 0$ (otherwise, the integral evaluates to $0$). This means that the middle expression of $y$ and the subsequent chain of differentiations (hence, also, the obtained expression for $y''$) are only valid for $s \neq 0$. To obtain a formula for all $s$, we must differentiate: $$y =...


0

Properties of wavelet: Wavelets have localized property w.r.t both frequency and space variable. It oscillates depending its vanishing moment. Application to signal processing: Fourier transform can not give information of signal with irregular structure. Actually Fourier transform gives the global information while wavelet transform gives the local ...


0

We may compute the Fourier transform of $g(x)=\log(x)\cdot\mathbb{1}_{x>0}$ as the following limit: $$ \widehat{g}(\xi)=\lim_{\lambda\to 0^+}\int_{0}^{+\infty}\log(x)e^{-i\xi x}e^{-\lambda x}\,dx=\frac{i\gamma}{\xi}-\frac{\pi}{2|\xi|}+\frac{i}{\xi}\log(|\xi|)\tag{1}$$ then compute the Fourier transform of $h(x) = e^{-x^2}$: $$ \widehat{h}(\xi) = \int_{-\...


1

For part (3), $$ ||g_n||_1=\int_{\mathbb{R}}\frac{|\sin(2\pi x)\sin(2\pi nx)|}{\pi^2x^2}\;dx=\frac{2n}{\pi}\int_{\mathbb{R}}\frac{|\sin(x)\sin(\frac{x}{n})|}{x^2}\;dx$$ $$ =\frac{4n}{\pi}\int_0^{\infty}\frac{|\sin(x)\sin(\frac{x}{n})|}{x^2}\;dx\geq \frac{4n}{\pi}\int_0^{n}\frac{|\sin(x)\sin(\frac{x}{n})|}{x^2}\;dx$$ Now $\sin t\geq t-\frac{t^3}{6}\geq \frac{...


2

I suggest you start from $F(-\nu)$ : $$F(-\nu) = \int_{-\infty}^\infty f(t) e^{j2 \pi \nu t} \;dt \;=\; \int_{\infty}^{-\infty} f(-u) e^{-j2 \pi \nu u} \,\;(-du) \;=\; \int_{-\infty}^{\infty} -f(u) e^{-j2 \pi \nu u} \,\;du \;=\; -F(\nu)$$ where I used the change of variables $u=-t$.


1

We consider the case $n=2$; the general case is the same except for notation. Define two partial Fourier transform operators $G_1$ and $G_2$ by $$G_1f(s,y)=\frac{1}{\sqrt{2\pi}}\int f(x,y)e^{-isx}\,dx$$and $$G_2f(x,t)=\frac{1}{\sqrt{2\pi}}\int f(x,y)e^{-ity}\,dy.$$Define $G_1^*$ and $G_2^*$ similarly, without the minus signs in the exponent. The one-...


1

Make the change of variables $x=t^2$ in both integrals in $\hat f(\omega)$ and use parity to extend the limits of integration to $-\infty$ and $\infty$. The things will then become much more clear.


1

\begin{align*} R\left(\omega\right) & =\sum_{k=-9}^{9}\left(10-\left|k\right|\right)e^{-ik\omega}=\left(\frac{\sin\left(5\omega\right)}{\sin\left(\omega/2\right)}\right)^{2} \end{align*}


4

First, note that we have$$|\mathcal{F}f(\xi)| = \left|\int_{\mathbb{R}^d} f(x)e^{-ix \cdot \xi}dx\right| \le \int_{\mathbb{R}^d} |f(x)|\,dx = 1 = \mathcal{F}f(0),$$where the final equality comes from the fact that $f \ge 0$. Since $f$ is real-valued, we may decompose $\mathcal{F}f(\xi)$ into its real and imaginary parts as$$\mathcal{F}f(\xi) = \int_{\mathbb{...


2

You should define all your terms. I presume $\cal F$ is the Fourier transform. The standard formula is that $$\cal F f(x) = \int_{\mathbb R} e^{-2\pi i x y} f(y) \, dy.$$ Since $2\pi i x y$ is purely imaginary or zero, $|e^{-2\pi i xy}| = 1$. Apply the triangle inequality to get $$|\cal F f(x)| \le \int_{\mathbb R} |e^{-2\pi i x y} f(y)| \, dy = \int_{\...


1

The function $F$ is continuous and of bounded variation because $$ V_{a}^{b}(F) = \int_{a}^{b}|f(t)|dt. $$ Therefore the Fourier series for $F$ converges pointwise everywhere to $F$ (in fact it must converge uniformly.) In particular, $$ F(0) = \lim_N\sum_{n=-N}^{N}\hat{F}(n)=2\lim_N\sum_{n=1}^{N}\frac{\hat{f}(n)}{in}. $$


1

Two more solutions. One in the spirit of this answer: \begin{align} \int_{-\infty}^\infty\frac{\sin^2x}{x^2}\mathrm e^{\mathrm itx}\,\mathrm dx &= \int_{-\infty}^\infty\frac{\sin^2x\cos tx}{x^2}\,\mathrm dx \\ &= \frac14\int_{-\infty}^\infty\frac{\cos(t+2)x-2\cos tx+\cos(t-2)x}{x^2}\,\mathrm dx \\ &= \frac14\int_{-\infty}^\infty\frac{(t+2)\sin(t+...


1

It's very atypical to be able to compute the eigenfunctions of a general integral operator explicitly. What is special about convolution operators is that they exhibit translation symmetry, which is why periodic functions "diagonalize" them. Wavelets are known to almost diagonalize certain kinds of nonstationary operators - this is discussed in Meyer's ...


1

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1

Another approach. Assume without loss of generality that $a > 0$, $b > 0$. Then \begin{align} (f_a * f_b)(x) &= \int_{-\infty}^{\infty}\frac{1}{(x-t)^2+a^2}\frac{1}{t^2+b^2}dt \\ &=\int_{-\infty}^{\infty}\frac{1}{2ia}\left[\frac{1}{t-x-ia}-\frac{1}{t-x+ia}\right]\frac{1}{2ib}\left[\frac{1}{t-ib}-\frac{1}{t+ib}\right]dt. \end{align} The ...


1

Following up on my comment, suppose we switch to coordinates $(x',y')=(x,y-mx)$. Then the stripe becomes $X'=\{(x',y')|-\frac12 t\leq y' \leq \frac12t\}$ and the Fourier transform takes the form $$\mathcal{F}[f](u,v) = \frac{1}{2 \pi}\int_{-\infty}^{\infty} f(x',y') e^{i(ux'+v(y'+mx'))}dx'dy'=\frac{1}{2 \pi}\int_{-\infty}^{\infty} f(x',y') e^{i(u+mv)x'+...


0

I see no reason to do anything substantially different from what rschwieb suggested first in his answer. Perhaps these things have a different name already, but I do not think there is any merit in introducing completely new names for things that are (sort of) well-known. Let $(A,\cdot)$ be a semigroup and $x,a\in A$. Then $x$ is an $n$-th root of $a$, if $...


2

One of the basic properties of Fourier series is localization: If $f=0$ in a neighborhood of $t$ then the Fourier series for $f$ converges to $0$ at $t$. Hence if $f=g$ in a neighborhood of $t$ then the two Fourier series either both converge at $t$ or both diverge at $t$. So: Given $f_1$ with a divergent series at $t_1$ and $f_2$ with a divergent series at ...


4

By Parseval's theorem we have $$\begin{align} \int_{-\infty}^{\infty}\frac{\sin^{4}\left(x\right)}{x^{4}}dx= & \pi\int_{-\infty}^{\infty}\frac{\sin^{4}\left(\pi t\right)}{\left(\pi t\right)^{4}}dt \\ = & \pi\int_{-1}^{0}\left(1+x\right)^{2}dx+\pi\int_{0}^{1}\left(1-x\right)^{2}dx \\ = & \frac{2}{3}\pi \end{align} $$ and now observe that $$\...


2

Fourier transform of $f_a$ is: $$\hat f_a = \frac 1a\sqrt{\frac\pi 2}e^{-a|\omega|} \tag 1$$ The fourier transform of a convolution is a regular product: $$\mathscr F\Big[ f_a\star f_b \Big] = \hat f_a\cdot\hat f_b = \frac 1a\sqrt{\frac\pi 2}e^{-a|\omega|} \cdot \frac 1b\sqrt{\frac\pi 2}e^{-b|\omega|} = \frac 1{ab}\frac\pi 2e^{-(a+b)|\omega|} \tag 2$$ The ...


2

The minimum set of conditions needed can be expressed using the Lebesgue integral. Essentially you need $f$ to be absolutely continuous with derivative $f' \in L^1$. Absolutely continuous means $f$ is the integral of its derivative: $$ f(y)-f(x)=\int_{x}^{y}f'(t)dt. $$ This all works out very nicely using the Lebesgue integral. For the Riemann ...


0

I am also learning this stuff at the moment so I'll let you know what I think. I don't see how you would use closed graph theorem, in fact, you would need either $\mathcal{S}=\mathcal{S}(\mathbb{R}^d)$ to be a Banach space or compact Hausdorff. I do not believe either conditions are satisfied ($\mathcal{S}$ is normable or compact, at least I don't think so). ...


0

The shift is defined by $g_a(x) = f(x-a)$. Then you write $$F[g_a](\xi) = \int_{\Bbb R}g_a(x)\exp(-ix\xi)dx = \int_{\Bbb R}f(x-a)\exp(-ix\xi)dx.$$ Conceptually, you first apply the shift and then apply the Fourier transform, but you can apply the shift only to the function, there is no sense in applying it to the exponent.


0

Let $g_0 = \frac{1}{2\pi}\int_{0}^{2\pi}g(x)dx$ and $f_0=\frac{1}{2\pi}\int_{0}^{2\pi}f(x)dx$. Then \begin{align} \frac{1}{2\pi}\int_{0}^{2\pi}f(x)g(nx)dx &=\frac{1}{2\pi}\int_{0}^{2\pi}f(x)(g(nx)-g_0)dx+f_0g_0 \\ &=\frac{1}{2\pi}\sum_{k=1}^{n}\int_{\frac{2\pi}{n}(k-1)}^{\frac{2\pi}{n}k}f(x)(g(nx)-g_0)dx+f_0g_0 \\ &=\frac{1}{...


2

Cosine is an even function; sine is odd. Using negative indices for these functions adds no information because, for example, $$5\sin 2x+ 4\sin(-2x) = \sin 2x$$ On the other hand, $e^{-2ix}$ is not just a constant multiple of $e^{2ix}$. And we need both of these to express $\sin 2x$, as in $$\sin 2x = \frac1{2i}(e^{2ix} - e^{-2ix})$$ A more precise ...


0

After a simple classification I have figured it out. We could solve it by induction on the degree of the polynomial $P(x)$. If $deg P(x)=1$, then $c_1 \in \mathbb{R} \setminus \mathbb{Q}$ and apparently $<c_1n+c_0>$ is equidistributed in $[0,1)$. Assume $deg P(x)=m$ and at least one of its coefficient $c_1,……,c_m$ is irrational, then $<P(n)>$ ...


1

You are trying to compute an NTT of length $n = 2$. You chose the modulus $p = 5$ with a primitive root $w = 2$, which are appropriate so far. The primitive root $w$ has order $p - 1 = 4$, which means $2^4 \equiv 1 \mod 5$. But what you want is an $n$th root of unity $a$, where $a^n = a^2 \equiv 1 \mod 5$. The easiest way to obtain this is to compute $a = w^...


3

We suppose the choice $m = j$, $C = j$ does not work for any $j$ and derive a contradiction. There thus exist Schwartz functions $\phi_j$ such that $|u(\phi_j)| > j\|\phi_j\|_j$, so if we define$$\psi_j = {1\over{j\|\phi_j\|_j}}\phi_j,$$it is clear that $|u(\psi_j)| > 1$ while $\|\psi_j\|_j = 1/j$. To show that $\psi_j \to 0$ in $\mathcal{S}(\mathbb{R}^...


2

We would still use the fast Fourier transform (FFT), but with interpolation. As pointed out, the signal can be written as$$s(t) = \int_{-\infty}^\infty d\nu\,\hat{s}(\nu)e^{i2\pi\nu t} = -\int_{-\infty}^\infty d\lambda {{\hat{s}(1/\lambda)}\over{\lambda^2}} e^{i2\pi t/\lambda},$$or$$s(t) = \int_{-\infty}^\infty {{d\lambda}\over{\lambda}} S(\lambda)e^{i2\pi t/...


3

This is a basic question about the overloaded usage of the word "spectrum," as all three terms are called spectra. Only the eigenvalues of a linear operator and the spectrum of a ring are directly related; Fourier coefficients are separate. As such, we are only going to explain that connection. I am not sure if there is an "explanation for dummies," in the ...


4

The Fourier series of a periodic function can be thought of as its decomposition into eigenfunctions for the translation operator $f(x) \mapsto f(x + t)$ on periodic functions. Say we're talking about $\mathbb{C}$-valued functions with period $2 \pi$: then the eigenfunctions are $e^{inx}, n \in \mathbb{Z}$ with eigenvalues $e^{int}$ (for translation by $t$). ...


2

If $\|g_n - f\|_{\infty} \to 0$ when $n \to \infty$, then $$\int_0^1 |g_n-f|^2 \leq \int_0^1 \|g_n - f\|_{\infty}^2 = \|g_n - f\|_{\infty} ^2 \to 0$$ when $n \to \infty$, so that $\|g_n-f\|_{L^2(0,1)} \to 0$ when $n \to \infty$.


1

Your approach is correct. Under the given assumptions, the function $$g(y) = \frac{f(x-y)-f(x)}{\sin(\pi y)} $$ is continuous and $1$-periodic, with $g(0)=-f'(x)/\pi$. In particular, it is integrable which is enough to apply the Riemann-Lebesgue lemma. More generally, this argument works for Hölder continuous $f$: an estimate of the form $|f(x-y)-f(x)|\le ...


2

The Hilbert transform $\mathcal{H}$ is sometimes said to be an anti-involution, as $\mathcal{H(H(u))}=-u$ (see Hilbert, inverse transform). I see this as a sub-case for your question only. I have witnessed $n$-idempotence too, so you could call it $4$-idempotent function.



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