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1

This is not a complete answer but I want show that there is nothing mysterious going on here. We want to prove that: $$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) \sim \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(s)\sum\limits_{d|n}\frac{\mu(d)}{d^{(s-1)}}$$ The Dirichlet inverse of the Euler totient function is $$a(n)=\sum\limits_{d|n} ...


0

Since you're referring to signals here, it seems appropriate to consider this question from the viewpoint of an electrical engineer. If we impose some restrictions on what kind of functions can be considered a "signal," then all periodic signals have a Fourier series. The function should be piecewise continuous. the function should be be bounded. These ...


1

A bounded periodic integrable function F will certainly "have" a Fourier series, but the sum of the series can fail to be equal to F at some points, even if F is continuous.


1

Well, we saw the other day that the observation that $1/N=N^2/N$ leads to a counterexample. Nothing like making a blithering idiot of oneself for motivation.... We will take $2\pi$ and perhaps other absolute constants to have the value $1$. The letters $I$ and $J$ will always denote intervals; as usual, $|I|$ is the length of $I$. Lightbulb: If $I$ and $J$ ...


0

$$\kappa(s) = \pi^{-s/2} \Gamma(s/2) \zeta(s) = \int_0^\infty x^{s-1} \sum_{n=1}^\infty e^{- \pi n^2 x^2} dx$$ (it is what Mike calls the the Gamma-completed Riemann zeta function. it is useful because $\kappa(s) = \kappa(1-s)$) you know that $\displaystyle f(x) = \sum_{n=1}^\infty e^{- \pi n^2 x} = \sum_{m=1}^\infty \frac{(-1)^{d_p(m)}}{e^{\pi mx}-1}$ ...


3

Let $\mathbf A = \left\{a_{ij}\right\}_{i,j=0}^n \in \mathbb C^{(n+1) \times (n+1)}$ be defined as $$ a_{ij} = c_{i+j \operatorname{mod} (n+1)}, $$ so $$ \mathbf A = \begin{pmatrix} c_0 & c_1 & \dots & c_n\\ c_1 & c_2 & \dots & c_0\\ \vdots & \vdots & \ddots & \vdots\\ c_n & c_0 & \dots & c_{n-1} \end{pmatrix}. ...


1

If $f$ and $f'$ are in $L^{2}$, then $ff'$ is absolutely integrable because $$ 2|ff'| \le |f|^{2}+|f'|^{2}. $$ Therefore, in this case, the following has finite limits as $x\rightarrow\pm\infty$: $$ f^{2}(x)-f^{2}(0)=2\int_{0}^{x}f(t)f'(t)dt. $$ In this case, the limits $\lim_{x\rightarrow\pm\infty}f^{2}(x)=L_{\pm}$ must be $0$ because, ...


3

This is called an anticirculant matrix, which is a special case of Hankel matrix. The eigenvalue/eigenvector formula for circulant matrix does not apply.


1

I am not familiar with your context, but I may be able to help out understand the definition and implications on $\lim_{x \rightarrow \pm \infty} f(x)$. A function is square integrable on $(-\infty, \infty)$, hence the name, if $$ \int_{-\infty}^\infty f(x)^{\color{red}{2}} \mathrm{d}x \lt \infty $$ (… and the above makes actually sense, ie. $f$ is ...


3

To derive Nash's inequality from $$\|f\|_2^2\leq 2\lambda \|f\|_1^2+\frac{1}{4\pi^2\lambda^2}\|f'\|_2^2\tag{1}$$ you should choose the value of $\lambda$ that minimizes the right hand side. To this end, consider the function $$g(\lambda)=2\lambda \|f\|_1^2+\frac{1}{4\pi^2\lambda^2}\|f'\|_2^2$$ of a real variable $\lambda$. Then ...


1

For the first one, $$\int_{|\xi|\ge \lambda} |\widehat{f}(\xi)|^2 d\xi = \frac1{\lambda^2} \int_{|\xi|\ge \lambda} |\lambda \widehat{f}(\xi)|^2 d\xi\le \frac1{\lambda^2} \int_{\mathbb{R}} |\xi \widehat{f}(\xi)|^2 d\xi = \frac1{c\lambda^2} \|f'\|_2^2,$$ where the last equality follows by Plancherel's theorem. For the second one, first use Plancherel's ...


1

Exponential with purely imaginary argument has absolute value equal to 1 so it is sufficient that $f$ vanishes at infinity.


0

Read my other answer! This one's wrong. EDIT: My answer was totally wrong. Sorry - will delete after everyone's had a chance to get a good laugh... WRONG: It's false. Let $$a=\frac\pi{2N}.$$For $0\le j<N$ set $$I_j=[2ja,(2j+1)a].$$ A fun way to calculate $\hat f(n)$ is to note that $f+\tau_af=\chi_{[0,\pi]}.$ Except that only works for odd $n$. Odd ...


0

It's a way of gathering things that sum to a constant. For example, \begin{align} \sum_{n=0}^{\infty}a_{n}z^{n}\sum_{n=0}^{\infty}b_{n}z^{n} & = \sum_{n=0}^{\infty}\left(\sum_{j+k=n}a_{j}b_{k}\right) z^{n} \\ & = \sum_{n=0}^{\infty}\left(\sum_{j=0}^{n}a_{j}b_{n-j}\right)z^{n}. \end{align} A Fourier transform can be viewed as an integral ...


0

$$ \mathfrak{F}(\int_0^x f(y)g(x-y)dy)=\mathfrak{F}(f(t))\times\mathfrak{F}(g(t))=0$$ In your case $g(x)=e^{-x^2/2}$. $$\mathfrak{F}(g(t))=e^{-x^2/2}>0$$ $$\mathfrak{F}(f(t))=0$$ $$f(t)=0$$


2

Set $g(x):=e^{-x^{2}/2}$ and recall that $$\widehat{g}(\xi)=\int_{\mathbb{R}}e^{-y^{2}/2}e^{-2\pi i y\xi}\mathrm{d}y=e^{-\xi^{2}/2},\qquad\forall \xi\in\mathbb{R}$$ Let $\left\{f_{n}\right\}$ be a sequence of functions in $L^{1}(\mathbb{R})\cap L^{2}(\mathbb{R})$ converging to $f$ in $L^{2}$ and pointwise a.e. Applying the convolution theorem to $f_{n}$ ...


5

Note that $|x|f(x)=\text{sgn}(x)xf(x)$. Therefore, the convolution theorem shows that $$ \mathcal{F}(|x|f(x))(\omega)=\mathcal{F}(\text{sgn}(x))(\omega)\ast\mathcal{F}(xf(x))(\omega)=\mathcal{F}(\text{sgn}(x))(\omega)\ast \frac{d}{d\omega} \mathcal{F}(f)(\omega)$$ I don't know how much you know about Fourier transforms, but the Fourier transform of ...


0

The way that Fourier came up with the Fourier transform and its inverse is through a contorted limiting process of the Fourier series expansion on $[-c,c]$ as $c\rightarrow\infty$. Most things that Fourier did could be fully justified once Math had evolved enough, but his arguments leading to the Fourier transform were horrible and cannot be made rigorous, ...


1

The Fourier Transform of a spatial variable is no different mathematically from a Fourier Transform of a temporal variable. The mathematics is agnostic to parameter interpretation. For the Fourier Transform pair for the time-frequency domain are often written $$F(\omega) = \mathscr{F}(f)(k) = \int_{-\infty}^{\infty} f(t) e^{i \omega t} \, dt$$ $$f(t) = ...


2

The Fourier inversion theorem has two different forms: $$f(t) \sim \sum_{n=-\infty}^{\infty} \left(\int_{-\pi}^{\pi} f(x)e^{-inx}\,dx\right) e^{int}$$ and $$f(t) \sim \int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty} f(x)e^{-i\omega x}\,dx\right)e^{i\omega t}\,d\omega.$$ You seem to be a bit mixed up on that point since at one point you're referencing ...


0

Of course, there are many relations between these two functions, but possibly the most interesting one is the fact that this function will represent a fixed-time evolution of the original function according to the solution of the Schrödinger's IVP $$ \begin{cases} \partial _t u = i \Delta u \\ u_0 = g \end{cases}$$ (This can be seen by observing that ...


2

One interesting generalization follows Fourier's original analysis for Ordinary Differential Equations on $\mathbb{R}$. It's easiest to break first into ODEs on a half line $[0,\infty)$ for Sturm-Liouville problems $$ Lf=-\frac{d^{2}f}{dx^{2}}+q(x)f(x) = g(x),\\ \cos\alpha f(0)+\sin\alpha f'(0) = 0. $$ where ...


1

As Qiaochu Yuan pointed out, both correspondences (circle $\to$ integers and line $\to$ line) are special cases of the Pontryagin duality for locally compact abelian group. But perhaps an informal description can be helpful too. The Fourier transform decomposes a function (thought of as a "signal") into waves of different frequencies. In general, a signal ...


1

As Ian pointed out, the Fourier transform of an $L^1$ function is continuous. For continuous functions, being zero a.e. and being zero everywhere are equivalent; so it makes more sense to use the shorter version, without a.e.


0

As commenters said, the answer is negative: e.g., $f(x) =|x|^{-1} e^{-|x|}$ is a nonintegrable function such that $|f(x)\sin x|$ is globally integrable. More generally, the following are equivalent: $f$ is integrable whenever $fg$ is; $\operatorname{ess\,inf}|g|>0$ Indeed, if 2) fails then the sets $A_n = \{x: 2^{-n}\le |g(x)|<2^{1-n}\}$ have ...


1

Let $f(x,y)$ be the function supported on $[-1,1]\times[-1,1]$ and there given by $\min\left(1-|x|,1-|y|\right)$. We have: $$\widehat{f}(u,v) = \frac{2 v \cos v \sin u-2 u\cos u\sin v}{\pi u^3 v-\pi u v^3}$$ and you just need to adjust the above transform by scaling and translating.


0

Your argument is solid, here are the ingredients, which for the most part you already have. From smoothness to decay The Fourier coefficients of a $C^\infty$ function decay faster than any power of $n$. Indeed, fix an integer $k$; the Fourier coefficients of the $k$th derivative of $f$ are bounded, hence $|\hat f(n)| = O(n^{-k})$. And this is valid for ...


1

What follows isn't a solution, but some observations--perhaps trivial--that may help somebody else obtain the $O(N/k)$ bound. Lemma. For any disjoint intervals $\left\{I_{j}=(a_{j},b_{j})\right\}_{j=1}^{N}\subset\mathbb{T}$, $$\sum_{n>k}\left|\widehat{f}(n)\right|^{2}\lesssim N^{2}/k$$ where $f:=\sum_{j=1}^{N}\chi_{I_{j}}$. Proof. By ...


0

(Very handwavy) I think Euler's formula was known to Fourier: $e^{\iota \phi} = \cos \phi + \iota \sin \phi$, and the isomorphism between $\mathbb{R}^2$ and $\mathbb{C}$. So there seems to be a "primitive" quality to trigonometric functions ("trascendental", rather). Wikipedia seems to hint that Fourier's result (expansion of arbitrary functions in sine ...


0

Remember that orthogonal means nothing if you do not specify the scalar product (norm) you're using to define it. There are other sets of functions that are orthogonal and complete but they are so under different norms. Sine and cosine are only special in a Hilbert Space where the norm is the one defined by Fourier.


0

compare it with the FFT of $\theta (0.3-t)$ and of $$H(t) = \theta (0.3-t)[\exp(2i \pi \ 30 \ t)+\exp(2 i \pi \ 100 \ t) + 3.0]$$ use a window (hanning) if you understand what's happening, you will understand nearly everything of the FFT.


2

The coefficients of the complex Fourier series for $e^x$ are given by $$\begin{align} c_n&=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^xe^{-inx}dx\\\\ &=\frac{1}{2\pi}\left.\frac{e^{(1-in)x}}{1-in}\right|_{-\pi}^{\pi}\\\\ &=\frac{(-1)^n\sinh(\pi)}{\pi(1-in)} \end{align}$$ Thus, the complex Fourier Series for $e^x$ is given by $$\bbox[5px,border:2px solid ...


0

Looks like the Andrews Curve was thought of for this purpose. Eg, http://sfb649.wiwi.hu-berlin.de/fedc_homepage/xplore/tutorials/mvahtmlnode9.html


11

I'm hardly qualified to answer this question, but you might find the following references useful. Let's start with the two classical examples of the Fourier transform: the Fourier transforms for $L^2$ functions on the line and the circle. Generalizing the notion of space The line and the circle are both topological abelian groups, and the Fourier ...


2

Another "intuitive" thing convolution does is smoothing. If you convolve a function $f$ with Dirac's delta function $\delta$ (which isn't a function is the sense of a mapping), you get $f$, and if you convolve $f$ with a smooth approximation to $\delta$, you get a smooth approximation to $f$. The convolution will be at least as smooth as the approximation ...


1

Concerning the second question, here's the (small)${}^{[1]}$ result mentioned in comments. Assume that the function $f\colon \mathbb{R}\to \mathbb{C}$ is integrable on the real line and satisfies the estimate \begin{equation} \left\lvert f(x)\right\rvert\le C e^{-a \lvert x \rvert} \end{equation} where $a>0$. (The value of $C>0$ is not important). ...


0

Hint: For 1), try a scaling argument. i.e. $g(x) = a f(bx) $ to find the pair $(\alpha, \beta)$. For 2), use Plancherel's Theorem. For 3), use something like $$ ab \leq \lambda a^2 + b^2 / \lambda $$ for $\lambda >0, a >0$. Now use 2) and 3) for 4).


1

First, let $y=\frac{x-x_0}{a}$. Then, $$\begin{align} \frac{N}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-(x-x_0)^2/a}e^{ik_0x}e^{-ikx}dx&=\frac{Na}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-y^2}e^{-i(k-k_0)(ay+x_0))}dy\\\\ &=\frac{Na}{\sqrt{2\pi}}e^{-i(k-k_0)x_0}\int_{-\infty}^{\infty}e^{-y^2}e^{-i(k-k_0)ay}dy\tag 1 \end{align}$$ Now, let's complete ...


0

Answered in comments: The series converges in $L^1$. The map $g \mapsto \int g(x)e^{2\pi i \xi\cdot x}\,dx$ is continuous for every $\xi$. That allows the interchange. ...we have $$\int \left(\lim_{N\to\infty} \sum_{n=1}^N f_n(x)\right)e^{2\pi i\xi\cdot x}\,dx = \lim_{N\to\infty} \sum_{n=1}^N \int f_n(x)e^{2\pi i\xi \cdot x}\,dx$$ by continuity of the ...


1

Well, there's a lot of creative answers but it seems that no one bothered to put here the "follow your nose" one, so I'll add it here for completeness. Recall the identity: $$\sin^3x = \frac{3 \sin x - \sin(3x)}{4}.$$ So we have: $$\int_{-\infty}^{+\infty} \left(\frac{\sin x}{x}\right)^3\,{\rm d}x = \int_{-\infty}^{\infty} \frac{3 \sin x - \sin ...


1

If you expand the exponential in the integral, only the constant term survives, yielding $R^2/2$, so the limit is $1/2$. This should in fact work for any $\gamma\gt0$, though one might have to argue more carefully in that case why the operations are allowed.


1

We have that $f*G_\chi =G_\chi$ if and only if $f$ is of the form $$f(m)=\delta(m)+\sum_{r:\ (r,N)>1} c_r e^{2\pi m r /N},$$ where $\delta$ is the delta function and the $c_r$ can be equal to any complex numbers. We can expand $G_\chi * f(m)$ as a double sum $$G_\chi * f(m)=\sum_{k\in\mathbb{Z}_{N}}\sum_{r\in\mathbb{Z}_{N}}\chi(r)e^{2\pi ...


0

If $n$ is large enough, linear regression should do the trick. In other words, you can jut consider the residuals of the linear regression.


4

Let's first prove that the Carleson operator is bounded on $L^p$ if Conjecture 9.11 holds for $d=1$. $$ \int_{-N}^{N} \hat f(y) e^{ixy} \, dy $$ is the inner product of $y \mapsto e^{i x y} \hat f(y)$ and $y \mapsto I_{|y|< N}$. By Parseval's Theorem this is equal to the inner product of the inverse Fourier transforms of these two functions, namely $z ...


3

There is indeed a connection, and it is a very useful one. Namely, the Parseval identity states that the $L^2$ energy (squared norm) of $f$ is the same as that of $\hat f$: $$ \int|f(x)|^2=\int|\hat f(k)|^2. $$ If you want to study the $L^2$ norm of the gradient instead, using $\widehat{\nabla f}(k)=2\pi ik\hat f(k)$ with the Parseval identity tells you ...


2

My convention for the Fourier transform is $$\widehat{f}(\xi)=\int_{\mathbb{R}}f(x)e^{-2\pi i x\xi}\mathrm{d}x,\qquad \xi\in\mathbb{R} \tag{1}$$ You can adjust to your own convention using the scaling properties of the Fourier transform. As Chester noted in the comments, you have a product of two functions $\sin t^{2}$ and $e^{-2\left|t-2\right|}$. However, ...


0

The frequency spectrum gives you the amplitudes for every frequency from DC to daylight that are required in order to reconstruct the original function. The only complication is that the frequency phase shift has to be accounted for as well. That's why you need sin and cosine function: $$ \sin(\omega t + \phi) = \cos(\phi)\sin(\omega ...


1

If $f \in L^{2}[0,\infty)$, then $\mathscr{L}\{f\}$ is holomorphic in the right half plane where $\Re s > 0$. and the Laplace transform is square integrable on all vertical lines in the right half plane, with $$ \frac{1}{2\pi}\int_{-\infty}^{\infty}|\mathscr{L}\{f\}(v+iw)|^{2}dw \le \int_{0}^{\infty}|f(t)|^{2}dt,\;\;\; 0 < v < \infty. $$ The ...


1

There is another easy way of calculating the Fourier Transform instead of going the direct route. In order to use that route, you will need to understand the concept of duality. The duality is a concept relating $t$ to $\lambda$. Basically, if $F(\lambda)$ is the Fourier Transform of $f(t)$, then $2\pi f(-\lambda)$ is the Fourier Transform of $F(t)$. If ...


4

METHOD 1: There are several ways to go here. If we are permitted use of Generalized Functions, then we can proceed as follows. Let $I(\lambda)$ be the integral $$I(\lambda)=\int_{-\infty}^{\infty}e^{i\lambda\tau}\frac{\sin a\tau}{\tau}d\tau\tag 1$$ Taking the derivative with respect to $\lambda$ of $(1)$ gives $$\begin{align} ...



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