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0

For every $a\in\mathbb R$, the complex exponential function $e^{iat}$ is positive-definite because $$ \sum_{k,l} e^{ia(t_k-t_l)}z_k\bar z_l = \left|\sum_{k} e^{iat_k} z_k\right|^2 \ge 0 $$ And $\cos^2 t$ can be written as a combination of complex exponentials with positive coefficients: $$ \cos^2 t = \frac12+\frac12 \cos 2t = ...


1

We have: $$ \frac{\pi}{2}\,a_n=\int_{0}^{\pi}\sqrt{1-k^2\sin^2\theta}\cos(2n\theta)\,d\theta =\frac{k^2}{4n}\int_{0}^{\pi}\frac{\sin(2\theta)\sin(2n\theta)}{\sqrt{1-k^2\sin^2\theta}}\,d\theta.\tag{1}$$ If we set: $$ b_m = \int_{0}^{\pi}\frac{\cos(2m\theta)}{\sqrt{1-k^2\sin^2\theta}}\,d\theta,\qquad c_m = ...


2

For the $s=0$ case, let $s_{n}=\sum_{k=1}^{n}c_{k}$ for $n \ge 1$ and let $s_{0}=0$. Then $s_{n}-s_{n-1}=c_{n}$ for all $n \ge 1$ and $\lim_{n} s_{n}=0$ by assumption. So the following are absolutely convergent for $0 \le r < 1$: $$ \begin{align} \sum_{n=1}^{\infty}r^{n}c_{n} & = \sum_{n=1}^{\infty}r^{n}(s_{n}-s_{n-1}) \\ & = ...


0

This is not true in general. Consider $f(x) = e^{ix}$ which is $2\pi$-periodic. Define $g(x) = f(-x)$ which is also $2\pi$-periodic. Observe that \begin{align*} \widehat f(k)&= \frac{1}{2\pi}\int_0^{2\pi} f(x) e^{-ikx} \, dx = \frac{1}{2\pi} \int_0^{2\pi} e^{ix}e^{-ikx} \, dx = \frac{1}{2\pi} \int_0^{2\pi} e^{e(1 - k)x} \, dx \\ &= \begin{cases} 1 ...


2

If we take out the $2 \pi$ for simplicity's sake and rescale time such that $f_m/f_c=r$ and $f_c=1$, then we are basically considering $$s(t)=\sin((1+\sin(rt))t) = \sin(t+t\sin(rt))$$ Now consider $r=1$ and going forward in time by $2 \pi$: $$s(t+2 \pi)=\sin(t+2 \pi + (t+2\pi) \sin(t+2\pi)) \\ = \sin(t + (t+2 \pi) \sin(t+2 \pi)) \\ = \sin(t + (t + 2 \pi) ...


1

Suppose the Fourier series of a periodic function $f$ with period $2\pi$ is $$ a_0+\sum_{k=1}^\infty \left(a_k \cos kx+b_k \sin kx\right) $$ and the Fourier series of $g$ is $$ q_0+\sum_{k=1}^\infty \left(q_k \cos kx+r_k \sin kx\right). $$ Then the Fourier series of $mf+ng$, where $m$ and $n$ are constants, is $$ (ma_0+nq_0)+\sum_{k=1}^\infty \Big((ma_k + ...


1

Yes, the Fourier Series is linear. The coefficients $a_k$ and $b_k$ are defined in terms of integrals and integrals are linear. For example, for constants $\lambda$ and $\mu$ and function $\mathrm{f}$ and $\mathrm{g}$ we have $$\frac{1}{L}\int_{-L}^L \left(\lambda\mathrm{f}(x)+\mu\mathrm{g}(x)\right)\cos\left(\frac{\pi n x}{L}\right) \mathrm{d}x=$$ ...


2

If you have a linear combination of functions, the resulting Fourier series is the corresponding linear combination of the Fourier series of the functions. So yes, it is linear. The key is that it is linear in the coefficients, even though the series is not linear in $x$.


2

For a function $f(x)=\frac{1}{\pi}\frac{a}{a^2+x^2}$ (also kwnown as Lorentzian function) the Fourier transform is: $$ \int_{-\infty}^{\infty}\frac{1}{\pi}\frac{a}{a^2+x^2}e^{-itx}dx=\frac{a}{\pi}\int_{-\infty}^{\infty}\frac{1}{(a+ix)(a-ix)}e^{-itx}dx $$ Now we will use Cauchy's integral formula. Although we only need to solve our integral for the real axis, ...


1

In your Calculation, x shouldn't be the lower border in the integral. It must be H(x)*x. Because, if x < 0, the integral is cut, then the lower border is 0. And in the end it is (x-2H(x)*x) = -|x| ... :)


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The important thing to remember is that you are dealing with multiple convolutions gaussian filter. If you have any probability background this is no different to multiple draws from a gaussian distribution and thus the same laws apply. Theorem: The normal distribution is stable. Specifically, suppose that X has the normal distribution with mean μ∈ℝ and ...


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You should look at how to compute a Fourier series on an arbitrarily lattice. Your data grid needs to be skew-shifted slightly and zero padded so that your data grid corresponds to the non-orthogonal lattice vectors of a triangular lattice. An ordinary 2D FFT on this grid of data produces the Fourier series coefficients corresponding to the data, where each ...


2

Do not use the triangle inequality to bring the absolute value inside the integral — it will not help, as $\lvert e^{-inx}\rvert =1$. Instead, use integration by parts. $$ \int_{ - \pi }^\pi nf (x)e^{- inx} dx = \left[i f(x)e^{- inx}\right]^\pi_{-\pi} -i \int_{[-\pi,\pi]} f^\prime(x) e^{- inx}dx $$ For the first term, leverage the periodicity of $f$. For ...


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Edit: This is the answer for the first version of the question. If period means the smallest period the answer is: no. Constant functions are counterexamples.


2

There is no Schwartz space, just $C^\infty(\mathbb T^n)$. Indeed, $\mathcal S(\mathbb R^n)$ is an intermediate class between all smooth functions and compactly supported smooth functions. But on $\mathbb T^n$, everything is compactly supported. Since the Fourier transform of a function on $\mathbb T^n$ is a function on $\mathbb Z^n$, it does not make ...


3

We wish to evaluate $$\int_{-\infty}^{\infty} e^{-|x|^{\alpha}}e^{-2\pi ixy}\,dx.$$ Making use of evenness, we get $$2\int_0^{\infty} e^{-x^{\alpha}} \cos(\pi xy)\,dx.$$ Expanding $\cos$ in a power series, we have $$2\sum_{n=0}^{\infty}\frac{(-1)^n\pi^{2n} y^{2n}}{(2n)!}\int_0^{\infty} e^{-x^{\alpha}} x^{2n}\,dx.$$ So we need only to evaluate the ...


1

We denote by $\phi$ the characteristic function of the random variable $X1_A/\mathbb{P}(A)$. Then we have $$\phi(t)=\mathbb E\left(\mathbb 1_A\exp\left(it\frac X{\mathbb{P}(A)}  \right)\right)+1-\mathbb{P}(A).$$ I'm not sure that this can be simplified more or written as a function of the characteristic function of $X$ and the indicator function of $A$. ...


1

Let us first establish the case $I=\left[-1,1\right]$. This yields $\lambda\left(I\right)=2$ and $c\left(I\right)=0$. Now, we use a dyadic partitition of $\mathbb{R}$, i.e. we write \begin{eqnarray*} \int\frac{\left|f\left(x\right)\right|}{\left(1+\frac{\left|x-c\left(I\right)\right|}{\lambda\left(I\right)}\right)^{N}}\,{\rm d}x & \lesssim & ...


2

This equation \begin{equation}f'(x)=2f(2x+1)-2f(2x-1), \quad f(0) = 1, \tag{$*$} \end{equation} has a finite solution which is also known as the $\mathrm{up}(x)$ or $\mathrm{hut}(x)$ function. It has compact support $\mathrm{supp}\,\mathrm{{up}}(x)=[-1,1]$ and its Fourier transform is $\hat{f}(t)=\prod\limits_{k=1}^{\infty}\mathrm{sinc}{(t\cdot ...


0

Now that I've had some time away from the problem I realize I was over-thinking this quite badly ... There is no need to do any kind of logs or anything. To simplify, I let $S = e^{-jw}$ Thus $H(w) = \dfrac{1}{(1-\frac{1}{4}e^{-jw})(1-\frac{1}{3}e^{-jw})}$ becomes $H(w) = \dfrac{1}{(1-\frac{1}{4}S)(1-\frac{1}{3}S)}$ Which I solved in the usual way.


2

If it was true, the Fourier transform of an even $L^1(\mathbb{R})$ function would be always increasing on $\mathbb{N}$. This is clearly false, just by considering $f(x)=e^{-x^2}$. Notice that neither the opposite inequality holds, since nothing forces the Fourier transform to be monotonic someway.


1

The operator $A$ is from $\mathbb{R}^k$ into itself, not into $\mathbb{R}^1$. If it is invertible, the change of variable $y=Ax$ gives $$\begin{align} \hat g(\xi)=\int_{\mathbb{R}^k}f(Ax)e^{ix\cdot\xi}\,dx&=\frac{1}{|A|}\int_{\mathbb{R}^k}f(y)e^{i(A^{-1}y)\cdot\xi}\,dy\\ ...


2

We have that: $$ g(t) = \int_{\mathbb{R}}\frac{\sin(\pi x)}{\pi x}e^{-itx}\,dx = \mathbb{1}_{(-\pi,\pi)}(t)+\frac{1}{2}\mathbb{1}_{\{\pi\}\cup\{-\pi\}}(t).\tag{1}$$ To prove such an identity, we may compute the inverse Fourier transform of $\mathbb{1}_{(-\pi,\pi)}$, or notice that: $$ g(t) = 2\int_{0}^{+\infty}\frac{\sin(\pi x)}{\pi x}\cos(tx)\,dx = ...


0

As indicated in the comments, there is no formula giving the Fourier coefficients of $\sqrt f$ in terms of the coefficients of $f$. Let $f(x)=1+t\,\cos x$ for $|t|\le1$. It is clear that $f(x)\ge0$ if $x\in[-\pi,\pi]$. For simplicity I will consider real instead of Fourier coefficients. Since $f$ is even, all sine Fourier coefficients vanish. The Fourier ...


2

$x/\sin x$ is not an integrable function, so that the Fourier transform is not defined. Moreover it is not locally integrable, so that the Fourier transform is not defined even as distribution.


0

(rather long to be a comment)Theorem:( J.L-. Lions 1951) If $u,v\in\mathcal{S}'({\mathbb{R}})$ with compact support then co.supp$(u\ast v)=$co.supp$(u)$+co.supp$(v)$, where "co" denotes the convex hull.


2

Such a function (at least with the additional properties you name) cannot exist: If the Fourier transform of $f$ is non-negative, then (by Fourier inversion) $$ |f(x)| = |\mathcal{F}^{-1}(\hat{f})(x)| = |\int \hat{f}(\xi) e^{2\pi i \langle x,\xi\rangle}\, d\xi | \leq \int |\hat{f}(\xi)|\, d\xi = \int \hat{f}(\xi)\, d\xi = f(0), $$ where the step before the ...


2

As you say, taking the Fourier transform yields $$\dot{\hat{u}}=\left(-\mathbf{k}^\mathsf{T}A\mathbf{k}+i\mathbf{k}\cdot\mathbf{b}+c\right)\hat{u}$$ and solving for $\hat{u}$ gives $$\hat{u}=\hat{\phi}(\mathbf{k})\exp\left[\left(-\mathbf{k}^\mathsf{T}A\mathbf{k}+i\mathbf{k}\cdot\mathbf{b}+c\right)t\right].$$ Letting $A=QDQ^\mathsf{T}$, we can rewrite the ...


1

You cannot have $\hat{f}=f$ unless you have normalized the Fourier transform so that it is isometric., meaning that $\|\hat{f}\|=\|f\|$, where $\|\cdot\|$ is the $L^{2}(\mathbb{R})$. That means $$ \hat{f}(s) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-ist}\,dt. $$


0

Proof Sketch: Just use the law of large of large numbers together with the continuous mapping theorem (under the assumption that the $X_1\in L^2k(\mathscr{A})$, for $k\in[1,\infty)$). Bonus: As a corollary if $X_1\in L^2k$ then $X_1\in L^j(\mathscr{A})$ for every $j\in [1,k]$, whence the binomial theorem and the continuous mapping theorem yields the ...


1

This is the lazy way, but a Google search for "proof of law of large numbers" comes up with a number of hits including this: https://www.math.ucdavis.edu/~tracy/courses/math135A/UsefullCourseMaterial/lawLargeNo.pdf which has proofs of both the weak and strong laws.


1

You are absolutely right: the Fourier transform of $h(-t)$ is $H(-f)$ (if $H(f)$ is the Fourier transform of $h(t)$). $H^*(f)$ is the Fourier transform of $h^*(-t)$. Of course, if $h(t)$ is real-valued we have $h(-t)=h^*(-t)$ and, consequently, $H(-f)=H^*(f)$. But in general you have the pairs $$h(-t)\Longleftrightarrow H(-f)\\ h^*(-t)\Longleftrightarrow ...


2

For any integer $k$, the numerator is $1 - 1 = 0$. If $k$ is also a multiple of $2N$ then the bottom is also $1-1 = 0$. This is indeterminate. (It seems their $N/2$'s should be $2N$'s.) Whenever $k$ is not a multiple of $2N$, the exponential on the bottom is not equal to one, so the expression on the bottom is nonzero, this whole expression is zero since ...


3

I am not completely sure at the moment how the supports of $\widehat{p_1 u} + \dots + \widehat{p_m u^m}$ and of $\widehat{u}$ could be connected, but if you consider $$ \widehat{u} = \delta_x, $$ i.e. $u = e^{2\pi i \langle x, \cdot\rangle}$, then $u^2 = e^{2\pi i \langle 2x, \cdot\rangle}$, i.e. $$ \widehat{u^2} = \delta_{2x}, $$ so that even ${\rm ...


0

The duality property is the statement that if $$\mathcal{F}(h(t)) = H(\omega)$$ then $$\mathcal{F}(H(t)) = h(-\omega)$$ To use the duality property to prove the statement you need to show that $$\mathcal{F}\left(i~\text{sign}(t)\right) = -\frac{1}{\omega\pi}$$ This can be done by a direct computation of $\mathcal{F}\left(i~\text{sign}(t)\right)$. To do ...


3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} ...


2

Definition of Fourier transform: \begin{align*} \tilde{f}(k) &= \int_{-\infty}^\infty e^{-ikx} f(x) dx \\ &= \int_{-\infty}^\infty f(x) \cos kx dx - i \int_{-\infty}^\infty f(x) \sin kx dx \\ \Rightarrow \tilde{f}^*(k) &= \int_{-\infty}^\infty f(x) \cos kx dx + i \int_{-\infty}^\infty f(x) \sin kx dx \\ &= \int_{-\infty}^\infty e^{ikx} f(x) ...


0

The coefficients are $$c_{n} = \frac{1}{4}\int_{0}^{2}e^{- i\frac{\pi n x}{2}}dx = \frac{i}{2 \pi n}(e^{- i\pi n}-1),$$ for $n \in \mathbb{Z}-\{0\}.$ For $n = 0:$ $$c_{0} = \frac{1}{4}\int_{0}^{2}dx \Rightarrow c_0 = \frac{1}{2}.$$ For $n$ odd: $$c_{n} = \frac{i}{2 \pi n}(e^{- i\pi n}-1)= -\frac{i}{\pi n}.$$ For $n$ even, and $n >0$: $$c_{n} = ...


2

When $t$ is positive, $U(t)$ is 1, and $U(-t)$ is zero, so you get only the first term, which equals $e^{-a|t|} = e^{-at}$. Now do the same thing when $t$ is negative... (I'm assuming here that your book/prof has the definition that $U(x) = 1$ for $x \ge 0$ and is zero otherwise.) (You might reasonably be worrying about the case $t = 0$, but if your ...


2

Yes, your proposition is true. By density of the continuous functions, there is a sequence of continuous $g_n$ with $g_n \to f$ in $L^1$. Replacing $g_n$ with $g_n^+$, we can assume $g_n \ge 0$. Passing to a subsequence, we can also assume $g_n \to f$ almost everywhere. Passing to a further subsequence, we can assume $\|g_n - f\|_{L^1} \le 2^{-n}$. ...


1

You use the density of smooth compactly supported functions in $L^1(\mathbb{R})$. Namely, if $f\in L^1(\mathbb{R})$. Then there is a sequence of smooth compactly supported functions $f_n$ such that $\|f-f_n\|_{L^1(\mathbb{R})}\to0$ as $n\to\infty$. This is exactly what lines 6-7 of the solution say.


5

This is about the first thing proved in any text on the Fourier transform: $$ |\hat f(\xi)|=\Bigl|\int_{\mathbb{R}^n}e^{ix\xi}\,f(x)\,dx\Bigr|\le\int_{\mathbb{R}^n}|f(x)|\,dx=\|f\|_1. $$


0

decompose the exponential function in terms of besselfunction as well as F, then integrate using the usual orhogonality relations.


1

It would be useful but not necessary. One does not need complex analysis even to consider the complex form of the Fourier series. Complex analysis may help with advanced aspects of Fourier series, for example finding certain series coefficients and an alternate way of solving Laplace's equation in two dimensions. (One way using a Fourier series ...


1

Well, Fourier series are more conveniently expressed in complex form as series $$\sum_{n=-\infty}^\infty c_ne^{inx}\;,\;\;n\in\Bbb N\;,\;\;c_n\in\Bbb C$$ I think complex analysis is way too much for the above, as things can be put in real form in a more or less easy way, but this may heavily depend on how deep that course of your intents to dig in these ...


3

The sinc function is the inverse Fourier transform of the characteristic function $\chi_{[-1,1]}$, at least up to a constant: $$ \begin{align} \chi_{[-1,1]}^{\vee}(x) & = \frac{1}{\sqrt{2\pi}}\int_{-1}^{1}e^{ixs}ds \\ & = \left.\frac{1}{\sqrt{2\pi}}\frac{e^{isx}}{ix}\right|_{s=-1}^{1} \\ & = ...


0

Because the sampling rate is less than two samples per cycle, the input will be aliased down to $26-44.1=-18.1$ kHz. If the wave is a sine wave, the minus sign just inverts the signal. If $t$ is a multiple of $1/(44.1$ kHz), $\sin (2\pi t 26$ kHz$)=\sin (-2\pi t 18.1$ kHz). If you have a large duration, you will see (almost all the) energy in frequencies ...


1

The question is not well-formed, but I will take a shot in an effort to help form it. The answer depends on how you extend $f(x)=-x^2+\pi x$, $0< x< \pi$ to $-\pi< x< 0$. If you do an even extension, $$f_e(x):=\begin{cases}f(-x), &-\pi<x<0,\\ f(x), &0<x<\pi,\end{cases}$$ then the full Fourier series will end up being just a ...


1

$M(\mathbb{R})\subset L^1(\mathbb{R})$. This post shows the existence of a continuous function $f$ with compact support (hence $f\in M(\mathbb{R})$) with $\hat f\notin L^1(\mathbb{R})$, and hence $\hat f\notin M(\mathbb{R})$. An explicit example of such an $f$ is $$ f(x)=\bigl(1-\log(1-x^2)\bigr)^{-\alpha}\cdot\mathbb{1}_{[-1,1]},\quad0<\alpha\le1. $$ ...


1

The smoother the function $f$ is, the smaller is $\hat f$ at $\infty$. Take a discontinuous function $f$ like the characteristic function of the interval $[-1,1]$, for which $$ \hat f(\xi)=\text{some constant}\cdot\frac{\sin \xi}{\xi}. $$ Note (added after ZHANG's comment): The example above is not continuous, so it is not an answer to the OP question.



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