New answers tagged

1

Take any nontrivial, nonnegative, symmetric function $g \in C_c^\infty (\Bbb{R}^d)$. If we let $h := \mathcal{F}^{-1}(g)$, then $h$ is real-valued (why?) and $$ h(0) = \int g(x) \, dx > 0, $$ since $g \geq 0$ and $g \not \equiv 0$. By continuity of $h$ and by rescaling (i.e., replace $g$ by $Cg$ for some large $C>1$), we get $h \geq 1$ on $B_{2\delta} ...


0

Your approach is perfectly appropriate. There was, however, a flaw in the execution of that way forward. Note that we have $$\begin{align} F(k)&=\langle H(a-|x|),e^{-ikx}\rangle\\\\ &=\frac{1}{-ik}\langle H(a-|x|),\frac{de^{-ikx}}{dx}\rangle\\\\ &=\frac{1}{-ik}\langle \text{sgn}(x)H'(a-|x|),e^{-ikx})\\\\ &=-\frac{1}{ik}\langle ...


1

$L^1$ is dense in $L^2$. For example, if $f \in L^2$, then $f_R=\chi_{[-R,R]}f \in L^1\cap L^2$ where $\chi_{[-R,R]}$ is the characteristic function of the finite interval $[-R,R]$; and $\lim_{R\rightarrow\infty}\|f-f_R\|_{L^2}=0$. The operators $\tau_y$, multiplication by $e^{-2\pi j\xi y}$ and the Fourier transform are isometric on $L^2$. Therefore, $L^2$ ...


3

The following can be found in a table of transforms: Let $f$ have the Fourier transform $\hat{f}$. $x \mapsto f(ax)$ has Fourier transform $k \mapsto {1 \over |a|}\hat{f}({k \over a}) $. $x \mapsto f(x-a)$ has Fourier transform $k \mapsto e^{-iak}\hat{f}(k) $. $x \mapsto e^{i a x} f(x)$ has Fourier transform $k \mapsto \hat{f}(k-a)$. The first two can be ...


0

\begin{align*} \int_{-\pi}^{\pi}|D_m(t)|\,dt&=2\int_{0}^{2\pi}\left|\frac{\sin((m+1/2)t)}{\sin(t/2)}\right|\,dt\\ &\geq2\int_{0}^{2\pi}\frac{|\sin((m+1/2)t)|}{t/2}\,dt\\ &=4\int_0^{(m+1/2)\pi}\frac{|\sin s|}{s}\,ds\\ &\geq4\sum_{j=1}^m\int_{(j-1)\pi}^{j\pi}\frac{|\sin s|}{s}\,ds\\ &\geq4\sum_{j=1}^m\int_{(j-1)\pi}^{j\pi}\frac{|\sin ...


0

Many trigonometric problems may be easily solved if you replaced the trig with exponentials. For example, $\sum_{n=0}^x\cos(n)$ can be changed into the "real" part of $\sum_{n=0}^xe^{ni}$, which is then found to be a very simple geometric sum. We can do the exact same thing with sine. Also, many calculus related trig problems can be easily solved as ...


0

I would say that the double approximation $$ \forall x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],\qquad 1-\frac{x^2}{2}\leq\cos(x)\leq e^{-x^2/2}$$ is everything we need. By approximating $\varepsilon+\cos(x)$ with $(1+\varepsilon)\,e^{-\frac{x^2}{2(1+\varepsilon)}} $ we have that $$ \int_{-\pi/2}^{\pi/2}\left(\varepsilon+\cos x\right)^{2k}\,dx \approx ...


0

It is just a convolution identity: $$\sum_{n\in\mathbb{Z}}\text{sinc}(A-n)\text{sinc}(n-B) = (\widehat{u}*\widehat{u})(A-B)=\left(\widehat{u\cdot u}\right)(A-B),$$ but since the inverse Fourier transform $u$ of the $\text{sinc}$ function is just the indicator function of an interval, $u^2=u$, hence the RHS of the previous line equals ...


0

Hint: This is apparently just Gaussian integrations in the complex plane. For $t>0$, we have: $$ u(x,t)~=~\frac{2}{\pi} \int_0^\infty \mathrm{d}k~ \exp\left[-k^2t\right]G_s(k)\sin kx $$ $$~=~\frac{1}{\pi} \int_{\mathbb{R}} \mathrm{d}k~ \exp\left[-k^2t\right]\sin 2k \sin kx $$ $$ ~=~ \frac{1}{2\pi} \int_{\mathbb{R}} \mathrm{d}k~ ...


2

I don't know if this has been considered resolved yet, but combining hints from @user1952009: Let $$f(t)=e^{-at}, -\pi<t<\pi$$ and have period $2\pi$. Then we want to find the Fourier series of $f(t)$ $$f(t)=\sum_{n=-\infty}^{\infty}a_ne^{int}$$ Then ...


1

Define the Fourier Transform of $f(x)$ as $$F(k)=\int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx$$ Then, the inverse Fourier Transform of $F(k)$ is given by $$f(x)=\frac1{2\pi}\int_{-\infty}^\infty F(k) e^{ikx}\,dk$$ We seek to find the inverse Fourier Transform for $F(k)=\frac{e^{-s|k|}}{k}$ where $s\in \mathbb{R}$ and $s>0$. ...


0

If by DC component you mean the overall integral of the function (i.e. the amount of the curve above the x axis minus the amount below it) then both are correct in that an odd function is also symmetrical about the y axis. As pointed out, no function is symmetrical over the x axis except $f(x)=0$ but what I assume you meant was it is the same for the ...


3

The dc component of $x\mapsto f(-x)$ is the same as for $f$. The dc componet of $x\mapsto -f(x)$ is the negative of the dc component of $f$. Hence for odd $f$, the dc component equals its own negative and must be zero. Also not that no function (except constant zero) is symmetric to the $x$-axis.


1

I like the analogy with traffic lights. With Fourier analysis you can tell the light has been an exact green frequency but you don't have information of the time it occurred. With wavelets you can tell the light was kind of green around a certain time. You sacrifice knowing the exact frequency and instead get approximations of the time and frequency.


5

If you can get $s(a) =\sum \dfrac1{a^2+n^2} $, then $s(a/2) =\sum \dfrac1{a^2/4+n^2} =4\sum \dfrac1{a^2+4n^2} =4\sum \dfrac1{a^2+(2n)^2} $, so $s(a)-\frac14 s(a/2) =\sum \dfrac1{a^2+n^2}-\sum \dfrac1{a^2+(2n)^2} =\sum \dfrac1{a^2+(2n+1)^2} $.


1

$\hat{f}(2x + 3) = \int_{-\infty}^\infty e^{- i \omega x} f(2 x+3) dx$ Now, $u=2 x + 3$ gives $du = 2 dx$ and $x = \frac{u-3}{2}$. Plugging this into the integral, you get $\hat{f}(2x + 3) = \frac{1}{2} \int_{-\infty}^\infty e^{- i \omega (\frac{u-3}{2}) } f(u) du = \frac{1}{2} \int_{-\infty}^\infty e^{- i \frac{\omega}{2} (u-3) } f(u) du = \frac{1}{2} ...


1

You are right exactly and you are almost done. Fix $1 < r < p < \infty$. Note that $\mathcal{M}$ is bounded $L^{p/r} \to L^{p/r}$. We have $$ \| \mathcal{M}_r f \|_p = \| \mathcal{M}(\lvert f \rvert^r) \|_{p/r}^{1/r} \leq C \| \lvert{f}\rvert^r \|_{p/r}^{1/r} = C \| f \|_p $$ so that $\mathcal{M}_r$ is bounded $L^p \to L^p$. Along with boundedness ...


1

Let $\lambda > 0$, define $\delta_\lambda f(x)=f(x/\lambda)$ we show that $\widehat{\delta_\lambda f}(\xi)=\lambda^n \widehat{f}(\lambda \xi)$. Place $z=x/\lambda$, i.e. $dx=\lambda^n dz$, obviously here we are in $\mathbb{R}^n$, then $\displaystyle \widehat{\delta_\lambda f}(\xi) := \int_{\mathbb{R}^n} e^{-2 \pi i x \cdot \xi} f(x/\lambda) dx = ...


0

Figured it out $$\tilde{u}(x,t) = \exp(-k^2t)\int_0^{\infty}\exp(k^2\tau)f(k,\tau)d\tau = \int_0^{\infty}\exp(k^2(\tau-t))f(k,\tau)d\tau $$ Which means that $$ u(x,t) = \int_0^{\infty} \int_\mathbb{R}\frac{\exp[-(x-k)^2/4(t-\tau)]}{\sqrt{4\pi(t-\tau)}}h(k,\tau)dkd\tau $$


1

Suppose $(e^{-x^2})^{(n)} =p_n(x) e^{-x^2} $. Then $\begin{array}\\ (e^{-x^2})^{(n+1)} &=(p_n(x) e^{-x^2})'\\ &=p_n'(x) e^{-x^2}-p_n(x)(2x) e^{-x^2}\\ &=e^{-x^2}(p_n'(x) -2xp_n(x))\\ \end{array} $ so if we define $p_0(x) = 1$ and $p_{n+1}(x) =p_n'(x) -2xp_n(x) $, then $(e^{-x^2})^{(n)} =p_n(x) e^{-x^2} $. Looking at this recurrence, we see ...


0

An exponential always beats a polynomial in the end... $|x| \le e^{|x|}$ so $|x|^n < e^{n |x|} < e^{x^2}$ if $|x| > n$, therefore $\left|x^n e^{-x^2}\right| < 1$ there. Since the continuous function $x^n e^{-x^2}$ is also bounded on the finite interval $[-n,n]$, we conclude that $x^n e^{-x^2}$ is bounded on $\mathbb R$. Take a linear ...


2

Hint. One may recall that $$ \cos nx=\frac{e^{inx}+e^{-inx}}2,\quad \sin nx=\frac{e^{inx}-e^{-inx}}{2i} $$ giving $$ \begin{align} f(x) &= a_0 + \sum_{n=1}^N (a_n \cos{nx} + b_n \sin{nx})\\\\ &=a_0 + \sum_{n=1}^N \frac{(a_n -ib_n)}2e^{inx}+ \sum_{n=1}^N \frac{(a_n +ib_n)}2e^{-inx} \end{align} $$ Can you take it from here?


3

I am wondering what if we just define it as $|f(x)|\le \frac{A}{x^2}$? The function $ x \mapsto \frac1{x^2}$ is not in $L^1(\mathbb{R})$ whereas $ x \mapsto \frac1{1+x^2}$ is in $L^1(\mathbb{R})$.


0

Using the reverse triangle inequality the way you did, too much stuff goes into the subtracted term. You need a more careful estimate, namely $$ \left|\sum_{|\alpha|\le m} a_\alpha(x)\xi^\alpha\right|\geq \frac{\gamma}{2}|\xi|^m - B \tag{1} $$ where $\gamma$ is from the definition of ellipticity and $B$ is a constant. When $(1)$ is applied to estimate ...


1

$$\begin{align} \widehat f(n)&=\int_{\mathbb{T}}f(x)\,e^{-inx}\,dx\\ &=\int_{\mathbb{T}}f\Bigl(x+\frac{\pi}{n}\Bigr)\,e^{-in\bigl(x+\tfrac{\pi}{n}\bigr)}\,dx\\ &=-\int_{\mathbb{T}}f\Bigl(x+\frac{\pi}{n}\Bigr)\,e^{-inx}\,dx. \end{align}$$ Averaging the two expressions for $\widehat f(n)$ we get $$ |\widehat ...


2

$\int \rho_{n,\epsilon}(x)dx = \int \rho_{\epsilon}(x)dx=\epsilon^{d}\int\rho(x)dx$ and the latter integral is considered of limit $\frac{1}{2^{j}}\searrow 0$ of Riemann summation of which the partition is equilateral cube of length $\frac{1}{2^{j}}$. That is, $$\int \rho(x)dx=lim_{j\searrow ...


1

In this case the Haar measure is simply the Lebesgue measure. So you have $$ \hat \mu(\varepsilon)=\int_{S^2} e^{-i\varepsilon x}dx. $$


0

You can use polar coordinates. Define $r=\sqrt{x^2+y^2}$ and $\omega=\sqrt{\omega_x^2+\omega_y^2}$. Then define the angles $\theta$ and $\alpha$ such that $(x,y)=(r\cos\theta,\,r\sin\theta)$ and $(\omega_x,\omega_y)=(\omega\cos\alpha,\omega\sin\alpha)$. Let us compute $$G(b,\omega,\alpha)=\int_0^{2\pi}\mathrm d\theta\int_0^\infty \mathrm e^{-br-\mathrm ...


0

Q1: $$ \mathcal{F}^{-1}[g](x) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \cos (\sqrt{ k^2 d -a^2 }) e^{i k x} d k $$ I see no reason to think it may be written any simpler than that. Also note that the integrand involves cosine of an imaginary argument, at least on part of the range.


3

$\newcommand\ip[2]{\left\langle #1,#2\right\rangle}$ Leaving out many technical details, just illustrating what that "duality argument" is: Let's define $$\ip fg=\int fg.$$ Suppose that $K$ is a kernel and define $\tilde K(t)=K(-t)$. Then, assuming you can justify the Fubini (typically by restricting to some nice subspace of $L^p$), you see that ...


0

$$ \mathcal F\{f(x)\}=\hat{f}(\lambda)= \int_{-\infty}^\infty f(x) e^{-i\lambda x}\, dx $$ The Fourier transform of $f(x)=e^{-a|x|}$ is $\hat{f}(\lambda)=\frac{2a}{a^2 + \lambda^2} $ and then for the duality property $\mathcal F\{\hat{f}(x)\}= 2\pi f(-\lambda)$ we have $$\mathcal F\left\{\frac{2a}{a^2 + x^2} \right\}=2\pi e^{-a|-\lambda|}$$ and for the ...


2

Change $u u_x$ to $(u^2/2)_x$. The Fourier transform of this term is $i k F(u^2/2)$. That is you apply fft directly to $u^2/2$, this is how you handle nonlinear terms in spectral method - there is really no trick. Below is what I wrote in matlab (use the simplest Euler forward step, so mind the time step size), it solves $x$ in $[0, 2\pi]$ with initial ...


0

The integral equals $\epsilon^n\widehat{f}(\epsilon z) e^{-ia\epsilon z}$, and thus the sum can be viewed as a Riemann sum for $\int_{\mathbb R^n}\widehat{f}(t)e^{-iat}\, dt=f(a)$.


1

Following David C. Ullrich's suggestion in the comments, let $g=f-\hat f(0)$. It is fairly straightforward to show $\hat g(n)=0$ for all $n\in\mathbb Z$, so by the usual uniqueness theorem, $g=0$ almost everywhere. This implies $f=\hat f(0)$ almost everywhere, i.e. $f$ is constant almost everywhere.


1

This doesn't answer your question, but circumvents it...The trick for Vicious Burgers Eq is to apply the Cole-Hopf Transformation: $$ u = - 2D \frac{1}{\phi} \frac{ \partial \phi }{ \partial x} $$ you'll obtain $$ \frac{ \partial }{ \partial x} \left ( \frac{1}{\phi} \frac{ \partial \phi }{ \partial t} \right) = D \frac{ \partial }{ \partial x} \left ( ...


1

Mr. Moonshine, there is no contradiction at all, you just messed it up. Let me explain. As Tryss and John Martin pointed out, the unilateral Fourier transform doesn't indeed has the property $$\mathcal{F}\left[f^{(n)}\right](\omega)=(2\pi i \omega)^n\mathcal{F}(\omega)$$ but rather $$\mathcal{F}\left[f^{(n)}\right](\omega)=(2\pi i ...


0

How about this (you can add the prefactors...) : $A\equiv f(uu_x) = \int u u_x \exp(I k x)$ and we do integration by parts: $A=f(uu_x) = u^2\exp(Ikx)\vert_{-\infty}^\infty-\int(u_x+Iku)u\exp(I k x)=-\int uu_x e^{Ikx}-\int u^2e^{Ikx}=-A-f(u^2)$. Hence, $A=-1/2f(u^2)$ (assuming your function goes to zero at infinities).


0

You may compute the Fourier transform directly. Since $f \in L^1(\mathbb{R})$, for $\xi \neq 0$, we have $$ \hat{f}(\xi) = \int_{-1}^1\, (1-|x|)\cdot e^{-2\pi ix\xi}\, dx = \left( \frac{\sin \pi \xi}{\pi \xi}\right)^2. $$ For $\xi=0$, $e^{-2\pi ix\xi}=1$, so $\hat{f}(0)=\int_{-1}^1\, (1-|x|)\, dx= 1$.


3

Actually, your heart is in the right place; you just have the wrong inner product in mind. Instead of taking the inner product of $f$ and $g$ to be$$\int_0^\infty f(t)g(t)\,dt,$$you should take it to be the average value of $f(t)g(-t)$, in an appropriate sense; this will be proportional to$$\int_{-\infty}^\infty f(t)g(-t)\,dt,$$and thus you can use this in ...


0

You've applied the duality property wrongly. What you should get is $$\mathcal{F}\{it X(t)\}=-2\pi\frac{dx(-\nu)}{d\nu}\tag{1}$$ Note that $X(t)$ is the Fourier transform of $x(\nu)$, not the other way around, as you supposed in your equation. Let $$\tilde{X}(\nu)=\mathcal{F}\{X(t)\}\tag{2}$$ From the duality property we know that $$\tilde{X}(\nu)=2\pi ...


1

My starting point is, of course, your work. $$ \big| f(x+h) - f(x)\big| ~=~ \left| \frac{1}{2\pi} \int_{-\infty}^\infty \widehat{f}(\xi) e^{i(x+h)\xi}\; d\xi ~-~ \frac{1}{2\pi} \int_{-\infty}^\infty \widehat{f}(\xi) e^{ix\xi}\; d\xi \right| \\=~\frac{1}{2\pi} \left|\int_{-\infty}^\infty \widehat{f}(\xi) \big(e^{i(x+h)\xi} - e^{ix\xi}\big) \; d\xi \right| ...


1

Calling $\mathcal{L}(f(t)) =F(s)$ and observing that $$ \frac{d^n}{ds^n}\mathrm e^{-st}= (-1)^nt^n \mathrm e^{-st} $$ we have $$F^{(n)}(s)= \int_0^\infty f(t) \left(\frac{d^n}{ds^n}e^{-st}\right)\mathrm dt =\int_0^\infty f(t) \left((-1)^nt^n \mathrm e^{-st}\right)\mathrm dt= (-1)^n \int_0^\infty t^nf(t)e^{-st}\ dt$$ that is $$(-1)^nF^{(n)}(s)= \int_0^\infty ...


1

Second(*) inequality, follows from, $K=\bigcup\limits_{n\in\mathbb{N}_0}(\mathfrak{P}^{-j}+\mathfrak{p}^{-j}u(n))$. For third (**), use Parseval's identity.


0

If $D$ is the differentiation operator, then your equation becomes $$ (D+\omega^2)y = \sin(nt). $$ The annihilator of the right side is $(D+n^2)$. So $y$ must be a solution of $$ (D+\omega^2)(D+n^2)y = 0. $$ The general solution of this equation is $$ y(t)=A\sin(\omega t)+B\cos(\omega t)+E\cos(nt)+F\sin(nt). $$ ...


1

As an additional exercise, you might like to try the following: Show that for any $x\in[0,2\pi]$ the set $\{x+2n\pi\alpha\mod2\pi:n\in\mathbb Z\}$ is dense in $[0,2\pi]$ Show that if $f\equiv c_1$ on $C_1$ and $f\equiv c_2$ on $C_2$, where $C_1$ and $C_2$ are both dense subsets of $[0,2\pi]$, then $c_1=c_2$ (Hint: use Riemann integrability) Conclude that ...


5

Compute the $n$th Fourier coefficient of $f$: $$\begin{aligned} a_n &= \frac{1}{2\pi}\int_0^{2\pi}f(x) e^{-inx} dx \\ &= \frac{1}{2\pi}\int_0^{2\pi}f(x + 2\pi \alpha) e^{-inx} dx \\ &= \frac{1}{2\pi}\int_0^{2\pi}f(x)e^{-in(x - 2\pi \alpha)} dx \\ &= e^{in2\pi \alpha} \frac{1}{2\pi}\int_0^{2\pi}f(x) e^{-inx} dx \\ &= e^{in2\pi \alpha} a_n ...


1

Complex conjugation, that is $\overline{x+yi}=x-yi$. With $x,y\in\mathbb{R}$


0

HINT: $$\left|\int_a^b f(x)\,dx\right|\le\int_a^b \left|f(x)\right|\,dx$$ and $|e^{ix}|=1$ for real $x$.


0

Just insert the absolute value into the integral. You have $$ \left| \hat{f}_n(k) - \hat{f}(k) \right| = \left| \int_0^1 (f(x) - f_n(x)) e^{-2\pi i k x} \, dx \right| \leq \int_0^1 \left|f(x) - f_n(x)\right| | e^{-2 \pi i k x}| \, dx = \int_0^1 |f(x) - f_n(x)| \, dx = ||f - f_n||_{L^1}. $$


0

HINT The function $f(x)$ is a triangular function and can be seen as the convolution of two rectangual functions and then the Fourier transform is the product of two sinc functions.



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