Tag Info

0

Pick a basis frequency which is relatively prime to your sine and your sine can be "arbitrarily well" described in $L^2$ (I think it is by Plancherel's theorem), but probably not in a finite number of terms. I can expand on this later when I get to my more computational computer. Illustration: We use the FFT, cycle through frequencies (linearly) between ...

0

Not possible. But some approximation is possible like: $$\cos(f_1+\Delta f)+\cos(f_1-\Delta f)=2\cos f_1 \cos \Delta f$$ If $\Delta f$ is small, then the overall waveform or graph would look like that it's oscillating at $f_1$ but the average of the waveform would be fluctuating at much slower frequency so may not bother. If the receiving device (either ...

0

The r.h.s. can be writen as $$\frac{1}{i\omega}F(\omega)+\pi F(0)\delta (\omega)=\left[\frac{1}{i\omega}+\pi\delta (\omega)\right] F(\omega)\equiv G(\omega)F (\omega)$$ where $$G(\omega)=\frac{1}{i\omega}+\pi\delta (\omega)$$ whose Inverse Fourier Transform is (http://mathworld.wolfram.com/FourierTransformHeavisideStepFunction.html) $$g(t)=2\pi ... 1 One way to look at it is to note that$$\int_{-\infty}^tf(\tau)d\tau=(f\star u)(t)\tag{1}$$where \star denotes convolution, and u(t) is the unit step function. Consequently, the Fourier transform of (1) is$$\mathcal{F}\left\{\int_{-\infty}^{t}f(\tau)\, d\tau\right\}=F(\omega)U(\omega)\tag{2}$$where F(\omega) and U(\omega) are the Fourier ... 0 FTT transforms for multiplications are only useful for very big numbers. The complexity hides a big constant factor. If you have to write algorithms for cryptography involving huge numbers, use it. But for small integers, you will waste time 1 The bra-ket notation is really an adaptation of Fourier's original ideas from more than a century before Dirac. There are ways of viewing this notation in terms of cyclic/irreducible representations associated with selfadjoint operators. However, the Laplace transform is associated with a non selfadjoint operator, which is not appropriate for study using the ... 0 The expressions are complicated. For instance$$\begin{align} a_0^{(3)}&=\frac1\pi\int_{-\pi}^\pi \Bigl(\frac{a_0^{(1)}}{2}+\sum_{n=1}^{\infty}a^{(1)}_n\cos(nx)+b^{(1)}_n\sin(nx)\,\Bigr)\Bigl(\frac{a_0^{(2)}}{2}+\sum_{n=1}^{\infty}a^{(2)}_n\cos(nx)+b^{(2)}_n\sin(nx)\Bigr)\,dx\\ ...

0

$$F[f](\xi) = F[(-\Delta+\lambda)((-\Delta+\lambda)^{-1}f)](\xi) = (|\xi|^2+\lambda)F[(-\Delta+\lambda)^{-1}f](\xi),$$ hence $$F[(-\Delta+\lambda)^{-1}f](\xi)=\frac{1}{|\xi|^2+\lambda}F[f](\xi)$$

2

Let $\varphi \in \mathcal D$, where $\mathcal D$ is the space of $\varphi \in C^\infty_0(\mathbb R)$ with the usual test function continuity notion. Consider $$\tilde \varphi(\xi) = \int_{\mathbb R} \frac{dx}{(2\pi)^{1/2}}e^{-i\xi x}\varphi(x)=\int_{\mathbb R} \frac{dx}{(2\pi)^{1/2}}\sum_{n=0}^\infty \frac{(-i\xi x)^n}{n!}\varphi(x);$$ since the integral ...

0

Let $0<s<1$ be fixed, and let $\alpha>0$ be such that $\alpha>2s$. Suppose $f\in C^{0,\alpha}(\mathbb{R}^{n})$ and $$|f(x)|\lesssim (1+|x|)^{\delta},\quad x\in\mathbb{R}^{n}$$ for $0<\delta<2s$. Then I claim that the integral above is absolutely convergent. Indeed, for \begin{align*} &\int_{1\geq ...

3

It is a typo. Dividing by $iw$ corresponds to the Fourier transform of $f(x)$ integrated. For example, in control engineering an integrator block is labeled $1/s$ where $s$ is the complex frequency variable of the Laplace transform, which becomes the Fourier transform for $s = iw$. Is it a first edition book? ...

0

You seem to think that the implied constants will be completely independent of our choice of Littlewood-Paley partition of unity. If you look at the proof of the standard Littlewood-Paley inequality, you'll see that the estimates for the vector-valued kernel depend on the LP partition of unity. For the upper LP estimate, we can say the following: for all ...

1

You know that $\bar{f}(t)$ corresponds to $\bar{F}(-\omega)$. There are two (reasonable) options to determine the Fourier transform of $e^{iat}\bar{f}(t)$. The first is to define a function $$g(t)=e^{-iat}f(t)\Longleftrightarrow G(\omega)=F(\omega+a)$$ from which $$\bar{g}(t)=e^{iat}\bar{f}(t)\Longleftrightarrow \bar{G}(-\omega)=\bar{F}(-\omega+a)$$ ...

0

Hint. Apply the complex form of differentiation under the integral sign theorem. If you define $$f_\alpha(z,t)=e^{-\vert t \vert^\alpha} e^{2\pi i z t},$$ you have $$F_\alpha(z)=\int_{-\infty}^{+\infty} f_\alpha(z,t) dt.$$ The partial derivative $$\frac{\partial f_\alpha}{\partial z}(z,t)=2\pi i t e^{-\vert t \vert^\alpha} e^{2\pi i z t}$$ is continuous on ...

1

Morera's Theorem and the related Cauchy's integral theorem and may be what you're seeking. $$\oint_\gamma f(z) dz = 0$$ If $f$ is continuous on an open set that contains $\gamma$, and satisfies the above integral, then $f$ is holomorphic on that set (analytic). By extension, if the above integral vanishes for every possible $\gamma$, then $f$ is analytic ...

1

The coefficient of the transformed function depends on your definition of the Fourier transform and the inverse transform. For example, I could define the forward transform and inverse respectively as $$F(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(x) e^{-ikx}\ dx$$ $$f(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty F(k) e^{ikx}\ dx$$ But this ...

1

Check out the Wikipedia page: There are several common conventions for defining the Fourier transform ... Wolfram Mathworld has at least one alternate definition. The problem is that different branches of mathematics and physics use different definitions for the Fourier Transform. Each definition gives a slightly different result. So perhaps we ...

1

The Fourier transform has different definitions in different settings. Some have a $\frac{1}{\sqrt{2\pi}}$, some have it without the square root, some have no such factor. The inverse transform is also defined with these $2\pi$ factors so that no matter which convention you use a consistent result is obtained. Partly it is a question of units, i.e. ...

0

Presumably, you mean a $C_{c}^{\infty}$ function by "bump function." I am going to change your notation for convenience: I will write $\widehat{\psi}$ where you have written $\psi$. For each $k$, set $\psi_{k}(x):=2^{kn}\psi(2^{k}x)$. For each $x$, define an operator-valued ($\mathbb{C}\rightarrow\ell^{2}(\mathbb{Z})$) by $\vec{K}(x)=(\psi_{k}(x))_{k}$. ...

1

Suppose $f$ is a periodic function on $\mathbb R$ with period $p$, with Fourier series $$f \sim \sum_{n\,\in\,\mathbb Z} c_n e^{2\pi inx/p}$$ The inner product $\langle f,g\rangle = \frac 1 p\int_0^p f(x)\,\overline{g(x)}\, dx$ (where $a\mapsto \overline{a}$ is complex conjugation) is used to determine the coefficients, which must satisfy $$c_n = ... 1 Partial work. Let define the following function:$$g(\xi)=\int_{\mathbb{R}}\exp\left(-\frac{x^2}{2a}\right)\exp(-i\xi x)\,\mathrm{d}x.$$Notice that:$$g'(\xi)=-i\int_{\mathbb{R}}x\exp\left(-\frac{x^2}{2a}\right)\exp(-i\xi x)\,\mathrm{d}x=-\frac{\xi}{a}g(\xi).$$I used derivation under the integral and then integration by parts. Moreover, using substitution, ... 4 There is a way to do this analytically using Fourier transforms. I will show this as soon as I can, but I promise you, it will be a lot messier than what I present below. GRAPHICAL METHOD To do this at the level of an undergrad signals class, you really need to draw a picture. But while it may be error-prone, there are checks along the way to see that ... 0 I cannot write a comment, so writing an answer. The function theta in the original question (and thus its Fourier transforms) is actually not the usual Heaviside step function, as$$\Theta(1)=0,\Theta(-1)=1.$$David C. Ullrich had already meantioned it, but I think it still causes confusion. The answer asked here is answered in detail in Graduate ... 3 F_{2l}(x) can be calculated in closed form for \text{Im}\ x<0 using the integral from Gradsteyn and Ryzhik$$ \int_0^1u^\lambda P_{2l}(u)du=\frac{(-1)^l\Gamma\left(l-\frac{\lambda}{2}\right)\Gamma\left(\frac{1}{2}+\frac{\lambda}{2}\right)}{2\Gamma\left(-\frac{\lambda}{2}\right)\Gamma\left(l+\frac{3}{2}+\frac{\lambda}{2}\right)},\quad \text{Re}\ ...

2

Think of a simple example where solutions are $$\phi_{n,m}(x,y)=\sin(n x)\sin(m y),\;\;\; \Omega = [0,\pi]\times[0,\pi].$$ The associated eigenvalues are $-(n^{2}+m^{2})$. The base eigenvalue is $\lambda=-2$, and this eigenspace has multiplicity $1$. The next is $\lambda=-5$, which is of multiplicity $2$. It's common to have multiplicity $1$ for the ...

0

The multiplication operation on $L^1(\mathbb{R})$ is convolution. If you choose $f \in L^1$ such that $f \ne 0$ and $\hat{f}(\xi)=0$ for a fixed, given $\xi\in\mathcal{R}$, then $h=f\star g = g\star f$ also has the property that $\hat{g}(\xi)=0$ (for this you need the Fourier transform and $\widehat{f\star g}=C\hat{f}\hat{g}$ where $C$ is a constant. And, if ...

2

This is simply a Bernoulli polynomial $\;B_2(x)\;$ up to a constant $\pi^2$ for $\,x \in (0,2\pi)$ : \begin{align} S&:=\sum_{k=1}^\infty \frac {\cos (kx)}{k^2}\\ &=\pi^2\;B_2\left(\frac x{2\pi}\right)\\ &=\pi^2\;\left(\left(\frac x{2\pi}\right)^2-\frac x{2\pi}+\frac 16\right)\\ &=\frac{\pi^2}6-\frac{\pi x}2+\frac{x^2}4\\ \end{align} On the ...

3

Hint: Notice that \begin{align} \frac{s^3}{(s^2+4)^2}&=\frac{s(s^2+4)-4s}{(s^2+4)^2}=\frac{s}{s^2+4}-\frac{4s}{(s^2+4)^2} \end{align} Also notice $\mathscr{L}\left\{\sin (2t)\right\}=\frac{2}{s^2+4}\,\,$ and $\,\,\frac{\mathrm d}{\mathrm ds}\left(\frac{2}{s^2+4}\right)=-\frac{4s}{s^2+4}$. Then use the fact that ...

0

METHOD $1$: We can proceed to evaluate the integral if we invoke Generalized Functions. To that end, we write the inverse Fourier Transform representation for $f$ as $$f(t)=\frac{1}{2\pi}\int_{- \infty}^{ \infty}\frac{e^{j\omega t}}{a+j\omega}\,d\omega \tag 1$$ Then, note that the derivative $f'$ is given by $$f'(t)=\frac{1}{2\pi}\int_{- \infty}^{ ... 2 Let S_0\subset L^1_0(\mathbb{R}^n) be the set of Schwartz functions with mean 0. S_0 is clearly dense (Edit: see the OP's comment below). Thus it suffices to approximate f\in S_0. Now let us try to finish your argument. Compute$$f-f_\delta=\mathcal{F}^{-1} ( \hat{f}\cdot \varphi(\xi/\delta) )=f*\psi_\delta,$$where \psi(x)=\hat{\varphi}(-x) ... 1 First, we find the solution of the non-difference equation:$$y'(t) + y(t) = \cos^2(\pi t) = \frac 12 \left( 1 + \cos 2\pi t \right)$$Homogeneous solution y_h(t) = Ce^{-t} Solution of the inhomogeneous equation y' + y = 1/2: \displaystyle y_1(t) = \frac 12 Solution of the inhomogeneous equation y' + y = \frac 12 \cos 2\pi t: \displaystyle ... 1 Expanding on Dr. MV's comment, consider a Fourier series for f on -L<x<L,$$f(x) = \sum_\limits{n=-\infty}^\infty c_n e^{i n \pi x/L}$$What's the period of the extended series? Now what if L=\pi as in your case - what's the period? When you restrict the sum's domain to [-L,L] you get f on that interval, which recreates the original ... 0 First write$$ \int_{0}^{\infty}e^{-\frac{4}{5}x}e^{-isx}dx =\frac{1}{\frac{4}{5}+is}. $$Then$$ \begin{align} \int_{0}^{\infty}x^{2}e^{-\frac{4}{5}x}e^{-isx}dx & =-\frac{d^{2}}{ds^{2}}\int_{0}^{\infty}e^{-\frac{4}{5}x}e^{-isx}dx \\ & = -\frac{d^2}{ds^2}\frac{1}{\frac{4}{5}+is} \\ & = ...

3

Note: You have the wrong Fourier coefficents on $[0,1]$. The coefficients should be $$\hat{f}(n)=\int_{0}^{1}f(t)e^{-2\pi in t}dt.$$ $\{ e^{2\pi in t}\}_{n=-\infty}^{\infty}$ is an orthonormal basis of $L^2[0,1]$. Every function $f \in L^2[0,1]$ is also in $L^1[0,1]$ because $$2|f| \le |f|^{2}+1.$$ If you have ...

2

Let $v(s)=\frac{1}{1+4s^2}$. Your equation can be written as: $$(u*v)(t) = \frac{1}{t^2+6t+10}\tag{1}$$ and by switching to Fourier transforms (defined through $\mathcal{F}(u)(\xi)=\widehat{u}(\xi)=\int_{-\infty}^{+\infty}u(x)e^{-2\pi i x \xi}\,dx$) we have: $$\widehat{u}(\xi)\cdot \frac{\pi}{2} e^{-\pi|\xi|}= \widehat{u}(\xi) \cdot \widehat{v}(\xi) = ... 1 I think you're not on the right track. You already pointed out some of the problems with your approach. Let me try to help you see how to tackle such a problem. First of all it helps to assume that f is a Schwartz function. In fact a bit more than your claim is true. Claim. If for some 1\le p,q\le \infty, there exists a constant ... 1 Observe what happens when you take the Fourier transform of a derivative:$$\begin{align}\widehat{\left(\frac{\partial u}{\partial x} \right)}(k) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \frac{\partial u}{\partial x} e^{-ikx}dx = - \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} u \frac{\partial}{\partial x} \left( e^{-ikx} \right) dx \\ &= ik ...

1

We will prove that if the sequence $(c_n)_{n \in \mathbb{Z}}$ is in $\ell^2(\mathbb{Z})$, then the series $$f(x) := \sum_{n \in \mathbb{Z}} c_n e^{2 \pi i n x}$$ converges to a function $f(x)$ that is in $L^2(0,1)$. First recall that $(e^{2\pi i n x})_{n \in \mathbb{Z}}$ is a complete orthonormal system here in $L^2(0,1)$. Then ...

0

Hint, too long for a comment. It's really bad practice to use the same variable in both the spatial and Fourier domain, and will lead to massive confusion. Switching to more usual notation, note that the result is $\frac{d^k}{d \omega^k} \hat{f}(\omega) = \mathcal{F}[{i^k x^k f(x)}](\omega)$, where $\hat{f}(\omega) = \mathcal{F}[f(x)](\omega)$. You know ...

0

Unfortunately, I do not believe that necessary and sufficient conditions for the absolute convergence of this integral are currently known. Your inversion formula is a special case of the Gil-Pelaez theorem, I found in this this paper which references Gil-Pilazez's paper However, I an unable to find an analytical result, only papers regarding numerical ...

0

Spectral theory gives a nice proof of the Plancherel Theorem. The operator studied is $L=\frac{1}{i}\frac{d}{dx}$ on $L^{2}(\mathbb{R})$. The resolvent of $L$ is an integral operator $$R(\lambda)f=(L-\lambda I)^{-1}f= \left\{\begin{array}{cc} i\int_{-\infty}^{x}e^{i\lambda(x-u)}f(u)du, & \Im\lambda > 0 \\ ... 0 The proof I came up with for the L^2 isometry of the Fourier transform on L^1(\Bbb R)\cap L^2(\Bbb R) hinges on the L^2 completeness of the Hermite-Gauss functions. It is well-known that these are eigenfunctions of the Fourier transform however the proofs using the Hermite-Gauss functions immediately jump to an L^2 theory and neglect the integral ... 1 It relies on the fact that (using your notation with "j" for what is usually written as "i")$$e^{jt}=\cos t + j\sin t$$From this it follows that$$e^{-jt}=\cos t - j\sin t$$so that$$\cos t = \frac12(e^{jt}+e^{-jt})$$1 I assume you are using j for the imaginary unit, which is more commonly written as i by English-speaking mathematicians (although electronics engineers like to use j, because they use i for current). The identity you seem to be unaware of is known as Euler's formula:$$e^{ix} = \cos x + i \sin x$$FWIW, that formula is valid for complex x as ... 1 Cauchy-Schwarz does the job:$$\begin{align} \int_\mathbb{R} |\hat{f}(k)|\,dk&=\int_\mathbb{R} |\hat{f}(k)|\,(1+k^2)^{1/2}\,(1+k^2)^{-1/2}\,dk\\ &\le\Bigl(\int_\mathbb{R} |\hat{f}(k)|^2\,(1+k^2)\,dk\Bigr)^{1/2}\Bigl(\int_\mathbb{R} (1+k^2)^{-1}\,dk\Bigr)^{1/2}\\ &=\sqrt{\pi}\Bigl(\int_\mathbb{R} |\hat{f}(k)|^2\,(1+k^2)\,dk\Bigr)^{1/2}. ...

0

Hint: you can Apply stone weirstrass, differentiable functions in $L^1_0$ with compact support are dense in continuous functions in $L^1_0$ which are dense in $L^1_0$.

0

This is actually much easier than I thought. In particular, the following example is compactly supported. First, I claim that the Hilbert transform of the function $$f:=\dfrac{1}{y(\log|x|)^{2}}\chi_{(0,1/2)}$$ is not locally integrable. Indeed, for $x<0$, \begin{align*} ...

0

Since $\mathbf P$ symmetric positive definite, $\mathbf P=\mathbf Q^T\,\mathbf D\,\mathbf Q$ where $\mathbf D$ is diagonal and $\mathbf Q$ is orthogonal. Then $$\mathbf{x}^T\, \mathbf{P}\, \mathbf{x}=(\mathbf{Q}\,\mathbf{x})^T\,D\,(\mathbf{Q}\,\mathbf{x}).$$ Make the change of variables $\mathbf y=\mathbf{Q}\,\mathbf{x}$, and take into account that since ...

1

$e^{i\theta} = \cos(\theta) + i \sin(\theta)$. With $\theta = 2\pi \cdot k$ for some integer k, $\sin(\theta)$ will always vanish, and $\cos(\theta)$ will equal 1. In your case, $k = -n\cdot j$. Since both $n$ and $j$ are integers, their product is an integer.

2

if $f$ is a smooth $L^1$ function, $\hat{f}$ is an $L^\infty$ function which decays faster than any polynomial at infinity This is not true as stated. Unquantified smoothness does not lead to quantitative conclusions. Indeed, the properties of $\hat f$ that you claimed imply $\hat f\in L^2$, hence $f\in L^2$. But a smooth $L^1$ function need not be in ...

Top 50 recent answers are included