# Tag Info

1

I'm aware of one necessary and sufficient condition. It comes out of the Spectral Theorem for selfadjoint operators. The domain of $L=\frac{1}{i}\frac{d}{dx}$ on $L^{2}[-\pi,\pi]$ consists of all periodic absolutely continuous functions on $[-\pi,\pi]$ with $f' \in L^{2}[-\pi,\pi]$. $L$ is selfadjoint on this domain. From the Spectral Theorem: $$... 1 As pointed out in my earlier comment, we need to assume s\in (0,1). Out of habit, I'm going using the convention \mathbb{T}\cong[0,1). You can modified everything to your own convention. The expression$$[f]_{W^{2,s}(\mathbb{T})}=\left(\iint_{\mathbb{T}\times\mathbb{T}}\dfrac{\left|f(x)-f(y)\right|^{2}}{\left|x-y\right|^{1+2s}}dxdy\right)^{1/2}$$... 0 Let's back up a tad and talk about Fourier series. Call \phi_k(t) = \exp(ikt). Let's work on L^2[0,2\pi]. We can take the Fourier expansion of some function:$$ x = \sum_{k\in\mathbb{Z}} a_k \phi_k $$where$$ a_k = \frac{\langle f, \phi_k \rangle}{\langle \phi_k, \phi_k \rangle} = \frac{1}{2\pi} \int_0^{2\pi} x(t) \phi_k(t)^*dt = ...

0

Schwartz space is topological vector space so what is really usually meant by "isomorphism" is a bijection that not only preserves algebraic, but also topological structure. That means that you also need to know that your mapping is homeomorphism. In this case, three-fold composition of Fourier transform ($\mathcal F$) is its inverse so it's sufficient to ...

1

Of course there are many functions with this property. An interesting way to get some: Say $\phi\in C^2_c(\Bbb R)$ (this is stronger than necessary) and the support of $\phi$ is contained in $[-\pi,\pi]$. Let $$f(x)=\frac1{\sqrt{2\pi}}\int e^{-itx}\phi(t)\,dt\quad(*).$$Then $$\int f(x)\,dx=\sum_{n\in\Bbb Z}f(n).$$First the inversion formula for the Fourier ...

0

In Fourier analysis, we mostly do not use $\cos$ and $\sin$ functions, but instead use $e^{i\xi x}$. The two approaches are equivalent, but with the complex exponentials, we have a huge advantage: If we change the phase (I.e. if we translate), we get $e^{i\xi (x+x_0)}=e^{i \xi x_0} e^{i \xi x}$, i.e. translation/change of phase corresponds to multiplying ...

-1

This is mechanized in Maple. with(OrthogonalExpansions): f := unapply(piecewise(and(x <= 0, x >= -1), 0, and(x > 0, x <= 1), sin(Pi*x)), x): FourierSeries(f(x), x = -1 .. 1, infinity, 'Coefficients'); $${\pi }^{-1}+\sum _{i=1}^{\infty } \left( \cases{-1/2\,{\frac {1}{\pi \, \left( i+1 \right) }}&i=1\cr -{\frac { \left( -1 ... 1 In the link I shared in the comments section you can see that the following holds: If f is absolutely continuous,$$ \left| \widehat f(n) \right| \le {K \over |n|}. $$If f is a BV function,$$ \left|\widehat f(n)\right|\le {\|f\|_{BV}\over 2\pi|n|}. $$If f \in C^p,$$ \left|\widehat{f}(n)\right|\le {\| f^{(p)}\|_1\over |n|^p}. $$If f\in C^p ... 2 Here's a partial answer, too long for a comment. I'm not sure about necessary and sufficient conditions for continuity or differentiability, but it's not too hard to find sufficient conditions. Certainly, if \sum |F(n)| < \infty then the series \displaystyle \sum F(n) e^{iun} converges absolutely and uniformly, by the Weierstrass M-test, so f is ... 2 First note that Theorem 4.1 holds for \psi\in L^2 by a simple approximation argument. Now assume f\in L^2, ||f||_2=1. Then, since |x|\le L_1/2 for x\in I_1,$$\left(\frac{L_1}{2}\right)^2\ge \left(\frac{L_1}{2}\right)^2\int_{I_1}|f|^2\ge\int_{I_1}x^2|f(x)|^2\ge\frac12\int_{\Bbb R}x^2|f(x)|^2.$$Similarly ... 0 When you solve the equation for X for \lambda =\alpha^2>0, this gives you$$ X=A\sin(\sqrt{\alpha}x)+B\cos(\sqrt{\alpha}x). $$Applying the initial conditions gives$$ 0=A\sin(\sqrt{\alpha}L) $$which implies that either A=0 or \sin(\sqrt{\alpha}L)=0. Note that A=0 gives a trivial result so we must have \sin(\sqrt{\alpha}L)=0, ie, ... 0 Just so that you have the identity \|f\|_2=\|\hat{f}\|_2, called Plancherel's theorem. 2 As discussed in the comments, probably the two answers differ by a factor of 2\pi because two different conventions for the definition of Fourier have been used. The second method probably uses$$ F(k)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\cos(ax)e^{-ikx} \, dx. $$There are several common conventions for the definition of the Fourier transform, in fact ... 1 Fourier series is not the same as Fourier transform. Fourier transform never requires periodicity: continuous distribution of frequencies ensures that the signal never repeats. In fact, Fourier transform is mostly used for nonperiodic signals. They do have to be L^2 integrable, which includes all finite signals. If you slacken the rules of convergence, you ... 0 There is a property of the Fourier Transform which explains this. It's the time scale property :$$ \mathscr{F}\{f(at)\} = \frac{1}{|α|} F\Big(j\frac{ω}{α}\Big) $$Where α is the scaling constant. For example considering the standard function: f(t)=e^{-kt}  and it's fourier transform which is : F(jω)=\frac{1}{jω + k}  if you scale that function, ... 3 To start with let's write (following the wikipedia convention)$$\nu(t):=\vartheta(0,it)=\sum_{n=-\infty}^{\infty}e^{-\pi n^2 t }=\sum_{n=-\infty}^{-1}e^{-\pi n^2 t }+1+\sum_{n=1}^{\infty}e^{-\pi n^2 t }=2\sum_{n=1}^{\infty}e^{-\pi n^2 t }+1$$Therefore:$$ I=\frac{1}{2}\int_{0}^{\infty}dt \left(\nu(t)-1\right)t^{s/2-1}=\sum_{n=1}^\infty\int_{0}^{\infty}dt ...

1

If $G$ is a locally compact group, then $G$ has a left invariant measure $$\mu(aX)=\mu(X)$$ for a measurable set $X$ and $a\in G$ called the left Haar measure. Using this, we can conclude the following: $$\int_G |f(y)|(\int_G |g(y^{-1}x)|dx)dy \\ =\int_G|f(y)|(\int_G|g(x)|dx)dy \\ =||f||_{L^1}||g||_{L^1}.~_{\square}$$

1

Let $\mathcal{F}\{u(x)\}(X)$ be the Fourier transform of $u(x)$. Then it's straightforward to show the inverse Fourier transform matches the original function in question. \begin{align*} &\mathcal{F}^{-1}\{F(X,Y)\}(x,y) \\ &= \frac{1}{(2\pi)^2}\iint_\limits{\mathbb{R}^2} F(X,Y) \exp(i(xX+yY)) dXdY \\ &= \frac{1}{(2\pi)^2} ...

2

Young's and Holder's inequalities make weak convergence easy to study, but the pointwise convergence depends on the behaviour of a maximal operator (the Carleson operator) for which it is not that easy to provide effective upper bounds. Carleson's greatest idea was probably to modify usual decomposition techniques in the Calderon-Zygmund theory in the ...

0

The coefficients of the complex Fourier Series are given by $$c_n=\frac{1}{T}\int_0^Ts(t)e^{i2\pi n t/T}dt$$ For $s(t)=1-e^{-2t}$ with a period $T=5$, we have \begin{align} c_n&=\frac{1}{5}\int_0^5 (1-e^{-2t})\,e^{i2\pi n t/5}dt\\\\ &=\delta_{n,0}+\frac12\,\frac{e^{10}-1}{e^{10}(-5+i\pi n)} \end{align} where $\delta_{n,m}$ is the Kronecker ...

0

Your computations aren't rigorous. Since $\partial_{j}^{d}$ is only homogeneous of degree $-d$, the integral expressions $$\int_{\mathbb{R}^{d}}\partial_{j}^{d}(\xi)\phi(\xi)e^{-2\pi ix\cdot\xi}d\xi$$ and $$\int_{\mathbb{R}^{d}}\partial_{j}^{d}m(\xi)(1-\phi(\xi))e^{-2\pi ix\cdot\xi}d\xi$$ a priori don't converge absolutely. In the first expression you have ...

1

The theorem is called Wiener-Khinchin Theorem. This is quite surprising a result. Not to reinvent the wheel, here is the relationship you are looking for with explanation on Wolfram.

2

By OP: I've finally found the source: E. M. Stein, G. Weiss, Introduction to Fourier Analysis on Euclidean Spaces, Princeton University Press, 1971. §1. Corollary 1.26 (p.15)

1

In A positive "Fourier transform" is integrable Zarrax gave a great answer to this question using mollifiers. He/she also provides a counterexample in the case $f$ is not continuous at $x=0$.

1

Maxima has the concept of nouns. If integrate fails for whatever reason, it returns the noun for what it would have done. Your integrates are failing and returning the noun output. You can also specify a function as a noun with a preceding single quote. Then you can set up values and variables before calling the function with ev and telling it to evaluate ...

2

Deciding if a function is odd or even is very useful for Fourier series. If $\mathrm{h}$ is an even function then $\displaystyle{\int_{-L}^L\mathrm{h}(x)~\mathrm{d}x = 2\int_0^L\mathrm{h}(x)~\mathrm{d}x}$ If $\mathrm{h}$ is an odd function then $\displaystyle{\int_{-L}^L\mathrm{h}(x)~\mathrm{d}x = 0.}$ We also need to remember that even $\times$ even = ...

0

For part 1), note that $f(x) = \frac{-1}{2} (e^{-x^{2}})'$ For Part 2), if $g(\lambda) = \frac{2}{1 + \lambda^{2}}$, then what will $2*g(2 \lambda)$ be?

1

To make Fourier Transforms easy there are a few simple rules like a rule for scaling, modulation, translation and so on. I will use these rules as we go on. They are fundamental when doing Fourier-transforms and you should be familiar with them. Here on the first page you can find many of them http://www.ibk.ethz.ch/ch/education/identmeth/fourier.pdf In the ...

1

A simple example of a family of functions with the properties you cite is $$G_{y,\alpha}(x)=e^{-\alpha(x-y)^{2}},\;\;\;\alpha > 0,y\in\mathbb{R}.$$ This function is analytic in $x$, and presumably that means you know that $(G_{y,\alpha}^{\wedge})^{\vee}=G_{y,\alpha}$ for all $\alpha > 0$, $y\in\mathbb{R}$. (Here I have used the ...

0

If you apply the change of variables $x = tL/\pi$ (or $t = \pi x/L$) to the integral over $[-\pi,\pi]$, you get $$\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\overline{g(t)}dt = \frac{1}{2L}\int_{-L}^{L} f(\pi x/L)\overline{g(\pi x/L)}dx.$$

0

Circulant Matrix will have a constant(same) diagonal entries.

2

As was said in comments, cosines and sines differ only by phase shift. The difference in their performance arises from their boundary behavior. On an interval $[0,\ell]$, the sine system $\left\{\sin \frac{2\pi n }{\ell}\right\}$ satisfies the Dirichlet boundary condition, attaining zero value at $0,\ell$. A generic function, e.g., one describing the ...

1

Try replacing $L$ with $\pi$ in the definition you have for the interval $[-L,L]$. Notice that you get back the original definition for $[-\pi,\pi]$. The generalization comes from wanting the cosine and sine parts to still be periodic over the interval $[-L,L]$, just as they were over $[-\pi,\pi]$. For the series on the interval $[0,2\pi]$, a useful trick ...

0

So what you do is to realize that you can replace $(k_x,k_y,k_z)\cdot(x,y,z)$ by $\kappa r \cos(\theta)$, where $\kappa = \sqrt{k_x^2 + k_y^2 + k_z^2}$ and $r = \sqrt{x^2+y^2+z^2}$ and $\theta$ is the angle between $(k_x,k_y,k_z)$ and $(x,y,z)$. Also, you have that the Fourier transform of the constant function is the Dirac delta function, so we can ...

1

For $x\ne 0,$ $$\hat f (x) =\int_\infty^\infty f(t)e^{-ixt}\, dt =1/(ix)\int_\infty^\infty f'(t)e^{-ixt}\, dt = 1/(ix)^2\int_\infty^\infty f''(t)e^{-ixt}\, dt.$$ I've integrated by parts twice; each time the boundary terms disappear because of compact support. The last expression is $1/x^2$ times a function in $C_0.$ We already know $\hat f$ is bounded. ...

3

It does work in a (very poor) sense, because $(-1)^n = -1$ for $n=1$, so in $$a_n = \frac{-2}{\pi(n^2-1)}[(-1)^n+1]$$ there are terms in the numerator and denominator that are zero. So if you prioritize the $(-1)^n+1 = 0$ part then we could interpret $a_1$ to be $0$. But it's more rigorous to say that $a_1$ is ill-defined (since $0/0$ is ill-defined), and ...

0

Hint?: This looks like the heat equation with added terms representing heat loss: https://en.wikipedia.org/wiki/Heat_equation#Derivation_in_one_dimension so you expect it to go to $0$. The second condition is strange to me - usually one deals with an initial non-negative temperature distribution at $t=0$.

0

[1] I assume that the term in which $t=n$,if it exists, equals $\sin t$...[2] Observe that $\sin (\pi t- \pi n)=(-1)^n \sin \pi t$ for integer $n$...[3] Assume uniform convergence. Then for some $N > |t|$ we have $|f_{N+1}(t)-f_N(t)| < 1/2$ for all $t$. But for $N>|t|$, we have $|f_{N+1}(t)-f_N(t)|=|[(2N+2) \sin \pi t ]/[\pi (t^2-(N+1)^2]|$ which ...

1

Consider any $h$ in $L^2(c,d)$. We note that for any $\epsilon > 0$, there exists an $N_0$ such that for $N > N_0$, $$\newcommand{\ip}[1]{\left\langle #1 \right\rangle} \int_a^b \left|h - \sum_{n=1}^N \ip{h,\phi_n}\phi_n\right|^2\,dx \leq \epsilon$$ We then note that $h \circ f \in L^2_{f'}(a,b)$, and that $$\int_c^d \left|h \circ f - ... 4 Ok, i will give a derivation of the leading order asymptotics for x\rightarrow\infty. For now, i also concentrate on the case where x>0, the opposite possibility should be doable with quite similiar techniques. We start from the complex function given by @Leucippus.$$ f(z,x)=\frac{e^{2ixz}}{\Gamma(\frac12+\frac{i}{\pi}z)\cosh^{3/2}(z)} $$We ... 2 Consider the integral:$$ I = \int_0^\infty \frac{\arctan(x)}{1+x^2} dx\,. $$By evaluating this integral through a change of variables:  u=\arctan x ,  and  du= {1\over 1+x^2}dx,$$ I=\int_0^ {\pi\over 2} u du= {\pi^2\over 8} \ $$Now,we notice that:$$ \arctan(x)=\int_0^1 \frac{x}{1+x^2y^2} dy\, $$Utilizing this fact, we can see:$$ I = ...

0

You've ignored the evaluation terms in the integration by parts: $$2\pi\hat{g}(0) = \int_{-\pi}^{\pi}g(t)dt=tg(t)|_{t=-\pi}^{\pi}-\int_{-\pi}^{\pi}tg'(t)dt \\ = \pi g(\pi)+\pi g(-\pi)-\int_{-\pi}^{\pi}tf(t)dt.$$ For $n \ne 0$, $$2\pi \hat{g}(n) = ... 0 Let the input signal be given by x[n], for n=0,1,\dots,N-1. We want to look at its N-point DFT (the FFT is just an efficient way to calculate the DFT). The k^{\mathrm{th}} sample of the DFT is given by$$ w[k] = \sum_n e^{-j2\pi k \cdot n / N} x[n] What you want to know is the frequency of this DFT bin. The frequency of the DFT bin is just the ... 1 \begin{align} & 1 - 2(1-\delta)\cos 2x + (1-\delta)^{2} \\ & = 1-2(1-\delta)\{1-2\sin^{2}x\}+(1-\delta)^{2} \\ & = (1-(1-\delta))^{2}+4(1-\delta)\sin^{2}x \\ & = \delta^{2}+4(1-\delta)\sin^{2}x \end{align} Therefore, 0 \le P_{\delta}(x) = \frac{1}{2\pi}\frac{\delta(2-\delta)}{\delta^{2}+4(1-\delta)\sin^{2}x}. $$So, ... 2 Recall that the FT in 3D is$$F(\mathbf{k}) = \int_{\mathbb{R}^3} d^3 \mathbf{r} \, f(\mathbf{r}) e^{i \mathbf{k} \cdot \mathbf{r}} $$Now, we have radial symmetry. Thus, F is radially symmetric and we may write$$\mathbf{k} \cdot \mathbf{r} = k r \cos{\theta} $$So we have$$F(k) = 2 \pi \int_0^{\infty} dr \, r \, e^{-a r} \, \int_0^{\pi} d\theta \, ...

3

Your function is odd and the interval is symmetric about zero, so the $a_n$, including $a_0$, should indeed be zero. But the $b_n$ should not be zero. Something that might help: because $f$ and $\sin$ are both odd, $\int_{-\pi}^\pi f(x) \sin(nx) dx = 2 \int_0^\pi x^2 \sin(nx)$. You can calculate that with integration by parts.

2

You can separate the dimensions. You start by trying to find a function $G(x,t)$ such that $G(x,t)$ satisfies the homogeneous heat equation in all of space and time, except formally at $x=t=0$; At $t=0$, $G(x,t)$ is zero forall non-zero values of $x$; For any nonzero real $(a,b)$, $\int_{-|a|}^{|b} G(x,0) dx = 1$. I'm not saying yet that this function ...

2

The Fourier transform of a nonzero function with compact support does not have compact support. So if something is "local in time" (i.e. it vanishes outside a finite time interval), then it is not "local in frequency" (i.e. it does not vanish outside any finite frequency interval). Similarly, the inverse Fourier transform of a nonzero function with compact ...

1

Let $\phi\geq 0$ be a Lebesgue integrable function with $\int_{\mathbb{R}}\phi=1$, and let $\mu$ be a (Borel) probability measure on $\mathbb{R}$. Define $\phi_{\delta}$ as above (I want to reserve the letter $\epsilon$ for later). Let $g$ be a continuous, bounded function on $\mathbb{R}$. Observe that ...

2

Given the two conditions listed in the proposed problem let $p = 2 z$ to obtain \begin{align} \frac{\psi(x)}{\sqrt{\pi}} = \int_{\Gamma} \, \frac{e^{2 i x \, z} \, dz}{\Gamma\left(\frac{1}{2} + \frac{i \, z}{\pi}\right) \, \left(\cosh(z)\right)^{\frac{3}{2}}}. \end{align} There is a branch cut due to the $(\cosh(z))^{3/2}$ term. The poles are determined by ...

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