# Tag Info

5

Knowing this series is convergent, you can find its sum using the formula $$\sum_{k=1}^{\infty} \frac{z^k}{k} = -\log{(1-z)}.$$ Then the sum is $$-\frac{1}{2}\left( \log{(1-e^{2\pi i \theta})}+\log{(1-e^{-2\pi i \theta})} \right),$$ and the limit of this as $\theta \to 0$ does not exist, since $\log{z}$ diverges as $z \to 0$.

4

Since $\hat{f}\in\ell^1\implies f\in L^\infty$ and $\hat{f}\in\ell^2\implies f\in L^2$, Riesz-Thorin interpolation guarantees that If $\hat{f}\in\ell^q$ where $\frac1p+\frac1q=1$ and $1\le q\le2$, then $f\in L^p$ Therefore, since the $\hat{f}$ you give above is in $\ell^q$ for all $q\gt1$, we have that $f\in L^p(\mathbb{T})$ for all $p\lt\infty$. ...

3

A correct way to do this with the formalism of the distributions: Let $T \in \mathcal S'$ , then $\hat{T}$,the Fourier transform of $T$ is defined by $$\forall \varphi \in \mathcal S, \langle \hat{T}, \varphi \rangle = \langle T, \hat{\varphi} \rangle$$ So the Fourier transform of $\delta_{x_0}$ is given by $$\forall \varphi \in \mathcal S, \langle ... 3 (need to fix the end. This is the right idea, but the end conclusion appears wrong.)$$-\log(1-z) = \sum_{n=1}^\infty \frac{z^n}{n}$$when |z|\leq 1 and z\neq 1. Then \sum_{n=1}^\infty \frac{\sin nx}{nx} is the imaginary part of:$$\frac{1}{x}\sum_{n=1}^\infty \frac{e^{inx}}{n}=-\frac{\log(1-e^{ix})}{x}$$When x=\frac{\pi}{4}, this is the ... 2 Write the Fourier transform as \mathcal{F}: \mathcal{S} \to \mathcal{S}. Observe that \mathcal{F}^{-1} \neq \mathcal{F} (assuming we choose the unitary definition), but that$$ \mathcal{F}^{-1} f(x) = \mathcal{F} f(-x) $$Therefore we conclude that$$ \mathcal{F}\mathcal{F}\mathcal{F}\mathcal{F} = \mathrm{Id}.$$This tells you that the eigenvalues must ... 2 The reason is as follows: by the Fourier inversion theorem, if f,\widehat{f}\in L^1, then we know that f agrees almost everywhere with a continuous function f_0. Moreover, f_0 = (f^{\vee})^{\wedge} so, since f^{\vee} is in L^1 (being that it is related to \widehat{f} via reflection), f_0 is bounded by the Riemann-Lebesgue Lemma. Hence f is ... 2 Just in order to fix notations, let:$$ \mathcal{F}(f)(\omega) = \widehat{f}(\omega) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f(x)\, e^{-i\omega x}\,dx.\tag{1} $$Assuming f_1(x)=e^{-|x|} and f_2(x) = \mathbb{1}_{[-1,1]}(x) we have:$$ \widehat{f_1}(\omega)=\sqrt{\frac{2}{\pi}}\frac{1}{\omega^2+1},\qquad ...

2

First, let me say that you can show this pretty easily be writing the convolution as a matrix vector product, with the matrix being a circulant one defined by one of the vectors. The result falls out due to the DFT diagnolizing circulant matrices. Anyway, you can also show this directly substituting the discrete convolution formula, and playing with ...

1

In view of this comment clarifying the question, we can give a counter example. Consider $$f = \exp( - |x|)$$ which is in any $L^p$. Its Fourier transform is well known to be (up to a constant depending on the dimension) $$(1 + |\xi|^2)^{-(n+1)/2}$$ This function is smooth and decays much faster (with all derivatives) than what you supposed. (In ...

1

The displacement $u(x,t)$ of the string is described by the wave equation for $t \ge 0$ and $0 \le x \le 1$: $$u_{tt} = c^{2}u_{xx},\\ u(x,0)=f(x),\;\;u_{t}(x,0)=0,\\ u(0,t)=u(1,t)=0.$$ The constant $c > 0$ is a physical constant associated with the properties of the string. You start by looking for separated solutions $T(t)X(x)$ with ...

1

You say you get to $$-\int_{-l}^l e^{ikx} \phi''(x) \, \mathrm{d} x$$ but then apply the Riemann-Lebesgue lemma.

1

Since the integrand is even, your integral is $2\int_0^\infty\dots dt$. For this MathWorld (38) gives a closed form in terms of Eulerian numbers $$\int_0^\infty \left(\frac{\sin x}{x}\right)^{2n} \mathrm{d} x = \frac{\pi}{2(2n-1)!} \genfrac{<}{>}{0}{}{2n-1}{n-1}$$

1

Here's a proof of Morlet's inversion formula. We define \begin{align*} C_\psi &:= \int_0^\infty \dfrac{\overline{\hat{\psi}(u)}}{u}\, \mathrm{d}u \end{align*} and assume that this quantity is finite. Assume that $x,\psi \in L^2(\mathbb{R})$. I make this assumption so that Parseval's identity holds. Then, using Parseval's identity and basic properties of ...

1

Certainly convergence of the Riemann sums does not follow from integrability and continuity of $g$. Consider this function $g$: Spikes of decreasing width. Area of the spike centered at $m\pi$ is $1/2^m$. We have: $\bullet$ $g$ is continuous $\bullet$ $\int_{-\infty}^{+\infty} |g(\xi)|\;d\xi = 1$ $\bullet$ $g(m\pi) = 1$ for $m=1,2,3,\dots$; ...

1

You've got to know a bit about generalized functions for this type of integral to make sense. The generalized function $1(x)$ can be defined by the sequence $1_n(x)=\exp(-x^2/n^2)$. The identification with $1$ comes from the limit, $$\lim_{n\rightarrow\infty}\int_{-\infty}^{\infty} f(x) 1_n(x) \mathrm{d}x = \int_{-\infty}^{\infty} f(x) \mathrm{d}x,$$ ...

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