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One of the basic properties of Fourier series is localization: If $f=0$ in a neighborhood of $t$ then the Fourier series for $f$ converges to $0$ at $t$. Hence if $f=g$ in a neighborhood of $t$ then the two Fourier series either both converge at $t$ or both diverge at $t$. So: Given $f_1$ with a divergent series at $t_1$ and $f_2$ with a divergent series at ...

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By Parseval's theorem we have \begin{align} \int_{-\infty}^{\infty}\frac{\sin^{4}\left(x\right)}{x^{4}}dx= & \pi\int_{-\infty}^{\infty}\frac{\sin^{4}\left(\pi t\right)}{\left(\pi t\right)^{4}}dt \\ = & \pi\int_{-1}^{0}\left(1+x\right)^{2}dx+\pi\int_{0}^{1}\left(1-x\right)^{2}dx \\ = & \frac{2}{3}\pi \end{align} and now observe that $$\... 2 Fourier transform of f_a is:$$\hat f_a = \frac 1a\sqrt{\frac\pi 2}e^{-a|\omega|} \tag 1$$The fourier transform of a convolution is a regular product:$$\mathscr F\Big[ f_a\star f_b \Big] = \hat f_a\cdot\hat f_b = \frac 1a\sqrt{\frac\pi 2}e^{-a|\omega|} \cdot \frac 1b\sqrt{\frac\pi 2}e^{-b|\omega|} = \frac 1{ab}\frac\pi 2e^{-(a+b)|\omega|} \tag 2$$The ... 2 The minimum set of conditions needed can be expressed using the Lebesgue integral. Essentially you need f to be absolutely continuous with derivative f' \in L^1. Absolutely continuous means f is the integral of its derivative:$$ f(y)-f(x)=\int_{x}^{y}f'(t)dt. $$This all works out very nicely using the Lebesgue integral. For the Riemann ... 0 I am also learning this stuff at the moment so I'll let you know what I think. I don't see how you would use closed graph theorem, in fact, you would need either \mathcal{S}=\mathcal{S}(\mathbb{R}^d) to be a Banach space or compact Hausdorff. I do not believe either conditions are satisfied (\mathcal{S} is normable or compact, at least I don't think so). ... 0 The shift is defined by g_a(x) = f(x-a). Then you write$$F[g_a](\xi) = \int_{\Bbb R}g_a(x)\exp(-ix\xi)dx = \int_{\Bbb R}f(x-a)\exp(-ix\xi)dx.Conceptually, you first apply the shift and then apply the Fourier transform, but you can apply the shift only to the function, there is no sense in applying it to the exponent. 0 Let g_0 = \frac{1}{2\pi}\int_{0}^{2\pi}g(x)dx and f_0=\frac{1}{2\pi}\int_{0}^{2\pi}f(x)dx. Then \begin{align} \frac{1}{2\pi}\int_{0}^{2\pi}f(x)g(nx)dx &=\frac{1}{2\pi}\int_{0}^{2\pi}f(x)(g(nx)-g_0)dx+f_0g_0 \\ &=\frac{1}{2\pi}\sum_{k=1}^{n}\int_{\frac{2\pi}{n}(k-1)}^{\frac{2\pi}{n}k}f(x)(g(nx)-g_0)dx+f_0g_0 \\ &=\frac{1}{... 2 Cosine is an even function; sine is odd. Using negative indices for these functions adds no information because, for example,5\sin 2x+ 4\sin(-2x) = \sin 2x$$On the other hand, e^{-2ix} is not just a constant multiple of e^{2ix}. And we need both of these to express \sin 2x, as in$$\sin 2x = \frac1{2i}(e^{2ix} - e^{-2ix})$$A more precise ... 0 After a simple classification I have figured it out. We could solve it by induction on the degree of the polynomial P(x). If deg P(x)=1, then c_1 \in \mathbb{R} \setminus \mathbb{Q} and apparently <c_1n+c_0> is equidistributed in [0,1). Assume deg P(x)=m and at least one of its coefficient c_1,……,c_m is irrational, then <P(n)> ... 1 You are trying to compute an NTT of length n = 2. You chose the modulus p = 5 with a primitive root w = 2, which are appropriate so far. The primitive root w has order p - 1 = 4, which means 2^4 \equiv 1 \mod 5. But what you want is an nth root of unity a, where a^n = a^2 \equiv 1 \mod 5. The easiest way to obtain this is to compute a = w^... 3 We suppose the choice m = j, C = j does not work for any j and derive a contradiction. There thus exist Schwartz functions \phi_j such that |u(\phi_j)| > j\|\phi_j\|_j, so if we define$$\psi_j = {1\over{j\|\phi_j\|_j}}\phi_j,$$it is clear that |u(\psi_j)| > 1 while \|\psi_j\|_j = 1/j. To show that \psi_j \to 0 in \mathcal{S}(\mathbb{R}^... 2 We would still use the fast Fourier transform (FFT), but with interpolation. As pointed out, the signal can be written as$$s(t) = \int_{-\infty}^\infty d\nu\,\hat{s}(\nu)e^{i2\pi\nu t} = -\int_{-\infty}^\infty d\lambda {{\hat{s}(1/\lambda)}\over{\lambda^2}} e^{i2\pi t/\lambda},$$or$$s(t) = \int_{-\infty}^\infty {{d\lambda}\over{\lambda}} S(\lambda)e^{i2\pi t/...

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This is a basic question about the overloaded usage of the word "spectrum," as all three terms are called spectra. Only the eigenvalues of a linear operator and the spectrum of a ring are directly related; Fourier coefficients are separate. As such, we are only going to explain that connection. I am not sure if there is an "explanation for dummies," in the ...

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The Fourier series of a periodic function can be thought of as its decomposition into eigenfunctions for the translation operator $f(x) \mapsto f(x + t)$ on periodic functions. Say we're talking about $\mathbb{C}$-valued functions with period $2 \pi$: then the eigenfunctions are $e^{inx}, n \in \mathbb{Z}$ with eigenvalues $e^{int}$ (for translation by $t$). ...

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If $\|g_n - f\|_{\infty} \to 0$ when $n \to \infty$, then $$\int_0^1 |g_n-f|^2 \leq \int_0^1 \|g_n - f\|_{\infty}^2 = \|g_n - f\|_{\infty} ^2 \to 0$$ when $n \to \infty$, so that $\|g_n-f\|_{L^2(0,1)} \to 0$ when $n \to \infty$.

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Your approach is correct. Under the given assumptions, the function $$g(y) = \frac{f(x-y)-f(x)}{\sin(\pi y)}$$ is continuous and $1$-periodic, with $g(0)=-f'(x)/\pi$. In particular, it is integrable which is enough to apply the Riemann-Lebesgue lemma. More generally, this argument works for Hölder continuous $f$: an estimate of the form $|f(x-y)-f(x)|\le ... 2 The Hilbert transform$\mathcal{H}$is sometimes said to be an anti-involution, as$\mathcal{H(H(u))}=-u$(see Hilbert, inverse transform). I see this as a sub-case for your question only. I have witnessed$n$-idempotence too, so you could call it$4\$-idempotent function.

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