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Yes, for the Mellin convolution $$[f(x) * g(x)](y)=\int\limits_0^\infty f(x)\, g\left(\frac{y}{x}\right)\,\frac{dx}{x}\tag{1}$$ one has $$\mathcal{M}_y[[f(x) * g(x)](y)](s)=\mathcal{M}_x[f(x)](s)\cdot ... 2 votes ### Do there exist mathematical transforms other than the Fourier Transform for which there exists some sort of a fast convolution theorem? The discrete Fourier transform, say in the form of the evaluation/interpolation ring isomorphism$$ \mathbb{C}[x]/(x^n-1)\cong\prod_i\mathbb{C}[x]/(x-\zeta^i)\cong\mathbb{C}^n $$takes cyclic ... 2 votes ### All DFT of binary numbers subsets of prime length are nonzero This DFT \hat S_k equals f(\zeta_k), where f(t) = x_p + \sum_{n=1}^{p-1} x_n t^n is a polynomial with integer coefficients and \zeta_k = \exp(2\pi i k/p) is a primitive pth root of unity. ... 2 votes ### How does this definition of Fourier transform in Fulton and Harris 3.32 relate to the usual notion of Fourier transform? This is a long story. This is the Fourier transform on a finite group. It's a generalization of the discrete Fourier transform, which it specializes to when G = \mathbb{Z}/n\mathbb{Z}. The sense in ... 1 vote Accepted ### Step in the proof that the function is nowhere differentiable in the Fourier Analysis textbook One way to see this is as follows, if g is assumed to be continuous on the compact interval [x_0 - \pi, x_0 + \pi]: Since g is differentiable at x_0, we have:$$\frac{g(x_0 - t) - g(x_0)}{-t} \...
In fact, we can show that $\|P_N'\|\leq 4\pi N\|P_N\|_{\infty}$. To prove this, we firstly claim that $P_N'(x)$ can be represented by  \frac{P_N'(x)}{2\pi iN}=((e^{-2\pi iN(\cdot)}P_N)*F_{N-1})(...