2
votes
Do there exist mathematical transforms other than the Fourier Transform for which there exists some sort of a fast convolution theorem?
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
$$\mathcal{M}_y[[f(x) * g(x)](y)](s)=\mathcal{M}_x[f(x)](s)\cdot ...
- 6,693
2
votes
Do there exist mathematical transforms other than the Fourier Transform for which there exists some sort of a fast convolution theorem?
The discrete Fourier transform, say in the form of the evaluation/interpolation ring isomorphism
$$
\mathbb{C}[x]/(x^n-1)\cong\prod_i\mathbb{C}[x]/(x-\zeta^i)\cong\mathbb{C}^n
$$
takes cyclic ...
- 9,559
2
votes
All DFT of binary numbers subsets of prime length are nonzero
This DFT $\hat S_k$ equals $f(\zeta_k)$, where $f(t) = x_p + \sum_{n=1}^{p-1} x_n t^n$ is a polynomial with integer coefficients and $\zeta_k = \exp(2\pi i k/p)$ is a primitive $p$th root of unity. ...
- 73.6k
2
votes
How does this definition of Fourier transform in Fulton and Harris 3.32 relate to the usual notion of Fourier transform?
This is a long story. This is the Fourier transform on a finite group. It's a generalization of the discrete Fourier transform, which it specializes to when $G = \mathbb{Z}/n\mathbb{Z}$. The sense in ...
- 409k
1
vote
Accepted
Step in the proof that the function is nowhere differentiable in the Fourier Analysis textbook
One way to see this is as follows, if $g$ is assumed to be continuous on the compact interval $[x_0 - \pi, x_0 + \pi]$:
Since $g$ is differentiable at $x_0$, we have: $$\frac{g(x_0 - t) - g(x_0)}{-t} \...
- 3,846
1
vote
Accepted
Trigonometric polynomial derivative upper bound
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})(...
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