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Given an "L" periodic Fourier series

$$\phi(x) = \sum_{1}^{\infty} \left( \frac{2}{l} \int_{0}^{l}\phi(x) sin\left(\frac{n \pi x}{l}\right)\, \mathrm dx \right) sin\left(\frac{n \pi x}{l}\right)$$

Is there a simple way to derive the Fourier Sine Transform? I've found derivations of $2 \pi$ periodic Sine transforms, but was hoping to better understand the "L" periodic case.

MarianD
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  • What is the Fourier sine transform to you, of what function ? Also rename your inner $x$ please. – reuns Apr 01 '19 at 02:09
  • Broadly, I understand the Fourier series to be an infinite series representation of a periodic function defined on its period. It is my understanding that the Fourier transform is taking the limit of this period to be infinite, allowing you to approximate a periodic functions. I recognize that the Fourier transform is usually represented in terms of complex exponential but I’m just trying to make the connection to a basic sine series more clear. –  Apr 01 '19 at 02:14

1 Answers1

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Let $f,f'$ be $L^1$ and $$\hat{f}(\xi) = \int_{-\infty}^\infty f(x)e^{-2i \pi \xi x}dx$$ For any $L$ let $$F_L(x)=\sum_n f(x+nL)$$ Then $F_L$ is $L$-periodic and $F_L,F_L'$ are integrable on one period and we have the uniformly convergent Fourier series $$F_L(x)= \sum_k \frac{1}{L}\hat{F}_L(k) e^{2i \pi xk/L}=\sum_k \frac{1}{L}\hat{f}(k/L) e^{2i \pi xk/L}$$ Locally uniformly $$f(x)=\lim_{L \to \infty}F_L(x)$$

And $$f(x)=\lim_{L \to \infty}F_L(x) = \lim_{L \to \infty}\sum_k \frac{1}{L}\hat{f}(k/L) e^{2i \pi xk/L} = \int_{-\infty}^\infty \hat{f}(\xi) e^{2i \pi \xi x}d\xi$$ where the series and the integral converge absolutely.

For the sine transform it is the same starting with $f$ odd, so that grouping the positive and negative frequencies together give series in $\sin$.

reuns
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  • Just to be clear, this is just a standard Fourier transform, correct? –  Apr 01 '19 at 03:38
  • @MarcoSun The Sine Fourier series/transform is just what you get from the standard Fourier series/transform when starting with $f$ odd (or odd-symmetrized) – reuns Apr 01 '19 at 21:02