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I am wondering if the function $\mathrm{sinc}(x)=\frac{\sin x}{x}$ can be represented in terms of Fourier series?

Thank you.

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    It's not periodic, so you'll need to restrict the domain somewhat. If the restricted domain is symmetric about the origin, your series becomes a cosine series since the sine cardinal is even.2011-12-17
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    Were you possibly thinking about the _Fourier transform_ of the sinc function which turns out to be a rect function ?2011-12-17
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    No, I am thinking about integral $\int_{-\pi}^{\pi}|sinc{x}|^ndx$.2011-12-17
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    I did not understand about the cosing sing. Could you elaborate, please. thank you!2011-12-17

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Well, in order to compute the Fourier series of any function, you need to specify the interval $(-\ell,\ell)$. But once you do, sure you can find the Fourier series of $f(x)={\sin x\over x}$ in the usual way.

For example, taking $\ell=\pi$, here is a plot of $f(x)$ (black) and the first three terms of its Fourier series (blue):

Mathematica graphics

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Yes, it can be represented in terms of the Fourier Series: $$\text {sinc} (x) =\lim_{k\to\infty}\prod_{n=1}^k\cos\left(\frac{x}{2^n}\right)= \lim_{k\to\infty}\frac{1}{2^{k - 1}}{\sum_{n = 1}^{2^{k-1}}\cos\left(\frac{(2n - 1)x}{2^k}\right)}.$$ This is because: $$\prod_{n=1}^k\cos\left(\frac{x}{2^n}\right)= \frac{1}{2^{k - 1}}{\sum_{n = 1}^{2^{k-1}}\cos\left(\frac{(2n - 1)x}{2^k}\right)}.$$