example of Fourier series

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Example:

Let $f\colon(-\pi,\pi)\to\mathbb{R}$ be the “identity” function, defined by

f(x)=x,\text{ for all }x\in(-\pi,\pi).

We will compute the Fourier coefficients for this function. Notice that $\cos(nx)$ is an even function, while $f$ and $\sin(nx)$ are odd functions.

$\displaystyle a_{0}^{f}$	$\displaystyle=$	$\displaystyle\frac{1}{2\pi}\int_{{-\pi}}^{{\pi}}f(x)dx=\frac{1}{2\pi}\int_{{-% \pi}}^{{\pi}}xdx=0$
$\displaystyle a_{n}^{f}$	$\displaystyle=$	$\displaystyle\frac{1}{\pi}\int_{{-\pi}}^{{\pi}}f(x)\cos(nx)dx=\frac{1}{\pi}% \int_{{-\pi}}^{{\pi}}x\cos(nx)dx=0$
$\displaystyle b_{n}^{f}$	$\displaystyle=$	$\displaystyle\frac{1}{\pi}\int_{{-\pi}}^{{\pi}}f(x)\sin(nx)dx=\frac{1}{\pi}% \int_{{-\pi}}^{{\pi}}x\sin(nx)dx=$
	$\displaystyle=$	$\displaystyle\frac{2}{\pi}\int_{{0}}^{{\pi}}x\sin(nx)dx=\frac{2}{\pi}\left(% \left[-\frac{x\cos(nx)}{n}\right]_{0}^{{\pi}}+\left[\frac{\sin(nx)}{n^{2}}% \right]_{0}^{{\pi}}=\right)=(-1)^{{n+1}}\frac{2}{n}$

Notice that $a_{0}^{f},a_{n}^{f}$ are $0$ because $x$ and $x\cos(nx)$ are odd functions. Hence the Fourier series for $f(x)=x$ is:

	$\displaystyle f(x)=x$	$\displaystyle=$	$\displaystyle a_{0}^{f}+\sum_{{n=1}}^{{\infty}}(a_{n}^{f}\cos(nx)+b_{n}^{f}% \sin(nx))=$
		$\displaystyle=$	$\displaystyle\sum_{{n=1}}^{{\infty}}(-1)^{{n+1}}\frac{2}{n}\sin(nx),\quad% \forall x\in(-\pi,\pi)$

For an application of this Fourier series, see value of the Riemann zeta function at $s=2$ .

Synonym:

example of Fourier coefficients

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