Find a Fourier series to represent $x-x^{2}$ from $x=-\pi$ to $x=\pi$. Hence show that, $$ \frac{1}{1^{2}}-\frac{1}{2^{2}}+\frac{1}{3^{2}}-\frac{1}{4^{2}}+\ldots \ldots=\frac{\pi^{2}}{12} . $$

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written 3.0 years ago by

prajapatijaimin • 3.3k

solution:

Let , $x-x^{2}=\frac{a_{0}}{2}+\sum_{n=1}^{\infty} a_{n} \cos n x+\sum_{n=1}^{\infty} b_{n} \sin n x$ By Euler's formulae, we have,

$a_{0}=\frac{1}{\pi} \int_{-\pi}^{\pi}\left(x-x^{2}\right) d x=\frac{1}{\pi}\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{-\pi}^{\pi}\\$

$=\frac{1}{\pi}\left[\left(\frac{\pi^{2}}{2}-\frac{\pi^{3}}{3}\right)-\left(\frac{\pi^{2}}{2}+\frac{\pi^{3}}{3}\right)\right]=-\frac{2 \pi^{2}}{3}\\$

$$ \begin{aligned} a_{n} &=\frac{1}{\pi} \int_{-\pi}^{\pi}\left(x-x^{2}\right) \cos n x d x \\\\ &=\frac{1}{\pi}\left[\left(x-x^{2}\right) \frac{\sin n x}{n}-(1-2 x)\left(-\frac{\cos n x}{n^{2}}\right)+(-2)\left(-\frac{\sin n x}{n^{3}}\right)\right]_{-\pi}^{n} \\\\ &=\frac{1}{\pi}\left[(1-2 \pi) \frac{\cos …

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