$ \text { Find the Fourier series of }f(x)=x \text { in the interval }(0,2 \pi) \text {. } $

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written 2.9 years ago by

prajapatijaimin • 3.3k

Solution:

The Fourier series of $f(x)$ with period $2 \pi$ is given by,

$f(x)=a_{0}+\sum_{n=1}^{\infty} a_{n} \cos n x+\sum_{n=1}^{\infty} b_{n} \sin n x \\$

$\begin{aligned} a_{0} &=\frac{1}{2 \pi} \int_{0}^{2 \pi} f(x) \mathrm{d} x \\\\ &=\frac{1}{2 \pi} \int_{0}^{2 \pi} x \mathrm{~d} x \\\\ &=\frac{1}{2 \pi}\left|\frac{x^{2}}{2}\right|_{0}^{2 \pi} \\\\ &=\frac{1}{2 \pi}\left(\frac{4 \pi^{2}}{2}\right) \\\\ &=\pi \\\\ a_{n} &=\frac{1}{\pi} \int_{0}^{2 \pi} f(x) \cos n x \mathrm{~d} x \\\\ &=\frac{1}{\pi} \int_{0}^{2 \pi} x \cos n x \mathrm{~d} x \\\\ &=\left.\frac{1}{\pi}\right|^{2} x\left(\frac{\sin n x}{n}\right)-(1)\left(-\frac{\cos n x}{n^{2}}\right)_{0}^{2 \pi} \\\\ &=\frac{1}{\pi}\left(\frac{\cos 2 n \pi}{n^{2}}-\frac{\cos 0}{n^{2}}\right)[\because \sin 2 n \pi=\sin 0=0] \\\\ &=0 \\\\ &=\frac{1}{\pi} \int_{0}^{2 \pi} f(x) \sin n x \mathrm{~d} x \\\\ & {[\because \sin n x \mathrm{~d} x} \\\\ & {[\cos 2 n \pi=\cos 0=1] } \\ \end{aligned}$

$\begin{aligned} &=\frac{1}{\pi}\left|x\left(-\frac{\cos n x}{n}\right)-(1)\left(-\frac{\sin n x}{n^{2}}\right)\right|_{0}^{2 \pi} \\\\ &=\frac{1}{\pi}\left[-2 \pi\left(\frac{\cos 2 n \pi}{n}\right)\right] \quad[\because \sin 2 \pi \\\\ &=-\frac{2}{n} \quad[\because \cos 2 n \pi=1] \\ \end{aligned}$

Hence,

$f(x)=\pi-2 \sum_{n=1}^{\infty} \frac{1}{n} \sin n x \\$

$x=\pi-2\left(\sin x+\frac{1}{2} \sin 2 x+\frac{1}{3} \sin 3 x+\cdots\right) \\$

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