If, $ f(x)=x, \quad 0<x<\frac{\pi}{2} $ $ =\pi-x, \quad \frac{\pi}{2}<x<\pi $ show that, (i) $f(x)=\frac{4}{\pi}\left[\sin x-\frac{\sin 3 x}{3^{2}}+\frac{\sin 5 x}{5^{2}}-\ldots \ldots\right]$

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written 2.9 years ago by

prajapatijaimin • 3.3k

Solution:

(i) For the half-range sine series,

Let, $f(x)=\sum_{n=1}^{\infty} b_{n} \sin n x \\$

Then, $b_{n}=\frac{2}{\pi} \int_{0}^{\pi} f(x) \sin n x d x=\frac{2}{\pi}\left[\int_{0}^{\pi / 2} x \sin n x d x+\int_{\pi / 2}^{\pi}(\pi-x) \sin n x d x\right] \\$

$\begin{aligned} &=\frac{2}{\pi}\left[x\left(-\frac{\cos n x}{n}\right)-1 \cdot\left(-\frac{\sin n x}{n^{2}}\right)\right]_{0}^{\pi / 2}+\frac{2}{\pi}\left[(\pi-x)\left(-\frac{\cos n x}{n}\right)-(-1)\left(-\frac{\sin n x}{n^{2}}\right]_{0}^{\pi}\right. \\\\ &=\frac{2}{\pi}\left[-\frac{\pi}{2 n} \cos \frac{n \pi}{2}+\frac{1}{n^{2}} \sin \frac{n \pi}{2}\right]+\frac{2}{\pi}\left[\frac{\pi}{2 n} \cos \frac{n \pi}{2}+\frac{1}{n^{2}} \sin \frac{n \pi}{2}\right] \\\\ &=\frac{2}{\pi}\left[\frac{2}{n^{2}} \sin \frac{n \pi}{2}\right]=\frac{4}{\pi n^{2}} \sin \frac{n \pi}{2} \end{aligned} \\$

When $n$ is even, $b_{n}=0$ .

$\therefore f(x)=\frac{4}{\pi}\left[\sin x-\frac{\sin 3 x}{3^{2}}+\frac{\sin 5 x}{5^{2}}-\ldots . .\right] \$

(ii) For the half-range sine series,

Let, $f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty} a_{n} \cos n x \\$

Then, $a_{0}=\frac{2}{\pi} \int_{0}^{\pi} f(x) d x=\frac{2}{\pi}\left[\int_{0}^{\pi / 2} x d x+\int_{\pi / 2}^{\pi}(\pi-x) d x\right] \\$

$\begin{aligned} &=\frac{2}{\pi}\left[\left|\frac{x^{2}}{2}\right|_{0}^{\pi / 2}+\left|\pi x-\frac{x^{2}}{2}\right|_{\pi / 2}^{\pi}\right] \\\\ &=\frac{2}{\pi}\left[\frac{\pi^{2}}{8}+\left(\pi^{2}-\frac{\pi^{2}}{2}\right)-\left(\frac{\pi^{2}}{2}-\frac{\pi^{2}}{8}\right)\right]=\frac{2}{\pi}\left[\frac{\pi^{2}}{4}\right]=\frac{\pi}{2} \end{aligned} \\$

$\begin{aligned} a_{n} &=\frac{2}{\pi} \int_{0}^{n} f(x) \cos n x d x=\frac{2}{\pi}\left[\int_{0}^{n / 2} x \cos n x d x+\int_{\pi / 2}^{\pi}(\pi-x) \cos n x d x\right] \\\\ &=\frac{2}{\pi}\left[x \cdot \frac{\sin n x}{n}-1 \cdot\left(-\frac{\cos n x}{n^{2}}\right)\right]_{0}^{\pi / 2}+\frac{2}{\pi}\left[(\pi-x) \cdot \frac{\sin n x}{n}-(-1)\left(-\frac{\cos n x}{n^{2}}\right)\right]_{\pi / 2}^{\pi} \\\\ &=\frac{2}{\pi}\left[\frac{\pi}{2 n} \sin \frac{n \pi}{2}+\frac{1}{n^{2}} \cos \frac{n \pi}{2}-\frac{1}{n^{2}}\right]+\frac{2}{\pi}\left[-\frac{\cos n \pi}{n^{2}}-\frac{\pi}{2 n} \sin \frac{n \pi}{2}+\frac{1}{n^{2}} \cos \frac{n \pi}{2}\right] \\\\ &=\frac{2}{\pi}\left[\frac{2}{n^{2}} \cos \frac{n \pi}{2}-\frac{\cos n \pi}{n^{2}}-\frac{1}{n^{2}}\right]=\frac{2}{\pi n^{2}}\left[2 \cos \frac{n \pi}{2}-\cos n \pi-1\right] \\ \end{aligned}$

$\begin{aligned} &\therefore a_{1}=0, a_{2}=\frac{2}{\pi \cdot 2^{2}}(2 \cos \pi-\cos 2 \pi-1)=\frac{-2}{\pi \cdot 1^{2}}, \\\\ &a_{3}=0, a_{4}=0, a_{5}=0, a_{6}=\frac{2}{\pi \cdot 6^{2}}(2 \cos 3 \pi-\cos 6 \pi-1)=\frac{-2}{\pi \cdot 3^{2}}, \\\\ &a_{7}=0, a_{8}=0, a_{9}=0, a_{10}=\frac{2}{\pi \cdot 10^{2}}(2 \cos 5 \pi-\cos 10 \pi-1)=\frac{-2}{\pi \cdot 5^{2}}, \ldots . . \end{aligned} \\$

Hence, $f(x)=\frac{\pi}{4}-\frac{2}{\pi}\left[\frac{\cos 2 x}{1^{2}}+\frac{\cos 6 x}{3^{2}}+\frac{\cos 10 x}{5^{2}}+\ldots \ldots\right] \\$

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