find forier series to half rang of f(x)=sinx ,0<x<π and prove that Σ


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Question Number 146758 by tabata last updated on 15/Jul/21

$f i n d f o r i e r s e r i e s t o h a l f r a n g o f$ $f (x) = s i n x, 0 < x < π a n d p r o v e t h a t$ $\underset{n = 1}{\sum^{\infty}} \frac{1}{4 n^{2} - 1} = \frac{1}{2}$
	Answered by Olaf_Thorendsen last updated on 15/Jul/21

	$S_{N} = \underset{n = 1}{\sum^{N}} \frac{1}{4 n^{2} - 1} = \frac{1}{2} \underset{n = 1}{\sum^{N}} (\frac{1}{2 n - 1} - \frac{1}{2 n + 1})$ $S_{N} = \frac{1}{2} \underset{n = 1}{\sum^{N}} (1 - \frac{1}{2 N + 1})$ $\Rightarrow S_{\infty} = \frac{1}{2}$
	Commented by tabata last updated on 15/Jul/21

	$a n d f o r i e r s i r ?$
	Answered by Olaf_Thorendsen last updated on 15/Jul/21

	$f (x) = \sin x, 0 < x < π$ $a_{0} = \frac{2}{π} \int_{0}^{π} f (x) d x = \frac{2}{π} \int_{0}^{π} \sin x d x$ $a_{0} = \frac{2}{π} {[- \cos x]}_{0}^{π} = \frac{4}{π}$ $a_{1} = \frac{2}{π} \int_{0}^{π} \sin x \cos d x = \frac{1}{π} \int_{0}^{π} \sin (2 x) d x$ $a_{1} = 0$ $n > 1 :$ $a_{n} = \frac{2}{π} \int_{0}^{π} f (x) \cos (\frac{n π x}{π}) d x$ $a_{n} = \frac{2}{π} \int_{0}^{π} \sin x \cos (n x) d x$ $a_{n} = \frac{2}{π} \int_{0}^{π} \frac{1}{2} [\sin ((n + 1) x) - \sin ((n - 1) x)] d x$ $a_{n} = \frac{2}{π} \int_{0}^{π} \frac{1}{2} [\sin ((n + 1) x) - \sin ((n - 1) x)] d x$ $a_{n} = \frac{1}{π} {[- \frac{\cos ((n + 1) x)}{n + 1} + \frac{\cos ((n - 1) x)}{n - 1}]}_{0}^{π}$ $a_{n} = \frac{1}{π} [- \frac{{(- 1)}^{n + 1}}{n + 1} + \frac{{(- 1)}^{n - 1}}{n - 1} + \frac{1}{n + 1} - \frac{1}{n - 1}]$ $a_{n} = - \frac{2}{π (n^{2} - 1)} [{(- 1)}^{n} + 1]$ $\sin x = \frac{a_{0}}{2} + a_{1} \cos x + \underset{n = 2}{\sum^{\infty}} a_{n} \cos (n x)$ $\sin x = \frac{2}{π} - \frac{2}{π} \underset{n = 2}{\sum^{\infty}} \frac{{(- 1)}^{n} + 1}{n^{2} - 1} \cos (n x)$ $n = 2 m$ $\sin x = \frac{2}{π} - \frac{4}{π} \underset{m = 1}{\sum^{\infty}} \frac{1}{4 m^{2} - 1} \cos (2 m x)$ $x = 0 :$ $0 = \frac{2}{π} - \frac{4}{π} \underset{m = 1}{\sum^{\infty}} \frac{1}{4 m^{2} - 1}$ $\underset{m = 1}{\sum^{\infty}} \frac{1}{4 m^{2} - 1} = \frac{1}{2}$
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