find-forier-series-to-half-rang-of-f-x-sinx-0-lt-x-lt-pi-and-prove-that-n-1-1-4n-2-1-1-2-

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}