written 7.8 years ago by | • modified 7.8 years ago |
Half range cosine series is given by
$a_{\theta} = \frac{1}{l}\int\limits_{0}^{l}f(x)dx$
$a_{n} = \frac{2}{l}\int\limits_{0}^{l} f(x)cos \frac{n \pi x}{l}dx , b_{n} = 0$
Here $f(x) = sinx \hspace{1cm} (0, \pi) i.e. 1 = \pi$
$\therefore f(x) = a_{0} + \sum a_{n} cos \frac{n\pi x}{l}$
$\therefore a_{0} = \frac{1}{\pi}\int\limits_{0}^{\pi}f(x) dx$
$\therefore a_{0} = \frac{1}{\pi} \int\limits_{0}^{\pi} sinxdx = \frac{1}{\pi}[-cosx]_{0}^{\pi} = \frac{2}{\pi}$
$\therefore a_{0} = \frac{2}{\pi}$
Now $a_{n} = \frac{2}{\pi}\int\limits_{0}^{\pi}sinx cosnxdx$
$a_{n} = \frac{2}{\pi}\int\limits_{0}^{\pi}[sin(1 + n)x + sin(1 - n)x]dx$
$a_{n} = \frac{1}{\pi} \int\limits_{0}^{\pi}[sin(1 + n)x + sin(1 - n)x]dx$
$a_{n} = \frac{1}{\pi}\bigg[\frac{-cos(1 + n)x}{(1 + n)} - \frac{cos(1 - n)x}{1 - n}\bigg]_{0}^{\pi}$
$a_{n} = \frac{1}{\pi}\bigg[\frac{-cos(\pi + \pi n)}{(1 + n)} - \frac{cos(\pi - \pi n)}{1 - n} + \frac{1}{1 + n} + \frac{1}{1 - n}\bigg]$
$a_{n} = \frac{1}{\pi}\bigg[\frac{cos\pi n}{(1 + n)} - \frac{cos\pi n}{1 - n} + \frac{2}{1 - n^2}\bigg]$
$a_{n} = \frac{1}{\pi}\bigg[cosn\pi * \frac{2}{1 - n^2} + \frac{2}{1 - n^2}\bigg]$
$\therefore a_{n} = 0$ if n is odd
$ = \frac{1}{\pi}\frac{4}{(1 - n^2)}$ if n is even
i.e.
$= \frac{1}{\pi}\frac{4}{(1 - n^2)}$ if n is even
$f(x) = sinx = \frac{2}{\pi} - \frac{4}{\pi}\bigg(\frac{cos2x}{3} + \frac{cos4x}{15} + \frac{cos6x}{35} +....\bigg).........(A)$
Now to deduce $\frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7}.......= \frac{1}{2}$
put x = 0 in equation (A) we get
$0 = \frac{2}{\pi} - \frac{4}{\pi}\bigg(\frac{1}{3} + \frac{1}{15} + \frac{1}{35} + ....\bigg)$
$\frac{2}{\pi} = \frac{4}{\pi} \bigg(\frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} +....\bigg)$
i.e.
$\frac{1}{2} = \frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + .........$ Hence proved