• Mar 21st 2011, 01:24 AM
Paymemoney
Hi

The following question i am unable to understand why the answer is:
http://img703.imageshack.us/img703/8...rierseries.png

Because this is what i got:

$f(t) ~ \frac{1}{\pi} + \frac{sin(t)}{2} - \frac{2}{\pi} \sum\limits_{n=1}^{\infty}\frac{cos(nt)}{n^2-1}$

Hope someone can help me
P.S
• Mar 21st 2011, 04:33 AM
tonio
Quote:

Originally Posted by Paymemoney
Hi

The following question i am unable to understand why the answer is:
http://img703.imageshack.us/img703/8...rierseries.png

Because this is what i got:

$f(t) ~ \frac{1}{\pi} + \frac{sin(t)}{2} - \frac{2}{\pi} \sum\limits_{n=1}^{\infty}\frac{cos(nt)}{n^2-1}$

Hope someone can help me
P.S

I get $\displaystyle{a_n=\frac{1}{\pi}\int\limits^\pi_0 \sin t\cos nt\,dt=\frac{1}{\pi}\int\limits^\pi_0\left[\sin(n+1)t-\sin(n-1)t\right]dt=}$

$\dispaystyle{\frac{1}{\pi}\left[-\frac{1}{n+1}\cos(n+1)t+\frac{1}{n-1}\cos(n-1)t\right]^\pi_0=}$

$\displaystyle{=\frac{(-1)^{n+1}-1}{\pi}\cdot\frac{2}{n^2-1}=\left\{\begin{array}{cc}0&\mbox{, if }n\mbox{ is odd}\\-\frac{4}{\pi(n^2-1)}&\mbox{ , if }n\mbox{ is even}\end{array}\right.}$ , so I think both

answers are wrong...or I am, of course.

Tonio