1. ## Fourier series problem

Here I got a question of Fourier series attached with this. Also I atemped some part of the question (that is attached here with) but I need to know is this answer is correct.. or not
and how I prove the part b) from the above result.

2. No, you should extend the domain of f to $\displaystyle [-\pi, \pi]$ such that the extension of f is even function .

3. How it is in even function.. please show me..

4. Try calculating $\displaystyle a_n$ again.

5. When now i calcalate this again as considering even function
then $\displaystyle b_n=0$

$\displaystyle a_0=\frac{2}{\pi}\int\limits_{0}^{\pi} f(x) dx$

$\displaystyle =\frac{2}{\pi}\int\limits_{0}^{\pi/2}(1)dx+\frac{2}{\pi}\int\limits_{\pi/2}^{\pi/2}0dx+\frac{2}{\pi}\int\limits_{\pi/2}^{\pi}(-1)dx$

$\displaystyle =(1-\frac{\pi}{2})$

$\displaystyle a_n=\frac{2}{\pi}\int\limits_{0}^{\pi}f(x)Cos nx dx$

$\displaystyle =\frac{2}{\pi}\int\limits_{0}^{\pi/2}1.Cos.nx.dx+\frac{2}{\pi}\int\limits_{\pi/2}^{\pi/2}.0.Cos.nx.dx+\frac{2}{\pi}\int\limits_{\pi/2}^{\pi}(-1).Cos.nx.dx$

$\displaystyle =\frac{2}{\pi}(\frac{Sinn(\pi/2)}{n})+\frac{2}{\pi}(\frac{-Sinn\pi}{n}+\frac{Sinn\pi}{2n})$

=0

Is this calculation is true? otherwise plz show me the incorrect place..