Fourier series - Integral

Question

Let $f$ be a complex-valued piecewise continuous function defined on the interval $[-\pi,\pi]$ and let \begin{equation} \frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left[a_{n}\cos(nx)+b_{n}\sin(nx) \right] \end{equation} denote its Fourier series on $[-\pi,\pi]$. Determine the value of \begin{equation} \frac{1}{\pi}\int_{-\pi}^{\pi}| f(x+\pi)-f(x)|^{2}dx \end{equation} in terms of $a_{n}$ and $b_{n}$.

I realize that this question will have to deal with Parseval's identity, however, I don't know how to use this fact to answer the question.

Answer 1

The question should have specified that $f$ is considered to be periodic with period $2\pi$ (otherwise your integrand is undefined for $x > 0$).

Start by substituting $x+\pi$ for $x$ in the Fourier series and simplifying.

Answer 2

i will try to use few term for demonstration,let us use $n=2$

so we have

$f(x)=a_0/2+a_1*cos(x+\pi)+b_1*sin(x+\pi)+a_2*cos(2*x+2*pi)+b_2*sin(2*x+2*\pi)$

now you see that $cos(x+2*\pi)=cos(x)$ and $sin(x+2*\pi)=sin(x)$

both are periodic with $2*\pi$ if we consider this fact we would have

$f(x)=a_0/2+a_1*cos(x+\pi)+b_1*sin(x+pi)+a_2*cos(2*x)+b_2*sin(2*x)$

if we subtract this from original function,we get

$f(x)=a_0/2+a_1*cos(x+\pi)+b_1*sin(x+\pi)+a_2*cos(2*x)+b_2*sin(2*x)$

-$(a_0/2+a_1*cos(x)+b_1*sin(x)+a_2*cos(2*x)+b_2*sin(2*x))$

can you square this ?please cancel also term $a_0/2$ and consider fact that $cos(x+\pi)=-cos(x)$ and $sin(x+\pi)=-sin(x)$