2
$\begingroup$

A periodic function f(x) is defined by:

$ f(n) = \begin{cases} {-x^2} & \textrm{ for - π < x ≤ 0} \\ x^2 & \textrm{ for 0 ≤ x < π } \\ \end{cases} \space , \space\space\space f(x+2π)=f(x)$

Determine the Fourier expansion for $f(x)$; $\space$ that is

$ f(x) = \frac{a_0}{2} + \sum\limits_{n=1}^\infty a_n\space cos(nx) + \sum\limits_{n=1}^\infty b_n\space sin(nx)$ .


I had first found the needed coefficients $ \space a_0 , \space a_n, \space b_n $ using their definitions:

$ a_0 = \frac{1}{\pi} \space $$\int_{-\pi}^{\pi} f(x) \space dx$$ $

$a_n = \frac{1}{\pi} \space $$\int_{-\pi}^{\pi} f(x) \space cos(nx)\space dx$$ $

$b_n = \frac{1}{\pi} \space $$\int_{-\pi}^{\pi} f(x) \space sin(nx)\space dx$$ $

I had to use integration by parts for $ a_n$ and $b_n$. However $a_0$ was more straightforward to solve by just applying the piecewise function to the integral (no further integration techniques needed).

I kept getting $a_0 = 0$, $\space$ $b_n=0$, $\space$ $a_n=0$. I cannot spot an error in my calculations, could the Fourier expansion for the above piecewise function really be equal to zero? If so, is there a quicker way of determining that instead of evaluating the integrals only to notice to get a zero answer?

$\endgroup$
3
$\begingroup$

Your function is odd and the interval is symmetric about zero, so the $a_n$, including $a_0$, should indeed be zero. But the $b_n$ should not be zero. Something that might help: because $f$ and $\sin$ are both odd, $\int_{-\pi}^\pi f(x) \sin(nx) dx = 2 \int_0^\pi x^2 \sin(nx)$. You can calculate that with integration by parts.

$\endgroup$
  • $\begingroup$ When I evaluated $ 2 $$\int_{0}^{\pi} x^2 \space sin(nx) dx$$ $ I kept getting $ 2$$\int_{-\pi}^{\pi} f(x) \space sin(nx) dx$ instead of $\int_{-\pi}^{\pi} f(x) \space sin(nx) dx$. Are you certain the relation you stated holds? $\endgroup$ – mnmakrets Aug 12 '15 at 4:05
  • $\begingroup$ @mnmakrets Yes. In detail, if $f$ and $g$ are both odd, then $\int_{-a}^a f(x) g(x) dx = \int_{-a}^0 f(x) g(x) dx + \int_0^a f(x) g(x) dx =- \int_a^0 f(-x) g(-x) dx + \int_0^a f(x) g(x) dx=-\int_a^0 f(x) g(x) dx + \int_0^a f(x) g(x) dx = 2 \int_0^a f(x) g(x) dx$. $\endgroup$ – Ian Aug 12 '15 at 10:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.