3
$\begingroup$

I'am a little confused. In my text book it is written that all odd function can be described by a sine series.

I have this following equation from an exercise:

$$A_{0}+\sum\limits_{n=1}^\infty \Big(A_{n} \cos(n \phi) + B_{n} \sin(n \phi)\Big)c^{n} = \sin\left(\frac{\phi}{2}\right)$$

It's a standard fourier series, where n and c is positive. Then it is written in the solution that $B_{n}c^{n} = 0$ because of symmetry reasons. I'am confused because then the fourier serie only have cosine term and the function on the right hand side is an odd function?!

$\endgroup$
  • $\begingroup$ You should indicate in which $\phi$-interval the equation is supposed to hold. It cannot be true for all real $\phi$. $\endgroup$ – Christian Blatter Dec 24 '12 at 10:15
3
$\begingroup$

Here's the idea: remember that you have to compute a bunch of integrals of the form $\int_{-\pi}^{\pi}f(t)\sin\;jt\;\mathrm dt$ $\int_{-\pi}^{\pi}f(t)\cos\;jt\;\mathrm dt$. Remember also that if a function $g(x)$ is odd, then $\int_{-c}^c g(t)\mathrm dt$ is zero (why?). Now, what happens if you try to expand an odd function?

As for your putative identity... the fact that there's no $c$ in the right hand side of your equation should make you doubtful that you wrote that one down correctly. Fourier series proper involve sines and cosines (or complex exponentials), not integer powers.

$\endgroup$

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.