Why does an odd function have only cosine terms in this Fourier series?

[sommerfugl]

Hello

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

I have this following equation from an exercise:

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

It's a standard Fourier serie, where n and c is positive. T
hen it is written in the solution that $$B_{n}c^{n} = 0$$ because of symmetry reasons. And I'am confused because then the Fourier serie only have cosine term and the function on the right hand side is an odd function?!

 

[mathman]

Off hand I would say you are right. For an odd function the A's should be 0, not the B's. Also if c is a constant, what is the point of cⁿ, since the direct calculation of the coefficients doesn't give them.

 

[AlephZero]

When you find the Fourier series, you are taking the function to be periodic with period $2\pi$.

I think your book is taking the function as $\sin \phi/2$ on the interval $[0, 2\pi]$ and extending it to be periodic for other values of $\phi$. That is an even function.

If you defined the function as $\sin \phi/2$ over the interval $[-\pi, \pi]$, that is a different function which is odd.

It could be that the book forgot to say which of these functions it is talking about.

I agree with #2, I don't see the purpose the $c^n$ (or for $c_n$, if there was a typo).