# Problem for Fourier series

1. Dec 6, 2005

### fufufuha

It's solution is
What I know is that insert 2n-1 instead of n
plz... explain to me
I don't know this reason

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Last edited: Dec 6, 2005
2. Dec 6, 2005

### Integral

Staff Emeritus
Please make an effort to provide a more complete statement of your problem.

3. Dec 6, 2005

### fufufuha

First I'm sorry that I can't write english well.
I have a problem's solution.
But I don't understand solution
For example Another problem's solution is n instead of 2n-1
But this problem's solution uses 2n-1 instead of n
I understood that 2n-1 is odd
but why do 2n-1 instead of n ?

4. Dec 6, 2005

### HallsofIvy

Staff Emeritus
When x= 1 or -1, 1-|x|= 0. Suppose you had just $\frac{n\pi}{2}x$. Then for even values of n, x= 1, you would have $cos(m\pi)$ (m= n/2) which is either -1 or 1. By restricting the numerator to be odd, you are making sure that you always have, for x= 1, $cos(\frac{\pi}{2})$, $cos(\frac{3\pi}{2})$, etc.

By the way, how does one do Fourier Series in precalculus?

Last edited: Dec 7, 2005