# Thread: Fourier Series- Concerning one of the coefficients.

1. ## Fourier Series- Concerning one of the coefficients.

Hello,

I'm working on the following functi(on

$\displaystyle f(x)= 0$ $\displaystyle x\epsilon[-\pi,0)$
$\displaystyle f(x)=1$ $\displaystyle x\epsilon[0,\pi)$

I was able to figure out the mean value by myself (which equals 1).

I got lost when it was time to figure out the cosine coefficient.
I was able to narrow it down to

$\displaystyle \frac{1}{n\pi} (sin(n\pi)-sin(0))$

after differentiating, but when I read my notes, it says the answer to that is 0.
I thought the answer would be

$\displaystyle \frac{sin(n\pi)}{n\pi}$

Can someone explain this to me?

2. Originally Posted by DarkMalice
Hello,

I'm working on the following functi(on

$\displaystyle f(x)= 0$ $\displaystyle x\epsilon[-\pi,0)$
$\displaystyle f(x)=1$ $\displaystyle x\epsilon[0,\pi)$

I was able to figure out the mean value by myself (which equals 1).

I got lost when it was time to figure out the cosine coefficient.
I was able to narrow it down to

$\displaystyle \frac{1}{n\pi} (sin(n\pi)-sin(0))$

after differentiating, but when I read my notes, it says the answer to that is 0.
I thought the answer would be

$\displaystyle \frac{sin(n\pi)}{n\pi}$

Can someone explain this to me?
The sine function has the property that it vanishes at multiples of $\displaystyle \pi$. In other words, $\displaystyle \sin n\pi = 0$

Problem solved!

3. Originally Posted by Opalg
The sine function has the property that it vanishes at multiples of $\displaystyle \pi$. In other words, $\displaystyle \sin n\pi = 0$

Problem solved!
Well duh! Wow for some reason i competely forgot pi equals 180 degrees. Associated it with 90 degrees and that's why I was messing up. Thank you! :P