Finding the Fourier Sine Series

1. Nov 3, 2013

trap101

Find the Fourier SIne Series for f(x) = x on -L < x < L (Full Fourier)

Ok, so my issue is in calculating the coefficients for the sine and cosine parts, more so an interpretation. So I have calulated the sine and cosine series to this point:

let An: Cosine series Bn: sine series:

An = l/(n$\pi$)) [ sin(n$\pi$ + sin(-n$\pi$)] - l/(n2$\pi$2) [sin(n$\pi$) - sin(-n$\pi$)]

Bn = l/(n$\pi$)) [ cos(n$\pi$ + cos(-n$\pi$)] - l/(n2$\pi$2) [cos(n$\pi$) - cos(-n$\pi$)]

My questions are these: How do I interpret the relationship between cos(n$\pi$) and cos(-n$\pi$)? and the same with the sine values?

My take is this: for the sine function, since all of the sine coeffcients have a $\pi$ in them then it doesn't matter what value I select for "n", they will all go to 0.

As for the cosine: Since cos(n$\pi$) and cos(-n$\pi$) end up going to the same value, just in opposite directions, I will have 2cos(ncos(n$\pi$) since cos(n$\pi$) = cos(-n$\pi$) in values.

2. Nov 3, 2013

LCKurtz

Yes. $\sin n\pi = 0$ for any integer $n$, so just leave those terms out of your series.

I think you mean $\cos(n\pi) + \cos(-n\pi) = 2\cos(n\pi)$ and $\cos(n\pi) - \cos(-n\pi) = 0$. Also sometimes you will see $\cos(n\pi)=(-1)^n$.

3. Nov 3, 2013

trap101

Yes that actually brings up anouther question I had with regards to $\cos(n\pi)=(-1)^n$

I noticed that they expressed a couple of final results using that expression of (-1)n. Why is that?

4. Nov 3, 2013

LCKurtz

I don't understand your question. They are equal.

5. Nov 3, 2013

trap101

It's ok I figured it out......there was another example with and extra (-1) that they didn't include initially so: (-1)(-1)n = (-1)n+1

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