Finding the Fourier Sine Series

[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 A_n: Cosine series B_n: sine series:



A_n = l/(n\pi)) [ sin(n\pi + sin(-n\pi)] - l/(n²\pi²) [sin(n\pi) - sin(-n\pi)]

B_n = l/(n\pi)) [ cos(n\pi + cos(-n\pi)] - l/(n²\pi²) [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.

 

[LCKurtz]

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 A_n: Cosine series B_n: sine series:



A_n = l/(n\pi)) [ sin(n\pi + sin(-n\pi)] - l/(n²\pi²) [sin(n\pi) - sin(-n\pi)]

B_n = l/(n\pi)) [ cos(n\pi + cos(-n\pi)] - l/(n²\pi²) [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.

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

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.

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 $\cos(n\pi)=(-1)^n$.

 

[trap101]

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 $\cos(n\pi)=(-1)^n$.

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?

 

[LCKurtz]

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?

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

 

[trap101]

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

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