erok81
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I'm not going to use the standard question as I've already solved it, this is more of a general question that doesn't fit into the three question format.
My question arises almost at the very end of the Fourier Series. It has been a looooong time since I've done summations so I am not sure if this is how they work.
Anyway...I have this:
[tex]\frac{p^2}{3}+\sum \frac{4p^2 cos(\pi n)}{\pi^2 n^2}[/tex]
The 4p^2/Pi^2 can come out to the front and the sign always seems to change when it does. So the final answer is this:
[tex]\frac{p^2}{3}-\frac{4p^2}{\pi^2}\sum \frac{cos(\pi n)}{\n^2}[/tex]
So, does the sign flip when you pull it out of the summation? Maybe one subtracts that from the sum and that is why the sign changes?
On a completely separate Fourier Series note. It seems whenever I have sin(pi*n) I can set that equal to zero for the series (at least I can get my answers to match that way). Is that a correct method? I know for any value of n it would be zero, but I want to make sure that is what is happening.
Hopefully that all makes sense. Thanks for the help.
My question arises almost at the very end of the Fourier Series. It has been a looooong time since I've done summations so I am not sure if this is how they work.
Anyway...I have this:
[tex]\frac{p^2}{3}+\sum \frac{4p^2 cos(\pi n)}{\pi^2 n^2}[/tex]
The 4p^2/Pi^2 can come out to the front and the sign always seems to change when it does. So the final answer is this:
[tex]\frac{p^2}{3}-\frac{4p^2}{\pi^2}\sum \frac{cos(\pi n)}{\n^2}[/tex]
So, does the sign flip when you pull it out of the summation? Maybe one subtracts that from the sum and that is why the sign changes?
On a completely separate Fourier Series note. It seems whenever I have sin(pi*n) I can set that equal to zero for the series (at least I can get my answers to match that way). Is that a correct method? I know for any value of n it would be zero, but I want to make sure that is what is happening.
Hopefully that all makes sense. Thanks for the help.
