Fourier Series/Summation Question

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Homework Help Overview

The discussion revolves around the manipulation of summations within the context of Fourier Series, specifically addressing the treatment of constants and the behavior of trigonometric functions in summations.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the process of pulling constants out of summations and question the implications of sign changes during this manipulation. There is also a discussion regarding the treatment of sin(pi*n) within the series.

Discussion Status

Some participants provide clarification on the summation limits and the correct handling of constants. There is an ongoing exploration of examples where sign changes may have been perceived, and one participant expresses uncertainty about their previous understanding.

Contextual Notes

Participants mention the summation limits and the potential confusion arising from different interpretations of the series. There is an acknowledgment of reliance on a solution manual, which may have influenced their reasoning.

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:

\frac{p^2}{3}+\sum \frac{4p^2 cos(\pi n)}{\pi^2 n^2}

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:

\frac{p^2}{3}-\frac{4p^2}{\pi^2}\sum \frac{cos(\pi n)}{\n^2}

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.:biggrin:
 
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Starting with this:
\frac{p^2}{3}+\sum \frac{4p^2 cos(\pi n)}{\pi^2 n^2}

and pulling out the constant from the summation, you should get this:
\frac{p^2}{3} + \frac{4p^2}{\pi^2}\sum \frac{cos(\pi n)}{n^2}

I'm assuming the summation is from n = 0 to infinity. There is no reason for the sign to change when you bring 4p2/(pi)2 out. Why would you think this?

Can you show an example of where you think this happened?
 
Mark44 said:
Starting with this:
\frac{p^2}{3}+\sum \frac{4p^2 cos(\pi n)}{\pi^2 n^2}

and pulling out the constant from the summation, you should get this:
\frac{p^2}{3} + \frac{4p^2}{\pi^2}\sum \frac{cos(\pi n)}{n^2}

I'm assuming the summation is from n = 0 to infinity. There is no reason for the sign to change when you bring 4p2/(pi)2 out. Why would you think this?

Can you show an example of where you think this happened?

The summation was from n=1 to infinity. I couldn't figure out how to add those with latex.

I agree with you, I didn't make sense but it seemed following the solution manual it's the only way I could get the correct answer.

Let me see if I can dig up the example that it happened with and I'll post it.
 
Here are your two summation expressions with the limits. Click either one to see how the limits should look.
\frac{p^2}{3}+\sum_{n = 1}^{\infty} \frac{4p^2 cos(\pi n)}{\pi^2 n^2}
= \frac{p^2}{3}+\frac{4p^2}{\pi^2}\sum_{n = 1}^{\infty} \frac{ cos(\pi n)}{ n^2}

The property used here is
\sum_{n = 1}^{\infty}k f(n) = k\sum_{n = 1}^{\infty}f(n)

Here k is a constant, so doesn't involve n.
 

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