How Do You Calculate the Sum of a Fourier Series at Specific Points?

ensten
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Homework Statement


<br /> h(x)=\left\{\begin{matrix}<br /> 9+2x , 0&lt;x&lt;\pi\\ -9+2x , -pi&lt;x&lt;0<br /> <br /> \end{matrix}\right.<br /> \\<br /> Find \ the \ sum \ of \ the \ Fourier \ series \ for \ x=\frac{3\pi}{2} and\ x=\pi<br /> \\<br /> The \ Fourier \ series \ is:<br /> \\<br /> h(x)=9+\pi + \sum_{n=1}^{inf} \frac{18-2(9+2\pi)(-1)^2}{n\pi}sin(nx)<br />
Also period is 2pi.

The Attempt at a Solution


I have already calculated the Fourier series. I however don't know how I can find the sum for the x's. I have searched the internet and looked in my book but I can't find any examples that help me. I'm stuck and don't know how to go from here.
 
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Can you simplify \sin(n \frac{3\pi}{2}) and \sin(n \pi)?
 
jbunniii said:
Can you simplify \sin(n \frac{3\pi}{2}) and \sin(n \pi)?

\sin(n \frac{3\pi}{2}) changes for different n's. Example: -1, 0 and 1. So I don't know how to simplify that. However \sin(n \pi) will always be 0 for all n.
 
\sin(n \frac{3\pi}{2}) is zero for even n. Try a change of variables so that you will only sum over the odd values of n.
 
jbunniii said:
\sin(n \frac{3\pi}{2}) is zero for even n. Try a change of variables so that you will only sum over the odd values of n.

I don't quite understand how to do that. I tried the sum function at wolframalpha and I get that it doesn't converge.
 
ensten said:

Homework Statement


<br /> h(x)=\left\{\begin{matrix}<br /> 9+2x , 0&lt;x&lt;\pi\\ -9+2x , -pi&lt;x&lt;0<br /> <br /> \end{matrix}\right.<br /> \\<br /> Find \ the \ sum \ of \ the \ Fourier \ series \ for \ x=\frac{3\pi}{2} and\ x=\pi<br /> \\<br /> The \ Fourier \ series \ is:<br /> \\<br /> h(x)=9+\pi + \sum_{n=1}^{inf} \frac{18-2(9+2\pi)(-1)^2}{n\pi}sin(nx)<br />
Also period is 2pi.

The Attempt at a Solution


I have already calculated the Fourier series. I however don't know how I can find the sum for the x's. I have searched the internet and looked in my book but I can't find any examples that help me. I'm stuck and don't know how to go from here.

You don't work this kind of problem by actually summing the series. You use the convergence theorem. Doesn't your text have a theorem something to the effect that for a periodic function f(x) satisfying the Dirichlet conditions the series converges to f(x) at points where f is continuous and to ##\frac {f(x^+) + f(x^-)} 2## at points ##x## where there is a jump discontinuity? So you can answer the question by examining the periodic extension of ##f(x)##. In fact, you don't even need to calculate the Fourier Series to answer the question.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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