# Finding the sum of an infinite series using Fourier

• John Jacke
In summary, the homework statement is trying to find the sum of (-1)3n+1/(2n-1)3. by using term-by-term integration on the cosine Fourier series x= L/2-4L/π2∑cos(((2n-1)πx)/L)/(2n-1)2. The Attempt at a Solution is when integrating and substituting Lx/2 for x's sine Fourier series I get ∑L2/π(2n-1)[(-1)n+1-4/π2(2n-1)2]sin(((2n-1)πx)/L) for the final

## Homework Statement

Trying to find the sum of (-1)3n+1/(2n-1)3. by using term-by-term integration on the cosine Fourier series x= L/2-4L/π2∑cos(((2n-1)πx)/L)/(2n-1)2.

Shown below

## The Attempt at a Solution

When integrating and substituting Lx/2 for x's sine Fourier series I get ∑L2/π(2n-1)[(-1)n+1-4/π2(2n-1)2]sin(((2n-1)πx)/L) for the final series.To find the sum of the series in class we would find Bn explicitly using the formula and then set it equal to this coefficient found by term by term integration. However when I do that they are the same except for that the only difference is that one term alternates and one does not in sign. So explicitly solving for Bn gives me what you see above except the first term doesn't alternate. So setting them equal cancels everything except those terms. Is this supposed to mean anything? I've gotten the correct answer but was off by a factor of 2. It should (I think) be π3/32.

One thing that I believe you are missing is the original ## f(x) ##, which is the periodic function that generates this series. Just a hint=I think I have this correct=is that ## f(x) ## is an even triangle shaped function. You need to integrate this ## f(x) ## on the left side of the equation. ( Unlike your post where you call it ## x ##, it s ## f(x) ##) . And also when you integrate the term ## L/2 ## from 0 to ## L/2 ##, you will get ## L^2/4 ##. ## \\ ## Editing: This one has a trick to it in the final integral that is used to evaluate the series=you would normally expect to evaluate this integral over the interval ## -L/2 ## to ## +L/2 ## or from ## 0 ## to ## L/2 ##. Try an ## L/4 ## limit instead. Also, in giving it in the functional form with an ## L ## in the amplitude, they could easily call this "1"" instead, and the same goes for the x limits on the triangle waveform. The waveform can go from ## -1/2 ## to ## +1/2 ## simplifying the algebra. (Additional editing, just so that I don't steer you in the wrong direction: I believe the function they gave you has a period equal to ##T=2L ##, but I'll let you verify that when you analyze the Fourier series for it.) ## \\ ## Editing: It is possible to use the waveform in the form ## y=x ## but that is only half of the even triangle. The other half ## y=-x ## needs to be included. In this case you could let the period go from -1 to 1. There are a couple of ways of choosing the even function triangle, but ## y=x ## by itself does not work. Looking over the form that they used, it looks like they may have ## f(0)=0 ## and ## f(x) ## increases as x increases. The identity can be proven with any of a couple of even triangle-shaped functions.(I succeeded at proving it, and I'm still working on figuring out the precise function that they supplied=additional edit=I figured out now, but I'll let you work out that part yourself.) It takes a little extra effort to figure out what the precise triangle is that they have for the function they gave you. It sort of works by trial and error. You try an even triangle function and compute its Fourier components and see if you have a match...Editing... Alternatively, a good computer graphics package would quickly graph out the function they gave you, but I don't have that resource presently available. ## \\ ## One additional suggestion: they could step you through the final integral instead of getting a zero=zero result if you integrate over the whole period of the function.

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