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## Homework Statement

From the Fourier series

[tex]\frac{1}{2} - \frac{1}{4}cos(x) + \sum\frac{(-1)^{n}}{1-n^{2}}cos(nx)[/tex]

of

(1/2)x*sinx on [tex][-\pi,\pi][/tex], find the function whose Fourier Series on [tex][-\pi,\pi][/tex] is

[tex]\frac{3}{4}sin(x) - \sum\frac{(-1)^{n}}{n-n^{3}}sin(nx)}[/tex]

Both sums go from n=2 to n=infinity. The latex stack was showing up weird.

## Homework Equations

## The Attempt at a Solution

Well, this seems like a fair bit of intuition. I tried to plug in F-Series formula for sin(x) and cos(x) into both F-Series but didn't get anywhere. I then realized that the second F-Series bears some resemblance to the integration of the first series. For example, the indefinite integral of the sum in the first F-Series is exactly the sum in the second series, aside from a constant. The integral of the -(1/4)cos(x) term yields (1/4)sinx, which is close and hence here has been everything I've tried.

Any suggestions to get me going in the right direction?

I also tried setting the a_n coefficient from the first F-Series equal to n*b_n of the second, and then I had the integrals equated but I can't really do much from there. Differentiating both integrals just gives 0= 0.

EDIT: Sorry, I was missing "sin(x)" in my second F-Series Formula.

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