Finding the function given a Fourier Series.

[Kizaru]

Homework Statement

From the Fourier series\frac{1}{2} - \frac{1}{4}cos(x) + \sum\frac{(-1)^{n}}{1-n^{2}}cos(nx)of (1/2)x*sinx on [-\pi,\pi], find the function whose Fourier Series on [-\pi,\pi] is \frac{3}{4}sin(x) - \sum\frac{(-1)^{n}}{n-n^{3}}sin(nx)}

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.

 

[Kizaru]

So no one has any ideas?
BTW this is a bump because I made a huge mistake on the latex for second F-Series.

 

[arithmetix]

well it looks to me like you could consider the Fourier series to the the resultant waveform whn a pure wave form is applied as the x in a f(x).
Imagine that that f(x)=x^2. Then the pure waveform is modulated by f(x) so that where we had sin(x) we now have (sin(x)). Now we apply the appropriate trig identity:
(sin(x))^2 = (sin(x))*(sin(x))= sin(a)*sin(B) where A=B (just so you recognise it!)
sin(A)*sin(B)=(1/2)(cos(A+B)-cos(A-B))
(sin(x))^2=(1/2)*(-cos(2*x)+cos(0)) (look out for that negative on the first cos!)
The Fourier series you quote is likely, I believe, to have resulted from such considerations at a higher power than 2. Such functions arise for instance when a small varying current passes through a silicon diode; the output voltage may well be represented by a function like this.

 

[Dick]

You were already on the right track. If you call your infinite series with cos(nx) C(x) and your infinite series with sin(nx) S(x) then the first relation is (1/2)-cos(x)/4+C(x)=(1/2)x*sin(x). The unknown function is 3*sin(x)/4-S(x). As you've noticed S'(x)=C(x). Solve the first relation for C(x), substitute C(x)=(S'(x)) and integrate to find S(x). Now you only have a constant of integration to find.

 

[Kizaru]

Hehe wow I feel pretty dumb, thanks for the help guys.