What is the half-range Fourier sine series for the function f(t) = t sin(t)?

NotStine
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Homework Statement



Question: Find the half-range Fourier sine series for the function f(t) = t sin(t)

Problem: According to all the examples I have gone through, they all have a limit when asking for the half-range. However, my teacher, in the question posted above, has not specified any limits. Is this a typing error? If not, can you please nudge me in the right direction.

Homework Equations





The Attempt at a Solution



None yet. I'm under the impression that question may have been typed wrong.
 
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I think -pi to pi are standard ...
Currently, it is an even function, I can suggest making it odd t*cos(t) and finding Fourier series from 0 to 2pi.
 
Ok here is what I gather so far:

I am looking for the sine half-series, which is bn.Sin(nt) from the Fourier series.

So,

bn = I{t.sin(t).sin(nt)} between 0 and 2pi

... which goes to ...

bn = I{t.sint(1+n)} between 0 and 2pi?

Is that correct?

EDIT: ... which gives me 0. I think I misunderstood.

Reading your suggestion again, you have changed t.sin(t) to t.cos(t)... Why is that? I can see we get an odd function (odd . even) but not sure how we came about the change...

Apologies in advance if I sound retarded, but 2 lectures on Fourier was no way near enough in my opinion.
 
Last edited:
Any ideas?
 
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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