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

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SUMMARY

The discussion focuses on finding the half-range Fourier sine series for the function f(t) = t sin(t). The user expresses confusion regarding the absence of specified limits for the function, which is typically required for half-range series. A suggestion is made to convert the function into an odd function by using t cos(t) to facilitate the calculation of the Fourier series from 0 to 2π. The user attempts to derive the coefficients bn using the integral I{t.sin(t).sin(nt)} over the interval from 0 to 2π, but encounters difficulties in understanding the transformation and its implications.

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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?
 

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