Soldato
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we've covered fourier series on the second year of my course, computer games technology. Afraid I can't help you with that though.. my memory fails 
Mathematics, help needed!
- Thread starter titaniumx3
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We did fourier series last year (first year Computer Science) - but I don't really have the energy to try and answer a proper question now!
The sketch is easy - it's just a "/\" around the origin. I'm sure you've got that already. Finding the fourier series shouldn't be too difficult - presumably you have notes with the fourier series equations somewhere. You'll need to do the integrals in two parts to take into account the two cases.
To prove it's absolutely convergent - I'm not quite sure. Maybe look through your notes for some clues about how to prove absolute convergence. I would have thought that for the second part you're expecting the even terms in the fourier series to come to zero and I would have expected the formula in the question to come out fairly neatly.
The sketch is easy - it's just a "/\" around the origin. I'm sure you've got that already. Finding the fourier series shouldn't be too difficult - presumably you have notes with the fourier series equations somewhere. You'll need to do the integrals in two parts to take into account the two cases.
To prove it's absolutely convergent - I'm not quite sure. Maybe look through your notes for some clues about how to prove absolute convergence. I would have thought that for the second part you're expecting the even terms in the fourier series to come to zero and I would have expected the formula in the question to come out fairly neatly.
Associate
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Damn
I did that last year, and I can't remember a thing
argh
I did that last year, and I can't remember a thing
argh
Express the Fourier transform in terms of sin and cos instead of using exponentials (it's fairly easy to switch between the two, since e^inx + e^-inx = 2cosx and e^inx - e^-inx = 2isinx). Clearly there will be no sin terms since this is an even function (extending it in the logical way.) Now you should get f(x) = pi/2 + 2/pi Sum[((1-(-1)^n)/n^2) cosnx, n=1 to inf] which is the same as the series given in the question. In fact I'm not sure if your transform is correct, but I'm afraid I don't have time to check. I did this sort of stuff last year, so I'm not too sure about the convergence, but since the Fourier coefficients look like 1/n^2 and |cosx| is bounded by 1 then the series is absolutely convergent by comparison to 1/n^2 (i think.) Hope that helps.
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^^ Yes you are right, I made a mistake whilst tediously drawing out the Fourier series with my mouse, lol. The denominator of the fraction should be n^2 not n.
Anyway, I totally forgot about the odd/even properties of the sine/cosine
, functions so when you expand exp(i*n*x), the even terms cancel out because of (-1)^n and the sine terms sum to zero, hence you eventually get the real series as required.
As for convergence, I'm still confused because in my notes there seem to be several methods to determine convergence, but I just used the fact that the series for 1/n^2 converges hence the fourier series terms also converge, using the comparision test and the fact that |cos(x)| is bounded by 1.
Thanks to everyone for the help!
Anyway, I totally forgot about the odd/even properties of the sine/cosine
, functions so when you expand exp(i*n*x), the even terms cancel out because of (-1)^n and the sine terms sum to zero, hence you eventually get the real series as required.As for convergence, I'm still confused because in my notes there seem to be several methods to determine convergence, but I just used the fact that the series for 1/n^2 converges hence the fourier series terms also converge, using the comparision test and the fact that |cos(x)| is bounded by 1.
Thanks to everyone for the help!
