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AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
Is there any general form for a Fourier series of a function f(x)? I'm not particularly wellversed in the subject, although I'll try to help... :)
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
My first observation of the problem is that f(x) goes through three cycles over the interval (3pi, 3pi)...
 one year ago

nissn Group TitleBest ResponseYou've already chosen the best response.0
the first thing I have to find is the fourier coefficient. I think it is 1/4?
 one year ago

nissn Group TitleBest ResponseYou've already chosen the best response.0
\[1/(2\pi)(\int\limits_{\pi}^{0}0 dx + \int\limits_{0}^{\pi}(1/\pi)x dx\]
 one year ago

nissn Group TitleBest ResponseYou've already chosen the best response.0
which is 1/4
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
Yep, that appears correct to me.
 one year ago

nissn Group TitleBest ResponseYou've already chosen the best response.0
\[a _{n}=1/\pi \int\limits_{\pi}^{0}0 dx + \int\limits_{0}^{\pi}(1/\pi)x dx\]
 one year ago

nissn Group TitleBest ResponseYou've already chosen the best response.0
I am not sure if I am doing it right on the last one
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
The resource I am checking indicate that the formula for a_n would be 1/pi * integral from pi to pi of f(x) cos(nx) dx
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
So... \( \displaystyle a_n = \frac{1}{\pi} \left( \int_{\pi}^{0} 0 \; \textrm{d}x + \int_{0}^{\pi} \frac{1}{\pi} x \cos nx \; \textrm{d}x \right) \)
 one year ago

nissn Group TitleBest ResponseYou've already chosen the best response.0
yeah so then it is correct
 one year ago

nissn Group TitleBest ResponseYou've already chosen the best response.0
so then it is 1/pi?
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
So I am finding: \( \displaystyle a_n = \frac{\pi n \sin n \pi + \cos \pi n  1}{\pi^2 n^2} \)
 one year ago

nissn Group TitleBest ResponseYou've already chosen the best response.0
and then I must do a \[b _{n}\]
 one year ago

nissn Group TitleBest ResponseYou've already chosen the best response.0
\[b _{n}=1/\pi(\int\limits_{\pi}^{0}0dx + \int\limits_{0}^{\pi}(1/\pi)x dx\]
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
My resource shows an additional sin(nx) there \( \displaystyle \frac{1}{\pi} \int_{0}^{\pi} \frac{1}{\pi} x \sin nx \; \textrm{d}x \)
 one year ago

nissn Group TitleBest ResponseYou've already chosen the best response.0
oh it's the same I just forgot to write the nx
 one year ago

nissn Group TitleBest ResponseYou've already chosen the best response.0
and that cos (pi/pi)
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
Sorry, I gotta go for school. I take the calculation of this integral to wolfram... Wolfram evaluates it out as: \( \displaystyle b_n = \frac{\sin \pi n  \pi n \cos \pi n}{\pi^2 n^2} \) http://www.wolframalpha.com/input/?i=integral+from+0+to+pi+of+1%2Fpi%5E2+x+sin%28n+x%29+dx
 one year ago

nissn Group TitleBest ResponseYou've already chosen the best response.0
okey. thank you
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
You're welcome! :)
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
and good luck, I think you're on the right track. :) I was using this as my resource: http://mathworld.wolfram.com/GeneralizedFourierSeries.html Just the end bit with the formula.
 one year ago

nissn Group TitleBest ResponseYou've already chosen the best response.0
thank you :)
 one year ago
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