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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... :)
 2 years 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)...
 2 years 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?
 2 years ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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