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find the fourier series of this

Mathematics
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of this
Is there any general form for a Fourier series of a function f(x)? I'm not particularly well-versed in the subject, although I'll try to help... :)
My first observation of the problem is that f(x) goes through three cycles over the interval (-3pi, 3pi)...

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Other answers:

yeah i know
the first thing I have to find is the fourier coefficient. I think it is 1/4?
\[1/(2\pi)(\int\limits_{-\pi}^{0}0 dx + \int\limits_{0}^{\pi}(1/\pi)x dx\]
which is 1/4
Yep, that appears correct to me.
\[a _{n}=1/\pi \int\limits_{-\pi}^{0}0 dx + \int\limits_{0}^{\pi}(1/\pi)x dx\]
I am not sure if I am doing it right on the last one
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
So... \( \displaystyle a_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} 0 \; \textrm{d}x + \int_{0}^{\pi} \frac{1}{\pi} x \cos nx \; \textrm{d}x \right) \)
yeah so then it is correct
so then it is 1/pi?
sin (1/pi)
So I am finding: \( \displaystyle a_n = \frac{\pi n \sin n \pi + \cos \pi n - 1}{\pi^2 n^2} \)
and then I must do a \[b _{n}\]
\[b _{n}=1/\pi(\int\limits_{-\pi}^{0}0dx + \int\limits_{0}^{\pi}(1/\pi)x dx\]
My resource shows an additional sin(nx) there \( \displaystyle \frac{1}{\pi} \int_{0}^{\pi} \frac{1}{\pi} x \sin nx \; \textrm{d}x \)
oh it's the same I just forgot to write the nx
and that -cos (pi/pi)
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
okey. thank you
You're welcome! :)
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.
thank you :)

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