so im trying to apply the fourier series to a saw function and following this website and if you look at eq 2, i dont understand the last step, i tried solving it myself but the second term disapears over there... i got:$$c_n=\frac{A(\pi2ni+1)}{4\pi^2n^2}$$ is it because only the complex part is used? and why isnt $c_0=0$?
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the +1 in cn is wrong and shouldnt be there, it should be cancel by a $e^{i 2 \pi n}$ – Olba12 Jun 11 '17 at 23:49
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You have $$ c_n = \frac1T \int_{0}^{T} f(t) e^{-i\frac{2\pi nt}{T}}dt \\ f(t) = \frac ATt $$
Then
$$ c_0 = \frac1T \int_{0}^{T} f(t) dt = \frac1T [\frac{At^2}{2T} ]_{0}^{T} = \frac1T(\frac{AT}2) = \frac A2 $$
For $c_n$ you should get arrive at
$$ -Ae^{i2\pi n}\frac{-2i \pi n + e^{i 2 \pi n } -1 }{4\pi^2 n^2} = \frac{iA}{2\pi n}. $$
Olba12
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ah right i forgot i should hav evaluated the integral for c0 before doing it... what about $c_n$? what am i missing? – wrong1man Jun 11 '17 at 23:30
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oh god im stupid.. i forgot "the lower integral", i subsrituted "t=T" in, but i forgot to subtract with "t=0" dumb me... thank you! – wrong1man Jun 12 '17 at 00:35
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the "i goes up" and the sign changes cause you multiply by $\frac{i}{i}$ correct? – wrong1man Jun 12 '17 at 00:41
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