I have to obtain Fourier Series of $$ F(x)=x(2\pi-x) $$ for $$ 0 < x < 2\pi $$ $$ f(x) = f(x+2\pi)$$ I am very confused what to do about the last condition: $$ f(x) = f(x+2\pi) $$ my calculated $a(0)$ is $-\frac{2}{3}\pi^2$ and my a(n) term is $ \frac{-2\left(\pi ^2n^2-2\right)\sin \left(\pi n\right)-4\pi n\cos \left(\pi n\right)}{\pi n^3} $ can someone confirm if these are correct?
Fourier Series with dc offset
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integration
definite-integrals
fourier-analysis
fourier-series
fourier-transform
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0If $n$ is an integer, then $\sin(n\pi)=1$. Similarly you can simplify $\cos(\pi n)$. I did not check if the integration was correct. – 2017-01-22