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I am having trouble converting the complex version of a Fourier Series into a purely real one. The function has value 1 for $0\leq x< \pi$ and 0 for $\pi\leq x< 2\pi$. Calculating the

Calculating the complex Fourier coefficients yields $C_n =\frac{e^{-i n \pi} -1}{-i n\sqrt{2\pi}}$

After obtaining the summation for the complex Fourier series by multiplying the coefficients by $e^{inx}$, applying Euler's formula, and splitting up the sum for even and odd n I obtain the following:

$2\sum _{n = -\infty, odd}^\infty \frac{\sin nx - i\cos nx}{n\sqrt{2\pi}}$

Should I just take the real part of the expression above or is there something I forgot to do ? (Calculations should be correct as I checked all my integrals on a calculator.) Thanks!

Clive
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  • First, I believe the Fourier series coefficient is actually $$C_n=-\frac{e^{-i \pi n}-1}{2 i \pi n}\tag{1}$$ and the Fourier series simplifies to $$\frac{1}{2}+\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^K \frac{\sin(n (\pi-x))+\sin(n x)}{n \pi}\right)\tag{2}$$ after combining the $n$ term with the $-n$ term and taking into account the special case of $C_0$. – Steven Clark Nov 28 '23 at 21:37
  • Thank you! Would you mind explaining how to obtain C0 ? – Clive Nov 29 '23 at 01:58
  • It's just the average $$C_0=\frac{1}{P}\int\limits_P f(x),dx$$ which in the case of your function can be evaluated as $$C_0=\frac{1}{2 \pi }\int_0^{\pi } 1 , dx=\frac{1}{2}$$ since your function is one for $0<x<\pi$ and zero for $\pi<x<2 \pi$. – Steven Clark Nov 29 '23 at 03:06

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