Using Complex Fourier Series to Find Real Coefficients

Question

I am about to go insane with this problem, so I really hope some kind, kind soul out there can help me.

I am trying to find the complex Fourier series of the following function and interval, and then use that to find the real Fourier series.

$I=(-\pi,\pi), f(x)=\exp(2x)$

I found the complex coefficient, $\mathbb{C}_n$, which I believe to be correct:

$\mathbb{C}_n=\frac{(-1)^n}{2\pi(2-ni)}(\exp(2\pi)-\exp(-2\pi))$

And here I am stuck, trying to find the Cos Coefficient, $a_n$:

$a_n=\mathbb{C}_n+\mathbb{C}_{-n}$ , which I have calculated about ten times to be:

$a_n=\frac{4*(-1)^n}{2\pi(4+n^2)}(\exp(2\pi)-\exp(-2\pi))$

Whereas WolframAlpha, when I plot in "FourierCosCoefficient[e^(2t),t,n]" yields a map of Tennessee half of the times, and the other times it yields:

$a_n=\frac{4(\exp(2\pi)(-1)^n-1}{\pi(n^2+4)}$

Answer 1

You may find out that wolfram alpha is assuming a different interval from which it is computing its fourier cosine coefficients. FourierCosCoefficient[e^(2t),t,n] seems to be integrating from 0 to pi.

I followed your work, and the sum you obtain for $c_n+c_{-n}$ is consistent. Furthermore, if you compute the cosine series coefficients using:

$$\frac{1}{\pi}\displaystyle\int_{-\pi}^\pi e^{2t}\ cos(nt)\ dt\ = \frac{4sinh(2\pi)(-1)^n}{\pi(n^2+4)}$$

... this is exactly the same as the answer you got for your cosine series.

Answer 2

This is not a complete answer, but I have extended your calculation a bit more closer to the answer given by wolfram alpha ,

$a_n={4{(-1)}^n\over 2\pi(4+n^2)}[e^{2\pi}- {1\over e^{2\pi} }]$

$={4{(-1)}^n\over \pi(4+n^2)}{e^{2\pi}\over 2}[{1-e^{-4\pi}}]$

Since $e^{-4\pi}$ is very close to $0$ and so cen be replaced by $0$.

$={4{(-1)}^n\over \pi(4+n^2)}{e^{2\pi}\over 2}$