I am looking at a certain derivation for the full Fourier series of $\phi(x)=\cosh{x}$ on the interval $(-l,l)$. The derivation uses the definition of hyperbolic cosine and the full Fourier series for $e^x$ as follows

$$\begin{align} \cosh{x}&=\frac{e^x+e^{-x}}{2} \\ &=\frac{1}{2}\left(\sum_{n=-\infty}^\infty (-1)^n \frac{l+in\pi}{l^2+n^2\pi^2}e^{\frac{in\pi x}{l}} \sinh{l}+\sum_{n=-\infty}^\infty (-1)^n \frac{l+in\pi}{l^2+n^2\pi^2}e^{-\frac{in\pi x}{l}} \sinh{l}\right)\\&= \frac{1}{2}\left(\sum_{n=-\infty}^\infty (-1)^n \frac{l+in\pi}{l^2+n^2\pi^2}e^{\frac{in\pi x}{l}} \sinh{l}+\sum_{n=-\infty}^\infty (-1)^{-n} \frac{l-in\pi}{l^2+n^2\pi^2}e^{\frac{in\pi x}{l}} \sinh{l}\right) \end{align}$$

My confusion is in seeing how the second sum on the middle line is apparently equal to the second sum on the last line. I have thought very long about this and may be missing something simple, but would really appreciate any help in seeing why this is true. Thanks in advance.

  • $\begingroup$ Change $n$ to $-n$. $\endgroup$ – Kavi Rama Murthy Sep 24 at 23:15
  • $\begingroup$ @KaviRamaMurthy thanks, this cleared it very well for me $\endgroup$ – mathim1881 Sep 25 at 16:07

This is similar to writing a full fourier series in complex form.

Follow this example:

$f(x) = \frac{a_0}{2}+\sum_{n=-\infty}^\infty (a_n cos(nx) b_n sin(nx)$

$= \frac{a_0}{2} + \sum_{n=-\infty}^\infty a_n \frac{e^{inx}+e^{-inx}}{2} + b_n \frac{e^{inx} -e^{-inx}}{2i}$

$=\frac{a_0}{2} + \sum_{n=-\infty}^\infty \frac{a_n-ib_n}{2} e^{inx} + \sum_{n=-\infty}^\infty \frac{a_n+ib_n}{2} e^{-inx}$

$=\sum_{n=-\infty}^\infty c_n e^{inx}$

Also, remember that $\cos(n\pi) =(-1)^n$ and De Moivre's Theorem $e^{ix}= \cos(x) + i \sin(x)$

so $(-1)^n e^{\frac{-in \pi x}{l}} = (-1)^n e^{\frac{-in \pi x}{l}} \frac{e^{\frac{in \pi x}{l}} }{e^{\frac{in \pi x}{l}} } $

Using this idea, we can apply it to the 2nd line posed in your question and should see the results follow.

Hope this helps.

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