\begin{align*} e^x &\sim A_0 + \sum\limits_{n=1}^\infty A_n \cos \frac{n \pi x}{L} \\ \end{align*}
The Fourier cosine series on the right will be even extended periodized version of $e^x$.
Now, I could solve for these coefficients directly,
\begin{align*} A_n &= \frac{2}{L} \int_0^L e^x \cos \frac{n \pi x} \, dx \\ \end{align*}
But this exercise asks for a different solution.
We can apply term by term differentiation to yield:
\begin{align*} e^x &\sim - \sum\limits_{n=1}^\infty \frac{n \pi}{L} A_n \sin \frac{n \pi x}{L} \\ \end{align*}
Setting the two definitions of $e^x$ equal yields the following which is equality for $[0,L]$:
\begin{align*} A_0 + \sum\limits_{n=1}^\infty A_n \cos \frac{n \pi x}{L} &\sim - \sum\limits_{n=1}^\infty \frac{n \pi}{L} A_n \sin \frac{n \pi x}{L} \\ \end{align*}
From there, how can I derive $A_n$?
EDIT: I found my original flaw, that last equation is only equality for $[0,L]$, not for $[-L,L]$. However, I'm still not sure how to calculate $A_n$ via this technique.
$f(x)=e^x$ is a mixed parity function which is neither even nor odd, so the fouries series will have both the compnents: $\sin (n x/L)$ and $\cos (nc/L)$.