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Let $f:[0, 2\pi] \to \mathbb{R}$ continuous function. Show that the Fourier series exponential form is equal to \begin{align} \sum_{n=-\infty}^{\infty}c_ne^{inx} \end{align} for a specific choice of $c_n$.

Now my attempt was basically taking the known methodology: \begin{align} f(x) = a_0 + \sum a_n \cos \left( nx \dfrac{\pi}{L} \right) + \sum b_n \sin \left( nx \dfrac{\pi}{L} \right) \end{align} and also \begin{align} &a_0 = \dfrac{1}{2L} \int_{-L}^{L}f(x) dx\\ &a_n = \dfrac{1}{L} \int_{-L}^{L}f(x)\cos \left( nx \dfrac{\pi}{L} \right) dx\\ &b_n = \dfrac{1}{L} \int_{-L}^{L}f(x)\sin \left( nx \dfrac{\pi}{L} \right) dx \end{align} and I am 100% sure we must use the identity $e^{ix} = \cos x+ i \sin x$ but I am not really sure how to piece this together.

Any thoughts?

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Use $c_{0} = a_{0}$ and solve the system

$c_{n}e^{inx}+c_{-n}e^{-inx} = a_{n}\cos\left(nx\right) + b_{n}\sin\left(nx\right)$

In the end you will get

$$ c_{n} = \frac{1}{2L}\int_{-L}^{L}f\left(x\right)e^{-inx}dx $$

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  • $\begingroup$ Many thanks. I did that on the closed segment $[0, 2\pi]$ but it's still okay. $\endgroup$ Dec 14, 2022 at 12:53

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