Fourier Series in Terms of e

Math Calculus

Proof

Let’s express a Fourier Series as:

$$\begin{gathered} v=\frac{2\pi nx}{P} \\ f(x)=\sum _{n=0}^{\infty }{A}_n\cos v+B_n\sin v \end{gathered}$$

We can deduce:

$$\begin{gathered} f(x)=\sum _{n=0}^{\infty }\frac{A_n{e}^{iv}+A_n{e}^{-iv}-i{B}_n{e}^{iv}+i{B}_n{e}^{-iv}}{2} \\ =\sum _{n=0}^{\infty }0.5(A_n+i{B}_n)e^{-iv}+0.5(A_n-i{B}_n)e^{iv} \\ =\sum _{n=0}^{\infty }\frac{e^{-iv}}{P}{\int }_{-P/2}^{P/2}f(x)(\cos v+i\sin v)dx+\frac{e^{iv}}{P}{\int }_{-P/2}^{P/2}f(x)(\cos -v+i\sin -v)dx \\ =\sum _{n=0}^{\infty }\frac{e^{-iv}}{P}{\int }_{-P/2}^{P/2}f(x)e^{iv}dx+\frac{e^{iv}}{P}{\int }_{-P/2}^{P/2}f(x)e^{-iv}dx \\ =\sum _{n=-\infty }^{\infty }\frac{e^{iv}}{P}{\int }_{-P/2}^{P/2}f(x)e^{-iv}dx \end{gathered}$$

Definitions

Definitions of $A_n$ and $B_n$:

Fourier Series Proof