How does one derive the complex form of the Fourier series?

Question

Specifically, I have gone from the Fourier Series in this form:

$$\sum\limits_{n=1}^{\infty} a_n\cos(nx) +b_n\sin(nx)$$

and I have taken it to this form:

$$\sum\limits_{n=1}^{\infty} \frac{(ib_n - a_n)e^{inx}}{2} + \sum\limits_{n=1}^{\infty} \frac{(ib_n + a_n)e^{-inx}}{2}$$

But how do I go from there to the far more familiar

$\sum\limits_{n = -\infty}^{\infty} C_ne^{inx}$

Answer 1

$$ \begin{align} \sum_{n=0}^\infty a_n\cos(nx)+b_n\sin(nx) &=\sum_{n=0}^\infty a_n\frac{e^{inx}+e^{-inx}}{2}+b_n\frac{e^{inx}-e^{-inx}}{2i}\\ &=\sum_{n=0}^\infty\frac{a_n-ib_n}{2}e^{inx}+\frac{a_n+ib_n}{2}e^{-inx}\\ &=\sum_{n=-\infty}^\infty c_ne^{inx} \end{align} $$ where $$ c_n=\left\{\begin{array}{l} \frac{a_n-ib_n}{2}&\text{if }n\gt0\\ \frac{a_{-n}+ib_{-n}}{2}&\text{if }n\lt0\\ a_0&\text{if }n=0 \end{array}\right. $$

Answer 2

Taking into account the missing term $a_0/2$ and correcting the errors in the coefficients, simply change $n$ to $-n$ in the second sum, and put

$$C_n=\frac{a_n-ib_n}{2}\qquad n>0$$

and

$$C_n=\frac{a_{-n}+ib_{-n}}{2}\qquad n<0$$

and

$$C_0=a_0/2.$$