How to get from $\sum_{n=0}^N (a_n \cos{nx} + b_n \sin{nx})$ to $\sum_{-N}^{N} c_n e^{inx}$?

Question

I'm currently reading Baby Rudin, and I'm in the section of Chapter 8 that covers Fourier series. There is one line that I just can't figure out for the life of me, and I can't find anything online about it. He defines trigonometric polynomials as finite sums of the form

$f(x) = a_0 + \sum_{n=1}^N (a_n \cos{nx} + b_n \sin{nx})$

but then he says that it can be written in the form

$f(x) = \sum_{-N}^{N} c_n e^{inx}$

I just can't figure out the steps here to get from one to the other, specifically why the range is now -N to N. Any pointers here would be super helpful.

In addition, if anyone has recommendations on resources to supplement this section, I would greatly appreciate those as well.

Answer 1

Hint. One may recall that $$ \cos nx=\frac{e^{inx}+e^{-inx}}2,\quad \sin nx=\frac{e^{inx}-e^{-inx}}{2i} $$ giving $$ \begin{align} f(x) &= a_0 + \sum_{n=1}^N (a_n \cos{nx} + b_n \sin{nx})\\\\ &=a_0 + \sum_{n=1}^N \frac{(a_n -ib_n)}2e^{inx}+ \sum_{n=1}^N \frac{(a_n +ib_n)}2e^{-inx} \end{align} $$

Can you take it from here?