Prove that $S_{n}(x) = \sum_{k=-n}^{n}c_{k}e^{ikx}$

Question

The problem is to prove that every trigonometric sum of the form $$S_{n}(x) := \frac{1}{2}a_{0} + \sum_{k=1}^{n}(a_{k}\cos kx + b_{k} \sin kx)$$ can be expressed as $$S_{n}(x) = \sum_{k=-n}^{n}c_{k}e^{ikx}$$ with $c_{k} := \frac{1}{2}(a_{k} - ib_{k})$ for $1 \leq k \leq n.$

Writing $e^{ikx} = \cos kx + i \sin kx$ did not help out, because I am stuck at dealing with $c_{k}$ for $-n \leq k \leq -1.$

Answer 1

Hint: Since $e^{ikx} = \cos kx + i \sin kx$ for all $k$, we have $$\cos kx = \frac{1}{2} (e^{ikx} + e^{i(-k)x}).$$ Now, write the analogous expression for $\sin kx$, substitute in your expression for $S_n(x)$, and solve for the $c_k$ by collecting like terms.

Answer 2

See this

$$ \sum_{k=-n}^{-1}c_{k}e^{ikx} = \sum_{k=1}^{n}c_{-k}e^{-ikx} $$