Fourier Series/Pi minus x over 0 to 2 Pi

Theorem
For $x \in \openint 0 {2 \pi}$:
 * $\displaystyle \pi - x = 2 \sum_{n \mathop = 1}^\infty \frac {\sin n x} n$

Proof
By definition of Fourier series:


 * $\displaystyle \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$

where:

for all $n \in \Z_{>0}$.

Thus:

For $n > 0$:

Now for the $\sin n x$ terms:

Finally: