I want to compute the sum of the fourier series for my 2pi periodic function $$y(x)=x^{2} $$ when $$x=6$$ its defined such that $$|x|\le \pi$$
I managed to compute the fourier series of my function and ended up with : $$y(x)=\frac{\pi^{2}}{3} + 2\sum_{n=-\infty}^\infty \frac{(-1)^{n}}{n^{2}}*e^{inx} $$
Therfor $$y(6)=\frac{\pi^{2}}{3} + 2\sum_{n=-\infty}^\infty \frac{(-1)^{n}}{n^{2}}*e^{in6} $$
But I have no clue how to evaluate this expression...any help?
The correct answer is $$(6-2\pi)^{2} $$
When dealing with even/odd functions, the real coefficients are particularly useful:
$$ f(x) \sim \frac{a_0}{2} + \sum_{n = 1}^{+\infty} \Big ( a_n \cos (nx) + b_n \sin (nx) \Big ) $$
where:
$$ b_n = 0 \qquad (\text{it's an even function}) $$ $$ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \ \mathrm dx = \frac{2}{\pi} \left. \frac{x^3}{3} \right |_{0}^{\pi} = \frac{2}{3} \pi^2 $$ $$ \begin{aligned} a_n &= \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \cos (nx) \ \mathrm dx \\ &= \frac{2}{\pi} \left [ \underbrace{\left. \frac{1}{n} x^2 \sin (nx) \right |_{0}^{\pi}}_{{} = 0, \ n \in \mathbb{N}} - \frac{2}{n} \int_{0}^{\pi} x \sin (nx) \ \mathrm dx \right ] \\ &= \frac{2}{\pi} \left [ \left. \frac{2}{n^2} x \cos (nx) \right |_{0}^{\pi} - \frac{2}{n^2} \underbrace{\int_{0}^{\pi} \cos (nx) \ \mathrm dx}_{{} = 0, \ n \in \mathbb{N}} \right ] \\ &= \frac{{4}}{n^2} (-1)^n \end{aligned} $$
Therefore:
$$ f(x) = \frac{\pi^2}{3} + 4 \sum_{n = 1}^{+\infty} \frac{(-1)^n}{n^2} \cos (nx) $$
In conclusion:
$$ f(6) = f(6 - 2 \pi) = (6 - 2\pi)^2 = \frac{\pi^2}{3} + 4 \sum_{n = 1}^{+\infty} \frac{(-1)^n}{n^2} \cos \big (n(6 - 2\pi) \big) $$ $$ \implies \sum_{n = 1}^{+\infty} \frac{(-1)^n}{n^2} \cos (6n) = \frac{1}{4} \left [ (6 - 2\pi)^2 - \frac{\pi^2}{3} \right ] $$