# Show that a the periodic function is even in a specific interval

I have just started to learn about Fourier series, and Even/Odd functions. I am supposed to show that the function below is even in the given period. I assumed that if I tried solving the $B_n$ it would be 0, but this was not the case. I thought that even functions would have a $B_n$ (sine series) of 0, I guess I have misunderstood something.

$$f(x) = \pi x -\frac{x^2}{2} - \frac{\pi^2}{3}, 0 \leq x \leq 2 \pi$$

The only additional information is that it is a 2$\pi$ periodic function. Could someone explain to me how to approach the problem, and clear up anything I might have misunderstood?

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By "even in the given period", I assume you mean symmetric around the midpoint of the interval, i.e. that $f(2 \pi - x) = f(x)$. Just plug in $2 \pi - x$ in place of $x$, and simplify. Or complete the square.
Your function $f$ is defined in $\left [0,2\pi \right ]$ and it's a parabola: it's easy to verify that its axis is $x=\pi$. Now, if you consider the interval $\left [ 0,\pi \right ]$ both $\{\cos (nx): n \in \mathbf{N}\}$ and $\{\sin (nx): n \in \mathbf{N}\}$ are c.o.n.s, so that you can expand $f$ for $0 \leq x \leq \pi$ in Fourier series. If you'll use the first c.o.n.s. your $\bar f(x)$ $x \in \mathbf{R}$ will be even; if you will use the second it will be odd.