Find the Fourier series of f(x) = x+ x2 in (−π,π) of periodicity 2π. Hence deduce that 1 π2

We are tasked with finding the Fourier series of $f(x) = x + x^2$ over the interval $(-\pi, \pi)$ with periodicity $2\pi$, and deducing that $\frac{1}{\pi^2} = \frac{1}{6} \sum_{n=1}^{\infty} \frac{1}{n^2}$.

Step 1: Fourier Series Expansion

For a $2\pi$-periodic function, the Fourier series is given by:

$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(nx) + \sum_{n=1}^{\infty} b_n \sin(nx)$ where the Fourier coefficients $a_0$, $a_n$, and $b_n$ are calculated as follows:

$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx$ $a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx$ [ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) , dx