In the Fourier series expansion of the periodic function $f(x) = x^2$ on the interval $[-\pi, \pi]$ with period $2\pi$:
$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(nx) + \sum_{n=1}^{\infty} b_n \sin(nx)$
The value of the Fourier coefficient $a_1$ is ________ (answer in integer).
Answer Key:
-4
Step-by-Step Solution & Explanation
Since $f(x) = x^2$ is an even function on $[-\pi, \pi]$, all sine coefficients $b_n = 0$.
The cosine coefficients are given by:
$a_n = \frac{2}{\pi} \int_0^\pi x^2 \cos(nx) \, dx$
Integrating by parts twice:
$\int x^2 \cos(nx) \, dx = \frac{x^2 \sin(nx)}{n} + \frac{2x \cos(nx)}{n^2} - \frac{2 \sin(nx)}{n^3}$
Evaluating between limits $0$ and $\pi$:
$a_n = \frac{2}{\pi} \left[ \frac{2\pi \cos(n\pi)}{n^2} \right] = \frac{4 (-1)^n}{n^2}$
For $n = 1$:
$a_1 = \frac{4 (-1)^1}{1^2} = -4$