The problem asks for the Fourier series expansion of $f(x) = x^2$ on the interval $[-\pi, \pi]$. Let’s solve it step by step:

Step 1: General Fourier Series Formula

The Fourier series of a function $f(x)$ on the interval $[-\pi, \pi]$ is: $f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty \left(a_n \cos(nx) + b_n \sin(nx)\right),$ where $a_0 = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \, dx,$ $a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos(nx) \, dx,$ $b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin(nx) \, dx.$

Step 2: Compute $a_0$

For $f(x) = x^2$: $a_0 = \frac{1}{\pi} \int_{-\pi}^\pi x^2 \, dx.$ Since $x^2$ is an even function, the integral simplifies to: $a_0 = \frac{2}{\pi} \int_0^\pi x^2 \, dx.$ $\int_0^\pi x^2 \, dx = \left[\frac{x^3}{3}\right]_0^\pi = \frac{\pi^3}{3}.$ Thus: $a_0 = \frac{2}{\pi} \cdot \frac{\pi^3}{3} = \frac{2\pi^2}{3}.$

Step 3: Compute $a_n$

For $n \geq 1$: $a_n = \frac{1}{\pi} \int_{-\pi}^\pi x^2 \cos(nx) \, dx.$ Since $x^2 \cos(nx)$ is an even function: $a_n = \frac{2}{\pi} \int_0^\pi x^2 \cos(nx) \, dx.$ Using integration by parts: $u = x^2, \quad dv = \cos(nx) \, dx, \quad du = 2x \, dx, \quad v = \frac{\sin(nx)}{n}.$ Integration by parts formula gives: $\int x^2 \cos(nx) \, dx = \frac{x^2 \sin(nx)}{n} - \int \frac{2x \sin(nx)}{n} \, dx.$ For the second term, we use integration by parts again: $u = x, \quad dv = \sin(nx) \, dx, \quad du = dx, \quad v = -\frac{\cos(nx)}{n}.$ This yields: $\int x \sin(nx) \, dx = -\frac{x \cos(nx)}{n} + \int \frac{\cos(nx)}{n} \, dx,$ and so on. After solving this, we get: $a_n = -\frac{4}{n^2} \cos(n\pi).$

Step 4: Compute $b_n$

Since $x^2$ is an even function, $b_n = 0$ because the integrand $x^2 \sin(nx)$ is odd: $b_n = \frac{1}{\pi} \int_{-\pi}^\pi x^2 \sin(nx) \, dx = 0.$

Step 5: Final Fourier Series

Substitute the computed coefficients into the Fourier series formula: $f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos(nx),$ where: $\frac{a_0}{2} = \frac{\pi^2}{3}, \quad a_n = -\frac{4}{n^2} \cos(n\pi), \quad b_n = 0.$ Thus: $f(x) = \frac{\pi^2}{3} - 4 \sum_{n=1}^\infty \frac{\cos(n\pi)}{n^2} \cos(nx).$ Note that $\cos(n\pi) = (-1)^n$, so: $f(x) = \frac{\pi^2}{3} - 4 \sum_{n=1}^\infty \frac{(-1)^n}{n^2} \cos(nx).$

Let me know if you'd like a deeper breakdown of any step!

Related Questions

1. What is the significance of the symmetry properties of $f(x)$ in Fourier series expansions?
2. Can you compute the Fourier series for $f(x) = x^3$ on $[-\pi, \pi]$?
3. How does the convergence of the Fourier series differ for discontinuous functions?
4. Why does $b_n$ vanish for even functions?
5. How would the Fourier series change if $f(x) = x^2$ were defined on a different interval?

Tip

Always identify the symmetry of a function before computing Fourier series coefficients; it simplifies the calculations significantly.