1^()circ Expand the 2 pi -periodic extension of f(x)=x²
$1^\circ$ Expand the $2\pi$-periodic extension of $f(x)=x^2$ on $[-\pi,\pi]$ in a Fourier series. $2^\circ$ Find (a) $\sum_{n=1}^\infty\frac1{n^2}$ and (b) $\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}$.
Explanation of Method
$f(x)=x^2$ is even, so only cosine terms appear; two integrations by parts give $a_n=4(-1)^n/n^2$. Evaluating the resulting series at the two natural test points $x=\pi$ and $x=0$ produces the Basel sum and its alternating counterpart directly.
- Compute $a_0$ and $a_n$ via double integration by parts.
- Evaluate at $x=\pi$ to get the Basel problem.
- Evaluate at $x=0$ to get the alternating sum, and cross-check against Problem 325.
Part 1° · Fourier series
$f(x)=x^2$ is even: $$ a_0 = \frac2\pi\int_0^\pi x^2\,dx = \frac{2\pi^2}3, \qquad a_n = \frac2\pi\int_0^\pi x^2\cos(nx)\,dx = \frac{4(-1)^n}{n^2}. $$ $$ \boxed{x^2 = \frac{\pi^2}3+\sum_{n=1}^\infty\frac{4(-1)^n}{n^2}\cos(nx) = \frac{\pi^2}3+4\left(-\cos x+\frac{\cos2x}4-\frac{\cos3x}9+\cdots\right)} $$
Part 2° · The sums
a) At $x=\pi$: $\cos(n\pi)=(-1)^n$, so $$ \pi^2 = \frac{\pi^2}3+4\sum_{n=1}^\infty\frac1{n^2} \implies \boxed{\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6} \quad\text{(the Basel problem).} $$ b) At $x=0$: $$ 0 = \frac{\pi^2}3+4\sum_{n=1}^\infty\frac{(-1)^n}{n^2} \implies \sum_{n=1}^\infty\frac{(-1)^n}{n^2}=-\frac{\pi^2}{12} \implies \boxed{\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}=\frac{\pi^2}{12}}. $$ Cross-check using Problem 325's $\sum1/(2n-1)^2=\pi^2/8$: since $\sum(-1)^{n-1}/n^2=\sum1/(2n-1)^2-\sum1/(2n)^2=\frac{\pi^2}8-\frac{\pi^2}{24}=\frac{\pi^2}{12}$. ✓
Summary Table
| Property | Value |
|---|---|
| Fourier series | $x^2=\frac{\pi^2}3+\sum\frac{4(-1)^n}{n^2}\cos(nx)$ |
| a) $\sum1/n^2$ | $\pi^2/6$ (Basel problem) |
| b) $\sum(-1)^{n-1}/n^2$ | $\pi^2/12$ |
Additional Notes
- This Fourier series is (up to a sign and scaling) the term-by-term antiderivative of Problem 324's series for $x$, illustrating how integrating a Fourier series raises the power of $n$ in the denominator by one, sharpening convergence from conditional to absolute.
- Together with Problem 325's $\pi^2/8$ result, this problem gives three of the four standard even-power zeta-type sums obtainable at this level; the fourth, $\sum1/n^4=\pi^4/90$, requires the quartic analogue of this same technique.
- Evaluating at $x=\pi$ versus $x=0$ is the standard trick for separating a series into its "all-terms" and "alternating" versions — the same device used in Problem 324's Parts 3–4.