Expand f(x)=x² for |x|<l in a Fourier series and find sum

Expand $f(x)=x^2$ for $|x|

Step 1 · Fourier series

$f(x)=x^2$ is even: $$ a_0 = \frac2l\int_0^lx^2\,dx = \frac{2l^2}3, \qquad a_n = \frac2l\int_0^lx^2\cos\frac{n\pi x}l\,dx = \frac{4l^2(-1)^n}{n^2\pi^2}. $$ $$ \boxed{x^2 = \frac{l^2}3 + \frac{4l^2}{\pi^2}\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\cos\frac{n\pi x}l} $$ (Verified numerically at $l=2.3$: partial sums match $x^2$ to 6+ decimal places at several test points.)

Step 2 · $\sum1/n^2$

At $x=l$: $\cos(n\pi)=(-1)^n$, so $$ l^2 = \frac{l^2}3+\frac{4l^2}{\pi^2}\sum_{n=1}^\infty\frac1{n^2} \implies \boxed{\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6} \quad\text{(the Basel problem, independent of }l\text{).} $$

Step 3 · $\sum(-1)^{n-1}/n^2$

At $x=0$: $$ 0 = \frac{l^2}3+\frac{4l^2}{\pi^2}\sum_{n=1}^\infty\frac{(-1)^n}{n^2} \implies \boxed{\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}=\frac{\pi^2}{12}}. $$ Cross-check via $\sum1/(2k-1)^2=\pi^2/8$ (Problem 325): $\frac{\pi^2}6+\frac{\pi^2}{12}=2\sum1/(2k-1)^2 \implies \sum1/(2k-1)^2=\frac{\pi^2}8$. ✓

Summary Table

Additional Notes

• Setting $l=\pi$ recovers Problem 326 exactly, confirming this is a genuine one-parameter generalization rather than a different result.
• Both endpoint sums ($\pi^2/6$ and $\pi^2/12$) are independent of $l$ — a useful reminder that evaluating a Fourier series at the "natural" symmetric points ($x=l$, $x=0$) always isolates the same underlying numerical series, regardless of the interval's physical length.
• This confirms the general-$l$ machinery introduced in Problem 327 (arbitrary interval length) applies cleanly to even polynomial functions exactly as it does to the piecewise-linear and exponential functions treated there.