= Solution
For the <half-range Fourier cosine series> on $(0,1)$, the coefficients are
$$
a_0=2\int_0^1x^2\,dx=\frac23,\qquad a_n=2\int_0^1x^2\cos(n\pi x)\,dx.
$$
Two integrations by parts, with $k=n\pi$, give
$$
\int_0^1x^2\cos(kx)\,dx=-\frac2k\int_0^1x\sin(kx)\,dx=\frac{2(-1)^n}{k^2}.
$$
Thus
$$
\boxed{x^2=\frac13+\frac4{\pi^2}\sum_{n=1}^{\infty}\frac{(-1)^n\cos(n\pi x)}{n^2}.}
$$
The even, $2$-periodic extension is continuous and piecewise continuously differentiable, so the <Fourier series> converges to it, including at $x=1$. Equivalently, the summable coefficient bound gives <uniform convergence> and permits taking the limit from the interior. Evaluating at $x=1$, where $(-1)^n\cos(n\pi)=1$, yields
$$
1=\frac13+\frac4{\pi^2}\sum_{n=1}^{\infty}\frac1{n^2},\qquad \boxed{\sum_{n=1}^{\infty}\frac1{n^2}=\frac{\pi^2}{6}.}
$$
This is the <Basel problem>.
Back to article page
