Fourier series of $F(x)=x^2$ and the Basel-sum identity: 04-BS-5, May 2015, Q2

Question 2: Fourier series of $F(x)=x^2$ and the Basel-sum identity (a) 15 marks; (b) 5 marks

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $F(x)=x^2$ on $0\lt x\lt2$, extended periodically with period $p=2$ (half-period $L=1$).

Find. (a) The Fourier series of $F$. (b) Use it to prove $\pi^2/6=\sum1/n^2$.

Approach. Compute $a_0,a_n,b_n$ by direct integration over one period, then evaluate the series at the jump point $x=0$, where it converges to the average of the left- and right-hand limits.

Constant term. $$a_0=\dfrac1L\int_0^{2}x^2\,dx=\int_0^2x^2\,dx=\dfrac83\ \Rightarrow\ \dfrac{a_0}{2}=\dfrac43$$
Cosine coefficients (integrate by parts twice). $$a_n=\int_0^2x^2\cos(n\pi x)\,dx=\dfrac{4}{n^2\pi^2}\qquad(n=1,2,3,\dots)$$
Sine coefficients. $$b_n=\int_0^2x^2\sin(n\pi x)\,dx=-\dfrac{4}{n\pi}\qquad(n=1,2,3,\dots)$$ Assembling (a): $$\boxed{F(x)\sim\dfrac43+\sum_{n=1}^{\infty}\left[\dfrac{4}{n^2\pi^2}\cos(n\pi x)-\dfrac{4}{n\pi}\sin(n\pi x)\right]}$$
Evaluate at the jump $x=0$ for part (b). The periodic extension has $F(0^+)=0$ and, by periodicity, $F(2^-)=4$; at a jump discontinuity the Fourier series converges to the midpoint $\tfrac12(0+4)=2$. Substituting $x=0$ into the series ($\cos0=1,\ \sin0=0$): $$2=\dfrac43+\sum_{n=1}^{\infty}\dfrac{4}{n^2\pi^2}\ \Longrightarrow\ \sum_{n=1}^{\infty}\dfrac1{n^2}=\left(2-\dfrac43\right)\dfrac{\pi^2}{4}=\dfrac23\cdot\dfrac{\pi^2}{4}$$ $$\boxed{\dfrac{\pi^2}{6}=\sum_{n=1}^{\infty}\dfrac1{n^2}}$$ (both sides $=1.644934\ldots$ to 4 decimal places).

Final results — Question 2
Quantity	Value
$a_0/2$	$4/3$
$a_n$	$4/(n^2\pi^2)$
$b_n$	$-4/(n\pi)$
Identity proved	$\pi^2/6=\sum_{n=1}^{\infty}1/n^2$

Topic: Fourier series and pointwise convergence at a jump.

Key relations: $a_0=\tfrac1L\int_0^{2L}F\,dx$, $a_n=\tfrac1L\int_0^{2L}F\cos(n\pi x/L)\,dx$, $b_n=\tfrac1L\int_0^{2L}F\sin(n\pi x/L)\,dx$; at a jump, the series converges to $\tfrac12[F(x_0^-)+F(x_0^+)]$.

Why this works: Fourier's theorem guarantees the series reproduces $F$ at every continuity point and the midpoint value at every jump; choosing $x=0$ (a jump of the periodic extension of $x^2$) turns the abstract series into a concrete numerical identity, because $\cos(n\pi\cdot0)=1$ kills the sine terms and isolates $\sum a_n$.

Common pitfall: evaluating at an ordinary interior point (say $x=1$) instead of the jump — that also gives a valid identity but not the classical Basel sum; forgetting the periodic extension's jump value $F(2^-)=4$ (not $F(0)=0$) when computing the average.

Source: Kreyszig, Advanced Engineering Mathematics, 10th ed., §11.1–11.2 (Fourier Series).

Question 2 of 7: Fourier series of $F(x)=x^2$ and the Basel-sum identity

Question 2: Fourier series of $F(x)=x^2$ and the Basel-sum identity (a) 15 marks; (b) 5 marks