Fourier Series and a Classical Identity: 04-BS-5, December 2015, Q2

Question 2: Fourier Series and a Classical Identity (A: 14 marks, B: 6 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. $f(x)$ is $2\pi$-periodic, piecewise linear on $(-\pi,0]$ and constant on $(0,\pi]$, as defined above.

Find. (A) The full Fourier series $a_0/2+\sum(a_n\cos nx+b_n\sin nx)$; (B) use it at $x=0$ to derive the Basel-type identity for $\pi^2/8$.

Approach. Compute $a_0$, $a_n$, $b_n$ from the Euler–Fourier formulas by splitting each integral at $x=0$; then evaluate the series at the continuity point $x=0$ (both branches agree there, $f(0)=\pi$) to extract the numerical identity.

Mean value $a_0$. $$a_0=\frac{1}{\pi}\int_{-\pi}^{0}(x+\pi)\,dx+\frac{1}{\pi}\int_{0}^{\pi}\pi\,dx = \frac{1}{\pi}\left(\frac{\pi^{2}}{2}\right)+\pi = \frac{3\pi}{2}\ \Rightarrow\ \frac{a_0}{2}=\frac{3\pi}{4}.$$
Cosine coefficients $a_n$. On $(0,\pi]$, $\int_0^\pi \pi\cos(nx)\,dx = \pi\sin(n\pi)/n = 0$. On $(-\pi,0]$, substituting $u=x+\pi$ (so $u:0\to\pi$) and using $\cos(nx)=\cos(nu-n\pi)=(-1)^n\cos(nu)$: $$\int_{-\pi}^{0}(x+\pi)\cos(nx)\,dx = (-1)^n\int_0^\pi u\cos(nu)\,du = \frac{1-(-1)^n}{n^{2}}.$$ Hence $a_n=\dfrac{1-(-1)^n}{\pi n^{2}}$, i.e. $$\boxed{a_n = 0\ (n\text{ even}), \qquad a_n=\frac{2}{\pi n^{2}}\ (n\text{ odd}).}$$
Sine coefficients $b_n$. The same substitution on the sine integral gives, after combining both branches, $$\boxed{b_n = \frac{(-1)^{n+1}}{n}, \qquad n=1,2,3,\dots}$$
Assemble the series (part A). $$f(x) = \frac{3\pi}{4} + \sum_{\substack{n=1\\ n\ \text{odd}}}^{\infty}\frac{2}{\pi n^{2}}\cos(nx) + \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin(nx).$$
Part B — evaluate at $x=0$. Both branches of $f$ agree at $x=0$: the left branch gives $f(0^-)=0+\pi=\pi$ and the right branch's limit is also $\pi$, so $f$ is continuous there and the Fourier series converges to $f(0)=\pi$. Setting $x=0$ in the series (every $\sin$ term vanishes): $$\pi = \frac{3\pi}{4} + \sum_{\substack{n=1\\ n\ \text{odd}}}^{\infty}\frac{2}{\pi n^{2}} \ \Rightarrow\ \sum_{\substack{n=1\\ n\ \text{odd}}}^{\infty}\frac{1}{n^{2}} = \left(\pi-\frac{3\pi}{4}\right)\frac{\pi}{2} = \frac{\pi}{4}\cdot\frac{\pi}{2}.$$ Re-indexing the odd integers as $n=2k-1$ gives the required identity: $$\boxed{\frac{\pi^{2}}{8} = \sum_{k=1}^{\infty}\frac{1}{(2k-1)^{2}}}$$

Final results — Question 2
Quantity	Result
$a_0/2$	$3\pi/4$
$a_n$	$0$ ($n$ even), $2/(\pi n^2)$ ($n$ odd)
$b_n$	$(-1)^{n+1}/n$
Identity proved	$\pi^{2}/8=\sum_{n=1}^{\infty}1/(2n-1)^{2}$

Topic: Fourier series of a piecewise-defined periodic function; using pointwise convergence at a continuity point to derive a numerical series identity.

Key relations: Euler–Fourier formulas $a_0=\frac1\pi\int_{-\pi}^{\pi}f\,dx$, $a_n=\frac1\pi\int_{-\pi}^{\pi}f\cos(nx)\,dx$, $b_n=\frac1\pi\int_{-\pi}^{\pi}f\sin(nx)\,dx$; Dirichlet convergence theorem (series converges to $f(x)$ at continuity points, to the midpoint of the jump at a discontinuity).

Why this works: Because $f$ happens to be continuous at $x=0$ (both one-sided values equal $\pi$), the Fourier series there converges exactly to $\pi$ rather than to a jump-average, which is what lets the series equation be read directly as a numeric identity for $\sum 1/(2n-1)^2$ — this is the standard textbook route to Basel-type sums, alongside the classic all-odd-square-reciprocal identity obtained from a pure square wave.

Common pitfall: Evaluating at a jump discontinuity instead of a continuity point (would require the Dirichlet midpoint rule, not the raw function value); also dropping the even-$n$ terms of $a_n$ without checking they are genuinely zero (they are, since $1-(-1)^n=0$ for even $n$).

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

Question 2 of 7: Fourier Series and a Classical Identity

Question 2: Fourier Series and a Classical Identity (A: 14 marks, B: 6 marks)