Fourier Series and a Classical Series Identity: 04-BS-5, May 2013, Q2

Question 2: Fourier Series and a Classical Series Identity (a) 14, (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. $f(x)=2\pi x-x^2$ on one period $0\le x\le2\pi$, extended periodically with period $p=2\pi$.

Find. (a) The Fourier series $\tfrac{a_0}2+\sum(a_n\cos nx+b_n\sin nx)$. (b) Use it to prove $\sum_{n=1}^{\infty}(-1)^{n+1}/n^2=\pi^2/12$.

Approach. (a) Integrate directly for $a_0$, and integrate by parts twice for $a_n$ (using $f$'s symmetry about $x=\pi$ to see $b_n=0$ without computing it). (b) Evaluate the series at a point where $f$ is known exactly and solve for the target sum.

Compute the constant term $a_0/2$. $$a_0=\frac1\pi\int_0^{2\pi}(2\pi x-x^2)\,dx=\frac1\pi\Big[\pi x^2-\tfrac13x^3\Big]_0^{2\pi}=\frac1\pi\Big(4\pi^3-\tfrac83\pi^3\Big)=\dfrac{4}{3}\pi^2$$ $$\boxed{\dfrac{a_0}{2}=\dfrac{2\pi^2}{3}}$$
Compute $a_n$ by integrating by parts twice. The boundary terms vanish because $\sin(2\pi n)=\sin0=0$; carrying out $\int_0^{2\pi}(2\pi x-x^2)\cos(nx)\,dx/\pi$ term by term, $$\boxed{a_n=-\dfrac{4}{n^2}}\qquad(n=1,2,3,\dots)$$
Compute $b_n$ using the midpoint symmetry, and assemble the series. Since $f(\pi+t)=2\pi(\pi+t)-(\pi+t)^2=\pi^2-t^2=f(\pi-t)$, $f$ is symmetric about $x=\pi$, so the sine coefficients vanish: $b_n=0$. Hence $$f(x)=\boxed{\dfrac{2\pi^2}{3}-4\sum_{n=1}^{\infty}\dfrac{\cos(nx)}{n^2}},\qquad 0\le x\le2\pi.$$
(b) Substitute $x=\pi$, a point where $f$ is known exactly. $f(\pi)=2\pi(\pi)-\pi^2=\pi^2$, and $\cos(n\pi)=(-1)^n$, so the series gives $$\pi^2=\dfrac{2\pi^2}{3}-4\sum_{n=1}^{\infty}\dfrac{(-1)^n}{n^2}$$
Solve for the target sum. $$4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}=\frac{2\pi^2}{3}-\pi^2=-\frac{\pi^2}{3}\ \Longrightarrow\ \sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}=-\frac{\pi^2}{12}$$ Multiplying by $-1$ turns $(-1)^n$ into $(-1)^{n+1}$: $$\boxed{\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n^2}=\dfrac{\pi^2}{12}}\qquad\blacksquare$$

Quantity	Result
$a_0/2$	$2\pi^2/3$
$a_n$	$-4/n^2$
$b_n$	$0$
Series	$f(x)=\tfrac{2\pi^2}3-4\sum\cos(nx)/n^2$
Identity proved	$\sum_{n=1}^{\infty}(-1)^{n+1}/n^2=\pi^2/12$

Concept & Theory

Topic. Fourier series of a function that is not obviously even/odd, exploited via its symmetry about the midpoint of the interval; using a convergent Fourier series to sum a numerical series.

Key relations. $a_n=\tfrac1\pi\int_0^{2\pi}f(x)\cos(nx)\,dx$, $b_n=\tfrac1\pi\int_0^{2\pi}f(x)\sin(nx)\,dx$; by Dirichlet's theorem, a convergent Fourier series reproduces $f$ exactly at any point of continuity.

Why this works. $f(x)=2\pi x-x^2$ is symmetric about $x=\pi$ (not about $x=0$), which is exactly the condition that kills every sine coefficient once the series is viewed as a function of $t=x-\pi$ — no integration of the $b_n$ integral is actually needed.

Common pitfall. Testing at $x=0$ first (the natural guess) only reproduces the already-known Basel sum $\sum1/n^2=\pi^2/6$; the alternating target identity needs the OTHER natural test point $x=\pi$, where $\cos(n\pi)=(-1)^n$ supplies the needed sign alternation.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 11 (§11.1–11.2 Fourier Series).

Question 2 of 7: Fourier Series and a Classical Series Identity

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