Fourier Series and a Classical Series Identity: 04-BS-5, May 2014, Q2

Question 2: Fourier Series and a Classical Series Identity (14+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. $F(x)$, periodic with period $p=2$ (half-period $l=1$): $F(x)=x$ on $(0,1]$, $F(x)=1$ on $(1,2]$, extended periodically.

Find. (A) The full Fourier series of $F(x)$. (B) The value of $\sum_{n=1}^{\infty}1/(2n-1)^2$, using the series at a convenient point.

F(x) over three periods: a unit ramp followed by a unit plateau, repeating with period 2. The dashed line marks the jump at the periodic wrap x=0 (equivalently x=2), which Part B exploits.

Approach. Compute $a_0,a_n,b_n$ from the Euler formulas over one period $[0,2]$ to get the series (Part A), then evaluate the series at $x=0$ — a jump point of the periodic extension — and invoke Dirichlet's theorem (the series there equals the average of the one-sided limits) to isolate the requested numeric sum (Part B).

Compute $a_0$. $$a_0=\int_0^2F(x)\,dx=\int_0^1x\,dx+\int_1^2 1\,dx=\tfrac12+1=\tfrac32,\qquad \frac{a_0}{2}=\frac34$$
Compute $a_n$. Integrating $x\cos(n\pi x)$ by parts on $(0,1]$ and $\cos(n\pi x)$ directly on $(1,2]$ and adding, $$a_n=\int_0^1x\cos(n\pi x)\,dx+\int_1^2\cos(n\pi x)\,dx=\boxed{\dfrac{(-1)^n-1}{\pi^2n^2}}=\begin{cases}-\dfrac{2}{\pi^2n^2} & n\text{ odd}\\[4pt]0 & n\text{ even}\end{cases}$$
Compute $b_n$. The same split for the sine integrals gives $$b_n=\int_0^1x\sin(n\pi x)\,dx+\int_1^2\sin(n\pi x)\,dx=\boxed{-\dfrac{1}{\pi n}}$$
Assemble the series — Part A answer. $$F(x)\sim\frac34-\frac{2}{\pi^2}\sum_{n\text{ odd}}\frac{\cos(n\pi x)}{n^2}-\frac1\pi\sum_{n=1}^{\infty}\frac{\sin(n\pi x)}{n}$$
Evaluate the Dirichlet value at $x=0$. $x=0$ is a jump point of the periodic extension: the right-hand limit is $F(0^+)=0$, and one period back the left-hand limit is $F(2^-)=1$. Dirichlet's theorem gives the series, at a jump, the midpoint value: $$\frac{F(0^+)+F(2^-)}{2}=\frac{0+1}{2}=\frac12$$
Solve for the target sum — Part B answer. At $x=0$, $\cos0=1$ and $\sin0=0$, so the series collapses to $a_0/2+\sum a_n$: $$\frac34-\frac{2}{\pi^2}\sum_{n\text{ odd}}\frac1{n^2}=\frac12\ \Longrightarrow\ \frac{2}{\pi^2}\sum_{n\text{ odd}}\frac1{n^2}=\frac14\ \Longrightarrow\ \boxed{\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}}$$

Fourier coefficients and the derived series identity
Quantity	Value
a₀/2	3/4
a_n, n odd	−2/(π²n²)
a_n, n even	0
b_n	−1/(πn)
Σ 1/(2n−1)²	π²/8

Topic: Fourier series of a piecewise ramp-plateau function, and harvesting a numeric series from the Dirichlet convergence theorem.

Key relations: Euler formulas $a_0,a_n,b_n$ on a period-$2l$ interval; Dirichlet's theorem — at a jump discontinuity the Fourier series converges to the average of the left- and right-hand limits.

Why this works: plugging a special value of $x$ into a convergent Fourier series turns an infinite trigonometric sum into an infinite numeric sum. Choosing $x=0$ here is deliberate: $\cos(n\pi\cdot0)=1$ kills the $n$-dependence inside the cosine terms and $\sin(0)=0$ removes the $b_n$ series entirely, leaving a single clean equation in $\sum 1/n^2$ (over odd $n$). This is the standard route to Basel-type identities from Fourier analysis.

Common pitfall: using the naive piecewise value $F(0)=0$ instead of the Dirichlet average of the periodic extension's two one-sided limits; also forgetting that $a_n=0$ for even $n$, so "$\sum_{n=1}^\infty a_n$" silently reduces to a sum over odd $n$ only — exactly the $\sum1/(2n-1)^2$ the question asks for.

Source: Kreyszig, Advanced Engineering Mathematics, Fourier series and Dirichlet's theorem (Ch. 11).

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

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