Fourier Series of a Piecewise-Linear Wave and a Series Identity: 04-BS-5, Undated paper, Q2

Question 2: Fourier Series of a Piecewise-Linear Wave and a Series Identity (20 marks: (a) 15, (b) 5)

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. One period of a $2\pi$-periodic wave built from two straight segments of different slope, together with the target series identity.

Given data
Quantity	Value
Period	$p = 2\pi$, so the harmonics are $\cos nx$, $\sin nx$
Branch on $-\pi \lt x \lt 0$	$f(x)=x+\pi$ (slope $1$, rises from $0$ to $\pi$)
Branch on $0 \lt x \lt \pi$	$f(x)=x/2$ (slope $\tfrac12$, rises from $0$ to $\pi/2$)
Discontinuity	at $x=0$ (and every $x=2k\pi$), from $\pi$ down to $0$

Find. The coefficients $a_{0}$, $a_{n}$, $b_{n}$ and the assembled series in part (a); then the numerical value of $\sum 1/(2n-1)^{2}$ extracted from that series in part (b).

Three periods of the given wave. Each period rises linearly from $0$ to $\pi$ on the left half and from $0$ to $\pi/2$ on the right half, so a jump of height $\pi$ occurs at every $x=2k\pi$; the dashed segments mark the jumps whose mean value $\pi/2$ is what part (b) exploits.

Approach. Integrate $f$ against $1$, $\cos nx$ and $\sin nx$ over one period, treating the two branches separately; then evaluate the resulting series at the discontinuity $x=0$, where Dirichlet's theorem replaces $f$ by the mean of its one-sided limits and every sine term vanishes.

Constant term. With $a_{0}=\dfrac{1}{\pi}\displaystyle\int_{-\pi}^{\pi}f(x)\,dx$, $$\int_{-\pi}^{0}(x+\pi)\,dx = \Big[\tfrac{x^{2}}{2}+\pi x\Big]_{-\pi}^{0} = \frac{\pi^{2}}{2},\qquad \int_{0}^{\pi}\frac{x}{2}\,dx = \frac{\pi^{2}}{4},$$ so $a_{0} = \dfrac{1}{\pi}\Big(\dfrac{\pi^{2}}{2}+\dfrac{\pi^{2}}{4}\Big) = \dfrac{3\pi}{4}$ and the constant term of the series is $\dfrac{a_{0}}{2} = \dfrac{3\pi}{8}$.
Cosine coefficients. Using $\int x\cos nx\,dx = \dfrac{x\sin nx}{n}+\dfrac{\cos nx}{n^{2}}$ and noting that $\int_{-\pi}^{0}\pi\cos nx\,dx = 0$, $$\int_{-\pi}^{0}(x+\pi)\cos nx\,dx = \frac{1-(-1)^{n}}{n^{2}},\qquad \int_{0}^{\pi}\frac{x}{2}\cos nx\,dx = \frac{(-1)^{n}-1}{2n^{2}} .$$ Adding them and dividing by $\pi$ leaves $$\boxed{\,a_{n} = \frac{1-(-1)^{n}}{2\pi n^{2}} = \begin{cases} \dfrac{1}{\pi n^{2}} & n \text{ odd} \\[4pt] 0 & n \text{ even}\end{cases}}$$
Sine coefficients. Using $\int x\sin nx\,dx = -\dfrac{x\cos nx}{n}+\dfrac{\sin nx}{n^{2}}$ together with $\int_{-\pi}^{0}\pi\sin nx\,dx = \dfrac{\pi}{n}\big((-1)^{n}-1\big)$, the left branch contributes $-\pi/n$ and the right branch $-\pi(-1)^{n}/(2n)$. Dividing by $\pi$, $$\boxed{\,b_{n} = -\frac{2+(-1)^{n}}{2n} = \begin{cases} -\dfrac{1}{2n} & n \text{ odd} \\[4pt] -\dfrac{3}{2n} & n \text{ even}\end{cases}}$$ The sine terms survive for every $n$, as they must: $f$ is neither even nor odd.
Assemble the series. Separating the odd harmonics $n=2k-1$ from the even ones $n=2k$, $$f(x) = \frac{3\pi}{8} + \frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\cos\big((2k-1)x\big)}{(2k-1)^{2}} - \frac{1}{2}\sum_{k=1}^{\infty}\frac{\sin\big((2k-1)x\big)}{2k-1} - \frac{3}{2}\sum_{k=1}^{\infty}\frac{\sin(2kx)}{2k} .$$ This completes part (a). The series converges to $f(x)$ at every point of continuity and to the mean of the one-sided limits at each jump.
Evaluate at the jump to obtain the identity (part b). At $x=0$ every sine term is zero, while the left- and right-hand limits of $f$ are $f(0^{-}) = \pi$ and $f(0^{+}) = 0$. Dirichlet's theorem therefore gives the series the value $\tfrac12\big(\pi+0\big) = \pi/2$, so $$\frac{\pi}{2} = \frac{3\pi}{8} + \frac{1}{\pi}\sum_{k=1}^{\infty}\frac{1}{(2k-1)^{2}} .$$ Subtracting and multiplying by $\pi$, $$\boxed{\;\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{2}} = \pi\left(\frac{\pi}{2}-\frac{3\pi}{8}\right) = \pi\cdot\frac{\pi}{8} = \frac{\pi^{2}}{8}\;}$$ as required. Numerically the left side is $1.2337$ and $\pi^{2}/8 = 1.2337$, which confirms the algebra.

Final results — Question 2
Quantity	Result
Constant term $a_{0}/2$	$3\pi/8 \approx 1.1781$
$a_{n}$	$1/(\pi n^{2})$ for odd $n$; $0$ for even $n$
$b_{n}$	$-1/(2n)$ for odd $n$; $-3/(2n)$ for even $n$
Series value at $x=0$	$\pi/2$ (mean of $\pi$ and $0$)
Series identity	$\sum_{n\ge1}1/(2n-1)^{2} = \pi^{2}/8 \approx 1.2337$

Topic. Fourier series of a $2\pi$-periodic piecewise-linear function, and the harvesting of a numerical series from it at a point of discontinuity.

Key relations. $a_{0}=\tfrac{1}{\pi}\int_{-\pi}^{\pi}f\,dx$, $a_{n}=\tfrac{1}{\pi}\int_{-\pi}^{\pi}f\cos nx\,dx$, $b_{n}=\tfrac{1}{\pi}\int_{-\pi}^{\pi}f\sin nx\,dx$; Dirichlet's convergence theorem, which sends the series to $\tfrac12[f(x^{-})+f(x^{+})]$ at a jump.

Why this works. The trigonometric system is orthogonal over any full period, so projecting $f$ onto each harmonic isolates one coefficient at a time; that is exactly what the three integrals do. Because $f$ is only piecewise smooth the series cannot converge to $f$ at the jump, and Dirichlet's theorem says precisely what it converges to instead. That substitute value is the lever for part (b): choosing the evaluation point at the jump kills all the sine terms and leaves a pure number on the left equal to a pure number plus a multiple of the wanted series on the right. The general result $\sum 1/(2n-1)^{2}=\pi^{2}/8$ is the odd-harmonic half of the Basel sum $\sum 1/n^{2}=\pi^{2}/6$.

Common pitfall. Evaluating the series at $x=0$ but setting the left side to $f(0)=0$ (or to $\pi$) instead of the mean $\pi/2$, which corrupts the identity by a factor; and dropping the $\pi$ in $x+\pi$ when integrating $\int_{-\pi}^{0}$, whose cosine contribution is zero but whose sine contribution is not.

Source. Kreyszig, Advanced Engineering Mathematics, 10th ed., §11.1 (Fourier series, Euler formulas) and §11.2 (convergence at jumps; functions of arbitrary period).

Question 2 of 7: Fourier Series of a Piecewise-Linear Wave and a Series Identity

Question 2: Fourier Series of a Piecewise-Linear Wave and a Series Identity (20 marks: (a) 15, (b) 5)