Fourier Series and Term-by-Term Differentiation: 04-BS-5, December 2016, Q2

Question 2: Fourier Series and Term-by-Term Differentiation (20 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 $-\pi\le x\le\pi$, extended with period $2\pi$; and $G(x)=x$ on the same interval, same period.

Find. The Fourier series of $F$, then (via differentiation of that result) the Fourier series of $G$.

Approach. $F$ is even, so its series has cosine terms only; compute $a_0,a_n$ by direct integration, then differentiate the resulting series term-by-term (legal here because $F$ is continuous and $F(-\pi)=F(\pi)$) to obtain the series for $F'(x)=2x$, and halve it to get $G(x)=x$.

Zero-th coefficient. $F$ even $\Rightarrow b_n=0$ for all $n$. $$a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}x^2\,dx=\frac{1}{\pi}\cdot\frac{2\pi^3}{3}=\frac{2\pi^2}{3}$$
General cosine coefficient (integrate by parts twice). $$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}x^2\cos(nx)\,dx=\frac{4(-1)^n}{n^2}$$
Assemble the series for $F$. $$\boxed{x^2=\frac{\pi^2}{3}+4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos(nx)},\qquad -\pi\le x\le\pi$$
Part (b) — differentiate term-by-term. $F$ is continuous with matching endpoint values, so its Fourier series may be differentiated term-by-term to give the series for $F'(x)=2x$: $$2x=-4\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\sin(nx)=4\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin(nx)$$ Dividing by $2$, $$\boxed{x=2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin(nx)=2\Bigl(\sin x-\frac{\sin2x}{2}+\frac{\sin3x}{3}-\cdots\Bigr)},\qquad -\pi\lt x\lt\pi$$ (the interval is open here because $G$'s periodic extension has a jump discontinuity at $x=\pm\pi$, where the series instead converges to the average $0$).

Function	Fourier series
$F(x)=x^2$	$\dfrac{\pi^2}{3}+4\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos(nx)$
$G(x)=x$	$2\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin(nx)$

Topic: Fourier series of an even function and term-by-term differentiation.

Key relations: $a_0=\frac1\pi\int F$, $a_n=\frac1\pi\int F\cos(nx)$; differentiating a Fourier series termwise reproduces the derivative's own Fourier series when legal.

Why this works: term-by-term differentiation of a Fourier series is valid when the function is continuous on the whole real line and its periodic extension has no jump — both hold for $F(x)=x^2$ since $F(-\pi)=F(\pi)=\pi^2$, so the boundary mismatch that would otherwise inject spurious terms is absent.

Common pitfall: assuming the same trick works in reverse — $G(x)=x$ itself has a jump at $\pm\pi$ ($G(-\pi^+)=-\pi\ne\pi=G(\pi^-)$), so its series may NOT be differentiated term-by-term to recover a convergent series for the constant function $1$.

Source: Kreyszig, Advanced Engineering Mathematics, Ch. 11 (Fourier Series).

Question 2 of 7: Fourier Series and Term-by-Term Differentiation

Question 2: Fourier Series and Term-by-Term Differentiation (20 marks)