Fourier Series of a Parabolic Arch and Its Derivative: 04-BS-5, December 2019, Q2

Question 2: Fourier Series of a Parabolic Arch and Its Derivative (a: 15 marks; b: 5 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 $0\le x\le2\pi$, period $p=2\pi$ (so half-period $L=\pi$); $G(x)=2\pi-2x$ on the same interval and period.

Find. (a) The Fourier series of $F(x)$. (b) The Fourier series of $G(x)$, obtained from part (a) rather than integrated from scratch.

Approach. Since $F$ is defined over a full period $[0,2\pi]$ (not a symmetric interval), compute $a_0,a_n,b_n$ directly by integration. Notice $F'(x)=2\pi-2x=G(x)$; since $F$'s periodic extension is continuous ($F(0)=F(2\pi)=0$), differentiate the series of $F$ term by term to get the series of $G$ — far faster than integrating $G$'s own coefficients from scratch.

Compute $a_0$. With $L=\pi$, $$a_0=\frac1\pi\int_0^{2\pi}(2\pi x-x^2)\,dx=\frac1\pi\left[\pi x^2-\frac{x^3}{3}\right]_0^{2\pi}=\frac1\pi\left(4\pi^3-\frac{8\pi^3}{3}\right)=\frac{4\pi^2}{3}\ \Rightarrow\ \frac{a_0}{2}=\frac{2\pi^2}{3}$$
Compute $a_n$ by parts (twice). $$a_n=\frac1\pi\int_0^{2\pi}(2\pi x-x^2)\cos(nx)\,dx$$ Splitting into the two terms and integrating by parts twice (using $\sin(2n\pi)=0,\ \cos(2n\pi)=1$ since $n$ is an integer), the $2\pi x$ piece and $-x^2$ piece combine and every boundary term with $\sin$ vanishes, leaving $$\boxed{a_n=-\dfrac{4}{n^2}}$$
Compute $b_n$. The same by-parts reduction on $\int_0^{2\pi}(2\pi x-x^2)\sin(nx)\,dx$ has every boundary term proportional to $\cos(2n\pi)-\cos(0)=1-1=0$ once the interior antiderivative terms are collected, giving $$\boxed{b_n=0}$$
Assemble $F(x)$. $$\boxed{F(x)=2\pi x-x^2=\dfrac{2\pi^2}{3}-4\sum_{n=1}^{\infty}\dfrac{\cos(nx)}{n^2}}$$ Check: $F(0)=F(2\pi)=0$ (periodic extension continuous, no jump), consistent with a pure-cosine-plus-constant series that is finite and single-valued at the shared endpoint.
Differentiate term by term for $G$. Because $F$ is continuous with a piecewise-continuous derivative, termwise differentiation of its Fourier series is valid: $$F'(x)=\frac{d}{dx}\!\left[\frac{2\pi^2}{3}-4\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}\right]=4\sum_{n=1}^{\infty}\frac{n\sin(nx)}{n^2}=4\sum_{n=1}^{\infty}\frac{\sin(nx)}{n}$$ Since $F'(x)=G(x)$ identically, $$\boxed{G(x)=2\pi-2x=4\sum_{n=1}^{\infty}\dfrac{\sin(nx)}{n}}$$

Quantity	Result
$a_0/2$	$2\pi^2/3$
$a_n$ (F, cosine coeff.)	$-4/n^2$
$b_n$ (F, sine coeff.)	$0$
$F(x)$	$\dfrac{2\pi^2}{3}-4\sum\cos(nx)/n^2$
$G(x)$	$4\sum\sin(nx)/n$

Topic: Fourier series on a full period $[0,2\pi]$; obtaining a companion series by termwise differentiation.

Key relations: full-range Fourier series $a_0=\frac1L\int_0^{2L}f\,dx$, $a_n=\frac1L\int_0^{2L}f\cos(n\pi x/L)dx$, $b_n=\frac1L\int_0^{2L}f\sin(n\pi x/L)dx$ (here $L=\pi$); termwise differentiation is valid when $f$ is continuous and $f'$ is piecewise smooth.

Why this works: $F$'s periodic extension has no jump discontinuity ($F(0)=F(2\pi)=0$), which is exactly the condition needed to differentiate its Fourier series term by term and still obtain a valid series for the derivative — here $G=F'$. This avoids a second, separate integration for $G$'s own coefficients.

Common pitfall: differentiating a Fourier series whose function is discontinuous at the period boundary (that operation is not legitimate there); also forgetting the chain-rule factor of $n$ that appears when a $\cos(nx)$ term is differentiated.

Source: Kreyszig, Advanced Engineering Mathematics, 10th ed., Ch. 11.1–11.3 (Fourier Series).

Question 2 of 7: Fourier Series of a Parabolic Arch and Its Derivative

Question 2: Fourier Series of a Parabolic Arch and Its Derivative (a: 15 marks; b: 5 marks)