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Past exam of the mathematics course of the University of Oxford

2023

Prelims

Mathematics V

{c}

Put $a=\pi/2$. The constant <Fourier coefficient> is
$$
a_0=\frac2\pi\left(\int_0^a x\,dx+\int_a^\pi a\,dx\right)=\frac{3\pi}{4}.
$$
For $n\geq1$, <integration by parts> gives
$$
\begin{aligned}
\int_0^a x\cos(nx)\,dx
&=\frac{a\sin(na)}n+\frac{\cos(na)-1}{n^2},\\
\int_a^\pi a\cos(nx)\,dx&=-\frac{a\sin(na)}n.
\end{aligned}
$$
The $n^{-1}$ terms cancel. Thus
$$
a_n=\frac{2}{\pi n^2}\left(\cos\frac{n\pi}{2}-1\right),\qquad
\boxed{f(x)=\frac{3\pi}{8}+\frac2\pi\sum_{n=1}^\infty
\frac{\cos(n\pi/2)-1}{n^2}\cos(nx)\quad(0\leq x\leq\pi).}
$$
This <Fourier cosine series> has $a_n=0$ when $n$ is divisible by four, $a_n=-4/(\pi n^2)$ when $n\equiv2\pmod4$, and $a_n=-2/(\pi n^2)$ when $n$ is odd.

Similarly,
$$
\begin{aligned}
\int_0^a x\sin(nx)\,dx&=-\frac{a\cos(na)}n+\frac{\sin(na)}{n^2},\\
\int_a^\pi a\sin(nx)\,dx&=\frac{a\cos(na)}n-\frac{a(-1)^n}n.
\end{aligned}
$$
Therefore
$$
b_n=\frac{2\sin(n\pi/2)}{\pi n^2}-\frac{(-1)^n}{n},\qquad
\boxed{f(x)=\sum_{n=1}^\infty\left(\frac{(-1)^{n+1}}n+
\frac{2\sin(n\pi/2)}{\pi n^2}\right)\sin(nx)\quad(0<x<\pi).}
$$
The <Fourier sine series> also equals $f(0)=0$ at zero, but has value zero at $\pi$, where $f(\pi)=\pi/2$.

Solution

Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics V / 4 / b / i / Solution

Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics V / 4 / b / i / Solution

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