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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2023

Paper 2

The mean of $f$ is zero. Its <Fourier cosine series> coefficients are
$$
\begin{aligned}
A_n
&=\frac1\pi\left[
\int_0^\pi\left(\frac\pi2-\theta\right)\cos(n\theta)\,d\theta
+\int_\pi^{2\pi}\left(\theta-\frac{3\pi}2\right)\cos(n\theta)\,d\theta
\right]\\
&=\frac{2(1-(-1)^n)}{\pi n^2}.
\end{aligned}
$$
Thus $A_n=0$ for <even> $n$ and $A_n=4/(\pi n^2)$ for <odd> $n$. Substitution into part (i) yields
$$
\boxed{
\phi(r,\theta)
=\log r-1
+\frac4\pi
\sum_{\substack{n\geq1\\ n\ {\rm odd}}}
\frac{\cosh(n(\log r-1))}{n^2\cosh n}\cos(n\theta)
}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 14A / b / ii / Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 14A / b / ii / Solution

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Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 14A / b / ii / Solution

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