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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2021

Paper 3

The function $f$ has zero mean and its cosine coefficients vanish:
$$
a_0=0,\qquad a_n=0.
$$
Its sine coefficients are
$$
b_n=\frac1\pi\left(\int_0^\pi\sin(n\theta)\,d\theta
-\int_\pi^{2\pi}\sin(n\theta)\,d\theta\right),
$$
so
$$
\boxed{
b_n=\begin{cases}
\dfrac4{\pi n},&n\text{ odd},\\
0,&n\text{ even}.
\end{cases}}
$$

Away from the corners, $F'=f$. Differentiating the <Fourier series> shows that $nB_n=a_n$ and $-nA_n=b_n$. Also,
$$
A_0=\frac1\pi\int_0^{2\pi}F(\theta)\,d\theta=\pi.
$$
Consequently
$$
\boxed{B_n=0,\qquad
A_n=\begin{cases}
-\dfrac4{\pi n^2},&n\text{ odd},\\
0,&n\text{ even},
\end{cases}}
$$
and
$$
F(\theta)=\frac\pi2-\frac4\pi
\sum_{r=0}^\infty\frac{\cos((2r+1)\theta)}{(2r+1)^2}.
$$

Evaluating the series for $f$ at $\theta=\pi/2$ gives the <Leibniz formula for π>,
$$
\boxed{\sum_{r=0}^\infty\frac{(-1)^r}{2r+1}=\frac\pi4}.
$$
Evaluating the continuous Fourier series for $F$ at $\theta=0$ gives
$$
0=\frac\pi2-\frac4\pi\sum_{r=0}^\infty\frac1{(2r+1)^2},
$$
and hence
$$
\boxed{\sum_{r=0}^\infty\frac1{(2r+1)^2}=\frac{\pi^2}{8}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 3 / 5A / Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 3 / 5A / Solution

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Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 3 / 5A / Solution

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