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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 <function> is odd, so only sine coefficients occur. For $n\geq1$,
$$
b_n=\frac2\pi\int_0^\pi(x^3-\pi^2x)\sin(nx)\,dx.
$$
Integration by parts gives
$$
\int_0^\pi x\sin(nx)\,dx=-\frac{\pi(-1)^n}{n},
$$
and
$$
\int_0^\pi x^3\sin(nx)\,dx
=-\frac{\pi^3(-1)^n}{n}+\frac{6\pi(-1)^n}{n^3}.
$$
The terms of order $1/n$ cancel, leaving
$$
b_n=\frac{12(-1)^n}{n^3}.
$$
Thus
$$
x^3-\pi^2x
=12\sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}\sin(nx),
\qquad -\pi<x<\pi.
$$

The <Parseval identity> gives
$$
\frac1\pi\int_{-\pi}^{\pi}(x^3-\pi^2x)^2\,dx
=144\sum_{n=1}^{\infty}\frac1{n^6}.
$$
Direct integration yields
$$
\frac1\pi\int_{-\pi}^{\pi}
(x^6-2\pi^2x^4+\pi^4x^2)\,dx
=\frac{16\pi^6}{105}.
$$
Therefore
$$
\boxed{\sum_{n=1}^{\infty}\frac1{n^6}=\frac{\pi^6}{945}},
$$
as recorded in the <Fourier series of x cubed minus pi squared x>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 3A / Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 3A / Solution

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Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 3A / Solution

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