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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2014

Paper 4

For real nonintegral $\alpha$, the <Parseval identity> and the coefficient formula give
$$
 \frac{\alpha^2\sin^2(\pi\alpha)}{\pi^2}
 \sum_{n\in\mathbb Z}\frac1{(n^2-\alpha^2)^2}
 =\frac1{2\pi}\int_{-\pi}^\pi\cos^2(\alpha x)dx
 =\frac12+\frac{\sin(2\pi\alpha)}{4\pi\alpha}.
$$
Solving for the sum yields
$$
 \boxed{\sum_{n\in\mathbb Z}\frac1{(n^2-\alpha^2)^2}
 =\frac{\pi^2}{2\alpha^2\sin^2(\pi\alpha)}
 +\frac{\pi\cot(\pi\alpha)}{2\alpha^3}.}
$$
At $\alpha=5/2$, the sine square is one and the cotangent is zero, giving \b[$2\pi^2/25$].

Solution

Past exam of the mathematics course of the University of Cambridge / 2014 / ib / Paper 4 / 17D / v / Solution

Past exam of the mathematics course of the University of Cambridge / 2014 / ib / Paper 4 / 17D / v / Solution

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Past exam of the mathematics course of the University of Cambridge / 2014 / ib / Paper 4 / 17D / v / Solution

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