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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2025

Paper 2

Integration by parts gives the recurrence
$$
\gamma(x,y)=y^{x-1}e^{-y}+(x-1)\gamma(x-1,y).
$$
Iterating $N$ times yields
$$
\gamma(x,y)=y^{x-1}e^{-y}
\sum_{n=0}^{N-1}(x-1)(x-2)\cdots(x-n)y^{-n}
+R_N,
$$
where the empty product for $n=0$ is one and
$$
R_N=(x-1)(x-2)\cdots(x-N)\gamma(x-N,y).
$$
For fixed $x$, the remainder has the order of the first omitted term as $y\to\infty$. Therefore
$$
\boxed{a_0(x)=1,\qquad
a_n(x)=\prod_{j=1}^n(x-j)\quad(n\geq1).}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 32D / a / Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 32D / a / Solution

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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 32D / a / Solution

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