Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 30D (Asymptotic Methods)

Table of contents
- a 30D
  - Solution a
- b 30D
  - Solution b

a

Solution

The phase function

cos t

attains its global maximum

1

on

[0, 3 π]

at the stationary endpoint

t = 0

and at the interior point

t = 2 π

. At either point,

cos (t_{0} + u) = 1 - \frac{u ^{2}}{2} + O (u^{4}) .

By Laplace's method at a nondegenerate stationary endpoint, the endpoint contributes

e^{x} \frac{π}{2 x} .

By Laplace's method, the interior maximum contributes

e^{x + 2 π} \frac{2 π}{x} .

The remaining parts of the interval are exponentially smaller. Summing the two contributions according to Laplace's method with multiple global maxima gives

I (x) \sim e^{x} \frac{π}{2 x} (1 + 2 e^{2 π}) .

b

Solution

u = lo g (1 + t), t = e^{u} - 1, d t = e^{u} d u .

Then

J (x) = \int_{0}^{l o g (1 + π)} g (u) e^{- xu} d u, g (u) = e^{u} [1 - cos (e^{u} - 1)] .

The Taylor series at the endpoint is

e^{u} - 1 = u + \frac{u ^{2}}{2} + \frac{u ^{3}}{6} + O (u^{4}),

and hence

g (u) = \frac{u ^{2}}{2} + u^{3} + O (u^{4}) .

The Watson lemma, together with

\int_{0}^{\infty} u^{n} e^{- xu} d u = \frac{n !}{x ^{n + 1}},

therefore yields

J (x) = \frac{1}{x ^{3}} + \frac{6}{x ^{4}} + O (x^{- 5}) .

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