Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 4 / 31B / a

Solution

Choose an analytic square-root branch of

f

w_{\pm} (z) \sim f (z)^{- 1/4} exp (\pm \int^{z} f (s) d s),

with successive inverse-power corrections; the two signs give an asymptotic basis in suitable sectors. Since

f = f_{0} + f_{1} / (2 f_{0} z) + \dots

, the exponent contains a linear term and a logarithm. For the specified

f

f = 1 + 1/ (2 z) + 7/ (8 z^{2}) + \dots

, explaining

e^{z} z^{1/2}

and

e^{- z} z^{- 1/2}

on the positive real axis.

To determine the requested coefficient from the differential equation itself, put

w_{1} = e^{z} z^{1/2} h (z)

. Substitution gives

h^{''} + (2 + 1/ z) h^{'} - \frac{9}{4 z ^{2}} h = 0.

For

h = 1 + a_{1} / z + \dots

, the order-

z^{- 2}

- 2 a_{1} - 9/4

. It must vanish, giving

a_{1} = - 9/8 .

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