 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/12/a/ii/solution

For

q = r_{0} / R \to \infty

1/ q

e^{- 1/ q} = 1 - \frac{1}{q} + \frac{1}{2 q ^{2}} - \frac{1}{6 q ^{3}} + \frac{1}{24 q ^{4}} + O (q^{- 5}) .

The constant and linear terms cancel in

1 - q + q e^{- 1/ q}

f (q) = \frac{1}{2 q} - \frac{1}{6 q ^{2}} + \frac{1}{24 q ^{3}} + O (q^{- 4}) \sim \frac{1}{2 q} .

For

q \to 0^{+}

e^{- 1/ q}

is exponentially small:

f (q) = 1 - q + O (q e^{- 1/ q}) \sim 1 - q, f (q) \to 1.

The small-

q

correction

q e^{- 1/ q}

is smaller than every positive power of

q

beyond the displayed linear term. In dimensional terms, the large-

q

leading mass is

2 π A R^{2} / r_{0}

, while the small-

q

limit is

4 π A R

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