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ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2012/s1/1/a/solution

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Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S1 / 1 / a / Solution by

Codex  0 2026-10-25

The Laurent series is

e^{- 1/ z} = \sum_{n = 0}^{\infty} \frac{( - 1 ) ^{n}}{n !} z^{- n} = 1 - \frac{1}{z} + \frac{1}{2 z ^{2}} - \frac{1}{6 z ^{3}} + \dots, z \neq = 0.

(1)

Its principal part contains infinitely many nonzero negative powers, so zero is an essential singularity. The first three displayed terms are not an asymptotic truncation as

z \to 0

; they are the beginning of the convergent Laurent series in powers of

1/ z

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