 Solution

ID: past-exam-of-the-short-options-paper-of-the-physics-prelims-of-the-university-of-oxford/2021/s01/1/b/solution

(2 + z)^{- 1}

converges for

∣ z ∣ < 2

, while the cancelled sine terms have an everywhere convergent Taylor series. Hence the requested Laurent series has no negative powers:

f (z) = n = 0 \sum \infty \frac{( - 1 ) ^{n} z ^{n}}{2 ^{n + 1}} + k = 1 \sum \infty \frac{( - 1 ) ^{k} z ^{2 k - 1}}{( 2 k + 1 )!}, 0 < ∣ z ∣ < 2.

For example,

f (z) = \frac{1}{2} - \frac{5 z}{12} + \frac{z ^{2}}{8} - \frac{13 z ^{3}}{240} + \frac{z ^{4}}{32} + O (z^{5}) .

The extension through zero also makes this a Taylor series. The pole at

- 2

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