 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/8/d/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 8 / d / Solution by

Codex  0 2026-10-18

For either exponent, the recurrence from part (c) implies

\frac{a _{n + 1} x ^{n + 1}}{a _{n} x ^{n}} = \frac{∣ x ∣}{∣ [ 2 ( n + 1 + σ ) - 1 ] ( n + σ ) ∣} ⟶ 0

(1)

for every finite complex

x

. The ratio test therefore gives

R = \infty for both power-series factors .

(2)

The factor multiplying

x^{σ}

is an entire function in each case. For

σ = 1/2

, this does not make the full Frobenius solution single-valued at zero: its square-root prefactor still requires a branch. The infinite radius refers to

\sum a_{n} x^{n}

, not to a Taylor expansion of the branched full solution.

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