 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-1/14e/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Since

1 - C = 1/2

z^{1/2} F (A - C + 1, B - C + 1; 2 - C; z) = z^{1/2} F (\frac{1 + k}{2}, \frac{1 - k}{2}; \frac{3}{2}; z) .

Under

z = sin^{2} x

, it must be a constant multiple of the independent elementary solution

sin (k x)

x \to 0

, both

z^{1/2}

and

x

are asymptotic to one another, while

s in (k x) \sim k x

. Matching the normalized hypergeometric factor, whose value at zero is one, gives

z^{1/2} F (\frac{1 + k}{2}, \frac{1 - k}{2}; \frac{3}{2}; z) = \frac{s i n ( k x )}{k} .

F (\frac{1 + k}{2}, \frac{1 - k}{2}; \frac{3}{2}; z) = \frac{sin ( k arcsin z )}{k z} = \frac{sin ( k x )}{k sin x} .

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