Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ii/paper-2/12e/solution
Past exam of the mathematics course of the University of Cambridge 2017 ii Paper 2 12E Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-05
Choose a branch of with and write . The Gauss hypergeometric equation isChanging to turns this into the same equation with third parameter . Its solution regular at is . Substitution of gives the Gauss hypergeometric equation for with parameters . Thus is also a solution. Their distinct nonintegrally separated characteristic exponents at a regular singular point imply linear independence near , soThis also follows from the exponent transformation in the Papperitz symbol.
Initially assume , so the Euler integral for the hypergeometric function applies. If , taking gives an integral evaluated by the beta function and determinesIndeed and there. To find , instead work in the nonempty parameter region , put and scale the integration variable by . The singular contribution isThe defining integral of the beta function evaluates this integral to . Since and , it determines . Analytic continuation in the parameters then yields the hypergeometric connection formulaThe usual power-series definition requires ; exceptional parameters need a separately specified analytic limit. Reciprocal gamma functions make vanishing coefficients at denominator poles meaningful. The branch must remain fixed when continuing in .
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