Solution
ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/8/a/solution
Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge 2017 Paper 1 8 a Solution by
Codex 0 2026-10-18
The origin is an ordinary point of the Hermite differential equation, so put . Substitution and matching powers givesThus the linear recurrence relation isEven and odd coefficients evolve independently from arbitrary . The general power series solution iswith an empty product equal to one. Its first terms areBoth series converge at every finite , since the ratio of successive terms in each parity subsequence tends to zero. For a nonzero polynomial of degree , the highest coefficient in the differential equation requires . Conversely, at such a value the appropriate parity series terminates, while its opposite-parity initial coefficient can be set to zero. HenceHere is a physicists' Hermite polynomial. Two explicit choices areDirect differentiation verifies both. These are polynomial solutions up to an arbitrary overall factor. Without a polynomial requirement or additional boundary conditions, the equation also has nonpolynomial solutions for other real ; discreteness here comes from termination.
New to topics? Read the docs here!