Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 8

The origin is an ordinary point of the Hermite differential equation, so put $y={\sum }_{n=0}^{\infty }{a}_n{x}^n$. Substitution and matching powers givesThus the linear recurrence relation isEven and odd coefficients evolve independently from arbitrary $a_0,a_1$. The general power series solution iswith an empty product equal to one. Its first terms areBoth series converge at every finite $x$, since the ratio of successive terms in each parity subsequence tends to zero. For a nonzero polynomial of degree $N$, the highest coefficient in the differential equation requires $\lambda =2N$. Conversely, at such a value the appropriate parity series terminates, while its opposite-parity initial coefficient can be set to zero. HenceHere $H_N$ 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 $\lambda$; discreteness here comes from termination.

Insert the Frobenius solution and its differentiated series into the equation. Collecting the terms before shifting an index givesThe coefficient of the lowest power $x^{\sigma -1}$ is $a_0\sigma (\sigma +\gamma -1)$. Because $a_0\ne 0$, the indicial equation givesFor the coefficient of $x^{n+\sigma -1}$ with $n\ge 1$, shift the index in the second sum. The hypergeometric Frobenius recurrence isWhenever the left coefficient is nonzero, division yieldsFor noninteger $\gamma$, the two independent local solutions can therefore be identified aswith a consistent local branch for the fractional power. The hypergeometric function coefficients follow by iterating the recurrence using rising factorials.

There is a parameter qualification to the PDF's statement for all real constants: a zero denominator cannot be divided through. At an undetermined coefficient at Frobenius resonance, use the undivided coefficient equation to test compatibility. For example, $\gamma =0,\sigma =0,n=1$ requires $0=\alpha \beta a_0$, so a pure exponent-zero series with $a_0\ne 0$ is impossible if $\alpha \beta \ne 0$. When $\gamma =1$ the two indicial roots coincide, and a second independent solution generally needs a logarithmic term. The indicial roots and undivided relation remain the valid coefficient conclusions in these exceptional cases.