Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 4 / 31A / Solution

Put $A=n^2-1/4$. Substitute $w=e^{sz}{\sum }_{k\ge 0}{c}_k{z}^{\rho -k}$ into $w^{\prime \prime }+(1-A/z^2)w=0$. The leading coefficient gives $s^2+1=0$, and the next gives $2s\rho c_0=0$. Thus $s=\pm i$, $\rho =0$. The coefficient of $e^{sz}{z}^{-k-2}$ then givesIn particular $c_1=sA{c}_0/2$ and $c_2=A(2-A)c_0/8$. The two Bessel inverse-power expansions at infinity start asTheir independent leading oscillations give two independent asymptotic solutions, represented by suitable multiples of $\sqrt{z}{H}_n^{(1)}(z)$ and $\sqrt{z}{H}_n^{(2)}(z)$ on the positive-real large-$z$ sector.

For the Liouville–Green approximation, $q=1-A/z^2$ giveswhere the integration constant is absorbed in the amplitude. Hence its two leading wave forms have first two termsExpanding one order further gives the same $A(2-A)/(8z^2)$ as the formal recurrence.

The infinite sums must be interpreted as asymptotic expansions, not literal convergent inverse-power series. For integer $n$, $A$ never equals $k(k+1)$, so the recurrence never truncates, and $|c_{k+1}/c_k|\sim k/2$. Thus the inverse-power series diverges at every finite nonzero $z$. Exact Hankel solutions have these asymptotic series; the printed “solutions of the form” uses that customary asymptotic meaning.