Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 14E (Further Complex Methods)

Put $y(x)=w(z)$ with $z={\sin }^{2}x$. The chain rule givesand henceAfter substitution and division by $4z(1-z)$, the ordinary differential equation becomesComparison with the Gauss hypergeometric equation gives, up to interchanging $A$ and $B$,

At $z=0$ and $z=1$, the coefficient of $w^{\prime }$ has at most a simple pole and that of $w$ has at most a double pole, so both are regular singular points. The regular singular point at infinity criterion gives the same conclusion at infinity. The corresponding Papperitz symbol isIndeed, the exponent pairs are $(0,1-C)=(0,1/2)$ at zero, $(0,C-A-B)=(0,1/2)$ at one, and $(A,B)=(k/2,-k/2)$ at infinity; their sum also satisfies the Fuchs relation.

The solution of $y^{\prime \prime }+k^{2}y=0$ with $y(0)=1$ and $y^{\prime }(0)=0$ is $y(x)=\cos (kx)$. Near $x=0$, the map $z={\sin }^{2}x$ has the local inverse $x=\arcsin \sqrt{z}$, so this solution is analytic in $z$, corresponds to exponent zero at $z=0$, and has value one there. Uniqueness of the normalized local solution of the Gauss hypergeometric equation therefore gives the hypergeometric cosine identityor, when $z={\sin }^{2}x$ on the chosen local branch,

Since $1-C=1/2$, the second local hypergeometric solution isUnder $z={\sin }^{2}x$, it must be a constant multiple of the independent elementary solution $\sin (kx)$. As $x\to 0$, both $z^{1/2}$ and $x$ are asymptotic to one another, while $sin(kx)\sim kx$. Matching the normalized hypergeometric factor, whose value at zero is one, givesConsequently the hypergeometric sine identity is