Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 1 / 8A / Solution

A Papperitz symbol records the three regular singular points of a second-order Fuchsian differential equation and the two local exponents at each. Exponents at infinity refer to solutions behaving as $z^{-\gamma }$, that is, to powers of the local coordinate $1/z$. Their sum satisfies the Fuchs relation: the six exponents add to one. The symbol represents the two-dimensional solution space, not one normalized solution by itself.

The specified exponents give the Gauss hypergeometric equationSet $d=c-a-b$ and let $v=F(c-a,c-b;c;z)$. Its equation isFor $u=(1-z{)}^{d}v$, differentiation gives $u^{\prime }=(1-z{)}^d[v^{\prime }-dv/(1-z)]$ and $u^{\prime \prime }=(1-z{)}^d[v^{\prime \prime }-2d{v}^{\prime }/(1-z)+d(d-1)v/(1-z{)}^2]$. Substitution into the first equation leaves $(1-z{)}^d$ times exactly the second equation: its $v^{\prime }$ coefficient is $c-(a+b+1+2d)z$, and its $v$ coefficient simplifies to $-(c-a)(c-b)$. Hence $u$ solves the same equation as $F(a,b;c;z)$.

The symbol gives the same check geometrically: multiplication by $(1-z{)}^d$ shifts the exponents at one from $\{0,-d\}$ to $\{d,0\}$ and those at infinity from $\{c-a,c-b\}$ to $\{b,a\}$; it does not change the exponents at zero. Choose the branch with $(1-0{)}^d=1$. Then $u$ is analytic at zero and $u(0)=1$. For $c\notin \{0,-1,-2,\dots \}$ the analytic recurrence uniquely determines this normalized solution, provingThe identity continues wherever both sides are defined with compatible branches. At exceptional parameters where the stated normalization does not uniquely define an analytic function, a parameter-continuation convention is needed; the generic proof does not silently assume uniqueness there. This is Euler's hypergeometric transformation.