Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 3 / 14E (Further Complex Methods)

Write the Papperitz symbol by listing pairs of exponents at $0,1,\infty$. For $F(a,c-b;c;w)$ they are $(0,1-c)$, $(0,b-a)$, and $(a,c-b)$. Under $w=z/(z-1)$ the points $0,1,\infty$ in the $z$-plane correspond respectively to $0,\infty ,1$ in the $w$-plane. Thus the pulled-back exponents are $(0,1-c)$ at zero, $(a,c-b)$ at one, and $(0,b-a)$ at infinity.

Multiplication by $(1-z{)}^{-a}$ shifts the exponents at one by $-a$ and those at infinity by $+a$, yieldingThis is the original Gauss hypergeometric equation. Both functions are analytic and normalized to one at zero, so uniqueness of that normalized local solution proves the Pfaff transformationContinue the identity using the branch $|\arg (1-z)|<\pi$. As usual, the parameters must admit the normalized hypergeometric solution; excluded singular parameter values are treated only when their limiting solution exists.

For $F(a,1+a-c;1+a-b;w)$, the exponents at $w=0,1,\infty$ are $(0,b-a)$, $(0,c-a-b)$, and $(a,1+a-c)$. The substitution $w=1/z$ exchanges zero and infinity, giving $(a,1+a-c)$ at $z=0$, $(0,c-a-b)$ at one, and $(0,b-a)$ at infinity. Multiplication by $(-z{)}^{-a}$ shifts the zero exponents by $-a$ and the infinity exponents by $+a$. The resulting Papperitz symbol is again $(0,1-c)$, $(0,c-a-b)$, $(a,b)$. Therefore $w_1$ satisfies the same hypergeometric equation, with the stated branch $|\arg (-z)|<\pi$.

First assume $a-b\notin \mathbb{Z}$ and nonsingular parameters, so the two infinity solutions $w_1$ and $w_2$, obtained by exchanging $a,b$, are independent. Write $F=A{w}_1+B{w}_2$. Since the power series at $1/z=0$ starts at one, $w_1\sim (-z{)}^{-a}$ and $w_2\sim (-z{)}^{-b}$.

In the parameter region $\Re (b-a)>0$, let $z\to -\infty$ in the Pfaff transformation. The $w_2$ term vanishes after multiplication by $(-z{)}^a$, while $z/(z-1)\to 1$ and $(-z)/(1-z)\to 1$. The supplied value at one of the Gauss hypergeometric function therefore givesAnalytic continuation in the parameters extends this coefficient to the nonresonant parameter domain. Exchange $a,b$ to determine $B$. Hence the hypergeometric connection formula at infinity isThe branch is $|\arg (-z)|<\pi$. The existence of $F(a,b;c;1)$ alone does not exclude integer $a-b$. At such resonant values the two separate displayed terms can be singular; the correct statement takes their combined parameter limit, which can contain logarithms. This qualification is essential, for example when $a=b$.