Paper 2, Section II, E

Further Complex Methods | Part II, 2017

The hypergeometric equation is represented by the Papperitz symbol

P{01∞00a1−cc−a−bb}P\left\{\begin{array}{ccc} 0 & 1 & \infty \\ 0 & 0 & a \\ 1-c & c-a-b & b \end{array}\right\}

and has solution y0(z)=F(a,b,c;z)y_{0}(z)=F(a, b, c ; z).

Functions y1(z)y_{1}(z) and y2(z)y_{2}(z) are defined by

y1(z)=F(a,b,a+b+1−c;1−z)y_{1}(z)=F(a, b, a+b+1-c ; 1-z)

and

y2(z)=(1−z)c−a−bF(c−a,c−b,c−a−b+1;1−z),y_{2}(z)=(1-z)^{c-a-b} F(c-a, c-b, c-a-b+1 ; 1-z),

where c−a−bc-a-b is not an integer.

Show that y1(z)y_{1}(z) and y2(z)y_{2}(z) obey the hypergeometric equation (∗)(*).

Explain why y0(z)y_{0}(z) can be written in the form

y0(z)=Ay1(z)+By2(z)y_{0}(z)=A y_{1}(z)+B y_{2}(z)

where AA and BB are independent of zz but depend on a,ba, b and cc.

Suppose that

F(a,b,c;z)=Γ(c)Γ(b)Γ(c−b)∫01tb−1(1−t)c−b−1(1−tz)−adtF(a, b, c ; z)=\frac{\Gamma(c)}{\Gamma(b) \Gamma(c-b)} \int_{0}^{1} t^{b-1}(1-t)^{c-b-1}(1-t z)^{-a} d t

with Re⁡(c)>Re⁡(b)>0\operatorname{Re}(c)>\operatorname{Re}(b)>0 and ∣arg⁡(1−z)∣<π|\arg (1-z)|<\pi. Find expressions for AA and BB.

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