Paper 1, Section II, A

Further Complex Methods | Part II, 2019

(a) Consider the Papperitz symbol (or P-symbol):

P{abcαβγzαβγ}(†)\tag{†} P\left\{\begin{array}{cccc} a & b & c & \\ \alpha & \beta & \gamma & z \\ \alpha^{\prime} & \beta^{\prime} & \gamma^{\prime} & \end{array}\right\}

Explain in general terms what this PP-symbol represents.

[You need not write down any differential equations explicitly, but should provide an explanation of the meaning of a,b,c,α,β,γ,α,βa, b, c, \alpha, \beta, \gamma, \alpha^{\prime}, \beta^{\prime} and γ.]\left.\gamma^{\prime} .\right]

(b) Prove that the action of [(za)/(zb)]δ[(z-a) /(z-b)]^{\delta} on ()(†) results in the exponential shifting,

P{abcα+δβδγzα+δβδγ}(‡)\tag{‡} P\left\{\begin{array}{cccc} a & b & c \\ \alpha+\delta & \beta-\delta & \gamma & z \\ \alpha^{\prime}+\delta & \beta^{\prime}-\delta & \gamma^{\prime} \end{array}\right\}

[Hint: It may prove useful to start by considering the relationship between two solutions, ω\omega and ω1\omega_{1}, which satisfy the PP-equations described by the respective PP-symbols () and ]

(c) Explain what is meant by a Möbius transformation of a second order differential equation. By using suitable transformations acting on ()(†), show how to obtain the PP symbol

P{0100az1ccabb}(*)\tag{*} P\left\{\begin{array}{cccc} 0 & 1 & \infty \\ 0 & 0 & a & z \\ 1-c & c-a-b & b \end{array}\right\}

which corresponds to the hypergeometric equation.

(d) The hypergeometric function F(a,b,c;z)F(a, b, c ; z) is defined to be the solution of the differential equation corresponding to ()(\star) that is analytic at z=0z=0 with F(a,b,c;0)=1F(a, b, c ; 0)=1, which corresponds to the exponent zero. Use exponential shifting to show that the second solution, which corresponds to the exponent 1c1-c, is

z1cF(ac+1,bc+1,2c;z).z^{1-c} F(a-c+1, b-c+1,2-c ; z) .

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