What is the cut structure of f(z) = F(a,b,c; z^(-2))
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Short working note by Phil, dated 2.18.10, on the cut structure of the hypergeometric function F(a,b,c;1/z^2). He argues the cuts lie along (-1,0) and (0,1) in the z plane because 1/z^2 exceeds 1 there. He uses Kummer's connection formulas (from Bateman) and power-function phase rules to show the values differ above and below the cut, checking with z=1/2.
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What is the cut structure of f(z) = F(a,b,c; 1/z2) PhL 2.18.10
Theory
First, here was my approach taken in my Legendre document:
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The F function has a cut for ξ = 1/z2 ≥ 1, according to our Section 5 above. It is useful to plot this function ξ = 1/z2,
This shows that as z runs ( 1→0), the argument ξ runs (1→∞). And:
This shows that as z runs (-1→0), the argument ξ runs (1→∞). I like to view this as two cuts in the z plane, the first is (0,1) and the second is (-1,0) [although one can of course draw them as one (-1,1)]. These two cuts will appear green below. [Note: The hypergeometric series F(a,b,c; 1/z2) diverges inside |z| = 1, but the hypergeometric function F(a,b,c; 1/z2) which continues this series has these green cuts. ]
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Now, the presence of a cut suggests that the function has different values on the two sides of the cut. But how would that arise? Consider some z between 0 and 1: R < 1
ξ = z-2 z = Re±iε ξ = R-2 e∓2iε = S e∓iε' S > 1 F(a,b,c,ξ)
So yes, ξ would indeed lie on the cut of F(a,b,c,ξ). For a point z in this range, we really have to use some continuation formula to see what is happening.
Kummer: The formulas starting with p 106 (25) all involve fixed phases and gamma functions, there are no functions of z appearing. We assume "general" values of a,b,c and avoid all gamma function poles. Our function F(a,b,c,ξ) is u1 on page 105. We would like to relate this to u3 and u4 which have 1/ξ. This would be a formula red 134 (I just added these letters in Bateman) which is p 107 (34). So think of this as
F(a,b,c,ξ) = A u3 + B u4 = A (-ξ)-aF(a,a+1-c; a+1-b; 1/ξ) + B A (-ξ)-bF(b,b+1-c; b+1-a; 1/ξ)
where the second term of course is the first with a↔b, as the symmetry of the LHS requires. Now set ξ = z-2 and we get
F(a,b,c, z-2) = A u3 + B u4 = A (-z-2)-aF(a,a+1-c; a+1-b; z2) + B A (-z-2)-bF(b,b+1-c; b+1-a; z2)
Consider one of these power factors:
(-z-2)-a = (-z2)a = (02-z2)a = (0-z)a (0+z)a
We have to say this has a cut to the left and a cut to the right, both branch points at 0. And the residual F functions have branch points for |z| > 1, so here is our picture:
It must be, then, that the outer cuts cancel, and that gives our green-cuts picture! But now we have a way to evaluate this thing say at z = Re±iε for R = 1/2. The residual F's are happy, and we have
(-z-2)-a = (0-z)a (0+z)a = (0- Re±iε)a (0+ Re±iε)a
Now using our theory of power functions "study of (1-z)^a and related v2.doc ", we have
(z-0)α0L = e±iαπ (0-z)α0R Im(z) 0
Then
(-z-2)-a = (0-z)aOR (0+z)aOL = (z-0)a0L e∓iaπ(0+z)aOL = e∓iaπ (z)2a
So we can rewrite our equation appropriate for points in the range z = (0,1)
F(a,b,c, z-2) = A e∓iaπ (z)2a F(a,a+1-c; a+1-b; z2) + B A e∓ibπ (z)2b F(b,b+1-c; b+1-a; z2)
For example, setting z = 1/2 gives:
F(a,b,c, 4) = A e∓iaπ (1/2)2a F(a,a+1-c; a+1-b; 1/4) + B A e∓ibπ (1/2)2b F(b,b+1-c; b+1-a;1/4)
and we see that indeed the two numbers will be different. So those green cuts are really there.