[math-fun] New(?) way to tabulate pFq identities
By a process not obvious how to automate, I extracted HypergeometricPFQ[{a, b, (2 a)/5 + (6 b)/5 + 1, 2 a + 2 b, a + 3 b}, {(2 a)/5 + (6 b)/5, a + (3 b)/2 + 1/2, a + (3 b)/2 + 1, 2 b + 1}, -(1/4)] == (4^-a Gamma[b + 1/2] Gamma[2 a + 3 b + 1])/( Gamma[a + b + 1/2] Gamma[a + 3 b + 1]) from a summation containing a messy, irreducible cubic factor, analogous to the sort used to discover recurrence relations via "creative telescopy". Punting the cubic cost two degrees of freedom, but there were hundreds of variants. The unpleasant choice seems to be between a useless messy cubic, and a nearly useless identity with only two degrees of freedom. Probably the identity you need was among the hundreds I threw away. Here's an interesting remedy. What you actually tabulate is, in this case, the fully general result in a somewhat intimidating form: ((-b - d) ((1/(-1 + b))(-1 + b - d) (-1 + a + b - d) (-1 + b + c - d) HypergeometricPFQ[{a, -1 + b, c, d, -1 - a + b + d, -1 + b - c + d}, {b/2 + d/2, 1/2 + b/2 + d/2, 1 - a + d, 1 - c + d, -a + b - c + d}, -(1/ 4)] - (19 + 5 b^2 - 7 c + b (-19 + 4 c - 11 d) + a (-7 + 4 b + c - 3 d) + 20 d - 3 c d + 5 d^2) HypergeometricPFQ[{a, b, c, d, -1 - a + b + d, -1 + b - c + d}, {b/2 + d/2, 1/2 + b/2 + d/2, 1 - a + d, 1 - c + d, -a + b - c + d}, -(1/ 4)] - 5 (-3 + b) (-2 + b) HypergeometricPFQ[{a, -1 + b, b, c, d, -1 - a + b + d, -1 + b - c + d}, {-3 + b, b/2 + d/2, 1/2 + b/2 + d/2, 1 - a + d, 1 - c + d, -a + b - c + d}, -(1/ 4)] + 3 (-2 + b) (-8 + a + 3 b + c - 3 d) HypergeometricPFQ[{a, -1 + b, b, c, d, -1 - a + b + d, -1 + b - c + d}, {-2 + b, b/2 + d/2, 1/2 + b/2 + d/2, 1 - a + d, 1 - c + d, -a + b - c + d}, -(1/ 4)]))/(a c (-1 + b + c) (1 - b + c - d) d) == -((Gamma[1 - a + d] Gamma[1 + b + d] Gamma[ 1 - c + d] Gamma[-a + b - c + d])/(a (-1 + b) c (-1 + b + c) Gamma[1 + d] Gamma[-1 - a + b + d] Gamma[ 1 - a - c + d] Gamma[b - c + d])) It's then fairly routine to determine whether your pFq[-1/4] is a special case. It's not pencil and paper, but many people now have access to computers. --rwg Count yourself lucky if you can read "Rauzy fractal" and not remember Them Moose Goosers.
On Tue, Jun 30, 2015 at 12:25 PM, Bill Gosper <billgosper@gmail.com> wrote:
By a process not obvious how to automate, I extracted HypergeometricPFQ[{a, b, (2 a)/5 + (6 b)/5 + 1, 2 a + 2 b, a + 3 b}, {(2 a)/5 + (6 b)/5, a + (3 b)/2 + 1/2, a + (3 b)/2 + 1, 2 b + 1}, -(1/4)] == (4^-a Gamma[b + 1/2] Gamma[2 a + 3 b + 1])/( Gamma[a + b + 1/2] Gamma[a + 3 b + 1])
from a summation containing a messy, irreducible cubic factor, analogous to the sort used to discover recurrence relations via "creative telescopy". Punting the cubic cost two degrees of freedom, but there were hundreds of variants. The unpleasant choice seems to be between a useless messy cubic, and a nearly useless identity with only two degrees of freedom. Probably the identity you need was among the hundreds I threw away.
Here's an interesting remedy. What you actually tabulate is, in this case, the fully general result in a somewhat intimidating
nauseating
form:
((-b - d) ((1/(-1 + b))(-1 + b - d) (-1 + a + b - d) (-1 + b + c - d) HypergeometricPFQ[{a, -1 + b, c, d, -1 - a + b + d, -1 + b - c + d}, {b/2 + d/2,
[This was so miscanonicalized that I'm vandalizing it here to make sure nobody wastes time trying to use it.]
It's then fairly routine to determine whether your pFq[-1/4] is a special
case.
I.e., you check your z argument = -1/4 and your parameters are congruent mod1.
It's not pencil and paper, but many people now have access to computers. --rwg
OK, here it is cleaned up. -(b - 2 d) (a + b - 2 d) (b + c - 2 d) HypergeometricPFQ[{a, -a + b, b - c, c, b - d, d}, {1/2 + b/2, 1 + b/2, 1 - a + b - c, 1 - a + d, 1 - c + d}, -(1/4)] + (b - d) (5 + 3 c + b (9 + 5 b + 4 c) + a (3 + 4 b + c - 7 d) - 18 d - 7 (3 b + c) d + 21 d^2) HypergeometricPFQ[{a, -a + b, b - c, c, 1 + b - d, d}, {1/2 + b/2, 1 + b/2, 1 - a + b - c, 1 - a + d, 1 - c + d}, -(1/4)] - 3 (5 + a + 3 b + c - 6 d) (b - d) (1 + b - d) HypergeometricPFQ[{a, -a + b, b - c, c, 2 + b - d, d}, {1/2 + b/2, 1 + b/2, 1 - a + b - c, 1 - a + d, 1 - c + d}, -( 1/4)] + 5 (b - d) (1 + b - d) (2 + b - d) HypergeometricPFQ[{a, -a + b, b - c, c, 3 + b - d, d}, {1/2 + b/2, 1 + b/2, 1 - a + b - c, 1 - a + d, 1 - c + d}, -(1/4)] == ( Gamma[1 + b] Gamma[1 - a + b - c] Gamma[1 - a + d] Gamma[1 - c + d])/ (Gamma[-a + b] Gamma[b - c] Gamma[d] Gamma[1 - a - c + d]) But maybe instead of a special case, you have a limiting case, say a -> oo : (b - 2 d) (b + c - 2 d) HypergeometricPFQ[{b - c, c, b - d, d}, {1/2 + b/2, 1 + b/2, 1 - c + d}, 1/ 4] - (3 + 4 b + c - 7 d) (b - d) HypergeometricPFQ[{b - c, c, 1 + b - d, d}, {1/2 + b/2, 1 + b/2, 1 - c + d}, 1/4] + 3 (b - d) (1 + b - d) HypergeometricPFQ[{b - c, c, 2 + b - d, d}, {1/2 + b/2, 1 + b/2, 1 - c + d}, 1/4] == ( Gamma[1 + b] Gamma[1 - c + d])/(Gamma[b - c] Gamma[d]) These are now 4F3[+1/4] instead of 6F5[-1/4] ! Also note the rhs of the a->oo limit is not the a->oo limit of the 6F5 rhs, but rather that limit transformed by Gamma[z]->1/Gamma[1-z] . I have only the vaguest notion why. Maybe there were illegal limit interchanges when I blew up my contour rectangles. I'm a big fan of handling all the contiguous cases, although it seems ambitious. I think for something like this you'd need a thirteen(?) dimensional system of 7x7 matrices, and seven-term recurrences. The chance that your 6F5[-1/4] will come out in Gammas seems mighty slim. Slightly better for certain 7F6[a+1,b,...g; a,h,...,i,j,k,L|-1/4] --rwg (Some of you may wonder about Mathematica's line-breaking. It's deliberately ghastly, to preclude premature end-of-expression on readback.)
On Tue, Jun 30, 2015 at 12:25 PM, Bill Gosper <billgosper@gmail.com> wrote:
By a process not obvious how to automate,
In very rough and somewhat gory form: gosper.org/hyper3by3.pdf (Pleasant introduction: gosper.org/pathi.pdf). Included: a triple 4F3[+1/4] = double 5F4 = single 6f5 (with conjugate quadratic surd parameters) all with three freedoms, as a subproblem. Probably nobody wants quadratic surd parameters, but there are numerous ways of rationalizing them by surrendering one freedom. So this is a compact way of transmitting them. A CAS could either exhaustively tabulate all the rational specializations, or just keep the general form, and check for a matching query by solving algebraic equations on the fly. This might be pushing the envelope when we get to cubics and beyond. --rwg I extracted
HypergeometricPFQ[{a, b, (2 a)/5 + (6 b)/5 + 1, 2 a + 2 b, a + 3 b}, {(2 a)/5 + (6 b)/5, a + (3 b)/2 + 1/2, a + (3 b)/2 + 1, 2 b + 1}, -(1/4)] == (4^-a Gamma[b + 1/2] Gamma[2 a + 3 b + 1])/( Gamma[a + b + 1/2] Gamma[a + 3 b + 1])
from a summation containing a messy, irreducible cubic factor, analogous to the sort used to discover recurrence relations via "creative telescopy". Punting the cubic cost two degrees of freedom, but there were hundreds of variants. The unpleasant choice seems to be between a useless messy cubic, and a nearly useless identity with only two degrees of freedom. Probably the identity you need was among the hundreds I threw away.
Here's an interesting remedy. What you actually tabulate is, in this case, the fully general result in a somewhat intimidating
nauseating
form:
((-b - d) ((1/(-1 + b))(-1 + b - d) (-1 + a + b - d) (-1 + b + c - d) HypergeometricPFQ[{a, -1 + b, c, d, -1 - a + b + d, -1 + b - c + d}, {b/2 + d/2,
[This was so miscanonicalized that I'm vandalizing it here to make sure nobody wastes time trying to use it.]
It's then fairly routine to determine whether your pFq[-1/4] is a special
case.
I.e., you check your z argument = -1/4 and your parameters are congruent mod1.
It's not pencil and paper, but many people now have access to computers. --rwg
OK, here it is cleaned up.
-(b - 2 d) (a + b - 2 d) (b + c - 2 d) HypergeometricPFQ[{a, -a + b, b - c, c, b - d, d}, {1/2 + b/2, 1 + b/2, 1 - a + b - c, 1 - a + d, 1 - c + d}, -(1/4)] + (b - d) (5 + 3 c + b (9 + 5 b + 4 c) + a (3 + 4 b + c - 7 d) - 18 d - 7 (3 b + c) d + 21 d^2) HypergeometricPFQ[{a, -a + b, b - c, c, 1 + b - d, d}, {1/2 + b/2, 1 + b/2, 1 - a + b - c, 1 - a + d, 1 - c + d}, -(1/4)] - 3 (5 + a + 3 b + c - 6 d) (b - d) (1 + b - d) HypergeometricPFQ[{a, -a + b, b - c, c, 2 + b - d, d}, {1/2 + b/2, 1 + b/2, 1 - a + b - c, 1 - a + d, 1 - c + d}, -( 1/4)] + 5 (b - d) (1 + b - d) (2 + b - d) HypergeometricPFQ[{a, -a + b, b - c, c, 3 + b - d, d}, {1/2 + b/2, 1 + b/2, 1 - a + b - c, 1 - a + d, 1 - c + d}, -(1/4)] == ( Gamma[1 + b] Gamma[1 - a + b - c] Gamma[1 - a + d] Gamma[1 - c + d])/ (Gamma[-a + b] Gamma[b - c] Gamma[d] Gamma[1 - a - c + d])
But maybe instead of a special case, you have a limiting case, say a -> oo :
(b - 2 d) (b + c - 2 d) HypergeometricPFQ[{b - c, c, b - d, d}, {1/2 + b/2, 1 + b/2, 1 - c + d}, 1/ 4] - (3 + 4 b + c - 7 d) (b - d) HypergeometricPFQ[{b - c, c, 1 + b - d, d}, {1/2 + b/2, 1 + b/2, 1 - c + d}, 1/4] + 3 (b - d) (1 + b - d) HypergeometricPFQ[{b - c, c, 2 + b - d, d}, {1/2 + b/2, 1 + b/2, 1 - c + d}, 1/4] == ( Gamma[1 + b] Gamma[1 - c + d])/(Gamma[b - c] Gamma[d])
These are now 4F3[+1/4] instead of 6F5[-1/4] ! Also note the rhs of the a->oo limit is not the a->oo limit of the 6F5 rhs, but rather that limit transformed by Gamma[z]->1/Gamma[1-z] . I have only the vaguest notion why. Maybe there were illegal limit interchanges when I blew up my contour rectangles.
I'm a big fan of handling all the contiguous cases, although it seems ambitious. I think for something like this you'd need a thirteen(?) dimensional system of 7x7 matrices, and seven-term recurrences. The chance that your 6F5[-1/4] will come out in Gammas seems mighty slim. Slightly better for certain 7F6[a+1,b,...g; a,h,...,i,j,k,L|-1/4] --rwg (Some of you may wonder about Mathematica's line-breaking. It's deliberately ghastly, to preclude premature end-of-expression on readback.)
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Bill Gosper