I said of
the general [3-balanced, well poised], noterminating case: hyper_f[4,3]([a,b,c,c-b-a+1/2],[c-a+1,c-b+1,b+a+1/2],1) =(b+a-1/2)!*(c-a)!*(c-b)! *(sqrt(%pi)*(c-2*b-2*a)!/((a-1/2)!*(b-1/2)!*(c-2*a)!*(c-2*b)!*(c-b-a)!) -hyper_f[4,3]([1,c-2*a+1,c-2*b+1,c-b-a+1],[c+1,c-2*b-2*a+2,c-b-a+3/2]) /((a-1)!*(b-1)!*((c+1)/2-b-a)*c!*(c-b-a+1/2)!)) 1 hyper_f ([a, b, c, c - b - a + -], 4, 3 2
1 [c - a + 1, c - b + 1, b + a + -]) = 2 1 (b + a - -)! (c - a)! (c - b)! (sqrt(%pi) (c - 2 b - 2 a)! 2 1 1 /((a - -)! (b - -)! (c - 2 a)! (c - 2 b)! (c - b - a)!) 2 2 - hyper_f ([1, c - 2 a + 1, c - 2 b + 1, c - b - a + 1], 4, 3 3 [c + 1, c - 2 b - 2 a + 2, c - b - a + -]) 2 c + 1 1 /((a - 1)! (b - 1)! (----- - b - a) c! (c - b - a + -)!)) 2 2
Notice the rhs 4F3 getting divided by infinity exactly when an upper parameter on the left is a nonpositive integer.
Gotcha! Notice the rhs 4F3 blowing up in revenge when c -> -n. The correct formula in this (terminating) case is 1 hyper_f ([a, b, - n, - n - b - a + -], 4, 3 2 1 [- n - a + 1, - n - b + 1, b + a + -]) 2 1 1 (- a - -)! (- b - -)! (- n - a)! (- n - 2 b - 2 a)! (- n - b)! 2 2 = -----------------------------------------------------------------. 1 sqrt(%pi) (- b - a - -)! (- n - 2 a)! (- n - 2 b)! (- n - b - a)! 2 Similar mischief attends c-b-a+1/2 = -n, when we have 1 hyper_f ([a, b, - n + b + a - -, - n], 4, 3 2 1 1 1 [- n + b + -, - n + a + -, b + a + -]) 2 2 2 1 1 1 1 sqrt(%pi) (b + a - -)! (- n + a - -)! (- n + b - -)! (n + b - a - -)! 2 2 2 2 = ---------------------------------------------------------------------. 1 1 1 1 1 (a - -)! (b - -)! (- n - -)! (- n + b - a - -)! (n + b + a - -)! 2 2 2 2 2 These were not easy, at least how I did them. Are they new? Also, does anyone know how to derive the closed form of a contiguous Dixon, e.g. 3F2[a,b,c;a-b+1,a-c+2], without using matrix products? The prospect of automating the 4F3, with four modes of termination and four- term contiguity relations, is somewhat daunting. But I'd love to get paid to do it! --rwg Tricky Nick was not a crook. PS, a goodie to stick in your share;functs.mac : /* all parts of xpr containing targ at depth dep (default 1) */ parts_with(xpr,targ,[dep]):=(dep:if dep='[] then 1 else dep[1], makelist(apply('part,cons(xpr,rest(l,-min(length(l),dep)))),l, inverse_part(xpr,targ)))$ E.g., if foo is the very first equation in this message, then (c93) parts_with(foo,hyper_f) (d93) [hyper_f , hyper_f ] 4, 3 4, 3 (c94) parts_with(foo,hyper_f,2) 1 (d94) [hyper_f ([a, b, c, c - b - a + -], 4, 3 2 1 [c - a + 1, c - b + 1, b + a + -]), 2 hyper_f ([1, c - 2 a + 1, c - 2 b + 1, c - b - a + 1], 4, 3 3 [c + 1, c - 2 b - 2 a + 2, c - b - a + -])] 2 (c95) parts_with(foo,c+1) 3 c + 1 (d95) [[c + 1, c - 2 b - 2 a + 2, c - b - a + -], -----] 2 2 One could also get d94 by parts_with(foo,d93[1]). PPS: Thanks to all who provided the ComplexExpand clue. But I guess nobody actually tried it, or they would have warned that it multiplies the mess from the sin nx/(2n choose n) series more than a hundredfold, leaving a bunch of Sins of Args of complex expressions. Can anyone tell me how to finish off, say, Sin[Arg[1 - E^(I*x)]]? E.g., (c120) sin(carg(1-%e^(%i*x))) x (d120) - cos(-) 2 PPS: Since there's never enough screenspace, I like Mma's small fonts, but is that an exponent or a speck of dust? I wish they'd steal the magnifying-glass idea from one of those document viewers.