New subject: [math-fun] hyjynx jinx

16 Dec 2005

      A datum, probably long known, apropos the computational complexity of
the Gamma fcn:
               3 1  6  9            5 1  6  8             1  6  10 
      hyper_2f1 (-, -, - ) hyper_2f1 (-, -, - ) hyper_2f1(-, -, -- )
                 7  7  7              7  7  7             7  7  7  

                                 7
                              1        %pi
                         588 (-)!  cos(---)
                              7        14
                 =       ------------------,
                                %pi

so (n/d)! is probably easy, but with effort proportional to d.  (Or
 maybe d^2).
-------
My POWERSERIES function got a double sum instead of Clausen's 4F3[z]
for not knowing the well-poised, 3-balanced (sesquiSaalschutzian?),
terminating 4F3[1] in the inner sum.  "SesquiPfaffian" is more just,
but also more confusing.  I got ambitious and derived the whole
4F3[a,b,c,d;e,f,g] (and 3F2[z]) matrix system, and extracted
the general, noterminating case:
                                  1
hyper_f    ([a, b, c, c - b - a + -], 
       4, 3                       2

                                            1
             [c - a + 1, c - b + 1, b + a + -], 1) = 
                                            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 + -], 1)
                                                      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.  Also
note the weird oo - oo when (c+1)/2 = a+b.  As usual, the
matrices provide the complete system of identities contiguous
to the above.

The algebra gets very heavy in transforming the general 4x4
system (which has eight dimensions), but there's a way to
exploit the symmetry in a,b,c,d and e,f,g.  You can actually
substitute a=b=c=d=S and later recover the original generality
with an extension of the following trick:  Supposing there were
only two variables, a and b:  Replace every f(S)^3 with
f(a) f(b) f((a+b)/2), then f(S)^2 with f(a) f(b) and finally
f(S) with f((a+b)/2).  This isn't rigorous, but elementary theology
seems to find the right symmetric functions.

A rigorous way to fight malignant algebroma is to get even more
ambitious and derive the q-extension of the 4F3 system.  Normally,
the addition of a new variable makes matters nonlinearly worse,
but q-extending

(n + d + c + b + a) (n + e + c + b + a) (n + e + d + b + a)

                               (n + e + d + c + a) (n + e + d + c + b)
to
      n + d + c + b + a        n + e + c + b + a
(1 - q                 ) (1 - q                 )

       n + e + d + b + a        n + e + d + c + a
 (1 - q                 ) (1 - q                 )

       n + e + d + c + b
 (1 - q                 )

actually *reduces* the expanded form from 247 terms to 31 terms,
with five 5-term exponents and 26 sixes.  All the nonlinear terms
(n^5, a^4, ...) disappear, but are recoverable with l'Hospital's
rule.  Unfortunately, 4F3 is just shy of the breakeven point.
But at least I now have the 4phi3 system too.
--rwg
(-: Mesotonic economist comes-into emoticons :-)
PS, I'm dabbling in sudoku as a programming exercise.  Our program
(brute-force backtrack) typically finds one or two "redundant" clues
in published puzzles, but of course removing them renders the puzzle
both unsymmetrical and insoluble without (perhaps humanly infeasible)
lookahead.

[math-fun] more hyjynx

R. William Gosper

R. William Gosper

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