Here's a nice test case: (2/9)! (1/3)!/((1/9)! (4/9)!) == 3^(2/3)/(2 (Cos[π/9] Sec[2π/9])^(1/3)) On Tue, Mar 25, 2014 at 7:27 PM, Bill Gosper <billgosper@gmail.com> wrote:
ITEM 1 (Schroeppel): (1/3)! and (2/3)! are interexpressible. (1/4)! and (3/4)! are interexpressible.
Thus these two pairs are of dimensionality one.
(1/10)! and (2/10)! are sufficient to express (N/10)! for all N. (1/12)! and (2/12)! are sufficient to express (N/12)! for all N. (1/3)! and (1/4)! are sufficient to express (N/12)! for all N.
Thus the three cases above are of dimensionality two.
PROBLEM: Find some order to this dimensionality business.
The reflection and multiplication formulas:
pi Z Z! (-Z)! = --------- sin(pi Z)
(N-1)/2 -NZ-1/2 (2 pi) N (NZ)! = Z! (Z-1/N)! (Z-2/N)! ... (Z-(N-1)/N)! --------------
I was under the misimpression that Mathematica had finally nailed this, but using
LR[L_, N_] := Block[{$MaxExtraPrecision = N}, LatticeReduce[ Transpose[Prepend[IdentityMatrix[Length[L]], Round[10^N*L]]]]]
(LR for Lattice Reduce or Linear Relation), which finds multiple solutions, unlike FindIntegerNullVector (PSLQ), to find relations among (k/12)! :
In[17]:= Join[{2, 3, 5}, (Range[6]/12)!]
Out[17]= {2, 3, 5, (1/12)!, (1/6)!, (1/4)!, (1/3)!, (5/ 12)!, Sqrt[\[Pi]]/2}
In[18]:= LR[Log[%], 69][[1 ;; 3]]
Out[18]= {{0, 6, -1, -2, 2, 3, -4, -6, 2, 3}, {-6, -2, 7, -4, 4, 0, -8, 0, 4, 0}, {-29420553, 655434586,
1447298892, -1344266042, -1749889224, 1253758671, 366295186, 83356101, -1110973039, -416024934}}
[...]
Out[24]= {75 (1/4)!^4 (1/3)!^6 == 8 π^(3/2) (1/12)!^2 (1/6)!^3 (5/12)!^2,
5 Sqrt[2] (1/4)!^2 == 3 3^(3/4) (1/12)! (5/12)!}
In[25]:= N[%]
Out[25]= {True, True}
Is this really that hard a problem?
Perhaps. --rwg