This Appell series is probably doable. Mma knows it for 4/3, 1/3, and -2/3 instead of 7/3. Do Appells obey recurrences similar to pFqs? I don't see them at http://functions.wolfram.com/HypergeometricFunctions/AppellF1/ but Wikipedia has four-termers, which may be just the ticket. There might even be a huge theory of these things, comparable to âFâ, including matrix products! A gold mine for thesis projects. (OtOH, Mathematica *still* lacks some of the âFâ contiguity machinery: In[168]:= Hypergeometric2F1[a, b, 1 - b + a, -1] Out[168]= (2^-a Sqrt[\[Pi]] Gamma[1 + a - b])/(Gamma[1/2 + a/2] Gamma[1 + a/2 - b]) In[170]:= Hypergeometric2F1[a, b, 1 - a + b, -1] // FunctionExpand Out[170]= Hypergeometric2F1[a, b, 1 - a + b, -1] ) --rwg On Sat, May 26, 2018 at 3:59 PM, Bill Gosper <billgosper@gmail.com> wrote:
Instead of just getting 4/Binomial[2/p,1/p] (or equivalent but messy đ functions) from Assuming[p > 0, 2 Integrate[(1 - Abs[x^p])^(1/p), {x, -1, 1}]], Assuming[p > 0, Area[ImplicitRegion[Abs[x^p] + Abs[y^p] < 1, {x, y}]]] does weird things that don't always work: 1/3 Area[ImplicitRegion[Abs[x]^(1/3)+Abs[y]^(1/3)<1,{x,y}]] 1/2 2/3 2/3 Area[ImplicitRegion[Abs[x]^(2/3)+Abs[y]^(2/3)<1,{x,y}]] 1 2 3/2 (2 2^(2/3) âĎ Gamma[2/3])/Gamma[1/6]-(2^(2/3) â(3 Ď) Gamma[5/6])/Gamma[-(2/3)] 2 Ď 5/2 - 5 â(2/(5+â5)) Ď Gamma[7/5])/(Gamma[-2/5] Gamma[4/5]))+(2^(1/5) âĎ Gamma[7/5])/Gamma[9/10] 3 3 3^(1/3) AppellF1[4/3,-(1/3),-(1/3),7/3 ,-(1/(-1+(-1)^(2/3))),1/(1+(-1)^(1/3))] 2 â3 (2^(1-1/Sqrt[3]) Sqrt[\[Pi]/3] Gamma[1/(2 Sqrt[3])])/Gamma[1/6 (3+Sqrt[3])] 4 (8 Gamma[5/4]^2)/âĎ 9/2 (2^(5/9) âĎ Gamma[11/9])/Gamma[13/18]+(Ď Csc[2 Ď/9] Gamma[11/9])/(Gamma[4/9] Gamma[7/9]) 5 (2 2^(3/5) âĎ Gamma[6/5])/Gamma[7/10]
Notice the AppellF1 for p=3, which must equal 4/Binomial[2/3,1/3], unbeknownst to FullSimplify. Is it beknownst to anybody? --rwg