Got it. Product[E^(1/12 - k^2/4 - Derivative[1, 0][Zeta][-1, 1])*k^(1/12 + k/2 + k^2/2)/ Hyperfactorial[k], {k, β}] == E^(3*Zeta[3]/(8*Pi^2) + Derivative[1, 0][Zeta][-1, 1]/2- 1/24) (2*Pi)^(1/24) == E^((3 Zeta[3])/(8 Pi^2)) (2 Pi)^(1/24)/βGlaisher Glaisher wins, this time. Hey, π(3) ! In[119]:= N@%[[2]] Out[119]= 0.997875917234445 An underestimate, for a change. N@%%[[1]] wimps out. In[120]:= 1/% Out[120]= 1.00212860409683 as predicted. βrwg On Mon, May 13, 2019 at 12:04 PM Bill Gosper <billgosper@gmail.com> wrote:
"Hyperfactorial" (= 1^1 2^2 ... n^n) has a particularly simple "Stirling's formula": In[54]:= Hyperfactorial@z | E^(1/12 - z^2/4 - Zeta'[-1]) z^(1/12 + z/2 + z^2/2)
(Zeta'[-1] and Zeta'[2] are connected
In[78]:= Zeta'[-1] // deGlaisher // Simplify // Distribute
Out[78]= 1/12 (1 - EulerGamma - Log[2 Ο)) + Zeta'[2]/(2 Ο^2)
via the zeta reflection formula.)
In[55]:= Table[% // Activate, {z, .0, 9}]
Out[55]= {1. | 0., 1. | 0.998755252375579, 4. | 3.99865591643225, 108. | 107.983585892636, 27647.9999999998 | 27645.6209013424, 8.63999999999998*10^7 | 8.63952270335811*10^7, 4.03107840000001*10^12 | 4.03092349178858*10^12, 3.31976639877119*10^18 | 3.31967257400642*10^18, 5.56964379417277*10^25 | 5.56952319419551*10^25, 2.1577941222941*10^34 | 2.15775718825365*10^34}
Unfortunately, an exact expression of the "figure of merit" infinite product of relative errors (~ 1.0021286) requires the "next" higher factorial
Wrong!
after "Hyperfactorial", 1^1^2 2^2^2 3^3^2 ... n^n^2, for which we still lack a consensual name. I've called it "Second Factorial", with Hyperfactorial being "First Factorial" and Factorial being "Zeroth Factorial", based on the dubious contention that the dot under the *' *in "*!*" is actually a tiny zero. Replacing that zero with 1, 2, ..., is typographically unattractive, but perhaps encircling those little numerals would help. βrwg
On Sun, May 12, 2019 at 10:10 AM Bill Gosper <billgosper@gmail.com> wrote:
n!/β(2Οn)e^n/n^n ~ 1 + O(1/n), whose infinite product over n blows up, giving an infinite figure of demerit. For convergence, we can improve Stirling's approximation: n! ~ β(2Ο(n+1/6)) n^n/e^n ~ n! (1+1/(144 n^2)+. . .). Then the infinite product of all the relative errors is In[86]:= Product[n!/β(2 Ο (n + 1/6)) n^n/E^n), {n, β}]
Out[87]= E^( 2 Zeta'[-1] - 1/12) β(β ! β(2Ο)))
Out[88]= 1.00781097654253
(BtW, Stirling's contribution was "only" the β(2Ο); de Moivre had already found the rest. This must have been nearly as stunning as Euler's π(2) = ΟΒ²/6.)
Stirling's approximation can be improved no end. Perhaps the next better is
E^-z β(2 Ο z) (1/(12 z) + z)^z ~ z! (1-1/(1440 z^3)+O[1/z]^4) whose "figure of merit" is Product[z!/(E^-z β(2 Ο z) (1/(12 z) + z)^z),{z,β}] == E^(2 Zeta'[-1]) ββ(2Ο)/(BarnesG[1 - I/(2β3)] BarnesG[1 + I/(2β3)]) ~ 1.001178221812, which seems to cry out for the BarnesG reflection formula. Which, unfortunately, only applies to G(z)/G(-z), not G(z)G(-z): E.g. BarnesG[1 - I/(2 β3)]/BarnesG[1 + I/(2 β3)] == (-1)^(1/24) E^(-((I PolyLog[2, E^(-(Ο/β3))])/(2 Ο))) (Ο Csch[Ο/(2 β3)])^(-(I/(2 β3))) βrwg