Searching for an asymptotic to exp[0.5]

[Leo.W]

Hi tommy
I read many of your posts, you're inspiring, respect!
I'm not here trying to disappoint you, please don't take my later words offensive or something  no offense at all

I came up with 2 ideas

First, that, exp^0.5 has no asymptotic to any combination of elementary functions,
I think it may be true because, the exp(z) and log(z) are elementary, too, they're only asymptotic to themselves, or adding descending terms like exp(z)~exp(z)+1, exp(z)~2sinh(z), etc.
I mean, you can never write an asymptotic of exp(exp(z)) not in the form that exp(exp(z)+h(z))+g(z) where h(z)~0 and g(z)=o(1) in small o notation, alternatively exp(exp(z)+h(z))*k(z)+g(z) or something, with h,j,g all elementary.
So this pattern may also apply to exp^0.5, just a guess lol
summary: exp^0.5~exp^0.5
And I read your post about using 2sinh(z) to approximate tetration, which is fantastic
So you may tell that if g(g(z))=2sinh(z), g(z)~exp^0.5(z) right?
Again if g(g(z))=exp(z)-1, will g(z)~exp^0.5(z)?
but since g(z) is not elementary... this is a loop though
so a question, if g(z)~exp^0.5(z), will g(g(z))~exp(z)?

Second, I wonder if you attempted finding the asymptotic of the Maclaurin Series of exp^0.5(z), it may give us some hints, assume \( exp^{0.5}(z)=\sum_{n\ge0}a_nz^n \), and asking an asymptotic of a_n, may help.
Indeed, asking the coefficients of the term is called Z-Transform in analytic theories, so I guess Z-transform may help, its inverse transformation is an integral, due to Cauchy's integral formula.
here's what I've found:
first generate the series, we can use Bell matrix to generate about correct 130 terms in 2 minutes, giving a full list of length 150

Code:

A={0.49856, 0.87634, 0.24755, 0.024572, -0.00095215, 0.00025335, \
0.000070930, -0.000048184, 2.6322*10^-6,
5.9669*10^-6, -1.3088*10^-6, -7.4742*10^-7, 2.6850*10^-7,
1.1251*10^-7, -4.8065*10^-8, -2.2028*10^-8, 8.1704*10^-9,
5.3099*10^-9, -1.2339*10^-9, -1.4183*10^-9, 1.0362*10^-10,
3.8903*10^-10, 3.5690*10^-11, -1.0434*10^-10, -2.9030*10^-11,
2.6037*10^-11, 1.3863*10^-11, -5.4973*10^-12, -5.5413*10^-12,
6.6662*10^-13, 1.9785*10^-12,
1.9479*10^-13, -6.3466*10^-13, -2.0676*10^-13, 1.7667*10^-13,
1.1330*10^-13, -3.7497*10^-14, -4.9753*10^-14, 2.0904*10^-15,
1.8867*10^-14, 3.7167*10^-15, -6.1889*10^-15, -2.9031*10^-15,
1.6492*10^-15,
1.5168*10^-15, -2.6212*10^-16, -6.5323*10^-16, -6.0239*10^-17,
2.4101*10^-16, 8.4076*10^-17, -7.4249*10^-17, -5.2086*10^-17,
1.6496*10^-17,
2.5018*10^-17, -3.1515*10^-19, -1.0167*10^-17, -2.4372*10^-18,
3.5095*10^-18, 1.8744*10^-18, -9.6023*10^-19, -1.0067*10^-18,
1.4189*10^-19, 4.4848*10^-19,
5.2808*10^-20, -1.7171*10^-19, -6.3925*10^-20, 5.5352*10^-20,
3.9864*10^-20, -1.3334*10^-20, -1.9715*10^-20, 8.6590*10^-22,
8.3933*10^-21, 1.5899*10^-21, -3.1262*10^-21, -1.3720*10^-21,
9.9393*10^-22, 7.9156*10^-22, -2.4138*10^-22, -3.8173*10^-22,
1.9922*10^-23, 1.6322*10^-22,
2.4944*10^-23, -6.3026*10^-23, -2.2692*10^-23, 2.1918*10^-23,
1.3517*10^-23, -6.6752*10^-24, -6.7915*10^-24, 1.6199*10^-24,
3.0858*10^-24, -1.8466*10^-25, -1.3073*10^-24, -1.1289*10^-25,
5.2539*10^-25, 1.1521*10^-25, -2.0279*10^-25, -7.0774*10^-26,
7.6010*10^-26, 3.6514*10^-26, -2.8008*10^-26, -1.7171*10^-26,
1.0301*10^-26, 7.6128*10^-27, -3.8536*10^-27, -3.2347*10^-27,
1.4963*10^-27, 1.3266*10^-27, -6.1243*10^-28, -5.2460*10^-28,
2.6499*10^-28, 1.9782*10^-28, -1.1963*10^-28, -6.9077*10^-29,
5.4911*10^-29, 2.0741*10^-29, -2.4784*10^-29, -4.0879*10^-30,
1.0546*10^-29, -6.5566*10^-31, -3.9676*10^-30, 1.3218*10^-30,
1.1431*10^-30, -8.8712*10^-31, -1.1506*10^-31,
3.8371*10^-31, -1.2479*10^-31, -8.3825*10^-32,
9.0056*10^-32, -2.0598*10^-32, -1.9684*10^-32,
2.0423*10^-32, -7.9655*10^-33, -5.8940*10^-34,
2.9461*10^-33, -2.2985*10^-33, 1.1897*10^-33, -4.7822*10^-34,
1.5768*10^-34, -4.3727*10^-35, 1.0324*10^-35, -2.0862*10^-36,
3.6089*10^-37, -5.3258*10^-38, 6.6558*10^-39, -6.9619*10^-40,
5.9902*10^-41, -4.1332*10^-42, 2.2005*10^-43, -8.4909*10^-45,
2.1140*10^-46, -2.5506*10^-48};

I took \( -ln\left |{a_{n}\right | \) and made a list plot, only concentrating on the 20th to 130th terms, we can see the plot is PRETTY asymptotic to a linear function, by calculation, mma deduced that MAYBE for large n, \( -ln\left |{a_{n}\right |\sim14.3+0.45n \), since then we MAY tell, that a boundary exists, by solving the sum of a_n*z^n, we arrive at
\( exp^{0.5}(z)\sim{g}(z)+\frac{7*10^{-11}z^{20}}{1-0.635z} \) where g(z) denotes the first 20 terms' summation, showing an asymptotic upper bound around z~0
This expression is quite not precise, though

Regards
Leo