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Im searching for an asymptotic to exp[0.5](x).

This is the continuation of the thread http://math.eretrandre.org/tetrationforu...hp?tid=854

where I and sheldon have gone off topic from the OP.

Thus we look for a function with " growth " = 0.5.

Or at least growth between 0 and 1.

Notice f^[n](x) = exp^[m](a^n ln^[m](x))

is not a solution for any m,a.
They all have growth 0.

2 candidates remain :

f1(z) = sum z^n/(n^2)!

f2(z) = ln(z)^e^ln^[3](z)^e^ln^[5]^e^...

We should investigate those perhaps ?

regards

tommy1729
(05/07/2014, 12:22 PM)tommy1729 Wrote: [ -> ]Im searching for an asymptotic to exp[0.5](x).

This problem will probably keep me occupied for a long time. I need to work on tools for figuring out the growth from the Taylor series, where growth is the limit of slog(f^n(x0))/n (I have some ideas).

I'm also thinking about what the Taylor series for an entire pseudo half iterate might look like, and what bounds can be put on the Taylor series coefficients. The pseudo half iterate of exp should converge with far fewer Taylor series terms than exp(z). half(10000)~=1E22, which is only 10000^5.5 Anyway, I expect this problem will keep me occupied for a long time, but that also means it might take awhile to make any real progress...

edit: Emperical testing suggests that an "entire" pseudo half iterate is very likely possible, with all positive Taylor series coefficients at z=0, and a probable growth value of 0.5, as defined by the "growth" equation. I can post the empirical results for the first 100 derivatives of such a conjectured asymptotic solution later. Each derivative is bounded to a maximum value by a particular value of half(z). I can post more later; still working on how to formalize the definition of the conjectured Taylor series.
- Sheldon
Perhaps this is trivial to Sheldon but some more comments.

Let p(x) be a polynomial.
The growth of f(x) and the growth of f(x) + p(x) are equal.

This could be intresting.

So the half-iterate of exp(x) and exp(x)+1 have the same growth rate.

SO growth works a bit like the concept of convergeance of a limit.
ONLY the " tail " of the expression matters.
Just like conv(lim a0 + a1 + a2 + a3 + ...) = conv(lim a2 + a3 + ...).

Hence when trying to find an asymptotic to exp^[0.5] we could as well use exp(x)-1+x = 0 + 2x + x^2/2 + ... .

This implies that we can play with the fixpoints and dummy variables.

So we can investigate C0 + C1 x + C2 x^2 + x^3/3! + x^4/4! + ...

I only see advantages for this.

However one disadvantage I see is that this might show that trying to find the half-iterate of exp(x) by taking the half-iterate of its truncated Taylor might not be such a good idea for studying the half-iterate for large x ( works fine for small x under some conditions ).
However thats a bit of topic here, but worth mentioning imho.

A further implication is that we can arrive at usefull results if we set f(0)=0.

I mean C0 + C1 x + C2 x^2 + x^3/3! + x^4/4! + ... with C0 = 0 and C1 > 1.

This allows us to use stuff like carleman matrices for example.

although C1,C2 appear as variables , 2 variables ( or finite ) are quite easy too handle.

This bring me to a few other remarks.

f(x) and f(a x) also have the same growth.
This implies that the Taylor coefficients t_n and a^n t_n imply the same growth rate !!

( Many conjectures about what gives an equal growth rate are possible , but its not immediate which ones are the most usefull )

These were algebraic ideas , but some calculus ideas occur too :

For instance use the Laplace transform instead of a Taylor series.
( with x = exp(-s) )
Then the theorems involving that transform can be used too !

For instance the Laplace transform of f ' (x) is very intresting.

It suggests that G(x) and G(x) * (ln(x)^k) have the same growth even when iterated.

HOWEVER one thing seems to put us on our feet again.

We want all derivatives to be nonnegative.
Its not clear those dummy variables can provide us with a solution that is both entire and has all Taylor coefficients nonnegative.

Why positive ?

Because then the Taylor is dominated by the largest coefficients.
Compare with exp vs 2sinh. The zero's coefficients of 2sinh do not effect its growth much. Because neither has negative coefficients !

This positivity removes unnessary up and down jumps in the sizes of the coefficients. So they are easier to approximate.

Also the positivity gives us insight about the functions values at complex imput... because of absolute convergeance !

I assume Sheldon was aware of all that.
But that information needed to be shared for all.

regards

tommy1729
(05/08/2014, 04:25 PM)sheldonison Wrote: [ -> ]
(05/07/2014, 12:22 PM)tommy1729 Wrote: [ -> ]Im searching for an asymptotic to exp[0.5](x).
...Emperical testing suggests that an "entire" pseudo half iterate is very likely possible, with all positive Taylor series coefficients at z=0, and a probable growth value of 0.5, as defined by the "growth" equation. .... Each derivative is bounded to a maximum value by a particular value of half(z)....still working on how to formalize the definition of the conjectured Taylor series.
- Sheldon

$\text{halfassym}(z) = \sum_{n = 1}^{\infty}\frac{x^n}{a_n!}$
Here, for a_n, factorial is extended to the reals with the gamma(n+1) function. I do not know what the limiting equation for a_n is as n gets arbitrarily large. These values were generated numerically. With 150 Taylor series terms, this series can accurately generate the half iterate for numbers up to 10^12.
Code:
a_1= 1.289368074687 a_2= 2.685542084449 a_3= 4.481892104368 a_4= 6.478743767633 a_5= 8.617330003873 a_6= 10.86883764614 a_7= 13.21549097355 a_8= 15.64480318533 a_9= 18.14765340676 a_10= 20.71664375389 a_11= 23.34605172725 a_12= 26.03107295368 a_13= 28.76759185629 a_14= 31.55218923626 a_15= 34.38183207053 a_16= 37.25395409452 a_17= 40.16617725981 a_18= 43.11650203072 a_19= 46.10306787913 a_20= 49.12423329725 a_21= 52.17844415943 a_22= 55.26438661508 a_23= 58.38072977033 a_24= 61.52640919181 a_25= 64.70028943211 a_26= 67.90141159249 a_27= 71.12888751647 a_28= 74.38183008946 a_29= 77.65944293902 a_30= 80.96100674887 a_31= 84.28582227410 a_32= 87.63323120070 a_33= 91.00261369743 a_34= 94.39338557884 a_35= 97.80499516680 a_36= 101.2369199152 a_37= 104.6886902658 a_38= 108.1598101190 a_39= 111.6498571303 a_40= 115.1583810162 a_41= 118.6850011605 a_42= 122.2293283987 a_43= 125.7909699804 a_44= 129.3696195111 a_45= 132.9648897794 a_46= 136.5764573911 a_47= 140.2040557334 a_48= 143.8473645074 a_49= 147.5060942647 a_50= 151.1799685187 a_51= 154.8686996662 a_52= 158.5720762206 a_53= 162.2898267064 a_54= 166.0217152588 a_55= 169.7675306959 a_56= 173.5270093037 a_57= 177.2999786775 a_58= 181.0862092231 a_59= 184.8855000278 a_60= 188.6976766463 a_61= 192.5225306431 a_62= 196.3598778123 a_63= 200.2095425885 a_64= 204.0713710077 a_65= 207.9451779799 a_66= 211.8308191885 a_67= 215.7281180480 a_68= 219.6369408364 a_69= 223.5570997849 a_70= 227.4885071866 a_71= 231.4309843808 a_72= 235.3843931661 a_73= 239.3486192792 a_74= 243.3235140576 a_75= 247.3089677325 a_76= 251.3048375095 a_77= 255.3110228135 a_78= 259.3273875110 a_79= 263.3538324564 a_80= 267.3902471867 a_81= 271.4365024525 a_82= 275.4924928105 a_83= 279.5581471603 a_84= 283.6333119804 a_85= 287.7179392030 a_86= 291.8118947335 a_87= 295.9151023881 a_88= 300.0274504396 a_89= 304.1488634944 a_90= 308.2792519888 a_91= 312.4185117624 a_92= 316.5665730911 a_93= 320.7233354589 a_94= 324.8887317451 a_95= 329.0626821729 a_96= 333.2451078801 a_97= 337.4359118041 a_98= 341.6350372412 a_99= 345.8424068745 a_100= 350.0579336860 a_101= 354.2815588969 a_102= 358.5132133215 a_103= 362.7528106159 a_104= 367.0003160828 a_105= 371.2556293444 a_106= 375.5186854508 a_107= 379.7894500858 a_108= 384.0678314842 a_109= 388.3537801607 a_110= 392.6472189586 a_111= 396.9481223641 a_112= 401.2563957391 a_113= 405.5719986392 a_114= 409.8948585083 a_115= 414.2249480026 a_116= 418.5621839981 a_117= 422.9065248225 a_118= 427.2579174535 a_119= 431.6162930147 a_120= 435.9816335167 a_121= 440.3538537394 a_122= 444.7329047543 a_123= 449.1187668605 a_124= 453.5113755327 a_125= 457.9106702550 a_126= 462.3166165089 a_127= 466.7291677505 a_128= 471.1482630381 a_129= 475.5738880517 a_130= 480.0059669786 a_131= 484.4444725900 a_132= 488.8893596420 a_133= 493.3405854525 a_134= 497.7981092466 a_135= 502.2618743290 a_136= 506.7318543022 a_137= 511.2080081242 a_138= 515.6902967113 a_139= 520.1786803772 a_140= 524.6731047065 a_141= 529.1735644829 a_142= 533.6799885996 a_143= 538.1923555703 a_144= 542.7106288953 a_145= 547.2347567376 a_146= 551.7647258028 a_147= 556.3004650155 a_148= 560.8419393211 a_149= 565.3890426629 a_150= 569.9416519995

One can compare this to Tommy's hypothetical candidate (quote below), and see that the (n^2)! in the denominator is growing much quicker than necessary, as compared to the empirical results. But one can also see that the denominator in the Taylor series coefficients for this asymptotic half iterate grow much faster than for the exp(z). The conjecture is as n gets arbitrarily large, for any z0>1, the slog(f^n(z0))/n~=0.5, and therefore this would be an entire function with half exponential growth.

tommy1729 Wrote:f1(z) = sum z^n/(n^2)!

The asymptotic half iterate function is defined such that all Taylor series coefficients are positive, and that the function is always less than but approaching the Kneser half iterate, for real(z)>0. The construction for the Taylor series I used is a somewhat complicated two stage process; I'll post more later. But the first stage is to note that if the Taylor series terms are all positive, than no individual Taylor series term can be larger than the desired sum, so we require that, $\forall {x>0} \; a_n x^n \lt \text{half}(x)$. This gives an upper bounds for the Taylor series coefficients. This function is always bigger than the Kneser half(z) function. The second stage is to scale down the terms by observing that for each value of half(z), one particular Taylor series term is largest contributer to the Taylor series sum. That determines for each term how much to scale that term down by. I think the resulting equation will always be less than the Kneser half iterate, but the ratio of this function over the Kneser half iterate will approach arbitrarily close to 1. Because the function is bounded to the right by the Kneser half iterate, we can safely say that this assymptotic half iterate is entire, assuming the construction works for arbitrarily large values of z and arbitrarily large Taylor series terms.

Here are some example of calculations using this assymptotic half Taylor series, as compared to the Kneser half iterate. It would probably make sense to set a_0 of the half iterate to sexp(-0.5), which is the half iterate of 0. I will half to generate a complex plan plot for this half iterate...
z, assymptotic_half, Kneser_half
0 0 0.4985632879411
1 1.126644950749 1.646354233751
10 58.93202104249 61.48617436731
100 187646.5930113 192708.5721853
1000 425414280682.2 432750850493.0
10000 9.638915213265 E21 9.750966938073 E21
100000 5.362748331798 E37 5.406412389290 E37
1000000 3.362567348729 E60 3.382539228002 E60
10000000 2.187210706560 E92 2.196854946875 E92
100000000 2.935957885769 E135 2.945782901678 E135
1000000000 3.788233214763 E192 3.798003781412 E192
10000000000 5.577154174589 E266 5.588492690694 E266
100000000000 3.101249943705 E361 3.106297055696 E361
1000000000000 6.614359301415 E480 6.622925643007 E480

One obvious questions from the Taylor series result, that I can't answer, because I have no idea how fast these functions grow as x goes to infinity, relative to exponentiaton. What is the "growth" of a functions like these, which should grow slower than exponentiation, but faster than any polynomial?
$\sum_{n = 1}^{\infty}\frac{x^n}{(2n)!}$
$\sum_{n = 1}^{\infty}\frac{x^n}{(4n)!}$
(05/10/2014, 12:14 PM)sheldonison Wrote: [ -> ]One obvious questions from the Taylor series result, that I can't answer, because I have no idea how fast these functions grow as x goes to infinity, relative to exponentiaton. What is the "growth" of a functions like these, which should grow slower than exponentiation, but faster than any polynomial?
$\sum_{n = 1}^{\infty}\frac{x^n}{(2n)!}$
$\sum_{n = 1}^{\infty}\frac{x^n}{(4n)!}$

I mentioned these before.

They are the " fake " exp(sqrt(x)) and exp(sqrt(sqrt(x))).

since a sqrt(x) is much closer to x than any positive iterate of a logaritm it follows that they both also have growth = 1.

These sums are related to the linear ordinary differential equations.
The first is cosh(sqrt(x)).
cosh(ln(x)) < 2cosh(sqrt(x)) < cosh(x).
SO growth = 1 follows.

---

Also like I said before the 0 terms do not change that much.
Maybe that makes more sense now.
---

Thanks for the data !

regards

tommy1729
(05/10/2014, 12:14 PM)sheldonison Wrote: [ -> ]One obvious questions from the Taylor series result, that I can't answer, because I have no idea how fast these functions grow as x goes to infinity, relative to exponentiaton. What is the "growth" of a functions like these, which should grow slower than exponentiation, but faster than any polynomial?
$\sum_{n = 1}^{\infty}\frac{x^n}{(2n)!}$
$\sum_{n = 1}^{\infty}\frac{x^n}{(4n)!}$

Wow -- both of these functions grow exactly exponentially, where growth is defined as slog(f^n)/n.

The first function is
$f = \sum_{n = 1}^{\infty}x^{n}{(2n)!} = \cosh(\sqrt{x})-1\approx \frac{1}{2}\exp(\sqrt{x})$

But If my math is correct then iterating f is the same as iterating $\exp(x/4)$, which is ..... the same as iterating an exponential!!!

The second function is
$f = \sum_{n = 1}^{\infty}\frac{x^{n}}{(4n)!} \approx \frac{1}{4}\exp(x^{0.25})$

Ok, the second function is a little more complicated, but if my math is correct, it is going to be the same as iterating $f(x)=\exp(\frac{x}{16})$, which .... drumroll ... is also exponential growth!

So, half exponential functions, especially the Taylor series of entire versions of half exponential functions need more study... Lets conjecture that I have a constructive definition of an entire half exponential Taylor series, for which I haven't given all of the details, but I have a pari-gp program. Then the Taylor series coefficients must eventually grow slower than all of these entire functions with exponential growth.... really interesting!!!
- Sheldon

Code:
a_1= 1.289368074687 a_2= 2.685542084449 a_3= 4.481892104368 a_4= 6.478743767633 a_5= 8.617330003873 a_6= 10.86883764614 a_7= 13.21549097355 a_8= 15.64480318533 a_9= 18.14765340676 a_10= 20.71664375389 a_11= 23.34605172725 a_12= 26.03107295368 a_13= 28.76759185629 a_14= 31.55218923626 a_15= 34.38183207053 a_16= 37.25395409452 a_17= 40.16617725981 a_18= 43.11650203072 a_19= 46.10306787913 a_20= 49.12423329725 a_21= 52.17844415943 a_22= 55.26438661508 a_23= 58.38072977033 a_24= 61.52640919181 a_25= 64.70028943211 a_26= 67.90141159249 a_27= 71.12888751647 a_28= 74.38183008946 a_29= 77.65944293902 a_30= 80.96100674887 a_31= 84.28582227410 a_32= 87.63323120070 a_33= 91.00261369743 a_34= 94.39338557884 a_35= 97.80499516680 a_36= 101.2369199152 a_37= 104.6886902658 a_38= 108.1598101190 a_39= 111.6498571303 a_40= 115.1583810162 a_41= 118.6850011605 a_42= 122.2293283987 a_43= 125.7909699804 a_44= 129.3696195111 a_45= 132.9648897794 a_46= 136.5764573911 a_47= 140.2040557334 a_48= 143.8473645074 a_49= 147.5060942647 a_50= 151.1799685187 a_51= 154.8686996662 a_52= 158.5720762206 a_53= 162.2898267064 a_54= 166.0217152588 a_55= 169.7675306959 a_56= 173.5270093037 a_57= 177.2999786775 a_58= 181.0862092231 a_59= 184.8855000278 a_60= 188.6976766463 a_61= 192.5225306431 a_62= 196.3598778123 a_63= 200.2095425885 a_64= 204.0713710077 a_65= 207.9451779799 a_66= 211.8308191885 a_67= 215.7281180480 a_68= 219.6369408364 a_69= 223.5570997849 a_70= 227.4885071866 a_71= 231.4309843808 a_72= 235.3843931661 a_73= 239.3486192792 a_74= 243.3235140576 a_75= 247.3089677325 a_76= 251.3048375095 a_77= 255.3110228135 a_78= 259.3273875110 a_79= 263.3538324564 a_80= 267.3902471867 a_81= 271.4365024525 a_82= 275.4924928105 a_83= 279.5581471603 a_84= 283.6333119804 a_85= 287.7179392030 a_86= 291.8118947335 a_87= 295.9151023881 a_88= 300.0274504396 a_89= 304.1488634944 a_90= 308.2792519888 a_91= 312.4185117624 a_92= 316.5665730911 a_93= 320.7233354589 a_94= 324.8887317451 a_95= 329.0626821729 a_96= 333.2451078801 a_97= 337.4359118041 a_98= 341.6350372412 a_99= 345.8424068745 a_100= 350.0579336860 a_101= 354.2815588969 a_102= 358.5132133215 a_103= 362.7528106159 a_104= 367.0003160828 a_105= 371.2556293444 a_106= 375.5186854508 a_107= 379.7894500858 a_108= 384.0678314842 a_109= 388.3537801607 a_110= 392.6472189586 a_111= 396.9481223641 a_112= 401.2563957391 a_113= 405.5719986392 a_114= 409.8948585083 a_115= 414.2249480026 a_116= 418.5621839981 a_117= 422.9065248225 a_118= 427.2579174535 a_119= 431.6162930147 a_120= 435.9816335167 a_121= 440.3538537394 a_122= 444.7329047543 a_123= 449.1187668605 a_124= 453.5113755327 a_125= 457.9106702550 a_126= 462.3166165089 a_127= 466.7291677505 a_128= 471.1482630381 a_129= 475.5738880517 a_130= 480.0059669786 a_131= 484.4444725900 a_132= 488.8893596420 a_133= 493.3405854525 a_134= 497.7981092466 a_135= 502.2618743290 a_136= 506.7318543022 a_137= 511.2080081242 a_138= 515.6902967113 a_139= 520.1786803772 a_140= 524.6731047065 a_141= 529.1735644829 a_142= 533.6799885996 a_143= 538.1923555703 a_144= 542.7106288953 a_145= 547.2347567376 a_146= 551.7647258028 a_147= 556.3004650155 a_148= 560.8419393211 a_149= 565.3890426629 a_150= 569.9416519995

It seems 0.5 n (ln(n)-1) < a_n < 2 n (ln(n)-1).

Hence the new conjecture is 1/(n ln(n)) ! as Taylor coefficients.

regards

tommy1729
(05/10/2014, 11:48 PM)sheldonison Wrote: [ -> ]
(05/10/2014, 12:14 PM)sheldonison Wrote: [ -> ]One obvious questions from the Taylor series result, that I can't answer, because I have no idea how fast these functions grow as x goes to infinity, relative to exponentiaton. What is the "growth" of a functions like these, which should grow slower than exponentiation, but faster than any polynomial?
$\sum_{n = 1}^{\infty}\frac{x^n}{(2n)!}$
$\sum_{n = 1}^{\infty}\frac{x^n}{(4n)!}$

Wow -- both of these functions grow exactly exponentially, where growth is defined as slog(f^n)/n.

The first function is
$f = \sum_{n = 1}^{\infty}x^{n}{(2n)!} = \cosh(\sqrt{x})-1\approx \frac{1}{2}\exp(\sqrt{x})$

But If my math is correct then iterating f is the same as iterating $\exp(x/4)$, which is ..... the same as iterating an exponential!!!

The second function is
$f = \sum_{n = 1}^{\infty}\frac{x^{n}}{(4n)!} \approx \frac{1}{4}\exp(x^{0.25})$

Ok, the second function is a little more complicated, but if my math is correct, it is going to be the same as iterating $f(x)=\exp(\frac{x}{16})$, which .... drumroll ... is also exponential growth!

So, half exponential functions, especially the Taylor series of entire versions of half exponential functions need more study... Lets conjecture that I have a constructive definition of an entire half exponential Taylor series, for which I haven't given all of the details, but I have a pari-gp program. Then the Taylor series coefficients must eventually grow slower than all of these entire functions with exponential growth.... really interesting!!!
- Sheldon

Nice to see you agree.
But I kinda said all those things recently.

regards

tommy1729
(05/10/2014, 11:58 PM)tommy1729 Wrote: [ -> ]
(05/10/2014, 11:48 PM)sheldonison Wrote: [ -> ].....
Wow -- both of these functions grow exactly exponentially, where growth is defined as slog(f^n)/n.
....
$f = \sum_{n = 1}^{\infty}x^{n}{(2n)!} = \cosh(\sqrt{x})-1\approx \frac{1}{2}\exp(\sqrt{x})$
....

Nice to see you agree.
But I kinda said all those things recently.

Agreed; and I'm sure I probably realized that before posting, my reply was also only a few minutes after your reply Anyway sometimes I need time to think it all through. Now I need to switch from numerical approximations mode to theoretical mode, the end goal is to come up with an equation like the one you conjectured, which I am not yet able to do.

(05/10/2014, 11:58 PM)tommy1729 Wrote: [ -> ]Hence the new conjecture is 1/(n ln(n)) ! as Taylor coefficients.

We desire an asymptotic under-approximation of the half exp(x) with all positive Taylor series coefficients at x=0, based on the Kneser half exp(x) function, which is not entire. So we need
$f(x) = \sum_{n = 1}^{\infty} a_n x^n \; < \; \exp^{0.5}(x)$
$\forall {x \gt 0 } \; \; a_n x^n \; < \; \exp^{0.5}(x)$
$\forall {x \gt 0 } \; \; \log(a_n x^n) \; < \; \log(\exp^{0.5}(x))$
$\forall {x \gt 0 } \; \; \log(a_n) + n\log(x) \; < \; \log(\exp^{0.5}(x))$
$\forall {x \gt 0 } \; \; \log(a_n) \; < \; \log(\exp^{0.5}(x)) - n\log(x)$

For log(exp^0.5(x)), we can substitute exp^0.5(log(x)).
$\forall {x \gt 0 } \; \; \log(a_n) \; < \; \exp^{0.5}(\log(x)) - n\log(x)$
Next, replace x with log(x), and change the range from x>0, to all x
$\forall {x } \; \; \log(a_n) \; < \; \exp^{0.5}(x) - nx$

Conjecture, for each a_n, there is a particular value of x that most limits a_n. In other words, we conjecture that for each value of n, exp^0.5(x)-nx has one minimum value, where the derivative is zero.
$\frac{d}{dx} \exp^{0.5}(x) - nx = -n + \frac{d}{dx} \exp^{0.5}(x)=0$

At the minimum, the derivative will be equal 0, so defining $\text{dexphalf}(x)=\frac{d}{dx} \exp^{0.5}(x)$
Now define
$h_n = \text{dexphalf}^{-1}(n)$
$\log(a_n) \; < \; \exp^{0.5}(h_n) - n h_n$
$a_n \; < \; \exp(\exp^{0.5}(h_n) - n h_n))$

This is the first step of my generation of an entire approximation of f(x), and is probably the most important, by replacing the "<" with equal".

$a_n = \exp(\exp^{0.5}(h_n) - n h_n)$

Here is the first entire asymptotic approximation of halfx, where $a_0=\exp^{0.5}(0)$
$f_1(x) = \sum_{n=0}^{\infty} a_n x^n$

With this approximation f(x) will be an entire over approximation of the exp^0.5(x). To get a better approximation, use the same values from above, and scale the a_n values to generate b_n. Formally, we can get a better approximation by scaling, where we evaluate f1(x) at $\exp(h_n)$, and compare to exp^0.5(exp(h_n)). Empirically, the scale factor is approximately 1/sqrt(n).
$b_n=a_n\frac{f_1(\exp(h_n))}{\exp^{0.5}(\exp(h_n))}$

Then a much better entire approximation of exp^0.5(x) is:
$f_2(x) = \sum_{n=0}^{\infty} b_n x^n$

For values of x on the order 1E10, the ratio of f_2(x) to exp^0.5(x) is accurate to about 1 part in 10000. I think the ratio of f_2(x) to exp^0.5(x) goes to exactly 1 as x goes to infinity. By the way, this f_2(x) is a little bit more accurate than the results I posted earlier in this thread, which used rough approximations for the minimum of a_n, instead of derivatives. One can also scale twice (or more times though the lower derivatives probably start oscillating); scaling twice, I get accuracy of 0.2 parts per million for f_3(1E10)/exp^0.5(1E10). By scaling three times, I get an accuracy of 1 part per billion, for f_4(1E10)/exp^0.5(1E10).

I hope the next step is to come up with a recursive equation, given any particular value of a_n, to come up with another a_n, for another much larger value of n. Specifically, can we come up with an approximation for how fast the factorial in the denominator grows? Specifically, given n, I can generate a_m, where h_n is the location used to evaluate a_n, from above. Starting with n, and a_n, the first two equations just use a_n from above.

$h_n = \text{dexphalf}^{-1}(n)$
$a_n = \exp(\exp^{0.5}(h_n) - n h_n)$

Skipping a lot of algebra, this is what I got to. I can include the algebra later. Here, we are generating a fractional Taylor series coefficient m, from the value n above, which would exactly match the equation above.

$m(n) = \frac{n \times \exp(\exp^{0.5}(h_n))}{\exp(h_n)}$
$\log(a_m) = \exp(\exp^{0.5}(h_n))\times(-n+1)$

I haven't been able to do anything useful in terms of limiting behavior (yet), but I only got this recursive relationship cleaned up a few minutes ago, this morning; I expect the relationship will be there! If I apply these equations to n=6, I get h_6=5.546380530883 and a_6=0.0000001055905600243, and m=700791.2, and log(a_m)= -149682094.7, which is correct. Then a_m = 1/9906980.030456! which is 1/(m*14.137)!, which actually matches Tommy's conjecture reasonably well since log(m)=13.4599.

Hopefully, the equations I just posted don't have too many typos.... If there are typos, I can say that I have a working pari-gp program that implements these mathematical equations (without typos).

The goal is to use this recursive relationship to prove something about f(n) where a_n = 1/f(n)!. Also, the second set of equations allows generating accurate recursive coefficients for much much larger values of n than the original equation. This could be used to check Tommy's conjecture....
- Sheldon
I'm not sure if this will help, but I know some about taylor series and can do some fractional calculus (Where can't we do fractional calculus)

If $\phi$ is holo on $\Re(z) < 1$ and satisfies some fast growth at imaginary infinity and negative infinity. And if $\phi(z) \neq \mathbb{N}$ in the half plane. Then fix $1>\tau > 0$:

$\frac{1}{2\pi i}\int_{\tau-i\infty}^{\tau+i\infty}\frac{\pi}{\sin(\pi z)\G(\phi(z))} w^{-z}\,dz = \sum_{n=0}^\infty \frac{w^n}{(\phi(-n))!}$

Maybe that might help some of you? The unfortunate part is as $w \to \infty$ we're going to get decay to zero. I'm not sure about the iterates. We can also note this is a modified fourier transform and so we can apply some of Paley Wiener's theorems on bounding fourier transforms from the original functions. I.e: We can bound the taylor series by the function in the integral. Therefore maybe if we get very fast decay to zero we can talk about asymptotics of $1/\exp^{0.5}(x)$.