Searching for an asymptotic to exp[0.5]

[sheldonison]

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 \( 1/\exp^{0.5}(x) \).

Hey James,

Actually, the half iterate is really well behaved in the complex plane, if you follow the well behaved branch for real(z)>0, especially as compared to Tetration, so something like that should work. There are singularities at L, and at z+pi i, I need to remember where the singularities are; Mike has done a nice complex plane plot where if you put the branches at L, The singularity at L, L* are the only two singularities you see.
- Sheldon

[JmsNxn]

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 \( 1/\exp^{0.5}(x) \).

Hey James,

Actually, the half iterate is really well behaved in the complex plane, if you follow the well behaved branch for real(z)>0, especially as compared to Tetration, so something like that should work. There are singularities at L, and at z+pi i, I need to remember where the singularities are; Mike has done a nice complex plane plot where if you put the branches at L, The singularity at L, L* are the only two singularities you see.
- Sheldon

Welllll!!! If that's so I have a different proposal for you.

Take \( \phi(z)=\frac{1}{\exp^{0.5}(z+1)} \). If it is holo in \( \Re(z) > -1 \) (i.e. the half iterate of the exponential has no zeroes) and If we can show \( |\phi(z)| < C e^{\alpha|\Im(z)| + \rho|\Re(z)|} \) for \( 0< \alpha < \pi/2 \) and \( \rho \ge 0 \) THEN!

\( \vartheta(w) = \sum_{n=0}^\infty \frac{w^n}{n!\exp^{0.5}(n+1)} \)

we have in \( 0 < \Re(z) < 1 \)

\( \phi(-z) = \frac{1}{\G(z)} \int_0^\infty \vartheta(-w)w^{z-1}\,dw \)

Transforming the mellin transform into a fourier transform is easily done by looking at the inverse mellin transform. Then we can talk about some bounded conditions. I suggest if you want to know what I'm talking about and where this comes from, to read the paper I posted in the other thread on tetration. It's all rigorously shown, and is quite simple and elegant, but remarkably powerful so far.


Tell me what you think, maybe we can do this with \( \exp^{\ell}(z+1) \) as well, with \( \ell \in (0,1) \)?

[tommy1729]

Good work guys.
Well perhaps not completely correct but I think going in a good direction.

I have the feeling that you guys are looking for Carlson's theorem.

This creates uniqueness and a newton series.

And by that we switch from the cardinality of functions to that of integers.

Afterthat we can give an integral representation of it.

Ok, maybe Im going to fast now , some more details :

Let a(x) and b(x) be entire functions that are asymptotic to exp^[0.5](x) for x > -1.
Also a(n) = b(n) for integer n.

Since a(n) and b(n) are entire and grow slower than exp , Carlsons theorem applies and

a(x) - b(x) = 0

!!!

So we can make a newton series.

And an integral representation.

And that might help in sheldon's wanted iterated algorithm ... to prove my conjecture for the coefficients.

regards

tommy1729

[sheldonison]

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

I used my new code algorithm, from post#9 in math equations earlier, to generate coefficients for ultra large values of n; these are unscaled values of a_n; f(x) using a_n Taylor series is on over approximation of exp^{0.5}, see post #9 and/or below for scaling equation.

\( \text{dexphalf}(x)=\frac{d}{dx} \exp^{0.5}(x) \;\; h_n = \text{dexphalf}^{-1}(n) \)

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

If the conjecture is correct, then a_n ~= n ln(n)^2, since ln(n!) ~= n ln(n)-n. But the pattern does not work, since k from the third column grows arbitrarily large, instead of converging to k=2.

\( \ln(a_n) = -n (\ln(n))^k \)
\( k_n=\frac{\ln(-\ln(a_n))-\ln(n)}{\ln(\ln(n))}\;\; \) these are minimum possible values k_n, scaling with f2 very slightly increases k_n

Code:

n                  ln(a_n)                  k_n   ln(a_n) = -n(ln(n))^k_n
10.0000000        -43.23747618988           1.755474309098
100.000000        -1702.383415564           1.856110652219
10000.0000        -808235.9469405           1.978208308724
100000000.        -51574683648.64           2.143700628920
1.00000000E16     -4.843045366425E19        2.352700385134
1.0000000E32      -7.278717641299E36        2.603697260942
1.0000000E64      -1.897211261800E70        2.895281429488
1.0000000E128     -9.271029181788E135       3.226308077405
1.0000000E256     -9.170982024460E265       3.595948495976
1.0000000E512     -1.983493098320E524       4.003740240479
1.0000000E1024    -1.014825960853E1039      4.449613625772
1.0000000E2048    -1.333046642964E2066      4.933883260692
1.0000000E4096    -4.898918992448E4117      5.457211597134
1.0000000E8192    -5.514831365014E8217      6.020558126848
1.0000000E16384   -2.093094173164E16414     6.625126235356
1.0000000E32768   -2.964298724943E32803     7.272315222911
1.0000000E65536   -1.743728846523E65577     7.963680815351
1.0000000E131072  -4.770569826246E131119    8.700904681531
1.0000000E262144  -6.838009537646E262198    9.485771998591
1.0000000E524288  -5.820676972960E524350   10.32015555316
1.0000000E1048576 -3.356313289241E1048647  11.20600483435

For comparison, for one of our "fake" half exp functions, if we try a_n=1/(4n)! for n=1E1048576, we get k=1.0943, so this is a very good test for detecting "fake" half exp functions, where ultimately k=1. For the conjectured a_n=1/(n ln(n))!, we would expect k=2.000000386, for the last term. So it would be nice to understand k for our entire function which is conjectured to converge to a half exponential growth. For an earlier conjecture, a_n=(n^2)!, k_n for the last term would be 164282.152, so that too gives a very different value.

Let a(x) and b(x) be entire functions that are asymptotic to exp^[0.5](x) for x > -1.
Also a(n) = b(n) for integer n.

Since a(n) and b(n) are entire and grow slower than exp , Carlsons theorem applies and

a(x) - b(x) = 0

!!!

So we can make a newton series.

I wasn't focused at all on uniqueness, just the fact that I wanted the asymptotic function to be entire. I don't claim that the asymptotic is equal to exp^0.5(x) anywhere, but I conjecture that the ratio, for the second function, which has been scaled approaches 1. So would Carlson's theorem allow us to generate an entire function that exactly equals exp^0.5(x) at the whole numbers, and also be entire? I don't know if a pure Newton series would have all positive derivatives, or whether or not that matters, or when an infinite Newton series would be entire.... It looks like Carlson's theorem requires that you start with an entire function, so I'm not sure it helps to turn a non-entire half iterate into an asymptotic entire function, and the existing conjectured asymptotic entire function already has a Taylor series, so we don't need Newton.

\( f_1(x) = \sum_{n=0}^{\infty} a_n x^n\;\; \) using a_n from above

\( b_n=a_n\frac{f_1(\exp(h_n))}{\exp^{0.5}(\exp(h_n))};\; \) using h_n from above, scaling is conjectured to allow the ratio to approach arbitrarily close to 1, and would be approximately \( b_n \approx a_n/\sqrt{n}\; \; \ln(b_n) \approx \ln(a_n) - 0.5\ln(n) \)

\( f_2(x) = \sum_{n=0}^{\infty} b_n x^n \)

\( \lim_{x \to \infty} \; \frac{f_2(x)}{\exp^{0.5}(x)}=1 \)

[tommy1729]

OK I do it slightly different.

I call the conjecture : the tommy-sheldon conjecture.

I feel the need to mention sheldon and I assume it will become a theorem very soon.

Let n be an integer > 3.

Let x be a real number > e^e.

A few words about this : first n > 3, this is justified because of the fact that only " the tail " matters as explained before.

Also x > 1 is important to me because the behaviour of x^n is very much influended by if x > 1 or not.
Also when we substitute x with ln(x) we need x > e for similar reasons.
Since the logic used has to do with approximations and inequalities we take the exp once more and get x > e^e just to be " safe ".

Also x and n are not to small to avoid fixpoint issues.

Let f(x) be the function we look for : the "fake" (entire) exp^[1/2](x) with all derivatives a_n > 0.

Now clearly for all such x and n we have :

f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ... a_n x^n + ...

and it follows that

a_n x^n < f(x)

both sides are > 0 hence we can take a logaritm on both sides without any issues.

ln(a_n) + n ln(x) < ln(f(x))

Now since x is sufficiently large we can approximate ln(f(x)) with f(ln(x)).
( you can compare this to how 2sinh(e^e) is close to e^(e^e) but sinh(0) is not close to exp(0) )

Thus we rewrite :

ln(a_n) + n ln(x) < f(ln(x))

Now we can substitute x with ln(x).
this is valid because x is sufficiently large : ln(x) > e.
(And e > 1 thus still justifying both f(ln(x)) = ln(f(x)) and the power remark about x^n )

( note Im not sure about sheldon's derivative , afterall substiting x with ln(x) and then taking the derivative without taking into account the substitution you just made ... in other words D f( ln(x) ) = f ' (ln(x) D ln(x) = f ' (ln(x)) / -x. However if you substitute ln(x) = y then you get f ' (y) dy ... where dy = dx/-x BUT (!) if you substitude ln(x) = x you get f ' (y) dx AND NOT f ' (y) / -x = f ' (ln(x)) / -x.
Although the division by x might not make a big difference in the conclusion , it is fundamentally wrong logic even if it leads to a correct result. And even if the result was correct , I think it created more confusion than clarity , sheldon even added a conjecture for it , WHEREAS with my method that is not neccessary. SO I believe this step is dubious and confused sheldon himself. Hence perhaps not a big issue I felt the need to communicate this. IN PARTICULAR BECAUSE THIS MIGHT WORK NOW BUT NOT IN A GENERALIZED CASE , HENCE THE MISTAKE MUST BE NOTED DUE TO DANGER OF BEING USED WRONGLY BY " UNEXPERIENCED MATHEMATICIANS ".
No offense to sheldon for all clarity. Btw writing 1/-x is also not very good , the irony ... but its valid for nonzero real x. Its better to say -1/x then 1/-x )



We continue

ln(a_n) + n ln(x) < f(ln(x))

ln(a_n) < f(ln(x)) - n ln(x)

As said we can substitute ln(x) = x.
( or ln(x) = y if you want , but it does not matter here because I do not take a derivative )

ln(a_n) < f(x) - n x

---
Now you see the issue with small n.
Say n = 1 or 2.
LHS = negative ... But RHS could be both negative or positive for small x and the inequality places a question mark on the " tail " argument and strong bounds on the a_n.
---

Since exp is a strictly rising function on the reals :

a_n < exp(f(x) - n x)

Now the trick is to write x as function of n.
AND NOT VICE VERSA !
ALSO NO DERIVATIVE !

Lagrange multiplication is also dubious here for similar reasons as mentioned above.

Let x = g(n).
Although the cardinality of the reals =/= the cardinality of the positive integers this is the way to go.
Differentiating this is (thus) also silly.


a_n < exp(f(x) - n x)

a_n < exp( f(g(n)) - n g(n) )

1/a_n > exp( n g(n) - f(g(n)) )

Now let g(n) = ln^[a](n).

Then clearly

n g(n) - f(g(n)) = n ln^[a](n) - f(ln^[a](n))

Remember f is close to exp^[1/2] since x is large enough.

=> n ln^[a](n) - (ln^[a-(1/2)](n))

Now this function is clearly maximized when a = 1/2.

=> n ln^[1/2](n) - n = n ( ln^[1/2](n) - 1 ).

SO our estimate is :

1/a_n > exp(n ( ln^[1/2](n) - 1 ) )

A better estimate is very very likely :

1/a_n > exp(n * ln^[1/2](n) )

The fact that a = 1/2 is a mini theorem.

( note the sequence is related to a discrete set and carlson might apply here to improve )

the conjecture is

the tommy sheldon conjecture

1/a_n = O( exp(n * (ln^[1/2](n))^B ) )

For some real B > 0.

The selfreference here is quite huge.

This also might be improved by recursion if we replace ln^[1/2](x) in the estimate or conjecture with the f(x).

Notice it is easy to prove 1/a_n = O( exp(n * (n^C) ) ) is false for any real C > 0.
Thereby destroying an earlier guess.


It very nice to know that this can be generalized to finding " fake " exp^[1/3] or other half-iterates of other functions etc etc.


I considered Collatz again... which gave me a headache.


regards

tommy1729

[sheldonison]

\( \text{dexphalf}(x)=\frac{d}{dx} \exp^{0.5}(x) \;\; h_n = \text{dexphalf}^{-1}(n) \)

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

I have an improvement in the equation for a_n
\( a_n = \frac{\exp(\exp^{0.5}(h_n) - n h_n)}{\sqrt{2 \pi \frac{d^2}{dx^2}\exp^{0.5}(h_n) }} \; \; a_0 = \exp^{0.5}(0) \)
edit: many changes made; more improvements below, stronger theoretical basis.

I implemented this Gaussian approximation with pari-gp, and get the following Taylor series for exp^{0.5}(x), which is accurate to 0.5E-4, for x in the range of 10^12. I like this solution best so far, because it starts out more accurate than my previous approximation was after scaling, and so it doesn't require any scaling, which might be harder to work with theoretically. If I replace the Gaussian estimation with a numeric integral, accuracy improves dramatically, so that the error term for x=10^12 improves to 1E-28.

The a_n coeffients can be thought of as either the upper bounds, from post#9, or equivalently, as calculating a Cauchy integral around the unit circle at a radius of exp(half(h_n)), for the x^n term. The radius is chosen to be the "best fit" for the x^n term. The d^2(exp^{0.5}) term comes from the fact that the nth term Cauchy unit circle integral can be approximated by an increasingly accurate Gaussian error function! As we go around that unit circle at exp(half(h_n)), where the x^2 Taylor series coefficient at half(h_n) gives the multiplier for that Gaussian error function which turns out to be the envelope for the Cauchy unit circle integral for calculating the a_n coefficient!! See the derivation of the Gaussian approximation below; we can treat Pi as reasonably close to infinity, and the higher order terms x^3, x^4 etc, become increasingly irrelevant.

The conjecture is that the ratio for the Gaussian approximation to the Kneser half iterate approaches 1, as x goes to infinity. I would still like to find an approximation for a_n as n to infinity, that can be expressed in terms of primitive functions like exp and gamma ....

Code:

Numeric integral for entire half Taylor series; this is the most accurate version
       0.4985632879411144346796190925
+x^ 1* 0.8710111130401411916658936025
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+x^119*2.406086562703090259789810633 E-952
+x^120*7.353576929945984825943727706 E-964
+x^121*2.062996044866367254648463530 E-975
+x^122*5.315484268416505872622525458 E-987
+x^123*1.258522567866236954382862472 E-998
+x^124*2.739546898498989761925977413 E-1010
+x^125*5.485531653181015540252830319 E-1022
+x^126*1.010882286100850364148000228 E-1033
+x^127*1.715303357604997360378961019 E-1045
+x^128*2.681349668109586179961519367 E-1057
+x^129*3.863215420318792302247556914 E-1069
+x^130*5.132575320637659521483538205 E-1081
+x^131*6.290973901451826499430290031 E-1093
+x^132*7.117034858844710792253182914 E-1105
+x^133*7.434963186913381616699570660 E-1117
+x^134*7.175535883707921136701383295 E-1129
+x^135*6.400595385142868802218878845 E-1141
+x^136*5.279212684551819192356274183 E-1153
+x^137*4.028009534268085274427661553 E-1165
+x^138*2.844276211365647877674429606 E-1177
+x^139*1.859508005039463803122793853 E-1189
+x^140*1.126035533626274114177059774 E-1201
+x^141*6.318505947802323797222722837 E-1214
+x^142*3.286724100651648951381637663 E-1226
+x^143*1.585531100527665483167915782 E-1238
+x^144*7.096152409897556355115618752 E-1251
+x^145*2.947675400321429892810975447 E-1263
+x^146*1.136880407531572331200690980 E-1275
+x^147*4.072828729047575634268426013 E-1288
+x^148*1.355781226961727800775025176 E-1300
+x^149*4.195262231654102499876908828 E-1313
+x^150*1.207165060933645530811450970 E-1325
+x^151*3.231263707211930861107870173 E-1338
+x^152*8.048865420729444233874864997 E-1351
+x^153*1.866422498230798292333708820 E-1363
+x^154*4.030445298086469258814308037 E-1376
+x^155*8.108058640520878451816788866 E-1389
+x^156*1.520032260920878101686380056 E-1401
+x^157*2.656503670689144383021561731 E-1414
+x^158*4.329499475128981565124536774 E-1427
+x^159*6.582349863516344984855565238 E-1440
+x^160*9.338668990553129883640703208 E-1453
+x^161*1.236781366859692389297678707 E-1465
+x^162*1.529489353130209794964280670 E-1478
+x^163*1.766793050594006027434806539 E-1491
+x^164*1.906993632878006517639988197 E-1504
+x^165*1.923863751515074897909419206 E-1517
+x^166*1.814665656457886373509359798 E-1530
+x^167*1.600849411868116034890257045 E-1543
+x^168*1.321200755442445354926834067 E-1556
+x^169*1.020429875486358289345036855 E-1569
+x^170*7.377746585147470422447048797 E-1583
+x^171*4.994814787113353646500336149 E-1596
+x^172*3.167361931007179005199974312 E-1609
+x^173*1.881849643653924045494284771 E-1622
+x^174*1.047866808778147260099053198 E-1635
+x^175*5.469970010811979820209209681 E-1649
+x^176*2.677589772452488813768643305 E-1662
+x^177*1.229432523769072880965526166 E-1675
+x^178*5.296476577303116987951193452 E-1689
+x^179*2.141460568563315391686316082 E-1702
+x^180*8.128147745800165253090960394 E-1716
+x^181*2.896998396963448961657560720 E-1729
+x^182*9.698300513209975888283389430 E-1743
+x^183*3.050339590552378843333948030 E-1756
+x^184*9.016111786083733120644214602 E-1770
+x^185*2.505077839762538013844591105 E-1783
+x^186*6.544330883519596549898149683 E-1797
+x^187*1.607909371744948895134530835 E-1810
+x^188*3.716373541858010751920509452 E-1824
+x^189*8.082520263145333164606871495 E-1838
+x^190*1.654440197610463210395411899 E-1851
+x^191*3.188147258600726509182831325 E-1865
+x^192*5.785140801075633800421509267 E-1879
+x^193*9.887398433329089225495816951 E-1893
+x^194*1.592010837665992733213999094 E-1906
+x^195*2.415507012130397350124062001 E-1920
+x^196*3.454382188898361752203626924 E-1934
+x^197*4.657285603659041661380254890 E-1948
+x^198*5.921005627893362617368461077 E-1962
+x^199*7.099978465660990342355250150 E-1976
+x^200*8.031848704412873538835534480 E-1990

Evaluation and comparison of Gaussian approximation, with Kneser half iterate, and numeric integration.

Code:

x             Gauss series at x   Gaussian/Kneser Half     numeric intetgral/Kneser          
1/2           1.009600147857       1.007947076464           0.9984707461893          
1             1.664777638208       1.011190425535           0.9987841217590          
2             3.475227097782       1.012802437685           0.9991092778066          
4             9.622858431184       1.011659759461           0.9988701672778          
8             37.09830694534       1.008919521341           0.9985195257893          
16            209.6008686180       1.006329241170           0.9986170527351          
32            1830.759692420       1.004559841264           0.9990581577384          
64            26151.47799104       1.003455005276           0.9994965790411          
128           648661.8576542       1.002721431338           0.9997710270601          
256           29786305.89243       1.002192879433           0.9999063463259          
512           2711120116.146       1.001793913224           0.9999651058788          
1024          525959987977.7       1.001485094862           0.9999884061876          
2048          2.349112269692 E14   1.001241871705           0.9999967699125          
4096          2.621071536336 E17   1.001047620358           0.9999993881637          
8192          7.965701394346 E20   1.000890626833           1.000000035772            
16384         7.224766966614 E24   1.000762421236           1.000000113394            
32768         2.153638714039 E29   1.000656756479           1.000000074344            
65536         2.335868262933 E34   1.000568948079           1.000000035914            
131072        1.026088749416 E40   1.000495432712           1.000000014689            
262144        2.043386486924 E46   1.000433465457           1+0.00000000532128446
524288        2.076940129986 E53   1.000380907824           1+0.00000000174075232
1048576       1.220321423874 E61   1.000336076709           1+0.00000000051912754
2097152       4.723250515159 E69   1.000297635079           1+1.418025757259 E-10    
4194304       1.381073893616 E79   1.000264511776           1+3.555589248262 E-11    
8388608       3.521531572583 E89   1.000235841927           1+8.189070349476 E-12    
16777216      9.099903639076 E100  1.000210922178           1+1.731794371046 E-12    
33554432      2.788647458979 E113  1.000189176699           1+3.359170814484 E-13    
67108864      1.194430932904 E127  1.000170131136           1+5.966788610686 E-14    
134217728     8.489581185954 E141  1.000153392461           1+9.685500427077 E-15    
268435456     1.197833913454 E158  1.000138633266           1+1.433153817130 E-15    
536870912     4.044763362506 E175  1.000125579421           1+1.927446011318 E-16    
1073741824    3.972335392963 E194  1.000114000308           1+2.348062402030 E-17    
2147483648    1.390244357146 E215  1.000103701046           1+2.580693625797 E-18    
4294967296    2.142544151526 E237  1.000094516252           1+2.546713291110 E-19    
8589934592    1.812291692626 E261  1.000086305016           1+2.243166560231 E-20    
17179869184   1.058083505201 E287  1.000078946824           1+1.749951927294 E-21    
34359738368   5.411414353099 E314  1.000072338235           1+1.196160689628 E-22    
68719476736   3.105872410678 E344  1.000066390156           1+7.045019712667 E-24    
137438953472  2.587656256198 E376  1.000061025608           1+3.465186873779 E-25    
274877906944  4.088298834066 E410  1.000056177870           1+1.299802399317 E-26    
549755813888  1.616444697079 E447  1.000051788938           1+1.206903030891 E-28    
1099511627776 2.132798432202 E486  1.000047808251           1-9.985173164859 E-29

The numeric integral is more complicated than the standard Gaussian error integral, and is derived below. The value h_n was chosen so that a1=n, and then the Cauchy integration multiplier for x^-n exactly cancels the a1 term..... Lets go back to the general case for a Cauchy integral used to calculate the Taylor series coefficients, where we have f(x) instead of exp^{0.5}(x)...

\( a_n = \frac{1}{2\pi} \oint \frac{f(x)}{x^{n+1}} \; \; \) The only practical way to calcute this integral is by using a circle centered on the origin;
\( a_n = \frac{1}{2\pi} \int_{-\pi i}^{\pi i} \exp(-nx)f(\exp(x)) dx \;\; \) radius=1 Cauchy integral formula. but we can use any radius
\( a_n = \frac{1}{2\pi r^n} \int_{-\pi i}^{\pi i} \exp(-nx) f(r\,\exp(x)) dx \; \; \) Now, for an arbitrary radius of r,
\( a_n = \frac{1}{2\pi r^n} \int_{-\pi i}^{\pi i} \exp(-nx) \exp^{0.5}(r\,\exp(x)) dx \; \; \) Now, replace f(x) with exp^{0.5}(x)
\( a_n = \frac{1}{2\pi r^n} \int_{-\pi i}^{\pi i} \exp(-nx) \exp(\exp^{0.5}(\ln(r\,\exp(x)))) dx\;\;\; \exp^{0.5}(x) = \exp(\exp^{0.5}(\ln(x))) \)
\( a_n = \frac{1}{2\pi r^n} \int_{-\pi i}^{\pi i} \exp(-nx) \exp(\exp^{0.5} ( \ln ( r ) + x ) ) dx \)
\( a_n = \frac{1}{2\pi r^n} \int_{-\pi i}^{\pi i} \exp(-nx + \exp^{0.5} ( \ln ( r ) + x ) ) ) dx \)

A couple of quick comments on exp^{0.5}. It doesn't follow the Cauchy integral rules, because the circle of radius r usually circles the singularities that are near the origin, at L,L*, and at +/- pi i -0.37. So its not a legal circular integral; there will be a discontinuity at -r, due to the branch singularities. But lets ignore that somewhat inconvenient fact, and evaluate the integral anyway. Now, we are free to choose any value of r we want. So let us choose the value of r, which has derivative for
\( \frac{d}{dx}\exp^{0.5}(\ln( r ) + x )=n \).

The \( \ln ( r ) = h_n \) is exactly the value I posted earlier in this post and is that radius. Now, we have the the following.

\( \exp^{0.5}(h_n+x)) - nx = \exp^{0.5}(h_n) + b_2 x^2 + b_3 x^3 + b_4 x^4 +
...\;\; \) Notice the b1 term conveniently cancels out!
\( a_n = \frac{1}{2\pi \exp(n h_n)} \int_{-\pi i}^{\pi i} \exp( \exp^{0.5}(h_n) + b_2 x^2 + b_3 x^3 + b_4 x^4 +
... ) dx \;\; \) substituting ln( r)=h_n and substituting the equation above
\( a_n = \frac{\exp( \exp^{0.5}(h_n) - n h_n)}{2\pi } \int_{-\pi i}^{\pi i} \exp ( b_2 x^2 + b_3 x^3 + b_4 x^4 +
... ) dx \;\; \) this a_n is the most accurate entire asymptotic Taylor series!
\( a_n \approx \frac{\exp( \exp^{0.5}(h_n) - n h_n)}{2\pi } \int_{-i\infty}^{i\infty} \exp ( b_2 x^2 ) dx\;\; \) The less accurate Gaussian approximation
Here b2 is the second derivative from the beginning of this post... and this is the Gaussian integral approximation for a_n from the beginning of the post, which works very well. The Gaussian approximation is derived as an approximation of the much more accurate numeric integral. But the Gaussian integral itself works very well, and better than any approximation I had before this post.

[tommy1729]

Ok after spending some time looking at the posts I take back most of my criticism. I did not have much time too think or read about this lately.

Btw sheldon edited his posts a few times, which makes them better naturally.

If we are talking in the context of growth approximation , rather than approximation of exp^[1/2] then it turn out my ideas and sheldons are similar and consistant with eachother and by themselves.

the tommy sheldon conjecture

1/a_n = O( exp(n * (ln^[1/2](n))^B ) )

For some real B > 0.

The conjecture made by sheldon that motivates his use of taking the derivative is true.

Hence h_n is justified.

The main difference between sheldon and me is then that i take ln^[1/2] or its " fake " and sheldon uses the h_n.

Although the h_n is more accurate ( since his derivative was justified ) it does not say how fast h_n " grows ".

So its basicly

ln[1/2](n) VS h_n

( I do not yet understand where his second derivative comes from , is this the same logic that goes from a_n to b_n ?? Also his integrals need more study and/or explaination. Forgive me if im a bit behind )

Some remarks first :

Its not so clear that

growth ( f(x) ) <=> growth ( exp^[1/2] ) <=> growth ( a_n )

is an equivalence relationship.
Just to avoid jumping into conclusions.

More importantly , we have encountered h_n before on this forum !
I do not know exactly where and how but I remember it.
That might be usefull.

Another remark of yet to determine value :
We could use the difference operator too :

a_n < exp( exp^[0.5](t_n) - n t_n )

where t_n is the inverse of exp^[1/2](x) - exp[1/2](x-1).

---

I Consider it important we only looked at

a_n < stuff

But a good boundary means

stuff < a_n < stuff

and the stuff < a_n is not adressed yet ...

---

BUT LETS CONTINUE :

ln[1/2](n) VS h_n

take the inverse on both sides

exp^[1/2](n) VS D exp^[1/2](n)

Now the key is too notice :

(exp^[1/2](n))^(1-o(1)) < D exp^[1/2](n) < (exp^[1/2](n))^(1+o(1))

Hence 50 % of the tommy sheldon conjecture has been proven, more specific :

1/a_n < O( exp(n * (ln^[1/2](n))^(1+o(1)) ) )

( B = 1 for the < part )

...

So it comes down (in part) to finding asymptotics to D exp^[1/2](n).

We know D exp^[1/2](n) = exp^[1/2](n) / ( n^( 1-o(1) ) )

from which the above follows easily.

regards

tommy1729

[sheldonison]

Btw sheldon edited his posts a few times, which makes them better naturally.

I've gone through at least 4 approximation versions, for the coefficients of an entire asymptotic half exponential function. Each one is better, more accurate, and more robust than the one before it. Sometimes, the next version shows up before I'm done documenting the previous version...

( I do not yet understand where his second derivative comes from , is this the same logic that goes from a_n to b_n ?? Also his integrals need more study and/or explaination. Forgive me if im a bit behind )
...But a good boundary means

stuff < a_n < stuff

and the stuff < a_n is not adressed yet ...

The \( \frac{d^2}{dx^2}\exp^{0.5}(h_n) \) shows up in the denominator of the Gaussian approximation, which is version III, and is very accurate, and is robustly shown (some details left out) to be an approximation of version IV (also in post#16), which involves integrals, and converges to the Kneser half iterate extremely extremely well, accurate to 28 decimal digits for half iterates O(1E12). Gaussian can be shown to always give a slight over approximation for the a_n (compared to the Integral). Multiplying Gaussian by any constant<1 will eventually give an under approximation for all a_n bigger than some particular value of n, assuming the Taylor series for Kneser half exp is eventually well behaved for large enough hn(n), (needs a definition). So that might rigorously address "stuff < a_n".

More importantly , we have encountered h_n before on this forum !
I do not know exactly where and how but I remember it.
That might be usefull.

h_n first shows up in the version II approximation, in post #9 of this thread. I'm unaware of h_n before that...

ln[1/2](n) VS h_n
....
Hence 50 % of the tommy sheldon conjecture has been proven, more specific :

1/a_n < O( exp(n * (ln^[1/2](n))^(1+o(1)) ) )

\( \ln(\frac{1}{a_n}) < n\exp^{0.5}(\ln(n)) \)
Tommy, empirical tests for your conjecture look remarkably good for this last equation, for n=10^1000000, this approximation is accurate to 99.8%, though only 60% accurate for n=100000000, so convergence is a little slow, and it appears to always be an under-approximation.

So, we have this nice asymptotic approximation, f=exp^0.5(x). Lets take the best one, version IV. What would f(f(z))/exp(z) look like???? Its an entire function, since f is entire, and since exp(z) has no zeros. The answer is really neat, and counter intuitive! I'll have to post some complex plane graphs tomorrow.
- Sheldon

[tommy1729]

To justify a bit more what I said notice that :

Let D be the differential operator.

\( \lim_{n->\infty} \frac{H_n}{\ln^{[1/2]}(n) } -> 1 \)

This follows from

\( \lim_{n->\infty} \frac{D \exp^{[1/2]}(n)}{\exp^{[1/2]}(n) } -> 1 \)

Now D exp^[1/2](x) is finally smaller than A * exp^[1/2](x) for any A > 1 BECAUSE

exp^[1/2](x) < exp(A x)

Take the logarithmic derivative on both sides :

( D exp^[1/2](x) ) / exp^[1/2](x) < A

QED.

Ofcourse we know D exp^[1/2](x) > exp^[1/2](x) / x from the simple consideration of a Taylor series.

This proves the previous post of me was correct.


regards

tommy1729

[tommy1729]

So, we have this nice asymptotic approximation, f=exp^0.5(x). Lets take the best one, version IV. What would f(f(z))/exp(z) look like???? Its an entire function, since f is entire, and since exp(z) has no zeros. The answer is really neat, and counter intuitive! I'll have to post some complex plane graphs tomorrow.
- Sheldon

I just posted why the tommy sheldon conjecture should hold.
So yes our f(x) is becoming very good.

But maybe we can learn from the classic masters something too ?

I mean surely you thought of the bell numbers right ?

Afterall if one wants boundaries on

e Bell(n)/n!

or equivalently

ln( Bell(n)/n!) + 1

Then one could use in a similar fashion

c_n x^n < exp(exp(x))

and repeat what we did here.

But the Bell numbers have been studied in different ways too , going from number theory and combinatorics to other uses of calculus and dynamics etc.

Or could the classic masters have learned from us ?
Maybe our methods give better approximations of bell numbers.
I recall that also Gottfried has studied the bell numbers in detail with matrices.

Maybe we can generalize the bell numbers in " tetration style ".


Another thing :

In NKS Wolfram mentions cellular automatons and recursions that grow at unknown growth rates.

I believe these relate to asymptotic function related to tetration and exp^[1/2] ...

Although it does not follow clearly from the equations , not even the recursive ones.

That might be an active research field.

As for f(z) and f(f(z))/exp(z) being entire ...

I repeat my questions/remarks :

What does the weierstrass product of those look like ??

Do we have f(z) = exp(g(z)) ( 1 + a_1 z)(1 + a_2 z) ... ?

or is there no exp(g(z)) term ??

Well it seems the g(z) is absurd.

HOWEVER we know g(z) could be a fake log !!
and so could ( 1 + a_1 z)(1 + a_2 z) ... be a fake 1/ln(x) !

So its not directly Obvious.

Notice f(z) = exp(g(z)) ( 1 + a_1 z)(1 + a_2 z) ...
implies f(z) has no positive real zero's but an infinite amount of negative zero's ( the a_n ).

Also f(z) = exp(g(z)) ( 1 + a_1 z)(1 + a_2 z) ... could be wrong !
The weierstrass product form could be more complicated !

Is it ?

Similar questions for f(f(z)/exp(z).

Since f(f(z))/exp(z) must be close to 1 for positive real z , it follows it must go to oo for other part of the complex plane.

( its clear from absolute convergeance that f(q z) does not grow faster than f(q) for z nonreal and q a positive real )

I bet this is what sheldon's plots will show us beautifully.

And there is also still carlson's theorem in the game

regards

tommy1729