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05/12/2014, 03:56 PM
(This post was last modified: 05/12/2014, 04:00 PM by sheldonison.)
(05/12/2014, 03:48 PM)JmsNxn Wrote: 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 is holo on and satisfies some fast growth at imaginary infinity and negative infinity. And if in the half plane. Then fix :
Maybe that might help some of you? The unfortunate part is as 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 .
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
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05/12/2014, 05:06 PM
(This post was last modified: 05/12/2014, 05:10 PM by JmsNxn.)
(05/12/2014, 03:56 PM)sheldonison Wrote: (05/12/2014, 03:48 PM)JmsNxn Wrote: 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 is holo on and satisfies some fast growth at imaginary infinity and negative infinity. And if in the half plane. Then fix :
Maybe that might help some of you? The unfortunate part is as 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 .
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 . If it is holo in (i.e. the half iterate of the exponential has no zeroes) and If we can show for and THEN!
we have in
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 as well, with ?
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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
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05/13/2014, 04:23 AM
(This post was last modified: 05/14/2014, 12:02 AM by sheldonison.)
(05/10/2014, 11:56 PM)tommy1729 Wrote: ...
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.
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.
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.
(05/12/2014, 11:35 PM)tommy1729 Wrote: 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 nonentire 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.
using a_n from above
using h_n from above, scaling is conjectured to allow the ratio to approach arbitrarily close to 1, and would be approximately
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05/13/2014, 11:44 PM
(This post was last modified: 05/13/2014, 11:51 PM by tommy1729.)
OK I do it slightly different.
I call the conjecture : the tommysheldon 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 halfiterates of other functions etc etc.
I considered Collatz again... which gave me a headache.
regards
tommy1729
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05/14/2014, 05:54 AM
(This post was last modified: 02/18/2016, 06:21 PM by sheldonison.)
(05/13/2014, 04:23 AM)sheldonison Wrote:
I have an improvement in the equation for a_n
edit: many changes made; more improvements below, stronger theoretical basis.
I implemented this Gaussian approximation with parigp, and get the following Taylor series for exp^{0.5}(x), which is accurate to 0.5E4, 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 1E28.
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
+x^ 2* 0.2538290681564146773775546123
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+x^ 5* 0.000006634106505118055642333198577
+x^ 6* 0.00000003523478404131539039591224937
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+x^155*8.108058640520878451816788866 E1389
+x^156*1.520032260920878101686380056 E1401
+x^157*2.656503670689144383021561731 E1414
+x^158*4.329499475128981565124536774 E1427
+x^159*6.582349863516344984855565238 E1440
+x^160*9.338668990553129883640703208 E1453
+x^161*1.236781366859692389297678707 E1465
+x^162*1.529489353130209794964280670 E1478
+x^163*1.766793050594006027434806539 E1491
+x^164*1.906993632878006517639988197 E1504
+x^165*1.923863751515074897909419206 E1517
+x^166*1.814665656457886373509359798 E1530
+x^167*1.600849411868116034890257045 E1543
+x^168*1.321200755442445354926834067 E1556
+x^169*1.020429875486358289345036855 E1569
+x^170*7.377746585147470422447048797 E1583
+x^171*4.994814787113353646500336149 E1596
+x^172*3.167361931007179005199974312 E1609
+x^173*1.881849643653924045494284771 E1622
+x^174*1.047866808778147260099053198 E1635
+x^175*5.469970010811979820209209681 E1649
+x^176*2.677589772452488813768643305 E1662
+x^177*1.229432523769072880965526166 E1675
+x^178*5.296476577303116987951193452 E1689
+x^179*2.141460568563315391686316082 E1702
+x^180*8.128147745800165253090960394 E1716
+x^181*2.896998396963448961657560720 E1729
+x^182*9.698300513209975888283389430 E1743
+x^183*3.050339590552378843333948030 E1756
+x^184*9.016111786083733120644214602 E1770
+x^185*2.505077839762538013844591105 E1783
+x^186*6.544330883519596549898149683 E1797
+x^187*1.607909371744948895134530835 E1810
+x^188*3.716373541858010751920509452 E1824
+x^189*8.082520263145333164606871495 E1838
+x^190*1.654440197610463210395411899 E1851
+x^191*3.188147258600726509182831325 E1865
+x^192*5.785140801075633800421509267 E1879
+x^193*9.887398433329089225495816951 E1893
+x^194*1.592010837665992733213999094 E1906
+x^195*2.415507012130397350124062001 E1920
+x^196*3.454382188898361752203626924 E1934
+x^197*4.657285603659041661380254890 E1948
+x^198*5.921005627893362617368461077 E1962
+x^199*7.099978465660990342355250150 E1976
+x^200*8.031848704412873538835534480 E1990
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 E10
4194304 1.381073893616 E79 1.000264511776 1+3.555589248262 E11
8388608 3.521531572583 E89 1.000235841927 1+8.189070349476 E12
16777216 9.099903639076 E100 1.000210922178 1+1.731794371046 E12
33554432 2.788647458979 E113 1.000189176699 1+3.359170814484 E13
67108864 1.194430932904 E127 1.000170131136 1+5.966788610686 E14
134217728 8.489581185954 E141 1.000153392461 1+9.685500427077 E15
268435456 1.197833913454 E158 1.000138633266 1+1.433153817130 E15
536870912 4.044763362506 E175 1.000125579421 1+1.927446011318 E16
1073741824 3.972335392963 E194 1.000114000308 1+2.348062402030 E17
2147483648 1.390244357146 E215 1.000103701046 1+2.580693625797 E18
4294967296 2.142544151526 E237 1.000094516252 1+2.546713291110 E19
8589934592 1.812291692626 E261 1.000086305016 1+2.243166560231 E20
17179869184 1.058083505201 E287 1.000078946824 1+1.749951927294 E21
34359738368 5.411414353099 E314 1.000072338235 1+1.196160689628 E22
68719476736 3.105872410678 E344 1.000066390156 1+7.045019712667 E24
137438953472 2.587656256198 E376 1.000061025608 1+3.465186873779 E25
274877906944 4.088298834066 E410 1.000056177870 1+1.299802399317 E26
549755813888 1.616444697079 E447 1.000051788938 1+1.206903030891 E28
1099511627776 2.132798432202 E486 1.000047808251 19.985173164859 E29
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)...
The only practical way to calcute this integral is by using a circle centered on the origin;
radius=1 Cauchy integral formula. but we can use any radius
Now, for an arbitrary radius of r,
Now, replace f(x) with exp^{0.5}(x)
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
.
The is exactly the value I posted earlier in this post and is that radius. Now, we have the the following.
Notice the b1 term conveniently cancels out!
substituting ln( r)=h_n and substituting the equation above
this a_n is the most accurate entire asymptotic Taylor series!
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.
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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](x1).

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))^(1o(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^( 1o(1) ) )
from which the above follows easily.
regards
tommy1729
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05/15/2014, 06:15 PM
(This post was last modified: 05/15/2014, 10:42 PM by sheldonison.)
(05/14/2014, 11:42 PM)tommy1729 Wrote: 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...
Quote:( 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 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".
Quote: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...
Quote: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)) ) )
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 underapproximation.
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
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To justify a bit more what I said notice that :
Let D be the differential operator.
This follows from
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
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(05/15/2014, 06:15 PM)sheldonison Wrote: 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
