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+--- Thread: Searching for an asymptotic to exp[0.5] (/showthread.php?tid=863)
RE: Searching for an asymptotic to exp[0.5] - sheldonison - 10/01/2015
(09/29/2015, 12:28 PM)tommy1729 Wrote: I think exp^[3] might fail tpid 17.
But i guess i need asymptotics for the inverse and derivatives ...
(09/30/2015, 12:25 PM)tommy1729 Wrote: I considered exp(exp(x)) ... If my calculations are correct, the fake coëfficiënts and the derivatives match very well.
In fact the correcting factor is sqrt( 2 pi n ) (1 + log(n)^C).
The Gaussian approximation for a_n convergence is slower than for exp(x). I printed the coefficients and the ratio over the correct coefficients for \( f(x)=\exp(\exp(\exp(x))) \) up to n=100000. Notice the ratio for n=100000 is about 1/100000 vs for exp(x) it would be 1/(12*100000); convergence is slower, but still pretty good. All of the Gaussian approximations are over-approximations of the correct values of a_n; since the g''' reduces the exact integral as compared with the Gaussian approximation. Though one can imagine that perhaps for some functions the Gaussian approximation could be an under-approximation, due to g'''', and other higher even derivatives
For the highest derivative I calculated, \( n=100000\;\; g''(h_n) ~= 21.6 \cdot h_n \), so the correcting factor for \( a_n \) is approximately \( 1/\sqrt{2\pi \cdot 21.6 \cdot h_n} \)
Code:
n ratio fake(a_n)/a_n fake(a_n) integration(a_n) formal power series(a_n)
1 1.15723940985387 47.6708060587910 41.1935556747161 41.1935556747161
2 1.09160388348818 106.083596475327 97.1814025948145 97.1814025948145
3 1.06898921269797 210.627119908936 197.033905868277 197.033905868277
4 1.05666212689989 382.207817995256 361.712422793656 361.712422793656
5 1.04867080297229 646.826803671434 616.806343647696 616.806343647696
6 1.04298401653642 1035.13141987781 992.471028765439 992.471028765439
7 1.03868925837461 1581.79911314889 1522.88001478340 1522.88001478340
8 1.03530915542802 2324.59752888503 2245.31727233106 2245.31727233106
9 1.03256648485568 3303.12956524844 3198.95097671131 3198.95097671131
10 1.03028799721656 4557.31116284059 4423.33713986059 4423.33713986059
11 1.02835933568202 6125.65311242020 5956.72436654410 5956.72436654410
12 1.02670169498546 8043.43440974102 7834.24674277465 7834.24674277465
13 1.02525883461920 10340.8641944161 10086.1010363855 10086.1010363855
14 1.02398943694716 13041.3319542067 12735.8071125100 12735.8071125100
15 1.02286239855711 16159.8416414750 15798.6466843152 15798.6466843152
16 1.02185381394878 19701.7152521193 19280.3657266644 19280.3657266644
17 1.02094497801528 23661.6362600396 23176.2110295482
18 1.02012102477752 28023.0843635677 27470.3527159234
19 1.01936997635353 32758.1917261114 32135.7235213982
20 1.01868206390559 37828.0287629513 37134.2837017470
21 1.01804923336751 43183.3059512592 42417.6990030404
22 1.01746477945636 48765.4583716498 47928.3994456354
23 1.01692224628397 54508.0627487729 53600.9697597178
24 1.01643597047673 60338.5233957121 59363.8089204820
25 1.01596323715813 66179.9541423454 65140.9868320276
26 1.01551188867641 71953.1782153540 70854.2209831832
27 1.01509167840415 77578.7670479892 76424.8947662030
28 1.01470415091722 82979.0418156307 81776.0418053547
29 1.01434378855430 88079.9676237661 86834.2266019472
30 1.01400369621485 92812.8790957969 91531.2604520085
31 1.01367956763000 97115.9869213079 95805.7022429442
32 1.01337016704666 100935.627007882 99604.1056694341
33 1.01307576287902 104227.226526516 102881.986926900
34 1.01279649575461 106955.973702653 105604.499396962
35 1.01253168208835 109097.190115455 107746.813668894
36 1.01227998862538 110636.415062925 109294.211980315
37 1.01203994540756 111569.220878326 110241.915456957
38 1.01181034360760 111900.785707435 110594.670146729
39 1.01159035416808 111645.256050902 110366.123649459
40 1.01137943310853 110824.935317003 109578.028113050
41 1.01117715812270 109469.336780894 108259.307559422
42 1.01098310615209 107614.139842365 106445.028051436
43 1.01079680921432 105300.087504960 104175.308301416
44 1.01061777085196 102571.860792664 101494.206177202
45 1.01044550585280 99476.9626243454 98448.6134283820
46 1.01027957304367 96064.6397358890 95087.1870859465
47 1.01011958833353 92384.8668237719 91459.3416295928
48 1.00996521991299 88487.4124140736 87614.3213990070
49 1.00981617441352 84421.0012454865 83600.3680555582
50 1.00967218272330 80232.5833724650 79463.9933569487
100 1.00578453692724 187.156981791671 186.080591737324
200 1.00341299501394 1.74510894750342 E-8 1.73917315818616 E-8
300 1.00249185889752 1.16810780755257 E-21 1.16520428289282 E-21
400 1.00198840302522 1.11337311130476 E-36 1.11116367010162 E-36
500 1.00166677366322 5.49504600939546 E-53 5.48590225200280 E-53
600 1.00144175611352 2.86723734773177 E-70 2.86310944218981 E-70
700 1.00127462579363 2.47946198504931 E-88 2.47630562203056 E-88
800 1.00114510607748 4.83232538668920 E-107 4.82679819074619 E-107
900 1.00104149537643 2.65156196967975 E-126 2.64880325333832 E-126
1000 1.00095654201347 4.84563528819201 E-146 4.84100466384362 E-146
2000 1.00054343575353 4.44050028010782 E-361 4.43808846415907 E-361
3000 1.00038874082092 1.26848456971695 E-595 1.26799164960216 E-595
4000 1.00030596222398 7.79342676222892 E-842 7.79104299738634 E-842
5000 1.00025385632356 2.84197593479513 E-1096 2.84125466433174 E-1096
6000 1.00021781130899 5.16340561932230 E-1357 5.16228121609325 E-1357
7000 1.00019128096097 7.80231979579972 E-1623 7.80082764599120 E-1623
8000 1.00017087568485 6.81651188733738 E-1893 6.81534731019828 E-1893
9000 1.00015465721180 1.41352928698151 E-2166 1.41331070828795 E-2166
10000 1.00014143364941 2.03221558548151 E-2443 2.03192820246049 E-2443
11000 1.00013043013483 4.69049191953417 E-2723 4.68988021782504 E-2723
12000 1.00012112023776 3.41159235878872 E-3005 3.41117919595346 E-3005
13000 1.00011313328559 1.35925315762782 E-3289 1.35909939824744 E-3289
14000 1.00010620042585 4.70370063713979 E-3576 4.70320115517424 E-3576
15000 1.00010012172476 2.08773817808774 E-3864 2.08752917106664 E-3864
16000 1.00009474529279 1.65935538247062 E-4154 1.65919818125315 E-4154
17000 1.00008995359053 3.15216667696599 E-4446 3.15188315375939 E-4446
18000 1.00008565418274 1.84141780147091 E-4739 1.84126008984270 E-4739
19000 1.00008177333722 4.12970846278349 E-5034 4.12937079035335 E-5034
20000 1.00007825149451 4.32839677072003 E-5330 4.32805809370588 E-5330
40000 1.00004302510030 2.00939366799719 E-11449 2.00930721735263 E-11449
60000 1.00003024269340 1.05575695570235 E-17808 1.05572502773402 E-17808
80000 1.00002352422901 1.90343379776152 E-24314 1.90338902200228 E-24314
100000 1.00001934754707 8.30238861997517 E-30925 8.30222799222832 E-30925RE: Searching for an asymptotic to exp[0.5] - tommy1729 - 10/01/2015
In post 150 I wrote the formula for the Tommy-Sheldon iterations.
I edited that with the dot product, to avoid confusion.
It is currently the most accurate general method without An integral.
Since edits are not marked on this forum I needed to say this.
The Tommy-Sheldon iterations might be associated with Some superfunction.
The method together with series reversion gives a simple algorithm to make a fake; simple enough for a basic program such as excel or C+ dos type.
----
But the main reason I make a post is to ask
When - for what f(x) - does fake(a_n)/a_n converge to its limit ( 1 ? ) quadratically ??
Where fake means the gaussian fake or the Tommy-Sheldon iterations.
Originally I thought this could be easily answered by a type of contour integral ...
Regards
Tommy1729
RE: Searching for an asymptotic to exp[0.5] - tommy1729 - 10/02/2015
In principle we can test how good our fake approximates the original for large x.
Lets assume our convergeance is slower then exp^[m] for positive integer m.
Fake[ exp^[m](x) ( f(x) - fake(f(x)) ] exp(- exp^[m-1](x)).
So now we can use this to get a better asymptotic.
---
Another idea is to consider
F(x) = Sum a(n) x^n
A(n) = Sum b(n) x^n (1)
Now compute the fake a(n) from (1).
Call them a&(n).
Now consider fake f(x) versus Sum a&(n) x^n.
Regards
Tommy1729
RE: Searching for an asymptotic to exp[0.5] - tommy1729 - 10/03/2015
If the conditions for the derivatives are not met we can still use fake function theory to find entire asymptotics.
For instance like the previous post.
Or like considered before asymptotic to sqrt(x)
Given by
Fake( exp(x) sqrt(x) ) exp(-x).
Maybe I mentioned this before but I think we should consider
Fake( exp(x) + sqrt(x) ) - sqrt(x).
And in general
Fake(a * b) / a versus Fake(a + b) - a.
Regards
Tommy1729
RE: Searching for an asymptotic to exp[0.5] - tommy1729 - 10/06/2015
Fake(a + b) - a seems to be always better then fake ( a b ) / a.
I can prove this for most functions.
Regards
Tommy1729
RE: Searching for an asymptotic to exp[0.5] - tommy1729 - 10/06/2015
Since the beginning of this thread i noticed a problem that has not been adressed.
The problem occurs with all methods.
I mentioned it but I guess it was not clear enough.
Maybe because I anticipated the problem before we had the actual methods.
So now that we have a dozen methods or so , it is time to bring it Up again.
This might also explain why MSE and MO users are ignoring this ( mick's Posts ).
In a way it resembles irrationality issues ; the irrationallity of a Sum of fractions is independant of the first n fractions. And by induction n+1 and by further induction ... You get the idea.
This looks cryptic so
Lets give the clarifying example :
Fake ( exp^[1/2] + 10 x) is off by a factor n^10 or n^10.5 depending on the method used.
Check it yourself.
Fake ( exp(x) + x^2 ) is even worse.
Regards
Tommy1729
-----
This appears to be a mistake and is considered for delete !
RE: Searching for an asymptotic to exp[0.5] - sheldonison - 10/08/2015
(10/07/2015, 10:17 PM)tommy1729 Wrote: As promised in the binairy partition function thread :
http://math.eretrandre.org/tetrationforum/showthread.php?tid=911&pid=8070#pid8070
I Will explain the connection between fake function theory and the binary partition function.
Lets use Jay's function J(x).
It is clear that J should be a good fake for the binary part function.
Also fake(J(x)) should be close to J(x).
Notice J(x) satisfies all neccessary conditions , even those for conjecture B and TPID 17.
All conditions for any method used so far.
J(x) grows much slower then exp but much faster then exp.
The growth of J(x) is 0. ( this is how tetration relates )
J(x) is close to exp( ln^2 (x) ).
But we do not even need this here.
J ' (x) = t(x) = J(x/2).
....
Remember, we're interested in the function g(x) = ln(f(exp(x)). Here, g(x)=ln(J(exp(x))). Then g'(h_n)=n. That's why you're getting nonsense results. If you do it correctly, using the approximation function:
\( f(x)=\exp\( (\ln(x))^2 \) \;\;\; g(x) = \ln (f(\exp(x))) \)
\( g(x)=x^2 \;\;\; g'(x)=2x \;\;\; g''(x)=2 \;\;\; h_n=\frac{n}{2} \)
\( a_n \approx \frac{\exp(g(h_n) - n h_n)}{\sqrt{2 \pi g''(h_n) }} \;\;\; \approx \frac{ \exp(-\frac{n^2}{4}) }{ \sqrt{4 \pi} } \)
Then you'll find the entire "fake function" approximation gives ideal results, using the Gaussian approximation. Of course, the partition function itself is non-analytic, so there's a limit to how well you can approximate it. I haven't done the calculations for the partition function itself in a while, but I did them once; I could post the Taylor series again. I find this particular function with \( g(x)=x^2 \) to be really interesting for fake functions precisely because it has such a nice closed form, and because the Gaussian approximation for this function is exactly correct! It also turns out this function has an infinite converging Laurent series, and an exactly defined error term too! See very closely related post#85 http://math.eretrandre.org/tetrationforum/showthread.php?tid=863&pid=7413
RE: Searching for an asymptotic to exp[0.5] - tommy1729 - 10/08/2015
Im on medication since post 153.
Seems to have its effect
My apologies for the last 2 or few more posts.
However the Tommy-Sheldon iterations and quadratic convergeance question still seem solid.
And I still wonder about the zero's of J(x).
Are they like - 2^n (n-1)/4n ?
I should probably post less until im cured.
Sorry.
Regards
Tommy1729
RE: Searching for an asymptotic to exp[0.5] - tommy1729 - 10/08/2015
(10/08/2015, 03:41 AM)sheldonison Wrote: \( f(x)=\exp\( (\ln(x))^2 \) \;\;\; g(x) = \ln (f(\exp(x))) \)
\( g(x)=x^2 \;\;\; g'(x)=2x \;\;\; g''(x)=2 \;\;\; h_n=\frac{n}{2} \)
\( a_n \approx \frac{\exp(g(h_n) - n h_n)}{\sqrt{2 \pi g''(h_n) }} \;\;\; \approx \frac{ \exp(-\frac{n^2}{4}) }{ \sqrt{4 \pi} } \)
Yes but this exp ( - n^2 / 4 ) is far from Jay's 2 ^ ( - n (n-1) ) / n !
Its a different base ; exp(- 1/4) =\= 2^(-1).
So this is the worst fit , rather then the best ?
It seems to disprove the conjectures ?!
Or do i need less or more medication ?
Regards
Tommy1729
RE: Searching for an asymptotic to exp[0.5] - tommy1729 - 10/08/2015
Sheldon , in your link you apparantly considered similar things.
But what is that about Laurent series ?
You mention Laurent and then you drop the negative terms ??
Or Maybe it related to the fake ln in the MSE threads.
I assume that.
Anyway.
As Said in the previous post , we seem to have a base problem.
So I reconsider.
I believe exp( ln ^ u ) ~ J is optimal for u = 2.
And i wonder about fake ( d(x) ) = J(x).
So i consider,
a > 1
Fa = exp( ln^a(x) ).
Ga = x^a
Ga ' = a x^(a-1)
Ga ' ^[-1] = (x/a)^(1/(a-1))
So a_n = exp( (n/a)^(a/(a-1)) - n (n/a)^(1/(a-1)) ).
So we get
1 + 1/(a-1) = 2
=> a = 2.
I assume t ' (x) = t(x/w) gives
t(x) ~ exp( ln(x) ^a(w) )
Where 1 + 1/(a(w) -1) = w.
Or t(x) ~ exp( ln(x) ^(1 + 1/(a-1)) )
( im running out of time to decide ).
So does a better estimate for J give a better fake !?
Or not ?
Regards
Tommy1729