# Tetration Forum

Full Version: Searching for an asymptotic to exp[0.5]
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(10/08/2015, 08:41 AM)tommy1729 Wrote: [ -> ]
(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

It is a different base; I interpolated $f(x) = \exp(\ln(x)^2)$ as per your earlier post. I leave it as a home work problem for you to generate the fake function for $f(x) \approx \exp$$\frac{\(\ln(x)$$^2}{2\ln(2)}\)\;\;\; g(x)\approx \frac{x^2}{2\ln(2)}$, which should be closer to Jay's series, in the sense that I think as x gets arbitrarily large, $g(x) \approx \frac{x^2}{2\ln(2)}$ for J(x). But this is a very rough order of magnitude approximation of J(x). If you like, I can post the results for the fake function for Jay's series exactly, but it would be a numerical Taylor series, not a nice closed form function.

Ok, here are the numerical approximations for the ratio of the Gaussian approximation for fake_a_n/a_n, for J(x). I would expect that as n gets arbitrarily large, this would go to a constant, but not get to exactly 1, unlike the case for the fake function for exp(z) or exp^{0.5}(z), where the ratio gets arbitrarily close to 1. The error term is going to be the ratio of the Gaussian approximation over the exact integral, which is complicated, but for this function, we think g'' goes to a constant, instead of getting arbitrarily small. So the error term involves the integral going from -infinity to infinity instead of the exact value of the integral from -pi to pi. Also the exact a_n integral includes all of the derivatives of g(z), not just the g'' approximation with all higher derivatives zero.

My "guess" for the ratio as n goes to infinity: 1.00000000002528325425; this is the limiting ratio as n goes to infinity for the fake function for $f(x) = \exp$$\frac{\(\ln(x)$$^2}{2\ln(2)}\)$. Now take the entire function, fake(x)=~f(x); and generate fake2(x) from fake(x), and that is how you get this limiting ratio. More later, only if there is interest.

(10/08/2015, 12:26 PM)tommy1729 Wrote: [ -> ]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 ??
If one includes the full Laurent series with negative terms, then there is a closed form for fake(x)-f(x), which might help with proofs. The negative Laurent series terms, $\frac{a_n}{x^n}$ terms cause a singularity at zero, but otherwise quickly become insignificant; elsewhere the full Laurent series converges. One might be able to prove the "guessed ratio" above.
Code:
n  ratio of fake_a_n/a_n for J(x) 1 1.048770528303 2 1.013850038715 3 1.005968616357 4 1.003144747621 5 1.001868901379 6 1.001204340672 7 1.000822844603 8 1.000587713520 9 1.000434687401 10 1.000330708190 11 1.000257534636 12 1.000204519322 13 1.000165153630 14 1.000135302082 15 1.000112249218 16 1.000094160750 17 1.000079766664 18 1.000068168339 19 1.000058717592 20 1.000050938776 21 1.000044477428 22 1.000039065682 23 1.000034498530 24 1.000030617249 25 1.000027297641 26 1.000024441575 27 1.000021970809 28 1.000019822446 29 1.000017945527 30 1.000016298466 31 1.000014847101 32 1.000013563191 33 1.000012423249 34 1.000011407636 35 1.000010499843 36 1.000009685927 37 1.000008954052 38 1.000008294132 39 1.000007697532 40 1.000007156832 41 1.000006665634 42 1.000006218397 43 1.000005810309 44 1.000005437176 45 1.000005095334 46 1.000004781568 47 1.000004493054 48 1.000004227300 49 1.000003982105 50 1.000003755518 51 1.000003545807 52 1.000003351428 53 1.000003171005 54 1.000003003307 55 1.000002847233 56 1.000002701791 57 1.000002566091 58 1.000002439331 59 1.000002320786 60 1.000002209802 61 1.000002105785 62 1.000002008196 63 1.000001916547 64 1.000001830392 65 1.000001749326 66 1.000001672978 67 1.000001601010 68 1.000001533112 69 1.000001469001 70 1.000001408415 71 1.000001351117 72 1.000001296886 73 1.000001245519 74 1.000001196830 75 1.000001150646 76 1.000001106809 77 1.000001065171 78 1.000001025596 79 1.000000987958 80 1.000000952139 81 1.000000918031 82 1.000000885534 83 1.000000854553 84 1.000000825000 85 1.000000796795 86 1.000000769861 87 1.000000744128 88 1.000000719529 89 1.000000696003 90 1.000000673491 91 1.000000651940 92 1.000000631299 93 1.000000611520 94 1.000000592559 95 1.000000574374 96 1.000000556926 97 1.000000540177 98 1.000000524094 99 1.000000508642 100 1.000000493793 101 1.000000479516 102 1.000000465784 103 1.000000452571 104 1.000000439854 105 1.000000427608 106 1.000000415814 107 1.000000404449 108 1.000000393494 109 1.000000382932 110 1.000000372744 111 1.000000362915 112 1.000000353428 113 1.000000344269 114 1.000000335424 115 1.000000326879 116 1.000000318622 117 1.000000310641 118 1.000000302925 119 1.000000295462 120 1.000000288242 121 1.000000281255 122 1.000000274493 123 1.000000267945 124 1.000000261604 125 1.000000255462 126 1.000000249511 127 1.000000243743 128 1.000000238151 129 1.000000232729 130 1.000000227471 131 1.000000222370 132 1.000000217420 133 1.000000212616 134 1.000000207953 135 1.000000203425 136 1.000000199028 137 1.000000194756 138 1.000000190606 139 1.000000186573 140 1.000000182653 141 1.000000178842 142 1.000000175136 143 1.000000171532 144 1.000000168026 145 1.000000164616 146 1.000000161296 147 1.000000158066 148 1.000000154921 149 1.000000151859 150 1.000000148878 151 1.000000145973 152 1.000000143144 153 1.000000140388 154 1.000000137702 155 1.000000135084 156 1.000000132532 157 1.000000130044 158 1.000000127618 159 1.000000125251 160 1.000000122943 161 1.000000120692 162 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1.000000010154 371 1.000000010073 372 1.000000009993 373 1.000000009913 374 1.000000009834 375 1.000000009756 376 1.000000009679 377 1.000000009603 378 1.000000009528 379 1.000000009453 380 1.000000009379 381 1.000000009306 382 1.000000009234 383 1.000000009162 384 1.000000009091 385 1.000000009021 386 1.000000008952 387 1.000000008883 388 1.000000008815 389 1.000000008748 390 1.000000008681 391 1.000000008615 392 1.000000008550 393 1.000000008485 394 1.000000008421 395 1.000000008358 396 1.000000008295 397 1.000000008233 398 1.000000008172 399 1.000000008111 400 1.000000008051

I Will Try to finish my homework today but im only at 2% of my potential.

Homework sounds a bit ... Oh well.

Anyway exp( (ln(x)^2) / 2 ln(2) ) is indeed a better estimate.

Why are you so carefull with the exact integral ?
Cant we simply express it with erf or similar special functions ?
I havent investigated it yet, but you seem to hold back or so.
Not sure about the how and why here.

How you computed your limit is a mystery to me.

Something to do with iterating the gaussian method ?

Reminds me of the Tommy-Sheldon iterations.

I Will now take my medication and do my other homework first ( number theory ).

Im not a student though.
There is no PhD in fake function theory yet.

But thanks for the post.

Regards

Tommy1729

I edited post 150 , where I mentioned the tommy-sheldon iterations.

Typo's , mistakes and confusion should be gone.

Although the convergeance conjecture disagrees with sheldon's recent not 1 ratio ... Maybe ...

Things should be clear now.

Regards

Tommy1729
Dont you just hate it when you Try to prove

D < 1.

And on day 1 you prove D < ln(n).

Day 2 : D < ln^[2] (n)

Day 3 : D < ln^[3] (n)

Etc

Oh Well

Regards

Tommy1729
I was thinking about g ' ' (h_n).

To avoid it being too small for the gaussian method , I invented the Tommy-Sheldon iterations.

However perhaps computing a fake g works better.

Although computing a fake g with the gaussian could result in " the same problem with g_2 ".
( the Tommy-Sheldon iterations do not have that problem )

Or maybe combining those.

Hmm.

Regards

Tommy1729
(10/08/2015, 11:08 PM)tommy1729 Wrote: [ -> ]I edited post 150 , where I mentioned the tommy-sheldon iterations.

Typo's , mistakes and confusion should be gone.

Although the convergeance conjecture disagrees with sheldon's recent not 1 ratio ... Maybe ...

Things should be clear now.

Regards

Tommy1729

Start with $f(x) = \exp$$\frac{\(\ln(x)$$^2}{2\ln(2)}\)\;\;\; g(x) = \frac{x^2}{2\ln(2)}$

g''(x) for f(x) is conjectured to approach exactly g''(x) for J(x), as x gets arbitrarily large, which is the initial reason for choosing this particular f(x). But f(x) is interesting on its own.

I generate the fake function for f(x) using the Gaussian method. I assume that this is what Tommy means in post#150 by F_2(x).
> F_2(x) = F_1(x) • sqrt( 2 pi G_1 '' (h_n) )

$g'(h_n) = n\;\;\;$ where for f(x) $g(x) = \frac{x^2}{2\ln(2)}\;\;\;$ optionally $a_0 = f(0)\;\;\; a_n = \frac{\exp(g(h_n) - n h_n)}{\sqrt{2 \pi g''(h_n) }}\;\;\; f_2(x) = \sum_{n=0}^{\infty} a_n x^n$
Is this the same as Tommy's F_2? Tommy has the g'' in the numerator which is a typo. It looks like Tommy's F_3 would be the fake function for F_2(x)? I'm not sure if that's what Tommy intended or not.

$f_2(x) = \sum_{n=0}^{\infty} \frac{\exp(-n^2\cdot \ln(2)/2)}{\sqrt{2\pi/\ln(2)}}\;\;\;$ f2(x) is the fake Gaussian approximation for f(x)

Now, starting with $f_2(x)\;\;\;$ let's generate $g_2(x)=\ln(f_2(\exp(x)))$
From there, we can generate a new set of $h_n$ values, which should be nearly identical to the original set of h_n values, and a new set of $b_n$ values which should be nearly identical to the $a_n$ values.

I did this numerically. I think I might be able to generate a closed form equation for this new ratio result, using the equations from post#85. I guess this ratio not going to a limiting value of 1 is a contradiction for your conjecture. Its actually rather interesting, especially if you consider the ratio for non integer values of a_n; n=20.5; vs the equation above. I can post more later if interested; it turns out we have a sine wave oscillating around $g'' = \frac{1}{\ln(2)}$ Anyway, assuming g''(x) for f(x) is asymptotically the same as g''(x) for J(x) as x gets arbitrarily large, I expect the limiting ratio for J(x) to be the same as the limiting ratio below, as n gets arbitrarily large.
Code:
ratio of b_n over a_n where f2(x) is the function from above 1 1.13160761703913345046 2 1.03756115378093045262 3 1.00584054797835817399 4 1.00042570875240853058 5 1.00001412678446418263 6 1.00000021835990293497 7 1.00000000163026326669 8 1.00000000003096900943 9 1.00000000002529304393 10 1.00000000002528326249 11 1.00000000002528325425 12 1.00000000002528325425 13 1.00000000002528325425 14 1.00000000002528325425 15 1.00000000002528325425 16 1.00000000002528325425 17 1.00000000002528325425 18 1.00000000002528325425 19 1.00000000002528325425 20 1.00000000002528325425 21 1.00000000002528325425 22 1.00000000002528325425 23 1.00000000002528325425 24 1.00000000002528325425 25 1.00000000002528325425 26 1.00000000002528325425 27 1.00000000002528325425 28 1.00000000002528325425 29 1.00000000002528325425 30 1.00000000002528325425
(10/09/2015, 08:15 AM)sheldonison Wrote: [ -> ]
(10/08/2015, 11:08 PM)tommy1729 Wrote: [ -> ]I edited post 150 , where I mentioned the tommy-sheldon iterations.

Typo's , mistakes and confusion should be gone.

Although the convergeance conjecture disagrees with sheldon's recent not 1 ratio ... Maybe ...

Things should be clear now.

Regards

Tommy1729

Start with $f(x) = \exp$$\frac{\(\ln(x)$$^2}{2\ln(2)}\)\;\;\; g(x) = \frac{x^2}{2\ln(2)}$

g''(x) for f(x) is conjectured to approach exactly g''(x) for J(x), as x gets arbitrarily large, which is the initial reason for choosing this particular f(x). But f(x) is interesting on its own.

I generate the fake function for f(x) using the Gaussian method. I assume that this is what Tommy means in post#150 by F_2(x).
> F_2(x) = F_1(x) • sqrt( 2 pi G_1 '' (h_n) )

$g'(h_n) = n\;\;\;$ where for f(x) $g(x) = \frac{x^2}{2\ln(2)}\;\;\;$ optionally $a_0 = f(0)\;\;\; a_n = \frac{\exp(g(h_n) - n h_n)}{\sqrt{2 \pi g''(h_n) }}\;\;\; f_2(x) = \sum_{n=0}^{\infty} a_n x^n$
Is this the same as Tommy's F_2? Tommy has the g'' in the numerator which is a typo. It looks like Tommy's F_3 would be the fake function for F_2(x)? I'm not sure if that's what Tommy intended or not.

$f_2(x) = \sum_{n=0}^{\infty} \frac{\exp(-n^2\cdot \ln(2)/2)}{\sqrt{2\pi/\ln(2)}}\;\;\;$ f2(x) is the fake Gaussian approximation for f(x)

Now, starting with $f_2(x)\;\;\;$ let's generate $g_2(x)=\ln(f_2(\exp(x)))$
From there, we can generate a new set of $h_n$ values, which should be nearly identical to the original set of h_n values, and a new set of $b_n$ values which should be nearly identical to the $a_n$ values.

I did this numerically. I think I might be able to generate a closed form equation for this new ratio result, using the equations from post#85. I guess this ratio not going to a limiting value of 1 is a contradiction for your conjecture. Its actually rather interesting, especially if you consider the ratio for non integer values of a_n; n=20.5; vs the equation above. I can post more later if interested; it turns out we have a sine wave oscillating around $g'' = \frac{1}{\ln(2)}$ Anyway, assuming g''(x) for f(x) is asymptotically the same as g''(x) for J(x) as x gets arbitrarily large, I expect the limiting ratio for J(x) to be the same as the limiting ratio below, as n gets arbitrarily large.
Code:
ratio of b_n over a_n where f2(x) is the function from above 1 1.13160761703913345046 2 1.03756115378093045262 3 1.00584054797835817399 4 1.00042570875240853058 5 1.00001412678446418263 6 1.00000021835990293497 7 1.00000000163026326669 8 1.00000000003096900943 9 1.00000000002529304393 10 1.00000000002528326249 11 1.00000000002528325425 12 1.00000000002528325425 13 1.00000000002528325425 14 1.00000000002528325425 15 1.00000000002528325425 16 1.00000000002528325425 17 1.00000000002528325425 18 1.00000000002528325425 19 1.00000000002528325425 20 1.00000000002528325425 21 1.00000000002528325425 22 1.00000000002528325425 23 1.00000000002528325425 24 1.00000000002528325425 25 1.00000000002528325425 26 1.00000000002528325425 27 1.00000000002528325425 28 1.00000000002528325425 29 1.00000000002528325425 30 1.00000000002528325425

Wait.

If a_n ~~ exp(g(h_n) - n h_n)

Is An underestimate ( as you say )

And the gaussian is beter then

Gaussian > exp( g(h_n) - n h_n).

Right ?

So gaussian can not be exp ... / sqrt( 2 pi g" (h_n)).

It has to be exp( g(h_n) - n h_n) * sqrt( 2 pi g " (h_n) ).

Or am I crazy today ?

Regards

Tommy1729
(10/09/2015, 11:56 AM)tommy1729 Wrote: [ -> ]So gaussian can not be exp ... / sqrt( 2 pi g" (h_n)).

It has to be exp( g(h_n) - n h_n) * sqrt( 2 pi g " (h_n) ).

Start with $\int_{-\infty}^{+\infty}\exp$$\frac{-g''x^2}{2}$$ = \sqrt{\frac{2\pi}{g''}}$

I think your just missing a little bit of algebra. We are interested an approximation for
$a_n \; = \; \frac{1}{2\pi i} \int_{h_n-\pi i}^{h_n+\pi i} \exp $$g(x) - n\cdot x$$ \;\;\;$ This is an exact equation if $g(x)=\ln(f(\exp(x)))$

$a_n \approx \; \frac{1}{2\pi i} \cdot \exp $$g(h_n) - n\cdot h_n$$ \cdot \int_{h_n- i\infty}^{h_n+ i\infty} \exp$$\frac{g''x^2}{2}$$ \;\;\;$ see post#16 for more details on getting to this step

$a_n \approx \; \exp $$g(h_n) - n\cdot h_n$$ \cdot \frac{1}{2\pi} \cdot \sqrt{\frac{2\pi}{g''}}$

$a_n \approx \; \exp $$g(h_n) - n\cdot h_n$$ \cdot \sqrt{\frac{1}{2\pi g''}}$

$a_n \approx \frac{\exp(g(h_n) - n h_n)}{\sqrt{2 \pi g''(h_n) }}$

All of the numerical approximations I have posted, use this version of the Gaussian approximation, and get excellent results. In most cases, the Gaussian approximation is also an over-approximation of an entire function with all positive derivatives.
(10/10/2015, 03:08 AM)sheldonison Wrote: [ -> ]
(10/09/2015, 11:56 AM)tommy1729 Wrote: [ -> ]So gaussian can not be exp ... / sqrt( 2 pi g" (h_n)).

It has to be exp( g(h_n) - n h_n) * sqrt( 2 pi g " (h_n) ).

Start with $\int_{-\infty}^{+\infty}\exp$$\frac{-g''x^2}{2}$$ = \sqrt{\frac{2\pi}{g''}}$

I think your just missing a little bit of algebra. We are interested an approximation for
$a_n \; = \; \frac{1}{2\pi i} \int_{h_n-\pi i}^{h_n+\pi i} \exp $$g(x) - n\cdot x$$ \;\;\;$ This is an exact equation if $g(x)=\ln(f(\exp(x)))$

$a_n \approx \; \frac{1}{2\pi i} \cdot \exp $$g(h_n) - n\cdot h_n$$ \cdot \int_{h_n- i\infty}^{h_n+ i\infty} \exp$$\frac{g''x^2}{2}$$ \;\;\;$ see post#16 for more details on getting to this step

$a_n \approx \; \exp $$g(h_n) - n\cdot h_n$$ \cdot \frac{1}{2\pi} \cdot \sqrt{\frac{2\pi}{g''}}$

$a_n \approx \; \exp $$g(h_n) - n\cdot h_n$$ \cdot \sqrt{\frac{1}{2\pi g''}}$

$a_n \approx \frac{\exp(g(h_n) - n h_n)}{\sqrt{2 \pi g''(h_n) }}$

All of the numerical approximations I have posted, use this version of the Gaussian approximation, and get excellent results. In most cases, the Gaussian approximation is also an over-approximation of an entire function with all positive derivatives.

Hmm.

The reason I got confused is

1) i incorrectly thought 1/n! ~ sqrt(2 pi n) (e/n)^n.
See the fake exp.

2) I assumed the gaussian was bigger then S9 because

2 a) the Gaussian is better then S9.

2 b) You Said in post 9 that the method ( S9 ) always gives An underestimate.

So apparantly 2 b) is false.

This was not mentioned before !

Im sorry. Big misunderstanding.

Did I understand my misunderstanding Well ?

Since 2 b) is false, does this imply there exist functions equal to their fake , independent of the method for fake ?
I guess so.
So we could solve S9( f(x) ) = f(x) ?

Intresting.

Regards

Tommy1729

Despite my recent confusions and mistakes , I edited the Tommy-Sheldon equations.

I still believe they are the best so far.

Regards

Tommy1729
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