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tetration base sqrt(e)
#1
tetration base sqrt(e).

I had some doubts on the title, since Im trying to explain some basic principles here.

But I thought it would be nice to unify things , so that things can be explained more clearly.

The main thing here is the concept of formula's for parabolic fixpoints and the general understanding of them. (beyond what has been written before and is well known )

Also the CONNECTIONS BETWEEN DYNAMICS AND CALCULUS play an important role here.

As explained in the thread " using sinh " http://math.eretrandre.org/tetrationforu...hp?tid=424


We can compute the tetration for any base b > 1 from the superfunction of 2sinh( ln(b) x) = 2sinh_b (x).

IF b > exp(1/2) then we have a hyperbolic fixpoint so

2sinh^[z]( ln(b) x ) = lim 2sinh_b^[n] ( (2 ln(b) )^(z-n) x )

But if b = exp(1/2) we have a parabolic fixpoint ( and we can easily see the formula above fails. )

Let c = exp(1/2).

2sinh_c^[z](x) = lim 2sinh_c^[n] ( g(z-n) x ).

Just as in the hyperbolic case where (2 ln(b))^(z-n) is " only " an approximation of arc2sinh_b^[n-z] it works nonetheless , and similarly for g(z-n).

So we need to understand the asymptotic behaviour of arc2sinh_c^[n] for large n.

Let x_(n+1) = arc2sinh_c (x_n)
The Taylor series for arc2sinh_c(x) starts like x + ...

Therefore

x_(n+1) = x_n + ...

Which leads towards

x_(n+1) - x_n = "arc2sinh_c - id"(x_n)

where "arc2sinh_c - id"(x) is ofcourse arc2sinh_c(x) - x.
( id comes from id(x) = x )

Now x_(n+1) - x_n tends to x_n ' as n gets large.


therefore we arrive at the differential equation ( Like I said its related to calculus ! )

d f(n)/dn = arc2sinh_c( f(n) ) - f(n).

The solution is

f^[-1](n) = integral dn /( arc2sinh_c(n) - n)

f(n) = (integral dn /( arc2sinh_c(n) - n))^[-1]

So our final solution is

2sinh_c^[z](x) = lim 2sinh_c^[n] ((integral d(z-n) /( arc2sinh_c(z-n) - (z-n)) )^[-1]
x ).

So this finally gives us tetration base sqrt(e).

If we were to truncate the inverse integral by say a truncated Taylor we would arrive at some of the familiar formula's for parabolic fixpoint expansion.

It is intresting to note that integrals like

dx/(exp(x) - x - 1)

dx/(2sinh(x) - x)

dx/(arc2sinh(x) - x)

dx/(sin(x) - x)

have no closed form.

This thread gives somewhat of a reason for that.
But more investigation is needed.
AND this also leads us to calculus ; differential Galois theory.
( or is that considered algebra ? ).

Using the ideas above we can express another often considered superfunction , the one of sin(x) or arcsin(x).

sin^[z](x) = arcsin^[n] ( sqrt ( 3 / (n-z) ) )

We found that by truncating the Taylor for sin up to O(x^4) BEFORE solving for the differential equation.
The inverse integral is then NOT truncated.

NOTICE 3 things !

1) sin^[-7](3/4) is problematic ! we enter the complex numbers.

2) arcsin^[n] grows superexponentially.

3) The switch from slowly to superexponential is made by going around a singularity/branch.
The singularity/branch created by arcsin is REPRESENTED by the singularity/branch of the sqrt !! We take sqrt's of negative numbers whenever arcsin iterations become nonreal !

Disclaimer : this is sketchy to avoid too long posts and may contain typo's or sign mistakes.
Also this is NOT the full story.

regards

tommy1729

" Truth is that what does not go away when you stop believing in it "
tommy1729
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#2
Questions :

A)
Does this agree with Levy's formula ?

Levy_f(x) = lim ( f^[n+1](x) - f^[n](u) )/( f^[n+1](u) - f^[n](u) )

B)
And can we , if not published already , extend Levy's method with a differential equation to speed up the convergeance ?

I think so.

I was thinking about introducing bring radicals to consider iterations of any entire g(x) = x - x^5 + ... .

C)

I know Levy and others published similar looking formula's , maybe that helps.

regards

tommy1729
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#3
W.J. Thron, Sequences generated by iteration., Trans. Am. Math. Soc. 96 (1960), 38-53
(English).

This is relevant.

W.J Thron proves that the integral can be truncated to A x^B under some conditions that come from this theorem :

(Thron 1960 Theorem 3.1.). Let f be analytic at 0 with powerseries expansion of the following form :

f(x) = x - a_m x^m + a_(m+1) x^(m+1) + ...

with a_m < 0.

then lim n -> oo
n^(1/(m-1)) h^[n](x) = (- a_m(m - 1))^(1/(m-1))

From which it follows.

regards

tommy1729
Reply


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