# Tetration Forum

Full Version: Interesting value for W, h involving phi,Omega?
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Ivars Wrote:I was studying the graph of selfroot of Lambert function:

W(x)^(1/W(x)).

It has maximum value at x=(e*(e^e)) and is e^(1/e) ; so

W(e*(e^e)) =W(e^(e+1))= e

If we use this, we can find :

h( (e^(-(e^e)))^e))= h(1,28785E-1

ln((e^(-(e^e))))^e)= e*ln(e^(-e^e)) = -e*e^e

so :

h( (e^(-(e^e)))^e))= -W(-ln((e^(-(e^e)))^e))/ln((e^(-(e^e)))^e)=

-W(e*e^e)/-e*e^e = -e/-e*e^e = 1/e^e= e^(-e)

So:

h( (e^(-(e^e)))^e))= h(1,28785E-1= e^(-e) = 0,065988036

and second superroot of ((e^(e^e))^e) = ln(((e^(e^e))^e))/W(ln ((e^(e^e))^e)) = e*(e^e)/W(e*(e^e)) = e*(e^e)/e = e^e= 15,15426224

Ssroot( ((e^(e^e))^e)= ssroot(7,76487E+17) = e^e = 15,15426224

For these values, h(1/a) = 1/ssroot(a)

I wonder are there any similar relations for W((1/e)*(e^e)) =W(e^(e-1))= W(5.574..) = 1.3894..

Ivars
Finally I understood defintion of Omega via e secondsuperroot of e :

Ssroot (e) = ln(e) / W(ln(e)) = 1/Omega

So e= (1/Omega)^(1/Omega)

1/e = (1/Omega)^(-1/Omega) = (Omega)^(1/Omega)

Which was known, of course.

Ivars
Sorry Henryk I had to add short version here as it belong Interesting things/ Omega thread if someone is looking here for more interesting values....

I constructed the following function (basically using only real part of module of each iterate) to see what happens when its iterated:
$f(x) = \ln(x) \text{ if } x>0$
$f(x) = \ln(-x) \text{ if }x<0$

Since $\ln(\Omega=0,567143..) = - \Omega$ it is obvious that $\Omega$ is the only starting point that will not move since every iteration will return to $\Omega$ as argument for $\ln$ and ${1/\Omega}$ will be next one to converge to $\Omega$ after first iteration.

So $\Omega$ may play a special role in this iterations, and it does.Firstly, all iterations from 1 to $\infty$ pass via this point $x=\Omega$.

Then I found $f(x)$ 8500 Excel precision iterates in the region x = ]0.001:0.001:2.71[ (for a starter, it can be done for negative /positive x outside this region as well), via formula:

$f^{\circ n}(x) = \ln(f^{\circ n-1}(x)\text{ if } f^{\circ n-1}(x)>0$
$f^{\circ n}(x) = \ln(-f^{\circ n-1}(x)\text{ if } f^{\circ n-1}(x)<0$

So first conclusion after some numerical modelling and graphing visible in

Limit of mean value of iterations of ln(mod(Re(ln(x))) = -Omega constant

from all this is :

$\lim_{n\to\infty}\frac{\sum_{n=1}^\infty f^{\circ n}(x)}{n}= -\Omega=-0.567143..$

Second, that iterations are bounded with span of values depending on the choice of x steps : size of step and linearity/nonlinearity.

Also:

$\lim_{n\to\infty}\frac{\sum_{n=1}^\infty f^{\circ n}({1/x})}{n}= +\Omega=+0.567143..$

Also obviously:

$f^{\circ n}(\Omega)= -\Omega=ln(\Omega)=-0.567143..$ for all positive n

$f^{\circ n}({1/\Omega})= +\Omega=ln({1/\Omega}) = +0.567143..$ for all positive n

Ivars
We can create the following expression:

$\lim_{n\to\infty}\frac{\sum_{n=1}^\infty f^{\circ n}(x)}{n}-f^{\circ n}(\Omega)= 0$

$\lim_{n\to\infty}\frac{\sum_{n=1}^\infty f^{\circ n}(x)}{n}-{n/n}*f^{\circ n}(\Omega)= 0$
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