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+--- Thread: Interesting value for W, h involving phi,Omega? (/showthread.php?tid=110)
RE: Interesting value for W, h involving phi,Omega? - Ivars - 03/16/2008
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) = 0,065988036and 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
RE: Interesting value for W, h involving phi,Omega? - Ivars - 03/17/2008
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
RE: Interesting value for W, h involving phi,Omega? - Ivars - 03/27/2008
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
RE: Interesting value for W, h involving phi,Omega? - Ivars - 04/04/2008
We can create the following expression:
\( \lim_{n\to\infty}\frac{\sum_{n=1}^\infty f^{\circ n}(x)}{n}-f^{\circ n}(\Omega)= 0 \)
\( \lim_{n\to\infty}\frac{\sum_{n=1}^\infty f^{\circ n}(x)}{n}-{n/n}*f^{\circ n}(\Omega)= 0 \)