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 Interesting value for W, h involving phi,Omega? Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/16/2008, 08:14 AM (This post was last modified: 03/19/2008, 07:33 AM by Ivars.) 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 Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/17/2008, 07:23 PM (This post was last modified: 03/25/2008, 06:20 PM by 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 Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/27/2008, 08:02 AM (This post was last modified: 04/02/2008, 06:42 AM by 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 Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 04/04/2008, 01:41 PM (This post was last modified: 04/10/2008, 07:16 AM by Ivars.) We can create the following expression: $\lim_{n\to\infty}\frac{\sum_{n=1}^\infty f^{\circ n}(x)}{n}-f^{\circ n}(\Omega)= 0$ $\lim_{n\to\infty}\frac{\sum_{n=1}^\infty f^{\circ n}(x)}{n}-{n/n}*f^{\circ n}(\Omega)= 0$ « Next Oldest | Next Newest »

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