Limit of mean of Iterations of f(x)=(ln(x);x>0,ln(-x) x<0) =-Omega constant for all x
#3
The character of convergence of the mean to

\( -\Omega \)

\( f(x) = \ln(x) \text{ if } x>0 \)
\( f(x) = \ln(-x) \text{ if }x<0 \)

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

can be seen from this graph where first 150 iterations of 100 points x=0.01 step 0.01 = 0.99 are plotted on top of each other.

\( f(x)_n = \ln(f(x)_{n-1}) \text{ if } f(x)_{n-1}>0 \)
\( f(x)_n = \ln(-f(x)_{n-1}) \text{ if } f(x)_{n-1} <0 \)


   

The limited spread (max-min) of each iteration is also visible with such linear choice and size of steps.

I guess this is somehow related to tree functions (which are related to Lambert function) and iterated logarithms. Just to remember always that studying \( \Omega \) constant is studying Lambert Wo(1)= \( \Omega \) so generalizations are possible if things are positioned correctly. So far I do not see them, but I see what to read next.

Ivars
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Messages In This Thread
RE: Limit of mean of Iterations of f(x)=(ln(x);x>0,ln(-x) x<0) =-Omega constant for all x - by Ivars - 03/26/2008, 11:51 AM

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