Limit of mean of Iterations of f(x)=(ln(x);x>0,ln(-x) x<0) =-Omega constant for all x

[bo198214]

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

This can be easier expressed as \( f(x)=\ln(|x|) \)

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

That seems indeed true. Basicly it has to do with that \( \Omega \) is a (the only) fixed point of \( x\to\ln|x| \). Proof:
\( x=\ln|x| \)
\( e^x=|x| \) this can not be satisfied for \( x>0 \) as \( e^x \) has no fixed point. So we consider the case \( e^x=-x \), which is equivalent to \( 1=(-x)e^{-x} \) which is by your definition \( -x=\Omega \).

However that fixed point is repelling \( |\ln(|x|)'(-\Omega)|=|\frac{1}{x}(-\Omega)|=|-\frac{1}{\Omega}|>1 \), thatswhy \( f^{\circ n}(x) \) does not converge. On the other hand the iterated value is always thrown back into the negative domain by \( \log(x) \) for positve \( x \). And there it oscillates around the fixed point until it is again repelled into the positive domain. So most time it is oscillating around \( -\Omega \), thatswhy the average is just this.