Limit of mean of Iterations of f(x)=(ln(x);x>0,ln(-x) x<0) =-Omega constant for all x
#5
Yes!

I learned how to iterate something.

THANKS. I wonder what I can do next with this. I have some ideas, like:

There must exist x for which iteration will never reach infinity. Question is- which numbers are those, and which are ones that will have infinite oscillating iterations having the average \( -\Omega \). It seems these numbers will make holes in a real line if that is considered as input /output line from this iteration.

The more iterations are done, the less points will be left from starting input interval [0:1] . Guess how many points will reach infinite oscillations? (or may be how many will NOT reach).

I think that ratio of finishing points at infinity to starting is \( \Omega \) again. I have no proof yet, nor even numerical test.

I think for any interval of input x ]-infinity<xo:x1<infinity[ the share of points reaching infinite iterations will be \( {\Omega/(x1-xo)} \)

And most likely I am wrong as such function ln(mod(ln(x)) can be constructed for any base,and each will have points with infinite oscillations, and points that will terminate to 1 and 0 after finite number of iterations, so if my conjecture is true, all other bases vs . e will have only \( 1-\Omega \) points left on unit interval of real axis that can be infinitely iterated.

e.g for base 4 similar fixed point is 1/2. f(f(f(............1/2) = 1/2; points that will attract to this fixed point are 2, -2, -1/2, etc..


Just to note down an idea for further investigation, for this average of log(log(mod(x)), may be for b>1, fixpoint= ssroot(b) ;
for b <1 b = (1/fixpoint)^(1/fixpoint) but fixpoint = - 1/ssroot(b) = -h(b).

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 - 03/27/2008, 03:38 PM

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