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Limit of mean of Iterations of f(x)=(ln(x);x>0,ln(-x) x<0) =-Omega constant for all x
I obtained the distributions around of integer positive iterations of these functions using module each time argument gets negative.

The distributions are symmetric against ordinate axis, point x=0, but they overlap. I am not sure of the type of the distributions, nor how does standard accuracy of Excel influence them- obviously, the values float away from where they should be, but, since we obtain correctly the mean values may be the distribution principially is the same regardless of accuracy given enough iterations are made. They seemed to be log normal but they are not. Weibull is too difficult for me to compare to data set.

Anyway, here are histograms- probability density functions and data parameters from excel:


Mean -0,569047639
Standard Error 0,012051586
Median -0,372148702
Mode #N/A
Standard Deviation 1,206784434
Sample Variance 1,456562737
Kurtosis 2,838543218
Skewness -1,189194674
Range 11,75801951
Minimum -9,512327352
Maximum 2,245692157



Mean 0,568939972
Standard Error 0,011986562
Median 0,372020783
Mode #N/A
Standard Deviation 1,205767153
Sample Variance 1,454070305
Kurtosis 2,858638726
Skewness 1,190184673
Range 11,83018032
Minimum -2,251410718
Maximum 9,578769599

In principle, all parameters are the same, just the ones who can, change sign. Obviously, location parameter needs to be added (its NOT 0) and mean and maximum does not coincide.
the value of median is Interesting. Could it be ? Probably not.

The distirbution looks a little like this:
log Weibull Ditribution

But ... I have a feeling that this distribution has to be iterated (tetrated?) as well to obtain some limit distribution, since I have been applying ln many times to those variables x which were quite regularly distributed at the beginning and obtained some peculiar chaos.


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 - 04/06/2008, 11:45 AM

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