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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 Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 04/06/2008, 11:45 AM (This post was last modified: 04/06/2008, 02:30 PM by Ivars.) I obtained the distributions around $+-\Omega$ of integer positive iterations of these functions using module each time argument gets negative. $\lim_{n\to\infty}\frac{\sum_{n=1}^\infty f^{\circ n}(x)}{n}= -\Omega=-0.567143..$ $\lim_{n\to\infty}\frac{\sum_{n=1}^\infty f^{\circ n}({1/x})}{n}= +\Omega=+0.567143..$ 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 $+-\Omega$ 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: ln(mod(x)):     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 ln(mod(1/x)):     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 $+- 1/e$? 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. Ivars « Next Oldest | Next Newest »

 Messages In This Thread Limit of mean of Iterations of f(x)=(ln(x);x>0,ln(-x) x<0) =-Omega constant for all x - by Ivars - 03/25/2008, 10:36 PM RE:Limit of mean of Iterations of f(x)=(ln(x);x>0,ln(-x) x<0) =-Omega for all x - by Ivars - 03/25/2008, 10:45 PM 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 RE: Limit of mean of Iterations of f(x)=(ln(x);x>0,ln(-x) x<0) =-Omega constant for all x - by bo198214 - 03/27/2008, 09:29 AM RE: Limit of mean of Iterations of f(x)=(ln(x);x>0,ln(-x) x<0) =-Omega constant for all x - by Ivars - 04/02/2008, 09:44 PM 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 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 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, 06:14 PM 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, 06:55 PM RE: Limit of mean of Iterations of f(x)=(ln(x);x>0,ln(-x) x<0) =-Omega constant for all x - by Ivars - 03/10/2009, 03:34 PM RE: Limit of mean of Iterations of f(x)=(ln(x);x>0,ln(-x) x<0) =-Omega constant for all x - by tommy1729 - 03/29/2015, 08:02 PM

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