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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 03/25/2008, 10:36 PM (This post was last modified: 03/26/2008, 04:08 PM by Ivars.) I was little nervous that in Henryk's formula for tetration with real/complex heights ( see here:Post ) any time logarithm takes negative value, iteration has to be stopped or complex values of logarithm used. Especially for all numbers < 1 iteration stops without even beginning. So I constructed the following function to see what happens when its iterated: Since it is obvious that is the only starting point that will not move since every iteration will return to as argument for and will be next one to converge to after first iteration. So may play a special role in this, and as we will see later, it does.Firstly, all iterations from 1 to pass via this point . Then I found iterates in the region x = ]0.001:0.001:2.71[ (for a starter, it can be done for negative /positive x>1 as well), via formula: It is clear that starting with all relatives to e like e, 1/e, e^(1/e), e^-e, e^e ; ((e^-e)^-e)^-e.......... etc iteration stops rather soon via sequence f(f(e) or f(f(f(1/e). The questions remains, if iterations are infinite, are there many numbers left on real axis which are not in some way related to iterated e-powers of smaller e parent? In case of infinite logarithmic iteration, some numbers would start at infinitesimal seed of e-sequence relative or parent, while some trees will stop at finite number of steps. Excuse me for obvious choice of names. Are there any numbers left? Probably yes. Well, this is how first 5 iterations look like: And iterations 6-10 obviuously increase the frequency of spikes. I did 8522 iterations in Excel with ordinary precision and what happens is interesting: The limit of average when n-> infinity of all points per iteration seems to converge to . The same limit value is reached as average value of iteration at each point x (Average of 8522 iterations in this case) This means that iterated sequences are More negative than positive, and in fact, the maximum postive value at any iteration over all points is always smaller than maximum negative, and their ratio (negative lower bound/positive upper bound) in limit tends to something very close to This last value has to be checked with better precision iteration, but about convergence to for averages mentioned above I have no doubt as it is logical. So first conclusion from all this is very interesting: This was so long, excuse me for mistakes but I hope tex and pictures explain. Ivars Attached Files Image(s) « 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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