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 Zeta iterations Balarka Sen Junior Fellow Posts: 25 Threads: 7 Joined: Feb 2013 02/24/2013, 08:00 AM Consider the (real) attractive fixed point of zeta z = -0.29590500557... In A069857, the author claims that this is also the limit of zeta^[m](z) for complex values of z as m tends to be large (or perhaps infinity, not sure if it exists). Can we show that the limit exists and -0.29590500557 IS the limit? Balarka . Gottfried Ultimate Fellow Posts: 757 Threads: 116 Joined: Aug 2007 02/24/2013, 11:10 AM (02/24/2013, 08:00 AM)Balarka Sen Wrote: Consider the (real) attractive fixed point of zeta z = -0.29590500557... In A069857, the author claims that this is also the limit of zeta^[m](z) for complex values of z as m tends to be large (or perhaps infinity, not sure if it exists). Can we show that the limit exists and -0.29590500557 IS the limit? Balarka . I don't have a final answer, but the following observation are perhaps helpful to find such an answer. A fixpoint is attracting, if the absolute value of its derivative is smaller than 1. So to see, whether z is attracting also for the complex plane its absolute value must be smaller in any direction of h in the derivative-formula (zeta(z+h/2)-zeta(z-h/2))/h . To do this in Pari/GP h can be defined as h=exp(I*phi)*1e-20 for varying phi. For all phi tested the absolute of the derivative at z was smaller than 1. However - a heuristic is no proof... Gottfried Gottfried Helms, Kassel Balarka Sen Junior Fellow Posts: 25 Threads: 7 Joined: Feb 2013 02/24/2013, 11:21 AM (This post was last modified: 02/24/2013, 11:24 AM by Balarka Sen.) Gottfried Wrote:to see, whether z is attracting also for the complex plane its absolute value must be smaller in any direction of h in the derivative-formula (zeta(z+h/2)-zeta(z-h/2))/h . Yes, it's obviously a question whether it's actually a fixed point in the whole complex plane but that wasn't the thing I've asked, actually. I want to know why zeta^[m](x) for complex x's tends to that unique fixed point for large m's? Is it just a mathematical coincidence or has a reason behind it? Thank you for your time, Balarka . Gottfried Ultimate Fellow Posts: 757 Threads: 116 Joined: Aug 2007 02/24/2013, 11:30 AM (This post was last modified: 02/24/2013, 11:31 AM by Gottfried.) (02/24/2013, 11:21 AM)Balarka Sen Wrote: want to know why zeta^[m](x) for complex x's tends to that unique fixed point for large m's? Is it just a mathematical coincidence or has a reason behind it? Thank you for your time, Balarka .Hi Belarka - possibly I just misunderstand your question. But if not: just because the absolute value of the derivative at z is smaller than 1 is that reason - at least as far as I understand that thing (I may be wrong, though). Gottfried Gottfried Helms, Kassel Balarka Sen Junior Fellow Posts: 25 Threads: 7 Joined: Feb 2013 02/24/2013, 11:36 AM Gottfried Wrote:just because the absolute value of the derivative at z is smaller than 1 is that reason I see, but how did you derived that? Is it very straightforward from the definition of attractive fixed point but I cannot see it? I am having a feeling that I possibly asked a pretty stupid question Gottfried Ultimate Fellow Posts: 757 Threads: 116 Joined: Aug 2007 02/24/2013, 12:28 PM (02/24/2013, 11:36 AM)Balarka Sen Wrote: I see, but how did you derived that? Is it very straightforward from the definition of attractive fixed point but I cannot see it?Hmm, perhaps you'll find better/more complete/more helpful explanations when googling for "basins of attraction" or "petals of attraction". Did you try our hyperoperation-wiki already? Perhaps it is also helpful to remember, that the function zeta1(s) = zeta(s)+1/(1-s) is entire (and has a power series which is easily derivable with Pari/GP). Then also the derivative must be entire; and also the derivative of the remaing part zeta2(s) = -1/(1-s) is zeta2(s)'=1/(1-s)^2 which is finite only except at s=1. But well, I think I'd stop here and confirm my much freehanded assumtions myself now... Gottfried Gottfried Helms, Kassel tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 02/24/2013, 07:54 PM (02/24/2013, 11:36 AM)Balarka Sen Wrote: Gottfried Wrote:just because the absolute value of the derivative at z is smaller than 1 is that reason I see, but how did you derived that? Is it very straightforward from the definition of attractive fixed point but I cannot see it? I am having a feeling that I possibly asked a pretty stupid question I wanted to comment on that to give some kind of intuitive explaination. For simplicity recenter the fixpoint at 0. Now if f(z) is analytic and f(0)=0 and there is no fixpoint near 0 then consider writing f(z) = 0 + a z + b z^2 + c z^3 + ... Now from calculus you know that if z gets very small ( where very depends on the size of the taylor coefficients and the remainder ) then f(z) gets well approximated by f(z) = a z + O(z^2) Now if a is 1 then f(z) approximates id(z) and hence the fixpoint 0 is " neutral " also called " invariant " or " parabolic ". Now clearly if a = i ( imaginary unit) then by f(z) = a z + O(z^2) we get a simple rotation but not a strong attraction or repellation. This already hints that looking at the complex number a in polar coordinates makes more intuitive sense. To visualize the 'small' z take a disk around z=0 with a small radius. Now because f(z) approximates a z , AND the theta of a determines how much we rotate , the absolute value must determine if it is attracting or repelling. By attracting we mean that any point in the small disk must be mapped closer to the fixpoint , and by repelling we mean that any point must be mapped further away from the fixpoint. For instance if f(z) is of the form 0 + 0.5 z + 0.0001 z^2 + ... then it is clear that f(z) in the small radius ( = z small => higher powers of z have lesser influence ) behaves like f(z)^[t] = 0.5^t z which is clearly attracting. By analogue if a = 2 it is clearly repelling. if a = 2 i , then that is equivalent to repelling + rotating. This more or less concludes my personal nonformal intuitive explaination ( for complex values ) and I hope it is clear and helpfull to you. Btw , Note that I did not discuss what happens when a = 0. I only gave a sketch of how and why when the absolute value of a satisfies 0 < abs(a) < oo. When a=0 many formula's and properties fail anyway and for the case a=0 you need to understand more of the Fatou-flower. I could say alot more but I do not want to confuse you with harder stuff and you can read up on the subject too. Hence I did not want to post more than that what occurs sponteanously to me. Together with some basic complex analysis and basic real calculus I think this will get you much further in the subjects and discussions about tetration and their cruxial details. regards tommy1729 bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 02/24/2013, 08:36 PM Perhaps formal derivation: Attractive fixpoint $z_0$ means $\left|\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}\right|=c<1$. By continuity of the left side in $z$ there is a disk $D_\eps(z_0)$ with center $z_0$ such that $\left|\lim_{z\to z_0}\frac{f(z)-z_0}{z-z_0}\right|=d<1$ for all $z\in D_\eps(z_0)$. ($c) So for any $z\in D_\eps(z_0)$ we have $|f(z)-z_0|, particularly $f(z)$ is again in $D_\eps(z_0)$. $|f(f(z))-z_0| and so on $|f^{[n]}(z)-z_0|. The right sight converges to 0 for $n\to\infty$, hence by definition of convergence: $f^{[n]}(z)\to z_0$ for $n\to\infty$. tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 02/24/2013, 09:13 PM How many fixpoints does zeta have actually ? If a fractal converges to a single fixpoint then that is the only fixpoint. So I guess the question is how many fixpoints does zeta have ? afterall if x is distinct from y and zeta(x)=x and zeta(y)=y then taking zeta zeta zeta ... zeta(y) = y ( and not x ! ). attractive or not is another matter. Notice that zeta(x+yi) for x >> y gives 1 and zeta(...zeta(1)...) = x I also note that zeta(z) = 1 + 1/2^z + ... so for Re(z) > 1 and abs(z) = oo we get the same limit as for zeta (...zeta (1)...). regards tommy1729 Gottfried Ultimate Fellow Posts: 757 Threads: 116 Joined: Aug 2007 02/25/2013, 12:17 AM (02/24/2013, 09:13 PM)tommy1729 Wrote: How many fixpoints does zeta have actually ?Because the bernoulli-numbers grow faster than their index, they have alternating signs and finally, because the bernoulli-numbers are interpolated by the zeta at negative arguments we have, that the negative part of the x=y-diagonal is crossed infinitely many times. So we have (at least countably) infinitely many real fixpoints. Gottfried Gottfried Helms, Kassel « Next Oldest | Next Newest »

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