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Zeta iterations  Balarka Sen  02/24/2013 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 . RE: Zeta iterations  Gottfried  02/24/2013 (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? 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 derivativeformula (zeta(z+h/2)zeta(zh/2))/h . To do this in Pari/GP h can be defined as h=exp(I*phi)*1e20 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 RE: Zeta iterations  Balarka Sen  02/24/2013 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 derivativeformula (zeta(z+h/2)zeta(zh/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 . RE: Zeta iterations  Gottfried  02/24/2013 (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?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 RE: Zeta iterations  Balarka Sen  02/24/2013 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 RE: Zeta iterations  Gottfried  02/24/2013 (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 hyperoperationwiki already? Perhaps it is also helpful to remember, that the function zeta1(s) = zeta(s)+1/(1s) 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/(1s) is zeta2(s)'=1/(1s)^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 RE: Zeta iterations  tommy1729  02/24/2013 (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 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 Fatouflower. 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 RE: Zeta iterations  bo198214  02/24/2013 Perhaps formal derivation: Attractive fixpoint means . By continuity of the left side in there is a disk with center such that for all . () So for any we have , particularly is again in . and so on . The right sight converges to 0 for , hence by definition of convergence: for . RE: Zeta iterations  tommy1729  02/24/2013 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 RE: Zeta iterations  Gottfried  02/25/2013 (02/24/2013, 09:13 PM)tommy1729 Wrote: How many fixpoints does zeta have actually ?Because the bernoullinumbers grow faster than their index, they have alternating signs and finally, because the bernoullinumbers are interpolated by the zeta at negative arguments we have, that the negative part of the x=ydiagonal is crossed infinitely many times. So we have (at least countably) infinitely many real fixpoints. Gottfried 