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 2 real fixpoints again ....... tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 01/15/2016, 01:29 PM (This post was last modified: 01/16/2016, 08:35 PM by tommy1729.) Consider a function f that is strictly increasing , u-shaped and real-analytic for real > 0. With fixpoints at 0 with derivative 0 and fixpoint at 1 with derivative 2. No other fixpoints. Then the superfunction starting at $x_o >0$ is given by : Lim n to oo $f^{[-n]} ( x_o ^{2^{(n+z + \theta(z) )}} ).$. Where $\theta$ is a 1 periodic real function. How about other cases ? Like also 2 fixpoints but at other positions with other derivatives. This solution was based on x^2 because that has a closed form. For instance $sinh ( m. arcsinh(x) ) ^{[z] }$ also has a closed form for positive integer m, just like x^2. A second question is how to get An analytic half-iterate out of this. 1) we need such a closed form 2) the super needs to be analytic 3) we need analytic continuation from x_a to x_b , where Small fixpoint < x_a < Large fixpoint < x_b. Leading to question 3,5 : the Roc of the Taylor series at expansion points and the agreement of 2 taylors when expanded at different points. In the Bummer thread this has been adressed a Tiny bit , but that was for Natural iteration ( theta = 0 , koenigs etc ) Here the attention Goes to the right theta , the correct helping function with a closed form like x^2 , and the questions above. Also Sheldon has investigated the natural method with An added theta_1 and theta_2 for resp. Each fixpoint , then setting them equal with the equation super_1 = super_2 and hoping to find the correct thetas like that so that the equality is fullfilled and analytic in An interval containing both fixpoints. But this is different thus. Also Bo asked questions about supers with closed forms. Sorry for the late edit. Regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 01/16/2016, 08:41 PM (This post was last modified: 01/16/2016, 08:42 PM by tommy1729.) I severely edited the OP. Sorry for the initial poor post and the late edit of it. [lack of time] I used Tex , keep in Mind that ^[] means composition. I think tex was necc to avoid confusion for Some people, however Im not very satisfied with the way x_o ^2 is displayed. The 2 is hanging to low , way to low for a power , even too low for An upper index. Looks like a multiplication :s Regards Tommy1729 marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 01/16/2016, 11:38 PM (This post was last modified: 01/16/2016, 11:40 PM by marraco.) tommy1729: If you find Latex inconvenient, you may find this useful: http://math.eretrandre.org/tetrationforu...p?tid=1050 I have the result, but I do not yet know how to get it. sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 01/17/2016, 12:51 AM (This post was last modified: 01/17/2016, 01:43 AM by sheldonison.) (01/15/2016, 01:29 PM)tommy1729 Wrote: Consider a function f that is strictly increasing , u-shaped and real-analytic for real > 0. With fixpoints at 0 with derivative 0 and fixpoint at 1 with derivative 2. Having a derivitive=0 at x=0 is the same as a super-attracting fixed point at x=0. The Op is refering to the classic example: $f(x)=x^2$ whose superfunction is the same from both fixed points. The fixed point at 1 is repelling with a period of 2, and there is also a super attracting fixed point at infinity. $f^{\circ z}(0.5) = 2^{(-2^z)}$ For f(x)=x^2, this is the same as the Schroeder function solution from the fixed point of 1. For all other super-attracting x^2 functions, $f(x)=x^2+\sum_{n=3}^{\infty}a_n\cdot x^n\;$ one can probably prove that the two superfunctions, one from the superattracting fixed point, and the other from the repelling fixed point, will always be different. One could look at Miller's book for the general form for putting the super-attracting fixed point function in congruence with the unit circle. There will be a fractal Julia set for the basin of attraction for the super-attracting fixed point, and the repelling fixed point of 1 will be on the Julia set boundary. We can always take the Abel function of the other fixed point's Superfunction, to generate a theta mapping. But theta mappings give us an infinite number of half iterates, abel functions, and super functions possible, all valid, besides the two formal solutions at the fixed points. It sort of reminds me of the sexp/slog sqrt(2) case, except without a uniqueness criteria. The sqrt(2) case has a uniqueness criteria when viewed as an extension of tetration. - Sheldon tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 01/18/2016, 02:58 AM (This post was last modified: 01/21/2016, 11:46 PM by tommy1729.) The " helping function " from the OP was x^2. Notice it is hard or even impossible to get a new helping function from An old one : Let g(x) be analytic with fixpoints 0,1. Then g^[-1]( g(x)^2 ) Has the same fixpoints with the same derivatives at them. So that does not work. Formula's (addition , product ) for trig and hyperbolic trig function also do not give the Desired results. Also the selfreference is not in our favour : This type of lim form for a super requires a helper , but that helper needs a super ... With in the case of using this type of lim , again a helper. --- So , putting this aside for a while , Maybe relaxing " closed form " helps. Like Sums , integrals , products. Not many product forms are discovered or written for functional equations and it seems rather hard. Gottried solved F(x+1) = e F(x) with the pxp(x). Since this OP is more focused on x^2 I Will solve T(x^2) = 2 T(x). For x>0. The aim is to generalize these products to get more helpers. Or show it can not be done, which would be hard ; a very general statement is always hard to disprove without An example. For x>0 and the product over n > 0 : T(x) = (x-1) $\Pi$ 2 (x^{2^(-n)} + 1)^{-1}. I call pxp the superproduct and T(x) the abelproduct. You can easily prove that T(x^2) = 2 T(x) by simply plugging in x^2. In fact T(xy) = T(x) + T(y) for Re(x,y) > 0. [ Re(x,y) = shorthand for Re(x),Re(y). ] Actually T(x) is just ln(x) for x>0. [yes , because we have a product for ln for positive x , this relates to fake function theory ... Notice we also have x^{1/2}, x^{1/4} etc in the Abelproduct , for which we have fakes as well. But that is a bit off topic here. In a way trivial because of the connection between products , zero's and fake function theory. However this is yet another way to arrive at a fake ln ( 3 or 2 ways , depending on how you count equivalent ) , so it is worth a comment imho. ] I already expressed the aim with this , but without Trying hard , a logical guess seems to be finding Abelproduct to solve 2 k(x) = k(a x^2 + b x + c). Notice a product that contains (a x^2 + b x + c)^[m] for integer m is acceptable. I consider a new thread , but this connects so strongly. Otherwise it might seem a curious coincidence of little importance or potential. Im also intrested in C^oo solutions. I was not able to find nowhere analytic C^oo Abelproducts , but then again i lack time. Have to go. Regards Tommy1729 sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 01/18/2016, 09:05 AM (This post was last modified: 01/18/2016, 09:26 AM by sheldonison.) (01/17/2016, 12:51 AM)sheldonison Wrote: $f(x)=x^2+\sum_{n=3}^{\infty}a_n\cdot x^n\;$ ... one from the superattracting fixed point ...One could look at Miller's book for the general form for putting the super-attracting fixed point function in congruence with the unit circle. tommy1729 Wrote:The " helping function " from the OP was x^2. Notice it is hard or even impossible to get a new helping function from An old oneTommy, the super-attracting fixed point at zero is actually an easy problem. You can get the Abel function from the formal Böttcher's function for the super-attracting fixed point. Among my pari-gp scripts, are solutions for the Böttcher function and its inverse for any arbitrary super-attracting fixed point series, and the corresponding Abel function and Superfunction for that arbitrary f(z) series: $f(z)=z^2+\sum_{n=3}^{\infty}a_n\cdot z^n$ $B(f(z)) = B(z)^2\;\;$ definintion of the Böttcher function for f(z) $\alpha(z) = \ln_2(-ln_2(B(z)))\;\;$ the Abel function for f(z) - Sheldon tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 01/18/2016, 10:02 AM Yes , but the Superattracting fixpoint was just picked to make the example easy. The actual goal is to get helpers for a pair of fixed points that are not superattractive. I prefer to talk with math formula instead of code. Boetcher does ring a few bells , although I assume i can learn more. Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 01/19/2016, 01:24 PM The Böttcher function does not really help much. It is just the double exponential of the Abel function. Or the exponential of A generalized Koenigs function. Consider j(x) Abel_1(j(x)) = Abel_1(x) + 1. A general Abel_2 is of the form Abel_2(x) = Abel_1(x) + invar(x). Where invar is the invariant : Invar(j(x)) = Invar(x). Invar relates trivially to the theta function used for going from Super_1 to Super_2. The general Böttcher is then basically a double exponential of this general Abel. So basically we just solved Abel Or Schröeder equation , and reached nothing more than the usual Koenigs and the understanding of the theta / invar. Anything beyond is conjectural in nature , and does not occur in the works of Böttcher. I assume your computation is just like this and thus not so helpfull ( in the context of this thread ). This is the reason i did not mention Böttcher before. His work seems not to help with matters like radius of convergeance , area of agreement with functional equation , agreement on fixpoints etc etc. Sorry Im skeptical. As often imho An example of a mathematician named after his least contribution ; His other work is imho more intresting. ( other examples are the forgotten work of Cantor , theorem named after people who did not prove it or conjectured it first etc ) We wrote alot about the theta wave and the invar. It seems logical to assume that a crucial invariant for base sqrt 2 should be of the form ( for validity in 2 < x < 4 ) : Sum Q(g^[k](x)) Where K runs over the integers and Q(2) = Q(4) = 0. Regards Tommy1729 sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 01/19/2016, 01:55 PM (This post was last modified: 01/19/2016, 05:38 PM by sheldonison.) tommy1729 Wrote:Yes , but the Superattracting fixpoint was just picked to make the example easy. The actual goal is to get helpers for a pair of fixed points that are not superattractive. (01/19/2016, 01:24 PM)tommy1729 Wrote: The Böttcher function does not really help much. .... His work seems not to help with matters like radius of convergeance , area of agreement with functional equation , agreement on fixpoints etc etc. ....It seems the super-attracting case is not essential to the discussion. For an attracting and a repelling fixed point, both fixed points have their own Schroeder function solutions, which don't agree with each other except in special cases (see http://math.stackexchange.com/questions/...-iteration for details). The biggest obstacle, is that there is no unique Abel/super function solution. For example, tetration b=sqrt(2) is no longer real valued, when viewed as a complex plane extension of tetration for real valued bases greater than exp(1/e). So tetration gives us two additional solutions to b=sqrt(2), which are complex conjugates of each other, depending on whether you rotate clockwise or counter-clockwise around the singularity at eta=exp(1/e). The result of this "tetration" theta mapping, is that the upper half of the complex plane looks like one of the fixed points, and the lower half of the complex plane looks like the other fixed point. There are an infinite number of other possible theta mappings of the two Abel or super-functions; some of them behave in other cool/interesting ways; but none of them seems particularly unique. There is also Gottfried's AIS sum solution. But there is no uniqueness, except when both fixed points are repelling, and then Henryk Trappmann's uniqueness proof is valid. - Sheldon tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 01/22/2016, 12:00 AM It seems hard to get Abelproducts ... Regards Tommy1729 « Next Oldest | Next Newest »

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