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 using sinh(x) ? tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 11/28/2012, 08:07 PM (This post was last modified: 11/28/2012, 08:11 PM by tommy1729.) Another thing I would like to talk about is the following property. I will start with an example. Let x be a real. Consider $f(n,x)=sin(n^2 x)/n$. If n goes to oo the limit gives $f(oo,x)=f(x)=0$. Now intuitively one would expect about the derivative with respect to x that $D f(x) dx =$ lim n-> oo $D f(n,x) dx = 0.$ Clearly $D f(x) dx = D 0 dx = 0.$ However $D sin(n^2 x)/n = n cos(n^2 x)$ so lim n-> oo $D f(n,x) dx$ =/= 0 ! This is an important and classic lesson in math. So many of our iterations used here require formal and carefull analysis. If you are not convinced notice F(n,x) = g^[n](Q(n,x)) IS something that occurs in the majority of limits related to tetration including e.g. tommysexp , basechange , interpolation methods , matrix methods , iterations to improve on riemann mappings , ... ! I do not know if $D^m F(n,x) dx = D^m F(x) dx$ has been proven here for any method ? Also note that I took x to be a real. So this is about Coo functions defined over the reals but for analytic functions defined over the complex number the issue is even bigger. Also note that in my example both f(n,x) and f(x) were Coo , in fact even analytic. So being analytic or Coo does not solve this issue. I think we need to work on this. --- For those clueness on how to such things , I point out that putting boundaries on values or signs and using the intermediate value theorem frequently does miracles. tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 12/17/2012, 04:34 PM I was wondering about the following conjectured uniqueness criterion for real-analytic half iterates ( easily generalized to all iterates btw) : (based upon my sinh method) for all real x: (D is the derivative operator) D exp^[1/2](x) > 0 D^2 exp^[1/2](x) > 0 exp^[1/2](x) - ln(ln(2sinh^[1/2](exp(exp(x))))) > 0. It seems intuitive , especially considering how the sinh method works. However tetration is sometimes (or often) counterintuitive. A stronger conjecture would be : D exp^[1/2](x) > 0 D^2 exp^[1/2](x) > 0 ln(ln(2cosh^[1/2](exp(exp(x))))) > exp^[1/2](x) > ln(ln(2sinh^[1/2](exp(exp(x))))) Although this requires some concept of cosh^[1/2]. ( I once considered a method based upon cosh but made a mistake ) Im not even sure about the uniqueness of these uniqueness criterions. For instance how this might be related to D^n exp^[1/2](x) > 0 for integer n with n>1 and/or curvature and/or length conditions. Intuitively it seems like f(x) = exp^[1/2](x) computed by the sinh method gives the shortest path/length between f(a) and f(b) for sufficiently large real a and b but im not even sure if that is a selfconsistant statement for ANY exp^[1/2](x). Also clearly shortest length and smallest curvature relate alot. Recently I posted a new thread that involved the point w := f ' (w) = 1 which might relate to uniqueness. Clearly there is still work. And I must admit its not all clear to me yet. Tetration is calculus on drugs regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 02/10/2015, 12:02 AM To extend the idea to smaller bases see this link : http://math.eretrandre.org/tetrationforu...21#pid7621 regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 04/22/2015, 12:27 PM Numerical test Use the Taylor expansions at 3,4,5 and see if they agree on the values of f(3),f(4),f(5) for say f= semi-exp. Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 04/23/2015, 04:49 PM Recently i started 2 threads about analytic like 2sinh type functions. Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 10/26/2015, 12:56 AM The 2sinh method is for bases larger than exp(1/2). There are easy analogues for bases > eta. For instance replace 2sinh with t(x) = exp(x) - 1/2 [ exp(-x) + exp( (3 - 2e) x) ]. Notice t(x) is close to 2sinh , is bijective from R to R and has Taylor t(x) = 0 + e x + O(x^2). Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 11/26/2015, 11:56 PM There are many ways to solve f(f(x)) = exp(x) or g(g(x)) = 2sinh(x). If g < f then g is uniquely so. Also using the 2sinh method on g gives us f. NOTICE it is impossible to directly solve for g in the 2sinh method. So we must use g < f. This implies a bijection between g and f , by the 2sinh or equivalently the inequality. Since there exist both analytic half iterates for both 2sinh and exp , for Every f there is a g and vice versa , we get the next BIG conclusion ; There MUST exist analytic solution of type 2sinh method. --- That is nice. But one wonders if f and g related as above must 1) both be analytic or both not. ? In other words is the bijection actually univalent ? I kinda repeat myself , but I assumed it belongs here better , and it was not understood by most. Regards tommy1729 sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 11/27/2015, 03:41 PM (This post was last modified: 11/27/2015, 07:34 PM by sheldonison.) (11/26/2015, 11:56 PM)tommy1729 Wrote: There are many ways to solve f(f(x)) = exp(x) or g(g(x)) = 2sinh(x). If g < f then g is uniquely so. Also using the 2sinh method on g gives us f. .... But one wonders if f and g related as above must 1) both be analytic or both not. If g=2sinh(z), and f=exp(z), One can start with the analytic Abel function $\alpha_g(z)\;\;$ and the superfunction for g. $\alpha^{-1}_g(z)$. For example, the abel and superfunction for 2sinh(z) can be developed using Koenig's method. Now this gives us $g^{[0.5]}(z) = \alpha_g^{-1}(\alpha_g(z)+0.5)$ We wish to generate the superfunction for f(z) from $\alpha_g\;\alpha_g^{-1}$, with the equation below. In the case at hand this limit for the superfunction of f converges at the real axis, but it is not analytic; it does not converge in the complex plane. So far, this seems always be the case for arbitrary exponentially growing functions, f and g. $\alpha^{-1}_f(x)\; = \; \lim_{n \to \infty} \; f^{[-n]}( \alpha^{-1}_g(x+n))\;\;\;$the superfunction for f(z) So the resulting half iterate of f developed in this way is not analytic either, even thought it converges at the real axis $f^{[0.5]}(z) = \alpha_f^{-1}(\alpha_f(z)+0.5)$ - Sheldon tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 11/28/2015, 01:21 PM I think you are confused. Sorry. But it seems you are not agreeing with what I said ... For dubious reasons. Again the 2sinh starts at real x > 0. Then we take the Limit of the continuations of the n th step iteration. NOT the continuation of the Limit of the n th step iteration. Afterwards we can take analytic continuation of THAT limit. Depending on the super/Abel you use for 2sinh this works infinitely AND fails infinitely often. AND .. Not or. Regards Tommy1729 sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 11/29/2015, 03:03 PM (This post was last modified: 11/29/2015, 03:03 PM by sheldonison.) Given half iterates, f(f(x))=exp(x), and g(g(x))=2sinh(x), which uniquely satisfy the Op's relationship, then the obvious conjecture from my post would be that only one of the two (f,g) can be analytic. If one of them is analytic, then the other is infinitely differentiable and nowhere analytic, so that the other does not converge to its Taylor series anywhere. This is kind of an interesting conjecture; that Tommy might be inclined to agree with based on earlier posts in this thread (see Tommy's post#61 for example). As far as "dubious reasons" go, I was merely answering Tommy's question, "both be analytic or both not.", as well as searching for cool mathematical truths. - Sheldon « Next Oldest | Next Newest »

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