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 using sinh(x) ? tommy1729 Ultimate Fellow Posts: 1,365 Threads: 333 Joined: Feb 2009 06/23/2010, 09:31 PM (06/23/2010, 09:11 PM)tommy1729 Wrote: to go towards the taylor series you simply take the derivates you need of $\ln^{[n]} (\2sinh^{[z]}(exp^{[n]}(x)))$ and create the taylor series with real coefficients. then just plug in complex Z instead of x and your done. however the question is , what is the radius of that taylor series expanded at x = 0 or even elsewhere. if that radius is nonzero , we could probably extend further by using analytic continuation , or mittag-leffler expansion. and that is the way we go to the complex plane. perhaps an intresting note is that if the ROC is large , we could check if that taylor series has the same period as exp(x) ( 2pi i ). if it is indeed large enough and 1) the period is indeed 2pi i then the limit formula might hold for the complex plane and be equal to the taylor of the limit formula for the reals. 2) the period is not 2pi i then the limit formula will NOT hold for the complex plane and NOT be equal to the taylor of the limit formula for the reals. also , if the radius is 0 everywhere , despite unlikely , then if the limit converges for all complex , then the function must have some local or global fractal or semi-fractal properties. regards tommy1729 perhaps constructing a fourier series is 'better' , assuming we have the period property of course. 'better ' in the sense of potentially easier to compute numerically , easier to compute 'symbolicly' ( four coeff ) and easier to prove related statements conjectures and properties. this makes me doubt if taylor = four , if the period really exists and if the result only converges for x > 0 ... for reasons not yet explained ... just saying that fourier expansion might be intresting imho. if valid... slightly off topic but i often think there should be a new type of series expansion designed for tetration ... regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,365 Threads: 333 Joined: Feb 2009 06/23/2010, 10:44 PM also worth mentioning i think : let the base be a^(1/b) > sqrt(e) so that we can compute the superfunction of f(x) = a^(1/b)^x with my method. then consider t(x) = b(x + c) and its inverse m(x) = x/b - c m(f(t(x))) = m(a^(x+c)) = (1/b) a^(x+c) - c = (a^c / b) a^x - c if a , b and c are chosen such that (a^c / b) a^x - c > x we can compute the superfunction of (a^c / b) a^x - c by computing m(f^[z](t(x))). regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,365 Threads: 333 Joined: Feb 2009 06/24/2010, 12:29 PM (06/23/2010, 10:44 PM)tommy1729 Wrote: also worth mentioning i think : let the base be a^(1/b) > sqrt(e) so that we can compute the superfunction of f(x) = a^(1/b)^x with my method. then consider t(x) = b(x + c) and its inverse m(x) = x/b - c m(f(t(x))) = m(a^(x+c)) = (1/b) a^(x+c) - c = (a^c / b) a^x - c if a , b and c are chosen such that (a^c / b) a^x - c > x we can compute the superfunction of (a^c / b) a^x - c by computing m(f^[z](t(x))). regards tommy1729 if (a^c / b) a^x - c = x we get the intresting case of yet another fixpoint. associating functions without fixpoint with functions with 1 or more fixpoints is a intresting but complicated idea ... it raises questions. can we determine the number of superfunctions by that ? can we define uniqueness in some way ? for instance we associate g , f , k with g no fixpoint , f one fixpoint and k 2 fixpoints. how many solutions ? 1 ? 2 ? 3 ? 4 ? 5 ? 6 ? oo ? keep in mind that solution might be equal to eachother. e.g. expanding f at its fixpoint = expanding k at its second fixpoint. the key might be to notice that if a function with no fixp has the same superf as the associated with 1 , it may be unique. tommy1729 Ultimate Fellow Posts: 1,365 Threads: 333 Joined: Feb 2009 06/24/2010, 07:22 PM (06/24/2010, 12:29 PM)tommy1729 Wrote: (06/23/2010, 10:44 PM)tommy1729 Wrote: also worth mentioning i think : let the base be a^(1/b) > sqrt(e) so that we can compute the superfunction of f(x) = a^(1/b)^x with my method. then consider t(x) = b(x + c) and its inverse m(x) = x/b - c m(f(t(x))) = m(a^(x+c)) = (1/b) a^(x+c) - c = (a^c / b) a^x - c if a , b and c are chosen such that (a^c / b) a^x - c > x we can compute the superfunction of (a^c / b) a^x - c by computing m(f^[z](t(x))). regards tommy1729 if (a^c / b) a^x - c = x we get the intresting case of yet another fixpoint. associating functions without fixpoint with functions with 1 or more fixpoints is a intresting but complicated idea ... it raises questions. can we determine the number of superfunctions by that ? can we define uniqueness in some way ? for instance we associate g , f , k with g no fixpoint , f one fixpoint and k 2 fixpoints. how many solutions ? 1 ? 2 ? 3 ? 4 ? 5 ? 6 ? oo ? keep in mind that solution might be equal to eachother. e.g. expanding f at its fixpoint = expanding k at its second fixpoint. the key might be to notice that if a function with no fixp has the same superf as the associated with 1 , it may be unique. however i believe in ' conservation of fixed points ' for analytic solutions. for instance (a^c / b) a^x - c = x has no solution in nonnegative real x and real a,b,c > 1. thus we linked the superfunction of a function without a real fixpoint to another without a real fixpoint. a similar thing happens with bases below eta. we then linked the superfunction of a function with 2 real fixpoints to another with 2 real fixpoints. as you can see the amount of fixpoints remains constant hence ' conservation of fixed points '. also intresting might be another example of substitution exp(x^2 / 2) ^[z] => sqrt( exp^[z] (x^2)) and notice exp(x^2 / 2) = sqrt(exp(x^2)) = sqrt(e)^(x^2) which leads me to the ' gaussian question ' exp(- x^2 ) ^[z] = ?? however we have a fixpoint there maybe intresting for statistics and combinatorics ... regards tommy1729 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/26/2010, 12:04 PM (06/24/2010, 07:22 PM)tommy1729 Wrote: as you can see the amount of fixpoints remains constant hence ' conservation of fixed points '. I am not really sure how you come to this conlcusion. Take Base function: b*x and its superfunction: b^x: The base function has 1 fixed point (0), while the superfunction has fixed points depending on the value of b. Either two real, 1 real, or no real fixed point. I would "conservation of fixed points" rather to non-integer iterates, as the demand that the non-integer iterates should have the same fixed points as the base function. PS: Again the reminder: Dont quote whole posts! Always pick the particular quote you are replying to, or no quote at all, if it is a general reply. People are able to read the previous post, no need to repeat it for them. tommy1729 Ultimate Fellow Posts: 1,365 Threads: 333 Joined: Feb 2009 06/26/2010, 09:19 PM no quote well Bo , i meant the conservation of the amount of important fixed points when linking one superfunction to another. e.g. the link between e^x - 1 and eta^x , both have only 1 important fixpoint. the link didnt change the amount of important fixpoints. sheldonison Long Time Fellow Posts: 638 Threads: 22 Joined: Oct 2008 06/27/2010, 03:31 AM (This post was last modified: 06/29/2010, 02:02 AM by sheldonison.) speaking of fixed points, the superfunction of 2sinh has an attracting fixed point of 0 + 1.895494239i, on the imaginary axis. I'm not sure what the exact region of convergence is (probably fractal), but any time real(superfunction(z))=0, then the superfunction(z+1, +2, +3 ... +n) will converge to this attracting fixed point. This is because if real(z)=0, then real(sexp2(z)) will also equal zero, which helps in understand why there is an attracting fixed point. Regions where real(SuperFunction(z))=0 occur whenever imag(SuperFunction(z))=i*0.5*pi/ln(2), or i*(0.5+n)*pi/ln(2). I started to sketch out where the contours are. The SuperFunction(i*0.5*pi/ln(2)+x) converges to this fixed attracting point as x (real valued) increases. Other imaginary values of z close to 0.5*pi/ln(2) also converge to the attracting fixed point. This adds complication to the "base change" converting the super function of sexp2 to sexp_e in those regions where real(superfunction(z)) approaches zero, since 2sinh has this attracting fixed point, but exp_e doesn't. Moreover, such regions approach the real axis as z increases. - Sheldon bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/27/2010, 09:16 AM (06/21/2010, 09:14 PM)tommy1729 Wrote: (06/09/2010, 12:18 PM)tommy1729 Wrote: $\operatorname{TommySexp_e}(z,x)= \lim_{n \to \infty } \ln^{[n]} (\2sinh^{[z]}(exp^{[n]}(x)))$ Ok, first let us verify that it is indeed an iteration of exp, i.e that $f^z(x)=\text{TommySexp_e}(z,x)$ indeed satisfies: $f^{v+w}(x)=f^{v}(f^{w}(x))$ and $f^1(x)=\exp(x)$. neglecting some rules of properly evaluating limits we get $f^v(f^w(x))=\lim_{n\to\infty} \ln^{[n]} (2\sinh^{[u]}(\exp^{[n]}(\ln^{[n]}(2\sinh^{[v]}(\exp^{[n]}(x))))))=\lim_{n\to\infty} \ln^{[n]}(2\sinh^{u+v}(\exp^{[n]}(x))=f^{u+v}(x)$. and $f^1(x)=\lim_{n\to\infty} \ln^{[n]}(2\sinh(\exp_^{[n]}(x))=\exp(x)$ because towards infinity $2\sinh$ gets arbitrarily close to $\exp$. Basically thats the iteration equivalent of the Abel function Lévy proposes: $\beta(x) = \lim_{n\to\infty} \alpha(\exp^{[n]}(x)) - \alpha(\exp^{[n]}(x_0))$ where $\alpha$ is the Abel function of $2\sinh$ (or in Lévy's case $\exp(x)-1$). The superfunction $\sigma$ is then (the inverse of $\beta$): $\lim_{n\to\infty} \alpha(\exp^{[n]}(x)) - \alpha(\exp^{[n]}(x_0))=y$ $\sigma(y)=x=\lim_{n\to\infty} \log^{[n]}(\alpha^{-1}(y+\alpha(\exp^{[n]}(x_0))))$ $\sigma(y)=\lim_{n\to\infty} \log^{[n]}(2\sinh^{[y]}(\exp^{[n]}(x_0))))$ which is the same as Tommy's superfunction. We can do something similar with not only $2\sinh$ or $\exp(x)-1$ but with any function that does not deviate too much from exp at infinity (i.e. all functions $h$ such that $\log^{[n]}(h(\exp^{[n]}(x)))\to \exp(x)$). Quote:as for the ROC i assume a plot for increasing n says more than a thousand words. plot n = 1 -> 100. z = 1/2 $\ln^{[n]} (\2sinh^{[z]}(exp^{[n]}(x)))$ I think you were not attentive when proposing to compute $\exp^{[100]}$, already $\exp^{[6]}(0)$ can not be computed in even sage's multiple precision arithmetic. tommy1729 Ultimate Fellow Posts: 1,365 Threads: 333 Joined: Feb 2009 06/28/2010, 11:23 PM (06/27/2010, 03:31 AM)sheldonison Wrote: speaking of fixed points, the superfunction of 2sinh has an attracting fixed point of 0 + 1.895494239i, on the imaginary axis. I'm not sure what the exact region of convergence is (probably fractal), but any time real(superfunction(z))=0, then the superfunction(z+1, +2, +3 ... +n) will converge to this attracting fixed point. This is because if real(z)=0, then real(sexp2(z)) will also equal zero, which helps in understand why there is an attracting fixed point. Regions where real(SuperFunction(z))=0 occur whenever imag(SuperFunction(z))=i*0.5*pi/ln(2), or i*(0.5+n)*pi/ln(2). I started to sketch out where the contours are. The SuperFunction(i*0.5*pi/ln(2)+x) converges to this fixed attracting point as x (real valued) increases. Other imaginary values of 0 close to 0.5*pi/ln(2) also converge to the attracting fixed point. This adds complication to the "base change" converting the super function of sexp2 to sexp_e in those regions where real(superfunction(z)) approaches zero, since 2sinh has this attracting fixed point, but exp_e doesn't. Moreover, such regions approach the real axis as z increases. - Sheldon no , we get rid of the fixed point of 2 sinh by the iterates of exp in my formula. regards tommy1729 sheldonison Long Time Fellow Posts: 638 Threads: 22 Joined: Oct 2008 06/29/2010, 03:18 AM (This post was last modified: 06/29/2010, 05:05 PM by sheldonison.) (06/28/2010, 11:23 PM)tommy1729 Wrote: (06/27/2010, 03:31 AM)sheldonison Wrote: ....This adds complication to the "base change" converting the super function of sexp2 to sexp_e in those regions where real(superfunction(z)) approaches zero, since 2sinh has this attracting fixed point, but exp_e doesn't. Moreover, such regions approach the real axis as z increases. - Sheldon no , we get rid of the fixed point of 2 sinh by the iterates of exp in my formula. regards tommy1729How is the fixed point removed? Starting with your equation for TommySexp, only interested in the "z" component, then let x=0, $\operatorname{TommySexp_e}(z)= \lim_{n \to \infty } \ln^{[n]} (\operatorname{2sinh}^{[z]}(\exp^{[n]}(0)))$ This function gives the exact same results for all values of n as the following equivalent equation: $\operatorname{TommySexp_e}(z)= \lim_{n \to \infty } \ln^{[n]} (\operatorname{2sinh}^{[z+\operatorname{SuperFunction^{-1}(\exp^{[n]}(0))]}} )$ Let $k=\operatorname{SuperFunction^{-1}(\exp^{[n]}(0))-n$, then the following equation is also exactly equivalent: $\operatorname{TommySexp_e}(z)= \lim_{n \to \infty } \ln^{[n]} (\operatorname{SuperFunction(z+k+n))$ As n goes to infinity, k converges quickly to the approximate value, k=0.067838366 (for n=0: k=-0.0734181, for n=1: k=0.0663658, for n>=2: k=0.067838366) Next, this equation can be used to compute TommySexp at z=0.5i*pi/ln(2), in the region where the attracting fixed point is, for increasing values of n $\ln^{[n]}(\operatorname{SuperFunction}(0.5i*\pi/\ln(2)+k+n))$ as n increases. My guess is that everywhere in the region around the i=0.5*pi/ln(2) value, where the Superfunction converges to the attracting fixed point, TommySexp will converge to the repelling fixed point of e^z. with all derivatives going to zero. For example, consider TommySexp for n=0, n=2, and n=10 at i0.5pi/ln(2). For n=0, real(x)=0, img(f)=i0.905, and for real(x)=1,img(f)=i1.572. This is a well defined analytic function. The function flattens out for n=2, and by the time n=10, it is converging towards the fixed point of "e"=0.318+i1.337 (hence my guess that all derivatives go to zero). Notice how the graph of TommySexp, for n=0, n=2 and n=10 flattens out and converges to the fixed point of "e" as n increases. - Sheldon « Next Oldest | Next Newest »

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