using sinh(x) ? - Printable Version +- Tetration Forum ( https://math.eretrandre.org/tetrationforum)+-- Forum: Tetration and Related Topics ( https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1)+--- Forum: Mathematical and General Discussion ( https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3)+--- Thread: using sinh(x) ? ( /showthread.php?tid=424) |

RE: using 2* sinh(x) ! - tommy1729 - 06/23/2010
(06/23/2010, 09:11 PM)tommy1729 Wrote: to go towards the taylor series you simply take the derivates you need of 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 RE: using sinh(x) ? - tommy1729 - 06/23/2010
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 RE: using sinh(x) ? - tommy1729 - 06/24/2010
(06/23/2010, 10:44 PM)tommy1729 Wrote: also worth mentioning i think : 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. RE: using sinh(x) ? - tommy1729 - 06/24/2010
(06/24/2010, 12:29 PM)tommy1729 Wrote:(06/23/2010, 10:44 PM)tommy1729 Wrote: also worth mentioning i think : 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 RE: using sinh(x) ? - bo198214 - 06/26/2010
(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. RE: using sinh(x) ? - tommy1729 - 06/26/2010
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. attracting fixed point of 2sinh - sheldonison - 06/27/2010
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 RE: using 2* sinh(x) ! - bo198214 - 06/27/2010
(06/21/2010, 09:14 PM)tommy1729 Wrote:(06/09/2010, 12:18 PM)tommy1729 Wrote: Ok, first let us verify that it is indeed an iteration of exp, i.e that indeed satisfies: and . neglecting some rules of properly evaluating limits we get . and because towards infinity gets arbitrarily close to . Basically thats the iteration equivalent of the Abel function Lévy proposes: where is the Abel function of (or in Lévy's case ). The superfunction is then (the inverse of ): which is the same as Tommy's superfunction. We can do something similar with not only or but with any function that does not deviate too much from exp at infinity (i.e. all functions such that ). Quote:as for the ROC i assume a plot for increasing n says more than a thousand words. I think you were not attentive when proposing to compute , already can not be computed in even sage's multiple precision arithmetic. RE: attracting fixed point of 2sinh - tommy1729 - 06/28/2010
(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. no , we get rid of the fixed point of 2 sinh by the iterates of exp in my formula. regards tommy1729 RE: attracting fixed point of 2sinh - sheldonison - 06/29/2010
(06/28/2010, 11:23 PM)tommy1729 Wrote:How is the fixed point removed? Starting with your equation for TommySexp, only interested in the "z" component, then let x=0,(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. This function gives the exact same results for all values of n as the following equivalent equation: Let , then the following equation is also exactly equivalent: 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 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 |