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 using sinh(x) ? tommy1729 Ultimate Fellow Posts: 1,352 Threads: 327 Joined: Feb 2009 02/28/2010, 12:22 AM im thinking about using asymt of exp(x) to solve tetration. perhaps not the best one , but sinh(x) comes to my mind. if we take the half-iterate of sinh(x) by using taylor series , we get a good approximation of the half-iterate of exp(x) for x large. if this good approximation is analytic in say [e^e,e^e^e^e] we could use that interval and take logs or exp of it to compute the half-iterate for [-oo,+oo] up to a relatively high precision. in fact , i assume , we can choose our precision if the good approximation is entire , by using bigger intervals and taking logs of it ( as somewhat done above ). we might even consider the mittag-leffler expansion to avoid problems with non-analytic issues ( thus taking mittag-leffler expansion of the formal taylor series ) thats basicly the idea , but as said maybe sinh(x) isnt the best , on the other hand its probably the easiest. furthermore some want - and me too actually - that half-iterates have non-negative (2+n)th derivatives and strictly positive zeroth , first and second derivatives. that last restriction troubles me , especially when i try to get closer to exp(x) than sinh(x) by using function that have non-negative derivatives ... getting conditions on a particular n'th derivate isnt hard but the whole problem is more troublesome ( maybe someone knows a solution to this ? ) i didnt mention that all the above ofcourse is about functions going from R -> R , its obvious , but i just mention it to avoid potential confusion. also every approximation of exp(x) should equal x at x = 0 only and be larger than id(x). hope its clear. maybe you considered this once too ? greetings my fellow euh ... tetrationalists regards tommy1729 ps : showing that all derivatives are positive can sometimes be easy , but in general its hard , see for example " li's criterion " for the RH. maybe its easy here and i missed a trivial thing ... ( im getting old ? :p ) tommy1729 Ultimate Fellow Posts: 1,352 Threads: 327 Joined: Feb 2009 03/06/2010, 12:35 PM (02/28/2010, 12:22 AM)tommy1729 Wrote: im thinking about using asymt of exp(x) to solve tetration. perhaps not the best one , but sinh(x) comes to my mind. if we take the half-iterate of sinh(x) by using taylor series , we get a good approximation of the half-iterate of exp(x) for x large. no replies ... is there a problem with the half-iterate of sinh(x) ? does its taylor series radius = 0 ? is the inner radius of the mittag-leffler = 0 ? ... bo198214 Administrator Posts: 1,373 Threads: 89 Joined: Aug 2007 03/07/2010, 11:19 AM (This post was last modified: 03/07/2010, 11:22 AM by bo198214.) Quote:is there a problem with the half-iterate of sinh(x) ? does its taylor series radius = 0 ? is the inner radius of the mittag-leffler = 0 ? Its not really a problem, but indeed the convergence radius of the fractional iterates would be 0. This is usually the case with parabolic ( f'(p)=1, f(p)=p ) development. But its not really a problem as you can use the Levy iteration formula (which though may be too slow) or the Ecalle formulas for parabolic iteration. tommy1729 Ultimate Fellow Posts: 1,352 Threads: 327 Joined: Feb 2009 03/09/2010, 01:31 PM ok , i found my silly mistake !! i wrote sinh(x) but i meant 2*sinh(x) !! so in the OP replace sinh(x) with 2*sinh(x). and then we dont have a parabolic fixpoint ! and we do have an asympt. sorry ! regards tommy1729 bo198214 Administrator Posts: 1,373 Threads: 89 Joined: Aug 2007 03/10/2010, 11:38 AM (03/09/2010, 01:31 PM)tommy1729 Wrote: so in the OP replace sinh(x) with 2*sinh(x). and then we dont have a parabolic fixpoint ! and we do have an asympt. And what do you do with this asymptote then? tommy1729 Ultimate Fellow Posts: 1,352 Threads: 327 Joined: Feb 2009 03/11/2010, 12:50 AM (03/10/2010, 11:38 AM)bo198214 Wrote: (03/09/2010, 01:31 PM)tommy1729 Wrote: so in the OP replace sinh(x) with 2*sinh(x). and then we dont have a parabolic fixpoint ! and we do have an asympt. And what do you do with this asymptote then? i take the half-iterate of it by using taylor series expanded at 0. f(x) is a taylor series , f(f(x)) = 2*sinh(x). f(x) is then a good asympt for the half iterate of exp(x) for large x. let g(g(x)) = exp(x) then g(x) = lim k-> oo log log log log ...( k times ) (f( e^e^e^...( k times ) ^ x)) something like that i think. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,352 Threads: 327 Joined: Feb 2009 03/14/2010, 01:07 AM (03/11/2010, 12:50 AM)tommy1729 Wrote: (03/10/2010, 11:38 AM)bo198214 Wrote: (03/09/2010, 01:31 PM)tommy1729 Wrote: so in the OP replace sinh(x) with 2*sinh(x). and then we dont have a parabolic fixpoint ! and we do have an asympt. And what do you do with this asymptote then? i take the half-iterate of it by using taylor series expanded at 0. f(x) is a taylor series , f(f(x)) = 2*sinh(x). f(x) is then a good asympt for the half iterate of exp(x) for large x. let g(g(x)) = exp(x) then g(x) = lim k-> oo log log log log ...( k times ) (f( e^e^e^...( k times ) ^ x)) something like that i think. regards tommy1729 all clear ? tommy1729 Ultimate Fellow Posts: 1,352 Threads: 327 Joined: Feb 2009 04/24/2010, 11:47 PM i think it works for the real line. however in the neighbourhood of the 2 fixed points of exp(x) it will fail. i think so because for fixed point y : log log f exp exp y => log log f(y) => equals y only if f(y) = y so i assume a "button-like" region of correctness. bo198214 Administrator Posts: 1,373 Threads: 89 Joined: Aug 2007 05/31/2010, 05:52 AM (This post was last modified: 05/31/2010, 07:03 AM by bo198214.) Actually Paul Lévy [1] showed how to obtain an iteration of e^x if we have an iteration of e^x-1. Say $\beta$ is an Abel function of e^x-1, then $\alpha(x)-\alpha(x_0)=\lim_{n\to\infty} \beta(\exp^{[n]}(x)) - \beta(\exp^{[n]}(x_0))$ is an Abel function of $e^x$. This should also work for beta being the Abel function of $2\sinh(x)$. This approach is actually equivalent to the "change of base" approach we considered here on the forum, also Walker [2] used a similar method. (But I am in the moment to lazy to detail how exactly they imply each other.) It is still open whether it is analytic, but it is proven to be infinitely differentiable in [2]. [1] Lévy, P. (1927). Sur l'itération de la fonction exponentielle. C. R., 184, 500–502. [2] Walker, P. L. (1991). Infinitely differentiable generalized logarithmic and exponential functions. Math. Comput., 57(196), 723–733. tommy1729 Ultimate Fellow Posts: 1,352 Threads: 327 Joined: Feb 2009 06/01/2010, 10:58 PM (This post was last modified: 06/01/2010, 11:00 PM by tommy1729.) (05/31/2010, 05:52 AM)bo198214 Wrote: Actually Paul Lévy [1] showed how to obtain an iteration of e^x if we have an iteration of e^x-1. Say $\beta$ is an Abel function of e^x-1, then $\alpha(x)-\alpha(x_0)=\lim_{n\to\infty} \beta(\exp^{[n]}(x)) - \beta(\exp^{[n]}(x_0))$ is an Abel function of $e^x$. This should also work for beta being the Abel function of $2\sinh(x)$. This approach is actually equivalent to the "change of base" approach we considered here on the forum, also Walker [2] used a similar method. (But I am in the moment to lazy to detail how exactly they imply each other.) It is still open whether it is analytic, but it is proven to be infinitely differentiable in [2]. [1] Lévy, P. (1927). Sur l'itération de la fonction exponentielle. C. R., 184, 500–502. [2] Walker, P. L. (1991). Infinitely differentiable generalized logarithmic and exponential functions. Math. Comput., 57(196), 723–733. but is that equivalent to my solution ? do we get the same half-iterate for exp(x) ? thanks for the post and references. regards tommy1729 « Next Oldest | Next Newest »

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