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f(f(x)) = exp(x) + 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: f(f(x)) = exp(x) + x (/showthread.php?tid=403) |
f(f(x)) = exp(x) + x - tommy1729 - 12/12/2009 let f(x) be Coo. f(f(x)) = exp(x) + x what is f(x) ? does the (only) fixpoint at - oo help ? can f(x) be entire ? is there a solution f(x) such that f(x) has no fixpoint apart from - oo ? can f(x) be expressed in terms of tetration ? is there a solution f(f(x)) = exp(x) + x with f(x) E Coo , f(x) mapping all reals to reals and f(x) having no fixpoint apart from - oo ? does " a solution f(f(x)) = exp(x) + x with f(x) E Coo , f(x) mapping all reals to reals and f(x) having no fixpoint apart from - oo " imply that all derivatives are strictly positive reals ? can f(x) be expressed in terms of pentation ? does the substitution y = 1/x help ? ( trying to 'move the fixpoint' but problems occur e.g. exp(1/x) has a singularity at 0 ! ... on the other hand perhaps considering a certain angle towards the singularity it might work to give a real-analytic solution ? ) does the strategy lim n-> oo f(f(x)) = exp(x) + x + 1/n work ? how about the carleman matrix method ? this seems like a difficult problem ... regards tommy1729 added exponential iterates - bo198214 - 12/12/2009 Indeed So one could think that one can not apply regular iteration. However as Tommy already suggested there is a fixed point at (complex) infinity. Well, the function is not analytic there, but for regular iteration it suffices that the function has an asymptotic powerseries development (approaching the fixed point in some sector) or even merely that the function is asymptotically real differentiable. This condition is met, the asymptotic derivatives for So we can apply the limit formula for regular iteration with multiplier 1 (derived from Lévy's formula for the Abel function): (*) This formula converges only very slowly, but it suffices to get 2 or 3 digits to plot a graph of Another way is to consider the analytic conjugate The conjugation formula shows also how to compute the inverse of RE: added exponential iterates - mike3 - 12/12/2009 I'm curious: where are the branch points for the positive-real-count fractional iterations such as RE: added exponential iterates - tommy1729 - 12/13/2009 (12/12/2009, 12:25 PM)bo198214 Wrote: Indeed that is intresting , though i dont know why you call it 'analytic conjugate' ? funny thing is i considered the above things in reverse order :p back to business , i wanted to point out that perhaps the above approach might also work for real-iterates of exp(z) ?!? basicly bo replaced the fixpoint from - oo to 0 , and i consider the possibility of doing the same with the 2 complex fixpoints of exp(z) towards 0. ( if succesfull i bet we arrive at kouznetsov's solution but with more proven properties ) so we look for strictly increasing functions T(x) resp f(x) with T(x) = f(x) ° exp(x) ° f°-1(x) with a fixpoint at 0. ( i didnt check but ) f(x) might just be simple like f(x) = (x - fp1)^a (x - fp2)^a realpoly(x) note fp1 and fp2 are the solutions of exp(z) = z and are eachother conjugate ( which is very important here ) of course also the correct branches of f°-1(x) resp f(x) are important. regards tommy1729 RE: added exponential iterates - mike3 - 12/14/2009 Most likely such a thing will turn out as equivalent to the usual regular iteration of exp from the repelling fixed point. This does not give real values for tetration at the real axis and does not have singularities (i.e. cannot satisfy RE: added exponential iterates - BenStandeven - 12/14/2009 Using the simplest conjugation function produces something of a mess; to simplify, I'll consider the conjugate F of exp(pi/2 x) instead; the fixed points are +/-I, so we can conjugate with w=z^2 + 1. The inverse function is z=sqrt(w-1); then I get: F(x) = exp(pi/2 * sqrt(x-1))^2 + 1 = exp(pi * sqrt(x-1)) + 1. Note that this function has a branch point at 1, but the conjugation function moves [0, oo] to [1, oo], so we must have two different functions near zero, one for each choice of branch cut. They will presumably yield different regular iteration functions, which correspond to the two entire regular iteration functions for exp. RE: f(f(x)) = exp(x) + x - bo198214 - 12/14/2009 (12/13/2009, 06:26 PM)tommy1729 Wrote: basicly bo replaced the fixpoint from - oo to 0 , and i consider the possibility of doing the same with the 2 complex fixpoints of exp(z) towards 0. Yes, I was also playing with this idea. However when you map a simply connected region with two fixed points on its boundary biholomorphically and real-analytically such that both fixed points go to 0, then you probably have a branch cut on the real axis (consider a sickle with the two fixed points at its ends, if you bent the fixed points to 0, there will be a slit or overlapping on the real axis up to the point where one part of the boundary intersects the real axis). (12/14/2009, 07:18 AM)BenStandeven Wrote: Using the simplest conjugation function produces something of a mess; to simplify, I'll consider the conjugate F of exp(pi/2 x) instead; the fixed points are +/-I, so we can conjugate with w=z^2 + 1. The inverse function is z=sqrt(w-1); then I get: F(x) = exp(pi/2 * sqrt(x-1))^2 + 1 = exp(pi * sqrt(x-1)) + 1. Very good example, it also illustrates my point above. The function F is non-real on (0,1) it has a cut from 1 to -oo (as you described) RE: f(f(x)) = exp(x) + x - tommy1729 - 12/14/2009 (12/14/2009, 09:52 AM)bo198214 Wrote:(12/13/2009, 06:26 PM)tommy1729 Wrote: basicly bo replaced the fixpoint from - oo to 0 , and i consider the possibility of doing the same with the 2 complex fixpoints of exp(z) towards 0. still i believe in the basic idea ... didnt say it would be easy ![]() regards tommy1729 |