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 f( f(x) ) = exp(x) solved ! ! ! tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 02/11/2009, 05:32 PM bo198214 Wrote:Existence was shown by Kneser. Uniqueness is a mostly unresearched yet. Sure is that it is not unique by analyticity alone. well , i assume the existence of tetration ( knesers exp(F(x)) = F(x+1)) leads to existence of f(f(x)) = exp(x). however , im not sure that f(f(x)) = exp(x) is not unique by analyticity alone ? care to explain ? regards tommy1729 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/11/2009, 06:04 PM tommy1729 Wrote:well , i assume the existence of tetration ( knesers exp(F(x)) = F(x+1)) leads to existence of f(f(x)) = exp(x). Yes, as we assume $F$ to be strictly increasing we can take the inverse function $F^{-1}$. Then you can verify that $f(x):=F\left(\frac{1}{2}+F^{-1}(x)\right)$ satisfies $f(f(x))=\exp(x)$ Quote:however , im not sure that f(f(x)) = exp(x) is not unique by analyticity alone ? care to explain ? Now if we have one analytic solution $F$ then we have a lot of other analytic solutions given by $F_\theta(x):=F(x+\theta(x))$ for any 1-periodic analytic function $\theta$ (prove that $F_\theta(z+1)=\exp(F_\theta(x))$ too!) If we make the amplitude of those $\theta$ sufficiently small, then $x+\theta(x)$ is strictly increasing and $F_\theta$ is too. Finally: $f_\theta(x):=F_\theta\left(\frac{1}{2}+F^{-1}_\theta(x)\right)$ is another analytic solution of $f(f(x))=F(x)$. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 02/11/2009, 08:05 PM bo198214 Wrote:Quote:however , im not sure that f(f(x)) = exp(x) is not unique by analyticity alone ? care to explain ? Now if we have one analytic solution $F$ then we have a lot of other analytic solutions given by $F_\theta(x):=F(x+\theta(x))$ for any 1-periodic analytic function $\theta$ (prove that $F_\theta(z+1)=\exp(F_\theta(x))$ too!) If we make the amplitude of those $\theta$ sufficiently small, then $x+\theta(x)$ is strictly increasing and $F_\theta$ is too. Finally: $f_\theta(x):=F_\theta\left(\frac{1}{2}+F^{-1}_\theta(x)\right)$ is another analytic solution of $f(f(x))=F(x)$. i had expected such a reply. but i disagree. let F( real ) map to reals. and let f( real ) map to reals. assuming those are satisfied , i feel that F(x+1) = exp(F(x)) should also satisfy F(x+1/2) = f(F(x)) where f(f(x)) = exp(x) , if it wants to be tetration. in general F(x+a) = f_a(F(x)) where f_a satisfies f_a(((... a times ...(x)))) = exp(x) should be satisfied. furthermore the inverse of F might be multivalued !! but that doesnt mean f(x) is all the possible results of F(1/2 + invF(x)) taking that into account , its probably clear that i will only accept examples of 2 distinct analytic solutions f(x) that map all reals to a subset of reals and satisfy f(f(x)) = exp(x). im not trying to be difficult. but this is important. working with F(x) seems like overkill , instead i focus on f : f(f(x)) = exp(x). ( as an analogue : there are multiple functions that satisfy f(x+1) = e*f(x) but that doesnt mean there are multiple functions that are a solution to exp( log(x) + 1 ) ( being e*x ) ) regards tommy1729 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/11/2009, 10:56 PM I dont really see what you mean. tommy1729 Wrote:taking that into account , its probably clear that i will only accept examples of 2 distinct analytic solutions f(x) that map all reals to a subset of reals and satisfy f(f(x)) = exp(x). And I gave you exactly those two distinct analytic functions: $f_1(x)=F(\frac{1}{2}+F^{-1}(x))$ and $f_2(x)=F_\theta(\frac{1}{2}+F^{-1}_\theta(x))$ Quote:( as an analogue : there are multiple functions that satisfy f(x+1) = e*f(x) but that doesnt mean there are multiple functions that are a solution to exp( log(x) + 1 ) ( being e*x ) ) Let one solution be $f(x)=\exp(x)$ and let the other solution be some other solution be $f_2=\exp\left(x+\frac{1}{2\pi}\sin(2\pi x)\right)$. Then $f(1+f^{-1}(x))=ex=f_2(1+f_2^{-1}(x))$ but $f\left(\frac{1}{2}+f^{-1}(x)\right)=e^{\frac{1}{2}}x\neq f_2\left(\frac{1}{2}+f_2^{-1}(x)\right)$ andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 02/14/2009, 04:18 AM bo198214 Wrote:the equation $f(f(x))=\exp(x)$ was not yet really solved, though we collected several approaches on that question on the forum.Agreed. I should have said "has several approaches" instead of "solved". bo198214 Wrote:What can be solved by regular iteration is $f(f(x))=\exp(x)-1$, though there are convergence issues, I think this is what you refer to. Well, because $F(x) = w^x - 1$ and $G(x) = w^{x/w} = e^x$ are topologically conjugate (where $w = -W(-1)$), I kind of skipped a step and thought of F (which has a fixed point of 0) and G (which has a fixed point of w) at the same time. Also, I do understand the general idea of tommy's method, to make a function with a fixed point at 0, as a limit of a function with a fixed point not at zero. It makes sense. I just need to sit down at look over the math for a week or so, to convince myself that it all works, and that it gives new insight to the problem. Andrew Robbins tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 02/17/2009, 12:30 AM [/quote] Well, because $F(x) = w^x - 1$ and $G(x) = w^{x/w} = e^x$ are topologically conjugate (where $w = -W(-1)$), I kind of skipped a step and thought of F (which has a fixed point of 0) and G (which has a fixed point of w) at the same time. Also, I do understand the general idea of tommy's method, to make a function with a fixed point at 0, as a limit of a function with a fixed point not at zero. It makes sense. I just need to sit down at look over the math for a week or so, to convince myself that it all works, and that it gives new insight to the problem. Andrew Robbins [/quote] ive been absent for a few days , but im thinking ( working ? ) on related ideas. such as convergeance acceleration. or a zero at other points then 0. i dont think my fixed point at 0 gives a convex solution to tetration ... i will post a conjecture soon. regards tommy1729 « Next Oldest | Next Newest »

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