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+--- Thread: f( f(x) ) = exp(x) solved ! ! ! (/showthread.php?tid=228)
RE: f( f(x) ) = exp(x) solved ! ! ! - tommy1729 - 02/11/2009
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
RE: f( f(x) ) = exp(x) solved ! ! ! - bo198214 - 02/11/2009
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 \( f(f(x))=F(x) \).
RE: f( f(x) ) = exp(x) solved ! ! ! - tommy1729 - 02/11/2009
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
RE: f( f(x) ) = exp(x) solved ! ! ! - bo198214 - 02/11/2009
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
\( f\left(\frac{1}{2}+f^{-1}(x)\right)=e^{\frac{1}{2}}x\neq f_2\left(\frac{1}{2}+f_2^{-1}(x)\right) \)
RE: f( f(x) ) = exp(x) solved ! ! ! - andydude - 02/14/2009
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
RE: f( f(x) ) = exp(x) solved ! ! ! - tommy1729 - 02/17/2009
[/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