f( f(x) ) = exp(x) solved ! ! !
#11
bo198214 Wrote:Pretending that there are no open questions does not help here.
Especially for you tommy I list them again:

  1. Do the approximations of \( \exp^{1/2} \) converge? I.e. we have approximations of \( \exp \) these are the functions \( g_n(x)=\exp(x)-\exp(-n x^2) \). These converge pointwise to \( \exp \) (except at 0). Then we take the regular iteration of these functions \( f_n={g_n}^{1/2} \). The question is whether they converge to anything.
  2. If they converge to \( f \), is \( f \) a differentiable or even analytic function?
  3. If they converge to \( f \), does \( f \) really satisfy \( f(f(x))=\exp(x) \)?
As I already told some pictures would make a good start if proofs could not provided yet.

question # n will be answered by "#n )"

1) yes they do converge.

because , clearly g_n does.

f_n(f_n) = g_n

its clear that f_n cannot be to far away from f_(n-1) and f_(n+1).

e.g. because both must be strictly rising for x > 1 and both are larger than the line y = x for x > 1.

in fact that distance ( not near 0 at least ) is bounded by

O ( exp(x) - g_n(x) )


2) since they converge , f is analytic.

because both sequences f_n and g_n are continu and analytic.

think of it like this :

( disregarding the neigbourhood at 0 for convenience )

suppose the opposite :

f_n is not analytic for some n.

since g_n is analytic and regular iterations of strictly rising functions preserve analytic properties(*). ( * if we also have f(0) = 0 which we do , we can construct taylor series , which are always analytic were they converge )

thus for all finite n , f_n is analytic.

thus only candidate for not being analytic is f_oo.

but that is absurd.

since lim n-> oo f_oo(x) - f_n(x) = 0.

if a function is not analytic it cannot have a non-positive real distance from an an analytic function everywhere !

3) yes f(f(x)) = exp(x) is really satisfied , at least for x = / = 0.

this is because lim n -> oo of g_n =

for x =/= 0 => exp(x)

for x = 0 => 0.


----------

there are still other questions than those 3.

for instance , is my method equivalent to another method ?

since my solution is analytic it can only be equivalent to other ( potentially ) analytic methods of course ...

another remark is , that my method applies to base e tetration.

it relates to other bases too , but how is not exactly clear yet.

a 'change of base formula' is not yet found.

also intresting are series and integrals of tetration.

etc etc etc

however im working on those too.

( it seems i will probably agree with most that has been written about it here )

one of the problems encountered in f(f(x)) = a^x

is that i might have an annoying zero : a^x = x. for a real x.

( so that my 'bumpy method' does not apply , at least not without a modification )

regards

tommy1729

" Statisticly , i dont exist " tommy1729
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Messages In This Thread
f( f(x) ) = exp(x) solved ! ! ! - by tommy1729 - 02/05/2009, 12:09 AM
RE: f( f(x) ) = exp(x) solved ! ! ! - by andydude - 02/05/2009, 09:17 PM
RE: f( f(x) ) = exp(x) solved ! ! ! - by bo198214 - 02/06/2009, 03:22 AM
RE: f( f(x) ) = exp(x) solved ! ! ! - by bo198214 - 02/06/2009, 06:02 PM
RE: f( f(x) ) = exp(x) solved ! ! ! - by bo198214 - 02/07/2009, 12:37 PM
RE: f( f(x) ) = exp(x) solved ! ! ! - by bo198214 - 02/08/2009, 12:15 PM
RE: f( f(x) ) = exp(x) solved ! ! ! - by tommy1729 - 02/08/2009, 03:59 PM
RE: f( f(x) ) = exp(x) solved ! ! ! - by bo198214 - 02/08/2009, 10:08 PM
RE: f( f(x) ) = exp(x) solved ! ! ! - by andydude - 02/14/2009, 04:18 AM

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