f(f(x)) = exp(x) + x - Printable Version

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f(f(x)) = exp(x) + x - Printable Version

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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 $F(x)=e^x+x$ (which I will call here added exponential) is a very interesting function as it has no complex fixpoints. If there was some complex fixpoint $z$ then $e^z+z=z$ , means $e^z=0$ which is never true.
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 $z\to -\infty$ of $F$ are: $F'(-\infty)=\lim_{z\to -\infty} e^z+1=1$ and all higher derivatives are $F^{(n)}(-\infty) = 0$ . I.e. $F$ has asymptotically the same derivatives as the identity function.

So we can apply the limit formula for regular iteration with multiplier 1 (derived from Lévy's formula for the Abel function):
(*) $F^{[t]}(z) = F^{[n]}( t F^{[-n+1]}(z) + (1-t) F^{[-n]}(z) )$
This formula converges only very slowly, but it suffices to get 2 or 3 digits to plot a graph of $f=F^{[0.5]}$ (blue):
[attachment=662]

Another way is to consider the analytic conjugate $M = \exp \circ F \circ \log$ . $M(x)=\exp(x+\log(x))=e^x x$ is the (what I call) multiplied exponential, which moves the fixed point at $-\infty$ of $F$ to the fixed point 0 of $M$ . This function is well regularly iterable at 0 (with powerseries and limit formula) at then $F^{[t]}=\log\circ M^{[t]} \circ\exp$ .

The conjugation formula shows also how to compute the inverse of $F(x)=x+e^x$ which is needed e.g. in the limit formula. The inverse is $F^{[-1]} = \log\circ M^{[-1]}\circ \exp$ , i.e.

$F^{[-1]}(x) = \log(W(e^x))$ , where $W$ is the Lambert function.

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 $F^{1/2}(z)$ on the complex plane? I tried a graph on the complex plane for z in the square from -5-5i to +5+5i and I noticed two cutlines (at $t \pm i \pi$ it appears (where t is real)), but they seem to go across the entire plane. So are these "real", i.e. are there branch points on them (or outside the graphing square), or are they just an artifact of the algorithm and the fractional iterate is actually entire (i.e. we could analytically continue it out of the region near 0 into the whole plane)?

RE: added exponential iterates - tommy1729 - 12/13/2009

(12/12/2009, 12:25 PM)bo198214 Wrote: Indeed $F(x)=e^x+x$ (which I will call here added exponential) is a very interesting function as it has no complex fixpoints. If there was some complex fixpoint $z$ then $e^z+z=z$ , means $e^z=0$ which is never true.
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.

(((snip part of quote)))

Another way is to consider the analytic conjugate $M = \exp \circ F \circ \log$ . $M(x)=\exp(x+\log(x))=e^x x$ is the (what I call) multiplied exponential, which moves the fixed point at $-\infty$ of $F$ to the fixed point 0 of $M$ . This function is well regularly iterable at 0 (with powerseries and limit formula) at then $F^{[t]}=\log\circ M^{[t]} \circ\exp$ .

The conjugation formula shows also how to compute the inverse of $F(x)=x+e^x$ which is needed e.g. in the limit formula. The inverse is $F^{[-1]} = \log\circ M^{[-1]}\circ \exp$ , i.e.

$F^{[-1]}(x) = \log(W(e^x))$ , where $W$ is the Lambert function.

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 $F(-1) = 0$ ) so it doesn't really seem to make sense for a generalization of tetration of base e to real or complex towers.

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.

( 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)

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.

( 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)

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)

still i believe in the basic idea ...

didnt say it would be easy $Wink$

regards

tommy1729