Fixed point and fractional iteration: if $F(k)=k$ then $F^{1\over n}(k)$ is another fixed point of $F$

Question

My knowledge of the fixed points and iteration equals zero, same for the notation and terminology but I really need to know if this deduction has trivial errors or is really as nice as it seems.

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I would like to prove the following:

Notation 1 - Given a function $f:A\rightarrow A$ define the set $Fix(f)\subseteq A$ as the set of $f$'s fixed points

$Fix(f):=\{\phi:f(\phi)=\phi\}$

Notation 2 - Given a function $f:A\rightarrow A$ and the definition of function composition $\circ$ define the function $f^n$ by recursion

$i)$ $f^0:=\operatorname{id}_A$

$i)$ $f^{n+1}:=f\circ f^n$

Definition 1 - Given a function $F:X\rightarrow X$, the " $1\over n$-iterate" of $F$ is a function $\Psi:X\rightarrow X$ with this property

$\forall x(x\in X) (\Psi^n(x)=F(x))$

I guess that we can write $\Psi=F^{1\over n}$

To Prove - If $k\in X$ is a fixed point of $F$ and exists a fucntion $\Psi$ such that $\Psi^n=F$ then $F^{1\over n}(k)$ is a fixed point of $F$

$$k\in Fix(F)\implies \forall n(n\ge1)(F^{1\over n}(k)\in Fix(F))$$

Proof 1 - For a fixed $n\gt 1$ define $\lambda:=\Psi(k)=F^{1\over n}(k)$

$\lambda:=\Psi (k)=\Psi(\Psi^n(k))$ because $k=\Psi^n(k)$

$\lambda=\Psi^n(\Psi(k))$ because iterates commute

$\lambda=\Psi^n(\lambda)$ because $\Psi(k)=\lambda$ by definition

Since $\Psi^n=F$ by definiton we conclude that $\lambda=F(\lambda)$ and thus

$$\lambda\in Fix(F)$$

Anyways this proof seems weird to me... I feel like there is something missing: I want to prove that for every natural number (greater than zero), if $F^{1\over n}$ exists, $F^{1\over n}(k)$ is a fixed point so maybe I need to use induction but I really don't know how I could do it

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Questions

$1)$ - Is this proof correct? If yes and it is a known result, can you add some info about it?

$2)$ - Is it possible to use induction for the proof? Or it is useless?

$3)$ - If the proof is correct, is this a result that can be strengthened? In fact it seems to me that the real generalized result would be something like $$k\in Fix(F)\implies \forall q(q\in\Bbb Q\land 0\lt q \lt 1)(F^{q}(k)\in Fix(F))$$

Answer 1

I think the main thing missing from the proof is a description of functions admitting an $n^{\rm th}$ root. There is nothing strange (that I can see) in the steps of your proof but it begs the question - for what $F$ is the set $\{ \Psi : \Psi^n = F \}$ nonempty?

Starting with $X = \mathbb{R}^k$, I believe any linear transformation admits a ${1\over n}$-iterate $\Psi$, but that still leaves the field very open.

Answer 2

I do not believe there is a short answer that will adress all your questions.

However consider the analytic solutions and the complex numbers.

Then we might have $f(z_0)=z_0$ where $z_0$ is a complex number and $f$ is an analytic function.

Now consider the function $F(z+1) = f(F(z))$.

Now your question is equivalent to $if$ $F(q)=z_0 $ where $q$ is finite , is it Always true that $F(q + {1\over m} + n) = F(q+{1\over m})$ for positive integers $n,m$ ?

And the answer is NO !

( In fact this has been mentioned and investigated by tommy1729 , Gottfried and probably all frequent posters at the tetration forum )

Here is the proof :

$F(q + {1\over m} + n) = F(q+{1\over m})$ implies that $F(q + {a\over m} + an) = F(q+{a\over m})$ for positive integer $a$.

( Notice the subtle $an$ term , not just $n$ ; the $a$ th iterate of $F({1\over m} + n)$ => $F({a\over m} + an)$. You might want to think about this. )

As a consequence since $q$ is finite and $an$ is an integer ... and $f$ is analytic then by analytic continuation $F$ ... and the fact that the rationals are dense in the reals ... $F$ is periodic with period $an$.

But superfunctions tend not to be periodic.

( superfunction means inverse abel function for those unfamiliar with the term )

Notice that $F(z+\theta(z))$ where $\theta$ is a one-periodic function satisfies the same functional equation but the same problem/solution (proof steps above) holds ... unless of course ( the "new" ) $q$ is no longer finite.

I can find you some related threads on the tetration forum if you like.

You know , periodicity is an important concept the re-occurs often. This is not Always clear from (basic) traditional education apart from trigonometry.

Guess that gives you some insight why your idea is a bit too optimistic.

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(ordinary reader may stop reading , this is a specialized comment ) As for tetration The kneser solution has the primary fixpoints of exp at +/- oo $i$. And there at oo $i$ the function is indeed flat and periodic. Hence there it is true what you assume ... by " design of the superfunction " , however I assume the fixpoints have no finite $q$ for the kneser function.

Formalizing that might be difficult. I considered thinking about univalent zones , but since exp is a chaotic map that seems difficult.

Maybe that would make a good question at the tetration forum.

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Answer 3

I think the Op's proof is correct assuming both functions are entire. However, even if $F(x)$ is entire, the fractional iterate is in general not entire. The Op's result does not hold if the fractional iterate is not entire

$$F^{o\frac{1}{n}}(x)\;\;\;\;h(x)=F^{o \frac{1}{2}}(x)\;\;\;\;h(h(x))=F(x)$$

If the half iterate is not entire then there are points for which $F(x)=h(h(x))$ is only by analytic continuation, and not by direct computation. So definition (1), $\forall x\;\;h(h(x))=F(x)$ does not hold $\forall x$, since the half iterate has multiple values depending on the path, unless the half iterate is also entire, which pretty much rules out all non-trivial half iterates!

I wanted to generate a counter example that was as simple as possible. Lets start with a function that has three fixed points, $(0,\pm 1)$, and we will develop the half iterate at the fixed point at the origin. I also wanted to avoid the parabolic case see Will Jagy's half iterate of sin(x) post which occurs when the first derivative of the fixed point=1, so I chose a positive 1st derivative at the fixed point at the origin of 4. Avoiding the parabolic case gives guaranteed convergence so then the formal half iterate would have guaranteed convergence near the origin, and would have a first derivative of 2.

So here is my counter example, for $F(x)=4x-3x^3$, for which the half iterate is $h(x)$, and $F(x)$ has fixed points of $(0,\pm 1)$. For all points within some radius of convergence, $h(h(x))=F(x)$, but eventually we get to the singularity of $h(x)$ which gives it a defined radius of convergence. But weird stuff happens when the radius of convergence of $h(x)$ is larger than $h(1)$ where 1 is one of the other fixed points. So, below, I post the formal half iterate of $F(x)$, which has a radius of convergence of $\frac{16}{9}\approx 1.78$. Here, is my counter example, where all three fixed points for $F(x)$ has $h(x)$ within its analytic radius of convergence, and even $h(1), h(-1)$ are within the radius of convergence so the half iterate seems to be completely unambiguous at these points. And yet this leads to a clear counter example.

$$F(1)=1$$ $$h(1)=1.66125776701924932137$$ $$h(1.66125776701924932137)=1$$ $$F(h(1))=F(1.66125776701924932137)=-7.10907369782592055937<>h(1)$$

However, even though the radius of convergence of the Taylor series of $h(x)>h(1)$, when you look at a number 1.2399067, $h(1.2399067)=16/9$, so $h(h(x))$ has a smaller radius of convergence is 1.2399067. Of course, this smaller radius of convergence has a singularity that cancels by analytic continuation since $F(x)$ is entire. And this is what allows weird stuff to happen... where in the complex plane, $h(x)$ at the fixed point of $F(x)$ has multiple values depending on the path, even though $F(x)$ is entire and is always well defined independent of the path. So $h(x)$ can only by fully defined by path dependent analytic continuation in the complex plane.

{h(x)=
+x   *2 
-x^ 3*3/10 
-x^ 5*27/850 
-x^ 7*243/44200 
-x^ 9*4391901/3862196000 
-x^11*4097709/15835003600 
-x^13*263696194479501/4216940633698000000 
-x^15*4352793841907459397/276378289132566920000000 
-x^17*0.00000408866926284292783744 
-x^19*0.00000108632410179569368855 
-x^21*0.000000293954426198467149790 
-x^23*0.0000000807297320769806555906 
-x^25*0.0000000224441951265113300999 
-x^27*0.00000000630439537551479828510 
-x^29*0.00000000178645419922952969101 
-x^31*0.000000000510065370119009481553 
-x^33*1.46596264762260914617 E-10 
-x^35*4.23776939074721938452 E-11 
-x^37*1.23135265881044955235 E-11 
-x^39*3.59433071863758569107 E-12  ....
}

Besides the formal Taylor series, one may generate the half iterate by using the identity:: $$h(x) = F \circ h \circ F^{-1} = \lim_{n \to \infty} F^{n} \circ h \circ F^{-n} = \lim_{n \to \infty} F^{n}(2\cdot F^{-n}(x)) $$

The $h = F \circ h \circ F^{-1}$ equation also shows that the half iterate radius of convergence is tied to the radius of convergence of $F^{-1}(x)$, which is $\frac{16}{9}$. This may be calculated from where $\frac{d}{dx}F(x)=0$ which is at $x=\pm \frac{2}{3}$ where $F(\pm \frac{2}{3})=\pm\frac{16}{9}$. Here is a graph of h(x), at the real axis, from -16/9 to +16/9, which is out to the radius of convergence, where the derivative of h(x) goes to infinity. The singularity is cancelled out when iterating $h(h(x))$ since the derivative of $h(x)=0$ where $h(x)=16/9$.

One more image showing $h(x), h^{o2}(x)=F(x), h^{o3}(x), h^{o4}(x)$ from 0 to 1. Odd iterations are in purple, and even iterations are in red. Notice that $h^{o3}(1)$=-7.10907 as expected, as opposed to $h(1)$=1.6612577. We see that $h(1)$ has multiple values, depending on how many times we have iterated $h(x)$. But F(x) is entire, so no matter how many times we iterate F(1)=1. Also notice that this still contradicts the Op's proof, since for analytic functions, the half iterate is multiple valued, apparently infinitely valued in this case, depending on the path in the complex plane.