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.

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


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


$1)$ - 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 $$k\in Fix(F)\implies \forall q(q\in\Bbb Q\land 0\lt q \lt 1)(F^{q}(k)\in Fix(F))$$

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    $\begingroup$ I don't know whether this is relevant, but I've one time stumbled on the problem of self-intersecting flows in the iteration of the exponentiation, which means in some mysterious way, that the same complex point could be arrived by different iteration-heights (and so should be a fixed point) but also been left on different pathes. I do not yet understand that phenomen (if it were one at all, I still scratch my head). It was for instance discussed on this post in the tetration-forum math.eretrandre.org/tetrationforum/… and other places. $\endgroup$ Jan 9, 2015 at 11:18
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    $\begingroup$ MphLee, sorry, it's just over my head at this days; I just wanted to give that reminder of an older discussion (which also had let me headscratching until today... ;-)) I hope someone else shall help out here. $\endgroup$ Jan 11, 2015 at 20:35
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    $\begingroup$ One more reminder. Of course, if we have $F$ has a fixpoint, say $F(t_0)=t_0$ and $\Psi$ as some $1/n$ iterate of $F$ with the property that $t_1 = \Psi(t_0) \ne t_0$ but $\Psi^{on}(t_0)=t_0$ then what you say should be obvious, and a set of points $t_0,t_1,...,t_n=t_0$ occurs. But even more should be expected for the generalization of $\Psi$ for continuous iterations: then instead of a set of points $t_0,t_1,...,t_n=t_0$ we should have a continuous curve (if in the complex plane). I considered one time a bizarre effect when then curve became fractal (and possibly some of (....) $\endgroup$ Jan 11, 2015 at 22:04
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    $\begingroup$ (...) your mentioned/assumed properties possibly do no more hold. I'd marked that discussion with the term "equator" and made some pictures for the tetration-forum, where also Sheldon made some nice comments. Perhaps this is interesting as well. See math.eretrandre.org/tetrationforum/… for instance, but all posts around it are related. Closing remark: sorry that the discussion of tetration comes so dominant here, I found it surely interesting for the more general case of iterated functions as properly adressed by your question. $\endgroup$ Jan 11, 2015 at 22:08
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    $\begingroup$ Definition (1) does not hold $\forall x$ if the half iterate is not entire... I will post exactly an example for which this is the case. Then, we can have the case where $h^{o2}(x)=F(x)$ within some radius of convergence. But call one of the other fixed points $L_1$, $h(L_1)=z \land $h(z)=L \land F(z)<>z$ $\endgroup$
    – Sheldon L
    Jan 13, 2015 at 12:04

3 Answers 3


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.

(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.

  • $\begingroup$ Tommy considered tetration for other numbers then the complex ones. But they suffered the same problem. Im not sure about non-complex numbers and iterations of other functions ( not exp ). $\endgroup$
    – mick
    Jan 10, 2015 at 23:05
  • $\begingroup$ mhh,$x_0$ is the fixed point of $f$, consider the $x_0$-based superfunction of $f$, yeah, it is the function that satiefies $F(0):=x_0$ and $F(z+1)=f(F(x))$ (we have that $F(z)=f^{\circ z}(x_0)$). Since $x_0$ fixed point implies that $F(n)=F(n+1)$ for all the naturals, my question is equivalent to aks if $F(n)=F(n+1)$ implies $F({1\over m})=F({1\over m}+1)$... while the third question asks if it is possible to conclude that $F(z+1)=F(z)$ always.Said that, thank you for the answer, I really should study alot more about periodicity and fixed points since now I'm very confused ;D. $\endgroup$
    – MphLee
    Jan 11, 2015 at 12:33
  • $\begingroup$ anyways I don't get the proof, can you rephrase it in a simpler way? I don't get some steps: I don't know nothing about analycity and the fact that the rationals are dense in the reals so I can't get the implication. $\endgroup$
    – MphLee
    Jan 11, 2015 at 20:36
  • $\begingroup$ I've edited your question, tell me if the edit is correct. $\endgroup$
    – MphLee
    Jan 12, 2015 at 8:34
  • $\begingroup$ Btw I don't get how you introduce the term $a$ $\endgroup$
    – MphLee
    Jan 12, 2015 at 8:38

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.

  • $\begingroup$ Ye, I didn't write it clear. The existence of $\Psi$ is given before. So the theorem I want to prove is of the form: If $\Psi$ exists and if a fixed point $k$ exists then... Anyways the question about the set $\{ \Psi : \Psi^n = F \}$ is interesting and I'd like to open a new question about it, if you are not interested in it. $\endgroup$
    – MphLee
    Jan 9, 2015 at 8:00
  • $\begingroup$ Assuming you have a $\Psi$, then the proof is very clear. You use the assumption on $k$, associativity of function composition (I'm sure in some esoteric setting this could fail, but by our everyday definition function evaluation is associative), then the definition of $\lambda$. This seems airtight. $\endgroup$
    – Titus
    Jan 9, 2015 at 8:28
  • $\begingroup$ As to your other questions: 2) you don't need induction for the proof since you are assuming a $\Psi$ exists for each integer $n$. 3) I'm not familiar with your notation $\wedge$, but if $\Psi$ is an $n^{\rm th}$ `root' of $F$, then any function in $\{ \Phi : \Phi^n = F^k ~~ \forall ~~ x \in X \}$ will also fix $k \in Fix(F)$, so $k$ is also a fixed point of $F^{k/n}$. $\endgroup$
    – Titus
    Jan 9, 2015 at 8:33
  • $\begingroup$ $\land$ means "and"... it means that q is ration AND the disequality holds. I don't get the answer to the 3rd question. Anyways can you glue all the comments and add it inside your answer, adding a proof to your last statement? $\endgroup$
    – MphLee
    Jan 9, 2015 at 9:51

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.

+x   *2 
-x^ 3*3/10 
-x^ 5*27/850 
-x^ 7*243/44200 
-x^ 9*4391901/3862196000 
-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 $h(x)=16/9$. graph of half iterate

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. graph of h, h^2, h^3, h^4

  • $\begingroup$ Damn, I see it but I can't believe it (hah just kidding...) actually I see it but I can't understand why O.O. I need time to even understand what I don't understand. Thank you for the answer. I will comment soon. $\endgroup$
    – MphLee
    Jan 13, 2015 at 15:29
  • $\begingroup$ @mick, MphLee; I really like your question, btw, it just took me awhile to realize that MphLee's proof was correct iff the fractional iterate was entire ... and then I got really interested. I wanted to post as simple and counterexample as I could. I probably should've skipped the Taylor series, and stuck with the limit definition of $h(x)$, but I generated the formal Taylor series solution of $h(x)$ first ... $\endgroup$
    – Sheldon L
    Jan 13, 2015 at 15:40
  • $\begingroup$ Do this actually implies that the existence of a function with $\Psi^n=F$ is equivalent to/implies having $\Psi$ and $F$ entire? Because the existence of a function with $\Psi^n=F$ already implies that $\Psi^n(x)=F(x)$ for all the $x$ in $dom(F)=dom(\Psi)=X$(domain of F).. and the fact that $\Psi^n$ exists means automatically that $cod(F)=\Psi^n[X]\subseteq \Psi^{n-1}[X]\subseteq...\Psi^2[X]\subseteq\Psi[X]\subseteq X$. (continue) $\endgroup$
    – MphLee
    Jan 13, 2015 at 18:33
  • $\begingroup$ (continue) Could this mean that this assumption is enough strong to imply entireness? But Entireness uses alot of structures such as $X=\Bbb C$ and holomorphy... while my deduction should run on a really wider class of maps over a generic set $X$...So at this point I wonder what are the real implication of such general assumption... $\endgroup$
    – MphLee
    Jan 13, 2015 at 18:39
  • $\begingroup$ Note 2 In the costruction using the limit you do use the abel function and the super function... anyways how do we know that is legit to do that?What if the superfunction or the abel function are not invertible? the formula will only hold on a subset of the complex where it gets invertible... is this related with the failure of the definition (1) adn thus with the failure of the theorem? $\endgroup$
    – MphLee
    Jan 13, 2015 at 18:41

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