I found the following equation on some web page I cannot remember, and found it interesting:


Out of curiosity I tried to solve it, but realized that I do not have a clue how to approach such an iterative equation except for trial and error. I also realized that the solution might not be unique, from the solution of a simpler problem

$$f(f(x)) = x$$

which has, for example, solutions $f(x) = x$ and $f(x) = \frac{x+1}{x-1}$.

Is there a general solution strategy to equations of this kind? Can you perhaps point me to some literature about these kind of equations? And what is the solution for $f(f(x))=\cos(x)$ ?

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    $\begingroup$ Are you looking for continuous solutions or arbitrary maps? $\endgroup$ Commented Mar 9, 2010 at 17:49
  • 7
    $\begingroup$ There is a collection of references and links for the general problem of solving $f(f(x))=g(x)$ for $f$, given $g$, at reglos.de/lars/ffx.html $\endgroup$ Commented Mar 9, 2010 at 23:02
  • 1
    $\begingroup$ After Sergey Ivanov's solution, I wonder how many discontinuity points a solution of $f(f(x)) = \cos(x)$ must have? $\endgroup$
    – Anonymous
    Commented Mar 10, 2010 at 1:22
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    $\begingroup$ Sergei's impossibility argument only applies to real functions, right? (Because it uses monotonicity.) Cosine is a perfectly healthy complex function, so you could ask the same question in the complex setting. As far as I can tell no one has answered that version of the question conclusively yet. $\endgroup$ Commented May 29, 2012 at 3:48
  • 2
    $\begingroup$ @Noah: for complex solution, see my reply mathoverflow.net/questions/17605/how-to-solve-ffx-cosx/… $\endgroup$ Commented May 29, 2012 at 11:57

12 Answers 12


The half-iterate of a function can be found by expressing its superfunction in a form of Newton series:

$$f^{[1/2]}(x)=\sum_{m=0}^{\infty} \binom {1/2}m \sum_{k=0}^m\binom mk(-1)^{m-k}f^{[k]}(x)$$

Where $f^{[k]}(x)$ means k-th iterate of $f(x)$ This series converges if two criteria are met:

  1. The superfunction(flow) of f(x) grows not faster than an exponent

  2. Runge phenomenon does not appear.

There is a number of strategies to combat Runge phenomenon which are outside of this answer's scope. It is worth noting though that trying to find a half iterate of the function $f(x)=\cos x$ leads to this Runge swamp and one needs to employ one of the mentioned techniques to achieve convergence.

Opposite case is with the function $f(x)=\sin x$. The superfunction is limited by $\pm 1$ and the series converges without any problem.

Below is a plot of half-iterate of $\sin x$, obtained with this formula. It is periodic with the same period as $\sin x$. The blue curve is the half-iterate, and the red curve is the half-iterate, repeated twice, and we can see that it is indeed very similar to sine function.


This plot is made from the first 50 terms of the above series.

This formula for the half-iterate can be used to find not only half-itertes but any real (or even complex!) iterate of a function by substituting the needed value instead of 1/2.

The formula can be also written in the following forms:

$$f^{[s]}(x)=\lim_{n\to\infty}\binom sn\sum_{k=0}^n\frac{s-n}{s-k}\binom nk(-1)^{n-k}f^{[k]}(x)$$

$$f^{[s]}(x)=\lim_{n\to\infty}\frac{\sum_{k=0}^{n} \frac{(-1)^k f^{[k]}(x)}{(s-k)k!(n-k)!}}{\sum_{k=0}^{n} \frac{(-1)^k }{(s-k) k!(n-k)!}}$$

There are also some other formulas giving the same result.

  • $\begingroup$ This is a truly great answer. But you use"Newton series" and "superfunction" in a way that will be misleading. By the "Newton series" you mean expansion of (1+x)^(1/2) in powers of x, not "Newton's iteration method", which is how some people read it. By "Superfunction" you mean an abstract iteration operator which when applied to functions composes them with other functions. This formula is a little hard to make precise. $\endgroup$
    – Ron Maimon
    Commented Jul 30, 2011 at 6:19
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    $\begingroup$ Newton series is the following series: $$f(x) = \sum_{k=0}^\infty \binom{x-a}k \Delta^k f\left (a\right)$$ It is equivalent to the above formula (just expand the deltas). Wikipedia: en.wikipedia.org/wiki/Finite_difference#Newton_series $\endgroup$
    – Anixx
    Commented Jul 30, 2011 at 6:47
  • $\begingroup$ A frustrating aspect of this construction is $f^{[0]}(x) = x.$ The infinite sum of all the coefficients of $x$ is 0. This must be true, $\sin x$ is periodic. The partial sum of a finite number of terms gives a nonzero coefficient for $x.$ This nonzero coefficient is necessary for uniform convergence on, say, the interval $(0,\pi - \varepsilon), \; \varepsilon > 0.$ On the other hand, by the time we have, say, $x > 2 \pi,$ the nonzero coefficient on $x$ itself (where all the other terms are periodic functions of $x$) makes matters worse, the partial summed function increases without bound. $\endgroup$
    – Will Jagy
    Commented Jul 30, 2011 at 20:22
  • $\begingroup$ One can take the sum not from k=0 but from k=1. The limit will be the same, and the coefficient of x will be zero. But this will also converge very slowly. $\endgroup$
    – Anixx
    Commented Jul 30, 2011 at 21:26
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    $\begingroup$ @Anixx Are you aware of any other similar formula, possibly with faster convergence? Computing large factorials is a bit of a problem numerically... $\endgroup$
    – Semola
    Commented Jan 23, 2019 at 22:15

There are no continuous solutions. Since the cosine has a unique fixed point $x_0$ (such that $\cos x_0=x_0$), it should be a fixed point of $f$. And f should be injective and hence monotone (increasing or decreasing) in a neighborhood of $x_0$. Then $f(f(x))$ is increasing in a (possibly smaller) neighborhood of $x_0$ while the cosine is not.

As for discontinuous ones, there are terribly many of them ($2^{\mathbb R}$) and you probably cannot parametrize them in any reasonable way. You can describe them in terms of orbits of iterations of $\cos x$, but I doubt this would count as a solution of the equation.

UPDATE: Here is how to construct a solution (this is technical and I might overlook something).

Let X be an infinite set and $g:X\to X$ is a map, I am looking for a sufficient conditions for the existence of a solution of $f\circ f=g$. Define the following equivalence relation on $X$: $x$ and $y$ are equivalent iff $g^n(x)=g^m(y)$ for some positive integers $m$ and $n$. Equivalence classes will be referred to as orbits (the term is wrong but I don't know what is a correct one). Two orbits are said to be similar is there is a bijection between them commuting with $g$. If $Y$ and $Z$ are two similar orbits, one can define $f$ on $Y\cup Z$ as follows: on $Y$, $f$ is that bijection to $Z$, and on $Z$, $f$ is the inverse bijection composed with $g$.

So if the orbits can be split into pairs of similar ones, we have a desired $f$. Now remove from the real line the fixed point of $\cos$ and all its roots ($\pi/2$ and the like). Then, if I am not missing something, in the remaining set $X$ all orbits of $\cos$ are similar, so we can define $f$ as above. Define $f$ so that $0$ has a nonempty pre-image (that is, the orbit containing $0$ should be used as $Z$ and not as $Y$). Finally, map the fixed point of $\cos$ to itself, and the roots of $\cos$ to some pre-image of $0$.

EDIT (by BP July 17, 2023): When removing the fixed point $t$ of $\cos$ above, one also needs to remove the grand orbit of $t$ under $\cos$ and define $f$ separately on this grand orbit, which is $\pm t + \mathbb{Z} \pi$. One possibility is to set $$f(\pm t + n\pi) = \begin{cases} t, &\text{ if $n=0$;}\\ -t, &\text{ if $n$ is even and nonzero; and}\\ t+2\pi, &\text{ if $n$ is odd.} \end{cases}$$

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    $\begingroup$ Im confused as to why x_0 must be a fixed point of f. Can't it be an involution (order two) point of f? $\endgroup$
    – Ben Weiss
    Commented Mar 9, 2010 at 21:06
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    $\begingroup$ @Ben: then there will be two fixed points of $f\circ f$. $\endgroup$ Commented Mar 9, 2010 at 21:09
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    $\begingroup$ since $f$ is continuous, if it has no fixed point you have e.g. $f(x)>x$ for all $x$, and thus $g(x)>x$ also. $\endgroup$
    – Homology
    Commented Apr 30, 2010 at 10:20
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    $\begingroup$ See my new answer. $\endgroup$
    – Anixx
    Commented May 29, 2012 at 3:10
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    $\begingroup$ Often the term "grand orbit" is used for the equivalence class you describe, to differentiate from the usual (forward) orbit, and the backward orbit. $\endgroup$ Commented Apr 28, 2020 at 13:06

There are a truly enormous number of solutions, if one only wants the solution to work on an interval. Indeed, one can find solutions to $f(f(x)) = g(x)$ for any function $g$ defined on an interval.

Specifically, I claim that if $g:[a,b] \to \Bbb R$, then there are $2^{|\Bbb R|}$ many functions $f$ from $\Bbb R$ to $\Bbb R$ with $f(f(x)) = g(x)$ for all $x$ in $[a,b]$.

One such solution $f$ is obtained as follows. First choose a $z$ such that $[a,b]$ and $[a + z, b + z]$ are disjoint. Now let $f(x) = x + z$, for $x$ in $[a,b]$, and $f(x) = g(x - z)$, for $x$ in $[a + z, b + z]$. Thus, $f(x)$ first translates $x$ to another interval, when $x$ is in $[a,b]$, and then $f$ computes $g$ of the reverse translate, when $x$ is not in $[a,b]$. So $f(f(x)) = g(x)$.

When $g$ is continuous, then this function $f$ will be continuous also, and can be made total by linearly extending.

More generally, if $h$ is bijection of $[a,b]$ with another interval $[a',b']$ disjoint from $[a,b]$, then let $f(x) = h(x)$ for $x$ in $[a,b]$, and $f(x) = g\big(h^{-1}(x)\big)$ for $x$ in $[a',b']$. It follows that $f(f(x)) = g(x)$. And since there are $2^{|\Bbb R|}$ many such functions $h$, there are similarly many functions $f$ satisfying the equation.

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    $\begingroup$ OK but I don't see how to generalise this trick so that it works for all x in the reals. $\endgroup$ Commented Mar 9, 2010 at 15:33
  • $\begingroup$ I've just asked an explicit question about this: mathoverflow.net/questions/17614/solving-ffxgx $\endgroup$ Commented Mar 9, 2010 at 15:43
  • $\begingroup$ Can you please show us a plot of a solution for cosine on interval say $[-\pi/2,\pi/2]$? $\endgroup$
    – Anixx
    Commented Aug 23, 2011 at 23:44
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    $\begingroup$ Sorry if I'm misunderstanding this: You said "When g is continuous, then this function f will be continuous also", but the answer with the most upvotes claims that there is no continous solution. $\endgroup$
    – Ovi
    Commented Jun 25, 2017 at 4:34
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    $\begingroup$ The solution in my answer only works on an interval, and this is an important difference. $\endgroup$ Commented Jun 25, 2017 at 4:46

Near a fixed point of $\cos(x)$ use the method of Schröder (1871) ...

Best historical reference: Daniel S. Alexander, A History of Complex Dynamics from Schröder to Fatou and Julia (1994).


It seems that contrary to some other answers a continuous solution can be constructed.

First of all we interpolate with Newton series the flow of function $\cos(\cos(z))$:

$$\phi_{1/2}(x,z)=\cases { \arccos^{[x]}(z), & \text{if } x < 0 \cr \cos^{[x]}(z), & \text{if } x \ge 0 } $$

$$\phi_{1}(x,z)=\sum_{m=0}^\infty \binom{x/2+1}{m} \sum_{k=0}^m (-1)^{k-m} \binom{m}{k} \phi_{1/2}(k-1,z)$$

We interpolate from the first integer point where the value is real, i.e. from x=-1.

We now obtain the approximation of the other half-flow of cos x by taking arccos on the above function:


We know that the flow of cos(x) should coincide with the first function in even integers and with the second function in odd integers.

So we make a stub of the flow following this knowledge (we also want that its absolute value to be monotonous).

$$\phi(x,z)=\frac{1}{2} \left((-1)^{x+1}+1\right) (\phi_{1}(x,z)-\text{FP})+\frac{1}{2} \left((-1)^x+1\right) (\phi_{2}(x,z)-\text{FP})+\text{FP}$$

where FP is the cosine fixed point.

This function coincides with the flow in integer points but still disagrees in between. To get a real flow we have to take a limit of repeated arccosine on the our stub:

$$\Phi(x,z)=\lim_{n\to\infty} \arccos^{[n]} (\phi(x+n,z))$$

Numerically this limit converges quite fast. If the limit exists, it by definition, satisfies the equation

$$ \cos(\Phi(x,z))=\Phi(x+1,z)$$

so it is the true flow.

The above can be illustrated by the graphic:

enter image description here

Here upper semi-flow (flow of cos(cos z)) ) is blue, lower semi-flow is red, real part of the flow is yellow, imaginable part of the flow is green. All flows are taken as point z=1.

Following this we can build a graphic of half-iterate of cosine $\Phi(1/2,z)$:

enter image description here

Here blue is the real part and red is the imaginary part.

We can verify that the half-iterate repeated twice $\Phi(1/2,\Phi(1/2,z))$ (blue) follows cosine (red) quite well at positive half-periods, and anywhere the cone is positive (that is, on the imaginary axis as well):

enter image description here

I think this coincodes with the answer by Gerald Edgar above. A modified function, iterated twice gives cosine in all real axis:

enter image description here

This is a true half-iterate of cosine, which works on the whole real axis, producing exactly cosine:

enter image description here

But as has been noted by Joel David Hamkins above, there is infinite number of such solutions, none of which work for the whole complex numbers.

This function can be considered though as the true solution on the complex plane if interpreted as a multi-valued function. To do this, take the function on the each interval and analytically extend it to the whole complex plane.

A mathematica notebook that produces the above is as follows:

$PlotTheme = None; 
f[x_, z_] := If[x >= 0, Nest[Cos, z, 2*x], Nest[ArcCos, z, -2*x]]
n := 30
s := 15
Ni[x_, z_] := 
 Sum[Binomial[x + 1, m]*
   Sum[(-1)^(k - m)*Binomial[m, k]*f[k - 1, z], {k, 0, m}], {m, 0, n}]
Semi2[x_, z_] := Ni[x/2, z]
Semi1[x_, z_] := ArcCos[Semi2[x + 1, z]]
FP := Evaluate[N[FixedPoint[Cos, 1.]]]
a := 21
Flow2[x_, z_] := 
 FP + (Semi2[x, z] - FP)*(((-1)^x + 1)/2) + (Semi1[x, z] - 
     FP)*(((-1)^(x + 1) + 1)/2)
FL[x_, z_] := Nest[ArcCos, Flow2[x + a, z], a]
Plot[{Semi1[x, 1], Semi2[x, 1], Re[FL[x, 1]], Im[FL[x, 1]]}, {x, -5, 
  5}, AspectRatio -> Automatic, PlotRange -> 3]
Plot[{Re[FL[0.5, x]], Im[FL[0.5, x]]}, {x, -5, 5}, 
 AspectRatio -> Automatic, PlotRange -> 3]
Plot[{Re[FL[0.5, FL[0.5, x]]], Cos[x]}, {x, -5, 5}, 
 AspectRatio -> Automatic, PlotRange -> 3]
HalfCos[z_] := 
 If[Im[z] == 0, Sign[Re[Cos[z]]]*FL[0.5, z], Sign[Re[z]]*FL[0.5, z]]
Plot[{Re[HalfCos[x]], Im[HalfCos[x]]}, {x, -5, 5}, 
 AspectRatio -> Automatic, PlotRange -> 3]
Plot[{Re[HalfCos[HalfCos[x]]], Cos[x]}, {x, -5, 5}, 
 AspectRatio -> Automatic, PlotRange -> 3]
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    $\begingroup$ I can't follow this answer... What's the meaning of the two colors in the plots? Is this function real or complex? And doesn't "follow quite well, at least at positive half periods" mean that it doesn't actually equal cos? $\endgroup$ Commented May 29, 2012 at 3:37
  • $\begingroup$ The function is complex. I added description to the plot. $\endgroup$
    – Anixx
    Commented May 29, 2012 at 3:53
  • $\begingroup$ The difference in sign in negative half-periods is due the fact that arccos is not the true inverse function but just only one branch. So the first iterate becomes $\cos(\arccos(x))$. One have to use multi-valued arccos to get the correct sign. $\endgroup$
    – Anixx
    Commented May 29, 2012 at 3:56
  • $\begingroup$ Are you claiming that your formula works for all complex numbers z? $\endgroup$ Commented May 29, 2012 at 3:58
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    $\begingroup$ @NoahSnyder I've found this same function independently written much differently. I have a proof around here somewhere, but it definitely works for $z$ in the immediate basin about the geometrically attracting fixed point on the real positive line. Every expression I've found once it hits the julia set just turns to bleh though--diverges as the terms $\phi(k,z)$ grow super exponentially with $k$. As does this one. $\endgroup$
    – user78249
    Commented Oct 18, 2016 at 18:54

This does not directly answer the question, which concerned iterative roots on the real axis. However, I have also seen complex solutions mentioned in some of the discussions, so it seems relevant to mention the following.

Baker (in "The iteration of entire transcendental functions and the solution of the functional equation f{f(z)} = F(z)") proves the following.

Theorem 1. If $F(z)$ is an entire function of finite order bounded on some continuous curve $\Gamma$ which extends to infinity, then the functional equation $f\{f(z)\} = F(z)$ has no entire solution.

Recall that finite order means that $\log_+ \log_+ |F(z)| = O(\log|z|)$ as $z\to\infty$. Of course, $\cos$ has order $1$, and is bounded on the real axis. So the equation $f(f(z)) = \cos(z)$ has no analytic solution defined on the entire complex plane.

Baker's argument is beautifully simple: Firstly, it is well-known that the composition of two functions of positive order cannot have finite order, so any such solution $f$ would have order zero. Now, either $f$ is itself bounded on $\Gamma$, or $f(\Gamma)$ is unbounded and $f$ is bounded on $f(\Gamma)$. In either case, $f$ is bounded on an unbounded connected set. But it is well-known that this is impossible for functions of order less than $1/2$.

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    $\begingroup$ Baker, I.N. The iteration of entire transcendental functions and the solution of the functional equation $f\{f(z)\}=F(z)$. Math. Ann. 129, 174–180 (1955). doi.org/10.1007/BF01362363 $\endgroup$ Commented Apr 28, 2020 at 14:11

About literature related to the topic of this question:

In the answer to the question f(f(x))=exp(x)-1 and other functions “just in the middle” between linear and exponential. one can find an interesting link with many references related to the problem:

There is also Kindermanns PhD thesis about finding solutions to iterative functional equations using a neural network (in german only)

which might be helpful.


Joel's answer made me think a bit and I believe I found an interesting solution for $f(x)$ :

$$ f(x) = \begin{cases} ix & \text{if } \mathrm{Im}(x) = 0, x\neq 0 \\ \cos(ix) & \text{if } \mathrm{Re}(x) = 0,x \neq 0 \\ 2\pi i & \text{if } x = 0 \end{cases}$$

It is of course a bit of a trick (reminds me of Wick Rotation), but I it works for all $x \in \mathbb R$, because

$$f(f(x)) = \cos(i(ix))=\cos(-x) = \cos(x)$$

Update: Added the case $x=0$. For this we have

$$f(f(0)) = \cos(i(2\pi i))=\cos(-2\pi) = \cos(0)$$

  • 7
    $\begingroup$ This is basically defining two functions $f$, $g$ such that $f(g(x)) = cos(x)$. Obviously you can choose $g(x)=x$ and $f(x)=cos(x)$; in your case you added a complication by requiring that the functions coincide at zero, which seems quite arbitrary. $\endgroup$ Commented Feb 19, 2014 at 18:23

If you assume, that $f$ can be written as a power series, lets say $f(x)=\sum_i a_ix^i$ (and the power series converges everywhere absolutely), then one can write down a power series for $f\circ f$, where the i-th coefficient is a polynomial in the lower coefficients. Comparing the coefficients of this power series to the one of $cos$, we have to solve a system of algebraic equations (which might lead to the disambiguity). This is just a first idea from a non expert.

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    $\begingroup$ "polynomial in the lower coefficients" works only if you expand about a fixed point of f ... in your case, only if f(0)=0 . $\endgroup$ Commented Mar 9, 2010 at 16:33

Henrik's idea is good, but it doesn't quite work for $\cos x$ since the power series must have a non-zero leading term, and so can't be substituted into a power series. To fix this, consider a power series about a root of $\cos x$, for example $f(x) = \sum_i a_i (x-\pi/2)^i$ with $a_0=0$.

  • 4
    $\begingroup$ No, you must use a fixed point of cos, not a zero of cos. $\endgroup$ Commented Mar 9, 2010 at 16:34

No answer, just a comment to illustrate the functionality of the Schröder-function which is not familiar for several participiants here plus a (futile) attempt to a real-to-real solution (but possibly reflecting parts of the Kneser-method)

1 - On the Schröder-mechanism

I applied the Schröder-mechanism which runs into a power series involving complex terms. Here is a graph for iterates in steps of 1/60 from a couple of starting points in the interval $0 \ldots \pi/2$ Near that two borders, the Schröder-function is difficult to handle and the iterates in that neighbourhoods are questionable.
For starting points below $\pi/4$ :

For starting points above $\pi/4$ :

Of course, the spirals/trajectories can be continued infinitely towards the fixpoint.

2 - On an attempt for a real-to-real solution

I investigated the possibility to find a real-to-real solution based on finding solutions for the polynomials of order $t$ as truncations of the power series of the $\cos()$ observing the characteristics when $t \to \infty$ . The heuristics suggest that we approach a formal power series whose coefficients blow up without bound (as some not yet estimated function of $t$) when increasing $t$ except the first terms which decreases, but I've no idea of the limit somewhere between zero and $0.5$
So it is surely hopeless to assume a meaningful solution this way.

Here is the attempt to a solution; it might illustrate the occuring problems well.

Let $$f_t(x) = \sum_{k=0}^{t-1} c_k x^k = \sum_{k=0}^{t-1} { ( (î x)^k + (-î x)^k) \over 2 \cdot k!} $$ the polynomial of degree $t-1$ of the $t$ leading terms of the power series for $\cos(x)$. Then we seek the polynomial $$ g_t(x) = \sum_{k=0}^{t-1} a_k { x^k } $$ such that $$g_t(g_t(x)) = f_t(x) + O(x^{t})$$

This process is interesting because in the case of the half-iterate of the $\exp()$ function it seems very likely that this process approximates well the famous Kneser's real-to-real solution (which was mentioned here in MO too). The machinery in Kneser's solution is highly intransparent and Kneser himself did not give an explicite way how to find the power series, however participants in the tetrationforum developed such explicite solutions (or at least asymptotic approximations) which give explicite power series to arbitrary many terms and arbitrary precision.

I found an iterative method to approximate the coefficents in $g_t(x)$ for each $t$ to arbitrary accuracy. The basic principle is the Newton-rootfinding algorithm applied to the (truncated) Carlemanmatrix $F_t$ assigned to $f_t(x)$ finding the (truncated) Carlemanmatrix $G_t$ and from this the assigned function $g_t(x)$ which gives indeed $G_t^2 = \hat F_t$ (where $\hat F_t$ is no more Carleman) . The key ingredient is here, that the Newton-iteration has a restriction to make sure, $G_t$ becomes a true (truncated) Carlemanmatrix - so we might introduce the name "restricted Newton squarerroot finding algorithm on Carlemanmatrices" (I have explained this a bit more in a recent MSE-answer and this on the half-iterate of the $\exp()$ where also the Kneser-solution was posted.)

The results are the following polynomials $g_t(x)$ which produce perfectly $g_t(g_t(x)) = f_t(x) + O(x^{t})$ that means they reproduce perfectly the $t$-leading coefficients of the $\cos()$-function. Here are the coefficients for the odd $t$ from $t=3$ to $t=21$ (columnwise):

   x         t=5         t=7         t=9        t=11        t=13        t=15        t=17        t=19        t=21
   0  0.71233691  0.69301041  0.67288261  0.65596547  0.64204889  0.63051446  0.62082937  0.61258889  0.60549199
   1   1.6102585   2.7287951   4.0085148   5.3987263   6.8710340   8.4067966   9.9930135   11.620254   13.281463
   2  -3.5729667  -10.358626  -21.756970  -38.394902  -60.724230  -89.082264  -123.72870  -164.86856  -212.66714
   3   3.6948540   21.217052   67.801784   162.22728   325.77604   581.54895   954.00757   1468.6586   2151.8298
   4  -1.4832464  -24.599214  -132.45279  -450.65133  -1181.2198  -2612.5304  -5126.5486  -9204.2529  -15429.785
   5           .   15.320488   166.15792   860.15843   3049.8356   8544.1701   20359.207   43097.180   83346.945
   6           .  -4.0249864  -130.81135  -1142.7490  -5750.4976  -20980.263  -61813.612  -156251.66  -351914.80
   7           .           .   59.142056   1043.7623   7979.3068   39297.326   146369.32   448528.28   1189365.8
   8           .           .  -11.778174  -627.54139  -8088.3168  -56426.961  -273197.44  -1033454.9  -3267909.5
   9           .           .           .   224.39055   5842.4420   61818.541   403310.68   1925686.7   7371868.9
  10           .           .           .  -36.268480  -2855.3854  -50872.280  -469386.59  -2908877.0  -13728482.
  11           .           .           .           .   848.08908   30498.821   426187.16   3553907.6   21143786.
  12           .           .           .           .  -115.82787  -12594.714  -295982.27  -3485952.6  -26885414.
  13           .           .           .           .           .   3207.6064   152012.62   2708168.3   28072241.
  14           .           .           .           .           .  -380.22736  -54457.346  -1629854.6  -23835563.
  15           .           .           .           .           .           .   12160.303   733291.62   16205217.
  16           .           .           .           .           .           .  -1275.3747  -232281.71  -8615795.2
  17           .           .           .           .           .           .           .   46234.919   3452645.4
  18           .           .           .           .           .           .           .  -4352.9127  -981161.26
  19           .           .           .           .           .           .           .           .   176317.73
  20           .           .           .           .           .           .           .           .  -15070.867

The coefficients in the polynomials show a clear growth with the degree $t$ and also suggest, that the "naive" extrapolation of the final series would in the leading terms look roughly like a geometric series with some function of $t$ as quotient.
Of course such an extrapolated series is divergent for all $|x|>0$ , and I assume, that a Kneser-like solution is accordingly impossible.

That polynomials $g_t(x)$ are also actually not very useful; while they reproduce the leading terms of the $\cos()$ well, the remaining data in $g_t(g_t(x))$ is much garbage. Here is an example for $t=5$ where $g_5(g_5(x)) = f_5(x) + O(x^5)$ but the $O(x^5)$ -part is really large (and grows with higher polynomials degrees $t$):

     g_5(g_5(x))  = f_5(x) + O(x^5) = 
        -2.2E-44 *  x      // nonzero because of stopping the Newton-iterations
     -0.50000000 *  x^2
        +2.2E-43 *  x^3    // nonzero because of stopping the Newton-iterations
      +0.0416667 *  x^4
      +46.474309 *  x^5
      -292.63771 *  x^6
      +946.43908 *  x^7
      -2017.5754 *  x^8
      +3098.6620 *  x^9
      -3562.7024 * x^10
      +3107.2484 * x^11
      -2045.3815 * x^12
      +992.01697 * x^13
      -336.46574 * x^14
      +71.533667 * x^15
      -7.1790424 * x^16

I like this question so I thought I'd bump it and give my two cents.

We'll denote $I$ as the immediate basin of the fixed point $x_0$ on the real positive line. This is the largest connected set about $x_0$ wherein $\lim_{n\to\infty} \cos^{\circ n}(x) \to x_0$. And we'll denote $\Psi$ as the Schroder function of $\cos$ about $x_0$--the function which linearizes $\cos$ i.e: $\Psi(\cos(x)) = -\sin(x_0)\Psi(x)$.

Naturally in a neighborhood of $x_0$ there are two half iterates

$$f_{01}(x) = \Psi^{-1}(\sqrt{-\sin(x_0)}\Psi(x))$$

for both branches of $\sqrt{}$ ($f_0$ and $f_1$ denoting different branches.). Now, these functions can be lifted to functions $f_{01} : I \to I$. The formula is a little cumbersome but essentially is the following (the proof is exhausting and my own so I'll leave it out).


$$\vartheta(x,t) = \sum_{n=0}^\infty \cos^{\circ 2(n+1)}(x)\frac{t^n}{n!}$$

and for $0 < \Re(z) < 1$

$$\phi(x,z) = \frac{1}{\Gamma(1-z)}\int_0^\infty \vartheta(x,-t)t^{-z}\,dt$$

which satisfies $\phi: I\times\mathbb{C}_{0 < \Re(z) < 1} \to I$ in $x$ and locally about $x_0$ looks like

$$\phi(x,z) = \Psi^{-1}(\sin(x_0)^{2z}\Psi(x))$$

choose $z_0$ and $z_1$ so that $\sin(x_0)^{2z_{01}} = \sqrt{-\sin(x_0)}$ for each branch of $\sqrt{}$ and voila $\phi(z_0,x) = f_0(x)$ and $\phi(z_1,x) = f_1(x)$. We've analytically continued $f_0$ and $f_1$ to the immediate basin. These are also the only analytic solutions to the equation

$$g : I \to I$$ $$g(g(x)) = \cos(x)$$

They are NOT real to real, which is all to do with the multiplier $-\sin(x_0)$ which is negative; there are no real square roots of negative numbers = there are no real composite square roots of functions with negative multipliers. Similarly for the function $h(h(h(x))) = \cos(x)$ there are 3 solutions, and only one of them is real to real (just like there are three cube roots of negative one and only one of them is real). Just as well, there are four solutions to $q(q(q(q(x)))) = \cos(x)$ and none are real to real (there are no real fourth roots of $(-1)$). So on and so forth.

Now I don't have a rigorous proof of the following, but it seems obvious enough that $f_0$ and $f_1$ are defined on their maximal domain. $\partial I$ is part of the julia set, and the function $\vartheta(x,t)$ diverges on the julia set because $\cos^{\circ 2(n+1)}(x)$ grows super exponentially with $n$ and the factorial no longer does its job. I think this is good intuition in inferring that $f$ has no extension to a larger domain. I could be wrong though--it'd be nice to see $f_{01} : \mathbb{C} \to \mathbb{C}$.

  • $\begingroup$ see mathoverflow.net/questions/45608/… and my self-answer I'm afraid the web page with those references is down but I have pdfs at home for the shorter pieces. Especially recommend I. Noel Baker $\endgroup$
    – Will Jagy
    Commented Nov 2, 2016 at 22:31
  • $\begingroup$ @WillJagy I've devoured everything there already unfortunately. I actually have my own formula for the half iterate of $\sin$ on the real line. By taking the uniform convergence of $f_n(x) = (1-1/n)\sin(x)$ to $\sin(x)$ on $\mathbb{R}$ one can show $f_n^{\circ t}(x) \to f^{\circ t}(x)$ given the above representation with $t \in \mathbb{R}^+$. It produces the exact graph you posted too. I was unable to show analycity though, only continuity. $\endgroup$
    – user78249
    Commented Nov 2, 2016 at 23:00

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