Parabolic Formal Powerseries

[bo198214]

In this post the question was raised whether \(f(z)=-z+z^2\) can have an analytic half iterate at 0.
It was found out that it does not even have a formal powerseries solution.
So that sparked the question which powerseries \(f\) have formal power series iterates and which not.
And I think I found a criterion.
So lets start with a powerseries \(f\) with \(f_1=1\) and consider the condition \(g\circ f = f\circ g\) then there is the equation system
\begin{align}
g_1 {f^1}_1 &= f_1 {g^1}_1 \\
g_1 {f^1}_2 + g_2 {f^2}_2 &= f_1 {g^1}_2 + f_2 {g^2}_2\\
g_1 {f^1}_3 + g_2 {f^2}_3 + g_3 {f^3}_3 &= f_1 {g^1}_3 + f_2 {g^2}_3 + f_3 {g^3}_3 \\
g_1 {f^1}_4 + g_2 {f^2}_4 + g_3 {f^3}_4 + g_4 {f^4}_4 &= f_1 {g^1}_4 + f_2 {g^2}_4 + f_3 {g^3}_4 + f_4 {g^4}_4 \\
\dots
\end{align}
(For notation see this post )
With \(f_1=1\) and \({h^k}_k = {h_1}^k\) and removing equal elements on both sides these are equivalent to
\begin{align}
g_1 f_2   &= f_2 {g_1}^2 \\
g_1 f_3 + g_2 {f^2}_3   &= f_2 {g^2}_3 + f_3 {g_1}^3\\
g_1 f_4 + g_2 {f^2}_4 + g_3 {f^3}_4 &=  f_2 {g^2}_4 + f_3 {g^3}_4 + f_4 {g_1}^4\\
\dots
\end{align}
From the first line we can see either \(g_1={g_1}^2\) or \(f_2=0\).
If \(f_2=0\) then on the second line we can see: either \(g_1={g_1}^3\) or \(f_3=0\),
if \(f_3 = 0\) then on the third line we can see: either \(g_1={g_1}^4\) or \(f_4=0\),
and so on (knowing that \({f^m}_n\) is a sum of products of \(f_k\) with maximum index \(k=n-m+1\), so this sum is also 0 if the previous \(f_k=0\)).

So we found out that \({g_1}^{v+1} = g_1\) is mandatory for \(v={\rm valit}[f]\) - valit is the maximum index up to which f is equal to the identity powerseries (e.g. \({\rm valit}[z+z^n+O(z^{n+1})]=n-1\)). 

This means if \(g\circ f=f\circ g\), \(f_0=0,f_1=1\) has a solution g, then must
\[ g^{{\rm valit}[f]} = 1 \]
(I didn't double check but vice versa all the solutions with \({g_1}^{{\rm valit}[f]}=1\) do exist.)

For the case \(h(z)=-z+z^2\) one would take \(f(z)=h(h(z))=z-2z^3+z^4\) (because then \(f_1=1\)) , this has \({\rm valit}[f] = 2\) so we can not take the 4th roots of \(f\) with \(g_1=\pm i\) but we can only take the second roots with \(g_1=\pm 1\).

However for the function \(h(z)=-z+z^4+z^5\), \(f\) would be \(f(z)=h(h(z))=z+z^5 + O(z^6)\). This has \({\rm valit}[f] = 4\) so here we *can* take all the 4th roots with \(g_1=\pm 1,\pm i\), particularly the two half iterates of h!
I will post the series after I updated my formal powerseries iteration algorithm, and check whether it is true what I claimed

[JmsNxn]

Bo, I don't have the time to get into it now, but this seems very similar to the problem:

\[
\text{The half iterate of}\,\,-z + \lambda z^2\,\,\text{has no power series at 0 for all}\,\,\lambda \neq 0\\
\]

Which is absolutely provable. And for higher powers of \(z\), it's reducible to this.

It can quite literally just become a question of \(y = z^n\) and there are \(n\) branches--in no world is \(y\) holomorphic at \(0\).



-----------------------------------------

I love how you reduce everything into taylor coefficients. I reduce everything into integrals or something like that. I love the taylor coefficient approach

[bo198214]

\[
\text{The half iterate of}\,\,-z + \lambda z^2\,\,\text{has no power series at 0 for all}\,\,\lambda \neq 0\\
\]

Yes, this follows from \(g_1^{{\rm valit}[f]} = 1\), the reasoning is like that:
First we integer iterate to make the first coefficient 1 (because this is much easier to handle than any primary roots).
\begin{align}
a(z)&=-z+\lambda z^2\\
f(z)&=a(a(z)) = z - \lambda z^2 + \lambda (-z + \lambda z^2) = z - 2\lambda z^3 + \lambda^2 z^4\\
{\rm valit}[f] &= 2
\end{align}
If you now have an iterational square root \(h\circ h = a\) then follows  \({h_1}^2 = -1\). But surely also \(h\circ f=f\circ h\) and hence \({h_1}^{2} = 1\) which is a contradiction.

But whom did you quote above?

Which is absolutely provable. And for higher powers of \(z\), it's reducible to this.

It can quite literally just become a question of \(y = z^n\) and there are \(n\) branches--in no world is \(y\) holomorphic at \(0\).

It is not generally true for higher powers, it really has to do with the valit. For example if you have
\begin{align}
a(z)&=-z+z^5\\
f(z)&=a(a(z))=z-2z^5 + O(z^6)\\
{\rm valit}[f]=4
\end{align}

Then you can take 4 4th iterational roots of f with first coefficients 1,i,-1,-1. Two of these roots are square roots of \(a\) corresponding to \(h_1=i,-i\):
\begin{align}
h^{|1}(x) & = x - \frac{1}{2} x^{5} - \frac{5}{8} x^{9} - \frac{25}{16} x^{13} - \frac{605}{128} x^{17} + O(x^{21})\\
h^{|2}(x) & = i x - \frac{1}{2} i x^{5} - \frac{5}{8} i x^{9} - \frac{25}{16} i x^{13} - \frac{605}{128} i x^{17}+ O(x^{21})\\
h^{|3}(x) & = - x + \frac{1}{2} x^{5} + \frac{5}{8} x^{9} + \frac{25}{16} x^{13} + \frac{605}{128} x^{17}+ O(x^{21})\\
h^{|4}(x) & = -i x + \frac{1}{2} i x^{5} + \frac{5}{8} i x^{9} + \frac{25}{16} i x^{13} + \frac{605}{128} i x^{17}+ O(x^{21})\\
\end{align}
(Unfortunately I use the indices already to address the coefficients, so I could not call it \(h_1,h_2,h_3,h_4\), so called it in that strange way)
In this simple case they are just the same function (i.e. the default regular iteration \(f^{\circ 1/4}\)) applied to x,ix,-x,-ix.
But this is different when you have more complicated \(a\).

For even powers we have: if \(a(z)=-z+z^{2n}\), (i.e. \({\rm valit}[a]=2n-1\)) then \({\rm valit}[a\circ a]=4n-2\).
Means we never get \({\rm valit}[a\circ a] = 4n\) so we never have those \(\pm i\) roots.

I love how you reduce everything into taylor coefficients. I reduce everything into integrals or something like that. I love the taylor coefficient approach

I think this is the most philosophical thing about complex function theory I learned, 
that the whole global behaviour of a holomorphic function is just given by the Taylor series at one point (because from there you just continue it anywhere you want).
But really I spend some time with formal powerseries - and built a sage-python library for all those iteration stuff (Though not with those Brent-optimizations, just the naive approach with recurrence formulas)
Up to now I only considered the standard case \(f_1=1\), and always wondered why the unit roots are not considered.
It really costed me some grey hairs to fiddle myself through the unit root case.
But now I am close to finish that case too.

[bo198214]

Just to test my sage-lib:
let \(a(x)=-x+x^4+x^5\) then \(f(x)=a(a(x))=x-2x^5+O(x^6)\), \({\rm valit}[f]=4\) and we have again the 4 4th roots:
\begin{align}
f^{\circ \frac{1}{4}|1}&=x - \frac{1}{2} x^{5} -  x^{7} + \frac{1}{4} x^{8} - \frac{5}{8} x^{9} + \frac{3}{2} x^{10} - \frac{17}{2} x^{11} - \frac{17}{16} x^{12} - \frac{209}{16} x^{13} + 22 x^{14} - \frac{429}{8} x^{15} + \frac{147}{8} x^{16} - \frac{31917}{128} x^{17} + \frac{2407}{16} x^{18} - \frac{4057}{8} x^{19}+O(x^20)\\
f^{\circ \frac{1}{4}|2}&=i x + 2 i x^{3} + \left(-\frac{1}{2} i + \frac{1}{2}\right) x^{4} + \frac{11}{2} i x^{5} + \left(-3 i + 4\right) x^{6} + \left(-\frac{15}{2} i - 1\right) x^{7} + \left(-16 i + \frac{101}{4}\right) x^{8} + \left(-\frac{1513}{8} i - 12\right) x^{9} + \left(\frac{43}{4} i + 101\right) x^{10} + \left(-\frac{18907}{12} i - \frac{211}{2}\right) x^{11} + \left(\frac{12329}{16} i + \frac{669}{8}\right) x^{12} + \left(-\frac{406943}{48} i - \frac{1119}{2}\right) x^{13} + \left(\frac{52139}{6} i - \frac{40595}{12}\right) x^{14} + \left(-\frac{2215209}{80} i - \frac{1901}{2}\right) x^{15} + \left(\frac{2798725}{48} i - \frac{1946047}{48}\right) x^{16} + \left(\frac{25537919}{640} i + \frac{130513}{6}\right) x^{17} + \left(\frac{7666655}{32} i - \frac{6096887}{20}\right) x^{18} + \left(\frac{3365536581}{2240} i + \frac{3905543}\\{12}\right) x^{19}+O(x^{20})\\
f^{\circ \frac{1}{4}|3}&=- x +  x^{4} + \frac{1}{2} x^{5} -  x^{7} + \frac{9}{4} x^{8} + \frac{5}{8} x^{9} + \frac{11}{2} x^{10} - \frac{21}{2} x^{11} + \frac{133}{16} x^{12} - \frac{367}{16} x^{13} + \frac{187}{2} x^{14} - \frac{699}{8} x^{15} + \frac{699}{4} x^{16} - \frac{82707}{128} x^{17} + \frac{17417}{16} x^{18} - \frac{11755}{8} x^{19}+O(x^{20})\\
f^{\circ \frac{1}{4}|4}&=-i x - 2 i x^{3} + \left(\frac{1}{2} i + \frac{1}{2}\right) x^{4} - \frac{11}{2} i x^{5} + \left(3 i + 4\right) x^{6} + \left(\frac{15}{2} i - 1\right) x^{7} + \left(16 i + \frac{101}{4}\right) x^{8} + \left(\frac{1513}{8} i - 12\right) x^{9} + \left(-\frac{43}{4} i + 101\right) x^{10} + \left(\frac{18907}{12} i - \frac{211}{2}\right) x^{11} + \left(-\frac{12329}{16} i + \frac{669}{8}\right) x^{12} + \left(\frac{406943}{48} i - \frac{1119}{2}\right) x^{13} + \left(-\frac{52139}{6} i - \frac{40595}{12}\right) x^{14} + \left(\frac{2215209}{80} i - \frac{1901}{2}\right) x^{15} + \left(-\frac{2798725}{48} i - \frac{1946047}{48}\right) x^{16} + \left(-\frac{25537919}{640} i + \frac{130513}{6}\right) x^{17} + \left(-\frac{7666655}{32} i - \frac{6096887}{20}\right) x^{18} + \left(-\frac{3365536581}{2240} i + \frac{3905543}{12}\right) x^{19}+O(x^{20})\\
a^{\circ \frac{1}{2}|1} &= f^{\circ \frac{1}{4}|2}\\
a^{\circ \frac{1}{2}|2} &= f^{\circ \frac{1}{4}|4}
\end{align}

[tommy1729]

Just to test my sage-lib:
let \(a(x)=-x+x^4+x^5\) then \(f(x)=a(a(x))=x-2x^5+O(x^6)\), \({\rm valit}[f]=4\) and we have again the 4 4th roots:
\begin{align}
f^{\circ \frac{1}{4}|1}&=x - \frac{1}{2} x^{5} -  x^{7} + \frac{1}{4} x^{8} - \frac{5}{8} x^{9} + \frac{3}{2} x^{10} - \frac{17}{2} x^{11} - \frac{17}{16} x^{12} - \frac{209}{16} x^{13} + 22 x^{14} - \frac{429}{8} x^{15} + \frac{147}{8} x^{16} - \frac{31917}{128} x^{17} + \frac{2407}{16} x^{18} - \frac{4057}{8} x^{19}+O(x^20)\\
f^{\circ \frac{1}{4}|2}&=i x + 2 i x^{3} + \left(-\frac{1}{2} i + \frac{1}{2}\right) x^{4} + \frac{11}{2} i x^{5} + \left(-3 i + 4\right) x^{6} + \left(-\frac{15}{2} i - 1\right) x^{7} + \left(-16 i + \frac{101}{4}\right) x^{8} + \left(-\frac{1513}{8} i - 12\right) x^{9} + \left(\frac{43}{4} i + 101\right) x^{10} + \left(-\frac{18907}{12} i - \frac{211}{2}\right) x^{11} + \left(\frac{12329}{16} i + \frac{669}{8}\right) x^{12} + \left(-\frac{406943}{48} i - \frac{1119}{2}\right) x^{13} + \left(\frac{52139}{6} i - \frac{40595}{12}\right) x^{14} + \left(-\frac{2215209}{80} i - \frac{1901}{2}\right) x^{15} + \left(\frac{2798725}{48} i - \frac{1946047}{48}\right) x^{16} + \left(\frac{25537919}{640} i + \frac{130513}{6}\right) x^{17} + \left(\frac{7666655}{32} i - \frac{6096887}{20}\right) x^{18} + \left(\frac{3365536581}{2240} i + \frac{3905543}\\{12}\right) x^{19}+O(x^{20})\\
f^{\circ \frac{1}{4}|3}&=- x +  x^{4} + \frac{1}{2} x^{5} -  x^{7} + \frac{9}{4} x^{8} + \frac{5}{8} x^{9} + \frac{11}{2} x^{10} - \frac{21}{2} x^{11} + \frac{133}{16} x^{12} - \frac{367}{16} x^{13} + \frac{187}{2} x^{14} - \frac{699}{8} x^{15} + \frac{699}{4} x^{16} - \frac{82707}{128} x^{17} + \frac{17417}{16} x^{18} - \frac{11755}{8} x^{19}+O(x^{20})\\
f^{\circ \frac{1}{4}|4}&=-i x - 2 i x^{3} + \left(\frac{1}{2} i + \frac{1}{2}\right) x^{4} - \frac{11}{2} i x^{5} + \left(3 i + 4\right) x^{6} + \left(\frac{15}{2} i - 1\right) x^{7} + \left(16 i + \frac{101}{4}\right) x^{8} + \left(\frac{1513}{8} i - 12\right) x^{9} + \left(-\frac{43}{4} i + 101\right) x^{10} + \left(\frac{18907}{12} i - \frac{211}{2}\right) x^{11} + \left(-\frac{12329}{16} i + \frac{669}{8}\right) x^{12} + \left(\frac{406943}{48} i - \frac{1119}{2}\right) x^{13} + \left(-\frac{52139}{6} i - \frac{40595}{12}\right) x^{14} + \left(\frac{2215209}{80} i - \frac{1901}{2}\right) x^{15} + \left(-\frac{2798725}{48} i - \frac{1946047}{48}\right) x^{16} + \left(-\frac{25537919}{640} i + \frac{130513}{6}\right) x^{17} + \left(-\frac{7666655}{32} i - \frac{6096887}{20}\right) x^{18} + \left(-\frac{3365536581}{2240} i + \frac{3905543}{12}\right) x^{19}+O(x^{20})\\
a^{\circ \frac{1}{2}|1} &= f^{\circ \frac{1}{4}|2}\\
a^{\circ \frac{1}{2}|2} &= f^{\circ \frac{1}{4}|4}
\end{align}

this is completely logical , almost trivial :

near fixpoint 0 , for small x and n > 2 ,  we have

f(x) = a x + b x^n + O(x^(n+1))

Notice the other fixed points are always a nonzero distance away from 0.

so if say a is 2 we are nearly solving

y = 2 x.

and hence the hyperbolic method works by the approximation y = 2x near 0.
AND we then get a unique solution.

Now if a  = 1 we get a different situation

f(x) = x + b x^n + ...

so we are nearly solving

y = x + b x^n

or

y = x ( 1 + b x^(n-1) )

y and x are close to 0 and close to eachother so

we are basicly solving

y = ( 1 + b x^(n-1) )

or

0 = 1 + b x^(n-1)

so we get (n-1) local solutions near the fixpoint 0.
and we can use those truncated (n-1) local solutions as asymptotics in our limit formulas for continu iterations , in the same way as we did with hyperbolic.

 ***

confused ? 
ok look at it geometrically

x + x^4 + ... = x ( 1 + x^3 + ...)  = id(x) + x^4 + ...

SO near the point 0 we get 3 solutions or said differently , 3 branches for the functional inverse.

so we have a riemann surface ( for the functional inverse ) with 3 layers near 0.

if it is analytic at 0 OR!! near 0 then by analytic continuation and the principles of riemann surfaces , the riemann surface will REMAIN to have 3 layers.

So for t element of ]0,1] we must also have 3 solutions for 

f^[t](x)  for x near 0.

***

yes your examples had -1 instead of 1 but its the same principle.

The higher terms are irrelevant because they vanish as x gets close to 0.

To avoid confusion i mean they vanish for a given analytic near 0 case f(x), I did mean vanish for the divergeant taylor or non-existant taylor of fractional derivates near 0. AND that is all that is required.


***

keep in mind , i said n > 2.

because n - 1 is 1 when n = 2 but that is invalid and absurd because then we get another linear.


Also keep in mind that I am considering x in the direction re(x) > 0.
This matters for the amount of petals.
for x = -1 we might not get solutions for half-iterates agreeing with x  = 0,1. but at x = -1 there might be a petal that works.

This suggests - by symmetry * re(x) > 0 or re(x) < 0 * -  that the number of petals when n > 2 is equal to 2*(n-1).


***

im wondering if real-analytic f(x) such that 

f(x) = x + b x^n + b_2 x^(n+1) + ...

where b_i are nonnegative.

and g(g(x)) = f(x) such that g(x) is the real half-iterate for x > 0 but not analytic at 0 ,

gives rise to the taylor coefficients g_m of g(x) growing as (m!)^(n-1) IFF some growth condition is given for b_i.



  


regards

tommy1729

[bo198214]

this is completely logical , almost trivial :

near fixpoint 0 , for small x and n > 2 ,  we have

f(x) = a x + b x^n + O(x^(n+1))

Notice the other fixed points are always a nonzero distance away from 0.

so if say a is 2 we are nearly solving

y = 2 x.

and hence the hyperbolic method works by the approximation y = 2x near 0.
AND we then get a unique solution.

...

Dear Tommy,
I think you heavily confusing stuff here.
We are not talking about taking inverses,
we talk about fractional iterates.
Let me give you a quick overview of the problem:
The regular iteration in the hyperbolic case has the form
\begin{align}
f(x) &= bx + O(x^2)\\
f^{\circ t}(x) &= b^tx +O(x^2)
\end{align}
So for t=1/q we have q fractional roots (with multiplicity) corresponding to \(b^{\frac{1}{q}}\).
In the case b=1 however the regular iteration look like this:
\begin{align}
f(x) &= x + cx^n + O(x^{n+1})\\
f^{\circ t}(x) &= x + t c x^n + O(x^{n+1})
\end{align}
So one would have only one q-th iterative root if 1 is the first coefficient.


So, then Leo came up with the example of \(f(x)=-x+x^2\) and could not find a
second iterative root for that and showed it can not exist.
Which then lead to the general question for which \(b=e^{2\pi i k/q}\) there would exist a
Q-th iterative root of 
\[f(x)=e^{\frac{2\pi i p}{q}}x + c x^n + O(x^{n+1})\]
and the answer is iff:

\[ {\rm valit}[f^{\circ q}] = Qqm \]

for some integer \(m\ge 1\). Then there would exist Q iterative Qth roots of f of the form 
\[ f^{\circ \frac{1}{Q}|k}(x)=e^{\frac{\frac{2\pi i p}{q}+2\pi i k}{Q}}x + O(x^2) , \quad 0\le k < Q\]

[tommy1729]

this is completely logical , almost trivial :

near fixpoint 0 , for small x and n > 2 ,  we have

f(x) = a x + b x^n + O(x^(n+1))

Notice the other fixed points are always a nonzero distance away from 0.

so if say a is 2 we are nearly solving

y = 2 x.

and hence the hyperbolic method works by the approximation y = 2x near 0.
AND we then get a unique solution.

...

Dear Tommy,
I think you heavily confusing stuff here.
We are not talking about taking inverses,
we talk about fractional iterates.
Let me give you a quick overview of the problem:
The regular iteration in the hyperbolic case has the form
\begin{align}
f(x) &= bx + O(x^2)\\
f^{\circ t}(x) &= b^tx +O(x^2)
\end{align}
So for t=1/q we have q fractional roots (with multiplicity) corresponding to \(b^{\frac{1}{q}}\).
In the case b=1 however the regular iteration look like this:
\begin{align}
f(x) &= x + cx^n + O(x^{n+1})\\
f^{\circ t}(x) &= x + t c x^n + O(x^{n+1})
\end{align}
So one would have only one q-th iterative root if 1 is the first coefficient.


So, then Leo came up with the example of \(f(x)=-x+x^2\) and could not find a
second iterative root for that and showed it can not exist.
Which then lead to the general question for which \(b=e^{2\pi i k/q}\) there would exist a
Q-th iterative root of 
\[f(x)=e^{\frac{2\pi i p}{q}}x + c x^n + O(x^{n+1})\]
and the answer is iff:

\[ {\rm valit}[f^{\circ q}] = Qqm \]

for some integer \(m\ge 1\). Then there would exist Q iterative Qth roots of f of the form 
\[ f^{\circ \frac{1}{Q}|k}(x)=e^{\frac{\frac{2\pi i p}{q}+2\pi i k}{Q}}x + O(x^2) , \quad 0\le k < Q\]

I am aware we are not talking About inverses.

But they matter.

Because when values are repeated in a neighbourhood ( we get multiple “ flows “ ) of a fix , we get multiple superfunctions adressing all those Locally multiple univalent neighbourhoods.
And thus we get also multiple fractional iterates. 

Just like exp(x) - 1 has 2 solutions that fundamentally disagree ( no 1 periodic function unites them ) 

Regards 

tommy1729

[bo198214]

Just like exp(x) - 1 has 2 solutions that fundamentally disagree ( no 1 periodic function unites them ) 

Two things:

1. There are two types of multiplicities here: One comes from the petals - for \(f(x)=x+a_{n+1}x^{n+1} + O(n+2)\) you always have 2n petals with (typically different) Abel functions, which then provide 2n solutions for all fractional/real/complex iterates. But all these solutions have multiplier 1, i.e. are of the form \(f^{\circ t}(x)=x+ta_{n+1}x^{n+1} + O(n+2)\) asymptotically at 0.
   
   However in this thread we are looking at fractional iterates with multiplier being an n-th root of unity! In the case \(f(x)=e^x-1\) is n=1, which means we can not take an iterational root with multiplier say -1, or i. We can do this for example for \(f(x)=x+x^3\), which has n=2, i.e. taking a half iterate with multiplier -1 is possible here:
   
   \[ f^{\circ \frac{1}{2}|2} = - x - \frac{1}{2} x^{3} + \frac{3}{8} x^{5} - \frac{9}{16} x^{7} + \frac{133}{128} x^{9} - \frac{507}{256} x^{11} + \frac{3399}{1024} x^{13} - \frac{6261}{2048} x^{15} - \frac{283347}{32768} x^{17} + \frac{4149645}{65536} x^{19} + O(x^{21}) \]
   
2. Of course they are united by a 1-periodic function - every two Abel functions of a function f that are suitably overlapping are "united" by a 1-periodic function.
    But I made some extra effort to even show that numerically, please have a look here.
   

PS: Calling something "almost trivial" with only a vague understanding of the problem bears a bad taste ...