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
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