09/03/2022, 02:05 PM
I think Leo has really found some nugget here.
I can verify \(h(h(z)))=-z+z^2\) does not have a formal powerseries solution. Remember writing \((f\circ g)_n = \sum_{m=1}^n f_m {g^m}_n\) where \(g^m\) is the (normal) power of \(g\) and \({g^m}_n\) the nth coefficient of that power and \({g^m}_n = \sum_{k_1+\dots+k_m=n} g_{k_1}\cdots g_{k_m}\).
So in our case h needs to satisfy the equations for at least the first 3 coefficients of h:
\begin{align}
h_1 {h^1}_1 &= -1\\
h_1 {h^1}_2 + h_2 {h^2}_2 &= 1\\
h_1 {h^1}_3 + h_2 {h^2}_3 + h_3 {h^3}_3 &= 0
\end{align}
which is equivalent to:
\begin{align}
h_1^2 &= -1\\
h_1 {h}_2 + h_2 h_1^2 &= 1\\
h_1 {h}_3 + h_2 2h_1 h_2 + h_3 h_1^3 &= 0
\end{align}
Putting the first equation in the 3rd we get \(2h_2^2h_1 = 0\) which means either \(h_1=0\) or \(h_2=0\), which contradicts the first two equations.
As we don't have formal powerseries iteration formulas for \(f_1^q=1\) for some \(q\ge 2\), let's have a look at \(f_1=1\), here we can always find an iterative square root \(h\) with \(h_1=1\) however only in certain cases a root with \(h_1=-1\), which can be seen from the equations indexed by n:
\begin{align}
\sum_{m=1}^n h_m {h^m}_n &= f_n\\
h_1 h_n + h_n h_1^n + \sum{m=2}^{n-1} h_m {h^m}_n &= f_n \\
h_n (h_1+h_1^n) &= f_n - \sum_{m=2}^{n-1} h_m {h^m}_n
\end{align}
Because with setting \(h_1=1\) the term \(h_1+h^n\) never vanishes, one can always calculate the next \(h_n\).
If however \(h_1=-1\) then e.g. at n=2 must
\[f_2 = h_1 {h}_2 + h_2 h_1^2 = -h_2 + h_2 = 0\]
This (using \(h_1=1\) and \(h_1=-1\)) would find both roots \(a(z)=-z+z^3\) and \(b(z)=z-z^3\) of \(a\circ a=b\circ b\).
But \(b\) would be the standard solution which generally always exists (\(b_1=1\)) and \(a\) would only be the side solution which in general not always exists (\(a_1=-1\)).
I think - though did not verify - that you can generalize this to q-th roots.
I.e. you can solve the equation \(h^{\circ q}=f\) (for \(f_1=1\)) always with initial setting \(h_1=1\), and occasionally with setting \(h_1=e^{2\pi i k/q}\) for \(2\le k<q\).
This then would mean that one can consider our original \(f(z)=-z+z^2\) a solution of \(h\circ h=f\circ f\) with initial setting \(h_1=-1\) and the square root of \(f\) would be the fourth root of \(f\circ f\) with initial settings \(h_1=i\) and \(h_2=-i\) - which we saw already does not exist, though.
I am still thinking how to express this with the Julia function/logit but didn't find something convincing.
For me there are these analogies of \(b^t=e^{t\log(b)}\) with \(f^{\circ t}={\rm expit}[t\;{\rm logit}[f]]\).
So if one wants to get the other roots of say \(b^{\frac{1}{2}}\) one would do something \(e^{\frac{1}{2}(\log(b)+2\pi i k)}=b^{\frac{1}{2}}e^{\pi i k}=\pm b^{\frac{1}{2}}\). I just don't see yet what would be the equivalent with expit & logit.
Btw. the logit of \(f(z)=-z+z^2\) does not exist - or rather is the 0-function according to my calculations.
Also it quite seems all the solutions of \(f\circ h=h\circ f\) are \(h=f^{\circ n}\), also according to my calculations
I can verify \(h(h(z)))=-z+z^2\) does not have a formal powerseries solution. Remember writing \((f\circ g)_n = \sum_{m=1}^n f_m {g^m}_n\) where \(g^m\) is the (normal) power of \(g\) and \({g^m}_n\) the nth coefficient of that power and \({g^m}_n = \sum_{k_1+\dots+k_m=n} g_{k_1}\cdots g_{k_m}\).
So in our case h needs to satisfy the equations for at least the first 3 coefficients of h:
\begin{align}
h_1 {h^1}_1 &= -1\\
h_1 {h^1}_2 + h_2 {h^2}_2 &= 1\\
h_1 {h^1}_3 + h_2 {h^2}_3 + h_3 {h^3}_3 &= 0
\end{align}
which is equivalent to:
\begin{align}
h_1^2 &= -1\\
h_1 {h}_2 + h_2 h_1^2 &= 1\\
h_1 {h}_3 + h_2 2h_1 h_2 + h_3 h_1^3 &= 0
\end{align}
Putting the first equation in the 3rd we get \(2h_2^2h_1 = 0\) which means either \(h_1=0\) or \(h_2=0\), which contradicts the first two equations.
As we don't have formal powerseries iteration formulas for \(f_1^q=1\) for some \(q\ge 2\), let's have a look at \(f_1=1\), here we can always find an iterative square root \(h\) with \(h_1=1\) however only in certain cases a root with \(h_1=-1\), which can be seen from the equations indexed by n:
\begin{align}
\sum_{m=1}^n h_m {h^m}_n &= f_n\\
h_1 h_n + h_n h_1^n + \sum{m=2}^{n-1} h_m {h^m}_n &= f_n \\
h_n (h_1+h_1^n) &= f_n - \sum_{m=2}^{n-1} h_m {h^m}_n
\end{align}
Because with setting \(h_1=1\) the term \(h_1+h^n\) never vanishes, one can always calculate the next \(h_n\).
If however \(h_1=-1\) then e.g. at n=2 must
\[f_2 = h_1 {h}_2 + h_2 h_1^2 = -h_2 + h_2 = 0\]
This (using \(h_1=1\) and \(h_1=-1\)) would find both roots \(a(z)=-z+z^3\) and \(b(z)=z-z^3\) of \(a\circ a=b\circ b\).
But \(b\) would be the standard solution which generally always exists (\(b_1=1\)) and \(a\) would only be the side solution which in general not always exists (\(a_1=-1\)).
I think - though did not verify - that you can generalize this to q-th roots.
I.e. you can solve the equation \(h^{\circ q}=f\) (for \(f_1=1\)) always with initial setting \(h_1=1\), and occasionally with setting \(h_1=e^{2\pi i k/q}\) for \(2\le k<q\).
This then would mean that one can consider our original \(f(z)=-z+z^2\) a solution of \(h\circ h=f\circ f\) with initial setting \(h_1=-1\) and the square root of \(f\) would be the fourth root of \(f\circ f\) with initial settings \(h_1=i\) and \(h_2=-i\) - which we saw already does not exist, though.
I am still thinking how to express this with the Julia function/logit but didn't find something convincing.
For me there are these analogies of \(b^t=e^{t\log(b)}\) with \(f^{\circ t}={\rm expit}[t\;{\rm logit}[f]]\).
So if one wants to get the other roots of say \(b^{\frac{1}{2}}\) one would do something \(e^{\frac{1}{2}(\log(b)+2\pi i k)}=b^{\frac{1}{2}}e^{\pi i k}=\pm b^{\frac{1}{2}}\). I just don't see yet what would be the equivalent with expit & logit.
Btw. the logit of \(f(z)=-z+z^2\) does not exist - or rather is the 0-function according to my calculations.
Also it quite seems all the solutions of \(f\circ h=h\circ f\) are \(h=f^{\circ n}\), also according to my calculations