07/17/2022, 08:41 AM
I still remember those times where Tetration was in the air and around the world people (including Daniel and me) were trying to figure out what the half-iterate of a function would be.
This can also be solved as formal powerseries:
\begin{align}
g\circ g &= f\\
\sum_{m=1}^n g_m {g^m}_n &= f_n\\
g_1 g_n + g_n g_1^n + \sum_{m=2}^{n-1} g_m {g^m}_n &= f_n \\
g_1 g_1 &= f_1\\
g_n &= \frac{1}{g_1+g_1^n}\left(f_n - \sum_{m=2}^{n-1} g_m {g^m}_n\right), n>1\\
\end{align}
But now compare that to
\begin{align}
g_n &= \frac{1}{{f_1}^n - f_1} \left(
f_n {g_1}^n - g_1 f_n + \sum_{m=2}^{n-1} f_m {g^m}_n - g_m{f^m}_n\right), n>1
\end{align}
and show that they are equal! (for \(g_1=\sqrt{f_1}\)) This seems ultra hard.
Luckily we don't need to do it "by hand" but know by the previous considerations that \(f^{\mathcal{R}\frac{1}{2}}\circ f^{\mathcal{R}\frac{1}{2}}=f\) and as the powerseries is determined by \(g\circ g=f\) we know that both must be equal.
Also it seems nearly impossible to prove \(f^{\mathcal{R}s+t} = f^{\mathcal{R}s}\circ f^{\mathcal{R}t}\) by arithmetical transformations.
This can also be solved as formal powerseries:
\begin{align}
g\circ g &= f\\
\sum_{m=1}^n g_m {g^m}_n &= f_n\\
g_1 g_n + g_n g_1^n + \sum_{m=2}^{n-1} g_m {g^m}_n &= f_n \\
g_1 g_1 &= f_1\\
g_n &= \frac{1}{g_1+g_1^n}\left(f_n - \sum_{m=2}^{n-1} g_m {g^m}_n\right), n>1\\
\end{align}
But now compare that to
\begin{align}
g_n &= \frac{1}{{f_1}^n - f_1} \left(
f_n {g_1}^n - g_1 f_n + \sum_{m=2}^{n-1} f_m {g^m}_n - g_m{f^m}_n\right), n>1
\end{align}
and show that they are equal! (for \(g_1=\sqrt{f_1}\)) This seems ultra hard.
Luckily we don't need to do it "by hand" but know by the previous considerations that \(f^{\mathcal{R}\frac{1}{2}}\circ f^{\mathcal{R}\frac{1}{2}}=f\) and as the powerseries is determined by \(g\circ g=f\) we know that both must be equal.
Also it seems nearly impossible to prove \(f^{\mathcal{R}s+t} = f^{\mathcal{R}s}\circ f^{\mathcal{R}t}\) by arithmetical transformations.