Maybe the solution at z=0 to f(f(z))=-z+z^2
#4
I want to complement something that I've missed
About the phenomenon that all functions, whether generalized by P method, simple iteration or limiting cases, would eventually goes to 0, I think I might give an explanation of this.
It's really fucking wierd, right? Why the fuck all such functions goes to 0 after so many iterations, it should just converge to an at-least-nonzero (continuous?) function. Let's still look at the case, denote \(F(z)=-z+z^2\) and \(f(f(z))=F(z)\), we took it for granted that \(f(z)=F(f(F^{-1}(z)))\) and this formula leads us to the 0 function.
The thing is, after each iteration, we'll jump from a branch cut of \(f\) to another totally different one branch, and this branch is fucking acting so holomorphic and it has a norm smaller than the initial branch, so it's now closer to 0. Or more briefly, denote \(f_1(z),f_2(z)\) are 2 branch cuts of the same function \(f\), we don't have \(f_1(z)=F(f_1(F^{-1}(z)))\), but we do have \(f_2(z)=F(f_1(F^{-1}(z)))\) where \(\|f_2(z)\|<\|f_1(z)\|\) most of the time.
Thus we always arrive at 0.

And in fact, we do have tons of different cuts, rather than \(f\) in the first post.

Now I'll show you computation stuffs.
The mathematica code(I can't afford it, I might just use a cloud or a cr*cked ver) is:
Code:
IF[z_] := 1/2 (1 - Sqrt[1 + 4 z])
F[z_] := -z + z^2
G[z_] := -I z - (1/2 - I/2) z^2 - I z^3 (I/2 + (2 I Log[z])/\[Pi]) -
  I z^4 ((-(1/2) + I/2) - I/\[Pi] + ((2 - 3 I) Log[z])/\[Pi]) -
  I z^5 ((5/8 - (3 I)/4) - (1 - (5 I)/4)/\[Pi] - (
     2 Log[z])/\[Pi]^2 - ((4 - (3 I)/2) Log[z])/\[Pi] - (
     6 Log[z]^2)/\[Pi]^2)
s = 100;
M[z_] := Nest[F, G[Nest[IF, z, s]], s]
Clear[L]
q = 2;
L[z_] := M[M[z]] - F[z]
Plot3D[0, {a, -q, q}, {b, -q, q}, PlotPoints -> 100,
PlotStyle -> Directive[RGBColor[1, 1, 1], Opacity[1]],
ColorFunction ->
  Function[{a, b, z},
   Hue[(Arg[L[N[a + b I]]])/(2 Pi), 1, 1 - (1/E)^Abs[L[N[a + b I]]]]],
  ColorFunctionScaling -> False, Mesh -> False, PlotRange -> {-1, 1},
BoxRatios -> {100, 100, .1}, ViewPoint -> Above]
Showing how close M(M(z)) in the code is to \(F(z)=-z+z^2\) in a field about \(\{z|z:\|z\|<.5, \arg(z)>-\tfrac{\pi}{2}\}\)

You can manipulate the s = 100; in the code, it's the number of iteration, s=200 gives a better half iterate etc.
Changing L[z_]:= M[M[z]] - F[z] to L[z_] := M[z] shows a graph about M(z), which is visualizing a descending absolute value of M(z) after each iteration, and as s goes bigger, the most part of the graph must be darker.

We can claim we found a good enough branch in this sense. The half iterates must violates \(f(F(z))=F(f(z))\) in some manner to prevent itself from being non-constructible.
Also I want to again emphasize that the deep rudiment of why we can't do a good calculation of this function is that \(1=e^{2\pi iz}\) doesn't always hold. This is the root of the crux.
Regards, Leo Smile
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Messages In This Thread
RE: Maybe the solution at z=0 to f(f(z))=-z+z^2 - by Leo.W - 01/20/2023, 01:35 PM

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