Using a family of asymptotic tetration functions...
#15
Hey James,
I've lately been considering an anologous familt of asymptotic tetration functions, satisfying the recurrence
\( f(z+1)=a^z*b^{f(z)} \) with arbitrary constant a and b
I simply take \( f(z)=g(a^z) \) and solve \( g(az)=z*b^{g(z)} \) term by term in series, which uses this code I wrote by wolfram mathematica 12, it iterates the recurrence to get the function converged
Code:
Clear[A, B, term, aa, IS, Z]
(* Solving \[Alpha] in coefficients *)
A = 1 + I;
B = 1/2;
term = 15;
aa[0] = 0;
aa[1] = 1/A;
IS = 1/A xx + Sum[aa[n] xx^n, {n, 2, term}];
Z = xx Series[Exp[IS Log[B]], {xx, 0, term}] - (IS /. xx -> A xx);
For[i = 2, i <= term, i++,
temp = Solve[Coefficient[Z, xx, i] == 0, aa[i]];
aa[i] = Simplify[temp[[1, 1, 2]]]]

Clear[\[Alpha], \[Beta], ff]
\[Alpha][z_] := Sum[aa[n] z^n, {n, 0, 15}]
\[Beta][z_] := Module[{x, q, o},
  x = N[z, 200];
  x = SetPrecision[x, 200];
  q = 0;
  While[Abs[x] > 10^-50,
   x = x/A;
   q = q + 1];
  o = \[Alpha][x];
  While[q > 0, o = x B^o; q = q - 1; x = A x];
  Return[o]] /; Abs[A] > 1
\[Beta][z_] := Module[{x, q, o},
  x = N[z, 200];
  x = SetPrecision[x, 200];
  q = 0;
  While[Abs[x] > 10^-50,
   x = x A;
   q = q + 1];
  o = \[Alpha][x];
  While[q > 0, o = Log[B, o/x]; q = q - 1; x = x/A];
  Return[o]] /; Abs[A] < 1
ff[z_] := \[Beta][A^z] /; Abs[A] > 1
ff[z_] := \[Beta][A^(z - 1)] /; Abs[A] < 1
I think maybe there's a relation between these functions, especially seeing how it diverges when a is close to 1, so I think, if this is correct:
\( \lim_{a\to1}f(z-g(a))\sim\mathrm{tet}_b(z) \) and g(a) is only determined by a, exploding to infinity when a is getting closer to 1
Also, these functions are multivalued(Taken the relation between f(z) and f(z-1)), maybe associated with Riemann surface?

Leo
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Messages In This Thread
RE: Using a family of asymptotic tetration functions... - by Leo.W - 08/05/2021, 04:51 PM

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