Using a family of asymptotic tetration functions...

[JmsNxn]

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

Hey, Leo

I'm sorry; I don't think I follow. Would you mind elaborating? What is \( b^f(z) \), particularly?

Regards, James



OHHHH WWAIT, nevermind, I get it. You meant to write \( b^{f(z)} \).  You are absolutely correct.

What you have constructed here; using Sheldon's idea of a modified Schroder function; you've made,

\(
f(s) = \Omega_{j=1}^\infty a^{s-j} b^z\,\bullet z\\
\)

This function will be holomorphic for \( |a| > 1 ,b \neq 0,s \in \mathbb{C} \). This is similar to how I constructed the \( \phi \) method, where I took \( a = b = e \). The conjecture that stands is that this can only construct a \( \mathcal{C}^\infty \) tetration on \( \mathbb{R}^+ \). And converges nowhere in \( \mathbb{C} \) when you apply iterated logs.

You're construction method is perfectly valid though; it's how Sheldon justified my method; both ways are equivalent; his is more hands on with taylor series though.

By this, I mean, you can construct a family of tetrations,

\(
F(a,b,s) = \lim_{n\to\infty} \log^{\circ n} f(s+n)\\
\text{for}\,\,a>1\,b > 0\,s \in \mathbb{R}\,s > R\,\,\text{for some}\,\,R > 0\\
b^{F(a,b,s)} = F(a,b,s+1)\\
\)

It will probably diverge in \( \mathbb{C} \) though. It's going to look like the \( \phi \) method.

---

I'd suggest looking at something that solves the asymptotic equation; and keep \( b > e^{1/e} \) and real.  In such a sense,

\(
g(b,\lambda, s) = \Omega_{j=1}^\infty \frac{b^z}{e^{\lambda(j-s)} + 1}\,\bullet z\\
\)

Which satisfies the equation,

\(
\log_b g(b,\lambda, s+1) = g(b,\lambda,s) - \log_b(1+e^{-\lambda s})\\
\)

Or fiddle with Tommy's gaussian approach. Much of this paper extends to all \( b > e^{1/e} \); I just kept it with \( e \) to keep it simpler. Theoretically the beta method works for all \( b > e^{1/e} \). Not too sure about complex values yet.

Regards, James