fa.functional analysis - What's a natural candidate for an analytic function that interpolates the tower function?

The question is often phrased, can tetration or iterated exponentiation be naturally extended to the real and complex numbers. Using the notation $^{1}a=a, ^{2}a=a^a, ^{3}a=a^{a^a}$, how do you compute a number like $^{.5}2$, and what are the properties of $^{x}e$ ?

The Derivatives of Iterated Functions

Consider the smooth function $f(z): mathbb{C} rightarrow mathbb{C}$ and its iterates $f^{;:t}(z), t in mathbb{N} $. The standard convention of using a coordinate translation to set a fixed point at zero is invoked, $f(0)equiv 0$, giving $f(z)=sum_{n=1}^{infty} frac{f_n}{n!} z^n$ for $0leq |z|< R$ for some positive $R$. Note that $f(z)$ is the exponential generating function of the sequence $f_0, f_1, ldots ,f_infty$, where $f_0=0$ and $f_1$ will be written as $lambda$. The expression $f_j^k$ denotes $(D^j f(z))^k |_{z=0}$ . Note: The symbol $t$ for time assumes $t in mathbb{N}$, that time is discrete. This allows the variable $n$ to be used solely in the context of differentiation. Beginning with the second derivative each component will be expressed in a general form using summations and referred to here as Schroeder summations.

The First Derivative

The first derivative of a function at its fixed point $Df(0)=f_1$ is often represented by $lambda$ and referred to as the multiplier or the Lyapunov characteristic number; its logarithm is known as the Lyapunov exponent. Let $g(z)=f^{t-1}(z)$, then

$ Df(g(z)) = f'(g(z))g'(z)$

$ = f'(f^{t-1}(z))Df^{t-1}(z) $

$ = prod^{t-1}_{k_1=0}f'(f^{t-k_1-1}(z))$

$ Df^t(0) = f'(0)^t $

$ = f_1^t = lambda^t $

The Second Derivative

$D^2f(g(z)) = f''(g(z))g'(z)^2+f'(g(z))g''(z)$

$= f''(f^{t-1}(z))(Df^{t-1}(z))^2+f'(f^{t-1}(z))D^2f^{t-1}(z) $

Setting $g(z) = f^{t-1}(z)$ results in

$ D^2f^t(0) = f_2 lambda^{2t-2}+lambda D^2f^{t-1}(0)$.

When $lambda neq 0$, a recurrence equation is formed that is solved as a summation.

$ D^2f^t(0) = f_2lambda^{2t-2}+lambda D^2f^{t-1}(0)$

$ = lambda^0f_2 lambda^{2t-2}$

$ +lambda^1f_2 lambda^{2t-4}$

$+cdots$

$+lambda^{t-2}f_2 lambda^2$

$+lambda^{t-1}f_2 lambda^0$

$ = f_2sum_{k_1=0}^{t-1}lambda^{2t-k_1-2} $

The Third Derivative

Continuing on with the third derivative,
$ D^3f(g(z)) = f'''(g(z))g'(z)^3+3f''(g(z))g'(z)g''(z)+f'(g(z))g'''(z)$

$ = f'''(f^{t-1}(z))(Df^{t-1}(z))^3 $

$ +3f''(f^{t-1}(z))Df^{t-1}(z)D^2f^{t-1}(z)$

$ +f'(f^{t-1}(z))D^3f^{t-1}(z)$

$ D^3f^t(0) = f_3lambda^{3t-3}+3 f_2^2sum_{k_1=0}^{t-1}lambda^{3t-k_1-5} +lambda D^3f^{t-1}(0) $

$ = f_3sum_{k_1=0}^{t-1}lambda^{3t-2k_1-3} +3f_2^2 sum_{k_1=0}^{t-1} sum_{k_2=0}^{t-k_1-2} lambda^{3t-2k_1-k_2-5} $

Note that the index $k_1$ from the second derivative is renamed $k_2$ in the final summation of the third derivative. A certain amount of renumbering is unavoidable in order to use a simple index scheme.

Iterated Functions

Putting the pieces together and setting the fixed point at $f_0$ gives,

$f^t(z) = sum_{j=0}^infty D^j f^t(f_0) (z-f_0)^j $

$ = f_0+lambda^t (z-f_0)+( f_2sum_{k_1=0}^{t-1}lambda^{2t-k_1-2}) (z-f_0)^2$

$+ (f_3sum_{k_1=0}^{t-1}lambda^{3t-2k_1-3} +3f_2^2 sum_{k_1=0}^{t-1} sum_{k_2=0}^{t-k_1-2} lambda^{3t-2k_1-k_2-5}) (z-f_0)^3+ ldots $

So far we have covered a decent amount of algebra, but still $t in mathbb{N}$. The equation $f^t(z)$ , $t in mathbb{N}$ is important because it is convergent when $f(z)$ is convergent.

Hyperbolic Fixed Points

When $lambda$ is neither zero nor a root of unity $lambda^t neq 1, t in mathbb{N}$, then the nested summations simplify to

$f^t(z)=f_0 + lambda ^t (z-f_0)+frac{lambda ^{-1+t} left(-1+lambda ^tright) f_2}{2 (-1+lambda )} (z-f_0)^2 $

$ + frac{1}{6} left(frac{3 lambda ^{-2+t} left(-1+lambda ^tright) left(-lambda +lambda ^tright) f_2^2}{(-1+lambda )^2 (1+lambda )}+frac{lambda ^{-1+t} left(-1+lambda ^{2 t}right) f_3}{-1+lambda ^2}right) (z-f_0)^3+ldots $

Hyperbolic Tetration

Let $a_0$ be a limit point for $f(z)=a^z$, so that $a^{a_0}=a_0$. Also $a_1=lambda$. This results in a definition for tetration of complex points for all except the set of points with rationally neutral fixed points. For the real numbers $a=e^{e^{-1}}approx 1.44467, a=e^{-e}approx 0.065988 $ have rationally neutral fixed points while $a=1$ is a superattractor. All other real values of $a$ are defined by hyperbolic tetration.

$ {}^t a = a_o + lambda ^tleft(1-a_oright)+frac{lambda ^{-1+t} left(-1+lambda ^tright) text{Log}left(a_oright){}^2}{2 (-1+lambda )}left(1-a_oright){}^2 $

$ + frac{1}{6}text{ }left(frac{3 lambda ^{-2+t} left(-1+lambda ^tright) left(-lambda +lambda ^tright)text{ }text{Log}left(a_oright){}^4}{(-1+lambda )^2 (1+lambda )}+frac{lambda ^{-1+t} left(-1+lambda ^tright) left(1+lambda ^tright)text{ }text{Log}left(a_oright){}^3}{(-1+lambda ) (1+lambda )}right)left(1-a_oright){}^3+ldots $

Summary

One issue that some researchers have with this approach is that it results in $^x e: mathbb{R} rightarrow mathbb{C} $.

Because this derivation is based on the Taylor series of $f^n(z)$, if $f(z)$ is convergent then $f^n(z)$ is convergent where $n in mathbb{N}$.