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 $0\leq |z|< R$ 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_2\lambda^{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_2\sum_{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_3\lambda^{3t-3}+3 f_2^2\sum_{k_1=0}^{t-1}\lambda^{3t-k_1-5} +\lambda D^3f^{t-1}(0) $

$ = f_3\sum_{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_2\sum_{k_1=0}^{t-1}\lambda^{2t-k_1-2}) (z-f_0)^2$

$+ (f_3\sum_{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 ^t\right) f_2}{2 (-1+\lambda )} (z-f_0)^2 $

$ + \frac{1}{6} \left(\frac{3 \lambda ^{-2+t} \left(-1+\lambda ^t\right) \left(-\lambda +\lambda ^t\right) 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 ^t\left(1-a_o\right)+\frac{\lambda ^{-1+t} \left(-1+\lambda ^t\right) \text{Log}\left(a_o\right){}^2}{2 (-1+\lambda )}\left(1-a_o\right){}^2 $

$ + \frac{1}{6}\text{ }\left(\frac{3 \lambda ^{-2+t} \left(-1+\lambda ^t\right) \left(-\lambda +\lambda ^t\right)\text{ }\text{Log}\left(a_o\right){}^4}{(-1+\lambda )^2 (1+\lambda )}+\frac{\lambda ^{-1+t} \left(-1+\lambda ^t\right) \left(1+\lambda ^t\right)\text{ }\text{Log}\left(a_o\right){}^3}{(-1+\lambda ) (1+\lambda )}\right)\left(1-a_o\right){}^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 $n \in \mathbb{N}$.