Superfunction

Superfunction comes from iteration of another function. Roughly, for some function $f$ and for some constant $t$, the superfunction could be defined with expression

${{S(z)} \atop \,}{= \atop \,}{{\underbrace {f{\Big (}f{\big (}...f(t)...{\big )}{\Big )}} } \atop {z\mathrm {~evaluations~of~function~} f\!\!\!\!\!\!}}$

then $S$ can be interpreted as superfunction of function $f$. Such definition is valid only for positive integer $z$. The most research and appllications around the superfunctions is related with various extensions of superfunction; and analysis of the existence, uniqueness and ways of the evaluation. For simple function $f$, such as addition of a constant or multiplication by a constant, the superfunction can be expressed in terms of elementary function. In particular, the Ackernann functions and tetration can be interpreted in terms of super-functions.

History

Analysis of superfunctions came from the application to the avaluation of fractional iterations of exponential. Superfunctions and their inverse functions allow evaluation of not onlu minus-first power of a function (inverse function), but also function in any real or even complex power. Historically, first function of such kind considered was ${\sqrt {\exp }}~$; then, function ${\sqrt {!~}}~$ was used as logo of the Physics department of the Moscow State University . That time, researchers did not have computational facilities for evaluation of such functions, but the ${\sqrt {\exp }}$ was more lucky than the $~{\sqrt {!~}}~~$; at least the existence of holomorphic function ${\sqrt {\exp }}$ has been demonstrated in 1950 by Belmuth Kneser .

Extensions

The recurrence above can be written as equations

$S(z\!+\!1)=f(S(z))~\forall z\in \mathbb {N} :z>0$

$S(1)=f(t)$.

Instead of the last equation, one could write

$S(0)=f(t)$

and extend the range of definition of superfunction $S$ to the non-negative integers. Then, one may postulate

$S(-1)=t$

and extend the range of validity to the integer values larger than $-2$. The following extension, for example,

$S(-2)=f^{-1}(t)$

is not trifial, because the inverse function may happen to be not defined for some values of $t$. In particular, tetration can be interpreted as super-function of exponential for some real base $b$; in this case,

$f=\exp _{b}$

then, at $t=0$,

$S(-1)=\log _{b}(1)=0$.

but

$S(-2)=\log _{b}(0)~\mathrm {is~not~defined}$.

For extension to non-integer values of the argument, superfunction should be defined in different way.

Definition

For complex numbers $~a~$ and $~b~$, such that $~a~$ belongs to some domain $D\subseteq \mathbb {C}$,
superfunction (from $a$ to $b$) of holomorphic function $~f~$ on domain $D$ is function $S$, holomorphic on domain $D$, such that

$S(z\!+\!1)=f(S(z))~\forall z\in D:z\!+\!1\in D$

$S(a)=b$.

Examples

Addition

Chose a complex number $c$ and define function $\mathrm {add} _{c}$ with relation $\mathrm {add} _{c}(z)=c\!+\!z~\forall z\in \mathbb {C}$ . Define function $\mathrm {mul_{c}}$ with relation $\mathrm {mul_{c}} (z)=c\!\cdot \!z~\forall z\in \mathbb {C}$.

Then, function $~\mathrm {mul_{c}} ~$ is superfunction ($~0$ to $~c~$) of function $~\mathrm {add_{c}} ~$ on $~\mathbb {C} ~$.

Multiplication

Exponentiation $\exp _{c}$ is superfunction (from 1 to $c$) of function $\mathrm {mul} _{c}$.

Abel function

Inverse of superfunction can be interpreted as the Abel function.

For some domain $E\subseteq \mathbb {C}$ and some $u\in E$,$v\in \mathbb {C}$,
Abel function (from $u$ to $v$ ) of function $F$ with respect to superfunction $S$ on domain $E\in \mathbb {C}$ is holomorphic function $A$ from $E$ to $D$ such that

$S(A(z))=z~\forall z\in E$

$A(u)=c$

The definitionm above does not reuqire that $A(S(z))=z~\forall z\in D$; although, from properties of holomorphic functions, there should exist some subset ${\mathcal {D}}\subseteq D$ such that $A(S(z))=z~\forall z\in {\mathcal {D}}$. In this subset, the Abel function satisfies the Abel equation.

Abel equation

The Abel equation is some equivalent of the recurrent equation

$F(S(z))=S(z\!+\!1)$

in the definition of the superfunction. However, it may hold for $x$ from the reduced domain ${\mathcal {D}}$.