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 evaluation of fractional iterations of functions. Superfunctions and their inverse functions allow evaluation of not only 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 Helmuth Kneser . Actually, for his proof, Kneser had constructed the superfunction of exp and corresponding Abel function ${\mathcal {X}}$, satisfying the Abel equation

${\mathcal {X}}(\exp(z))={\mathcal {X}}(z)+1$

the inverse function is the entire analogy of the super-exponential (although it is not real at the real axis).

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

$S(0)=t$

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

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

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

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

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 $t=1$,

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

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

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

$S(a)=b$.

Uniqueness

Holomorphism declared in the definition is essential for the uniqueness. If no additional requirements on the continuity of the function in the complex plane, the strip $\{z\in \mathbb {C} :-1<\Re (z)\leq 0\}$ can be filled with any function (for example, the Dirichlet function), and extended with the recurrent equation. A little bit more regular approach is fitting of the superfunction at the part of the real axis with some simple function (for example, the linear function), and following extension of this step-vice function to the whole complex plane.

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 $\mathrm {mul} _{c}$.

Quadratic polynomials

Let $H(z)=2z^{2}-1$. Then, $f(z)=\cos(\pi \cdot 2^{z})$ is a $(\mathbb {C} ,~0\!\rightarrow \!1)$ superfunction of $H$.

Indeed,

$f(z+1)=\cos(2\pi \cdot 2^{z})=2\cos(\pi \cdot 2^{z})^{2}-1=H(f(z))$

and

$f(0)=\cos(2\pi )=1$

In this case, the superfunction $f$ is periodic; its period

$T={\frac {2\pi }{\ln(2)}}\mathrm {i} \approx 9.0647202836543876194\!~i$

and the superfunction approaches unity also in the negative direction of the real axis,

$\lim _{x\rightarrow -\infty }f(x)=1$

The example above and the two examples below are suggested at

Rational function

In general, the transfer function $H$ has no need to be entire function. Here is the example with meromorghic function $H$. Let

$H(z)={\frac {2z}{1-z^{2}}}~\forall z\in D~$; $~D=\mathbb {C} \backslash \{-1,1\}$

Then, function

$F(z)=\tan(\pi 2^{z})$

is $(C,0\!\mapsto \!0)$ superfunction of function $H$, where $C$ is the set of complex numbers except singularities of function $F$. For the proof, the trigonometric formula

$\tan(2\alpha )={\frac {2\tan(\alpha )}{1-\tan(\alpha )^{2}}}~~\forall \alpha \in \mathbb {C} \backslash \{\alpha \in \mathbb {C} :\cos(\alpha )=0||\sin(\alpha )=\pm \cos(\alpha )\}$

can be used at $\alpha =\pi 2^{z}$, that gives

$H(F(z))={\frac {2\tan(\pi 2^{z})}{1-\tan(\pi 2^{z})}}=\tan(2\pi 2^{z})=F(z+1)$

Exponentiation

Let $b>1$, $H(z)=\exp _{b}(z)$, $C=\{z\in \mathbb {C} :\Re (z)>-2\}$. Then, tetration $\mathrm {tet} _{b}$ is a $(C,~0\!\rightarrow \!1)$ superfunction of $\exp _{b}$.

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 ${\mathcal {D}}$.