# Tetration Forum

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In this thread, there is mention of "Trappmann's Balanced Hyper-operator", and then there is a small section on it at the very end of the FAQ. I guess I finally have to learn about the Lambert W function...

We use a uniqueness condition on sexp : for x,y >=0 : sexp(x+yi) is real entire.

We could change the base e to base 2 or change tetration to pentation to generalize things.

Imho that is the way to do hyperoperation and I believe that answers almost all questions. ( I read your paper ).

Imho there are 2 big questions remaining :

1) informally speaking : what lies between tetration and pentation ?

Once again I mean the " half-super functions " as has been discussed on this forum before ( mainly by myself and James Nixon ).

Let S mean "superfunction of ..." and S^[-1] "is the superfunction of ..."

We have S^[-1](f(x)) = f ( f^[-1](x)+1)

examples :
S(exp(x)) = sexp(x)
S^[-1](sexp(x)) = sexp(slog(x)+1) = exp(x)

Question : if we say S^[a+b](f(x)) = S^[a](S^[b](f(x)) = S^[b](S^[a](f(x))

Then what is S^[1/2](f(x)) ? Or what is S^[1/2](exp(x)) ?

(Question 2 is still under investigation and not formulated yet)

regards

tommy1729
Quote:what lies between tetration and pentation ?
Do we know what lies between addition and multiplication, or multiplication and exponentiation? I would be happy to know those first. I assume they would be simpler to find, but I can also imagine that they would be equally difficult to find.

Quote:Question : if we say S^[a+b](f(x)) = S^[a](S^[b](f(x)) = S^[b](S^[a](f(x))
Then what is S^[1/2](f(x)) ? Or what is S^[1/2](exp(x)) ?
I found an answer to part of your question. By that I mean I was able to find S^[1/2](exp(x)):

By definition:
$S^{1/2}(S^{1/2}(e^x))=e^x$

So we are trying to find some function $f$ such that
$f(f(x))=e^x$

If we define $b$ such that
$f(x)=e^{bx}$

then
$f(f(x))=e^{be^{bx}}=e^x$

$\Rightarrow be^{bx}=x$

$\Rightarrow bx\cdot e^{bx}=x^2$

$\Rightarrow bx=W(x^2)$, where W is the Lambert W function

$\Rightarrow b=W(x^2)/x$

$\Rightarrow f(x)\equiv S^{1/2}(e^x)=e^{bx}=e^{W(x^2)}$

There is your half-superfunction of exp(x).
(03/22/2014, 12:06 AM)hixidom Wrote: [ -> ]
Quote:what lies between tetration and pentation ?
Do we know what lies between addition and multiplication, or multiplication and exponentiation? I would be happy to know those first. I assume they would be simpler to find, but I can also imagine that they would be equally difficult to find.

Quote:Question : if we say S^[a+b](f(x)) = S^[a](S^[b](f(x)) = S^[b](S^[a](f(x))
Then what is S^[1/2](f(x)) ? Or what is S^[1/2](exp(x)) ?
I found an answer to part of your question. By that I mean I was able to find S^[1/2](exp(x)):

By definition:
$S^{1/2}(S^{1/2}(e^x))=e^x$

So we are trying to find some function $f$ such that
$f(f(x))=e^x$

If we define $b$ such that
$f(x)=e^{bx}$

then
$f(f(x))=e^{be^{bx}}=e^x$

$\Rightarrow be^{bx}=x$

$\Rightarrow bx\cdot e^{bx}=x^2$

$\Rightarrow bx=W(x^2)$, where W is the Lambert W function

$\Rightarrow b=W(x^2)/x$

$\Rightarrow f(x)\equiv S^{1/2}(e^x)=e^{bx}=e^{W(x^2)}$

There is your half-superfunction of exp(x).

Sorry for not using tex before but

By definition:
$S^{1/2}(S^{1/2}(e^x))=S^{1/2+1/2}(e^x)=S(e^x)=sexp(x)$

Sorry.

regards

tommy1729
Ah. I see.

By half-superfunction, I thought you meant the superfunction of exp(x), S(x,n), evaluated at n=1/2.

But I guess you're talking about the superfunction of S(x,n), $S^m(x,n)$, evaluated at m=1/2.

Since S is a function of 2 variables, I guess I have to ask...
Is $S^2(x,n)\equiv S(S(x,n),n)$, $S(x,S(x,n))$, or $S(S(x,n),S(x,n))$?
Superfunction is a multivalued function defined over a set of functions not over a set of numbers:

$S(f)=F$ means that $S$ takes a function $f$ and gives a function $S(f)=F$ calles superfunction of $f$

such that $F$ satisfies

1) $F(x+1)=f(F(x))$

since there are infinite solution for $F$ (infinite superfunctions) that means that $S(f)=F$ is multivalued and then is not a function at all and we have to put some restrictions:
using Trapmann-Kouznetsov terminology used in their paper "5+ methods..." we call $S_u(f)=F_u$ the $u$-based superfunction of $f$ the function $F_u(x)$ that satifies two requirements

1) $F_u(x+1)=f(F_u(x))$

2) $F_u(0)=u$

and we have

$F_u(n)=f^{\circ n}(u)$

In this way we obtain uniqueness over the naturals: in fact superfunction is equivalent to the "definition by recursion" that is unique .
But is not over the reals... there we need more requirments.

Obviously this is still not enough to achieve the uniqueness of $F_u$ (iteration of $f$) that would mean having $S$ to be a function over a set of functions (not multivalued).

By the way I guess that Trapmann and Kouznetsov tried to find such additionals requirments but my math level is not enough to understand it.

Anyways we have that $S^{\circ -1}$ is a function and $S^{\circ 1/2}$ is the half superfunction.

example :
let define $add_b(x)=b+x$ and $mul_b(x)=bx$ we have

$S_0(add_b)=mul_b$ (multiplication is the 0-based superfunction of addition)

so we search for a $S^{\circ 1/2}$ such that

$S^{\circ 1/2}(S^{\circ 1/2}(add_b))=mul_b$

and that if $b[1,5]x=hyper-(1,5)_b(x)$ we should have (maybe...)

$S^{\circ 1/2}(add_b)=hyper-(1,5)_b$
and

$S^{\circ 1/2}(hyper-(1,5)_b)=mul_b$

I apologize if I did some mistakes.
So here is a link to the updated document. I've added a little bit on non-integer iteration of the [x] operator as well as [x] for non-integer x. I used the results to write matlab code that plots $[x]^n a$ over ranges of a, n, and x values. The plots are also in the document. There are still some limitations, but the expansion method (See http://arxiv.org/pdf/hep-th/9707206v2.pdf, pg.31) seems to work very well for x<3 and a,n<2. $[x]^n a$ blows up for larger values of a and/or n, as expected, and the expansion produces poor results for x>3, since I currently only know inverse operations for [1], [2], and [3], and so my expansions for non-integer x are limited to 4 terms.

Plot over a:
[attachment=1067]
Plot over n:
[attachment=1066]
Plot over x:
[attachment=1068]
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