Generalized arithmetic operator

[hixidom]

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...

[tommy1729]

https://sites.google.com/site/tommy1729/...e-property

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

[hixidom]

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.

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).

[tommy1729]

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.

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:
\( S^{1/2}(S^{1/2}(e^x))=S^{1/2+1/2}(e^x)=S(e^x)=sexp(x) \)

that is sufficient to see your answer is wrong ...
Sorry.

regards

tommy1729

[hixidom]

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 \( S(S(x,n),S(x,n)) \)?

[MphLee]

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.

[hixidom]

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:
   
Plot over n:
   
Plot over x: