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+--- Thread: Generalized arithmetic operator (/showthread.php?tid=840)
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RE: Generalized arithmetic operator - hixidom - 03/12/2014
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...
RE: Generalized arithmetic operator - tommy1729 - 03/21/2014
https://sites.google.com/site/tommy1729/extended-distributive-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
RE: Generalized arithmetic operator - hixidom - 03/22/2014
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))I found an answer to part of your question. By that I mean I was able to find S^[1/2](exp(x)):
Then what is S^[1/2](f(x)) ? Or what is 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).
RE: Generalized arithmetic operator - tommy1729 - 03/22/2014
(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))I found an answer to part of your question. By that I mean I was able to find S^[1/2](exp(x)):
Then what is S^[1/2](f(x)) ? Or what is 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
RE: Generalized arithmetic operator - hixidom - 03/22/2014
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 \( S(S(x,n),S(x,n)) \)?
RE: Generalized arithmetic operator - MphLee - 03/22/2014
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.
RE: Generalized arithmetic operator - hixidom - 06/11/2014
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: