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 Generalized arithmetic operator hixidom Junior Fellow Posts: 20 Threads: 3 Joined: Feb 2014 03/12/2014, 06:43 PM 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 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 03/21/2014, 10:31 PM 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 Junior Fellow Posts: 20 Threads: 3 Joined: Feb 2014 03/22/2014, 12:06 AM (This post was last modified: 03/22/2014, 12:09 AM by hixidom.) 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). tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 03/22/2014, 12:13 AM (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: $S^{1/2}(S^{1/2}(e^x))=S^{1/2+1/2}(e^x)=S(e^x)=sexp(x)$ that is sufficient to see your answer is wrong ... Sorry. regards tommy1729 hixidom Junior Fellow Posts: 20 Threads: 3 Joined: Feb 2014 03/22/2014, 12:42 AM (This post was last modified: 03/22/2014, 12:47 AM by 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 Fellow Posts: 95 Threads: 7 Joined: May 2013 03/22/2014, 09:38 AM (This post was last modified: 03/22/2014, 03:01 PM by 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. MathStackExchange account:MphLee hixidom Junior Fellow Posts: 20 Threads: 3 Joined: Feb 2014 06/11/2014, 05:10 PM (This post was last modified: 06/11/2014, 05:11 PM by 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: « Next Oldest | Next Newest »

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