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 Generalized arithmetic operator hixidom Junior Fellow Posts: 20 Threads: 3 Joined: Feb 2014 03/11/2014, 03:52 AM (This post was last modified: 03/11/2014, 03:54 AM by hixidom.) Hello forum. So I've finally hit a dead end in my hyperoperator theory. "Dead end" as in I fear I've discovered that there is, in general, no closed for for [x]a, where [x] is my hyperoperator (with argument x) and a is the argument. So this is what I've got so far: Amateur document (The real meat of the theory starts in Section 4) Crude summary Define a binary operation, $a[x]b$, where $a[1]b=a+b$, $a[2]b=a\cdot b$, $a[3]b=a^b$ and so on. My goal is to get to a point where I can evaluate $a[x]b$ for $x\in C$. Next, define a unary operation, $[x]a$. Denote iteration of the binary and unary operators via a superscript: e.g. $a[x]^3 b=((a[x]b)[x]b)[x]b$ $[x]^3 a=[x][x][x]a$ Next, establish the following axioms: A0: $a[x]^0 b=a$ A1: $[x]a\equiv a[x]a$ A2: $[x]^n [x]^m a=[x]^{n+m}a$ (perhaps this is not really an axiom based on the way I have defined iteration of the unary operator) A3: $[x]^n a=a[x]2^n$ From these "axioms" the following relations, among others, can be proven (see the document for derivations): $a[4]b$ is not tetration (sorry forum) $(a[x]b)[x]c=a[x](b\cdot c)$ $a[x]^n b=a[x]b^n$ $(a[x]b)[x]^{-1} b=a$ (definition of inverse) $[x]2=[x+1]2$ (2 is a fixed point, explaining why $2+2=2\cdot 2=2^2$, etc.) The problem: I find that $[x+y]a=f_y^{log_2 a}(a)$, where $f_y(a)$ is the yth super-iteration of $f_1(a)\equiv [x]a$. What do I mean by "super-iteration" and how do I come to that conclusion? It's in Section 4.3 of the document, but here it is again anyway: Given the axioms, we can find that $[x]^{log_2 a}a=[x+1]a$. Define $f_1(a)\equiv [x]a$. Now we can iterate both sides $log_2 a$ times to obtain $f_1^{log_2 a}(a)=[x]^{log_2 a}a=[x+1]a$. Now we must define $f_2(a)\equiv f_1^{log_2 a}(a)$, and repeat the $log_2 a$ iteration to obtain $f_2^{log_2 a}(a)=[x+1]^{log_2 a}a=[x+2]a$. Repeating this algorithm y times results in $f_y^{log_2 a}(a)=[x+y]a$. I know that the "super-iteration" method is necessary because I've tried other more naive derivations of $[x+y]a$ which turned out to be wrong when testing various combinations of x and y (e.g. $[x+y]a$ should be the same for (x,y)=(0,1) and (x,y)=(1,0)). So anyways, I'm putting this out here on the forum because super-iteration seems to be an unmanageable concept to me, and I have not been able to find a way to avoid it. I hope that someone with more experience in this area of mathematics will have some idea of how to manipulate the super-iteration into something simpler or how to relate $[x]a$ and $[x+y]a$ without the need for super-iteration. Ideally, the end result is a closed form expression for $[x]a$, where x can be any complex number. I'm using the term "super-iteration" because it is used on another thread on the forum (which, btw, is the only place on the internet where I could find information on anything similar to what I'm encountering). Thanks for reading. I know that this is my problem, but if anyone finds it interesting enough to think about, I would appreciate any comments on the validity of what I have so far and suggestions on how to proceed. Thanks again. hixidom JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 03/11/2014, 03:15 PM (This post was last modified: 03/11/2014, 03:17 PM by JmsNxn.) Well I'm not certain, but your operators seem to be like a modified lower hyperoperators. These operators satisfy the recursion: $(x[s]y)[s-1] x = x[s](y+1)$ where $x[0]y = x+y$ Check if this function works. I'm pulling it out of a hat but I have a lot of math behind it.: $\frac{2}{x[s]y} = \frac{1}{\Gamma(-s)} \sum_{n=0}^\infty \frac{(-1)^n}{(x[n]y)n!(n-s)} + \sum_{k=0}^\infty \frac{a_k}{\Gamma(k-s+1)}$ where $a_k = \sum_{n=0}^\infty \frac{(-1)^k}{x[n+k]y}$ This may or may not work. Depending on if 1/x[s]y is holomorphic or meromorphic and if (x[s]y)[s-1]x is holo as well with decent enough behaviour. hixidom Junior Fellow Posts: 20 Threads: 3 Joined: Feb 2014 03/11/2014, 06:24 PM Quote:$(x[s]y)[s-1]x=x[s](y+1)$ I can't seem to prove or disprove that statement in my system, except by the following discrepancy: Wikipedia has $a[4]b=a^{a^{b-1}}$, which doesn't agree with what I have. For example: According to Wiki, $a[4]4=a^{a^3}$ According to what I have, $a[4]4=[3]^2 a=[3][3]a=[3](a^a)=(a^a)^{(a^a)}$. So I don't think that that recursion relation results in the same set of hyperoperators. Do you think that the equation with the summation still applies? What exactly should I take from that equation? Is it to be evaluated numerically? Thanks for the response. hixidom MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 03/11/2014, 10:49 PM These are not the "natural" (left-associative) hyperoperations family, and seems that are different from lower hyperoperations too. Why you chose this kind of bracketing? MathStackExchange account:MphLee hixidom Junior Fellow Posts: 20 Threads: 3 Joined: Feb 2014 03/11/2014, 11:20 PM (03/11/2014, 10:49 PM)MphLee Wrote: These are not the "natural" (left-associative) hyperoperations family, and seems that are different from lower hyperoperations too. Why you chose this kind of bracketing? I'm not familiar with the difference between left- and right-associative hyperoperators. I would say that the unary version is definitely righ-associative, as $[x][y][z]a=[x]([y]([z]a))$ but perhaps that is not what you are referring to. If I had to write a recursion relation along the lines of what has been posted so far, I would say it is: $a[x]b=(a[x]\frac{b}{2})[x-1](a[x]\frac{b}{2})$ which is neither left- nor right-associative, if I understand the meaning of that phrase. The ones that have been posted so far are $a [x] b=a[x-1](a [x] (b-1))$, and $a [x] b=(a [x] (b-1))[x-1]a$. which are very similar, with the only difference being that the [x-1] operation on the right side is flipped. But neither of these definitions result in the same [4] operator that I have. \begin{align*}a[4]2&=[3]a&=a^a\\a[4]4&=[3]^2a&=(a^a)^{(a^a)}\\a[4]8&=[3]^3a&=\left ((a^a)^{(a^a)}\right )^{\left ((a^a)^{(a^a)}\right )}\end{align*} etc. JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 03/12/2014, 02:59 AM (This post was last modified: 03/12/2014, 03:02 AM by JmsNxn.) Yes I noticed that it wasn't exactly lower hyper operators, but it's closer to it than anything else. It depends on a lot of factors but the summation formula I gave may or may not work. I think the recursion is more aptly written by the general law: $a [x] \alpha + \beta = (a[x] \alpha)[x-1](a[x] \beta)$ So it's almost like it's necessarily satisfying the exponential law. Which is really cool now that I think about it. Let's see. I'm thinking there's trouble evaluating $a[n]k$ for arbitrary natural n and arbitrary natural k and complex a. So let's be a bit more modest with our findings. So let us fix a to N for some natural N $N [m] k$ is definitely calculatable for m, k natural as well. Now let us suppose that $1/N [m] s$ is holo in s for m greater than or equal to exponentiation Define: $2/N [m] s = \frac{1}{\Gamma(-s)} \sum_{k=0}^\infty \frac{(-1)^k}{n! N[m]k (k-s)} + \sum_{k=0}\frac{a_k}{\Gamma(k-s+1)}$ Where $a_k = \sum_{j=0}^\infty \frac{(-1)^j}{j!N[m](k+j)}$ Now do the same trick (hope fully you get what I mean so that I don't have to reexplain) so that we get: $2 / w[m]s = \frac{1}{\Gamma(-w)} \sum_{k=0}^\infty \frac{(-1)^k}{k!(k[m]s)(k-w)} + \sum_{k=0}^\infty \frac{b_k}{\Gamma(k-w+1)}$ And do the same trick one more time to get, $w + s \neq 0$, and $w \cdot s \neq 0$: $2 / w [ z] s = \frac{1}{\Gamma(-z)} \sum_{k=0}^\infty \frac{(-1)^k}{k!(w[k]s)(k-z)} + \sum_{k=0}^\infty \frac{c_k}{\Gamma(k-z+1)}$ This may or may not work depending on if the functions you defined are holo morphic. Mine are. They should obey the recursion by some fractional calculus theorems I've given, it's a little clunky. but Not very hard to show. They may not equal your operators though, you'll have to do the calculations :p I don't really want to spend the time to rigourously justify this. It's very time consuming and it requires a lot of finesse to pull off. But I can see the cauchy integrals coming out as they should after about 3 pages of equations for one identity. lol hixidom Junior Fellow Posts: 20 Threads: 3 Joined: Feb 2014 03/12/2014, 04:37 AM I'll try to work with the equation you provided, but it will take me a while to figure it out. Thanks for the suggestion! Regarding the relation provided: $a [x](\alpha + \beta) = (a[x] \alpha)[x-1](a[x] \beta)$ It is only true when $\alpha=\beta$, unfortunately, because $(a[x] \alpha)[x-1](a[x] \beta)\neq(a[x] \beta)[x-1](a[x] \alpha)$ in general. For example (x=4): $(a[4]2)[3](a[4]4)\neq(a[4]4)[3](a[4]2)\leftrightarrow (a^a)^{\left ((a^a)^{(a^a)}\right )}\neq\left ((a^a)^{(a^a)}\right )^{(a^a)$ MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 03/12/2014, 11:18 AM (This post was last modified: 03/13/2014, 08:40 AM by MphLee.) (03/11/2014, 11:20 PM)hixidom Wrote: I'm not familiar with the difference between left- and right-associative hyperoperators. I would say that the unary version is definitely righ-associative, as $[x][y][z]a=[x]([y]([z]a))$ but perhaps that is not what you are referring to. If I had to write a recursion relation along the lines of what has been posted so far, I would say it is: $a[x]b=(a[x]\frac{b}{2})[x-1](a[x]\frac{b}{2})$ which is neither left- nor right-associative, if I understand the meaning of that phrase. The ones that have been posted so far are $a [x] b=a[x-1](a [x] (b-1))$, and $a [x] b=(a [x] (b-1))[x-1]a$. which are very similar, with the only difference being that the [x-1] operation on the right side is flipped. But neither of these definitions result in the same [4] operator that I have. \begin{align*}a[4]2&=[3]a&=a^a\\a[4]4&=[3]^2a&=(a^a)^{(a^a)}\\a[4]8&=[3]^3a&=\left ((a^a)^{(a^a)}\right )^{\left ((a^a)^{(a^a)}\right )}\end{align*} etc. well , i think i got it. tell me if i'm right first lets notice that we have $a[4]2= {^{2}a}$ $a[4]4= {^{2}(^{2}a)}$ $a[4]8= {^{2}(^{2}(^{2}a))}$ so we have that if $t(a)=a^a={^2 a}$ (note that since $[3]=hyper 3$ we have that $[3]a=t(a)$) $a[4]2^k=t^{\circ k}(a)=[3]^{k}a$ and so $t(a[4]2^k)=a[4]((2^k) \times 2)=t^{\circ k+1}(a)=[3]^{k+1}a$ so the formulae that define all your hyperoperation family are Quote:1) $a[1]n:=a+n$ 2) $[s]a:=f_s(a)=a[s]a$ (that is A1) 3) $a[s+1]n=[s]^{\log_2(n)}a$ (that is A3) with this three difinitions (and if we already have defined the function iteration) we can get all the values of your hyperoperations. and if im right again we have too $a[s+1](2n)=(a[s+1]n)[s](a[s+1]n)$ I think i have already found this family somewhere, probably in the tetration FAQ pdf ----- I've jsut noticed that we can maybe write it as functional equation let $d(x):=2x$ Quote:1') $F_{1,a}(n):=a[1]n:=a+n$ 2') $f_s(a)=[s]a:=F_{s,a}(a)$ 3') $F_{s+1,a}(d(n))=f_s(F_{s+1,a}(n))$ the last can be rewritten in this way 3'') $F_{s+1,a}\circ d=f_s\circ F_{s+1,a}$ and we have this property $F_{s+1,a}\circ d^{\circ t}=f_s^{\circ t}\circ F_{s+1,a}$ and there is a name for this functional equation, I don't remember it but is a kind of abel functional equation i guess. I'm sure that someone here can help you more about this. Lee EDIT: BINGOOO found them. I knew it! bo198214 wrote about this family long ago (2008 ) BALANCED HYPEROPERATIONS http://math.eretrandre.org/tetrationforu...operations MathStackExchange account:MphLee hixidom Junior Fellow Posts: 20 Threads: 3 Joined: Feb 2014 03/12/2014, 05:54 PM (This post was last modified: 03/12/2014, 06:21 PM by hixidom.) Thanks Lee! That is very exciting to read. One of the last things he mentions is Quote:$x[k+1]y={f_k}^{\circ (\log_2 y)}(x)$. which is equivalent to what I have as $a[x+1]b=[x]^{log_2 b}a$ Somewhere in his OP he says it can be extended to real-argumented hyperoperators, but he never gets there as far as I can tell. He does plot a[4]a, which is really cool, and shows the fixed point at a=2. There's a section on "Balanced hyperoperators" in the Wikipedia page. Apprently they were first considered in this paper. MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 03/12/2014, 06:19 PM I know! is even what i said too but i use $s$ for the rank usually MathStackExchange account:MphLee « Next Oldest | Next Newest »

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