# Example $x$, $y$ and $z$ values for $x\uparrow^\alpha y=z$ where $\alpha\in \Bbb R-\Bbb N$

$\uparrow^n$ and $G(n,\cdot,\cdot)$ are notations for hyperoperation.

http://en.m.wikipedia.org/wiki/Hyperoperation

$n$ is the hyperoperations rank.

Can example $x$, $y$ and $z$ values be provided for either the following formula?

The formula can be notated:

$$z = x\uparrow^{-3}y$$

$$\text{or}$$

$$z = G(-1, x, y)$$

Or, alternatively, for this formula?

The formula can be notated:

$$z = x\uparrow^{-0.5}y$$

$$\text{or}$$

$$z = G(1.5, x, y)$$

I prepose a possible name for $\uparrow^{-0.5}$ to be addiplication, also resulting in the words addiply, addiplicative and addiplicativly.

• If you have time, I'm preparing a long answer including some interesting stuff about non-integer ranks. Commented Apr 19, 2015 at 11:57
• Yes please :) I'll check back later today, and I won't be marking answers for a few days. Commented Apr 19, 2015 at 12:07
• I tried to improve the look of the title, fell free to reset the edit. Commented Apr 19, 2015 at 15:33
• I strogly suggest to improve you question look in order to make it more easy to find. The fact that you use such long title with two really specific cases and the Goodstein's Notation, wich is really really uncommon, weakens the effectiveness of the title and limits the spread fo the question. continue Commented Apr 20, 2015 at 19:35
• Thanks, for persisting and explaining. Commented Apr 22, 2015 at 9:40

note for the unfortunate reader: my native language is not English so I apologize in advance for errors

the references link seems to be fixed now

# On non-integer rank Hyperoperations

When you ask if values of $G(n,-,-)$ can be provided if $n\in\Bbb Z$ or $n\in \Bbb R$ you are actually asking if is possible to define an extension of $G$ to the integers, rational or real numbers (or complex) satisfying the Hyperoperations recursion over all the domain.

## Brief introduction

Notation 1: With $G:\Bbb N\times\Bbb N\times\Bbb N\to\Bbb N$ we mean the Goodstein function [1] or equivalently the Hyperoperations sequence: let's work with $G$ (a $3$-ary function) as if it is an indexed family of binary functions $+_{s\in \Bbb N}:\Bbb N \times \Bbb N \to \Bbb N$ and call the index $s$ the rank of the hyperoperation.

$$m+_s n:=G(s,m,n)$$

Notation 2: The infix notation that I'm using ($+_s$) is not common at all, as the Goodstein $G$. Usually the most common notations for $+_s$ are $H_s$ (Wikipedia page prefix notation) or $[s]$ (Square bracket infix notation) or maybe $A_s(m)=2[s]m$ (comes from the Ackermann function and is used in some recursion papers about the Grzegorczyk hierarchy and in some recent works of D. Kouznetsov [6]) and the Knuth's Uparrow notation $\uparrow^n$ (that has a different indexing starting from $\uparrow^0=\times$) [2]

$$H_s(x,y)=x[s]y=x\uparrow^{s-2}y=G(s,x,y)=x+_s y$$

Definition 1 (Hyperoperations sequence): We define the indexed family $\{+_s\}_{s\in \Bbb N}$ recursively over the natural numbers ($b,n, s\in \Bbb N$)

$i)$ $b+_0n=n+1$

$ii)$ $b+_{s+1}0=b_{s+1}$

$iii)$ $b+_{s+1}(n+1)=b+_s(b+_{s+1}n)$

Where base values $b_{s+1}$ that give us the "natural/classical" hyperoperations sequence are the following

$iv)\,\,\,\, b_ {s+1}:= \begin{cases} b, & \text{if$s=0$} \\ 0, & \text{if$s=1$} \\ 1, & \text{if$s\gt 1$} \\ \end{cases}$

I will call the argument $b$ base, $s$ rank and $n$ exponent.

Remark 0 Note that this definition gives us the "standard" hyperoperations

$H_1(x,y)=x+_1y=x+y$

$H_2(x,y)=x+_2y=xy$

$H_3(x,y)=x+_3y=x^y$

$H_4(x,y)=x+_4y={}^{y}x$ (Tetration)

Terminology 1: About the use of the term rank for the argument $s$ I don't think that is official but at least it makes sense for various reasons. As far as I know the term was introduced by K. A. Rubtsov and G.F Romerio in different papers/reports for the first time since the year 2006 [3] (p.3) and the term has been widely adopted by the Tetration Forum in the subsequent years. Another good reason is that every Hyperoperation $H_n$ belongs to the class $\mathcal E^n$ of the Grzegorczyk Hierarchy [4], a Sub-recursive hierarchy that organizes the Primitive Recursive functions according to their growth rate. Since the position inside a hierarchy is usually denoted by the term rank (See $V_\alpha$ hierarchy of sets for example) it seems to me a perfect choice. About the base and exponent terms I'm not sure, Rubtsov and Romerio tried to introduce, and used systematically, the the terms hyperbase for $b$ and hyperexponent for $n$ together with an uniform terminology for right and left inverse hyperoperations [5].

Remark 1: This sequence follows from original Goodstein definition and it implies that the $0$-th rank hyperoperation trivially coincides with the successor function $a+_0 n=n+1$ and so does all the integers-negative ranks: you can find more about this in the good David K answer here, in my and Ibrahin Tencer's answers here [7] and on the Tetration Forum [8] Anyways it is possible to avoid the imposition of $a+_0 n=n+1$ with alternative definitions of the hyperoperations sequence that give us more freedom for the negative ranks: about this there is a large amount of work by Rubtsov and Romerio under the name of Zeration [3], [9], [10] and by Cesco Reale in [11]. The topic is quite controversial and not very well known so I suggest you those two thread on TetrationForum [12], [13]

Back to your question, Imho is important to notice that is unlikely that you can extend the rank to non-ingers values without finding a way to extend the base and the exponent too. If you look at the sequence $x_s:=2+_s n$, for a fixed $n$ , we have that $x_s\in \Bbb N$ if the rank is a natural number but if we want it to be continuous or analytic in the variable $s$, thus extending to $s \in \Bbb R$, is very likely that for most of the non-integers ranks $x_s$ will have non-integer values making the functions $+_s$ not closed on the naturals for most of the non-integers ranks even when the base and exponent are natural numbers.

Example: if $2+_{1.5}3=q$ and $q\in \Bbb R\setminus \Bbb Z$ then evaluating $2+_{1.5}4=q$ would need to know how to evaluate $2+_{0.5}q$

That's my impression.

## The higher-order function iteration approach

An interesting way to continue [14] is to look at some suitable space of binary functions $\mathcal H$ with $+_s\in\mathcal H$ together with a function $\Sigma:\mathcal H\to \mathcal H$ with the following property

$$\Sigma [+_s]=+_{s+1}$$

and continue investigating its dynamics because it turns out that the Hyperoperations are the natural iterations of this map $\Sigma$ applied to the $0$-th rank hyperoperation

$$\Sigma^{\circ n}[+_0]=+_n$$

The operator $\Sigma$ increases by one the rank of the hyperoperations so it's plausible to expect that the fractional/real/complex iteration of this map Is going to give us the fractional/real/complex rank hyperoperations:

$$\forall \sigma\in \Bbb C(\Sigma^{\circ \sigma}[+_0]=+_\sigma)$$

Terminology 2: The iteration we are talking about $-^{\circ n}$ can be defined recursively in the following way: given a function $f:X\to X$

$i)$ $f^{\circ 0}(\beta)=\beta$ or $f^{\circ 0}={\rm id}_X$

$ii)$ $f^{\circ n+1}(\beta)=f(f^{\circ n}(\beta))$ or $f^{\circ n+1}=f\circ f^{\circ n+1}$

My naive opinion is that if one is able to find a good space $\mathcal H$ such that $\Sigma$ is an operator then we could try to apply the powerful tools of operator theory.

This kind of point of view can be used for a larger class operation sequences. The idea of reducing the non-integer ranks problem ot a non-integer iteration problem is, again, as far as I know due to Henrik Trappmann [15](2008), the founder of the Tetration Forum . Some years later this idea was better developed by James Nixon (2011) with the concept of "meta-superfunctions" [16] who is still working on this point of view (see later).

To explain better this we have to first find what kind of map is or should be $\Sigma$, the map that increases the rank by one unit: the discourse is a bit long so I'll send you to one of my answer at MSE [17]. As you can read in my answer $\Sigma$ is closely related to the process of iterating a function, also called "taking the superfunction" in the case of $1$-ary functions and is also closely related to known problems as finding the solutions of Abel functional equations of the form $\chi(z)+1=\chi(f(z))$, Schröder's equations $s\cdot\Psi(z)=\Psi(f(z))$

A massive amount of research on finding/building unique superfunctions [20] was made and is still carried by D. Kouznetsov [18], [19] and you can find most of his work in this online enciclopedia [23].

Finding an unique $\beta$-based superfunction actually give us an "higher order" (because send functions to functions) function that maps the function $f(x)$ to the function $F(z)=f^{\circ z}(\beta)$

Definition 2 Given a function $f$ we define intuitively its $\beta$-based superfunction $F_\beta$ as the function that maps to every $z$ the $z$-th application of $f$ to $\beta$ $$F(z)=f^{\circ z}(\beta)$$

Proposition 1: The $\beta$-based superfunction of $f$ satisfies this equations

$i)$ $F_\beta(0)=\beta$

$ii)$ $F_\beta(z+1)=f(F_\beta(z))$

Definition 3 Given a suitable collection functions $H$ we define intuitively the $\beta$-based superfunction map as a function $\mathcal S:H\to H$ that maps to every $f$ its $\beta$-based superfunction $F_\beta$ $$\mathcal S_\beta:f\mapsto F_\beta$$ $$\mathcal S_\beta[F](z)=f^{\circ z}(\beta)$$

Is easy to see what this has to do with hyperoperations and with the $\Sigma$.

Lets define the sequence of "hyper-exponentiations"

Definition 4 We define the family $\{H_{b,s}\}_{b,s\in \Bbb N}$ of hyper-exponentiations as follows

$$H_{b,s}(n):=b+_s n$$

we have that $H_{b,1}(n)={\rm add}_b(n)=b+n$, $H_{b,2}(n)={\rm mul}_b(n)=bn$ and $H_{b,3}(n)=\exp_b(n)=b^n$

Proposition 2: From the definition 1 we have the following

$$H_{b,s+1}(n+1)=H_{b,s}(H_{b,s+1}(n))$$

In other words we have that the superfunction map is the map $\Sigma$ we are looking for because we have that every $(s+1)$-rank hyper-exp is the superfunction of the $s$-rank hyper-exp $$\mathcal S[H_{b,s}]=H_{b,s+1}$$

At this point it makes sens to ask if finding the non-integers iterations of $\mathcal S_\beta$ really gives us non-integer ranks hyper-exponentiations functions for some bases $b$

$$\mathcal S_{\beta=1}^{\circ \sigma}[H_{b,2}]=H_{b,2+\sigma}$$

Remark 2: In the equation above I've setted $H_{b,2}$ and $\beta=1$ because for all the natural ranks $s\ge 3$ we have that $H_{b,s}(0)=1$ by definition.

Question 1: The real question is: It is possible to find the unique... real... non-integer iterations of $\mathcal S$?

As far as I know the answer to this question is still unknown or at least unknown to me, an amateur mathematician, even if some hot posts by the user JmsNxn appeared in the last two months on the Tetration Forum [21], [22].

## References

• wow, nice work :) Commented Apr 19, 2015 at 18:47
• Do you agree with David Ks findings, that I hope I am expressing correctly here, that G(-∈ℕ, x, y) that is really G(-∈ℕ, x) = x + 1? Commented Apr 19, 2015 at 18:54
• @alan2here pls read my answer carefully, I explain it clearly and I give u links to other post about that and some of those are from me (see at tetration forum). Commented Apr 20, 2015 at 17:49
• Wow, what a tour-de-force! We seem to be in agreement on the negative integer ranks. It looks like the fractional definition is still an open question, but this answer has interesting links on that topic, and it seems that if you digest this answer you will be well-prepared (so far as MSE can provide, anyway) to read further. Commented Apr 20, 2015 at 18:17
• @DavidK yeah thx, You are right, I needed years to find, collect and understand all this informations. I'm an "active researcher" in this topic (I'd call it Hyperoperations Theory) since the year 2010 and this is the summary of my knowledge on the topic. Commented Apr 20, 2015 at 19:21

$$G(n,a,b) = \begin{cases} b + 1 & \text{if } n = 0 \\ a &\text{if } n = 1, b = 0 \\ 0 &\text{if } n = 2, b = 0 \\ 1 &\text{if } n \ge 3, b = 0 \\ G(n-1,a,G(n,a,b-1)) & \text{otherwise} \end{cases}$$ for non-negative integers $n$, $a$, and $b$.

If we can extend this definition to negative integers $n$, then presumably we still want it to satisfy the recurrence $G(n,a,b) = G(n-1,a,G(n,a,b-1))$ when $a\geq 0$ and $b > 0$. In particular, for any $a\geq 0$ and $b > 0$, $$G(0,a,b) = G(-1,a,G(0,a,b-1)) = G(-1,a,b)$$ since $G(0,a,b-1) = b.$ That is, $$G(-1,a,b) = G(0,a,b) = b+1.$$

We can take this even further: for an integer $k \geq 0$, suppose $G(-k,a,b) = b+1$ for all $b \geq k$. Then $G(-k,a,b-1) = b$ for all $b \geq k+1,$ and $$G(-(k+1),a,b) = G(-(k+1),a,G(-k,a,b-1)) = G(-k,a,b) = b+1.$$ By induction, $G(n,a,b) = b+1$ for all $n\leq 0$ and all $b\geq -n.$

If you assume from the beginning that $G(0,a,b) = b+1$ for any integer $b$, then you end up with $G(n,a,b) = b+1$ for any integer $b$ and any integer $n\leq 0.$

I do not have any good idea yet how to deal with non-integer values of $n$.

• I think that this can add stuff to your nice answer: math.stackexchange.com/questions/170398/… Commented Apr 19, 2015 at 9:25
• @MphLee Good reference. It confirms what I suspected: I'm not the first to come to this conclusion, even on math.SE. Commented Apr 19, 2015 at 13:04
• @alan2here I defined $G(0,a,b)$ to match $H_0(a,b)$ on the page you linked to. Whether you use $b+1$ or $a+1$, either way leads to the conclusion that Successor is the operation "before" Successor. As observed in MphLee's answer to an earlier question, what we have at that point is really just a unary operation on one of the last two parameters of $G$. Commented Apr 19, 2015 at 13:19
• As for awards, I think you can give ordinary votes and accept an answer without awarding the bounty immediately, but you might as well hold out for a bit to see if anyone comes up with an interpretation of the operation for $n=3/2$. I'd really like to see that too. Commented Apr 19, 2015 at 13:19
• @David K hey, look At my answer, ive added alot of info and references about the negative ranks problem. Commented Apr 20, 2015 at 17:46

I think you could extend to non-integers (and even complex numbers, and so on) if we shift the definition of $$G(n,a,b)$$. $$$$G_{+}(n,a,b) = \left\{\begin{matrix} ?\hfill & & & n=0\hfill \\ b+1\hfill & & & n=1,~~b=0 \\ 0+a\hfill & & & n=2,~~b=0 \\ 0\times a\hfill & & & n=3,~~b=0 \\ 0^a\hfill & & & n\ge4,~~b=0 \\ G(n-1,a,G(n,a,b-1)) & & & \text{otherwise} \\ \end{matrix}\right.$$$$ Extending to negative number would then give you something like this... I think. $$$$G_{-}(n,a,b) = \left\{\begin{matrix} ?\hfill & & & n=0\hfill \\ b-1\hfill & & & n=-1,~~b=0 \\ 0-a\hfill & & & n=-2,~~b=0 \\ 0\div a\hfill & & & n=-3,~~b=0 \\ \sqrt[a]{0},~~\log_{a}(0+1) \hfill & & & n\le-4,~~b=0 & ??? \\ G(n+1,a,G(n,a,b-1)) & & & \text{otherwise} & ???\\ \end{matrix}\right.$$$$ This makes sense to me since negative values of $$n$$ should gives you the (left-, right-)inverse elements of their respective $$n$$ operator. meaning. $$$$G_{-}(n,a,G_{+}(-n,a,b)) = \left\{\begin{matrix} ?\hfill & & & n=0\hfill \\ b=b+1-1\hfill & & & n=-1\\ b=b+a-a\hfill & & & n=-2 \\ b=\frac{ba}{a}\hfill & & & n=-3 \\ b=\sqrt[a]{b^a}\hfill & & & n\le-4 \\ \cdots\hfill & & & \text{otherwise} \\ \end{matrix}\right.$$$$ And trying to get the more convoluted left-inverse... $$$$G_{-}(n,G_{+}(-n,a,b),b) = \left\{\begin{matrix} ?\hfill & & & n=0\hfill \\ b-1\hfill & & & n=-1\hfill \\ -a=b-(b+a)\hfill & & & n=-2\hfill \\ \frac{1}{a}=\frac{b}{ba}\hfill & & & n=-3\hfill \\ ?=\sqrt[b^a]{b}\hfill & & & n\le-4\hfill \\ \cdots\hfill & & & \text{otherwise} \\ \end{matrix}\right.$$$$ ...and we have failed. But since generally speaking only have two inverses then maybe is time to introduce complex numbers. Since I don't want find convoluted real solution to the problem. $$$$Z(n,a,b) = \left\{\begin{matrix} \exp_{a}^{n}{b}\hfill & & & n\ge4\hfill \\ b\times a\hfill & & & n=3\hfill \\ b+a\hfill & & & n=2\hfill \\ b+1\hfill & & & n=1\hfill \\ ?\hfill & + & ?\hfill & n=0\hfill \\ b-1\hfill & + & i\cdot(a-1)\hfill & n=-1 \\ b-a\hfill & + & i\cdot(a-b)\hfill & n=-2 \\ b\div a\hfill & + & i\cdot(a\div b)\hfill & n=-3 \\ \text{ssrt}_{a}^{n}(b)\hfill & + & i\cdot\text{slog}_{a}^{n}(b) & n\le-4\\ \end{matrix}\right.$$$$ Letting $$$$a[n]b = \{Z(n,b,a) \mid \forall n \in \mathbb{Z}\}$$$$ $$n=-1,0,1$$ seem kinda wastefull you could probably get rid of them with impunity. Ok drum roll... $$$$a=\Re\{(a[n]b)[-n]b)\}~~~~~~\&~~~~~~b=\Im\{(a[n]b)[-n]b)\}$$$$ Lets try it out... $$$$a=(a+b)-b)~~~~~~\&~~~~~~b=(a+b)-a$$$$ And $$$$a=\sqrt[b]{a^b}~~~~~~\&~~~~~~b=\log_a(a^b)$$$$ But unfortunately, $$$$a=(a+1)-1~~~~~~\&~~~~~~b\ne(a+1)-1$$$$ Also there was no need to include the complex numbers I just needed a left and right inverse tuple. If it seems confusing its probably because I flip the order when defining $$a[n]b$$ so that the leftness and rightness stays consistent with the input. from here we now know what $$a[0]b$$ should be that's because 1) $$a[+0]b=a[-0]b$$ and should also follow $$$$a=(a\cdot b) \cdot b)~~~~~~\&~~~~~~b=(a\cdot b) \cdot a$$$$ not much of an involution expert but zero seems to fit the bill (:B). also this should hold. $$$$a[0](b+1) = a[-1](a[0](b))$$$$

OK BACK TO THE ORIGINAL QUESTION. What the heck is $$a[2.5]b$$. Well we would probably still need to use the pesky recursive formula. $$$$a[n]b = a[n-1](a[n](b-1))$$$$ The fact that we not able to get the right-inverse for $$n=1$$ suggests the Z(n,a,b) is not ready to answer the original question. We now need some other definition that has a right inverse for $$n=1$$ but I will do that in a follow up answer becasue its already getting too long.