What is this operator called?

Question

If $x \cdot 2 = x + x$

and $x \cdot 3 = x + x + x$

and $x^2 = x \cdot x$

and $x^3 = x \cdot x \cdot x$

Is there an operator $\oplus$ such that:

$x \oplus 2 = x^x$

and $x \oplus 3 = {x^{x^x}}$?

Also, is there a name for such a set of operators ops where...

Ops(1) is addition

Ops(2) is multiplication

Ops(3) is exponentiation

Ops(4) is $\oplus$

...and so on

Also, is there a branch of math who actually deals with such questions? Have these questions already been answered like 2000 years ago?

You may be interested in the Ackermann function. Makes for fun problems in recursion theory and such. — Jyrki Lahtonen, Jul 05 '16 at 09:48
FYI $(x^x)^x = x^{(x^2)} \ne x^{(x^x)}$ so the operation is not associative so the notation ${x^x}^x$ is ambiguous. We usually take it to mean $x^{(x^x)}$ because $(((x^x)^x)^x).... = x^{(x^m)}$ and can be expressed that way. — fleablood, Jul 05 '16 at 17:34
i think fleablood got the convention backward. $a^{b^c}$ is understood as $a^{(b^c)}$ because $(a^b)^c = a^{bc}$. — Anton Sherwood, Jul 06 '16 at 00:52

Mario Carneiro · Accepted Answer · 2016-07-05T08:51:25.360

55

This operation ${\rm Ops}(4)$ is called tetration, from the greek root tetra meaning four; it's also sometimes called a "power tower". There are also many further generalizations of this type of sequence; Knuth's up-arrow notation gives $a^{a^{a^a}}=a\uparrow\uparrow4$, so that $a\uparrow\uparrow n$ is the tetration operation. By adding more arrows you get pentation and so on, and the Conway chained arrow notation generalizes this still further.

FYI, for "to the power of-ation" the word you're looking for is exponentiation.

edited Jul 05 '16 at 08:51

answered Jul 05 '16 at 08:45

Mario Carneiro

27,438

6

Also that is similar to graham number. https://en.wikipedia.org/wiki/Graham%27s_number – Takahiro Waki Jul 05 '16 at 09:19
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Aha, and the answer to the second question is hyperoperation: https://en.m.wikipedia.org/wiki/Hyperoperation – uzilan Jul 05 '16 at 12:19
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Ah, the whims of the internet. This does not deserve to be my highest-voted answer. – Mario Carneiro Jul 06 '16 at 04:03
1

It's the bikeshed, Mario; you get used to it. – J. M. ain't a mathematician Jul 06 '16 at 12:36
The problem with this answer is that a↑↑n is not defined to be tetration operation. It is a consequence of a↑n being defined as $a^n$ and then further arrows is defined to be n number of iterations of itself minus one arrow on a. Much prefer @Curd 's answer – Ariana Jul 08 '16 at 17:59
@ArianaGrande It follows exactly the same definitional scheme as the OP used. Also I don't understand the distinction you are making - they have exactly the same expansion, only the notation is different. Anyway I'm not judging which is the "best" here, just giving some links to further information. – Mario Carneiro Jul 08 '16 at 22:14
@MarioCarneiro If you are complaining, I'll just give you a high five without any upvote. – Simply Beautiful Art Apr 20 '17 at 20:46
@Simply Or you could upvote one of my other posts in its stead! ;) – Mario Carneiro Apr 20 '17 at 23:21

score 9 · Answer 2 · answered Jul 05 '16 at 17:26

A more general function that combines all those operators has been defined by Ackermann:

$ \varphi(m,n,p) = \begin{cases} \varphi(m, n, 0) = m + n \\ \varphi(m, 0, 1) = 0 \\ \varphi(m, 0, 2) = 1 \\ \varphi(m, 0, p) = m &\text{ for } p > 2 \\ \varphi(m, n, p) = \varphi(m, \varphi(m, n-1, p), p - 1) &\text{ for } n > 0 \text{ and } p > 0. \end{cases} $

So for $p = 0, 1, 2$ you get

$\phi(m, n, 0) = m + n $
$\phi(m, n, 1) = m \cdot n $
$\phi(m, n, 2) = m ^ n $

and

$\phi(m, n, 3) = \overbrace{{{m ^ m} ^ m} ^ {...}}^{n}$

More generally see fast growing hierarchy. You can find an introduction to some of these topics in the articles cited in my post on Goodstein's Theorem. — Bill Dubuque, Jul 05 '16 at 23:53

Ariana · Answer 3 · 2016-07-06T02:39:17.190

It is known as a tetration, and it is normally written as $^na$ where n is the height of the power tower. It is the forth hyperoperation.

The zeroth hyperoperation is the successor function, and the first is the zeroth hyperoperation iterated, and so on

A more general way to define the nth hyperoperation is, using the notation, $H_n(a,b)$ where n is the nth hyperoperation,

${\displaystyle H_{n}(a,b)={\begin{cases}b+1&{\text{if }}n=0\\a&{\text{if }}n=1{\text{ and }}b=0\\0&{\text{if }}n=2{\text{ and }}b=0\\1&{\text{if }}n\geq 3{\text{ and }}b=0\\H_{n-1}(a,H_{n}(a,b-1))&{\text{if }n\in\mathbb{N},n>3}\end{cases}}}$

Some notations for hyperoperations are(for $H_n(a,b)$:

Square bracket notation: $a[n]b$
Box notation: $a{\,{\begin{array}{|c|}\hline {\!n\!}\\\hline \end{array}}\,}b$
Nambiar's notation : $a\otimes ^{n-1}b$
Knuth's up arrow notation: $a\uparrow^{n-2}b$
Goodstien's notation: $G(a,b,n)$
Conway's chained arrow notation: $a\rightarrow b\rightarrow (n-2)$
Bowers exploding array function: $\{a,b,n,1\}$
Original Ackermann function: ${\begin{matrix}\phi (a,b,n-1)\ {\text{ for }}1\leq n\leq 3\\\phi (a,b-1,n-1)\ {\text{ for }}n\geq 4\end{matrix}}$

Wow, Arianna Grande as a mathematician and stackexchange user. Exciting, — Chisko, Jul 06 '16 at 17:26
Is any of these 8 extending Peter Hurfords $C(a,\dots)$ (see Extending the Extensions)? — Raphael J.F. Berger, Mar 11 '18 at 21:47
@R_Berger The extension, I believe, grows faster than every function in the 8 listed — Ariana, Mar 12 '18 at 10:24

What is this operator called?

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