Talk:Knuth's up-arrow notation

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This could do with conversion to use Wikipedia:TeX notation

I'll get right on it! Dysprosia

Can someone move this to "Knuth's up-arrow notation", please?

If it's gonna be moved, I'd vote for Knuth arrow. BTW I'd removed Ackermann function since Ackermann page no longer refers to it. Kwantus

use Knuth arrow for relatively small numbers LMAO! =) Kwantus

It was either that or relatively small large numbers! :) (Maybe smaller magnitude...) Dysprosia 05:49, 30 Aug 2003 (UTC)

I tried to set it so that 3(arrow arrow)5 was shown as also being equal to 3^(3^7625597484987), but couldn't get the notation right. Can someone fix that? DS 14:15, 7 May 2005 (UTC)

Done. - Gauge 23:03, 2 October 2005 (UTC)

[edit] 3²⁷=27³

Why is $3^{3^3}$ = 3²⁷, not 27³? Thanks, --Abdull 23:12, 30 September 2005 (UTC)

27³ = (3³)³ = 3⁹ ≠ 3²⁷. Writing the exponentials as $3^{3^3}$ = pow(3, pow(3, pow(3, 1))), we see that we evaluate these functions from the inside out, so pow(3, pow(3, pow(3, 1))) = pow(3, pow(3, 3)) = pow(3, 27) = 3²⁷, rather than 27³. - Gauge 22:59, 2 October 2005 (UTC)

[edit] Study

As I said in the article on the closely related Hyper Operator, I think it would beinteresting to investigate the notation for a non-natural number of arrows, equivalent to: $a^{(n)}b|n\notin\mathbb{N}$ I've already defined over all integers n, but I think it would be interesting to expand this to all reals, or possibly even all complex numbers. He Who Is 01:02, 23 April 2006 (UTC)

Sounds like hard work! I hope you can do it. Kaimiddleton 07:54, 22 April 2006 (UTC)

[edit] Formal definition

Okay, what's the formal definition of $a \uparrow^n b$ when $n = 0$ ? It was originally written as 1 in the formal notation, but as $a b$ in the tables below. Now the line has been removed altogether, meaning the function is currently undefined for $n = 0$ . It should be either

$a\uparrow^n b= \left\{ \begin{matrix} a^b, & n \isin \{ 0, 1 \}; \\ a\uparrow^{n-1}(a\uparrow^n(b-1)), & \mbox{otherwise} \end{matrix} \right.$

$a\uparrow^n b= \left\{ \begin{matrix} ab, & n = 0; \\ a^b, & n = 1; \\ a\uparrow^{n-1}(a\uparrow^n(b-1)), & \mbox{otherwise} \end{matrix} \right.$ .

cBuckley (Talk • Contribs) 13:31, 21 May 2006 (UTC)

The article had b=0, not n = 0. As the article stands, the up-arrow is defined for n ≥ 1. Dysprosia 13:40, 21 May 2006 (UTC)

Okay, my mistake. So what's with the first line in each of the tables? cBuckley (Talk • Contribs) 14:26, 21 May 2006 (UTC)

I see no reason why it isn't defined for n = 0. The definition $a\uparrow^n b = a^{(n-2)} b$ gives a perfect definition for all n &ge -2; n $\in\mathbb{Z}$ He Who Is 20:24, 21 May 2006 (UTC)

Well, what would be the point of that? The arrow notation should be used when it should be used, so there's no practical need to make such a definition when one can resort to conventional notation. Dysprosia 22:35, 21 May 2006 (UTC)

I don't mean to say that there is a practical need for it, I merely mean to say that because of that definition, the Knuth up-arrow notation is defined for all n greater than of equal to -2.He Who Is 22:31, 22 May 2006 (UTC)

I think you mean to say that the up-arrow is defined for all n ≥ 2: as it stands, the hyper operator isn't defined for negative n. Dysprosia 22:38, 22 May 2006 (UTC)

Actually I did mean negative two, because the standard notation is defined for all n greater than or equal to zero (or depending on your point of view, all integers), and a $\uparrow$ ^n b = a^(n-2) b. But, as you said, anyone can resort to the standard notation if they have to, so there's no reson to argue this point.He Who Is 19:45, 23 May 2006 (UTC)