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)

Contents

[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)
Very big difference... 327 = 7,625,597,480,000 ≠ 273 = 19,683 Frozen Mists 21:59, 28 June 2007 (UTC)

Left-associative exponents and higher operators ARE used in some applications (although I can't site one offhand), but if you define higher operators based on left-association, you don't get essentially new operators, since in that case a^^b would = a^{(a-1)^b}, rather than a^a^...^a^a (b a's). —Preceding unsigned comment added by 24.165.184.37 (talk) 23:12, 11 March 2008 (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 ab 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.

or

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)

[edit] OR for citation

I understand that this is WP:OR, but it's the "proof" for the citation (or an estimate).

let's define T=SQRT(10). (T is approximately equal to 3.16, which is close enough to 3 for something like "moderately sized hard drives"). Using T as a base, 10 (in decimal) would be expressed as 100T. 10010 is expressed as 10000T. The amount of space on a hard drive (d) it would take to store a number (using ASCII) on a hard drive is one byte per digit. For the powers of 10, let N be the power. the desired number to store is 10N. Expressed in N, the number of bytes it takes to store a number is N+1. If Expressed as base T, it would be 2N+1. Let's Examine T7625597484987. (I know it isn't equal to 37625597484987, but comparatively, they're close). Such a beast would have 7625597484988 digits when expressed in Base T. As a result, it would have approximately 3812798742494 digits when expressed in base 10. assuming each digit takes one byte, the number can be stored in 3 812 798 742 494 bytes, or approximately 4 Terabytes (hard drive manufacturers terabyte). If using a binary storage mechanism (as might be used to most efficiently store any number, say as in a twos compliment storage), 3 decimal digits, can approximately be stored in 10 binary digits. at 8 bits to a byte, this would take approximately 1 588 666 142 705 bytes, or 1.5 terabytes. Moderately large hard drives these days are about 300gb, (large ones at 500gb), which is 12ish (ascii) or 5ish (binary) "moderately large hard drives"? McKay 06:38, 19 January 2007 (UTC)

[edit] Error in table?

It seems to me that there is an error in the first table under table of values. First, we have 2\uparrow\uparrow 6 = 2^{2^{65536}}\approx 10^{6.0 \times 10^{19,728}}

Then, in the very next table we have 2\uparrow\uparrow 7 = 2^{2^{2^{65536}}}\approx 10^{10^{6.0 \times 10^{19,728}}} .

Assuming the first one is correct, shouldn't this be 2\uparrow\uparrow 7 = 2^{2^{2^{65536}}}\approx 2^{10^{6.0 \times 10^{19,728}}} since we are simply looking at 2 raised to the power in the previous column? I'm not sure which of these is incorrect, but I thought I would just bring this to the attention of someone who is more certain. GreekHouse 02:11, 12 October 2007 (UTC)

Compared to the effect of the rounding error in the 6.0, replacing 2 by 10 at the bottom has a negligible effect: a factor a little more than 3 at the second level, i.e. a term 0.5 at the third level.--Patrick 06:15, 12 October 2007 (UTC)

[edit] Small issue on convention

This article starts with the example of multiplication. Although I think that the example of 3x2 is the correct one, it is not consistent with the general example ab (a copies of b instead of b copies of a). However the general trend in this article is of course b copies of a. So should we make the first example ba or change the explicit example to two copies of three, which is counter-intuitive? Paul (talk) 08:15, 25 January 2008 (UTC)