Talk:Conway chained arrow notation

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As it stands, this page is gibberish. I have no doubt that the equations are "correct" but they need explanation. Are the first 3 equations alternative schemes or are they all a single scheme defining one large number. -- SGBailey 22:07, 2003 Aug 27 (UTC)

They're supposed to make up a recursive definition. They're not correct, though. They don't cover the case with more than two arrows. Matthew Woodcraft

I understand Knuth's up-arrow notation which is explained. This "X>p>1=X>p" and "p>q=p^q" (using > for right arrow and ^ for exponent) explains nothing. I am inclined to empty the article unless someone can explain why I shouldn't. -- SGBailey 09:57, 2003 Aug 28 (UTC)

Fixed up the first part, however I don't get the chaining of 3 arrows as below. Dysprosia 10:06, 28 Aug 2003 (UTC)

Ach. Didn't see this page before. I sorta expected email. The eqns were correct, but although i specified, i guess i didn't make clear what X, p, and q meant. Okay, I tried working three 4-element chains, two fizzles and one probably explodes. You try it! =) Kwantus

Contents 1 Micropædia 1.1 More not-clearness 2 Resolve? 3 Why, when, and where? 4 New extreme-power notation proposed.

[edit] Micropædia

(Some things which don't belong in a main page IMO but are useful to understanding.--Kwantus)

• Every chain is equivalent to some shorter chain with the same head. That is, with any subchains X and Y, X→Y=X→n for some (usu very large) number n dependent in a nontrivial way on X and Y. This is a simple consequence of rules 1 & 2.
• A chain beginning with two 2s stands for 4. By rules 3 and 2, 4=2→2=2→2→1. By rule 1, 2→2→(n+1)=2→2→n. By induction, 4=2→2→n for all n. By the result above, all chains 2→2→Y=2→2→n for some n.
• A chain can be truncated ahead of any 1. By the result above, any chain X→1→Y=X→1→n for some n, and rule 1 instantly reduces that to X.

[edit] More not-clearness

" If, for convenience, we define f(n)=3→3→n then
3→3→64→2 = f⁶⁴(1) (see functional powers),

• now that doesn't follow. 3→3→64→2 = f(64→2) does however. Why should 64→2 become a superscript 64 and the 2 become a 1 inside the parentheses? If this is some special meaning of "functional powers, it isn't made clear by the link to function.
• The remaining lines of the paragraph will remain clouded in confusion until the above is sorted out.

G=f⁶⁴(4), and
3→3→65→2 = f⁶⁵(1) = f⁶⁴(27)
Since f is strictly increasing,
f⁶⁴(1) < f⁶⁴(4) < f⁶⁴(27)
which is the given inequality.

-- SGBailey 15:12, 2004 Jan 27 (UTC)

You're confusing 3→3→64→2 with 3→3→(64→2). Remember, "one must be careful to treat an arrow chain as a whole." Using the first rule, with X=3→3, p=64, and q=1, we get

3→3→64→2 = 3→3→(3→3→(...→(3→3)→1...)→1)→1, with 64 copies of 3→3 before all the 1s. Those 1s can be removed, leaving

3→3→(3→3→(...→(3→3)→1))...), again with 64 copies of 3→3, which is equal to f⁶⁴(1).

Factitious 11:02, September 5, 2005 (UTC)

[edit] Resolve?

"In this case the notation eventually resolves to being the leftmost number raised to some integer (usually enormous) power" This statement is pretty silly. I mean exponentiation just 'resolves' into repeated multiplication in the same way as multiplication resolves into an addition and addition really resolves into repeated incrementing (by 1). You might as well say that the notation resolves into a really long series of increments. Asteron 01:33, 25 January 2006 (UTC)

I disagree. It clarifies the structure of the result; the chained arrow notation is very hard to parse on first sight and statements like this make it easier to know what to expect. Also, it puts the chained arrow notation in its proper place as a generalization of simple exponentiation. For example, Knuth's up-arrow notation is also a generalization of simple exponentiation; would you say that it is silly to note that is a power of a? Obviously it would be ridiculous to describe either as "resolving" into repeated addition; the point is to relate them to the next most natural and already understood operation. Ryan Reich 02:00, 25 January 2006 (UTC)

[edit] Why, when, and where?

Why did Conway invent this notation?

When did Conway invent this notation?

Where (i.e., in what book or article) did Conway specify this notation? --Oz1cz 14:58, 3 June 2006 (UTC)

It's specified on page 61 of The Book of Numbers, but I don't know if that's where he first specified it. Factitious 22:30, 15 September 2006 (UTC)

[edit] New extreme-power notation proposed.

Let's consider the function Cw1 (x, n) (Conway's first sort extended function):

Cw1 (x) = x->x->x->...->x with n-1 arrows according to Conway chained arrow notation.

For example, Cw1 (3, 4) = 3->3->3->3; Cw1 (10, 5) = 10->10->10->10->10 etc.

Now, let's consider another function Cw2 (x) (Conway's second sort extended function):

Cw2 (x) = Cw1 (x, x).

For example, Cw2 (4) = Cw1 (4, 4) = 4->4->4->4; Cw2 (5) = Cw1 (5, 5) = 5->5->5->5->5 etc.

Now we can construct the Extreme-Power Notation:

x~1 = x;

x~2 = Cw2 (x)= x->x->x->...->x (with x-1 arrows) (incomparably larger than Graham's number!);

x~3 = Cw2 (x~2) = x~2 -> x~2 - x~2 -> ... -> x~2 (with (x~2)-1 arrows);

x~4 = Cw2 (x~3) = x~3 -> x~3 - x~3 -> ... -> x~3 (with (x~3)-1 arrows) etc.

...

x~ ~1 = x;

x~ ~2 = x~x;

x~ ~3 = (x~ ~2)~ ~2 = (x~x)~ ~2 = (x~x)~(x~x);

x~ ~4 = (x~ ~3)~ ~3 etc.

...

x~ ~ ~1 = x;

x~ ~ ~2 = x~ ~x;

x~ ~ ~3 = (x~ ~ ~2)~ ~ ~2 = (x~ ~x)~ ~ ~2 = (x~ ~x)~ ~(x~ ~x);

x~ ~ ~4 = (x~ ~ ~3)~ ~ ~3 etc.

...

Finally, lets say that Blt (x) = x~ ~ ~...~x (with x tildes (!)) (Beloturkin's function).

For example, 1 Beloturkin = Blt (G), where G is Graham's number;

1 Beloturkinion = Blt (Beloturkin);

1 Beloturkiniard = Blt (Beloturkinion).

Beloturkin 08:23, 15 September 2006 (UTC)