Talk:Pentation

Disputed names

Are tetration, pentation, hexation, etc., the commonly used names? I'm not convinced. — Arthur Rubin (talk) 19:27, 23 June 2009 (UTC)[reply]

Yes, they are the most commonly used names, except for "hyper-4", "hyper-5" and "hyper-6". --116.14.26.124 (talk) 07:32, 24 June 2009 (UTC)[reply]

Disputed

The claim that $e{\uparrow }{\uparrow }(-1.85)=-1.85$ depends on the specific, non-standard definition of ultra exponential function, which does not appear in tetration. — Arthur Rubin (talk) 14:01, 24 June 2009 (UTC)[reply]

This is obtained using Kneser tetration, which, in recent days, is seen as "canonical" tetration. Kwékwlos (talk) 10:59, 28 March 2023 (UTC)[reply]

Tetration links to ultra exponential function. --116.14.26.124 (talk) 04:54, 26 June 2009 (UTC)[reply]

Yes, it does. But there is not a claim that the the ultra exponential function is tetration, only that it's an example of it. — Arthur Rubin (talk) 16:09, 26 June 2009 (UTC)[reply]

negative results

Tetration reports that 1^^(-1) is undefined. I commented that 0^^(-1) = log₀0 = undefined, but that may be incorrect. It follows that (-1)^^^3 is undefined. — Arthur Rubin (talk) 19:47, 16 October 2009 (UTC)[reply]

My mistake, restoring. 1^^^-1 is undefined, however, so some of the negative exponents are more confused. — Arthur Rubin (talk) 20:26, 16 October 2009 (UTC)[reply]

Hexation

Let's extend this up by one level to hexation, which is merely repeated pentation. We know that for a^^^^b, where a and b are natural numbers, the following is a complete list of all values of a^^^^b that are writeable:

1^^^^n = 1
n^^^^1 = n
2^^^^2 = 4

These values hold for all operations throughout the sequence. However, I think the article is interesting when it talks about defining (-1)^^^n for different values. Can this be extended to hexation?? How do we define (-1)^^^^n for all integers n?? Georgia guy (talk) 13:38, 22 January 2012 (UTC)[reply]

This article is about the pentation function and nothing more. Hexation is not deemed notable enough to have an article on it's own, though it is briefly mentioned in the article on Hyperoperations, so if you feel that the function needs to be covered in more depth fee free to discuss it at Talk:Hyperoperation. Robo37 (talk) 18:04, 29 March 2012 (UTC)[reply]

Pentation bounding ELEMENTARY.

According to Elementary recursive, ELEMENTARY is bounded by tetration (unless I'm misreading this). The line suggesting that pentation is a (upper) bound would therefore be correct, but it would be misleading since it seems to suggest this is a notable property of pentation, and so tetration does not bound ELEMENTARY.

Can someone confirm whether ELEMENTARY is bounded above by tetration? Either way, this sentence needs to be clarified (it's not as clear as it should be to outsiders what is meant by "bounds"), or else removed if entirely inaccurate. TricksterWolf (talk) 17:08, 21 May 2012 (UTC)[reply]

a↑↑b?

So, a↑b is a^b. But what then is a↑↑b? The sentence represents the value obtained by repeatedly applying the function x → a↑x, for b repetitions, starting from the number 1, tells quite little. How is it marked in exponents? 85.217.15.230 (talk) 01:45, 11 January 2014 (UTC)[reply]

a↑↑1 = a

a↑↑2 = a^a

a↑↑3 = a^(a^a)

a↑↑4 = a^(a^(a^a))

etc. That's what the sentence in question means. —David Eppstein (talk) 02:26, 11 January 2014 (UTC)[reply]

Ok, now I get it. But what means: x → a↑x, for b repetitions? If b = 2, then it will be a^a, but what has the x got to do with that? 85.217.15.230 (talk) 13:40, 21 January 2014 (UTC)[reply]

It means "replace x by a↑x, then replace the new value of x by a↑x, etc., for a total of b replacements." —David Eppstein (talk) 17:07, 21 January 2014 (UTC)[reply]

If one writes: f(x)=a^x, then a↑↑b = f^b(1), where f^b represents the bth iterate of f. I don't know if this helps anyone, but it's the way I think of it. — Arthur Rubin (talk) 11:07, 22 May 2014 (UTC)[reply]

Examples with 2

All examples with 2 are very misleading. This is because for any hyperoperation $2[n]2=2[n-1]2=4$ . You cannot see which is which. All examples must start from 3. It is indeed more complex to understand that way and use notation, but that goes with the subject already. 158.248.76.166 (talk) 13:30, 25 July 2021 (UTC)[reply]

More notations

Just wanted to mention, You can represent pentation as X^^^Y! Thlvs (talk) 00:00, 9 February 2023 (UTC)[reply]

This is already in the article, when it talks about Knuth's up-arrow notation. Unless maybe you consider ^ to be a different symbol from

\uparrow

, I guess. —David Eppstein (talk) 02:10, 9 February 2023 (UTC)[reply]

I believe that based on the prevalence of Knuth's notation, using any other notation would be a poor choice for this article. AJRobbins (talk) 18:03, 7 September 2024 (UTC)[reply]

3^^^x

Using Kneser's tetration as a base and then using the fixed point at x ~ -1.87451 (where 3^^x = x), we have:

3^^^-2.0 = -1.00000

3^^^-1.5 ~ -0.50211

3^^^-1.0 = 0.00000

3^^^-0.5 ~ 0.48464

3^^^0.0 = 1.00000

3^^^0.5 ~ 1.67959

3^^^1.0 = 3.00000

3^^^1.5 ~ 9.68335

3^^^2.0 = 7625597484987.00000

3^^^2.5 ~ 10^10^10^10^10^10^(1.20513*10^21977) Kwékwlos (talk) 11:02, 28 March 2023 (UTC)[reply]

teory for negative results

as you know, there are adition, multiplication, exponenciation, etc... and in the other hand, we have subtraction, division, radiciation, etc... multiplying with negative values needs negative values ((x*y)-0) exponenciating with negative y, needs division ((x↑-y)↓1) what if tetration uses radiciation ((x↑↑-y)↓↓2), pentation uses tetrational pentation ((x↑↑↑-y)↓↓↓3) etc something like that would maybe work and keep the logic sorry if i have bad grammar, im portuguese