Talk:Tetration

The superimaginary?

Define a surreal constant s as ssrt(-1). Is it possible to break this down into real/imaginary parts, or would it turn into its own unique unit? ArtismScrub (talk) 21:34, 13 March 2018 (UTC)

$s=-1$, as $^{2}(-1)=(-1)^{-1}=-1.$. So it's in fact real(!). Alfa-ketosav (talk) 12:18, 1 July 2020 (UTC)

Open questions

The article has an outdated sentence: "In particular, it is not known whether either of ⁴π or ⁵e is an integer." In 2020, modern computer algebra systems are able to answer one of these questions, namely that ⁴π is not an integer: ${\begin{aligned}\pi ^{\pi ^{\pi ^{\pi }}}\approx 9.08\times 10^{666262452970848503}\end{aligned}}$ 213.41.135.21 (talk) 08:44, 12 July 2020 (UTC)

That doesn't prove it's not an integer; it only proves that if it is an integer it has 666,262,452,970,848,504 digits (that makes it big!) Georgia guy (talk) 11:40, 12 July 2020 (UTC)

But has it not been recently proved (by example of the first Skewes`s number) that the result of multiplying two transcendental numbers cannot be an integer? 213.41.135.21 (talk) 15:23, 12 July 2020 (UTC)

If that statement were true, it would prove that there are no transcendental numbers. Two transcendental numbers CAN make an integer when multiplied together; an example is pi times (1/pi) equals 1. Georgia guy (talk) 15:29, 12 July 2020 (UTC)

I put it a little inaccurately. I meant the exponentiation (1/pi)^(1/pi) is not an integer 213.41.135.21 (talk) 15:41, 12 July 2020 (UTC)

It clearly isn't; it's approximately 0.69462799224. Georgia guy (talk) 15:50, 12 July 2020 (UTC)

Euler paper on infinite height

How come that this paper is not mentioned: De formulis exponentialibus replicatis https://scholarlycommons.pacific.edu/cgi/viewcontent.cgi?article=1488&context=euler-works — Preceding unsigned comment added by Laczkó Dávid (talk • contribs) 19:07, 22 August 2020 (UTC)

Examples correct?

I think there is an error in the table of "Examples of tetration", I find it suspect that for x=3,...,9, when x^^4 = exp_10^3(z), then x^^5 = exp_10^4(z) (??) which is 10^exp_10^3(z) = 10^(x^^4), while x^^5 = x^(x^^4), and 10 ≠ x. — MFH:Talk 13:56, 17 February 2022 (UTC)

Wait... x^^5 = x^(x^^4) = 10^y with y = x^^4 lg x = exp_10^3(z) lg x = 10^t with t = exp_10^2(z) + lg lg x, and OK, lg lg x ~ -0.3 is small w.r.t. 10^10^z but not completely negligible for z = 1.09... but yes, it won't change the z-value within the given precision. This might merit a remark: For numbers x ∈ [2 .. 99] and n > 2, we have x^^n = exp_10^m(z) => x^^(n+1) = exp_10^(m+1)(z') with |z' - z| < 10^(-???). — MFH:Talk 15:18, 17 February 2022 (UTC)

"Misnomer"?

Is the following sentence (currently in the article) meaningful?:

"The term power tower is occasionally used, in the form "the power tower of order n" for ${\ \atop {\ }}{{\underbrace {a^{a^{\cdot ^{\cdot ^{a}}}}} } \atop n}$. This is a misnomer, however, because repeatedly raising to a power is not tetration (see below). Tetration is instead iterated exponentiation."

Tetration is, according to the "see below"-link, exactly that: a power tower. It is even specifically (and correctly) pointed out that exponentiation is not associative / right-assotiative, so a power tower of height n does the same as tetration by n. --Felix Tritschler (talk) 15:09, 19 February 2022 (UTC)

I took a whack at describing the iterated exponentiation process, using right-associativity as the starting point (the top-right), then descending down the tower, to the left. --Ancheta Wis   (talk | contribs) 16:28, 19 February 2022 (UTC)