# Talk:Tetration

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## 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The article has an outdated sentence: "In particular, it is not known whether either of 4π or 5e is an integer."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 (talkcontribs) 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)