Talk:Tetration/Archive 3

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Archive 3

"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)

Please add new sections below

I think,according to the definition,0↑↑(-2)=1≠-∞, 0↑↑(-3)=0.The reasons are that 0^0=1 and 0^1=0,so 0↑↑1=0,0↑↑2=1, 0↑↑↑3=0,…,and we can get that 0↑↑n=0(n is odd) or 1(n is even).Since -2 is even,we can say that 0↑↑(-2)=0.Besides,according to the formulas: ①loga[a↑↑(n+1)]=a↑↑n. So,if n is an odd, 0↑↑(n+1)=1 and log0[0↑↑(n+1)]=log01=0=0↑↑n Similarly,if n is an even, 0↑↑(n+1)=0 and log0[0↑↑(n+1)]=0↑↑n=1=log00. So,when we use this formula for calculating 0↑↑n,we should define that log01=0 and log00=1. So,0↑↑0=log0(0↑↑1)=log00=1, 0↑↑(-1)=log01=0,0↑↑(-2)=log00=1, 0↑↑(-3)=log01=0 — Preceding unsigned comment added by Constant numbers (talk • contribs) 06:21, 8 July 2022 (UTC)

I also hold the viewpoint that (-1)↑↑0=-1≠1,(-1)↑↑(-1)=-1≠0,(-1)↑↑(-2)=-1,(-1)↑↑(-3)=-1 and 1↑↑(-1)=1≠0,1↑↑(-2)=1≠-∞, 1↑↑(-3)=1. Since 1^1=1,(-1)^(-1)=-1,so we can get that 1↑↑n=1 and (-1)↑↑n=-1 for any n∈N+. So,according to the formula that loga[a↑↑(n+1)]=a↑↑n,we can get 1↑↑1=1=log1(1↑↑2)=log1(1), 1↑↑2=1=log1(1↑↑3)=log1(1),…, here,we also ought to define that log1(1)=1,so 1↑↑0=log1(1↑↑1)=log1(1)=1, 1↑↑(-1)=log1(1↑↑0)=log1(1)=1≠0, 1↑↑(-2)=log1[1↑↑(-1)]=1=log1(1) ≠-∞, 1↑↑(-3)=log1[1↑↑(-2)]=1=log1(1) Similarly,(-1)↑↑1=-1= log(-1)[(-1)↑↑2]=log(-1)(-1), (-1)↑↑2=-1=log(-1)[(-1)↑↑3] =log(-1)(-1),…,here,we also ought to define that log(-1)(-1)=-1,so (-1)↑↑0=log(-1)[(-1)↑↑1] =log(-1)(-1)=-1≠1, (-1)↑↑(-1)=log(-1)[(-1)↑↑0] =log(-1)(-1)=-1≠0, (-1)↑↑(-2)=log(-1)[(-1)↑↑(-1)] =-1=log(-1)(-1)≠-∞. (-1)↑↑(-3)=log(-1)[(-1)↑↑(-2)] =-1=log(-1)(-1).

Tetration to real heights

The linear approximation method isn't easily a good approximation of tetration if the base is lower than 2. I am suprised there is no mention of any kind of Kneser's method being used for real- and complex-valued tetration. Using Kneser, the values should be: e^^(pi/2) ~ 5.82366 (against 5.868...) and 0.5^^-4.3 = log_0.5(log_0.5(log_0.5(log_0.5(0.5^^-0.3)))) ~ -1.07191 - 3.10267i (against 4.03335...). In the quadratic approximation, 2^^0.5 ~ 1.45933..., but Kneser yields 1.45878. Kwékwlos (talk) 12:28, 16 February 2023 (UTC)

Usage in speech

What's the correct way to refer to the tetration operation when reading a mathematical expression out loud? For example, when reading the expression 3↑↑5, would you say, "three tetrated to five," or "three to the fifth tetration," or some such thing? — Preceding unsigned comment added by Mvrog (talk • contribs) 23:09, 27 April 2023 (UTC)

Tetration

What is 456 tetrated to 789? 2A02:C7C:5F3D:D500:7DA0:1FC2:12EF:D8B1 (talk) 21:14, 23 December 2023 (UTC)

456^^789 is congruent modulo 10^20 to 96042614856384249856. Moreover, 456^^789 is congruent modulo 10^790 to 456^^(789+c) for every positive integer c (the proof easily follows from my paper entitled "The congruence speed formula" (DOI: 10.7546/nntdm.2021.27.4.43-61)). --Marcokrt (talk) 14:29, 5 January 2024 (UTC)

Integer tetration peculiar property

In the "Properties" section of the Tetration page, I think that the constancy of the congruence speed should be mentioned since it is a peculiar property of hyper-4, it has been proven to hold (in radix-10, the well-known decimal numeral system) for any base that is not a multiple of 10 (see https://arxiv.org/pdf/2208.02622.pdf), and an explicit formula has also been given (see Equation 16 of https://nntdm.net/volume-28-2022/number-3/441-457/). Now, I am not going to edit the mentioned section since I received warnings in the past for this kind of stuff, but I will be glad to help you and provide proper references if someone thinks that such a result is worth mentioning. As a (trivial) special case, knowing that the constant congruence speed of the tetration base 3 is equal to 0 iff the hyperexponent is 1 and that it is 1 otherwise, we can state that Graham's number, G:=3^^b, is congruent modulo 10^(b-1) to 3^^c for any integer c=b+1,b+2,... and at the same time that the b-th rightmost digit of Graham's number is not the same of 3^^c for any integer c greater than b. Marcokrt (talk) 03:59, 5 January 2024 (UTC)

Added a short description (in parentheses) of a peculiar property characterizing integer tetration (i.e., tetration is the only hyperoperator having a constant congruence speed for nontrivial bases), providing a couple of references to the above-mentioned result since it is not easy to properly state it in less than a few lines.

In the above, I implicitly assumed radix-10, but it would be possible to derive analogous rules for any other square-free numeral system. Marcokrt (talk) 18:00, 7 January 2024 (UTC)

Unfortunately, somebody (anonimously) is trying to blank any contribution related to the discovery of the constancy of the congruence speed on the Web, deleting entire sections/pages with no reason (just as a personal attack against myself, I guess - see 16:18, 4 May 2024‎), and this occurred multiple times (see for instance https://en.wikipedia.org/w/index.php?title=Graham%27s_number&action=history - 16:23, 4 May 2024‎) on different platforms on the same couple of days, May 3rd and 4th, 2024, as you can see here (https://googology.fandom.com/wiki/Graham%27s_number?action=history - 16:43, 3 May 2024 ‎).

Now, I hope that this kind of vandalism can be prevented and forbidden in order to avoid losing the relevant information that these unknown people are trying to hide for some sort of personal reason and without providing any serious argument able to disprove peer-review results (and I am myself an endorser for the arXiv section math.NT who studyied this very specific topic for 10+ years, not some random amateur/mathematical crank... just to clarify the point). Thanks in advance for understanding!

Marco Marcokrt (talk) 22:28, 6 May 2024 (UTC)

The IP is correct, and was not commiting vandalism. This is a nonnotable result added by a person with a plain conflict of interest based on an unreliable journal. It never should have been added to the article in the first place. MrOllie (talk) 01:58, 7 May 2024 (UTC)

Rounding in the Examples table

I think the Examples table has an issue with spurious precision. Right now ${}^{6}2$ is stated to be 2.12004 × 10^{6.03123×1019,727}. I don't think the 2.12004 factor should be there because the rounding of 6.03123 has a much larger effect on the accuracy of the number.

The number was calculated by finding the common log of ${}^{6}2$. The integer part of the result starts with 60312260... and is 19,728 digits long. The fractional part starts with .3263437.... Then 10 is raised to the power of the integer and fractional parts to get the value of ${}^{6}2$. The fractional part gives us $10^{0.3263437...}=2.120038...$. This is where that factor comes from.

But remember that the integer part of the log is 19,728 digits long. It was rounded to 6.03123×10^19,727. Adding just one more digit of accuracy is a change of 4×10^19,721. The inaccuracy from this rounding vastly overshadows the 2.12 figure. It would be like saying the sun is 93 million miles and 4.2 inches away. If the 93 is rounded at all then the 4.2 is meaningless. Likewise, I believe the 2.12004 is meaningless.

Another way to calculate this would be as follows:

${}^{6}2=2^{{}^{5}2}\approx 2^{2.00353\times 10^{19,728}}=10^{\log _{10}{(2)}\times 2.00353\times 10^{19,728}}\approx 10^{0.603123\times 10^{19,728}}=10^{6.03123\times 10^{19,727}}$

I will be changing this as well as ${}^{4}4$, ${}^{4}5$, and ${}^{4}6$ for the same reason. It's quite possible that I made a mistake in my math or in my reasoning so if anyone wants to double check this it would be very helpful. Jak86 (talk)(contribs) 05:45, 23 April 2024 (UTC)

For ${}^{6}2$ the first few digits are 21200, we're just arguing over how many millions or billions of digits follow. The rounded part is in the integral power that ten will be raised to: it changes where the digits are, but not what they are. The number that has been rounded will always just be a one followed by some number of zeros: rounding changes the number of zeros. Adding one more digit of accuracy to the exponent of ten changes the number dramatically, but changes 2.12004 in no way at all. This is the magic of logarithms, and fundamentally how floating point math works. 66.113.23.42 (talk) 23:38, 20 July 2024 (UTC)

Open question?

The article states “It is not known whether ⁿq is rational for any positive integer n and positive non-integer rational q”. This doesn’t make any sense, as (if I’m understanding the concept of titration correctly) ¹0.5 would be equal to 0.5, and would fit all 3 criteria:

1. n is a positive integer
2. q is a position non-integer rational
3. ⁿq is rational

Is the article supposed to say “every positive integer n”? 203.220.166.72 (talk) 05:31, 14 September 2024 (UTC)

Tetration to infinite heights

The article should show the existence and the coordinates of the inflection point on the curve of y=x↑↑∞ within the domain interval [e⁻ᵉ , ᵉ√e]. It is somewhere near x=0.3944, y=0.5819, but greater precision should be used when presenting it. 50.110.99.89 (talk) 17:07, 18 December 2023 (UTC)

I used WolframAlpha to get those values at a much higher precision, so here it is:

x = 0.394416066798979621617062515433809095220247702822598132414641394633374582791277160634942

y = 0.581932705608592190120933697309893947888922734948425097472649678032890093766822476175890

I think that might be too much, but at least I helped you out with the precision. 107.9.41.132 (talk) 17:03, 5 October 2024 (UTC)

OEIS entry for 0.581932705608592190... —Quantling (talk | contribs) 17:35, 6 October 2024 (UTC)

Real heights

The section on real heights has serious issues. The continuity, differentiability, and regularity requirements do not seem to produce inconsistent results, contrary to the article (added a citation needed for this). In fact, [15], the Paulsen & Cowgill paper referenced in the complex heights section gives a result which fulfills all of these. This entire section should likely be reworked with this and subsequent papers in mind. 50.1.19.82 (talk) 19:42, 12 October 2024 (UTC)

"Application" section

The application section does not show an application, just a math problem contrived to illustrate tetration. To my (pretty extensive) knowledge tetration is not necessary or even useful to describe any physical phenomenon or scientific theory, nor particularly useful in a pure mathematics sense. I submit there is no application for tetration in the usual sense of the word "application" and that the section be deleted unless someone comes up with a sensible and legitimate real-world use. 2601:647:6480:B640:1D9F:27C1:933F:E72A (talk) 04:52, 9 January 2025 (UTC)

I agree that it's pure math, not anything applied. For pure math, it is a key step in constructing some ordinal numbers, for example in Cantor normal form, on the way to ε₀, which plays a key role in logic systems. Perhaps we should talk about some of this, though the section name might nonetheless need changing. —Quantling (talk | contribs) 13:41, 9 January 2025 (UTC)

the layeradd method for real heights

i discovered this myself btw also all slog() (they are all base 10) and 10^^ are linear approximation in the intermediate function x^^[3+~5] is ((10^)^5) x^^3 and x^^[5+~pi]=10^^(slog(x^^5)+pi) BUT x^^(I+F) where I is integer and F is in the range [0,1) is limit of log_x^k(x^^[(I+k)+~F]) where k goes to infinity and log_x^k is log base x iterated k times 2001:9E8:E1DE:7900:2EDE:7178:D4D6:8F96 (talk) 14:20, 4 May 2025 (UTC)

Which versions of specific tetration values should we show?

Please discuss here. —Quantling (talk | contribs) 16:18, 2 June 2025 (UTC)