Talk:Tetration/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  &#124; 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)