Talk:Tetration

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The content of Super-root was merged into Tetration. The former page's history now serves to provide attribution for that content in the latter page, and it must not be deleted as long as the latter page exists. For the discussion at that location, see its talk page.

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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)[reply]

Which versions of specific tetration values should we show?

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

Recursive formula for nth super roots, not just the 3rd

I did a bit of messing around with the recursive formula for the 3rd super root and how it was derived, and I found that other super roots can be calculated with a similar formula, ${\displaystyle x_{n+1}=\exp(W(W(x_{n}W({x_{n}}^{x_{n}}W(\cdots W(^{m-2}{x_{n}}\ln y))))))}$, which gives the solution to ${\displaystyle ^{m}{x}=y}$ when x₀ is set to 1 and as n approaches infinity. Although, it does get more complicated with higher values of m, and it only works for when m is a positive integer greater than 2. This is original research however, so this shouldn’t be included in the article. I’ll show how I derived it though. It goes something like this: ${\displaystyle {\begin{aligned}^{m}x=y\rightarrow x^{^{m-1}x}=y\rightarrow e^{^{m-1}x\ln {x}}=y\rightarrow ^{m-1}x\ln {x}=\ln y\rightarrow x^{^{m-2}x}\ln {x}=\ln y\\\rightarrow e^{^{m-2}x\ln x}\ln {x}=\ln y\rightarrow e^{^{m-2}x\ln x}({^{m-2}x})\ln {x}=^{m-2}x\ln y\\\rightarrow ({^{m-2}x})\ln {x}=W(^{m-2}x\ln y)\cdots \end{aligned}}}$Repeat the 4th to last steps shown here until you reach ${\displaystyle ^{1}x\ln(x)=x\ln(x)}$ on the left side.

${\displaystyle {\begin{aligned}x\ln x=W(xW(x^{x}\cdots W(^{m-2}x\ln(y))))\rightarrow \ln x=W(W(xW(x^{x}\cdots W(^{m-2}x\ln(y)))))\\\rightarrow x=\exp(W(W(xW(x^{x}\cdots W(^{m-2}x\ln(y)))))\end{aligned}}}$You can then replace the x on the left side with x_n+1, and all of the x’s on the right side with x_n and set x₀ to 1. 107.9.36.50 (talk) 02:02, 23 June 2025 (UTC)[reply]

lower tetration

can you add lower tetration which is n copies of a combined by exponentiation, left-to-right. 46.98.144.28 (talk) 18:14, 1 October 2025 (UTC)[reply]

I think that you are talking about $${\displaystyle \operatorname {LowerTetration} (a,n)=a^{\left(a^{n-1}\right)}}$$ for n > 0.

I don't see that naming this right-hand side to be "Lower tetration" adds much of anything to this article. Do you have sources that argue otherwise? —Quantling (talk | contribs) 19:40, 1 October 2025 (UTC)[reply]

i tried to write a reply with proof of existence but it got removed because it had a blacklisted link 46.98.144.28 (talk) 05:52, 2 October 2025 (UTC)[reply]

anyways, for example 2 lower tetrated to 4 is 256 46.98.144.28 (talk) 05:53, 2 October 2025 (UTC)[reply]

I agree that 2 "lower tetrated" to 4 is 256. Not coincidentally, it is the case that ${\displaystyle 2^{\left(2^{4-1}\right)}=256}$ too. To include something like this definition or formula in the article, we need a reliable source — like a well-known textbook — that has a similar discussion. —Quantling (talk | contribs) 15:02, 2 October 2025 (UTC)[reply]

Although ${\displaystyle 2\leq \alpha <\omega \land 4\leq \beta <\omega \implies \alpha ^{\alpha ^{(-1+\beta )}}<(\alpha \uparrow \uparrow \beta )}$, notice that ${\displaystyle 2\leq \alpha <\omega \land \omega <\beta \implies (\alpha \uparrow \uparrow \beta )=\omega <\alpha ^{\alpha ^{(-1+\beta )}}}$. So lower=tetration beats tetration in the long run. JRSpriggs (talk) 21:33, 29 December 2025 (UTC)[reply]

I am not at all sure of this but, ... maybe the formula LowerTetration(α, β) = α^αβ−1 applies only for β < ω. Directly from their definitions, does lower tetration beat tetration in the long run? —Quantling (talk | contribs) 22:05, 29 December 2025 (UTC)[reply]

Letting α = 2. We see ${\displaystyle (2^{2^{-1+\beta }})^{2}=2^{2^{-1+\beta }\cdot 2}=2^{2^{-1+\beta +1}}}$ holds even when ω≤β.

Perhaps we could get the best of both worlds by redefining tetration as follows:

${\displaystyle \alpha \uparrow \uparrow 0=1}$,

${\displaystyle \alpha \uparrow \uparrow (\beta +1)=\max(\alpha ^{(\alpha \uparrow \uparrow \beta )},{(\alpha \uparrow \uparrow \beta )}^{\alpha })}$,

${\displaystyle \alpha \uparrow \uparrow \lambda =\sup(\{\alpha \uparrow \uparrow \beta :\beta <\lambda \})}$ when λ is a limit.

OK? JRSpriggs (talk) 23:31, 29 December 2025 (UTC)[reply]

If there's a reasonable source that discusses mixing tetration and lower tetration in that way, we could discuss it in the article. Because I'm always fearful of the hyperoperation cases where either argument is 0, I would define tetration and lower tetration for the other cases with

$${\displaystyle \alpha \uparrow \uparrow 1=\alpha }$$

$${\displaystyle \alpha \uparrow \uparrow \beta =\sup _{0<\gamma <\beta }\alpha ^{\alpha \uparrow \uparrow \gamma }}$$

$${\displaystyle \operatorname {LowerTetration} (\alpha ,1)=\alpha }$$

$${\displaystyle \operatorname {LowerTetration} (\alpha ,\beta )=\sup _{0<\gamma <\beta }\operatorname {LowerTetration} (\alpha ,\gamma )^{\alpha }}$$

And if it is noteworthy, we could add

$${\displaystyle \operatorname {BothTetration} (\alpha ,1)=\alpha }$$

$${\displaystyle \operatorname {BothTetration} (\alpha ,\beta )=\max(\sup _{0<\gamma <\beta }\operatorname {BothTetration} (\alpha ,\gamma )^{\alpha },\sup _{0<\gamma <\beta }\alpha ^{\operatorname {BothTetration} (\alpha ,\gamma )})}$$

(We could carefully add the cases with 0 too.)

—Quantling (talk | contribs) 15:50, 30 December 2025 (UTC)[reply]

I agree that LowerTetration needs to start with β = 1 because otherwise 1^α = 1 and then you are stuck there.

However, BothTetration works fine starting with β = 0 and α = 0.

If 2 ≤ α, then max (1^α, α¹) = max (1, α) = α.

Also max (1¹, 1¹) = max (1, 1) = 1. So for α = 1, BothTetration is the constant function 1 of β.

And max (1⁰, 0¹) = max (1, 0) = 1. So for α = 0, BothTetration is also the constant function 1 of β.

Notice that BothTetration acts like LowerTetration when α = 0 or β = the successor of a limit ordinal, and it acts like Tetration otherwise.

When α = 0, ordinary Tetration gives 1 when β is even and 0 when β is odd; so it cannot be continuous.

OK? JRSpriggs (talk) 15:51, 31 December 2025 (UTC)[reply]

real and complex heights

Has anyone defined the value of tetration for a height z that is a complex number, perhaps restricted to those with a real part in [0, 1), as follows? For some large integer n compute ⁿa. Use z to adjust ⁿa to get a value that approximates ^(n+z)a. Iteratively apply log_a to this value n times, to get an approximation for ^za. Finally take the limit using this process as n → +∞.

For example, we could use a z-based adjustment that makes sense for the case that the limit ^∞x = y exists. In particular, starting with x^y = y, we can write

x^{y + ε} = x^y (x^ε) = y (1 + ε ln x + O(ε²)) = y + εy ln x+ O(ε²) = y + ε ln y + O(ε²).

Iterating, by taking x raised to that and repeating z times in total, gives a final value of y + ε (ln y)^z + O(ε²).

The big win is that (ln y)^z is easy to interpolate for non-integer values of z. We can use ε = ⁿx − y and then define

${\displaystyle {}^{z}x=\lim _{n\to +\infty }\log _{x}^{[n]}\left(y+\left({}^{n}x-y\right)\left(\ln y\right)^{z}\right)}$

where log_x^[n] means apply the logarithm (base x) for n times, and where y = ^∞x. We hope that the limit makes sure that the O(ε²) terms are negligible.

So, for the case that y exists, does this work for non-integer values of z, is it analytic, does it somehow extend to values of x for which the limit ^∞x does not exist, and/or is there a textbook that thinks any of this is notable? —Quantling (talk | contribs) 18:01, 17 November 2025 (UTC)[reply]

Undue weight in section Complex heights

This section appears to give undue weight to the work of Vincent Vey. A neutral reformulation and possible reduction would be welcome. Furthermore, the only reference currently provided by Vey appears to be a self-published PDF that reads more like a set of personal notes than a scientific article.Lonico978 (talk) 22:21, 27 December 2025 (UTC)[reply]