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1

Using the relation given in Wikipedia $$ssrt(a^ab^b)=\frac {\log(a^ab^b)}{W(\log(a^ab^b)}=\frac {a \log a + b \log b}{W(a \log a + b \log b)}$$ but as there is no nice formula for $W(x+y)$ this doesn't really get you where you want to go.


1

First, we have $$ e^{\textstyle 10^{10^{2.8}}} = 10^{\textstyle \log_{10}(e) \cdot 10^{10^{2.8}}} = 10^{\textstyle 10^{\left( 10^{2.8} + \log_{10}\log_{10}(e) \right)}} = 10^{\textstyle 10^{10^{A}}} $$ where $$A = \log_{10}(10^{2.8} + \log_{10}\log_{10}(e)) \approx \log_{10}(630.96 - 0.36) \approx 2.7998 $$ Now, setting $B=10^{10^A}$, do the same ...


1

In general, not necessarily. The left-hand side needs to be big enough. The smallest we can get is if we let $x \to -\infty$, which gives us $$ 10^{10^{10^{10^{x}}}} \to 10^{10^{10^{10^{-\infty}}}} \to 10^{10^{10^{0}}}=10^{10^{1}} = 10^{10} $$ so basically, we cannot reach down to $10^{10}$, but any number above that is large enough that we can find an $x$ ...


1

This is discussed on the Wikipedia page for tetration. Your question is somewhat ill-defined, in that we can trivially extend any function $f:\mathbb{N}\to\mathbb{N}$ to a new function $g:\mathbb{C}\to\mathbb{C}$ such that $g(z)=f(z)$ when $z\in\mathbb{N}$ and $g(z)=0$ otherwise. Now you might say that's silly, but now you need to tell me what critera a ...


1

This is an attempt at proving statement 1. Showing that the conclusion holds for any $y \in \mathbb{R}$ with $|y| > y^\star$ basically boils down to proving $\lim\limits_{|y| \to \infty}(x+yi)^{x+yi} = 0$ for fixed $x$. Without loss of generality we may assume $y >0$ since $\bar z^{\bar z} = \overline{z^z}$. We may also ignore the argument of $(x+yi)...


1

Just to make it possible to "close the case": The article is possibly Baker, I. N., and Rippon P. J. "A Note on Complex Iteration." The American Mathematical Monthly 92.7 (1985): 501-04. Web. This is online, for instance via JStor. (Access possible using an account of an affiliated university or organiszation)


0

Does ssrt(x) = x^^(1/2) actually no. because that means if a = n^^b then n = a^^(1/b) so let us assume that is true if x^^n = z , n aproaches to infinity. then x = z^^(1/n) = z^^(0) but we already know z^^(0) is always equal to 1 but x = z^(1/z) , (see previous topic) then z^^(0) = z^(1/z) and that is not true. hence the assumption has ...


3

Letting $M = 4\uparrow^3 4 \uparrow^3 4$, we have $$ 4\uparrow^3(M+1) = 4 \uparrow\uparrow (4 \uparrow^3 M) = 4\uparrow\uparrow N < N \uparrow\uparrow N$$ but $$ 4 \uparrow^3 (M+2) = 4\uparrow\uparrow (4 \uparrow\uparrow (4 \uparrow^3 M)) = 4\uparrow\uparrow (4 \uparrow\uparrow N) > 4 \uparrow\uparrow (2N) > (4 \uparrow\uparrow N) \uparrow\...



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