# Tag Info

## New answers tagged tetration

-1

I got a message from Tommy1729. He considered nonzero $T$ such that : $$T = \lim \frac{A^n}{(f(n,2) - \sqrt 2 - L \ln 2 ^n)}$$. Where $f(n,2)$ is the nth superroot of $2$ , $L$ is the constant from the Op and $A$ is a constant. If the limit $T$ does not exist at least the best fitting $A$ is considered. In other words : $$A^n$$ ~ $$\frac{1}{(f(n,2) ... 0 Instead of x^{f(x)} , I should have used (\sqrt2)^{f(x)}. Then we get  D (\sqrt 2)^{f(x)} / f ' (x) = \ln(\sqrt 2) (\sqrt 2)^{f(\sqrt 2)} f ' (\sqrt 2) / f ' (\sqrt 2) = \ln(2). Notice that  D (\sqrt 2)^{(\sqrt 2)^{f(x)}} / f ' (x) = ( \ln 2)^2  as it should. 2 The actual definition is$$a_{m,n}= \begin{cases} 1 & \text{if }n=0\\ \frac{1}{n!}& \text{if }m=1 \\ \frac{1}{n}\sum_{j=1}^n j a_{m,n-j} a_{m-1,j-1} & \text{otherwise} \end{cases} $$So to construct this array you can think of a_{m,n} as a function f(m,n) defined recursively as$$f(m,n)= \begin{cases} 1 & \text{if }n=0\\ ...

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You have: $${}^3(ab)$$ you already know that it is equal to: $$(ab)^{({}^2a)^b*({}^2b)^a}$$ from your second equation. So using the fact $$(xy)^z=x^z*y^z$$ you can determine $$a^{({}^2a)^b*({}^2b)^a}*b^{({}^2b)^a*({}^2a)^b}$$ expanding into exponentiation you get $$a^{(a^a)^b*(b^b)^a}*b^{(b^b)^a*(a^a)^b}$$ That and the power of power rule get you ...

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$$\mathcal L=\lim\limits_{n\to\infty}\frac{\sqrt[n]2_s-\sqrt2}{(\ln2)^n}\tag1$$ This limit is only possible if $$\lim\limits_{n\to\infty}\frac{\sqrt[n+1]2_s - \sqrt 2}{\sqrt[n]2_s - \sqrt 2}= \ln2$$ To show this , use l'hopital We get with $f(x) = x^{f(x)}$ : $\frac{D x^{f(x)}} {f ' (x)} = \frac{ \sqrt 2 ^2 2\ln(\sqrt2) }{2} = \ln2$. Qed This is part ...

-1

This might be relevant Limit involving tetration It might not be compatible with the Op conjecture ?

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$$\mathcal L=\lim\limits_{n\to\infty}\frac{\sqrt[n]2_s-\sqrt2}{(\ln2)^n}\tag1$$ Notice the resemblance with the Koenigs function https://en.m.wikipedia.org/wiki/Koenigs_function In fact it is a Koenigs function with the variable fixed to the value $2$. Since $1 < 2 < \exp(1/e)$ and the derivative is not $0$ or $1$ , the Koenigs function converges ...

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I have some observations, from where someone more experienced might be able to derive the proof - maybe this is helpful) Let the iterated functional root ("superroot of order") $B(z,n)$ (which finds the " B "ase of the powertower) be defined as $$\;^n b = z \qquad \to \qquad B(z,n) = b$$ For the following let us always denote $u$ for the ...

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I've just entered in this problem again and propose now to use the power series for the inversion $\;^3 W(x)= \text{reverse}(x \cdot \exp(x \cdot \exp(x)))$ using the Lagrange series-inversion. You'll get a series with a very limited radius of convergence; however it seems to be finite and not really zero. But the signs of the coefficients alternate, so ...

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I would suggest to study formally $f_n(x)$ : look at the values in 0 and 1, and the variations in between study $f_n$ recursively At this point prove that there is 1 root only $r_n$ prove that $r_n$ decreases show that the limit cannot be > 0.

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This relates to the deeper questions of where does the iterates of $a^z$ converge in the complex plane for infinite exponential towers at a fixed point $A$ such that $a^A=A$ and where is the onset for period $n$ behavior. Let $\Delta z$ be an infinitesimal. Since $a^A=A$, $a^{A+\Delta z}=A a^{\Delta z}=A + Ln A \ \! {\Delta z}$. Therefore \$A+\Delta z ...

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