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I was looking at a problem that deals with tetration. Here's the problem: $$\begin{split}x^{x^{x^{\ddots}}}=3 \\ x^3=3 \Rightarrow x= \sqrt[3]{3} \end{split}$$ However, I saw that the solution is $-\frac{3}{\ln(3)}W_{-1}\left(-\frac{3}{\ln(3)}\right)\approx 2.478$ and watched some videos which explain something like the convergence "test" of $e^{-e}$ and $e^e$ and more ones some Lambert $W_{-1}$ function but I don't undertand this one.

I try to solve that one by this logic: $$x^{x^{x^{\ddots}}}=n$$ So $$x^n=n\text{, which implies, $\forall n \in \mathbb{R}$,}\, x=\sqrt[n]{n}\text{.}$$

What's wrong with this logic?

Gabriel Benício
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  • Really nice and deep dive into this topic: https://arxiv.org/pdf/1908.05559.pdf – Benjamin Wang Mar 13 '24 at 01:30
  • @Benjamin, thanks for the article. It's will hep me a lot. – Gabriel Benício Mar 13 '24 at 02:21
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    Where did you get that "solution"? It's wrong. $-\frac{3}{\ln(3)}W_{-1}\left(-\frac{3}{\ln(3)}\right)$ is not $\approx 2.478$. In fact, is not even real. – jjagmath Mar 13 '24 at 02:48
  • To enhance on @jjagmath's comment $$ w=\exp(-LambertW(-\log(3),-1)) \ (result:) = 0.229750106592 + 1.26644774360 \cdot î \ \qquad \text{check:} \ w^{\frac1w} \ (result:) = 3.00000000000 $$ (computation by Pari/GP – Gottfried Helms Mar 16 '24 at 09:42

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The answer is simpler and no knowledge about the Lambert function is necessary.

The infinite exponential tower $$x^{x^{x^\cdots}}$$ is defined as the limit of the sequence $$x,x^x,x^{x^x},x^{x^{x^x}},\cdots$$ when this sequence converge. This happens to be for $x \in [e^{-e},e^{1/e}]$. So those are the only real values of $x$ for which the expression $$x^{x^{x^\cdots}}$$ makes sense. That means that $$f(x)=x^{x^{x^\cdots}}$$ defines a function $f:[e^{-e},e^{1/e}] \to \Bbb R$. The image of that function turn out to be $[1/e,e]$.

Since $3\not\in [1/e,e]$, that simply means that $$x^{x^{x^\cdots}} = 3$$ has no solution.

But your reasoning works if you choose a number in the image of the function. For any $y\in [1/e,e]$ the solution of $$x^{x^{x^\cdots}} = y$$ is in fact $x=\sqrt[y]y$.

jjagmath
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