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I was fiddling around with tetration and I stumbled across an interesting idea, ${^\infty}{\sqrt[x]{x}}$. I messed around with the concept a little bit and I had the following idea:

Let ${^\infty}y = x$, then $$ {^\infty}y = x = y^x $$ Now we have $$ x = y^x,\\ y = \sqrt[x]{x} $$ Thus $$ \sqrt[x]{x}^{\sqrt[x]{x}^{\sqrt[x]{x}^\cdots}} = x $$ I think it makes sense, but I'm not entirely sure if it's right or not. Plugging in some values for x and shoving it into a recursive function in python showed that it worked for $x = 2$, but at $x = 3$ it converged at $2.478051576300804$. I'm not sure if this is because my identity is false or because floating-point precision isn't precise enough. So is this statement true?

user3002473
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    To the person who down-voted, care to explain why? – user3002473 Jul 13 '14 at 17:13
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    I think people are downvoting because this is considered to be a "well-known" problem (which is arrogant if you ask me). I upvoted because the problem shows some critical thought on the part of OP, and it is an interesting question. – Ryan Jul 13 '14 at 17:25

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It is not only true, but mentioned explicitly in the Wikipedia article on the subject. :-) In fact, it has been known since the time of Euler. What you forgot to add is that infinite tetration only converges for $x\in\Big[e^{-e}~,~\sqrt[e]e\Big]$.

Lucian
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  • Oh so that's where the down-votes were coming from. Why does it converge for $x \in [e^{-e}, \sqrt[e]{e}]$? – user3002473 Jul 13 '14 at 17:19
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    Because $\sqrt[x]x$ has a global maximum in $x=e$. Also, notice that $e^{-e}=\sqrt[1/e]{1/e}$ . – Lucian Jul 13 '14 at 17:34