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This question already has an answer here:

I was reading Evaluating tetration to infinite heights (e.g., $2^{2^{2^{2^{.^{.^.}}}}}$). For what values of $x$ does tetration to infinite heights (i.e., $x^{x^{x^{x^{.^{.^{.}}}}}}$) converge?

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marked as duplicate by J. M. is a poor mathematician, Jim, Julian Kuelshammer, Davide Giraudo, Martin May 28 '13 at 6:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ what do you think? $\endgroup$ – Lost1 Apr 15 '13 at 14:20
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    $\begingroup$ Note that such a number $x=t^{t^{t^{.^{.^.}}}}$ (if it exists) will necessarily lie at a point of intersection of $y=x$ and $y=t^x$. $\endgroup$ – Cameron Buie Apr 15 '13 at 14:24
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    $\begingroup$ You express your expression in terms of the Lambert $W$ function as $ y={x^{x^{x^{.^{.^{.}}}}}} \implies y = -\frac{W(-\ln(x))}{\ln(x)} $ $\endgroup$ – Mhenni Benghorbal Apr 15 '13 at 14:28
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Look up "exponential tower" or "tetration". Calculating the expression involves Lambert's famous $W$ function with a value of $\frac{W(-\ln x)}{-\ln x}$.

Wikipedia has a nice discussion at http://en.wikipedia.org/wiki/Tetration.

As shown there, your expression converges for $e^{-e} \le x \le e^{1/e}$. As with at least 50% of the questions proposed here, this was shown by Euler.

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  • $\begingroup$ +1 :) Could you explain a bit about how you got to $e^{-e} \le x \le e^{1/e}$? $\endgroup$ – Yatharth Agarwal Apr 15 '13 at 15:56
  • $\begingroup$ It's in the Wikipedia article, a little ways in. $\endgroup$ – marty cohen Apr 16 '13 at 1:13
  • $\begingroup$ strange bound. thanks. $\endgroup$ – GA316 Apr 16 '13 at 4:07
  • $\begingroup$ I think Lambert functions explains the tetration's convergence very well. thanks. $\endgroup$ – GA316 Apr 16 '13 at 7:42
  • $\begingroup$ I see where the Wikipedia article states this interval, but I don't see where the article shows it. $\endgroup$ – Marcel Besixdouze Sep 26 '15 at 3:55

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