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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what do you think? – Lost1 Apr 15 '13 at 14:20
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7Note that such a number $x=t^{t^{t^{.^{.^.}}}}$ (if it exists) will necessarily lie at a point of intersection of $y=x$ and $y=t^x$. – Cameron Buie Apr 15 '13 at 14:24
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1You express your expression in terms of the Lambert $W$ function as $ y={x^{x^{x^{.^{.^{.}}}}}} \implies y = -\frac{W(-\ln(x))}{\ln(x)} $ – 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.
marty cohen
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+1 :) Could you explain a bit about how you got to $e^{-e} \le x \le e^{1/e}$? – Yatharth Agarwal Apr 15 '13 at 15:56
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I think Lambert functions explains the tetration's convergence very well. thanks. – GA316 Apr 16 '13 at 07:42
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I see where the Wikipedia article states this interval, but I don't see where the article shows it. – Marcel Besixdouze Sep 26 '15 at 03:55