# Values for which tetration to infinite heights (i.e., $x^{x^{x^{x^{.^{.^{.}}}}}}$) converges [duplicate]

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?

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

• what do you think? – Lost1 Apr 15 '13 at 14:20
• Note that such a number $x=t^{t^{t^{.^{.^.}}}}$ (if it exists) will necessarily lie at a point of intersection of $y=x$ and $y=t^x$. – Cameron Buie Apr 15 '13 at 14:24
• You 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

Look up "exponential tower" or "tetration". Calculating the expression involves Lambert's famous $W$ function with a value of $\frac{W(-\ln x)}{-\ln x}$.
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.
• +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