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For some values of $x$ the limit of infinite tetration converges. For example when $x=\sqrt{2}$ this is fixed point

$$\lim_{n\rightarrow \infty} \sqrt{2} \uparrow \uparrow n = \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\cdot^{\cdot^{\cdot}}}}} = \sqrt{2}$$

This limit converges for all $e^{-e} \le x \le e^{1/e}$. What is the minimum of this function?

$$f(x) = \frac{\lim_{n\rightarrow \infty} \sqrt{x} \uparrow \uparrow n}{\sqrt{x}} $$

Numerically the answer looks to have a single minima at $f(x\approx 0.579690102929) = 0.687413037245$, is there an exact solution?

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  • $\begingroup$ I think you shouldn't have $\sqrt 2$ in your definition of $f$. Am I wrong ? $\endgroup$ – Watson Feb 4 '16 at 21:05
  • $\begingroup$ @Watson you're correct, thanks, and fixed! $\endgroup$ – Hooked Feb 4 '16 at 21:13
  • $\begingroup$ The key, why $x=\sqrt{2}$ works is not because it is a root, but an exponential of the form $x=b^{1\over b}$ at $b=2$. Thus a better generalization is for instance $ ^3 \sqrt{3}$ etc. After that try to find the number $b$ where $x=2 = b^{1 \over b}$ and look what happens when you evaluate x^b,x^x^b,x^x^x^b,... resp 2^b,2^2^b, 2^2^2^b,... $\endgroup$ – Gottfried Helms Feb 5 '16 at 22:24

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