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Given that $k,x \in \mathbb{R}^+$ and $k > 1$, the function $f$ defined by $$f(x) = k^{k^{\frac{-1}{x}}}$$, generates a sequence $x_0,f(x_0), f(f(x_0)), \cdots$ is observed to converge to a fixed point independent of $x_0$. (Proof)

I was wondering if $f$ is a contraction on the metric space of $\mathbb{R}^+$ equipped with absolute distance.

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  • $\begingroup$ Yes, given you exclude $0$ from the domain. $\endgroup$ – Michael Angelo Feb 21 '16 at 8:27
  • $\begingroup$ @fawningflagellum Can you help me prove it? $\endgroup$ – Agnishom Chattopadhyay Feb 21 '16 at 10:10
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If $f(x) = k^{k^{\frac{-1}{x}}} = e^{\ln(k)k^{\frac{-1}{x}}} $, so that

$\begin{array}\\ f'(x) &=f(x)(\ln(k)k^{\frac{-1}{x}})'\\ &=f(x)(\ln(k)e^{\ln(k)\frac{-1}{x}})'\\ &=f(x)\ln(k)((\ln(k)\frac{-1}{x})'e^{\ln(k)\frac{-1}{x}})\\ &=f(x)\ln^2(k)\frac{1}{x^2}k^{\frac{-1}{x}}\\ &=\dfrac{k^{k^{\frac{-1}{x}}}\ln^2(k)k^{\frac{-1}{x}}}{x^2}\\ \end{array} $

The condition for a contraction mapping is $|f'(x)| < 1 $. In this case, The set of $x$ for which this is true depends on numerical computation, since it appears highly unlikely that an analytic solution if possible.

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