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This is related: Do these points trace out a function? $ P(2^{2^s},2^{2^{-s}}) $

What function does $P(2^{2^{2^s}},2^{2^{2^{-s}}})$ trace out?

I tried going through the answer that was given in the previous post but could not figure it out for this extension of the problem. I don't understand how to find the function.

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    $\begingroup$ For fixed $s$, write $x = 2^{2^{2^s}}, y = 2^{2^{2^{-s}}}$. Then $\log_2(y) = f(\log_2(x))$, where $f$ is as in the previous question. So $y = 2^{f(\log_2(x))}$. $\endgroup$ – Ben FL Nov 16 '19 at 22:23
  • $\begingroup$ I don't see how that works $\endgroup$ – Ultradark Nov 17 '19 at 18:13
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Let's split this into a few steps:

  • We can get from $x:=\color{red}{2^s}$ to $y:=\color{red}{2^{-s}}$ with $\color{red}{y=1/x}$.
  • We can get from $x:=\color{orange}{2}^{\color{red}{2^s}}$ to $y:=\color{orange}{2}^{\color{red}{2^{-s}}}$ with $\color{red}{\log_2y=1/\log_2x}$, i.e. $\color{orange}{y=2^{1/\log_2x}}$.
  • We can get from $x:=\color{limegreen}{2}^{\color{orange}{2}^{\color{red}{2^s}}}$ to $y:=\color{limegreen}{2}^{\color{orange}{2}^{\color{red}{2^{-s}}}}$ with $\color{orange}{\log_2y=2^{1/\log_2\log_2x}}$, i.e. $\color{limegreen}{y=2^{2^{1/\log_2\log_2x}}}$.
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  • $\begingroup$ It looks like the right answer but it's not working when I actually plot it $\endgroup$ – Ultradark Nov 17 '19 at 18:27
  • $\begingroup$ @Ultradark Could you edit your question to show how you plotted it and what happened, and why you feel it's wrong somehow? $\endgroup$ – J.G. Nov 17 '19 at 18:35
  • $\begingroup$ Oh I got it now $\endgroup$ – Ultradark Nov 17 '19 at 18:37

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