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

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.

• 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 $y = 2^{f(\log_2(x))}$. – Ben FL Nov 16 '19 at 22:23
• I don't see how that works – Ultradark Nov 17 '19 at 18:13

## 1 Answer

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