Transcendental function #1

Euler showed that $\displaystyle \lim_{n \ \to \ \infty} \underbrace{x^{x^{\cdot ^{\cdot ^{x}}}}}_{\text{n times}}$ converges for $\displaystyle \large \frac{1}{e^e} \leq x \leq e^{\frac1e}$ , which bounds $\displaystyle \large y = \underbrace{x^{x^{\cdot ^{\cdot ^{x}}}}}_{\text{n times}}$ to $\displaystyle \large \frac 1e \leq y \leq e$ . The limit, should it exist, is a positive real solution of the equation $\large y = x^y$ . Thus, $\large x = y^{\frac 1y}$ .

The limit defining the infinite tetration of $\large x$ fails to converge for $\large x > e^{\frac1e}$ because the maximum of $\large y$ is $\large e$ . Thus, we can rule out $\boxed{1}$ and $\boxed{32}$ .

$\displaystyle \boxed{2}$ is not a solution to $\displaystyle \large y = x^y$ :

$\displaystyle \begin{aligned} \sqrt[3]{3}^{\sqrt{3}} = 3^{\frac{\sqrt{3}}{3}} = 3^{\frac{1}{\sqrt{3}}} = \sqrt[\sqrt{3}]{3} \neq \sqrt{3} \\ \\ \\ \end{aligned}$

The rest converge given that they are solutions to $\displaystyle \large y = x^y$ .

$\displaystyle \boxed{4} \ :$

$\displaystyle \begin{aligned} \sqrt[4]{4}^{\sqrt[4]{4}^{\sqrt[4]{4}^{\cdots}}} & = 2 \\ \sqrt[4]{4} ^{\ 2} & = 2 & \ \ \ \small \color{#3D99F6} 4^{\frac 24} = 4^{\frac12} \\ \sqrt{4} & = 2 \end{aligned}$

$\displaystyle \boxed{8} \ :$

$\displaystyle \begin{aligned} \sqrt[3]{2.25}^{\sqrt[3]{2.25}^{\sqrt[3]{2.25}^{\cdots}}} & = 1.5 \\ \sqrt[3]{2.25} ^{\ 1.5} & = 1.5 & \ \ \ \small \color{#3D99F6} 2.25^{\frac{1.5}{3}} = 2.25^{\frac12} \\ \sqrt{2.25} & = 1.5 \end{aligned}$

$\displaystyle \boxed{16} \ :$

$\displaystyle \begin{aligned} \sqrt[5]{6.25}^{\sqrt[5]{6.25}^{\sqrt[5]{6.25}^{\cdots}}} & = 2.5 \\ \sqrt[5]{6.25} ^{\ 2.5} & = 2.5 & \ \ \ \small \color{#3D99F6} 6.25^{\frac{2.5}{5}} = 6.25^{\frac12} \\ \sqrt{6.25} & = 2.5 \end{aligned}$

$\displaystyle \boxed{64} \ :$

$\displaystyle \begin{aligned} \sqrt[11]{51.53632}^{\sqrt[11]{51.53632}^{\sqrt[11]{51.53632}^{\cdots}}} & = 2.2 \\ \sqrt[11]{51.53632} ^{\ 2.2} & = 2.2 & \small \color{#3D99F6} 51.53632^{\frac{2.2}{11}} = 51.53632^{\frac15} \\ \sqrt[5]{51.53632} & = 2.2 \\ \\ \\ \\ \end{aligned}$

Hence, our answer is $\displaystyle 4 + 8 + 16 + 64 = 92$ .