More Fun With Power Towers

[Necessity] Fix a number $a>1$ . Let us first characterize a necessary condition for the number $a$ , so that the limit $a^{a^{\cdot ^{\cdot {^\cdot} }}}$ exists in $\mathbb{R}$ .

If the limit does exist, then there must exist a real number $x$ , such that the following equation has a real solution $a^x=x$ Now, consider minimizing the convex function $f(x)=a^x-x, x\in \mathbb{R}$ . At the minima $x^*$ , the first derivative must vanish, thus we must have $a^{x^*} \ln(a)=1$ Hence, $x^*=-\frac{\ln(\ln(a))}{\ln(a)}$ Hence the minimum value of $f(x)$ is $a^{x^*}-x^*=\frac{1+\ln(\ln(a))}{\ln(a)}$ . For the equation $a^x=x$ to have a real solution, this minimum value must be non-positive. Thus, the limit above does not exist if $1+\ln(\ln(a))>0$ i.e., $a>e^{1/e}$ Hence, for the limit to exist, we must have $a\leq e^{1/e}$ .

[Sufficiency] Now, we show that for any $1<b< e$ , the mapping $g:[1,b] \to [1,b]$ , given by $g(x)={b^{x/b}}$ , is a well-defined contraction mapping. This can be directly verified by taking derivative of $g(x)$ and showing that it is strictly less than unity in the compact interval $[1,b]$ .

Taking $a=b^{1/b}$ , it follows by Banach's fixed point theorem , that the iteration $g^m(a)=a^{{g^{m-1}}(a)}$ converges, since the point $a$ belongs to the domain (i.e., $1\leq a=b^{1/b} \leq b$ ). Thus the supremum of all numbers $a$ for which the limit exists is $e^{1/e}$ with the corresponding supremum value of the limit being $e$ .

$e$ is where ${ y }^{ \frac { 1 }{ y } }$ has the maximum value. See graphic.

Note that $\dfrac { d }{ dy } \left( { y }^{ \frac { 1 }{ y } } \right) ={ y }^{ \frac { 1 }{ y } -2 }\left( 1-Log\left( y \right) \right)$

If ${ a }^{ { a }^{ { a }^{ ... } } }=b$

then $\frac { W\left( -\ln { a } \right) }{ -\ln { a } } =b$

by identity, $\frac { W\left( -\ln { a } \right) }{ W\left( -\ln { a } \right) { e }^{ W\left( -\ln { a } \right) } } =b$

$\frac { 1 }{ { e }^{ W\left( -\ln { a } \right) } } =b$

${ e }^{ -W\left( -\ln { a } \right) }=b$

$W\left( -\ln { a } \right) =-\ln { b }$

$-\ln { a } =-\ln { b } { e }^{ -\ln { b } }$

$\ln { a } =\frac { \ln { b } }{ b }$

So, $a=\sqrt [ b ]{ b }$

We want $a$ to the maximum so $\sqrt [ b ]{ b }$ has to be maximum; and for a function to be maximum (or minimum), its derivative has to be 0. The derivative of $\sqrt [ b ]{ b }$ is ${ -b }^{ \frac { 1 }{ b } -2 }\left( \ln { b-1 } \right)$

So, ${ -b }^{ \frac { 1 }{ b } -2 }\left( \ln { b } -1 \right) =0$

that means either ${ -b }^{ \frac { 1 }{ b } -2 }=0$ or $\ln { b } -1=0$

The first one is not possible, even if $b$ is 0, so

If $\ln { b } -1=0$

then $\ln { b } =1$

thus $b=e$

Thus the maximum is $e$