# Is there an exact value to the minimum of the infinite tetration of $x$?

Is there an exact value to the minimum of the function $$f_k(x)=\underbrace{x^{x^{x^{.^{.^.}}}}}_{2k\,\text{times}}$$ as $k\to\infty$, where $k=1,2,3,\cdots$?

This visualisation in Desmos shows that the minimum is around $0.3792$.

I have specifically stated $2k$ as the function is strictly increasing for odd tetrations.

I am certain that it will not tend to $0$ since for every $x^x$ I add to the tetration, it decreases by $0.0004$, then $0.0003$ and $0.0002$.

Differentiation looks extremely tedious to do...

So is there a rigorous approach to find the exact value, if there exists one, of $\min(f_\infty(x))$?

If not, is there a method to find a lower bound?

• Note that $e^{1/e} = 1.44466786101...$ Mar 24, 2018 at 22:50
• math.stackexchange.com/questions/2490769/… My answer shows a detailed graph and equations for the fixed points of $b^z$ and $b^{b^z}$ in the neighborhood of $b=\exp(-e)$ Mar 25, 2018 at 17:54

Claim: The minimum $1/e \approx 0.36788$ is achieved at $x = e^{-e} \approx 0.065988$.
We proceed by first establishing that the minimum is upper bounded by $1/e$ and then show that it is also lower bounded by $1/e$.
Proof of upper bound. Let $y_{0}=1$ and define the iteration $$y_{n+1}=x^{y_{n}}.$$ Note that $$y_{1}=x,\qquad y_{2}=x^{x},\qquad\text{etc.}$$ tends to the infinite tetration. It is known that this iteration converges (possibly to $\infty$) for all $x\geq e^{-e}$. Define $y_{\infty}$ as the function which associates to each $x \geq e^{-e}$ the limit $\lim_{n\rightarrow\infty}y_{n}$. Note that if $y_\infty(x)$ is finite, then $$y_{\infty}(x)=\lim_{n\rightarrow\infty}y_{n+1}=\lim_{n\rightarrow\infty}x^{y_{n}}=y^{y_{\infty}(x)}$$ and hence $$(y_{\infty}(x))^{1/y_{\infty}(x)}=x.$$ Solving for $y_{\infty}(x)$, $$y_{\infty}(x)=\frac{W(\log(1/x))}{\log(1/x)}.$$ Let $x_0=1$ and define also the iteration $$x_{n+1}=x^{x^{x_{n}}}.$$ Since it is monotone, this iteration converges (possibly to $\infty$) for all $x\geq0$. Define $x_{\infty}$ as the function which associates to each $x \geq 0$ the limit $\lim_{n\rightarrow\infty}x_{n}$. Since $x_{\infty}=y_{\infty}$ on $[e^{-e},\infty)$, $$\min_{x\geq0}x_{\infty}(x)\leq y_{\infty}(e^{-e})=1/e.$$
Proof of lower bound. Fix $x$ and let $a=x_{\infty}(x)$ for brevity. Then, $$x^{x^{a}}=a\implies x=a^{1/(x^{a})}\implies x^{a}=a^{a/(x^{a})}\implies(x^{a})^{x^{a}}=a^{a}.$$ Therefore, $$a^{a}\geq\min_{u\geq0}u^{u}=e^{-1/e},$$ which in turn implies $a\geq1/e$.
• $$b=\exp(-e) \approx 0.0659880358453$$ $$\lim_{n \to \infty} b \uparrow \uparrow n = \exp(-1)\approx 0.36787944117$$ This is the Op's minimum; for b in the range of 0..1. I will post links or answer with more detail later. Mar 25, 2018 at 15:39
• See this related answer for the fixed points of $f(z)=b^z$ in the neighborhodd of $b=\exp(-e)$ math.stackexchange.com/questions/2490769/… See the graph and note that there is a parabolic fixed at $b>\exp(-e)$, and for $b_2>b$ the fixed points become complex conjugate pairs. Mar 25, 2018 at 15:52
• Thank you for your contributions, parsiad and SheldonL. I have repetitively added more terms to the function and it does look like it converges to $1/e$! Mar 25, 2018 at 19:22