I recently saw a YouTube video here and it got me thinking, does anyone on this site have any insight on how we would tackle (change to base ten, evaluate, compute, or even put limits on approaching) some things like this?


  1. $\displaystyle e^{e^{e^e}}$
  2. $\displaystyle \pi^{\pi^{\pi^\pi}}$
  3. $\displaystyle \sqrt{2}^{\sqrt{2}^{\sqrt{2}^\sqrt{2}}}$
  4. $\ln(2)^{\ln(2)^{\ln(2)^{\ln(2)}}}$

What methods are proposed to even look at these things? Would it be a good idea to change to complex for pi? Do all towers of root 2 just simplify to 2?

note: Apologies for being new and not knowing how to format the text correctly.

Edit: Things I am thinking, for

$$\displaystyle \sqrt{2}^{\sqrt{2}^{\sqrt{2}^\sqrt{2}}} = \sqrt{2}^{\sqrt{2}^{\sqrt{2}\cdot \sqrt{2}}} = \sqrt{2}^{\sqrt{2}^{2}} = \sqrt{2}^{2} = 2 $$

This gets me thinking that regardless of the height (greater then 3) of the tower of it will always be 2.

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    $\begingroup$ $\sqrt{2}^{\sqrt{2}}<\sqrt{2}^2=2$. $\endgroup$ – J.G. Mar 2 at 21:37
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    $\begingroup$ What does "tackle some things like this" mean? Compute it? $\endgroup$ – Winther Mar 2 at 21:39
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    $\begingroup$ I would be surprised if there is any known "method" to look at these. $\endgroup$ – GEdgar Mar 2 at 21:49
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    $\begingroup$ Regarding your Edit: $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^\sqrt{2}}} = 1.840{\dots}$, which is not $2$. $\endgroup$ – Eric Towers Mar 2 at 23:17
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    $\begingroup$ What do you mean by solve, OneCold? It doesn't make any sense to "solve" a number. $\endgroup$ – Gerry Myerson Mar 3 at 5:47

The question is ambiguous, but we could interpret it as one about computability and decidable equality. Let $\mathbb{Q}$ be the set of rational numbers, $\mathbb{Q}^+$ be the set of positive rational numbers, and $\mathbb{R}$ be the set of real numbers. A computable real number is a real number $x \in \mathbb{R}$ such that there exists a computable function $f_x : \mathbb{Q}^+ \rightarrow \mathbb{Q}$ such that

$$ |f_x(\varepsilon) - x| \leq \varepsilon $$

That is, $f_x$ is a computer program such that, for any desired rational error tolerance $\varepsilon > 0$, $f_x$ yields a rational number within $\varepsilon$ of $x$. For an implementation of this concept, see Toward an API for the real numbers (summary).

Now note that

\begin{align} a^{a^{a^a}} &= \exp a^{a^a} \log a \\ &= \exp (\exp a^a \log a) \log a \\ &= \exp (\exp (\exp a \log a) \log a) \log a \end{align}

So we can reduce any power tower to exponentials and logarithms, which are computable. Finally, note that the constants $e, \pi, \sqrt{2}, \ln 2$ are computable. Therefore, all the numbers listed in the question are computable.

In fact, we have something even stronger:

Can we decide equality for recursive reals computed from integer constants by a combination of the following operations, which we will refer to as "calculator operations": (1) the four basic arithmetic operations, and square roots; (2) the sin, cos and tan trigonometric functions and their inverses; and (3) exponential and (natural) logarithm functions.

The answer turns out to be yes, as proven in the 1994 paper The identity problem for elementary functions and constants. Since $e = \exp 1$ and $\pi = 2 \sin^{-1} 1$, we can decide equality between any of the numbers listed in your question and any given integer. Therefore, we can decide whether any of the numbers listed in your question is an integer.

For example, to decide whether $x = \pi^{\pi^{\pi^\pi}}$ is an integer: Compute an integer upper bound $u \geq x$, say $u = 4^{4^{4^4}}$. Then, for all integers $k \in \{1, \dots, u\}$ (of which there are only finitely many), decide whether $x = k$. If equality holds for any $k$, $x$ is integer-valued. Otherwise, it is not integer-valued.


Welcome to StackExchange. There seems to be quite a few questions that you might be asking. First, using $I$ to represent an irrational number, let's look at converting $I^{I^{I^I}}$ to its decimal representation.

$$ I^{I^{I^I}} = I^{\left(I^{\left(I^I\right)}\right)} $$

As with all algebra, we need to work on the inside-most parentheses. Add elements of the Taylor ($I^1 = I)$ or MacLaurin ($I^0=1$) Series in order until they are smaller than the precision of your desired answer. Repeat as you work your way outwards. This, of course, is a naive implementation, but it will work. Taylor Series for $e^x$ are better known and easier to work with, so let's check that out:

$$ \begin{align*} e^x &= a^b \\ x &= \ln\left(a^b\right) = b \ln (a)\\ I^I &= e^{I \ln(I)} \text{ giving you more rational factors}\\ I^{\left(I^I\right)} &= e^{e^{I \ln(I)} \ln (I)}\\ I^{\left(I^{\left(I^I\right)}\right)} &= e^{e^{e^{I \ln(I)} \ln (I)}\ln(I)} \end{align*} $$

Still, to do a numerical approximation, there's a lot of work to use a Taylor Series. Another option would be to use Newton's method since you know the values at $0$ and $1$ and finding derivatives of the function seems doable. I believe this would be a much faster implementation than the naive Taylor. A more sophisticated Taylor is likely also available in which you find an approximation of the entire beast instead of each step individually. It sounds hard.

Next, can these towers have rational representations? It's possible with probability zero. There are a infinitely more irrational numbers than than rationals. Last, it appears your example with $\sqrt 2$ misunderstands exponentiation. $a^{b^c} = a^{\left(b^c\right)} \neq \left(a^b\right)^c \neq a^{(b\cdot c)}$.


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