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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?

examples:

  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.

PNT
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    $\sqrt{2}^{\sqrt{2}}<\sqrt{2}^2=2$. – J.G. Mar 02 '21 at 21:37
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    What does "tackle some things like this" mean? Compute it? – Winther Mar 02 '21 at 21:39
  • Sorry I am of the understanding that you can not compute pi^pi^pi^pi, it is too large, maybe some of the others we can. – OneCold Ruben Mar 02 '21 at 21:42
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    I would be surprised if there is any known "method" to look at these. – GEdgar Mar 02 '21 at 21:49
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    If there were a widely known method to approach whether any of these were an integer, it would be in the video or a top comment on that video. – Mark S. Mar 02 '21 at 21:54
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    Might be worth mentioning transcendental number theory as a field that studies this, Gelfond-Schneider as a theorem that at least can tell you $\sqrt{2}^{\sqrt{2}}$ is transcendental, and Schanuel's conjecture as something that can at least prove (1) is transcendental. (And maybe some of the others? Unsure). Also, might point out your examples 3 and 4 are, respectively, in $(1,2)$ and $(0,1)$. – Milo Brandt Mar 02 '21 at 22:26
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    Regarding your Edit: $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^\sqrt{2}}} = 1.840{\dots}$, which is not $2$. – Eric Towers Mar 02 '21 at 23:17
  • Might be worth mentioning that the video you cite asks whether a particular power tower is an integer. You don't mention any relation between this goal and your phrase "how we would tackle some things like this". "Tackle" in what sense? – Eric Towers Mar 02 '21 at 23:19
  • Good point, I will define what I mean when I say tackle in the original post. – OneCold Ruben Mar 03 '21 at 04:37
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    What do you mean by solve, OneCold? It doesn't make any sense to "solve" a number. – Gerry Myerson Mar 03 '21 at 05:47
  • If you are given a question, you give an answer, or a solution, you solve the problem. Is this a comment about me using the word solve when there is no x in the problem? Come on, set the problem to x then there x to solve for? Definition :solve /sälv/ verb find an answer to, explanation for, or means of effectively dealing with (a problem or mystery): – OneCold Ruben Mar 03 '21 at 18:26
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    I'm sorry, OneCold, but "solve $e^{e^{e^e}}$" makes no more sense than "solve $17$." In any event, I don't know what it means, and I can't help you if I can't understand your question. What do you actually want to know about $e^{e^{e^e}}$ and those other numbers? – Gerry Myerson Mar 04 '21 at 11:55
  • Seriously? Ok, how about change to the normal base ten we are familiar with looking at numbers as? Is familiar now ambiguity or subjective? I love you math people, I swear that some of you get your kicks telling people they asked their question wrong. Come on. You are telling me there is not way for a reasonable person to interpret the meaning of a question, What are you a computer? No. you are a human, capable of understanding meaning, not a computer. Also if you can't help, then don't, thanks. – OneCold Ruben Mar 05 '21 at 19:26
  • You probably can't change these numbers to base ten, since there's not enough paper/bits in the universe to write the digits. (Though it could be difficult to prove some of these aren't rational!) We can certainly get approximations, but those are inexact in a way which usually isn't what we're after in pure mathematics. The point, I think, isn't "the question is wrong", but "the question isn't precise enough to give useful answers". – aschepler Mar 05 '21 at 19:41
  • Ok, I give up, Please take the question down. It was not a worth use of time. – OneCold Ruben Mar 05 '21 at 19:51
  • @OneColdRuben This is like the $4$th question I've seen about this video lol – Some Guy Mar 05 '21 at 20:41
  • Also this is wrong $ \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$ – Some Guy Mar 05 '21 at 20:42
  • I'm sad to see you give up, OneCold. You could probably learn something, if you took our comments seriously, instead of trying to fight them. Anyway, I think your emphasis on irrational numbers is misplaced. $e^{e^{e^e}}$ is no harder to "tackle" than is $3.5^{3.5^{3.5^{3.5}}}$, or even $4^{4^{4^4}}$, which is a number with something like $10^{150}$ digits if you could write it out in base ten. I doubt we will ever know, say, the middle digit of that number, and there's no irrationality there. – Gerry Myerson Mar 05 '21 at 22:11
  • As for my insistence on careful phrasing of questions, it's when I see you write $\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2}}}=\sqrt2^{\sqrt2^{\sqrt2\cdot\sqrt2}}$ that I recognize, even if you don't, that your sloppiness and carelessness have real consequences and lead you to gross mistakes. If you take a little more care in how you express yourself, you won't make stupid errors like that one. – Gerry Myerson Mar 05 '21 at 22:15
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    So, after you gave up, you got three answers. Do you want to engage with these users who have tried to help you? Have they solved your question? Is there anything you want to ask them to clarify? – Gerry Myerson Mar 07 '21 at 11:51

2 Answers2

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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.

user76284
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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. $a^{b^c} = a^{\left(b^c\right)} \neq \left(a^b\right)^c \neq a^{(b\cdot c)}$.