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I just watched a video by blackpenredpen where he solved this equation. Here are the steps he took (The process transitions from one line to another. He didn't explicitly use an implies or iff sign, so I'll leave it as such):

$x^{x^{x^{x^{x^{...}}}}} = 2$

$x^2 = 2$

$\left[ \begin{array}{l} x = \sqrt{2} \\ x = -\sqrt{2} \end{array} \right. $

At this point, he just crossed out the $-\sqrt{2}$ option.

Could someone please explain why $-\sqrt{2}$ is not a solution? Thank you!

azimut
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ten_to_tenth
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    The power tower only converges when $e^{-e}\leq x\leq e^{\frac{1}{e}}$ – cansomeonehelpmeout Feb 19 '24 at 15:42
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    Did you try seeing what the values are if you repeatedly take $-\sqrt{2}$ to the power of $-\sqrt{2}$? It doesn't give you $2$. That value is actually $-\frac{2W(\ln{\sqrt{2}})}{\ln{2}}\approx -0.76666469596...$ where $W$ is the Lambert W function. – Warren Moore Feb 19 '24 at 15:55
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    If we're dealing with only real numbers, then non-integer exponents are not defined for negative numbers, so $-\sqrt2$ cannot be the base of an exponent. – Polygon Feb 19 '24 at 16:00
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    bprp is also silent on what his method would yield if the RHS were replaced by 4. – user43208 Feb 19 '24 at 17:03
  • @cansomeonehelpmeout Could you please explain why it is the case? I'm a first-year CS student so I'm not sure if I have the mathematical maturity to understand it. I'm asking just in case it's something within my knowledge. – ten_to_tenth Feb 19 '24 at 17:18
  • @WarrenMoore Thank you for the pointer! I'm not familiar with the Lamber W function so I didn't take that approach. I'll def come back to your comment when I understand that topic. – ten_to_tenth Feb 19 '24 at 17:20
  • @Polygon Thank you! I have a question. Since the tower continues to infinity and converges to 2, doesn't that mean the exponent of each $x$ in the tower is always 2, which is an integer? – ten_to_tenth Feb 19 '24 at 17:23
  • @ideals_go If you've got some time, you might get a lot out of this video. I don't think it directly answers your specific question, but it gives a great intuitive analysis of the power tower, and it could help you reason about it. – Polygon Feb 19 '24 at 19:16
  • @Polygon Thank you so much! It looks great!. I'm watching it right now! – ten_to_tenth Feb 19 '24 at 19:30
  • @WarrenMoore Yes, you are correct. The Lambert W is built into my WP-34s calculator. You can't get the physical machine any more, but there is a free iPhone emulator that runs at least fifty times faster. – richard1941 Feb 24 '24 at 03:45

1 Answers1

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Short answer

$x^{x^{x^{x^{x^{...}}}}}$ is not defined for $x < 0$.

Explanation

How is $x^{x^{x^{x^{x^{...}}}}}$ defined? It is the limit $\lim_{n\to\infty} a_n$ of the sequence \begin{align*}a_0 & = 1 \\ a_1 & = x \\ a_2 & = x^x \\ a_3 & = x^{x^x} \\ a_4 & = x^{x^{x^x}} \\ & \ldots \end{align*} Starting from $a_2$, these are powers whose exponent is not necessarily an integer. Such kinds of general powers are defined only if their base is positive, which for the above powers means $x > 0$.

(The general power $a^b$ is defined as $\exp(b\ln(a))$, and the expression $\ln(a)$ requires $a > 0$.)

Comment 1

In fact, the power $a^b$ can be defined for $a < 0$ in a meaningful way as long as $b$ is a rational number with an odd denomiator, see the comment of the user "Especially Lime" below. For example, $(-8)^{1/3} = -2$ since $(-2)^3 = -8$. However, this is still not enough for our case.

Comment 2

Actually, the condition $x \geq 0$ is not enough to ensure that $x^{x^{x^{x^\ldots}}}$ is well-defined. The reason is that we did not check if the limit $\lim_{n\to\infty}a_n$ actually exists. You can find this discussion in this answer. The result is that the domain of valid numbers $x$ is $$\frac{1}{e^e} \leq x \leq e^{\frac{1}{e}},$$ which approximates to $$0.066 < x < 1.445.$$ Strictly speaking, this discussion is also missing in the video linked in the question: To show that $x = \sqrt{2}$ is a solution of $x^{x^{x^{x^\ldots}}} = 2$, one has to ensure that the limit exists in the case $x = \sqrt{2}$. (Disclaimer: Since the calculation in the video doesn't use any implication signs, we don't know what he claims to have actually achieved.)

Comment 3

The discussion of the domain of $x$ leads to the range of $$x^{x^{x^{x^\ldots}}} = y$$ as $$\frac{1}{e} \leq y \leq e.$$ I find it quite surprising that the range has a finite maximum.

Comment 4

Within the above range of $y$, we can solve the equation $$x^{x^{x^{x^\ldots}}} = y$$ in general, in the same style as as it is done in the video linked in the question. We have $$x^y =x^{\left(x^{x^{x^\ldots}}\right)} = x^{x^{x^{x^\ldots}}} = y$$ and therefore $$x = y^{\frac{1}{y}}.$$ In the case $y = 2$ this reproduces the solution of the original problem as $x = 2^{\frac{1}{2}} = \sqrt{2}$.

azimut
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    Hi, thank you! I have a question. Is it necessary for $a_{n+1} = x^{a_{n}}$ to be defined as $a_{n+1} = e^{a_{n}ln(x)}$? For example, we don't rewrite $(-2)^2$ as $e^{2ln(-2)}$ because -2 < 0, but $(-2)^2$ is still a valid expression, so it seems a bit to me like we're imposing an expression that requires $x > 0$ to obtain $x > 0$. – ten_to_tenth Feb 19 '24 at 17:31
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    @ideals_go For powers $a^b$ with $b$ a non-negative integer, we can use the usual definition $a^b = a \cdot a \cdot \ldots \cdot a$ ($b$ times). But for a general real number $b$, this is not possible any more, and the common definition in analysis is $a^b = \exp(b \ln (a))$. – azimut Feb 19 '24 at 17:58
  • Thank you, someone also mentioned this in a comment above and I also had a question for him/her. The tower continues to infinity and converges to 2; doesn't that mean the exponent of each $x$ in the tower is always 2, which is an integer? – ten_to_tenth Feb 19 '24 at 18:02
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    @ideals_go On your other point: From a pedantic point of view, we are only allowed to write down expressions which are well-defined. So strictly speaking, the exercise should start with "Let $x > 0$ such that the limit $x^{x^{x^\ldots}}$ exists." – azimut Feb 19 '24 at 18:05
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    @ideals_go The symbol $x^{x^{x^\ldots}}$ is a bit misleading, look at the recursive definition of $a_{n+1}$. The sequence of exponents in $a_{n+1}$ converges to $2$, but it never equals $2$. – azimut Feb 19 '24 at 18:11
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    @ideals_go To make my last statement more concrete: $a_2 = x^x$. This expression is not well-defined for $x = -\sqrt{2}$, since the exponent is a non-integer $< 0$. – azimut Feb 19 '24 at 18:42
  • Thank you very much! I believe I get it now from your last two comments! – ten_to_tenth Feb 19 '24 at 19:14
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    If $a<0$ you can still define the power $a^b$ provided $b$ is a rational with odd denominator. However, that doesn't really help with $x^{x^{x^\cdots}}$ as $x^x$ will be irrational for non-integer rational $x$. – Especially Lime Feb 20 '24 at 08:14
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    @EspeciallyLime Thank you for this comment. You are right, but it gets messy, and as you say, it doesn't help in this case. I will add a remark in my solution. – azimut Feb 20 '24 at 08:18
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    @ideals_go I've now added some additional info; maybe that's also interesting for you. – azimut Feb 20 '24 at 09:07
  • @azimut thank you very much for your rewrite! As to your last point, I examined the function $f(x) = x^{\frac{1}{x}}$ and found that $f(x)$ increases from $0$ (at $x = 0$) to $e^{\frac{1}{e}}$ (at $x = e$) and decreases to $1$ at infinity, so that should mean as $\frac{1}{e^e} \leq y^{\frac{1}{y}} \leq e^{\frac{1}{e}}$, $y \in [\frac{1}{e}, \infty)$. I wondered if you or I might have made a mistake. Please check again! – ten_to_tenth Feb 20 '24 at 10:33
  • @ideals_go Where exactly do you think there is a mistake? I've added details to the last point, separating a comment 4. – azimut Feb 20 '24 at 13:15
  • In terms of tetration, $a_n$ may be denoted ${}^nx$ or $x\uparrow\uparrow n$. – J.G. Feb 20 '24 at 13:24
  • @azimut It appears I made a mistake somewhere in my reasoning. I was wondering why $1/e \leq y \leq e$. Here is what I did: $x = y^{\frac{1}{y}}$. I graphed this function, with $x$ on the vertical axis and $y$ on the horizontal axis, and found that as $x$ varies in its interval, $y$ varies in $[\frac{1}{e}, \infty)$, but yours is $[\frac{1}{e}, e]$, which should be right as the expression is known to never converge to a value larger than $e$, but I couldn't figure out what goes wrong in my reasoning. – ten_to_tenth Feb 20 '24 at 14:37