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$$x^{x^{x^{\cdot^{\cdot^{\cdot}}}}} =\quad 2$$ This equation has the answer $\sqrt{2}$ by taking $\log$ to both side. This answer is correct because I'd proven it by replacing $x$ with $\sqrt{2}$ then computed the expression repeatedly and the answer is close to $2$.

But now consider,$$x^{x^{x^{\cdot^{\cdot^{\cdot}}}}} =\quad 3$$ I tried using the same solving method and it gave me that $\sqrt[3]{3}$ is the answer but when I tried proving the answer by replacing $x$ with $\sqrt[3]{3}$ and computed it repeatedly, the answer is closer to $2.478052680288297$ and not $3$

I don't know why this happens, could someone explain the answer to $x^{x^{x^{\cdot^{\cdot^{\cdot}}}}} = 3$ ?
I don't think it's $\sqrt[3]{3}$

The first equation problem came from brilliant.org

offchan
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  • These are known as power towers or tetrations. Check out wikipedia for more information. – RghtHndSd Oct 21 '14 at 18:20
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    Just a small thing, doing convincing numerical calculations doesn't suffice as a proof – HBeel Oct 21 '14 at 20:00
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    Note: The fact that this is a limit can be noticed by using Knuth's up-arrow notation which is useful in expressions involving power towers such as this: $$x^{x^{x^{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}}} = ,,\lim_{n\to \infty} \left(x\uparrow\uparrow n\right)$$ – Nick Oct 21 '14 at 20:30
  • $\sqrt[3]{3}=x(say)$. Then $x^{x^{x^{x}}}=3$. So,when I tried proving the answer by replacing $x$ with $\sqrt[3]{3}$ and computed it repeatedly the answer is 3.How it becomes closer to 2.478052680288297 ? I can not understand the question? – Empty Oct 25 '14 at 14:42
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    Try replacing x with 3rd root of 3 but replace the last x first not the first one. so x^x^x^... becomes x^[x^(3rd root of 3)] = x^ (3rd root of 3 ^ 3rd root of 3) and so on. If you replace the first x first then the answer limit is infinity not 2.47... If you calculate correctly then it should become closer to 2.47 not 3 – offchan Oct 26 '14 at 06:51

3 Answers3

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Here's a question: if $x = \sqrt[3]{3}$, then what is the value of $y = x^{x^{x^{\dots}}}$?

As you say, if $y$ is a solution to this, then $y \ln x = \ln y$, so that $$ \frac{\ln y}{y} = \ln(x) = \frac{\ln(3)}{3} $$ Now, how can we "solve this" for $y$? As it ends up, there are two solutions for $y$. The answer you are getting is the second root, $y \approx 2.47805$.

We can, in fact, conclude that the equation $x^{x^{x^{\dots}}} = 3$ has no solution.

So, the salient question is how do we "choose" one value of $y$ over the other?

First of all, we should decide when this sequence converges at all. For a fixed $a$, define the function $f_a(x) = a^x$. Consider the following recursive definition of a sequence:
$$ x_0 = 1\\ x_n = a^{x_{n-1}} $$ If any one value for $y = a^{a^{a^{\dots}}}$ makes sense, it's the limit $\lim_{n \to \infty} x_n$. This has now become an analysis of fixed-point iteration.

It's relatively easy to show that we can guarantee that this sequence converges as long as near $x = 1$ (our starting point), $|f'(x)| < 1$. Because $f'(x) = \ln(a) a^x$, we find that $|f'(1)| < 1$ exactly when $a < e$. So, this sequence will necessarily converge for $a \in (e^{-1},e)$.

As it turns out, however, this is not the only situation in which the sequence converges. In fact, it was shown by Euler (see wiki page and comments below) that the sequence will converge for $e^{-e} < a < e^{1/e}$. Correspondingly, the limit of the sequence will necessarily lie in $(1/e,e)$. Because $3$ lies beyond these bounds, it cannot be the limit of such a sequence.

See also this wikipedia page.

Ben Grossmann
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  • Nice work! That is brilliant and I'm glad that 2.47 wan't popping out of nowhere. Wouldn't we say that both 2.47 and 3rd root 3 are solutions, as opposed to no solution? – nathan.j.mcdougall Oct 21 '14 at 18:30
  • And thus, we can now proof that it has no solution if $y > e$. – Irvan Oct 21 '14 at 18:31
  • You should be a math professor. Your knowledge is too much than I can perceive. – offchan Oct 21 '14 at 18:42
  • That's what the answer I'm looking for, at least it tells me that the number I'm getting is not a magical number that has no reason 2.47... and it tell me that this kind of equation isn't always that easy – offchan Oct 21 '14 at 18:50
  • I appreciate the compliment. Let me know if there's something I should clarify here. – Ben Grossmann Oct 21 '14 at 18:51
  • @ Omnomnomnom, We have, $e^{\dfrac{1}{e}}=1.444667861...$.So according to your answer the solution is impossible for $y=2$, which is contradictory....Please clear it to me.... – Empty Dec 08 '14 at 04:25
  • @Panja.S. I must have miscalculated – Ben Grossmann Dec 08 '14 at 04:53
  • Well, it's impossible for 2 as well. – Panglossian Oporopolist Dec 30 '15 at 05:27
  • @Kugelblitz that's not true: taking $x = \sqrt{2}$ works for $2$. – Ben Grossmann Dec 30 '15 at 13:43
  • Right... but going by your last statement (Which discuss the domain of y), it's contradictory. I'm perfectly aware that sqrt(2) satisfies y=2... it happens so that your answer's last line is incorrect; I made the previous comment just to highlight your fallacy. I presume you actually mean $x$ should lie between $e^{1/e}$ and $e^{-1/e}$, and not y,; see this: https://math.dartmouth.edu/~euler/docs/originals/E532.pdf – Panglossian Oporopolist Dec 30 '15 at 14:10
  • You can check Wolfram Mathworld or Wikipedia as well, the last line should read $e^{-1/e} \leq x \leq e^{1/e}$ :) – Panglossian Oporopolist Dec 30 '15 at 14:12
  • @Kugelblitz you'll have to forgive me, it's been more than a year since I've thought about this problem. In the future, when you see little mistakes like that, you should feel free to just edit the correction yourself. Thank you, though. – Ben Grossmann Dec 30 '15 at 14:17
  • No issues. I make so many more mistakes than you! I wished that I ask you to do the correction yourself because I thought correcting it myself is, somewhat, rude. Besides, it's not like you made a conceptual error... – Panglossian Oporopolist Dec 30 '15 at 14:36
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$$ x^{x^{x^{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}}}=y $$ $$ \ln{x^{x^{x^{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}}}}=\ln y $$ $$ x^{x^{x^{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}}}\ln{x}=\ln y $$ $$ y\ln{x}=\ln y $$ $$ \ln{x}=(\ln{y})/y $$ $$ \ln{x}=\ln({y^{1/y}}) $$ $$ x= y^{1/y} $$

Substitute in $2$ and $3$ to see why these are the solutions of $x$. They will always be the $y^{\text{th}}$ root of $y$.

Rustyn
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nathan.j.mcdougall
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  • I wrote a python script for this: $f(n)$ is after $n$ iterations. f(10) = 2.284634256264522, f(100) = 2.47803028602696, f(1000) = 2.4780526802882967, f(10000) = 2.4780526802882967, f(100000) = 2.4780526802882967, f(1000000) = 2.4780526802882967.

    It's surprising that it would eventually converge.

     – Irvan Oct 21 '14 at 18:12
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    Sorry if I oversee something trivial here, but what you have done seems to work wonderfully if there is a solution $x$, such that $x^{x^{x^{\cdots}}}=y$. I am, however, not entirely convinced this method immediately proves that such an $x$ necessarily exists? I mean, we know that if there is such an $x$, then it is $\sqrt[3]{3}$. But I'm not entirely clear as to why this series now has to converge. – Some Math Student Oct 21 '14 at 18:13
  • Wow, that is interesting, and actually rather unsettling. It does seem to be very much fixed on that value, and makes me wonder whether there is an error in the solution. :\

    Some Math Student: Good point. I will think about it and get back to you.

     – nathan.j.mcdougall Oct 21 '14 at 18:14
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    Note that $4^{\frac 14}=2^{\frac 12}$ – Mark Bennet Oct 21 '14 at 18:15
  • If you mean hundreds, what I mean by computing repeatedly is after 350 iterations the result is 2.47... and to the millionth iteration the last digit is still the same. I think it's never going to be any closer to 3 soon. But for sqrt(2), after only around 200 iterations the answer became closer to 2.0000 I wrote the program using c++ code. – offchan Oct 21 '14 at 18:16
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    I think you are right. But it's not diverging, which is the frustrating thing. If it's converging on something it should be 3rd root 3 >.< – nathan.j.mcdougall Oct 21 '14 at 18:18
  • It's not diverging, since $3^{1/3} < e^{1/e}$. – Irvan Oct 21 '14 at 18:23
  • Exactly, so what's the value it converging to? This is rather frustrating. – nathan.j.mcdougall Oct 21 '14 at 18:28
  • @nathan.j.mcdougall see my answer – Ben Grossmann Oct 21 '14 at 18:28
  • Nice. Now I can sleep well. – nathan.j.mcdougall Oct 21 '14 at 18:33
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There is one more observation which should be useful for you.
Begin again with the problem of iterated exponentiation to base $b=\sqrt2$. If you start with some value smaller than $x_0 \lt 4$ you arrive at $x_\infty=t_0=2$:

$ \quad \qquad \displaystyle x_0=1.1 \\ \quad \qquad x_k = b ^{x_{k-1}} \qquad \qquad \text{ where } b=\sqrt 2\\ \quad \qquad \lim_{k \to \infty} x_{k+1}=x_k=t_0=2 $

so we might call $t_0=2$ a "fixpoint" of that iterated operation, and because from all initial values (in a certain interval) it arrives at the same fixpoint, we call it an "attracting" fixpoint.

For instance

$ \quad \qquad \displaystyle x_0=5.3 \\ \quad \qquad x_k = b ^{x_{k-1}} \\ \quad \qquad \lim_{k \to \infty} x_{k+1}= \infty $

begins outside that interval and the iteration diverges to $\infty$

But if some value is a fixpoint for an operation, then it should also be the fixpoint of the reverse operation, thus we might look at

$ \quad \qquad \displaystyle x_0=1.1 \\ \quad \qquad x_k = \log_b (x_{k-1}) \qquad \qquad \text{where} \log_b(\cdot) = \log(\cdot)/ \log(b) \text{ and } b=\sqrt2\\ \quad \qquad \lim_{k \to \infty} x_{k+1} =x_k= t_{-1} \qquad \qquad \text{ (=some complex value) } $

But now if you begin at $x_0=5.3$ then we find convergence

$ \quad \qquad \displaystyle x_0=5.3 \\ \quad \qquad x_k = \log_b (x_{k-1}) \\ \quad \qquad \lim_{k \to \infty} x_{k+1} =x_k= t_1 = 4 $
$\qquad \qquad $ (and the same for any value $2 \lt x_0 \le 4$)

which is obviously "attracting" for the reverse operation in an interval in which the original operation has a divergent trajectory.

If in a third view we modify $x_0$ only the slightly least bit around $t_1$ then the original operation runs away from $4$, and so we might introduce the name "$t_1=4$ is a repelling fixpoint".

The same can now be applied to your other base where you find $b= 3^{1/3}$ . You can insert the exact value $x_0=3$ as initial value and shall find that it is a fixpoint. However it is a repelling one: if you let $x_0 = 3-\epsilon$ then the original operation iterates to the attracting fixpoint which is a bit larger than 2 and if you apply the reverse operation you'll find the repelling fixpoint $t_1 = 3$ which is what you wanted.

Gottfried Helms
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