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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 proved it by computing the equation repeatedly by replacing $x$ with $\sqrt{2}$ 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

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  • $\begingroup$ These are known as power towers or tetrations. Check out wikipedia for more information. $\endgroup$ – RghtHndSd Oct 21 '14 at 18:20
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    $\begingroup$ Just a small thing, doing convincing numerical calculations doesn't suffice as a proof $\endgroup$ – HBeel Oct 21 '14 at 20:00
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    $\begingroup$ 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: $$x^{x^{x^{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}}} = \,\,\lim_{n\to \infty} \left(x\uparrow\uparrow n\right)$$ $\endgroup$ – Nick Oct 21 '14 at 20:30
  • $\begingroup$ $\sqrt[3]{3}=x(say)$. 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? $\endgroup$ – Empty Oct 25 '14 at 14:42
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    $\begingroup$ 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 $\endgroup$ – off99555 Oct 26 '14 at 6:51
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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 show that this process 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.

The punchline is that we can guarantee that this sequence converges as long as near $x = 1$, $|f'(x)| < 1$. In fact, we have $$ f'(x) = \ln(a) a^x $$ so that $|f'(1)| < 1$ exactly when $a < e$. So, this sequence will only (necessarily) converge for $a \in (e^{-1},e)$.

More importantly, the only values the only value that this sequence can converge to are in $(e^{-1},e)$.

It follows that the equation $x^{x^{x^{\dots}}} = y$ can only be solved when $e^{-1/e} \leq x \leq e^{1/e}$, hence the impossibility of solution when $y = 3$.

See also this wikipedia page.

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  • $\begingroup$ 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? $\endgroup$ – nathan.j.mcdougall Oct 21 '14 at 18:30
  • $\begingroup$ And thus, we can now proof that it has no solution if $y > e$. $\endgroup$ – Irvan Oct 21 '14 at 18:31
  • $\begingroup$ You should be a math professor. Your knowledge is too much than I can perceive. $\endgroup$ – off99555 Oct 21 '14 at 18:42
  • $\begingroup$ 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 $\endgroup$ – off99555 Oct 21 '14 at 18:50
  • $\begingroup$ I appreciate the compliment. Let me know if there's something I should clarify here. $\endgroup$ – Omnomnomnom Oct 21 '14 at 18:51
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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 $y$.

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  • $\begingroup$ 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. $\endgroup$ – Irvan Oct 21 '14 at 18:12
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    $\begingroup$ 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. $\endgroup$ – Some Math Student Oct 21 '14 at 18:13
  • $\begingroup$ 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. $\endgroup$ – nathan.j.mcdougall Oct 21 '14 at 18:14
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    $\begingroup$ Note that $4^{\frac 14}=2^{\frac 12}$ $\endgroup$ – Mark Bennet Oct 21 '14 at 18:15
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    $\begingroup$ 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 >.< $\endgroup$ – nathan.j.mcdougall Oct 21 '14 at 18:18
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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.

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