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$\newcommand\iddots{\mathinner{ \kern1mu\raise1pt{.} \kern2mu\raise4pt{.} \kern2mu\raise7pt{\Rule{0pt}{7pt}{0pt}.} \kern1mu }}$ Out of boredom, I decided to recall the following equation:

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

Which, I simply rewrote like this:

$x^2 = 2$, and therefore $x = \sqrt{2}$. Then I took a look at the more general form:

$$x^{x^{{x}^{\iddots}}} = p.$$

I then concluded that $x = (p)^{\frac{1}{p}}$, and this was the solution set for all $p$. Then I thought for a while and determined that there are multiple values where this function equals the square root of $2$, so it's not injective. However, this led to an awkward statement based on the fact that $\sqrt{2} = (4)^{\frac{1}{4}}$. We deduce the following:

The equation $x^{x^{{x}^{\iddots}}} = 2$ and the equation $x^{x^{{x}^{\iddots}}} = 4$ have the same solution, namely $x = \sqrt{2}$. So we conclude that $2 = 4$.

I showed this to friend of mine, and he conjectured that there are no valid solutions for all $p > e$. To prove this, I tried writing the following:

I tried differentiating $f(x) = (x)^{(1/x)}$ and setting it equal to $0$ to find the maximum. I determined that there was a maximum at the point where $x = e$. Does this mean that the maximum possible solution $x$ for the infinite power is $e^{(1/e)}$ therefore showing that the maximum value for $p$ is indeed $p = e?$

Adrian Keister
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ra1nmaster
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    The Wikipedia article for $e$ talked about this very thing, I guess the proof is due to Euler that $e^{1/e}$ is the upper limit for $p$. – abiessu Sep 12 '13 at 23:36
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    There is a lower positive value too, of $x=e^{-e}$, if you want finite tetration to converge in the limit. – Henry Sep 12 '13 at 23:58

3 Answers3

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First you need to determine what $x^{x^{x^{\cdots}}}$ means.

A natural attempt to give meaning to this would be the limit (if it exists) of the sequence $$ 1, ~x,~x^x,~x^{x^x}, ~x^{x^{x^x}}, \ldots $$ which can also be written as a recurrence

$$ z_0=1 \qquad z_{n+1} = x^{z_n} $$

If the limit exists at all, it must be a fixed point of the map $z\mapsto x^z$, but that doesn't mean that every fixed point has to be a limit.

In the case $x=\sqrt 2$, we easily see that both $z=2$ and $z=4$ are fixed points. The question is just which (if any) of them is the limit of the sequence. Plotting the function and using some iteration theory we see that $z=2$ is a stable fixed point and must be reached when we start from $1$, wheras $z=4$ is not stable -- if $z_n$ is close to $4$, the next $z_{n+1}$ will be less close to $4$.

You have already proved that IF there is any $x$ such that $x^{x^{\cdots}}=4$ when $x$ must be $\sqrt 2$. However, since $x=\sqrt 2$ doesn't in fact give $4$, then the conclusion is simply that $x^{x^{\cdots}}$ never takes the value 4.

hmakholm left over Monica
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In addition to Henning's excellent answer, I would add this as well: we have that $y=2$ implies $x=\sqrt{2},$ and $y=4$ implies $x=\sqrt{2}.$ Schematically, we have $A\to C$ and $B\to C.$ This doesn't imply that $A$ and $B$ are equivalent. That would be like saying "If you're a dog, you're a mammal. If you're a cat, you're a mammal. Therefore, if you're a dog, you're a cat." In logic, this is the fallacy of the undistributed middle.

Adrian Keister
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$$\{x \text{ s.t } f(x) = y\} = \{x \text{ s.t } f(x) = z\} \implies y=z$$

A vacuous counter example would be the empty set, $x^2 = -1$ and $x^2 = -2$ have the same solution set in reals, but obviously $-1 \ne -2$.

Otherwise the statement is actually true. Let $q$ be some value in the inhabited set, then it must hold that $f(q) = y = z$.

I then concluded that x=p^{1/p}, and this was the solution set for all $p$.

This is the statement that isn't right for the reasons explained in the accepted answer.

DanielV
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