# What is wrong with this funny proof that 2 = 4 using infinite exponentiation?

Out of boredom, I decided to recall the following equation:

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

Which, I simply rewrote like this:

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

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

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 its not injective. However, this led to an awkward statemend based on the fact that $\sqrt{2} = (4)^{\frac{1}{4}}$. We deduce the following:

The equation $x^{x^{x\cdots}} = 2$ and the equation $x^{x^{x\cdots}} = 4$ 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 $p = e$?

• 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
• 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

## 1 Answer

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.