The Infinite Exponential Tower That Shoots Straight Up to the Sky

直上云霄的无穷指数方程 苏剑林 · 2010-02-06 · Mathematics exponent equation infinity

Browsing the web yesterday, I stumbled upon an interesting equation:

$$x^{x^{x^{\dots}}}=2$$

Don't rush ahead, dear reader — why not try solving it yourself first? more

Actually it's simple enough: the number inside the parentheses of $x^{x^{x^{\dots}}}=2$ is itself 2, so the original equation becomes $x^2 =2$, from which we get $x=\sqrt{2}$.

Simple, isn't it? ^_^ But wait, we're not done yet! Now try solving this equation: $x^{x^{x^{\dots}}}=4$

Well, applying the same method as before, you'll find that likewise $x=\sqrt{2}$.

Now we're left scratching our heads: is $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\dots}}}$ equal to 2 or equal to 4?

Actually, this number should be equal to 2. Here's a rather non-rigorous way to see it: in order to find an upper bound for this number (this term might not be quite the right one to use here), $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\dots}}}} \leq \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\dots^2}}}=2$, and so $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\dots}}}$ cannot possibly be greater than 2.

It seems that once infinity gets involved, all sorts of bizarre things become possible. Even a method this basic can end up producing "extraneous roots"...

By the way, readers might like to try using the method above to solve for $2^{2^{2^{\dots}}}$: let $2^{2^{2^{\dots}}}=k$, then $2^k =k$, and it turns out this equation has no solution at all!!! — Wait, actually there is one solution: $k\to\infty$, which means $2^{2^{2^{...}}}$ diverges.

So what, then, is the condition for convergence? From the method above it's not hard to see that: when $x=\sqrt[n]{n}$, $f(x)$ can take a definite value (though not necessarily equal to n), and this is precisely the condition for convergence. $\sqrt[n]{n}$ attains its maximum at n = 2. Therefore, the range over which $x^{x^{x^{\dots}}}$ converges is $x \in [0,\sqrt{2}]$