27

Solving $x^{x^{x^{.^{.^.}}}}=2\Rightarrow x^2=2\Rightarrow x=\sqrt 2$.

Solving $x^{x^{x^{.^{.^.}}}}=4\Rightarrow x^4=4\Rightarrow x=\sqrt 2$.

Therefore, $\sqrt 2^{\sqrt 2^{\sqrt 2^{.^{.^.}}}}=2$ and $\sqrt 2^{\sqrt 2^{\sqrt 2^{.^{.^.}}}}=4\Rightarrow\bf{2=4}$.

What's happening!?

user26486
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JSCB
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4 Answers4

33

Let's add the hypothesis that $x>0$ to the problem, so that it's clear your derivations are correct.

Pay attention to what you've proven:

  • If $x^{x^{\cdot^\cdot}} = 2$, then $x = \sqrt{2}$
  • If $x^{x^{\cdot^\cdot}} = 4$, then $x = \sqrt{2}$

This is very different from

  • If $x = \sqrt{2}$, then $x^{x^{\cdot^\cdot}} = 2$
  • If $x = \sqrt{2}$, then $x^{x^{\cdot^\cdot}} = 4$

Your argument that $2=4$ requires this latter pair of statements, but you haven't proven either of them; instead, what you've proven are the first pair of statements!

It's easy to get in the habit of forgetting about the direction you've argued a problem, and in many situations, arguments are reversible, making it hard to see why direction matters. But this is an example of the dangers of getting things wrong!


Incidentally, if $x = \sqrt{2}$ then $x^{x^{\cdot^\cdot}} = 2$ is correct, if we assume the usual meaning of infinite power towers as a limit of finite ones. If you're familiar with limits of sequences, then you can use an inductive proof to show that the sequence

$$ a_0 = \sqrt{2} \qquad \qquad a_{n+1} = \sqrt{2}^{a_n} $$

is strictly increasing and bounded above by $2$, and so the limit converges. And if $L$ is the limit, then because exponentiation is continuous, we can take the limit of the recursive relation to see that

$$L = \sqrt{2}^L$$

letting you complete the proof.

  • Your last statement $L=\sqrt2^L$ lets $L=2 \text{ or } L=4$, so doesn't look helpful for finding the limit. – Ruslan May 11 '13 at 13:06
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    Those are the only possibilities, and the information above let's you rule one of them out! The harder part is showing those are the only possibilities. –  May 11 '13 at 13:07
  • See the wikipedia entry on the [Lambert W function](http://en.wikipedia.org/wiki/Lambert_W_function), and note in particular [Example 3](http://en.wikipedia.org/wiki/Lambert_W_function), which gives the formula $z^{z^{z^\cdots}}=-W(\ln z)/\ln z$ whenever the left hand side converges. – Harald Hanche-Olsen May 11 '13 at 13:44
  • Then how to show $\sqrt 2^{\sqrt 2^{.^{.^.}}}$ is $2$ not $4$? – JSCB May 12 '13 at 03:12
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    @ᴊᴀsᴏɴ: If $a_n$ is an increasing sequence bounded by $2$.... –  May 12 '13 at 07:48
  • I get it, does it means there's no solution in $\mathbb R$ for $x^{x^{x^{.^{.^.}}}}=4$? – JSCB May 13 '13 at 08:15
9

You have merely shown that the equation $\sqrt 2^y = y$ has more than one solution.

Then you assumed that $x^{x^{x^\ldots}}$ somehow made sense and tried to talk about it as if it meant "the solution $y$ of $x^y = y$". Which of course is nonsense when the equation has several solutions.

mercio
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3

Your reasoning does not make sense because you did not specified what does $x^{x^{x\ldots}}$ mean. It is not a finitary operation so it is not clear what does that term denote. If it stands for an outcome of a certain limiting procedure then you are in trouble since that limit can be $1$ or $\infty$ for positive $x$. Thus with this interpretation your premises are false and you can deduce from them anything you want, for instance that $0=1$.

Godot
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2

In general, you can solve the equation in terms of the Lambert W function as

$$ y=x^{x^{x^{.^{.^.}}}} \implies \ln(y)=y\ln(x) \implies y = -\frac{W(-\ln(x))}{\ln(x)}. $$

Try to use this closed form to see what the problem is. Note this, if you ask maple to solve the equations

$$ -\frac{W(-\ln(x))}{\ln(x)}=2,\quad -\frac{W(-\ln(x))}{\ln(x)}=4, $$

you will get the same answers

$$ x=1,\sqrt{2} $$

More generally, the solution of

$$ -\frac{W(-\ln(x))}{\ln(x)}=a $$

is given by

$$ \left\{ x=1,x={{\rm e}^{{\frac {\ln \left( a \right) }{a}}}}\right\}. $$

Now, if you use the above solution with $a=2,4$, you will see why?

Mhenni Benghorbal
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