6
$\begingroup$

What is $x$ when is satisfies $x^{x^{x^{\dots}}}=2$ ?

I am really confused with this; the root is $\sqrt{2}$, but why does the equation $x^{x^{x^{\dots}}}=4$ have the same root?

$\endgroup$
  • $\begingroup$ Why does \udots not work? How can I write dots going from bottom left to top right? $\endgroup$ – FraGrechi Jul 10 '16 at 14:12
  • $\begingroup$ "How much" is not exactly correct wording in this case $\endgroup$ – Yuriy S Jul 10 '16 at 14:14
  • $\begingroup$ {\cdot^{\cdot^{\cdot}}} gives e.g. $ x^{x^{x^{\cdot^{\cdot^{\cdot}}}}} $ $\endgroup$ – Joffan Jul 10 '16 at 14:25
  • 1
    $\begingroup$ It's clearly a convergence issue, I believe its known that this tower only converges in a certain interval containing $2$ but not $4$. Something with $\exp(1/e)$ and $\exp(e)$ if I remember correctly. Hopefully someone more knowledgable will remember it exactly. $\endgroup$ – Myself Jul 10 '16 at 17:25
  • 2
    $\begingroup$ The true question is: why do you believe that $x^{x^{x^{\dots}}}=4$ has the same positive root? If you can clarify this, we shall be able to answer your question too. $\endgroup$ – Alex M. Jul 10 '16 at 17:53
8
$\begingroup$

Let's first start with defining what we mean by the expression $a = x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}$: the infinite power tower is the limit of the recursion

$$a_{n+1} = x^{a_n},~~~~~~~ a_0 = x$$

If $x>1$ then $x^x > x$ so $a_{n}$ is monotonically increasing and converges iff it's bounded above (this is the case only if $e^{-e} \leq x \leq e^{\frac{1}{e}}\approx 1.44$; see e.g. Wiki:Tetration). If $x=\sqrt{2}$ then it follows by induction that $a_n \leq 2$ since $a_0 = \sqrt{2} < 2$ and

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

Thus $a_n$ is bounded above by $2$ (and the recursion therefore converges). Taking the limit of the recursion we get that it satsify the equation $a=x^a$. As you have noted this equation has the two solutions $a=2$ and $a=4$ (which are also the only real solutions). Since the limit is well defined only one of these two solutions are valid and since $a\leq 2$ we have that the solution $a=4$ does not describe the limit of the recursion above.

Why is there an extra solution? Note that

$$\lim\limits_{n\to\infty} a_n = a \implies a = x^a$$

is a one way implication and it's not true that $a = x^a \implies \lim\limits_{n\to\infty} a_n = a$. This is similar to

$$x=1 \implies x^2 = 1 \implies x = \pm 1$$

In this simple example it's squaring that introduces an extra solution. This happens quite a lot when manipulating equations: if not all of the steps are two-way implications then we might introduce extra solutions and we must use some other means to determine which one is the right one to choose.


In general when the infinite tower converges the limit is given by

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

where $W$ is the (principal branch of the) Lambert function. The maximum value (for $x$ where it converges) is found for $x = e^{\frac{1}{e}}$ where we have $a = e < 4$. Thus $x^{x^{x^{\cdot^{\cdot^{\cdot}}}}} = 4$ has no solutions at all in real numbers so the manipulations you do to get $x = \sqrt{2}$ are not justified.

$\endgroup$
1
$\begingroup$

(I've answered this at an earlier similar question, see the link , the following is a copy&paste from there)


Also you should consider, whether you would better like to write $\small x$, $\small _bx $ , $\small _{_b}{_b}x $ ,$\small {_{...} } _{_b}{_b}x $ , because you always begin the evaluation at the top of the powertower and not at the bottom. And also then it is unambiguous to discuss $\small 2= 2 $, $\small 2 = _\sqrt22 $ , $\small 2= _{_\sqrt2}{_\sqrt2}2 $ and $\small 2= {_{...} } _{_\sqrt2}{_\sqrt2}2 $ as evaluated from the top. It is then also correct to write $\small 4= 4 $, $\small 4 = _\sqrt24 $ , $\small 4= _{_\sqrt2}{_\sqrt2}4 $ and $\small 4= {_{...} } _{_\sqrt2}{_\sqrt2}4 $ as a second solution. (This is clearly no standard notation, but I really do not know why this did not become standard)

[added] Then one could also write $\small 2= \lim {_{...} } _{_\sqrt2}{_\sqrt2}x \text{ for } -\infty \lt x \lt 4$ to note the convergence of all that initial values x, and because $\small x=\sqrt2 $ is in that range we can say $\small 2= \lim {_{...} } _{_\sqrt2}{_\sqrt2}\sqrt2 $


[not in the original answer:] The same occurs also for each of the infinitely many (complex) fixpoints of the equation $ \sqrt{2}^{z_k}=z_k$
For instance $z_2 \approx \small 0.145108802639 - 0.220160521524 î $ is such a complex fixpoint in a sense because it fits $z_2 = \exp ( (\log(b)-2 \pi î) \cdot z_2)$ where $b=\sqrt{2}$

$\endgroup$
-1
$\begingroup$

Assuming the limit exists, then since $$x^{x^{x^{\cdots }}} = 2$$ then it follows that $$x^2 = 2$$ which has two solutions: $x = \sqrt{2} $ and $x = -\sqrt{2}$. If you're working over the reals, then you'll want to ignore $-\sqrt{2}$.

Do note that all this operates under the assumption that the limit exists, you need to prove it does before the working above is valid.

$\endgroup$
  • 11
    $\begingroup$ Despite all the upvotes, this is not what the OP asked for. He already knows these solutions, he asks why they are the same when the right-hand side is $4$ instead of $2$. I always find it ridiculous when people vote without reading. $\endgroup$ – Alex M. Jul 10 '16 at 17:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.