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$\left(\sqrt{2}\right)^{\sqrt{2}^{{\,}^{\textstyle.^{\textstyle\,.^{\textstyle \,.}}}}}=x$

well if the infinite power tower of roots is $x$, then

$\left(\sqrt{2}\right)^x=x$

Now, I just tried out reasonable candidates, namely $2$ and $4$, which are the solution(s).

But how to do it without "brute force"?

Also, if I just type it in my calculator, $2$ seems to be the limit of this expression, so is $4$ wrong?

Alexis Olson
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SAJW
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    see http://math.stackexchange.com/questions/87870/are-these-solutions-of-2-xxx-cdot-cdot-cdot-correct – E.H.E Oct 15 '16 at 02:20

1 Answers1

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Let's interpret the infinite tower as the limit of the sequence $a_0 = \sqrt{2}, a_1 = \sqrt{2}^{a_0} = \sqrt{2}^{\sqrt2}, a_2 = \sqrt2^{a_1} = \sqrt2^{\sqrt2^{\sqrt2}}, \ldots$ . Indeed any limit $x$ must satisfy $x = \sqrt{2}^x$, which has the solutions $x=2$ and $x=4$.

To show that the limit must be $2$, you can prove that if $a_i \le 2$ then $a_{i+1} \le 2$. Therefore, by induction, $a_i \le 2$ for all $i$, so the limit must also be $\le 2$.

arkeet
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  • this way seems easy for me. But i might ask if i can come to the same result if i show that $f(a_n)$ grows very slowly and will never reach values above 2.? Or other way asked, would this be the actual way to prove the $a_i$ stuff? – SAJW Oct 15 '16 at 02:44