How can I prove
$$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}...}}}}=2$$
I don't know which method can be used for this?
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$\begingroup$ You can start by defining what $\sqrt2^{\sqrt2^\ldots}$ is. $\endgroup$– user2345215Commented Dec 22, 2014 at 13:09
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$\begingroup$ There are several things related to convergence of $x^{x^{x^{\cdots}}}$ i.e., an infinite exponentiation that you might want to check first. Firstly it only makes sense when $x \in [e^{-e},e^{1/e}]$, then you use monotone convergence to finish the task ! :) $\endgroup$– scionaCommented Dec 22, 2014 at 13:16
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2$\begingroup$ related question $\endgroup$– Ben GrossmannCommented Dec 22, 2014 at 13:16
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$\begingroup$ Wouldn't it also fast, if you just look at $ | 2 - a_n| < \epsilon $ and then show there is always an $ a_n $ in $ (a_n)_n $, so that this inequation is true $\forall \epsilon > 0$ ? Well, someone had still to show that this series is strictly monotone rising. Personally I find this approach more intuitive, though most proofs for this work in a very similar fashion. $\endgroup$– ImagoCommented Dec 22, 2014 at 14:36
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3 Answers
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We can define $x=\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}...}}}}$ as follows:
Let $x_1 = \sqrt 2$ and $x_{n+1} = (\sqrt 2)^{x_{n}}$
We can show $x_n \lt 2\ \forall n$ by induction, since if $y \lt 2$, then $(\sqrt 2)^y \lt 2$. And $x_n$ is clearly monotonically increasing, so $x_n \to x$.
But $$x_{n+1} = (\sqrt 2)^{x_{n}}$$ so taking limits, we get that $$x = (\sqrt 2)^{x}$$
Solving this, and using the fact that $x \le 2$ gives $x =2$.
answered Dec 22, 2014 at 13:14
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$\begingroup$ I probably miss something, but if the sequence $x_n$ is bounded above bu $x$ and monotone, it converges, but it doesn't necessarily converge to $x$ $\endgroup$– AlexCommented Dec 22, 2014 at 13:25
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2$\begingroup$ @Alex $x$ is defined as the limit of the $x_n$. This is the most natural way to make sense of the infinite series of powers. $\endgroup$ Commented Dec 22, 2014 at 13:27
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Step One. Define the recursive sequence $$ a_0=\sqrt{2}, \quad a_{n+1}=\sqrt{2}^{a_n},\,\,n\in\mathbb N. $$
Step Two. Show that $\{a_n\}$ is increasing (inductively), and upper bounded by $2$ (also inductively).
Step Three. Due to Step Two the sequence $\{a_n\}$ is convergent. Let $a_n\to x$. Clearly, $\sqrt{2}<x\le 2$. But $a_{n+1}=\sqrt{2}^{a_n}\to x$, as well. Hence $$ x=\lim_{n\to\infty}a_n=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\sqrt{2}^{a_n}=\sqrt{2}^x. $$ Thus $x$ satisfies $\sqrt{2}^x=x$.
Step Four. Show that $x<\sqrt{2}^x$, for all $x\in (\sqrt{2},2)$, and thus $x=2$.
answered Dec 22, 2014 at 13:20
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Define $x_1=\sqrt{2}$ and $x_{n+1}=\sqrt{2}^{x_n}$.
Prove by induction that $x_n \leq x_{n+1} \leq 2$. As the sequence is bounded and increasing, it is convergent, and the limit is between $x_1=\sqrt{2}$ and $2$.
Finish the proof by observing that $$\sqrt{2}^x=x$$ has an unique solution on the interval $[\sqrt{2}, 2]$.
For the last part, as well as for the monotony, you should study first the monotony/sign of $x-\sqrt{2}^x$ on $ [\sqrt{2}, 2]$
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$\begingroup$ Is there a way to solve $\sqrt{2}^x=x$ without making a function: $f(x)=x-\sqrt{2}^x$? $\endgroup$– shinzouCommented Dec 22, 2014 at 13:30