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This question already has an answer here:

$$\Large x^{x^{x^{x^{x^{.^{\,.^{\,.}}}}}}} = 2$$

What is $x$?

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marked as duplicate by Newb, vonbrand, user127.0.0.1, Ross Millikan, user61527 Feb 3 '14 at 5:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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What you have there is called an infinite tetration. For your case, $x^2 = 2 \implies x = \sqrt2$.

In general, for $y = \Large x^{x^{x^{.^{\,.^{\,.}}}}}$, Euler showed that it is necessary that $e^{-e} \leq x \leq e^{\frac{1}{e}}$ for convergence to occur for real $x$.

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  • $\begingroup$ what if you tetrate something infinitly, (infinite pentration)? $\endgroup$ – tox123 Aug 17 '15 at 18:08
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So, we have $$x^2=2$$

In general, if $$x^{x^{x^{\cdots}}}=y, x^y=y$$

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    $\begingroup$ This is a good answer if you have convergence. Convergence needs to be demonstrated to make this work. $\endgroup$ – Ross Millikan Feb 3 '14 at 4:25
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    $\begingroup$ @RossMillikan, if the Right Hand Side is finite, so should be the left, right? $\endgroup$ – lab bhattacharjee Feb 3 '14 at 4:30
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    $\begingroup$ It takes more than that. Think of $1-1+1-1\dots$ A similar argument would be $S=1-1+1-1\dots=1-S, S=\frac 12$ but I can make it come out other things. $\endgroup$ – Ross Millikan Feb 3 '14 at 4:35
  • $\begingroup$ Except that here every element of the sequence is identical. $\endgroup$ – JPi Feb 3 '14 at 4:55

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