# x raised to itself infinite number of times [duplicate]

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

What is $x$?

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

• See this thread: math.stackexchange.com/q/492109/264 – Zev Chonoles Feb 3 '14 at 4:23
• @ZevChonoles Thanks for the link. Apologies to anyone getting annoyed with dupes..it's quite hard searching for stuff here! – mchangun Feb 3 '14 at 4:25
• – Lucian Feb 3 '14 at 4:29

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 $x$.

• what if you tetrate something infinitly, (infinite pentration)? – tox123 Aug 17 '15 at 18:08

So, we have $$x^2=2$$

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

• This is a good answer if you have convergence. Convergence needs to be demonstrated to make this work. – Ross Millikan Feb 3 '14 at 4:25
• @RossMillikan, if the Right Hand Side is finite, so should be the left, right? – lab bhattacharjee Feb 3 '14 at 4:30
• 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. – Ross Millikan Feb 3 '14 at 4:35
• Except that here every element of the sequence is identical. – JPi Feb 3 '14 at 4:55