$$\Large x^{x^{x^{x^{x^{.^{\,.^{\,.}}}}}}} = 2$$
What is $x$?
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2See this thread: http://math.stackexchange.com/q/492109/264 – Zev Chonoles Feb 03 '14 at 04:23
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2@ZevChonoles Thanks for the link. Apologies to anyone getting annoyed with dupes..it's quite hard searching for stuff here! – mchangun Feb 03 '14 at 04:25
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See infinite tetration. – Lucian Feb 03 '14 at 04:29
2 Answers
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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 $x$.
Einsteinium
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Yiyuan Lee
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Yiyuan, This is the answer I needed. However if the RHS is 4, then the solution would be 4^(1/4) which is exactly √2 . Thus it seems that the same sequence of numbers has not one limit (2), but two limits: 2 and 4. Could you, please, explain this paradox? . – Dimitre Novatchev Mar 11 '24 at 17:15
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Yiyuan, I found the answer here: https://arxiv.org/pdf/1908.05559.pdf . the value of the RHS must be in the interval:
[1/e, e]. This means that only two natural numbers qualify as values of the RHS: 1 and 2. – Dimitre Novatchev Mar 11 '24 at 19:23 -
I edited the answer, adding this new information - hope you don't mind. – Dimitre Novatchev Mar 11 '24 at 19:32
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So, we have $$x^2=2$$
In general, if $$x^{x^{x^{\cdots}}}=y, x^y=y$$
lab bhattacharjee
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5This is a good answer if you have convergence. Convergence needs to be demonstrated to make this work. – Ross Millikan Feb 03 '14 at 04:25
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1@RossMillikan, if the Right Hand Side is finite, so should be the left, right? – lab bhattacharjee Feb 03 '14 at 04:30
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3It 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 03 '14 at 04:35
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