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Possible Duplicate:
Infinite tetration, convergence radius

Recently in this thread, Pseudo Proofs that are intuitively reasonable, I learned that $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}} = 2$$

The next natural question to ask is, what is the largest number $x$ such that $$f(x)=x^{x^{x^{x^{...}}}}$$ converges?

A short exercise in matlab coding suggests that either $f(1.5)$ diverges, or whatever it converges to is too large for my computer to handle. Thus the answer should be somewhere between 1.41 and 1.5.

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marked as duplicate by Aryabhata, Qiaochu Yuan Mar 26 '12 at 20:15

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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    $\begingroup$ It is not explicitly mentioned on the linked question, but the radius of convergence you are looking for is $e^{1/e} = \max_y y^{1/y} \approx 1.4446678$. $\endgroup$ – TMM Mar 26 '12 at 20:18
  • $\begingroup$ Thanks for the answer TMM. Sorry to ask a duplicate question - I searched for "power tower" before asking, but did not know of the term "tetration". $\endgroup$ – Nick Alger Mar 26 '12 at 20:22
  • $\begingroup$ your expression could be reduced to an infinite product of (1/2)ln(x). Now use the infinite product convergence theorems and try compute rhs. sorry i am int rush to get to work would try post more details later. $\endgroup$ – Comic Book Guy Mar 26 '12 at 20:46