# Tetration Forum

Full Version: Another proof of TPID 6
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(10/07/2009, 12:03 AM)andydude Wrote: [ -> ]Conjecture

$\lim_{n\to\infty} f(n) = e^{1/e}$ where $f(n) = x$ such that ${}^{n}x = n$

Discussion

To evaluate f at real numbers, an extension of tetration is required, but to evaluate f at positive integers, only real-valued exponentiation is needed. Thus the sequence given by the solutions of the equations
• $x = 1$
• $x^x = 2$
• $x^{x^x} = 3$
• $x^{x^{x^x}} = 4$
and so on... is the sequence under discussion. The conjecture is that the limit of this sequence is $e^{1/e}$, also known as eta ($\eta$). Numerical evidence indicates that this is true, as the solution for x in ${}^{1000}x = 1000$ is approximately 1.44.

lim n-> oo x^^n = n conj : any real x = eta

since (eta+q) ^^ n grows faster than n for any positive q , we can use the squeeze theorem

lim q -> 0 eta =< x <= eta + q

hence x = eta

see also http://en.wikipedia.org/wiki/Squeeze_theorem

QED

regards

tommy1729