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 Discussion of TPID 6 JJacquelin Junior Fellow Posts: 5 Threads: 1 Joined: Oct 2010 10/22/2010, 11:27 AM (This post was last modified: 10/22/2010, 11:32 AM by JJacquelin.) (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. I think that the conjecture is false. First, the numerical computation have to be carried out with much more precision. The solution for x in ${}^{1000}x = 1000$ is approximately 1.44467831224667 which is higher than e^(1/e) The solution for x in ${}^{10000}x = 10000$ is approximately 1.4446796588047 which is higher than e^(1/e) As n increases, x increasses very slowly. But, in any case, x is higher than e^(1/e) = 1.44466786100977 Second, on a more theoretical viewpoint, if x=e^(1/e), the limit of ${}^{n}x$ is e , for n tending to infinity. So, the limit isn't = n , as expected. « Next Oldest | Next Newest »

 Messages In This Thread Discussion of TPID 6 - by JJacquelin - 10/22/2010, 11:27 AM RE: Discussion of TPID 6 - by bo198214 - 10/24/2010, 07:44 AM RE: Limit of self-super-roots is e^1/e. TPID 6 - by sheldonison - 10/22/2010, 11:41 PM RE: Limit of self-super-roots is e^1/e. TPID 6 - by nuninho1980 - 10/23/2010, 07:57 PM

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