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 Logarithm reciprocal bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 07/20/2010, 04:13 AM (This post was last modified: 07/24/2010, 02:23 AM by bo198214.) Hello all you tetration brainies out there, looking somewhat deeper into the intuitive Abel function of f(x)=b*x (which is supposed to be log_b(x) however unproven until now), I found a somewhat direct expression of the coefficients, which boils down to the following challenging question: Let the sequence $(a_n)_{n\in\mathbb{N}}$ be defined recursively in the following way for $b>0$: $a_1 = \frac{1}{b-1}$ and $a_n = \frac{1}{1-b^n}\sum_{m=1}^{n-1} a_m \left(n\\m\right) (1-b)^{n-m} b^m$ for $n\ge 2$ Is $\lim_{n\to\infty} a_n = \frac{1}{\ln(b)}$? Does it converge? The following graph of the sequence for $b=2$, $1/\ln(b)\approx 1.442695$ leaves the question open:     (The messed up numbers on the left side are due to a bug in sage *sigh*) An equivalent slightly nicer formulation of the problem Let the sequence $(a_n)_{n\in\mathbb{N}}$ be defined recursively in the following way for $b>0$: $a'_1 = 1$ and $a'_n = \frac{1}{1-b^n}\sum_{m=1}^{n-1} a'_m \left(n\\m\right) (1-b)^{n-m} b^m$ for $n\ge 2$ Is $\lim_{n\to\infty} a'_n = \frac{b-1}{\ln(b)}$? edit: this can be found now as TPID 9 in the open problems thread. « Next Oldest | Next Newest »

 Messages In This Thread Logarithm reciprocal - by bo198214 - 07/20/2010, 04:13 AM RE: Logarithm reciprocal - by tommy1729 - 07/20/2010, 09:11 PM RE: Logarithm reciprocal - by bo198214 - 07/21/2010, 02:54 AM RE: Logarithm reciprocal - by bo198214 - 07/24/2010, 02:19 AM RE: Logarithm reciprocal - by tommy1729 - 07/24/2010, 09:33 PM RE: Logarithm reciprocal - by bo198214 - 07/24/2010, 11:10 PM RE: Logarithm reciprocal - by bo198214 - 08/11/2010, 02:35 AM RE: Logarithm reciprocal - by bo198214 - 07/25/2010, 03:39 PM RE: Logarithm reciprocal - by Gottfried - 07/26/2010, 12:14 PM RE: Logarithm reciprocal - by bo198214 - 07/26/2010, 03:56 PM RE: Logarithm reciprocal - by Gottfried - 07/26/2010, 05:08 PM

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