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.
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.
