08/03/2009, 08:37 AM
(08/03/2009, 12:41 AM)andydude Wrote: What's wrong with "super-logarithm"?
Well I thought this is obvious: The prefix "super" for a function f describes a function F with
F(x+1)=f(F(x)). But super logarithm is not a superfunction of the logarithm, but an inverse superfunction of the exponential.
Quote:Hmm... I have no idea how I would describe it without matrices...
Ah I see now, it is because of the composition, for which you need the Carleman matrix. I had in mind to use the direct formula for powerseries composition (which is also in the paper already)
\( (f\circ g)_n = \sum_{k=1}^n f_k (g^k)_n \)
If you apply this to the Abel equation \( \alpha \circ f = \alpha+1 \) you get an infinite linear equation system in \( \alpha_k \) which expresses exactly
\( \vec{\alpha} C[f] = \vec{\alpha} + (1,0,0,...) \). Which is a row vector equation; if you transpose you get the usual column vector equation.
Ya, think about it. I let it to you. The goal should be a motivated and understanding reader

Quote:However, it would be good to generalize a little, because the definition of the Abel matrix doesn't apply to doing intuitive iteration of (x -> a x) because you have to use different truncations for this case.
Of course it does not work out of the box for f(x)=ax because it has a fixed point at 0 (and the method does not work at fixed points).
You have to (linearly) conjugate f such that the fixed point is somewhere else. It is an open question whether the intuitive Abel function is indeed log_a (and we should put that into the paper as conjecture. If it is true it would boost the significance of this method.). I followed this idea here. I merely could (non-publicly) achieve that the solution of the infinite equation system does not depend on the fixed point determined through the conjugation.
