Thread Rating:
  • 0 Vote(s) - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Superlog determinant factored!
I will write the proof as a story, so as to make it more interesting, but first, some definitions.

Let be the nth truncated Abel matrix of f expanded at , and let . As I wrote about in my 2005 paper, each nth solution is unique iff is nonzero. These solutions can be used as the coefficients of the superlog as .

Let indicate that for all . In some textbooks, this is referred to as "f and g have contact of order n at 1". In order to keep the notation small, every occurance of this notation will refer to f and g as functions of b, not any other variable, and the expansion point will always be 1.



So one day, I was playing around with the determinants , and began differentiating them. I was bored, and stumbled upon the quotient:

and I noticed that the series expansions of this expression at b=1 increased with every approximation number n, so I then looked at the quotient:

and to my surprise, the series expansions did not change at all with n, except for the coefficients of , which are beyond the claim of . So after looking at the quotient a bit more, I saw that it was the logarithmic derivative of , so I integrated both sides.

The extra terms are to make up for the constant of integration, and for the integral of in the big summation on the right. So after some simplification, I realized that

or in other words, the can be factored into a function of b and a function of . Putting it all together:


Although it is not as exciting as I hoped at first, it is still interesting. The formula can be verified because it asserts something that is true for all n, and so if it is true for it will probably be true for more n. I have tried to use this formula to find information about base-e, but these functions don't converge out there. I'm thinking it has something to do with the fact that base-1 is divergent. I'm hoping that this kind of formula could help with the higher bases, but so far no such luck. I have also tried to find a similar formula for the same matrix with the first column replaced with (1, 0, 0, 0, ...) which represents the numerator of Cramer's rule for the first coefficient/derivative of superlog.

One final remark. Notice that n has been isolated. This makes it obvious that

diverges to infinity if and converges to 0 if , but is this only for bases near 1?

Andrew Robbins
OMG, I just realized:
(11/09/2009, 09:39 PM)andydude Wrote: OMG, I just realized:

So I guess I should rephrase the statement as:

which may provide a useful shortcut to showing whether or not intuitive/natural iteration even works. Suppose we make a Cramer's rule matrix with all the same entries of , except the first column is replaced with (1, 0, 0, 0, ...) so we can solve for . Let , then if the limit exists, which is a rational polynomial mess.

If we take into account the above formula, then

which may provide a clear path to answering the question of convergence.

Possibly Related Threads...
Thread Author Replies Views Last Post
  Branch points of superlog mike3 0 3,182 02/03/2010, 11:00 PM
Last Post: mike3
  Superlog with exact coefficients andydude 7 11,328 03/13/2009, 06:14 PM
Last Post: bo198214

Users browsing this thread: 1 Guest(s)