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An incremental method to compute (Abel) matrix inverses
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(07/09/2010, 06:31 AM)bo198214 Wrote: I fiddled a bit around with Gottfried's suggestion of LU decomposition of the Abel matrix (though in the end the formula is independent of the LU decomposition).
(...)
Hi Henryk -

I've also tried a similar thing: to compute the new column/row of the LU-factors by the old ones - optimally by reference to the previous row/column only... without success so far.
Triggered by your msg I looked at a symbolic representation, using the symbol "u" for the log of the base (which can be substituted by 1 if the base is e = exp(1)).
So B is the Bell-matrix for x->exp(u*x), B1 is truncate(B - I) ,
Then L (lower) , D (diagonal), U (upper) the inverses of the LU-factors of B1, such if it converges, B1^-1 = U*D*L , where for the slog we need only the first column of L.
Looking at the last column of U with the idea to compose it by the previous column only (or by some composition of some earlier columns) I notice, that the entries are polynomials in u and I didn't see yet a simple possibility to determine that polynomials based on that of the previous colums. The same is true analoguously for the n'th entry in D and for the n'th row in L.

If I determine the coefficients for the powerseries for the slog by U*D*L[,0], then I see, that the degree of the polynomials, by which the coefficients are defined, increase binomially (n,2) and in our context of tetration I can't remember any recursionformula for such a situation. (In my treatize about the symbolic description of the coefficients for the dxp-Bell-matrix I came across the same growthrate of the order of the polynomials and did not find a formula how to compute one coefficient only by its index and possibly some known constants)

It would be nice to find such a formula for "the last" column in U and "last row" in L by a short recursion - this would then also allow to dismiss that matrix-inversion(s) completely... but I think, your derivation is comparably complex (so that the inversion is even faster)?

Gottfried
ah - ps: I tried also to express it in terms of (u-1) or (u+1) instead of u - but with little progress.
Gottfried Helms, Kassel
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Messages In This Thread
RE: An incremental method to compute (Abel) matrix inverses - by Gottfried - 07/20/2010, 12:13 PM

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