10/29/2007, 11:30 AM
Jay ,
an approximate answer... I didn't get the meaning and the subsequent computing from your two-liner about how to compute slog when I read it. But anyway, I recognized that you are using that matrix, that I call B, however modified by subtracting something(like identity matrix) from the subdiagonal and use this in the matrix-root-solving formula.
Let the subdiagonal-detail aside, then an approximate answer may be, that B can be decomposed into a product of a factorial scaled Stirling-matrix 2'nd kind and a binomial-matrix:
B = S2 * P~
and your matrix-root-solving formula was something like
(B - D) *X = Y
where D means the subtraction in the subdiagonal, Y seems to be the second column of the identity-matrix and X are the terms, that you use.
Well, let D aside, then this is about
S2 * P~ * X = Y
and the root solving uses the inversion of B (well, actually B-D) , so
X = P~^-1 * S2^-1 * Y
S2 contains the coefficients of the exponential-series in its second column, and S2^-1 = S1 contains the factorial scaled Stirling numbers 1'st kind in the same column - which are just -1,1/2,-1/3,...
So it is no surprise for me to read, that for slog and sexp you get terms which approximate to that values. May be you can improve your observation by a binomial-transform of your terms, so computing
Z = P~ * X
may have a even clearer pattern.
Hope, I'm not completely out of the path...
Gottfried
an approximate answer... I didn't get the meaning and the subsequent computing from your two-liner about how to compute slog when I read it. But anyway, I recognized that you are using that matrix, that I call B, however modified by subtracting something(like identity matrix) from the subdiagonal and use this in the matrix-root-solving formula.
Let the subdiagonal-detail aside, then an approximate answer may be, that B can be decomposed into a product of a factorial scaled Stirling-matrix 2'nd kind and a binomial-matrix:
B = S2 * P~
and your matrix-root-solving formula was something like
(B - D) *X = Y
where D means the subtraction in the subdiagonal, Y seems to be the second column of the identity-matrix and X are the terms, that you use.
Well, let D aside, then this is about
S2 * P~ * X = Y
and the root solving uses the inversion of B (well, actually B-D) , so
X = P~^-1 * S2^-1 * Y
S2 contains the coefficients of the exponential-series in its second column, and S2^-1 = S1 contains the factorial scaled Stirling numbers 1'st kind in the same column - which are just -1,1/2,-1/3,...
So it is no surprise for me to read, that for slog and sexp you get terms which approximate to that values. May be you can improve your observation by a binomial-transform of your terms, so computing
Z = P~ * X
may have a even clearer pattern.
Hope, I'm not completely out of the path...
Gottfried
Gottfried Helms, Kassel

