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