I sincerely apologize, I need to fix two formulas in my post above.
The first is based on a transpose of the formula I actually gave, and I forgot to change the order or matrix multiplication when I took the transpose:
Furthermore, I could be mistaken and this formula may only apply for \( x_0 = 0 \). The second formula is actually the determinant of the Cramer's Rule matrix divided by the determinant of the truncated matrix, and I forgot about the determinant when I wrote it above:
Thats all. Sorry for the mistakes.
About the comment about the last equation, yes it's basically restating the infinite system of equations in a different form. I was hoping that by changing it from an iteration/power/derivatives to a combinatorial equation, that some properties could be found that would be helpful, but so far I have found nothing to be gained by putting it in combinatorial form.
Andrew Robbins
The first is based on a transpose of the formula I actually gave, and I forgot to change the order or matrix multiplication when I took the transpose:
\(
\mathbf{R}[f](x_0) = (\text{Pdm}[f](x_0) - \mathbf{I})\mathbf{J}
\)
\mathbf{R}[f](x_0) = (\text{Pdm}[f](x_0) - \mathbf{I})\mathbf{J}
\)
Furthermore, I could be mistaken and this formula may only apply for \( x_0 = 0 \). The second formula is actually the determinant of the Cramer's Rule matrix divided by the determinant of the truncated matrix, and I forgot about the determinant when I wrote it above:
\(
A_k = \lim_{n \rightarrow \infty} \frac{\det(\mathbf{R}[f](x_0)_{nk})}{\det(\mathbf{R}[f](x_0)_{n0})}
\)
A_k = \lim_{n \rightarrow \infty} \frac{\det(\mathbf{R}[f](x_0)_{nk})}{\det(\mathbf{R}[f](x_0)_{n0})}
\)
Thats all. Sorry for the mistakes.
About the comment about the last equation, yes it's basically restating the infinite system of equations in a different form. I was hoping that by changing it from an iteration/power/derivatives to a combinatorial equation, that some properties could be found that would be helpful, but so far I have found nothing to be gained by putting it in combinatorial form.
Andrew Robbins
