What is the Pade approximation of the matrix logarithm?

Question

I would like to use the Pade approximation in my numerical procedure and I would like to use it to approximate the logarithm of a matrix. However, I couldn't find the correct expression for it in the literature. I know the general expression of the Pade approximation but I do not know how to use it for my Matrix logarithm.

So suppose that I have a matrix $\textbf{X}$ that takes the following form

$\textbf{X}=\begin{pmatrix} 1.002 & 0 & 0 \\ 0 & 1.01 & 0 \\ 0 & 0 & 1.1 \end{pmatrix}$.

The expression for the Pade approximation reads

$(\textbf{A}+\textbf{A}^2/2).(\textbf{I}+\textbf{A}+\textbf{A}^2/6)^{-1}$

where $\textbf{A}=\textbf{X}^T.\textbf{X}-\textbf{I}$ and $\textbf{I}$ is a 3 by 3 identity matrix.

But upon using this formulation, I always get $1/0$ expression.

What is the missing part in my calculation?

Answer 1

I think you are making some mistake in computing. The following PARI/GP code (with a $1/2$ factor included)

default(realprecision, 7);     
X = matdiagonal([1.002, 1.01, 1.1]);
A = X~*X - matid(3);
L = 1/2 * (A+A^2/2) / (1+A+A^2/6);
print(L);

outputs

[0.001998003, 0, 0; 0, 0.009950331, 0; 0, 0, 0.09530948]

while the code

log([1.002, 1.01, 1.1])

outputs [0.001998003, 0.009950331, 0.09531018] and the two answers are close. Also note that the code

X = 1 + x; A = X*X - 1;
L = 1/2 * (A+A^2/2) / (1+A+A^2/6);
print(L + O(x^5));

outputs x - 1/2*x^2 + 1/3*x^3 - 1/4*x^4 + O(x^5) which is the first four terms of the power series expansion of $\,\log(1+x).$