Inverse of matrix expansion with negative exponents
$begingroup$
The answer to this question shows that if I have a real nonsingular matrix $M$, such that its Taylor expansion in $epsilon$ is
$$M(x+epsilon)= sum_{n=0}^infty M_n(x) epsilon^n $$
its inverse can be written as an expansion using the following formula
$$B = sum_{i=0}^infty b_nepsilon^n$$
$$b_0 = a_0^{-1},$$
$$b_1 = -a_0^{-1}a_1a_0^{-1}$$
$$b_2 = a_0^{-1}a_1a_0^{-1}a_1a_0^{-1} -a_0^{-1}a_2a_0^{-1}$$
$$b_3 = - a_0^{-1}a_1a_0^{-1}a_1a_0^{-1}a_1a_0^{-1} + a_0^{-1}a_1a_0^{-1}a_2a_0^{-1} + a_0^{-1}a_2a_0^{-1}a_1a_0^{-1} - a_0^{-1}a_3a_0^{-1}$$
My question is: how does this change if I also have negative powers of $epsilon$ in the expansion? i.e.
$$mathcal{M}(x)=sum_{n=-infty}^infty mathcal{M}_n(x)epsilon^n$$
matrices power-series taylor-expansion inverse matrix-calculus
asked Jan 16 at 15:11
usumdelphiniusumdelphini
323111
$endgroup$
add a comment |
$begingroup$
The answer to this question shows that if I have a real nonsingular matrix $M$, such that its Taylor expansion in $epsilon$ is
$$M(x+epsilon)= sum_{n=0}^infty M_n(x) epsilon^n $$
its inverse can be written as an expansion using the following formula
$$B = sum_{i=0}^infty b_nepsilon^n$$
$$b_0 = a_0^{-1},$$
$$b_1 = -a_0^{-1}a_1a_0^{-1}$$
$$b_2 = a_0^{-1}a_1a_0^{-1}a_1a_0^{-1} -a_0^{-1}a_2a_0^{-1}$$
$$b_3 = - a_0^{-1}a_1a_0^{-1}a_1a_0^{-1}a_1a_0^{-1} + a_0^{-1}a_1a_0^{-1}a_2a_0^{-1} + a_0^{-1}a_2a_0^{-1}a_1a_0^{-1} - a_0^{-1}a_3a_0^{-1}$$
My question is: how does this change if I also have negative powers of $epsilon$ in the expansion? i.e.
$$mathcal{M}(x)=sum_{n=-infty}^infty mathcal{M}_n(x)epsilon^n$$
matrices power-series taylor-expansion inverse matrix-calculus
asked Jan 16 at 15:11
usumdelphiniusumdelphini
323111
$endgroup$
add a comment |
$begingroup$
The answer to this question shows that if I have a real nonsingular matrix $M$, such that its Taylor expansion in $epsilon$ is
$$M(x+epsilon)= sum_{n=0}^infty M_n(x) epsilon^n $$
its inverse can be written as an expansion using the following formula
$$B = sum_{i=0}^infty b_nepsilon^n$$
$$b_0 = a_0^{-1},$$
$$b_1 = -a_0^{-1}a_1a_0^{-1}$$
$$b_2 = a_0^{-1}a_1a_0^{-1}a_1a_0^{-1} -a_0^{-1}a_2a_0^{-1}$$
$$b_3 = - a_0^{-1}a_1a_0^{-1}a_1a_0^{-1}a_1a_0^{-1} + a_0^{-1}a_1a_0^{-1}a_2a_0^{-1} + a_0^{-1}a_2a_0^{-1}a_1a_0^{-1} - a_0^{-1}a_3a_0^{-1}$$
My question is: how does this change if I also have negative powers of $epsilon$ in the expansion? i.e.
$$mathcal{M}(x)=sum_{n=-infty}^infty mathcal{M}_n(x)epsilon^n$$
matrices power-series taylor-expansion inverse matrix-calculus
asked Jan 16 at 15:11
usumdelphiniusumdelphini
323111
$endgroup$
The answer to this question shows that if I have a real nonsingular matrix $M$, such that its Taylor expansion in $epsilon$ is
$$M(x+epsilon)= sum_{n=0}^infty M_n(x) epsilon^n $$
its inverse can be written as an expansion using the following formula
$$B = sum_{i=0}^infty b_nepsilon^n$$
$$b_0 = a_0^{-1},$$
$$b_1 = -a_0^{-1}a_1a_0^{-1}$$
$$b_2 = a_0^{-1}a_1a_0^{-1}a_1a_0^{-1} -a_0^{-1}a_2a_0^{-1}$$
$$b_3 = - a_0^{-1}a_1a_0^{-1}a_1a_0^{-1}a_1a_0^{-1} + a_0^{-1}a_1a_0^{-1}a_2a_0^{-1} + a_0^{-1}a_2a_0^{-1}a_1a_0^{-1} - a_0^{-1}a_3a_0^{-1}$$
My question is: how does this change if I also have negative powers of $epsilon$ in the expansion? i.e.
$$mathcal{M}(x)=sum_{n=-infty}^infty mathcal{M}_n(x)epsilon^n$$
matrices power-series taylor-expansion inverse matrix-calculus
matrices power-series taylor-expansion inverse matrix-calculus
asked Jan 16 at 15:11
usumdelphiniusumdelphini
323111
asked Jan 16 at 15:11
usumdelphiniusumdelphini
323111
asked Jan 16 at 15:11
usumdelphiniusumdelphini
323111
asked Jan 16 at 15:11
usumdelphiniusumdelphini
323111
asked Jan 16 at 15:11
usumdelphiniusumdelphini
323111
323111
add a comment |
add a comment |
1 Answer
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$begingroup$
If you define $M(x)$ as a Cauchy principal value, that is, $M(x)=lim_{krightarrow +infty}sum_{n=-k}^k M_n(x)epsilon^n$, then
$M(x)=sum_{n=0}^{+infty}U_n(x)epsilon^n$ where $U_0=M_0,U_{2p}=M_{2p}+M_{-2p},U_{2p+1}=M_{2p+1}-M_{-2p-1}$.
answered Jan 27 at 11:18
loup blancloup blanc
24.1k21851
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
If you define $M(x)$ as a Cauchy principal value, that is, $M(x)=lim_{krightarrow +infty}sum_{n=-k}^k M_n(x)epsilon^n$, then
$M(x)=sum_{n=0}^{+infty}U_n(x)epsilon^n$ where $U_0=M_0,U_{2p}=M_{2p}+M_{-2p},U_{2p+1}=M_{2p+1}-M_{-2p-1}$.
answered Jan 27 at 11:18
loup blancloup blanc
24.1k21851
$endgroup$
add a comment |
$begingroup$
If you define $M(x)$ as a Cauchy principal value, that is, $M(x)=lim_{krightarrow +infty}sum_{n=-k}^k M_n(x)epsilon^n$, then
$M(x)=sum_{n=0}^{+infty}U_n(x)epsilon^n$ where $U_0=M_0,U_{2p}=M_{2p}+M_{-2p},U_{2p+1}=M_{2p+1}-M_{-2p-1}$.
answered Jan 27 at 11:18
loup blancloup blanc
24.1k21851
$endgroup$
add a comment |
$begingroup$
If you define $M(x)$ as a Cauchy principal value, that is, $M(x)=lim_{krightarrow +infty}sum_{n=-k}^k M_n(x)epsilon^n$, then
$M(x)=sum_{n=0}^{+infty}U_n(x)epsilon^n$ where $U_0=M_0,U_{2p}=M_{2p}+M_{-2p},U_{2p+1}=M_{2p+1}-M_{-2p-1}$.
answered Jan 27 at 11:18
loup blancloup blanc
24.1k21851
$endgroup$
If you define $M(x)$ as a Cauchy principal value, that is, $M(x)=lim_{krightarrow +infty}sum_{n=-k}^k M_n(x)epsilon^n$, then
$M(x)=sum_{n=0}^{+infty}U_n(x)epsilon^n$ where $U_0=M_0,U_{2p}=M_{2p}+M_{-2p},U_{2p+1}=M_{2p+1}-M_{-2p-1}$.
answered Jan 27 at 11:18
loup blancloup blanc
24.1k21851
answered Jan 27 at 11:18
loup blancloup blanc
24.1k21851
answered Jan 27 at 11:18
loup blancloup blanc
24.1k21851
answered Jan 27 at 11:18
loup blancloup blanc
24.1k21851
24.1k21851
add a comment |
add a comment |
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