Inverse of matrix expansion with negative exponents












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$begingroup$


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.



$$mathcal{M}(x)=sum_{n=-infty}^infty mathcal{M}_n(x)epsilon^n$$










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    0












    $begingroup$


    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.



    $$mathcal{M}(x)=sum_{n=-infty}^infty mathcal{M}_n(x)epsilon^n$$










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      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.



      $$mathcal{M}(x)=sum_{n=-infty}^infty mathcal{M}_n(x)epsilon^n$$










      share|cite|improve this question









      $endgroup$




      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






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      asked Jan 16 at 15:11









      usumdelphiniusumdelphini

      323111




      323111






















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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 $U_0=M_0,U_{2p}=M_{2p}+M_{-2p},U_{2p+1}=M_{2p+1}-M_{-2p-1}$.






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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 $U_0=M_0,U_{2p}=M_{2p}+M_{-2p},U_{2p+1}=M_{2p+1}-M_{-2p-1}$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              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 $U_0=M_0,U_{2p}=M_{2p}+M_{-2p},U_{2p+1}=M_{2p+1}-M_{-2p-1}$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                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 $U_0=M_0,U_{2p}=M_{2p}+M_{-2p},U_{2p+1}=M_{2p+1}-M_{-2p-1}$.






                share|cite|improve this answer









                $endgroup$



                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 $U_0=M_0,U_{2p}=M_{2p}+M_{-2p},U_{2p+1}=M_{2p+1}-M_{-2p-1}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 27 at 11:18









                loup blancloup blanc

                24.1k21851




                24.1k21851






























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