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How to prove e^At e^(-At_0) = e^A(t-t_0)?

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Avatar for sciencolic

Sciencolic (sciencolic)

I love Physics, Mathematics and Science.
Japan

  • Mathematics
    		• 597
  • Power series
    		• 0
  • Matrix (mathematics)
    		• 0
    26 Aug 10

    Given that A is a matrix, I would like to ask how to show:

    \f[e^{A t}e^{-A t_0}=e^{A\left(t-t_0\right)}\f]

    Thank you very much.

    2 comments >
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    Avatar for ross_tang

    Ross Tang (ross_tang)

    Degree in Physics and Mathematics, Master in Physics
    香港

  • Mathematics
    		• 809
    26 Aug 10

    Have you tried to expand it using power series of exponential and do the grouping? Anyway, here are the steps:

    \f[\begin{align*} e^{A t}e^{-A t_0}&=\sum _{k=0}^{\infty } \frac{1}{k!}A^kt^k\sum _{i=0}^{\infty } \frac{1}{i!}(-1)^iA^it_0{}^i \\ &=\sum _{i=0}^{\infty } \sum _{k=0}^{\infty } \frac{(-1)^i}{k!i!}A^{k+i}t^kt_0{}^i \\ \end{align*}\f]
    Now, let m = k + i, and try to change the summation variables:

    \f[\begin{align*} e^{A t}e^{-A t_0}&=\sum _{m=0}^{\infty } \sum _{k=0}^m A^m\frac{(-1)^{m-k}t^k t_0^{m-k}}{k!(m-i)!}\\ &=\sum _{m=0}^{\infty } \left(\sum _{k=0}^m \frac{(-1)^{m-k}t^kt_0^{m-k}}{k!(m-k)!}\right)A^m\\ &= \sum _{m=0}^{\infty } \frac{\left(t-t_0\right){}^mA^m}{m!}\\ &=e^{A\left(t-t_0\right)} \end{align*}\f]

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    Avatar for juaninf

    Juan Grados (juaninf)

    Master Student Scientific Computing at LNCC
    RJ, Brazil

    27 Aug 10

    mmm, where i can read about change of variables in summation inifinite,

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