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Exponentiation of a matrix

  1. Jan 3, 2014 #1
    I need to retrieve a finite rotation matrix (with cos and sin) from the exponentiation of the infinitesimal version of it.

    Suppose my infinitesimal matrix is ω. I then compute exp(ω).

    My guess would be

    [tex]\exp(\omega)=\sum_{k=0}\frac{\omega^{2k}}{2k!}+\sum_{k=0}\frac{\omega^{2k+1}}{(2k+1)!}[/tex]

    i.e. the even and odd contributions.

    The notes I'm reading suggest instead:

    [tex]\exp(\omega)=I+\sum_{k=1}\frac{\omega^{2k}}{2k!}+\sum_{k=1}\frac{\omega^{2k+1}}{(2k+1)!}[/tex]

    which looks weird to me; if I take the identity matrix I to be the k=0 contribution of the even part (ω^0=1), then I don't know where the term linear in ω is in the series any more. I think it's not there at all.

    Even more: I do need the k=0 contributions later on to retrieve the series expansion expressions for cos and sin.

    What do you think? Any comments?
     
  2. jcsd
  3. Jan 3, 2014 #2

    AlephZero

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    I agree. Replacing the ##\omega^0/0!## term by ##I## is fair enough, but the second sum should be from ##k = 0##.

    It's probably just a typo.
     
  4. Jan 3, 2014 #3

    HallsofIvy

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    ?? The "linear term in [itex]\omega[/itex]" is the term with [itex]\omega^1[/itex] which means it is the term in the second sum with k= 0: [itex]\frac{\omega^{2(0)+ 1}}{(2(0)+ 1)!}= \omega[/itex].

    The linear approximation to [itex]e^\omega[/itex] is [itex]I+ \omega[/itex].
     
  5. Jan 29, 2014 #4

    FactChecker

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    I agree. It's a typo. If you are doing this, don't forget that you can use the characteristic equation of the matrix ω to eliminate all the high powers.
     
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