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Thread: Proof with polynomial differential operators

  1. #1
    Junior Member
    Oct 2009

    Proof with polynomial differential operators


    Got a problem that I have no idea how to approach.
    The question is a follows:
    p(D) is a polynomial D operator of degree n>m. Suppose a is a m fold root of p(t)=0, but not a (m+1) fold root.
    Verify that \frac{1}{p(D)}e^{at}=\frac{1}{p^{(m)}(a)}t^me^{at} where p^{(m)}(t) is the m^{th} derivative of p(t).

    I can see some use for the exponential shift formula. That would make the left hand side equal e^{at}\frac{1}{p(D+a)}.
    From here, I can't see where to go.

    I'm thinking maybe there is a typo in the question. Maybe p^{(m)}(a) is supposed to be p^{(m)}(t). However, the above is copied directly from the source. I don't know what to make of it.

    Any input is greatly appreciated.
    Thanks in advance.
    Last edited by Silverflow; April 20th 2010 at 04:15 AM. Reason: Extra information
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