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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:
    $\displaystyle p(D)$ is a polynomial $\displaystyle D$ operator of degree $\displaystyle n>m$. Suppose $\displaystyle a$ is a $\displaystyle m$ fold root of $\displaystyle p(t)=0$, but not a $\displaystyle (m+1)$ fold root.
    Verify that $\displaystyle \frac{1}{p(D)}e^{at}=\frac{1}{p^{(m)}(a)}t^me^{at}$ where $\displaystyle p^{(m)}(t)$ is the $\displaystyle m^{th}$ derivative of $\displaystyle p(t)$.

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

    I'm thinking maybe there is a typo in the question. Maybe $\displaystyle p^{(m)}(a)$ is supposed to be $\displaystyle p^{(m)}(t)$. 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; Apr 20th 2010 at 04:15 AM. Reason: Extra information
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