Proof with polynomial differential operators

**Silverflow** · April 20th 2010, 04:10 AM

Heyya,

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.

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