# Thread: Proof with polynomial differential operators

1. ## Proof with polynomial differential operators

Heyya,

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.