Let $V$ be the vector space over $\mathbb{R}$ consisting of polynomials of degree less than or equal to $3$. Let $T : V \to V$ be the operator sending $f(t)$ to
$f(t + 1)$, and $D : V \to V$ the operator sending $f(t)$ to $df(t)/dt$. Then $T$ is a polynomial in $D$.
Here, the problem is that the image of $D$ i.e., $D(V)$ consists of polynomials of degree $2$ or less. But, image of $T$ contains polynomials of degree $3$. So, how can $T$ be a polynomial of $D$?
I think there is some problem with the question.
Thank you!
Best Answer
Here $D^{2}f(t)=f''(2),D^{3}f(t)=f^{(3)}(2),...$ Taylor's Theorem gives $f(t+1)=f(t)+\frac {f '(t)} {1!}+\frac {f ''(t)} {2!}+\frac {f ''(t)} {2!}+\frac {f^{(3)}(t)} {3!}$. Hence $T=I+\frac D {1!}+\frac {D^{2}} {2!}+\frac {D^{3}} {2!}=p(D)$ where $p(z)=1+\frac z {1!}+\frac {z^{2}} {2!}+\frac {z^{3}} {3!}$.