Recall the theorem on page 162:
Theorem: The differentiation operator $\frac{\mathrm{d}}{\mathrm{d}x}$ is a linear operator from $\mathbb{R}^n$ into $\mathbb{R}^n$, and for any $t\in\mathbb{R}$ \begin{equation}\mathrm{e}^{t\frac{\mathrm d}{\mathrm dx}}=H^t,\end{equation} where $H^t: \mathbb{R}^n\rightarrow\mathbb{R}^n$ is the operator of translation by $t$, i.e. $\left(H^tf\right)(x)=f(x+t)$.
Remark that $\mathbb{R}^n$ above represents a vector space of quasi-polynomials $\mathrm{e}^{\lambda x}p(x)$ (whose space is isomorphic to $\mathbb{R}^n$),and where $p(x)$ is a polynomial whose degree is less than $n$.
My question is how to understand the sign $\mathrm{e}^{t\frac{\mathrm d}{\mathrm dx}}$, and how this sign acts on the quasi-polynomial?
Thanks!
Consider that $$ (e^{t\frac{d}{dx}}f)(x)=\sum_{k=0}^\infty\frac{t^k}{k!}f^{(k)}(x) $$ is the Taylor series expansion of $f$ around $x$, and that for analytical functions this is exactly $f(x+t)$, the result of the translation operator. As the space of polynomials is a subset of the space of analytical functions, this shows that the claim is true. Using polynomials avoids any convergence considerations as the series involved will be finite sums.