Denote the action of a linear functional on a polynomial $p(x)$ as
$$\langle L \mid p(x) \rangle$$ and define a bilinear functional on the space of polynomials $P$ by
$$\langle f(t) \mid x^n \rangle=a_n$$
where $f(t)=\sum_{n\ge 0}a_n\frac{t^n}{n!}$ is an element of the algebra of formal power series $\mathcal{F}$. We can consider this bilinear since if $p(x)=\sum_{k=0}^n{c_kx^k}$ we also have
$$\langle f(t)\mid p(x)\rangle=\sum_{k=0}^n{c_k\langle f(t)\mid x^k\rangle}$$
There are a few basic results that can be shown, such as
$$\langle t^k \mid x^n\rangle = n!\delta_{n,k}$$
$$\langle e^{yt} \mid x^n \rangle=y^n \Rightarrow \langle e^{yt} \mid p(x)\rangle=p(y)$$
$$\langle e^{yt}-1 \mid p(x)\rangle=\langle e^{yt} \mid p(x)\rangle - \langle 1 \mid p(x) \rangle = p(y)-p(0)$$
The previous two functionals are referred to as the "evaluation functional" and the "forward difference functional", respectively. So extending, what can we say about
$$\langle e^{yt}-T_n(yt) \mid p(x)\rangle$$ where
$$T_n(yt)=\sum_{j=0}^n\frac{(yt)^j}{j!}$$
If $n=2$, we have
$$\langle e^{yt}-1-yt \mid p(x)\rangle=\langle e^{yt} \mid p(x)\rangle - \langle 1 \mid p(x) \rangle - \langle yt \mid p(x) \rangle$$
I think that $\langle yt \mid p(x)\rangle = yp'(0)$, but it doesnt seem right. I know that
$$\langle t \mid x^n \rangle = \delta_{n,1}$$ and so then if $p(x)=c_nx^n+...+c_1x+c_0$,
$$\langle t \mid p(x) \rangle=\sum_{k=0}^n{c_k\langle t \mid x^k \rangle}=\sum_{k=0}^n{c_k\delta_{k,1}}=c_1=p'(0)$$
So is it true that $\langle yt \mid p(x) \rangle = yp'(0)$? From there i think the arugment goes that increased subtraction of the power series terms will yield subtraction of the Taylor series of $p(x)$. Then i would need to extend this to
$$\left\langle y^k\frac{t^k}{k!} \mid p(x) \right\rangle$$