I am reading a paper on quadratic decomposition of Appell Sequences and would like to see if I can apply it to a particular Appell sequence that I am working with. However, my undergraduate Linear Algebra course didn't really go hard into linear functionals and inner products as much as I would like, so I'm just trying to get help notationally and get a little push in the right direction. I'm copying the first paragraph here.
Suppose we are looking at $P(\mathbb{C})$, the vector space of polynomials with coefficents in $\mathbb{C}$, and its dual space $P'$. The elements of $P'$ are called $forms$. The action of $u\in P'$ on $f\in P$ is denoted as $\langle{u,f}\rangle$. In particular, we denote by $(u)_n=\langle{u,x^n}\rangle, n\ge 0,$, the moments of $u$. Recall that a linear operator $T:P\rightarrow P$ has a transpose $T^t:P'\rightarrow P'$ defined by $$\langle{T^t(u),f}\rangle=\langle{u,T(f)}\rangle, u\in P', f\in P$$ For example, for any form $u$, any polynomial $g$, let $Du=u'$ and $gu$ be the forms defined as usually $$\langle{u',f}\rangle=-\langle{u,f'}\rangle, \langle{gu,f}\rangle=\langle{u,gf}\rangle$$ where $D$ is the differential operator. Thus, the differentiation operator $D$ on forms is minus the transpose of the differentiation operator $D$ on polynomials.
I have a sequence of polynomials generated by $$\frac{2e^x}{e^{2x}+1+2x}\cdot e^{xz}=\sum_{n=0}^\infty B_n(z)\frac{x^n}{n!}$$ associated with the defined polynomials above.
I understand that, for example, $\langle{u_0,B_0}\rangle=1$ and $\langle{u_0,B_k}\rangle=0$ for $k>0$. If my first few polynomials are $$\{1,z-1, z^2-2z+3,z^3-3z^2+9z-15,...\}$$ how do I use the above information to figure out my dual sequence?
Edit: I have updated the question to more appropriately convey my issues and I copied the first paragraph from the paper I'm wanting to work with to hopefully get some understanding and apply the concepts to my own Appell Sequence.
I'll demonstrate how you can calculate some of the moments of $\langle u_0, \cdot \rangle$, i.e. $\langle u_0, x^n \rangle$. To ease notation, let $$L_0(p) = \langle u_0, p \rangle$$ If we know that $L_0(B_n(x)) = \delta_0(n)$ then $$L_0(B_0) = L_0(1) = 1$$ By linearity, $ 0 = L_0(B_1) = L_0(1) - L_0(x) = 1 - L_0(x)$ so $$L_0(x) = 1$$ Again $0 = L_0(B_2) = L_0(x^2 - 2x + 3) = L_0(x^2) - 2L_0(x) + 3 L_0(1)$ so $$L_0(x^2) = -1$$ etc.