Subjects: Umbral calculus
Links: Delta Operators, Dual Vector Spaces

Let $P$ be the algebra of polynomials in the single variable $x$ over the field $C$ of characteristic zero. Let $P^{\ast }$ be the dual vector space of $P$. $$\langle L \mid p(x)\rangle$$borrowed notation from physics, to denote the action of the linear functional $L$ on a polynomial $p(x)$, and we recall that the vector space operations on $P^{\ast }$.

Since a linear functional is uniquely determined by its action on on a basis, $L$ is uniquely determined by the sequence of constant $⟨L\mid x^n⟩$

We denote $\mathcal{F}$ denote the algebra of formal power series in the variable $t$ over the field $C$. The formal power series $$f(t) = \sum_{k = 0}^\infty \frac{a_k}{k!}t^k$$defines a linear functional on $P$ by setting $$\langle f(t) \mid x^n\rangle =a_n$$
for all $n\ge 0$. In particular $$\langle t^k \mid x^n \rangle = n! \delta_{n, k}$$
Actually, any linear functional $L$ in $P^{\ast }$ has the form $$f_L(t) = \sum_{k = 0}^\infty \frac{\langle L \mid x^k\rangle}{k!} t^k$$then we get that $$\langle f_L(t) \mid x^n\rangle = \langle L \mid x^n\rangle$$
Th: The map $L\mapsto f_L(t)$ is a vector space isomorphism from $P^{\ast }$ onto $\mathcal{F}$.

Given this isomorphism between the vector space $P^{\ast }$ and the algebra $\mathcal{F}$. Then we can identify each linear functional with a formal power series and we can define an algebra structure on $P^{\ast }$, namely the algebra of formal power series. We shall call $\mathcal{F}$ the umbral algebra.

Prop: If $f(t),g(t)\in \mathcal{F}$, then $$\langle f(t)g(t) \mid x^n\rangle = \sum_{k = 0}^n {n \choose k} \langle f(t) \mid x^k\rangle \langle g(t) \mid x^{n-k}\rangle$$
Prop: If $f_1,\dots ,f_{,}\in \mathcal{F}$, then $$\langle f_1(t) \cdots f_m(t)\mid x^n\rangle = \sum {n \choose i_1, \dots, i_m} \langle f_1(t) \mid x^{i_1}\rangle \cdots \langle f_m(t) \mid x^{i_m}\rangle$$
where $i_1+\cdots +i_m=n$

Def: the order $o(f(t))$ of a power series $f(t)$ is the smallest integer $k$ for which the coefficient $t^k$ doesn't vanish. We take $o(f(t))=\mathrm{\infty }$ if $f(t)=0$.

We have a couple of things that are easy to see:

• $o(f(t)g(t))=o(f(t))+o(g(t))$
• $o(f(t)+g(t))\ge min\{o(f(t)),o(g(t))\}$
• The series has a multiplicative inverse if $o(f(t))=0$.

Def: if $o(f(t))=1$, then the formal power series $f(t)$ has a a compositional inverse $\overline{f}(t)$, satisfying $f(\overline{f}(t))=\overline{f}(f(t))=t$. A series for which $o(f(t))=1$ will be called a delta series.

The sequence $f(t{)}^k$ of powers of a delta series form a pseudobasis for $\mathcal{F}$. That is, for any series $g(t)$ in $\mathcal{F}$ there exists a unique sequence of constants $a_k$ for which $$g(t) = \sum_{k = 0}^\infty a_k f(t)^k$$

Prop: If $o(f(t))>\mathrm{deg}(p(x))$, then $⟨f(t)\mid p(x)⟩=0$

Prop: If $o(f_k(t))=k$ for all $k\ge 0$, then $$\left\langle \left.\sum_{k = 0}^\infty a_k f_k(t); \right|; p(x)\right\rangle = \sum_{k = 0}^\infty a_k \langle f_k(t) \mid p(x)\rangle$$for all $p\in P$, the second sum is finite

Prop: If $o(f_k(t))=k$ for all $k\ge 0$ and if $$\langle f_k(t) \mid p(x) \rangle = \langle f_k(t) \mid q(x)\rangle$$for all $k$, then $p(x)=q(x)$.

Prop: If $\mathrm{deg}{p}_k(t)=k$ for all $k\ge 0$ and if $$\langle f(t) \mid p_k(x) \rangle = \langle g(t) \mid p_k(x)\rangle $$for all $k$, then $f(t)=g(t)$

We actually have two nice things in particular $$\langle t^k \mid p(x) \rangle = p^{(k)}(0)$$for all $k$.

Def: When we are considering a delta series $f(t)\in \mathcal{F}$ as a linear functional we will refer to it as delta functional. Similarly, when we are considering invertible series as a linear functional, we use the term invertible functional.

Prop: The series $f(t)$ is a delta functional iff $$\langle f(t) \mid 1\rangle = 0 \qquad \langle f(t) \mid x\rangle \ne 0$$
Prop: The series $f(t)$ is an invertible functional iff $⟨f(t)\mid 1⟩\ne 0$.

Prop: If $f(t)\in \mathcal{F}$, then $$\langle f(t) \mid x p(x) \rangle = \langle \partial_t f(t) \mid p(x)\rangle$$ for all polynomials.

Prop: For any $f(t)\in \mathcal{F}$, and $p(x)\in P$, $$\langle f(t) \mid p(ax)\rangle = \langle f(at) \mid p(x)\rangle$$

Linear Operators

We are going to adopt a weird notation. Let $f(t)$ be a power series $$f(t) = \sum_{k = 0}^\infty \frac{a_k}{k!}t^k$$we can make it a linear operator on $P$ defined by $$f(t) x^n = \sum_{k = 0}^\infty {n \choose k} a_k x^{n-k}$$
Meaning that we use juxtaposition $f(t)p(x)$ to denote the action of the operator $f(t)$ on the polynomial $p(x)$.

As with linear functional, we note two elements $f(t),g(t)\in \mathcal{F}$ are equals as formal power series iff they are equal as linear operators. We can also conclude that $$[f(t) g(t)]p(x) = f(t)[g(t) p(x)]$$for all $f(t),g(t)\in \mathcal{F}$ and $p(x)\in P$, and so we can write $f(t)g(t)p(x)$ without ambiguity, Using the commutativity of the product in $\mathcal{F}$, we get that $$f(t)g(t)p(x) = g(t)f(t)p(x)$$for all $f(t),g(t)\in \mathcal{F}$ and $p(x)\in P$.

Def: When we think of a delta (or invertible) series as an operator, we shall refer to it as a delta (or invertible) operator.

Prop: If $o(f(t))>\mathrm{deg}p(x)$, then $f(t)p(x)=0$

Prop: If $o(f_k(t))=k$ for all $k\ge 0$, then $$\left[\sum_{k = 0}^\infty a_k f_k(t)\right] p(x) = \sum_{k = 0}^\infty a_k [f_k(t)p(x)]$$for all $p(x)\in P$, the second one being a finite sum

Prop: If $o(f_k(t))=k$ for all $k\ge 0$ and if $f_k(t)p(x)=f_k(t)q(x)$ for all $k$, then $p(x)=q(x)$

Prop: If $\mathrm{deg}{p}_k(x)=k$ for all $k\ge 0$ and if $f(t)p_k(x)=g(t)p_k(x)$ for all $k$, then $f(t)=g(t)$.

Prop: If $f(t),g(t)\in \mathcal{F}$ and $p(x)\in P$, then $$\langle f(t) g(t) \mid p(x) \rangle = \langle g(t) \mid f(t) p(x)\rangle$$