November 10, 2025 00:36 UTC (Last updated on November 10, 2025 00:41 UTC)

Tags: math

Introduction

I’ll start by rationalizing an example of “old” umbral calculus from Wikipedia. |\newcommand{pair}[2]{\langle{#1}\mid{#2}\rangle}| |\newcommand{bigpair}[2]{\left\langle{#1}\ \middle|\ {#2}\right\rangle}| |\newcommand{pseq}[2]{{#1}_{#2 \in \mathbb N}}| |\newcommand{ucomp}[2]{#1_n(\underline #2(x))}|

We know |(x+y)^n = \sum_{k=0}^{n {n \choose k} x}{n-k} y^k|. We then “infer” that |B_n(x+y) = \sum_{k=0}^n {n \choose k} B_{n-k}(x) y^k| where |B_n(x)| are the Bernoulli polynomials. Actually, the “old” style would be even more dubious. You’d have a “rule” like representing |B_n(x+y)| as |(b+y)^n|, then expand like usual and replace |b^k = (b + 0)^k| with |B_k(x)|. The variables like |b| were the “shadow” or “umbral” variables.

Rationalizing it using techniques I’ll describe below. Let |\varepsilon_y| be the linear operator on polynomials satisfying |\varepsilon_y p(x) = p(x + y)|. Since |D_x[\varepsilon_y p(x)] = \varepsilon_y D_x p(x)| for all |y| where |D_x| is differentiation by |x|, |\varepsilon_y| is induced from a formal power series. In particular, |\varepsilon_y = e^{yD_x}|.

Let |T| be the linear operator (the transfer or umbral operator) characterized by mapping |x^n| to |B_n(x)| representing a change of basis on the vector space of polynomials. We can apply |T| to the equation [ \varepsilon_y(x^n) = (x+y)^n = \sum_{k=0}^{n {n \choose k} x}{n-k} y^k ] to get via linearity [ T(\varepsilon_y(x^n)) = T((x+y)^n) = \sum_{k=0}^{n {n \choose k} T(x}{n-k}) y^k = \sum_{k=0}^n {n \choose k} B_{n-k}(x) y^k ]

The key property we then need is |T(\varepsilon_y(x^n)) = \varepsilon_y T(x^n) = B_n(x + y)| which can be reduced to |D_xT(x^n) = T(D_x x^n)|. This is just the statement that |D_x B_n(x) = nB_{n-1}(x) = nT(x^{n-1}) = T(nx^{n-1}) = T(D_x x^n)| using a well-known property of Bernoulli (and other) polynomials. In fact, this relation implies that |T| is itself induced by a formal power series and thus commutes with any other linear operator so induced.

Ultimately, the only properties we needed for this was that we had a (linear) change of basis from the monomial basis to the polynomial sequence, which we’ll have for any polynomial sequence whose |n|th element has degree |n|, and that the change of basis commuted with the |D_x| operator. The latter is more stringent, but there are various ways we can expand the scope of the argument.

First, |D_x x^n = \frac{c_n}{c_{n-1}} x^n| with |c_n = n!| completely characterizes (with |D_x x^0 = 0|) the differentiation operation on polynomials. Choosing a different sequence |c| leads to different notions of “differentiation”. This will change |\varepsilon_y| and lead to different formulas, but they will be structurally similar.

In a different direction, we can ask for formal power series and polynomial sequences that relate to each other the way |D_x| and |x^n| do. We say that a polynomial sequence |s_n(x)| is Sheffer for a pair |(g(t), f(t))| with |\deg g = 0| and |\deg f = 1| when |\pair{g(t)f(t)^k}{s_n(x)} = c_n\delta_k^n|. This has |g(t)f(t)^k| taking the place of |t^k| and |s_n(x)| taking the place of |x^n|. This would let us transfer identities involving the |s_n(x)| to any linear operator that commutes with |g(t)f(t)|. While changing |c| changes our notion of “differentiation”, using a Sheffer sequence allows us to consider other “differential operators” using the same notion of “differentiation”.

This is based primarily off of works by Steven Roman and Gian-Carlo Rota. It closely follows The Theory of the Umbral Calculus. I by Steven Roman (1982). See also The Umbral Calculus by Steven Roman (1984).

I’ll include proofs below to illustrate that each bit of reasoning is fairly straightforward.

• Conventions

• Formal Power Series

• Linear Functionals

• Properties

• Evaluation Functional

• Formal Derivative

• Linear Operators

• Evaluation Operator

• Characterizing Linear Operators Induced from Formal Power Series

• Polynomial Sequences

• Recurrence Formulas

• Transfer Formulas

• Umbral Composition and Transfer Operators

• Example: Chebyshev Polynomials

• Summary

Conventions

Fix a field |\mathbb K| of characteristic |0|. I’ll write |\mathscr F = \mathbb K[![t]!]| for the |\mathbb K|-algebra of univariate formal power series (with indeterminate |t|) and |P = \mathbb K[x]| for the |\mathbb K|-algebra of univariate polynomials (with indeterminate |x|). Further, |\mathrm{Hom}(X, Y)| will be the |\mathbb K|-vector space of |\mathbb K|-linear maps from a |\mathbb K|-vector space |X| to a |\mathbb K|-vector space |Y|. In particular, |X^* = \mathrm{Hom}(X,\mathbb K)| is the dual space of |X|.

The four main |\mathbb K|-vector spaces we’ll be focused on are |\mathscr F|, |P|, |P^*|, and |\mathrm{Hom}(P, P)|. The first two are additionally |\mathbb K|-algebras, and we’ll find that the third is as well and is, in fact, isomorphic to |\mathscr F| which is arguably a key enabling fact of umbral calculus. In particular, the |\mathbb K|-algebra structure induced on |P^*| via |\mathscr F| is called the umbral algebra.

Given the above, unsurprisingly, we’ll be talking a lot about formal power series and polynomials. To save a bit of space and typing for me, unless otherwise specified, if I say |f| is a formal power series or |f \in \mathscr F|, then that will also defined the sequence |\pseq{f_n}{n}| such that |f(t) = \sum_{k=0}^{\infty f_k t^k|. Generally, when I state something is a sequence it will be a function |\mathbb N \to X| for some |X| and the parameter will be written as a subscript. So the formal power series |f| also defines a sequence, also called |f|, from |\mathbb N \to \mathbb K|. (In this case, we could literally identify formal power series with these sequences.) Similarly, stating |p| is a polynomial or |p \in P| defines a sequence also called |p| such that |p(x) = \sum_{n=0}}{\deg p}p_n x^n| where |\deg p| is the degree of the polynomial, i.e. the largest value of |n| such that |p_n \neq 0|. These are also sequences |\mathbb N \to \mathbb K|, and we could identify polynomials with such sequences that are eventually always |0|. We also have the degree of a formal power series written |\deg f| which is the smallest |k| such that |f_k \neq 0|. Note that this is dual to the notion for polynomials. It’s clear that |\deg(fg) = \deg f + \deg g|.

Of course, sometimes I will explicitly state something like |f(t) = \sum_{k=0}^\infty a_k t^k| in which case the sequence |f| is not defined. Usually, this will arise with a formal power series expression, e.g. |(f \circ g)(t)| so there shouldn’t be any ambiguity. As is typical, I’ll often say “the formal power series |f(t)|” as opposed to “the formal power series |f|”. Finally, as has been illustrated, I’ll endeavor to use |k| as the indexing variable for formal power series and |n| for polynomials, but that won’t always be possible.

The Kronecker delta is written |\delta_k^n| and defined by |\delta_n^n = 1| and |\delta_k^n = 0| for |k \neq n|. This should typically be thought of as a way of forcing |n| and |k| to be equal, i.e. |f(n)\delta_k^n = f(k)\delta_k^n| and |\delta_{f(k)}^{n = \delta_k}{f^{-1}(n)}| for an invertible function |f|.

Formal Power Series

I’ll assume familiarity with the basic algebra of formal power series. This lecture gives a nice in-depth and more technical overview, though it goes far beyond what we’ll need. I’ll recall a few results and fix the terminology and notation that will be used in this article which largely follows Roman but there are many variations in the literature.

Theorem (id:wewp): |f \in \mathscr F| has a multiplicative inverse, written |f^{-1}| if and only if |\deg f = 0|.

Proof: (click to expand) |f(t)g(t) = \sum_{k=0}^{\infty c_n t^k| where |c_k = \sum_{i+j=k} f_i g_j|. Clearly, |c_0 = f_0 g_0| which makes it clear a multiplicative inverse to |f| can only exist if |f_0 \neq 0|. It also makes it clear that for |g| to be |f}{-1}|, |g_0 = 1/f_0|. A simple calculation shows that |g_k = f_0^{-1} \sum_{n=1} f_n g_{k-n}| for |k > 0| which gives a recurrence computing all the remaining coefficients of |f^{-1}|. |\square|

Thus, |f| being invertible will be synonymous with |f| having degree |0|.

It’s worth noting that |f/g| can be defined even when |g| isn’t invertible, e.g. |t/t = 1|. If |\deg f > \deg g|, then we can cancel out common factors of |t| until the denominator is invertible.

Suppose |g : \mathbb N \to \mathscr F|, which we’ll write as |g_k(t) \in \mathscr F|, such that |\deg g_k \to \infty| as |k \to \infty|. Given any |a : \mathbb N \to \mathbb K|, then |\sum_{k=0}^\infty a_k g_k(t)| is a well-defined element of |\mathscr F|. In particular, if |\deg g > 0|, then |g_k(t) = g(t)^k| satisfies the conditions. If we have |\deg g_k = k|, which is the case when |g_k(t) = g(t)^{k| for |g| with degree |1| for example, then |g| forms a pseudobasis1 for |\mathscr F| meaning for any |f \in \mathscr F|, there exists a unique sequence |a| such that |f(t) = \sum_{k=0}}\infty a_n g_k(t)|. A series |f| with |\deg f = 1| is called a delta series. Every delta series gives rise to a pseudobasis of |\mathscr F|.

If |f, g \in \mathscr F| and |\deg g > 0|, then we can thus form the composition |f(g(t)) = \sum_{k=0}^\infty f_k g(t)^k|. It’s clear that |\deg(f\circ g) = \deg f\deg g|.

Theorem (id:cjme): A series, |f|, has a compositional inverse, written |\bar f|, meaning |f(\bar f(t)) = t = \bar f(f(t)| if and only if |\deg f = 1|.

Proof: (click to expand) Suppose |g \in \mathscr F| such that |f(g(t)) = t|. By taking degrees, we immediately see that |f| (and |g|) need to be of degree exactly one to have any chance for |g| to be a compositional inverse to |f|. If |g(t)^{k = \sum_{n=0}}\infty b_{k,n} t^n|, then clearly we need |f_1 b_{1,1} = 1|. By induction, we have |b_{k+1,n} = \sum_{i+j=n} b_{1,i}b_{k,j}| but note that |b_{k,n} = 0| for all |n < k| so this sum always has |k \leq j < n| and thus |1 \leq i \leq n-k|. This leads to [\sum_{k=1}^{n f_k b_{k,n} = 0 = f_1 b_{1,n} + \sum_{k=2}n f_k b_{k,n}] where the last sum only involves |b_{1,i}| for |i < n|. Alternatively, simply note that if |f(t) = th(t)| and |g(t) = tk(t)| then, |h| and |k| have degree |0|, and |t = f(g(t)) = t k(t)h(g(t))|. So |k(t) = h(g(t)){-1} = h(tk(t))}{-1}|. |h| is invertible and |tk(t)| clearly has degree greater than |0| so the composition is well-defined. Unfolding this expression for |k(t)| shows that the |i|th coefficient only depends on earlier coefficients. |\square|

A useful result linking these two together is if |\deg f = 1|, then [ 1 = t’ = [\bar f(f(t))]’ = \bar f’(f(t))f’(t) ] In other words, |f’(t)^{-1} = \bar f’(f(t))|.

Linear Functionals

For |L \in P^*| and |p \in P|, we’ll write |\pair{L}{p(x)}| for the action of |L| on |p|. Any such |L| is uniquely defined by its values on |x^n| for all |n\in\mathbb N|.

If |c : \mathbb N \to \mathbb K\setminus {0}|, we can define for each |f \in \mathscr F| a linear functional which we’ll also write as |f| or |f(t)| via [\pair{f(t)}{x^n} = c_n f_n] Really, we should write something like |\pair{f(t)}{p(x)}_c| to indicate the dependence on |c|. This play on notation is unambiguous since |f(t) = g(t)| if and only if |\pair{f(t)}{x^n} = \pair{g(t)}{x^n}| for all |n|, i.e. |f| and |g| are equal as power series if and only if the induced linear functionals are equal.

Notable choices for |c| are:

• |c_n = n!| is the most traditional case.
• |c_n = 1|
• |c_n = 1/{\lambda \choose n}| for |\lambda| not a negative integer.

The definition of the linear functional induced by |f \in \mathscr F| implies that |\pair{t^k}{x^n} = c_n\delta_k^n|. This leads to [\bigpair{\sum_{n=0}^{\infty a_n t^n}{p(x)} = \sum_{n=0}}\infty a_n \pair{t^n}{p(x)} ] where the right-hand side is well-defined because only finitely many of the terms of the sum will be non-zero. (We can generalize to allow Laurent series with only finitely many negative powers on the left and Laurent series with only finitely many positive powers on the right.)

We can articulate L’Hôpital’s rule with this notation as: if |\deg f \geq \deg g > 0|, then [\pair{f(t)/g(t)}{x^0} = \pair{f’(t)/g’(t)}{x^0} ]

We can explicitly write the formal power series, |f_L|, corresponding to the linear functional, |L|, as [f_L(t) = \sum_{k=0}^{\infty \frac{\pair{L}{x^k}}{c_n}t^k] It is trivial to verify that the linear functional induced by |f_L| is |L|. This gives an isomorphism |\mathscr F \cong P}*| as |\mathbb K|-vector spaces. However, the algebra structure on |\mathscr F| then induces an algebra structure on |P^*|. We can compute [\pair{f(t)g(t)}{x^n}\rangle = \sum_{i+j=n}\frac{c_n}{c_i c_j}\pair{f(t)}{x^i}\pair{g(t)}{x^j}]

We’ll call |L| a delta/invertible functional when it corresponds to a delta/invertible power series.

Properties

If |\deg f > \deg p| then |\pair{f(t)}{p(x)} = 0|.

If |\deg p_n(x) = n| and |\pair{f(t)}{p_n(x)} = 0| for all |n \in \mathbb N|, then |f(t) = 0|.

|\pair{f(at)}{p(x)} = \pair{f(t)}{p(ax)}| [Proof: Follows immediately from |\pair{a^n t^n}{x^n} = a^n c_n = \pair{t^n}{a^n x^n}|. |\square|]

|p(x) = \sum_{n=0}^\infty \frac{\pair{t^n}{p(x)}}{c_n}x^n|

If |\deg f_k = k| and |\pair{f_k(t)}{p(x)} = 0| for all |k \in \mathbb N|, then |p(x) = 0|.

Evaluation Functional

We always have the evaluation functional |\varepsilon_y| for |y \in \mathbb K| defined by [\pair{\varepsilon_y(t)}{p(x)} = p(y)] Note that this definition doesn’t depend on the choice of |c|. We quickly compute |\pair{\varepsilon_y(t)}{x^n} = c_n y^n| so [ \varepsilon_y(t) = \sum_{k=0}^\infty \frac{y^k}{c_k}t^k]

When |c_n = n!|, then |\varepsilon_y(t) = e^{yt}|.

Formal Derivative

The formal derivative of |f \in \mathscr F|, written |\partial_t f(t)|, is defined as [\partial_t t^k = \begin{cases} \frac{c_k}{c_{k-1}}t^{k-1}, & k > 0 \ 0, & k = 0 \end{cases}] which leads to the key property |\pair{\partial_t f(t)}{p(x)} = \pair{f(t)}{xp(x)}|.

As an example, we immediately compute that |\partial_t\varepsilon_y(t) = y\varepsilon_y(t)|.

We will also use the ordinary derivative of formal power series which we’ll notate with |f’(t)|. The formal derivative and the ordinary derivative coincide when |c_n = n!| as suggested by the previous example.

Linear Operators

We’ve identified formal power series with linear functionals on |P|. Next, we want to identify them with linear operators on |P|. We’re clearly not going to get an isomorphism in this case as multiplication (i.e. composition) of linear operators doesn’t commute in general, while multiplication of formal power series does. Nevertheless, we will derive simple characterizations of which linear operators are of this form.

One of the most important properties we will want is the following adjointness property: [ \pair{f(t)g(t)}{p(x)} = \pair{f(t)}{g(t)p(x)} ] where |g(t) p(x)| is the action of the linear operator induced by |g(t)| on |p(x)|. We can derive what the induced linear operator must be to satisfy this property.

[\begin{align} c_n \delta_m^n & = \pair{t^m}{x^n} \ & = \pair{t^{m-k} t^k}{x^n} \ & = \pair{t^{m-k}}{t^k x^n} \ & = \bigpair{t^{m-k}}{\frac{c_n}{c_{n-k}}x^{n-k}} \end{align}] so |t^k x^n = \frac{c_n}{c_{n-k}} x^{n-k}| for |k \leq n| and |0| otherwise. Thus, [f(t) x^n = \sum_{k=0}^{n \frac{c_n}{c_{n-k}} f_k x}{n-k} ] which can be extended to all polynomials by linearity.

This should look familiar. |tx^n = \frac{c_n}{c_{n-1}}x^{n-1}| which is exactly the same formula as the one for |\partial_t| except this operates on polynomials while |\partial_t| operates on formal power series. In particular, when |c_n = n!|, |t| behaves exactly like the derivative of polynomials with respect to |x|, and we see that the formal power series pick out a special class of differential operators on polynomials.

A simple calculation shows that |(t^j t^k) x^n = t^j (t^k x^n)| which lifts to a general associativity law: |(f(t) g(t)) p(x) = f(t) (g(t) p(x))|. The adjointness property also immediately implies that the induced linear operators commute.

As before, we will say delta/invertible operator when the linear operator is induced by a delta/invertible formal power series.

Define |Dx^n = nx^{n-1}|, |D^{-1}x^n = \frac{1}{n+1}x^{n+1}|, and |x^{-1} x^n = \begin{cases}x^{n-1}, & n > 0 \ 0, & n = 0\end{cases}|. Then for various choices of |c|, |t| behaves as the following linear operators:

• |c_n = n!| implies |t = D|
• |c_n = 1| implies |t = x^{-1}|
• |c_n = (n!)^{m+1}| implies |t = (Dx)^m D|
• |c_n = 1/(-\lambda)_{(n)}| implies |t = -(\lambda + xD)^{{-1} x}{-1}|
• |c_n = 1/n| implies |t = x^{-1} D^{-1} x^{-1}|
• |c_n = 1/{-\lambda \choose n}| implies |t = -(\lambda + xD)^{-1} D|
• |c_n = 2^{2n}(1+\alpha)^{{(n)}/(1 + \alpha + \beta)}{(2n)}| implies |t = 4(1 + \alpha + \beta + 2xD)^{{-1} (2 + \alpha + \beta + 2xD)}{-1} x^{-1} (\alpha + xD)|
• |c_n = (1-q^n)^{-1} \prod_{k=1}^n (1-q^k)| implies |tp(x) = (p(qx) - p(x))/(qx - x)|

A linear operator |T| on |\mathscr F| is continuous if given a sequence of formal power series |\pseq{f_k}{k}| such that |\deg f_k \to \infty| as |k \to \infty|, we have |\deg T(f_k) \to \infty| as |k \to \infty|.

Theorem (id:fqys): If |T| is a continuous linear operator on |\mathscr F|, then [ T\left(\sum_{k=0}^{\infty a_k f_k(t)\right) = \sum_{k=0}}\infty a_k T(f_k(t)) ] for all sequences |\pseq{a_k}{k}| in |\mathbb K| and |\pseq{f_k}{k}| in |\mathscr F| for which |\deg f_k \to \infty| as |k \to \infty|.

Proof: (click to expand) By the assumptions, both |\pair{T\left(\sum_{k=0}^{\infty a_k f_k(t)\right)}{x^n}| and |\pair{\sum_{k=0}}\infty a_k T(f_k(t))}{x^n}| involve only finitely many terms of the sum. That is, for every |n| there is some |N| such that for all |m > N|, |\pair{T\left(\sum_{k=0}^\infty a_k f_k(t)\right)}{x^n} = \pair{T\left(\sum_{k=0}^{m a_k f_k(t)\right)}{x^n}| and |\pair{\sum_{k=0}}\infty a_k T(f_k(t))}{x^n} = \pair{\sum_{k=0}^{m a_k T(f_k(t))}{x^n}|. But [\begin{align} \bigpair{T\left(\sum_{k=0}}\infty a_k f_k(t)\right)}{x^n} & = \bigpair{T\left(\sum_{k=0}^m a_k f_k(t)\right)}{x^n} \ & = \bigpair{\sum_{k=0}^{m a_k T(f_k(t))}{x^n} \ & = \bigpair{\sum_{k=0}}\infty a_k T(f_k(t))}{x^n} \end{align}] so these two linear functionals agree on a pseudobasis and thus are the same which implies the formal power series are the same as well. |\square|

This can be cast as an instance of topological continuity, but I won’t describe that here.

Evaluation Operator

Unlike the linear functional case, the linear operator induced by the formal power series corresponding to the evaluation functional does depend on the choice of |c|. In general, we have: [\varepsilon_y(t) x^n = \sum_{k=0}^{n \frac{c_n}{c_{n-k} c_k} y^k x}{n-k} ]

For |c_n = n!|, |\varepsilon_y(t) p(x) = p(x + y)|. For |c_n = 1|, |\varepsilon_y(t) x^n = \frac{x^{n+1} - y^{n+1}}{x-y}|.

Characterizing Linear Operators Induced from Formal Power Series

Theorem (id:lgtb): Let |U| be a linear operator on |P|. There is an |f \in \mathscr F| such that |Up(x) = f(t) p(x)| for all |p \in P| if and only if |U| commutes with the operator |t|, i.e. |Utp(x) = tUp(x)| for all |p \in P|.

Proof: (click to expand) The only if direction is obvious. For the other direction, first note that |\deg Ux^n \leq n| because |t^k Ux^n = Ut^k x^n = 0| if |k > n| so |\pair{t^k}{Ux^n} = 0| for all |k > n|. Now define [f(t) = \sum_{k=0}^{\infty \frac{\pair{t^0}{Ux^k}}{c_k}t^k ] Then, [\begin{align} f(t) x^n & = \sum_{k=0}n \frac{\pair{t^0}{Ux^k}}{c_k}t^k x^n \ & = \sum_{k=0}n \frac{c_n}{c_k c_{n-k}} \pair{t^0}{Ux^k} x{n-k} \ & = \sum_{k=0}n \frac{\pair{t^0}{Ut{n-k} x^n}}{c_{n-k}} x{n-k} \ & = \sum_{k=0}n \frac{\pair{t^0}{t{n-k} Ux^n}}{c_{n-k}} x}{n-k} \ & = \sum_{k=0}^{n \frac{\pair{t}{n-k}}{Ux^n}}{c_{n-k}} x^{n-k} \ & = \sum_{k=0}^n \frac{\pair{t^k}{Ux^n}}{c_k} x^k \ & = Ux^n \end{align}] The last equality relies on the degree of |Ux^n| being less than or equal to |n|. |\square|

Corollary (id:viti): A linear operator on P has the form of |f(t)| for an |f \in \mathscr F| if and only if it commutes with any specific delta operator.

Proof: (click to expand) The sequence of powers of the formal power series associated with the specific delta operator form a pseudobasis which means we can write |t| as an infinite linear combination of them. Thus the linear operator commutes with |t| and we can apply the theorem. |\square|

Corollary (id:qllw): A linear operator on P has the form of |f(t)| for an |f \in \mathscr F| if and only if it commutes with |\varepsilon_y(t)| for all |y \in \mathbb K|.

Proof: |\varepsilon_y(t) - c_0^{-1} t^0| is a delta operator. |\square|

Polynomial Sequences

When we say |\pseq{p_n(x)}{n}| is a (polynomial) sequence, that will always mean that |\deg p_n(x) = n|.

Theorem (id:cgsr): Let |f| be a delta series and |g| be an invertible series, then there is a unique polynomial sequence |\pseq{s_n(x)}{n}| such that [\pair{g(t)f(t)^k}{s_n(x)} = c_n\delta_k^n ] holds for all |n,k \in \mathbb N|.

Proof: (click to expand)

Uniqueness follows easily by considering |\pair{g(t)f(t)^k}{s_n(x) - r_n(x)} = 0| where |\pseq{r_n(x)}{n}| is another sequence satisfying the same property.

For existence, we can just brute force it. If |g(t)f(t)^{k = \sum_{i=k}}\infty b_{k,i} t^i| and we set |s_n(x) = \sum_{j=0}^n a_{n,j} x^j|, then we want to solve for the |a_{n,i}| induced by the following triangular system of linear equations: [\begin{align} c_n\delta_k^n & = \pair{g(t)f(t)^{k}{s_n(x)} \ & = \bigpair{\sum_{i=k}}\infty b_{k,i} t^i}{\sum_{j=0}^n a_{n,j} x^j} \ & = \bigpair{\sum_{i=k}^n b_{k,i} t^i}{\sum_{j=0}^n a_{n,j} x^j} \ & = \sum_{i=k}^n \sum_{j=0}^n b_{k,i} a_{n,j} \pair{t^i}{x^j} \ & = \sum_{i=k}^n c_i b_{k,i} a_{n,i} \end{align}] |b_{k,k} \neq 0| since |\deg g(t)f(t)^k = k| and |c_n \neq 0| by assumption. Therefore, the diagonal entries of the triangular matrix corresponding to this system of linear equations are non-zero, and thus the matrix is invertible. |\square|

We’ll say |\pseq{s_n(x)}{n}| is the Sheffer sequence or is Sheffer for the pair |(g, f)|. When |g(t) = 1|, then we say that the corresponding Sheffer sequence is the associated sequence to |f|. When |f(t) = t|, then we say that the corresponding Sheffer sequence is the Appell sequence for |g|. Often I’ll use |\pseq{p_n(x)}{n}| for associated sequences, i.e. when |g(t) = 1|, and reserve |\pseq{s_n(x)}{n}| for the general case. The idea is that if |g(t)f(t)^k| takes the place of |t^k|, then |s_n(x)| takes the place of |x^n|. This is illustrated by the defining property and the following theorems.

Theorem (Expansion Theorem): Let |\pseq{s_n(x)}{n}| be a Sheffer for |(g, f)|. Then for any |h \in \mathscr F|, [ h(t) = \sum_{k=0}^\infty \frac{\pair{h(t)}{s_k(x)}}{c_k}g(t)f(t)^k ]

Proof: (click to expand) |\pseq{s_n(x)}{n}|, like any polynomial sequence, is a basis for |P|, so two linear functionals are equal if they agree on all |s_n(x)|. Clearly, [\begin{align} \bigpair{\sum_{k=0}^\infty \frac{\pair{h(t)}{s_k(x)}}{c_k}g(t)f(t)^k}{s_n(x)} & = \bigpair{\sum_{k=0}^n \frac{\pair{h(t)}{s_k(x)}}{c_k}g(t)f(t)^k}{s_n(x)} \ & = \sum_{k=0}^n \frac{\pair{h(t)}{s_k(x)}}{c_k}\pair{g(t)f(t)^k}{s_n(x)} \ & = \pair{h(t)}{s_n(x)} \end{align}] |\square|

Corollary (Polynomial Expansion Theorem): Let |\pseq{s_n(x)}{n}| be a Sheffer for |(g, f)|. Then for a |p \in P|, [ p(x) = \sum_{n=0}^\infty \frac{\pair{g(t)f(t)^n}{p(x)}}{c_n} s_n(x) ]

Proof: (click to expand) Choose |h = \varepsilon_y| in the Expansion Theorem, then apply it as a linear functional to |p|. |\square|

Theorem (Generating Function): |\pseq{s_n(x)}{n}| is Sheffer for |(g, f)| if and only if [ \frac{1}{g(\bar f(t))} \varepsilon_y(\bar f(t)) = \sum_{k=0}^\infty \frac{s_k(y)}{c_k} t^k ] for all |y \in \mathbb K|.

Proof: (click to expand)

For the forward implication, using the Expansion Theorem we have: [ \varepsilon_y(t) = \sum_{k=0}^\infty \frac{\pair{\varepsilon_y}{s_k(x)}}{c_k} g(t) f(t)^{k = \sum_{k=0}}\infty \frac{s_k(y)}{c_k} g(t) f(t)^k ] Substituting |\bar f(t)| for |t| and dividing both sides by |g(\bar f(t))| gives the result.

For the reverse implication, if |\pseq{r_n(x)}{n}| is the Sheffer sequence for |(g, f)| then we immediately get [ \sum_{k=0}^{\infty \frac{r_k(y)}{c_k} t^k = \frac{1}{g(\bar f(t))} \varepsilon_y(\bar f(t)) = \sum_{k=0}}\infty \frac{s_k(y)}{c_k} t^k ] and applying both sides to |x^n| then gives |s_n(y) = r_n(y)| for all |y \in \mathbb K|. |\square|

Theorem (Conjugate Representation): |\pseq{s_n(x)}{n}| is Sheffer for |(g, f)| if and only if [ s_n(x) = \sum_{k=0}^{n \frac{\pair{g(\bar f(t))}{-1}\bar f(t)^k}{x^n}}{c_k} x^k ]

Proof: (click to expand) Applying |\varepsilon_y| to both sides gives [\begin{align} s_n(y) & = \sum_{k=0}^{n \frac{\pair{g(\bar f(t))}{-1}\bar f(t)^{k}{x^n}}{c_k} y^k \ & = \bigpair{\sum_{k=0}n \frac{y^k}{c_k}g(\bar f(t)){-1}\bar f(t)k}{x^n} \ & = \bigpair{\sum_{k=0}\infty \frac{y^k}{c_k}g(\bar f(t))}{-1}\bar f(t)^{k}{x^n} \ & = \pair{g(\bar f(t))}{-1}\varepsilon_y(\bar f(t))}{x^n} \ \end{align}] but [ s_n(y) = \bigpair{\sum_{k=0}^\infty \frac{s_n(y)}{c_k}t^k}{x^n} ] so we can apply the Generating Function theorem. |\square|

Theorem (Multiplication Theorem): Let |\pseq{s_n(x)}{n}| be Appell for |g|, then [ s_n(\alpha x) = \alpha^n \frac{g(t)}{g(t/\alpha)} s_n(x) ] for |\alpha\neq 0|.

Proof: (click to expand) [\begin{align} \pair{t^k}{g(t/\alpha) s_n(\alpha x)} & = \pair{t^k g(t/\alpha)}{s_n(\alpha x)} \ & = \pair{(\alpha t)^k g(\alpha t/\alpha)}{s_n(x)} \ & = \alpha^k \pair{g(t) t^k}{s_n(x)} \ & = \alpha^k c_n \delta_k^n \ & = \alpha^n c_n \delta_k^n \ & = \alpha^n \pair{g(t)t^k}{s_n(x)} \ & = \pair{t^k}{\alpha^n g(t) s_n(x)} \end{align}] so |g(t/\alpha) s_n(\alpha x) = \alpha^n g(t) s_n(x)|. |\square|

Theorem (id:qqes): |\pseq{s_n(x)}{n}| is Sheffer for |(g, f)| if and only if |\pseq{g(t)s_n(x)}{n}| is the associated sequence for |f|.

Proof: Just apply adjointness to the definition. |\square|

Theorem (id:cutg): A sequence |\pseq{p_n(x)}{n}| is the associated sequence for |f| if and only if 1) |\pair{t^0}{p_n(x)} = c_0 \delta_n^0| for all |n \in \mathbb N|, and 2) |f(t) p_n(x) = \frac{c_n}{c_{n-1}}p_{n-1}(x)| for all |n \in \mathbb N|.

Proof: (click to expand)

|f(t)^{0 = t^0| implies the first condition. For the second condition, if |\pseq{p_n(x)}{n}| is associated to |f|, then for |k > 0| [\begin{align} \bigpair{f(t)}{k-1}}{\frac{c_n}{c_{n-1}}p_{n-1}(x)} & = \frac{c_n}{c_{n-1}}\pair{f(t)^{k-1}}{p_{n-1}(x)} \ & = c_n \delta_k^n \ & = \pair{f(t)^{k}{p_n(x)} \ & = \pair{f(t)}{k-1}}{f(t)p_n(x)} \end{align}] and |\pseq{f(t)^k}{k}| is a pseudobasis.

Conversely, assuming (1) and (2) hold, then [\begin{align} \pair{f(t)^k}{p_n(x)} & = \pair{t^0}{f(t)^k p_n(x)} \ & = \frac{c_n}{c_{n-k}} \pair{t^0}{p_{n-k}(x)} \tag{(2) k times} \ & = \frac{c_n}{c_{n-k}} c_0 \delta_{n-k}^0 \tag{(1)} \ & = c_n \delta_k^n \end{align}] |\square|

Theorem (id:hvdt): A sequence |\pseq{s_n(x)}{n}| is Sheffer for |(g, f)| for some invertible |g| if and only if |f(t) s_n(x) = \frac{c_n}{c_{n-1}}s_{n-1}(x)|.

Proof: (click to expand)

For the forward implication, simply apply the previous theorem to |\pseq{g(t)s_n(x)}{n}| which is associated to |f|, then apply |g(t)^{-1}| to the resulting recurrence equation.

For the reverse implication, let |\pseq{p_n(x)}{n}| be the associated sequence for |f| and |U| be the linear operator defined by sending |s_n(x)| to |p_n(x)|. Then we have [ Uf(t)s_n(x) = \frac{c_n}{c_{n-1}}Us_{n-1}(x) = \frac{c_n}{c_{n-1}}p_{n-1}(x) = f(t)p_n(x) = f(t)Us_n(x) ] Since |\pseq{s_n(x)}{n}| is a basis, we see that |U| commutes with a delta series and thus must be of the form |g(t)| for some |g| which is invertible because |U| preserves degree. Thus |\pseq{g(t)s_n(x)}{n}| is associated to |f| which is equivalent to |\pseq{s_n(x)}{n}| being Sheffer for |(g, f)|. |\square|

Corollary (id:tvdx): [ ts_n(x) = \sum_{k=0}^{n-1} \frac{c_n}{c_k c_{n-k}} \pair{t}{p_{n-k}(x)} s_k(x) ]

Proof: (click to expand) Start by expanding |ts_n(x)| via the Polynomial Expansion Theorem [\begin{align} ts_n(x) & = \sum_{k=0}^{\infty \frac{\pair{g(t)f(t)k}{ts_n(x)}}{c_k} s_k(x) \ & = \sum_{k=0}{n-1} \frac{\pair{t}{g(t)f(t)k s_n(x)}}{c_k} s_k(x) \ & = \sum_{k=0}{n-1} \frac{c_k}{c_{n-k} c_k} \pair{t}{g(t) s_{n-k}(x)} s_k(x) \tag{theorem (id:hvdt)} \ & = \sum_{k=0}}{n-1} \frac{c_k}{c_{n-k} c_k} \pair{t}{p_{n-k}(x)} s_k(x) \tag{theorem (id:qqes)} \end{align}] |\square|

Theorem (Sheffer Identity): A sequence |\pseq{s_n(x)}{n}| is Sheffer for |(g, f)| for some invertible |g| if and only if [ \varepsilon_y(t) s_n(x) = \sum_{i+j=n} \frac{c_n}{c_i c_j} p_i(y) s_j(x) ] for all |y \in \mathbb K| where |\pseq{p_n(x)}{n}| is associated to |f|.

Proof: (click to expand)

First, we’ll establish that [ \varepsilon_y(t) p_n(x) = \sum_{i+j=n} \frac{c_n}{c_i c_j} p_i(y) p_j(x) ]

Applying |f(t)^k| to both sides of the equation leads to [\begin{align} \pair{f(t)^k}{\varepsilon_y(t) p_n(x)} & = \pair{\varepsilon_y(t)}{f(t)^k p_n(x)} \ & = \frac{c_n}{c_{n-k}} \pair{\varepsilon_y(t)}{p_{n-k}(x)} \ & = \frac{c_n}{c_{n-k}} p_{n-k}(y) \ & = \sum_{i+j=n} \frac{c_n}{c_i c_{j-k}} p_i(y) \pair{t^0}{p_{j-k}(x)} \ & = \sum_{i+j=n} \frac{c_n}{c_i c_j} p_i(y) \pair{t^0}{f(t)^k p_j(x)} \ & = \sum_{i+j=n} \frac{c_n}{c_i c_j} p_i(y) \pair{f(t)^k}{p_j(x)} \ & = \bigpair{f(t)^k}{\sum_{i+j=n} \frac{c_n}{c_i c_j} p_i(y) p_j(x)} \end{align}] which is most easily read from outwards in.

Doing the same trick as the previous proof, we let |U| be a linear operator defined by sending |s_n(x)| to |p_n(x)|. If |\pseq{s_n(x)}{n}| is Sheffer for |(g, f)| for some |g|, we can choose |U = g(t)| and applying |U| to both sides of the above equation gives the forward direction since |g(t)| commutes with |\varepsilon_y(t)|. Conversely, if we assume the equation then we’ll see that |U| must commute with |\varepsilon_y(t)| which implies its of the form |g(t)|. |\square|

As an example, we see that the Bernoulli polynomials |B_n(x)| are Sheffer for |t| with |c_n = n!| for which the associated polynomials are |\pseq{x^n}{n}|. The Sheffer Identity is thus the one mentioned in the Introduction.

Theorem (id:jtbs): Let |\pseq{s_n(x)}{n}| be Sheffer for |(g, f)| and |\pseq{p_n(x)}{n}| be associated to |f|. For all |h, l \in \mathscr F|, [ \pair{h(t)l(t)}{s_n(x)} = \sum_{i+j=n} \frac{c_n}{c_i c_j} \pair{h(t)}{p_i(x)} \pair{l(t)}{s_j(x)} ]

Proof: (click to expand) Use the Expansion Theorem on |h| with respect to |\pseq{p_n(x)}{n}| and |l| with respect to |\pseq{s_n(x)}{n}|. |h(t)l(t)| will then be [ \sum_{i+j=k} \left(\frac{\pair{h(t)}{p_i(x)}}{c_i} f(t)^i\right) \left(\frac{\pair{l(t)}{s_j(x)}}{c_j} g(t)f(t)^j\right) = \left[\sum_{i+j=k} \frac{\pair{h(t)}{p_i(x)}\pair{l(t)}{s_j(x)}}{c_i c_j}\right] g(t)f(t)^k ] Applying this to |s_n(x)| gives the result. |\square|

See the paper for an interesting alternative proof of this result.

Recurrence Formulas

Given a linear operator |\mu| on |P|, the adjoint |\mu^*| is a linear operator on |\mathscr F| characterized by: [ \pair{\mu^* f(t)}{p(x)} = \pair{f(t)}{\mu p(x)} ]

We can readily compute that: [ \mu^* f(t) = \sum_{k=0}^\infty \frac{\pair{f(t)}{\mu x^k}}{c_k} t^k ]

Theorem (id:ahdu): The adjoint to a linear operator on |P| is continuous.

Proof: (click to expand) Let |\pseq{f_k(t)}{k}| where |\deg f_k \to \infty| as |k \to \infty|. Let |K_n| be an index such that |\deg f_{K_n} > \sup_{i=0}^{n \deg \mu x^i|. |\pair{f_{K_n}(t)}{\mu x^m} = 0| for all |m \leq n|, so, using the formula above, |\deg \mu}* f_{K_n} \geq n|. |\square|

In fact, a linear operator on |\mathscr F| is an adjoint to one on |P| if and only if it is continuous.

Theorem (id:fdui): If |T| is a continuous linear operator on |\mathscr F|, then there exists a linear operator |\mu| on |P| such that |T = \mu^*|.

Proof: (click to expand) Define |\mu x^n = \sum_{k=0}^\infty \frac{\pair{Tt^k}{x^n}}{c_k} x^k| which is well-defined because |T| being continuous means only finitely many |\pair{Tt^k}{x^n}| are non-zero. By construction, we see that |\pair{t^k}{\mu x^n} = \pair{Tt^k}{x^n}|. |\square|

If |\pseq{p_n(x)}{n}| is associated to |f|, then the umbral shift |\theta_f| associated to |f| is the linear operator on |P| defined by [ \theta_f p_n(x) = \frac{(n+1)c_n}{c_{n+1}} p_{n+1}(x) ] for all |n \in \mathbb N|. In the case where |p_n(x) = x^n| and |c_n = n!| the umbral shift is just multiplication by |x|. Since, famously, multiplication by |x| does not commute with differentiation, |Dx - xD = 1| as operators, the umbral shift isn’t induced as a linear operator by a formal power series. We’ll see that this is generally the case below.

A derivation on an algebra |A| is a linear operator |\partial| on |A| satisfying [ \partial(ab) = (\partial a)b + a\partial b ] for all |a, b \in A|.

Lemma (id:lxnr): A continuous linear operator |\partial| on |\mathscr F| is a continuous derivation if and only if |\partial 1 = 0| and for any delta series |f|, |\partial f(t)^{k = kf(t)}{k-1}g(t)| for all |k \in \mathbb N| for some |g|.

Proof: (click to expand) A continuous derivation satisfies |\partial h(t) = h’(t)\partial t| from which the result follows with |g(t) = f’(t)\partial t|. A continuous linear operator satisfying these laws satisfies |\partial h(t) = h’(t)f’(t)^{{-1}g(t)| which we can see by expanding |h(t)| in terms of the pseudobasis |\pseq{f(t)k}{k}| and using continuity to push the |\partial| into the sum. Given this [\begin{align} \partial (h(t)k(t)) & = (h(t)k(t))’f’(t){-1}g(t) \ & = k(t)h’(t)f’(t)}{-1}g(t) + h(t)k’(t)f’(t)^{-1}g(t) \ & = (\partial h(t))k(t) + h(t)\partial k(t) \end{align}] |\square|

Theorem (id:oqyq): An operator |\theta| on |P| is the umbral shift for the delta series |f| if and only if its adjoint |\theta^*| is a derivation on |\mathscr F| and [ \theta^* f(t)^{k = kf(t)}{k-1} ] for all |k \in \mathbb N|.

Proof: (click to expand)

If |\theta| is the umbral shift for |f| associated to |\pseq{p_n(x)}{n}|, then [\begin{align} \pair{\theta^* f(t)^{k}{p_n(x)} & = \pair{f(t)k}{\theta p_n(x)} \ & = \frac{(n+1)c_n}{c_{n+1}}\pair{f(t)k}{p_{n+1}(x)} \ & = (n+1)c_n \delta_k{n+1} \ & = kc_n \delta_k{n+1} \ & = kc_n \delta_{k-1}n \ & = \pair{kf(t){k-1}}{p_n(x)} \end{align}] We of course have |\pair{\theta}* t^0}{p_n(x)} = \pair{t^0}{\theta p_n(x)} = \frac{(n+1)c_n c_0}{c_{n+1}} \pair{f(t)^{0}{p_{n+1}(x)} = \frac{(n+1)c_n c_0}{c_{n+1}} \delta_0}{n+1} = 0| since |n+1| is never |0|. Since |\theta^*| is an adjoint, it’s continuous, and we can apply the previous lemma to conclude that it is a derivation.

If |\theta^*| is a derivation satisfying |\theta^* f(t)^{k = kf(t)}{k-1}| then we can rearrange the equations of the first result to get: [\begin{align} \pair{f(t)^{k}{\theta p_n(x)} & = \pair{\theta}* f(t)^{k}{p_n(x)} \ & = \pair{kf(t)}{k-1}}{p_n(x)} \ & = kc_n \delta_{k-1}^{n \ & = kc_n \delta_k}{n+1} \ & = (n+1)c_n \delta_k^{n+1} \ & = \frac{(n+1)c_n}{c_{n+1}}\pair{f(t)^k}{p_{n+1}(x)} \end{align}] |\square|

In the particular case where |f(t) = t|, the above states that |\theta_t^* t^k = kt^{k-1}|, i.e. |\theta_t^* f(t) = f’(t)|. (Not to be confused with |\partial_t| which is only the true derivative when |c_n = n!|.) We can easily compute that |\theta_t t = xD|. Notably, this does not depend on the choice of |c|.

Theorem (id:kscz): If |\pseq{s_n(x)}{n}| is Sheffer for |(g, f)|, then [ \theta_t s_n(x) = \sum_{k=0}^{n+1} \left[\frac{c_n}{c_k c_{n-k}} \pair{g’(t)}{s_{n-k}(x)} + \frac{kc_n}{c_k c_{n-k+1}} \pair{g(t)f’(t)}{s_{n-k+1}(x)} \right] s_k(x) ]

Proof: (click to expand) Start with the Polynomial Expansion Theorem. [\begin{align} \theta_t s_n(x) & = \sum_{n=0}^{{n+1} \frac{\pair{g(t)f(t)k}{\theta_t s_n(x)}}{c_k} s_k(x) \ & = \sum_{n=0}{n+1} \frac{\pair{\theta_t}*(g(t)f(t)^{k)}{s_n(x)}}{c_k} s_k(x) \ & = \sum_{n=0}}{n+1} \left[\frac{\pair{g’(t)f(t)^{k + kg(t)f(t)}{k-1}f’(t)}{s_n(x)}}{c_k}\right] s_k(x) \ & = \sum_{n=0}^{{n+1} \left[\frac{\pair{g’(t)}{f(t)k s_n(x)} + \pair{kg(t)f’(t)}{f(t){k-1} s_n(x)}}{c_k}\right] s_k(x) \ & = \sum_{n=0}}{n+1} \left[\frac{c_n}{c_k c_{n-k}}\pair{g’(t)}{s_{n-k}(x)} + \frac{kc_n}{c_{n-k+1}c_k}\pair{g(t)f’(t)}{s_{n-k+1}(x)}\right] s_k(x) \end{align}] |\square|

Lemma (id:golc): A surjective derivation on |\mathscr F| is continuous.

Proof: (click to expand) For any derivation |\partial|, we have |\partial 1 = \partial 1 + \partial 1| so |\partial 1 = 0|. We also have |\partial t^k = kt^{k-1}\partial t|. Since |\deg \partial t^0 = \infty| and |\deg \partial t^k = \deg \partial t + k - 1|, the only way to get something of degree |0| and have |\partial| be surjective is if |\deg \partial t = 0|. In general, if |\deg f = k| then |f(t) = t^k g(t)| where |\deg g = 0|. Therefore, |\deg \partial f(t) = \deg (t^k \partial g(t) + kt^{k-1}g(t)\partial t)| and the right-hand side has degree |k-1| implying |\partial| is continuous. |\square|

Theorem (id:gxkw): A surjective derivation on |\mathscr F| is adjoint to an umbral shift and vice versa.

Proof: (click to expand)

For the reverse direction, since |f| is a delta series in the previous theorem and |\theta^* f(t)^{k = kf(t)}{k-1}|, |\pseq{\theta^* f(t)^{k}{k}| is a pseudobasis and so |\theta}*| is surjective.

For the forward direction, if |\partial| is a surjective derivation, then it is an adjoint of a linear operator on |P| because it is continuous. Generally, we have |\partial f(t) = f’(t)\partial t| so we can solve |f’(t) = (\partial t)^{{-1}| with |f(0) = 0|, i.e. |\deg f = 1|. Then |\partial f(t)k = kf(t){k-1} f’(t)\partial t| and |f’(t) = (\partial t)}{-1}| so we end up with just |\partial f(t)^{k = k f(t)}{k-1}| as desired. |\square|

Lemma (id:woxc): If |f| and |g| are delta series, then [\theta_f^* = (\theta_f^* g(t))\theta_g^* ]

Proof: (click to expand) For any derivation |\partial|, we have |\partial a^k = (\partial a)ka^{k-1}| and |\theta_f^*| is a derivation. Therefore [ \theta_f^* g(t)^{k = (\theta_f}* g(t))kg(t)^{{k-1} = (\theta_f}* g(t)) \theta_g^* g(t)^k ] and |g(t)^k| for |k\in \mathbb N| is a pseudobasis so this suffices to show the operators are equal. |\square|

Theorem (id:dxii): If |\theta_f| and |\theta_g| are umbral shifts, then [ \theta_f = \theta_g \circ (\theta_g^* f(t))^{-1} ]

Proof: (click to expand) [\begin{align} \pair{t^k}{\theta_f p(x)} & = \pair{\theta_f^* t^k}{p(x)} \ & = \pair{(\theta_g^* f(t))^{-1}\theta_g* t^k}{p(x)} \ & = \pair{\theta_g^* t^k}{(\theta_g^* f(t))^{{-1} p(x)} \ & = \pair{t^k}{\theta_g (\theta_g}* f(t))^{-1} p(x)} \end{align}] |\square|

Theorem (id:ouma): If |\pseq{p_n(x)}{n}| is associated to |f|, then [ p_{n+1}(x) = \frac{c_{n+1}}{(n+1)c_n} \theta_t(f’(t))^{-1} p_n(x) ]

Proof: This is just the previous theorem applied to |p_n(x)| with |g(t) = t|. |\square|

Lemma (id:xnvj): Let |\theta_f| be the umbral shift for |f|. Then [ \theta_f^*(h(t)) = h(t) \theta_f - \theta_f h(t) ] for all |h \in \mathscr F|. The left-hand side is the linear operator on |P| induced by the formal power series that is the output of |\theta_f^*|.

Proof: (click to expand) [\begin{align} \pair{t^k}{\theta_f^*(h(t))x^n) + \theta_f h(t) x^n)} & = \pair{t^k}{\theta_f^*(h(t))x^n)} + \pair{t^k}{\theta_f h(t) x^n)} \ & = \pair{\theta_f^*(h(t)) t^k}{x^n)} + \pair{h(t)\theta_f^*(t^k)}{x^n)} \ & = \pair{\theta_f^*(h(t)) t^k + h(t)\theta_f^*(t^k)}{x^n)} \ & = \pair{\theta_f^*(h(t) t^k)}{x^n)} \ & = \pair{t^k}{h(t) \theta_f x^n)} \end{align}] |\square|

This lemma shows that no umbral shift has the form |g(t)| for a formal power series |g|.

Theorem (id:omlq): Let |\pseq{s_n(x)}{n}| be Sheffer for |(g, f)|. Then if |\theta_f| is the umbral shift for |f|, [ s_{n+1}(x) = \frac{c_{n+1}}{(n+1)c_n}(g(t)\theta_f^*(g(t)^{-1}) + \theta_f)s_n(x) ]

Proof: (click to expand) We have that |\pseq{g(t)s_n(x)}{n}| is associated to |f| leading to [ g(t)s_n(x) = \frac{c_{n+1}}{(n+1)c_n} \theta_f g(t)s_{n+1}(x) ] The previous lemma leads to [\begin{align} s_n(x) & = \frac{c_{n+1}}{(n+1)c_n} g(t)^{{-1}\theta_f g(t)s_{n+1}(x) \ & = \frac{c_{n+1}}{(n+1)c_n} g(t)}{-1}(g(t)\theta_f - \theta_f^*(g(t)))s_{n+1}(x) \ & = \frac{c_{n+1}}{(n+1)c_n} (\theta_f - g(t)^{-1}\theta_f*(g(t)))s_{n+1}(x) \ & = \frac{c_{n+1}}{(n+1)c_n} (\theta_f + g(t)\theta_f^*(g(t)^{{-1}))s_{n+1}(x) \end{align}] where the final equality comes from [ 0 = \theta_f}*(1) = \theta_f^*(g(t)g(t)^{{-1}) = g(t)}{-1}\theta_f^*(g(t)) + g(t)\theta_f^*(g(t)^{-1}) ] |\square|

Theorem (id:kusb): Let |\pseq{s_n(x)}{n}| be Sheffer for |(g, f)|. [ s_{n+1}(x) = \frac{c_{n+1}}{(n+1)c_n}\left(\theta_t - \frac{g’(t)}{g(t)}\right) \frac{1}{f’(t)} s_n(x) ]

Proof: (click to expand)

Starting from the previous theorem [ s_{n+1}(x) = \frac{c_{n+1}}{(n+1)c_n}(g(t)\theta_f^*(g(t)^{-1}) + \theta_f)s_n(x) ]

Theorem (id:dxii) gives |\theta_f = \theta_t f’(t)^{{-1}|. Since |\theta_f}* h(t) = h’(t)\theta_f^* t| for any |h|, we first note that using theorem (id:oqyq), |1 = \theta_f^* f(t) = f’(t) \theta_f^* t| or |\theta_f^* t = f’(t)^{{-1}|. Then [ g(t)\theta_f}*(g(t)^{{-1}) = -\frac{g(t) g’(t)}{g(t)2} \theta_f* t = -\frac{g’(t)}{g(t)} \theta_f}* t = -\frac{g’(t)}{g(t)} f’(t)^{-1} ] |\square|

Theorem (id:vxhh): Let |\pseq{s_n(x)}{n}| be Sheffer for |(g, f)|. If [ T = \left(\theta_t - \frac{g’(t)}{g(t)}\right)\frac{f(t)}{f’(t)} = \left(xD - \frac{tg’(t)}{g(t)}\right)\frac{f(t)}{tf’(t)} ] then [ Ts_n(x) = ns_n(x) ] In other words, |s_n(x)| is an eigenfunction for |T| with eigenvalue |n|.

Proof: (click to expand) The equality for |T| just involves inserting a |t/t| in the middle. [ Ts_n(x) = \left(\theta_t - \frac{g’(t)}{g(t)}\right)\frac{1}{f’(t)}f(t)s_n(x) = \frac{c_n}{c_{n-1}}\left(\theta_t - \frac{g’(t)}{g(t)}\right)\frac{1}{f’(t)}s_{n-1}(x) = n s_n(x) ] where [theorem (id:kusb