Let
$$\begin{aligned} \mathcal{F}= \Biggl\{ f(t)=\sum_{k=0}^{\infty }a_{k} \frac{t^{k}}{k!} \bigg| a_{k}\in \mathbb{C}_{p} \Biggr\} \end{aligned}$$
be the algebra of all formal power series in t with coefficients in \(\mathbb{C}_{p}\). Let \(\mathbb{P}=\mathbb{C}_{p}[x]\) be the ring of all polynomials in x with coefficients in \(\mathbb{C}_{p}\), and let \(\mathbb{P}^{*}\) denote the vector space of all linear functionals on \(\mathbb{P}\) (see [6]). Let \(\langle L|P(x)\rangle \) denote the action of the linear functional L on the polynomial \(P(x)\).
The formal power series
$$\begin{aligned} f(t)=\sum_{k=0}^{\infty }a_{k} \frac{t^{k}}{k!}\in \mathcal{F} \end{aligned}$$
(8)
defines the λ-linear functional on \(\mathbb{P}\) by setting
$$\begin{aligned} \bigl\langle f(t)|(x)_{n,\lambda } \bigr\rangle _{\lambda }=a_{n}\quad (n\ge 0), \bigl(\text{see [6, 11]}\bigr). \end{aligned}$$
(9)
Thus, by (8) and (9), we get
$$\begin{aligned} \bigl\langle t^{k}|(x)_{n,\lambda } \bigr\rangle _{\lambda }=n! \delta _{n,k}\quad (n,k \ge 0), \bigl(\text{see [5, 6]}\bigr), \end{aligned}$$
(10)
where \(\delta _{n,k}\) is the Kronecker symbol.
Here, \(\mathcal{F}\) denotes both the algebra of formal power series in t and the vector space of all λ-linear functionals on \(\mathbb{P}\), so an element \(f(t)\) of \(\mathcal{F}\) will be thought of as both a formal power series and a λ-linear functional. We shall call \(\mathcal{F}\) the λ-umbral algebra. The λ-umbral calculus is the study of λ-umbral algebra. The order \(o(f(t))\) of the power series \(f(t)(\ne 0)\) is the smallest integer k for which \(a_{k}\) does not vanish. If \(o(f(t))=0\), then \(f(t)\) is called an invertible series; if \(o(f(t))=1\), then \(f(t)\) is said to be a delta series (see [1, 4–6, 13]).
For \(f(t),g(t)\in \mathcal{F}\), with \(o(f(t))=1\) and \(o(g(t))=0\), there exists a unique sequence \(S_{n,\lambda }(x)\) \((\deg S_{n,\lambda }(x)=n)\) such that \(\langle g(t)(f(t))^{k}|S_{n,\lambda }(x)\rangle _{\lambda }=n!\delta _{n,k}, (n,k\ge 0)\). Such a sequence \(S_{n,\lambda }(x)\) is called the λ-Sheffer sequence for \((g(t),f(t))\), which is denoted by \(S_{n,\lambda }(x)\sim (g(t),f(t))_{\lambda }\) (see [6]). In [6], we note that
$$\begin{aligned} S_{n,\lambda }(x)\sim \bigl(g(t),f(t) \bigr)_{\lambda }\quad\Longleftrightarrow\quad \frac{1}{g(\bar{f}(t))}e_{\lambda }^{x} \bigl(\bar{f}(t) \bigr)=\sum _{n=0}^{ \infty }S_{n,\lambda }(x) \frac{t^{n}}{n!}, \end{aligned}$$
(11)
where \(\bar{f}(t)\) is the compositional inverse function of \(f(t)\) with \(f(\bar{f}(t))=\bar{f}(f(t))=t\).
By (8), (9), and (10), we easily get
$$\begin{aligned} f(t)=\sum_{k=0}^{\infty } \frac{\langle f(t)|(x)_{k,\lambda }\rangle _{\lambda }}{k!}{t^{k}},\qquad P(x)= \sum_{k=0}^{\infty }\frac{\langle t^{k}|P(x)\rangle _{\lambda }}{k!}(x)_{k, \lambda }\quad \bigl(\text{see [5, 6]}\bigr). \end{aligned}$$
(12)
The formal power series \(f(t)=\sum_{k=0}^{\infty }a_{k}\frac{t^{k}}{k!}\in \mathcal{F}\) defines the λ-differential operator \((f(t))_{\lambda }\) on \(\mathbb{P}\), which is given by
$$\begin{aligned} \bigl(f(t) \bigr)_{\lambda }(x)_{n,\lambda }=\sum _{k=0}^{n}\binom{n}{k}a_{k}(x)_{n-k, \lambda }\quad (n\ge 0), \end{aligned}$$
(13)
and by linear extension (see [5, 6, 11]).
For \(k\ge 0\), by (13), we easily get
$$\begin{aligned} \bigl(t^{k} \bigr)_{\lambda }(x)_{n,\lambda }= \textstyle\begin{cases} (n)_{k}(x)_{n-k,\lambda } & \text{if $k \le n$,} \\ 0 & \text{if $k>n$}, \end{cases}\displaystyle \quad\bigl(\text{see [6]}\bigr). \end{aligned}$$
(14)
Before proceeding further, we would like to say a little about the differences between the λ-umbral calculus and umbral calculus. Facts on umbral calculus are obtained from the corresponding ones on λ-umbral calculus by letting \(\lambda \rightarrow 0\) and then suppressing 0s from everywhere. Here we mention a few of those.
The formal power series
$$\begin{aligned} f(t)=\sum_{k=0}^{\infty }a_{k} \frac{t^{k}}{k!}\in \mathcal{F} \end{aligned}$$
defines the linear functional on \(\mathbb{P}\) by setting
$$\begin{aligned} \bigl\langle f(t)|x^{n} \bigr\rangle =a_{n}\quad (n\ge 0). \end{aligned}$$
(15)
In particular, we get
$$\begin{aligned} \bigl\langle t^{k}|x^{n} \bigr\rangle =n!\delta _{n,k}\quad (n,k\ge 0), \bigl(\text{see [5, 6]}\bigr), \end{aligned}$$
(16)
where \(\delta _{n,k}\) is the Kronecker symbol.
For \(f(t),g(t)\in \mathcal{F}\) with \(o(f(t))=1\) and \(o(g(t))=0\), there exists a unique sequence \(S_{n}(x)\) \((\deg S_{n}(x)=n)\) such that \(\langle g(t)(f(t))^{k}|S_{n}(x)\rangle =n!\delta _{n,k}, (n,k\ge 0)\). Such a sequence \(S_{n}(x)\) is called the Sheffer sequence for \((g(t),f(t))\), which is denoted by \(S_{n}(x)\sim (g(t),f(t))\). Moreover, we have
$$\begin{aligned} S_{n}(x)\sim \bigl(g(t),f(t) \bigr)\quad\Longleftrightarrow\quad \frac{1}{g(\bar{f}(t))}e^{x \bar{f}(t)}=\sum_{n=0}^{\infty }S_{n}(x) \frac{t^{n}}{n!}. \end{aligned}$$
(17)
By (15) and (16), we get
$$\begin{aligned} f(t)=\sum_{k=0}^{\infty }\frac{\langle f(t)|x^{k}\rangle }{k!}{t^{k}},\qquad P(x)= \sum_{k=0}^{\infty }\frac{\langle t^{k}|P(x)\rangle }{k!}x^{k} \quad\bigl(\text{see [5, 6]}\bigr). \end{aligned}$$
(18)
The formal power series \(f(t)=\sum_{k=0}^{\infty }a_{k}\frac{t^{k}}{k!}\in \mathcal{F}\) defines the differential operator on \(\mathbb{P}\), which is given by
$$\begin{aligned} f(t)x^{n}=\sum_{k=0}^{n} \binom{n}{k}a_{k}x^{n-k} \quad(n\ge 0), \end{aligned}$$
(19)
and by linear extension (see [5, 6, 11]).
In particular, for \(k\ge 0\), by (19) we get
$$\begin{aligned} t^{k}x^{n}=\textstyle\begin{cases} (n)_{k}x^{n-k} & \text{if $k \le n$}, \\ 0 & \text{if $k>n$}, \end{cases}\displaystyle \quad\bigl(\text{see [6]}\bigr), \end{aligned}$$
(20)
where \((x)_{0}=1, (x)_{n}=x(x-1)\cdots (x-n+1), (n\ge 1)\).
For further details on umbral calculus, we let the reader refer to [13, 14]. The next theorem is important for our discussion in the following and contains results not addressed in [6].
Theorem 1
Let \(\mathcal{E}_{\lambda }(t)=\frac{1}{\lambda }(e^{\lambda t}-1)\). Let \(f(t),g(t)\in \mathcal{F}\) with \(o(f(t))=1\) and \(o(g(t))=0\). Then we have the following:
-
(a)
\(S_{n,\lambda }(x) \sim (g(t), f(t))_{\lambda } \Longleftrightarrow S_{n, \lambda }(x) \sim (g(\mathcal{E}_{\lambda }(t)), f(\mathcal{E}_{\lambda }(t)))\).
-
(b)
For any \(P(x) \in \mathbb{P}=\mathbb{C}_{p}[x]\), we have
$$\begin{aligned} \bigl(e_{\lambda }^{y}(t) \bigr)_{\lambda }P(x)=P(x+y),\qquad \bigl\langle e_{\lambda }^{y}(t)|P(x) \bigr\rangle _{\lambda }=P(y). \end{aligned}$$
-
(c)
Let \(S_{n,\lambda }(x) \sim (g(t), f(t))_{\lambda }\). Then we have
$$\begin{aligned} \bigl(f(t) \bigr)_{\lambda }S_{n,\lambda }(x)=f \bigl( \mathcal{E}_{\lambda }(t) \bigr)S_{n, \lambda }(x)= nS_{n-1,\lambda }(x). \end{aligned}$$
-
(d)
For any \(f(t) \in \mathcal{F}\) and \(P(x) \in \mathbb{P}=\mathbb{C}_{p}[x]\), we have
$$\begin{aligned} \bigl(f(t) \bigr)_{\lambda }P(x)=f \bigl(\mathcal{E}_{\lambda }(t) \bigr)P(x). \end{aligned}$$
-
(e)
Let \(S_{n,\lambda }(x) \sim (g(t), f(t))_{\lambda }\), and let \(P_{n,\lambda }(x)=(g(t))_{\lambda }S_{n,\lambda }(x) \sim (1, f(t))_{ \lambda }\). Then we have
$$\begin{aligned} \biggl(\frac{1}{g(t)} \biggr)_{\lambda }P_{n,\lambda }(x)= \biggl( \frac{1}{g(\mathcal{E}_{\lambda }(t))} \biggr)P_{n,\lambda }(x)=S_{n, \lambda }(x). \end{aligned}$$
-
(f)
If \(S_{n,\lambda }(x) \sim (g(t), t)_{\lambda }\) (that is, \(S_{n,\lambda }(x)\) is a λ-Appell sequence for \(g(t)\)), then we have
$$\begin{aligned} \biggl(\frac{1}{g(t)} \biggr)_{\lambda } (x)_{n,\lambda }= \biggl( \frac{1}{g(\mathcal{E}_{\lambda }(t))} \biggr) (x)_{n,\lambda }=S_{n, \lambda }(x). \end{aligned}$$
Proof
(a) Observe first that \(\bar{\mathcal{E}}_{\lambda }(t)=\frac{1}{\lambda }\log (1+\lambda t)\). Then, from (17) and (11), we have
$$\begin{aligned} S_{n,\lambda }(x) \sim \bigl(g \bigl(\mathcal{E}_{\lambda }(t) \bigr), f \bigl(\mathcal{E}_{ \lambda }(t) \bigr) \bigr) &\Longleftrightarrow \quad\sum _{n=0}^{\infty }S_{n, \lambda }(x) \frac{t^{n}}{n!} =\frac{1}{g(\bar{f}(t))}e^{x \bar{\mathcal{E}}_{\lambda }(\bar{f}(t))}=\frac{1}{g(\bar{f}(t))}e_{ \lambda }^{x} \bigl(\bar{f}(t) \bigr) \\ & \Longleftrightarrow\quad S_{n,\lambda }(x) \sim \bigl(g(t), f(t) \bigr)_{ \lambda }. \end{aligned}$$
(b) It suffices to show these for \(P(x)=(x)_{n,\lambda }\). The second one is immediate from definition (9). The first one follows from (13) as follows:
$$\begin{aligned} \bigl(e_{\lambda }^{y}(t) \bigr)_{\lambda }(x)_{n,\lambda }&= \Biggl(\sum_{l=0}^{ \infty }(y)_{l,\lambda } \frac{t^{l}}{l!} \Biggr)_{\lambda }(x)_{n,\lambda } \\ &=\sum_{k=0}^{n}\binom{n}{k}(y)_{k,\lambda }(x)_{n-k,\lambda }=(x+y)_{n, \lambda }. \end{aligned}$$
(c) Here we show \((f(t))_{\lambda }S_{n,\lambda }(x)=nS_{n-1,\lambda }(x)\). Once we show (d), the middle equality will follow.
$$\begin{aligned} \bigl\langle g(t)f(t)^{k} | \bigl( f(t) \bigr)_{\lambda }S_{n,\lambda }(x) \bigr\rangle _{ \lambda }&= \bigl\langle g(t)f(t)^{k+1} |S_{n,\lambda }(x) \bigr\rangle _{\lambda }=n! \delta _{n,k+1} \\ &=n(n-1)!\delta _{n-1,k}= \bigl\langle g(t)f(t)^{k} |nS_{n-1,\lambda }(x) \bigr\rangle _{\lambda }. \end{aligned}$$
Now, the result follows from the uniqueness of λ-Sheffer sequence. (d) By linear extension, it is enough to show that \((t^{k})_{\lambda }(x)_{n,\lambda }= (\mathcal{E}_{\lambda }(t) )^{k}(x)_{n, \lambda }\) for \(0 \le k \le n\).
As \(\sum_{n=0}^{\infty }(x)_{n,\lambda }\frac{t^{n}}{n!}=e_{\lambda }^{x}(t)=e^{x \bar{\mathcal{E}}_{\lambda }(t)}\), \((x)_{n,\lambda } \sim (1, \mathcal{E}_{\lambda }(t))\), and hence \(\mathcal{E}_{\lambda }(t)(x)_{n,\lambda }=n(x)_{n-1,\lambda }\), by Theorem 2.3.7 in [13]. Now, from (14) we have
$$\begin{aligned} \bigl(t^{k} \bigr)_{\lambda }(x)_{n,\lambda }=(n)_{k}(x)_{n-k,\lambda }= \bigl( \mathcal{E}_{\lambda }(t) \bigr)^{k}(x)_{n,\lambda }. \end{aligned}$$
(e) From (a), we note that \(S_{n,\lambda }(x) \sim (g(\mathcal{E}_{\lambda }(t)), f(\mathcal{E}_{ \lambda }(t)))\), and \(P_{n,\lambda }(x)=g(\mathcal{E}_{\lambda }(t))S_{n,\lambda }(x) \sim (1, f(\mathcal{E}_{\lambda }(t))\). Then, from p.107 of [13], we have \(S_{n,\lambda }(x)=\frac{1}{g(\mathcal{E}_{\lambda }(t))}P_{n,\lambda }(x)= (\frac{1}{g(t)} )_{\lambda }P_{n,\lambda }(x)\).
(f) This follows from (e) by noting that \((x)_{n,\lambda } \sim (1,t)_{\lambda }\). □