# Linear differential equations for families of polynomials

## Abstract

In this paper, we present linear differential equations for the generating functions of the Poisson-Charlier, actuarial, and Meixner polynomials. Also, we give an application for each case.

## Introduction

As is well known, the Poisson-Charlier polynomials $$C_{k}(x;a)$$ are Sheffer sequences (see [14]) with $$g(t) = e^{a(e^{t}-1)}$$ and $$f(t) = a(e^{t}-1)$$, which are given by the generating function

\begin{aligned} C(x,t)=e^{-t}(1+t/a)^{x}=\sum_{n\geq0}C_{n}(x;a) \frac{t^{n}}{n!}\quad (a\neq0). \end{aligned}
(1)

They satisfy the Sheffer identity

$$C_{n}(x+y;a)=\sum_{k=0}^{n} \binom{n}{k} a^{k-n}C_{k}(y;a) (x)_{n-k},$$

where $$(x)_{n}$$ is the falling factorial (see [5]). Moreover, these polynomials satisfy the recurrence relation

$$C_{n+1}(x;a)=a^{-1}xC_{n}(x-1;a)-C_{n}(x;a)\quad \bigl(\mbox{see [5]}\bigr).$$

The first few polynomials are $$C_{0}(x;a) = 1$$, $$C_{1}(x;a) = -\frac{(a-x)}{a}$$, $$C_{2}(x;a) = \frac {(a^{2}-x-2ax+x^{2})}{a^{2}}$$.

The actuarial polynomials $$a_{n}^{(\beta)}(x)$$ are given by the generating function of Sheffer sequence

\begin{aligned} F(x,t)=e^{\beta t+x(1-e^{t})}=\sum_{n\geq0}a_{n}^{(\beta)}(x) \frac {t^{n}}{n!} \quad\bigl(\mbox{see [5]}\bigr), \end{aligned}
(2)

and the Meixner polynomials of the first kind $$m_{n}(x;\beta,c)$$ are also introduced in [5] as follows:

\begin{aligned} M(x,t)=\sum_{n\geq0}m_{n}(x; \beta,c)\frac {t^{n}}{n!}=(1-t/c)^{x}(1-x)^{-x-\beta}. \end{aligned}
(3)

In mathematics, Meixner polynomials of the first kind (also called discrete Laguerre polynomials) are a family of discrete orthogonal polynomials introduced by Josef Meixner (see [610]). They are given in terms of binomial coefficients and the (rising) Pochhammer symbol by

$$m_{n}(x,\beta,c) = \sum_{k=0}^{n} (-1)^{k}{n \choose k} {x\choose k}k!(x-\beta )_{n-k}c^{-k} \quad\bigl(\mbox{see [5]}\bigr).$$

Some interesting identities and properties of the Poisson-Charlier, actuarial, and Meixner polynomials can be derived from umbral calculus (see [1113]). Kim and Kim [12] introduced nonlinear Changhee differential equations for giving special functions and polynomials. Many researchers have studied the Poisson-Charlier, actuarial and Meixner polynomials in the mathematical physics, combinatorics, and other applied mathematics (for example, see [14, 15]).

In this paper, we study linear differential equations arising from the Poisson-Charlier, actuarial, and Meixner polynomials and derive new recurrence relations for those polynomials from our differential equations.

## Poisson-Charlier polynomials

Recall that the falling polynomials $$(x)_{N}$$ are defined by $$(x)_{N}=(x-1)\cdots(x-N+1)$$ for $$N\geq1$$ with $$(x)_{0}=1$$. For brevity, we denote the generating functions $$C(x,t)$$ and $$\frac{d^{j}}{dt^{j}}C(x;t)$$ by C and $$C^{(j)}$$ for $$j\geq0$$.

### Lemma 1

The generating function $$C^{(N)}$$ is given by $$(\sum_{i=0}^{N}a_{i}(N,x)(t+a)^{-i} )C$$, where $$a_{0}(N,x)=(-1)^{N}$$, $$a_{N}(N,x)=(x)_{N}$$, and

$$a_{i}(N,x)=(x-i+1)a_{i-1}(N-1,x)-a_{i}(N-1,x)\quad (1 \leq i\leq N-1).$$

### Proof

Clearly, $$a_{0}(0,x)=1$$. For $$N=1$$, by (1) we have $$C^{(1)}=(-1+x(t+a)^{-1})C$$, which proves the lemma for $$N=1$$ (here $$a_{0}(1,x)=-1$$ and $$a_{1}(1,x)=x$$). Assume that $$C^{(N)}$$ is given by $$(\sum_{i=0}^{N} a_{i}(N,x)(t+a)^{-i} )C$$. Then

\begin{aligned} C^{(N+1)}&= \Biggl(-\sum_{i=0}^{N} a_{i}(N,x)i(t+a)^{-i-1} \Biggr)C + \Biggl(\sum _{i=0}^{N} a_{i}(N,x) (t+a)^{-i} \Biggr) \bigl(-1+x(t+a)^{-1}\bigr)C\\ &= \Biggl(\sum_{i=1}^{N+1}(x-i+1)a_{i-1}(N,x) (t+a)^{-i} -\sum_{i=0}^{N} a_{i}(N,x) (t+a)^{-i} \Biggr)C. \end{aligned}

This shows that the generating function $$C^{(N+1)}$$ is given by

\begin{aligned} &\Biggl(-a_{0}(N,x)+\sum_{i=1}^{N} \bigl((x-i+1)a_{i-1}(N,x) -a_{i}(N,x) \bigr) (t+a)^{-i}\\ &\quad{}+(x-N)a_{N}(N,x) (t+a)^{-N-1} \Biggr)C. \end{aligned}

Comparing with $$C^{(N+1)}= (\sum_{i=0}^{N+1} a_{i}(N+1,x)(t+a)^{-i} )C$$, we complete the proof. □

In order to obtain an explicit formula for the generating function $$C^{(N)}$$, we need the following lemma.

### Lemma 2

For all $$0\leq i\leq N$$, the coefficient‘s $$a_{i}(N,x)$$ in Lemma 1 are given by

$$a_{i}(N,x)=(x)_{i}\binom{N}{i}(-1)^{N-i}.$$

### Proof

By Lemma 1 we have that

$$a_{i}(N+1,x)=(x-i+1)a_{i-1}(N,x)-a_{i}(N,x),\quad 0\leq i\leq N+1,$$

with $$a_{0}(0,x)=1$$ and $$a_{i}(N,x)=0$$ whenever $$i>N$$ or $$i<0$$. Define $$A_{i}(x;t)=\sum_{N\geq i}a_{i}(N,x)t^{N}$$. Then we have

$$A_{i}(x;t)=\frac{(x+1-i)t}{1+t}A_{i-1}(x)$$

with $$A_{0}(x;t)=\frac{1}{1+t}$$. By induction on i we derive that $$A_{i}(x,t)=\frac{(x)_{i} t^{i}}{(1+t)^{i+1}}$$. Hence, by the fact that $$\frac{1}{(1+t)^{i+1}}=\sum_{j\geq0}\binom{i+j}{i}(-1)^{j}t^{j}$$ we obtain that $$a_{i}(N,x)=(x)_{i}\binom{N}{i}(-1)^{N-i}$$, as required. □

Thus, by Lemmas 1 and 2 we can state the following result.

### Theorem 3

The linear differential equations

$$C^{(N)}= \Biggl(\sum_{i=0}^{N}(x)_{i} \binom{N}{i}(-1)^{N-i}(t+a)^{-i} \Biggr)C\quad (n=0,1,\ldots)$$

have a solution $$C(x,t)=e^{-t}(1+t/a)^{x}$$, where $$(x)_{i}=x(x-1)\cdots(x+1-i)$$ with $$(x)_{0}=1$$.

As an application of Theorem 3, we obtain the following corollary.

### Corollary 4

For all $$k,N\geq0$$,

$$C_{k+N}(x;a)=\sum_{i=0}^{N}\sum _{m=0}^{k}(x)_{i}\binom{N}{i} \binom{k}{m}(-1)^{N-i+m}(i+m-1)_{m}a^{-i-m}C_{k-m}(x;a).$$

### Proof

By (1) and Theorem 3 we have

$$C^{(N)}= \Biggl(\sum_{i=0}^{N}(x)_{i} \binom{N}{i}(-1)^{N-i}(t+a)^{-i} \Biggr) \sum _{\ell\geq0}C_{\ell}(x;a)\frac{t^{\ell}}{\ell!}.$$

Since $$\frac{1}{(1+t)^{i+1}}=\sum_{j\geq0}\binom{i+j}{i}(-1)^{j}t^{j}$$, we obtain

$$C^{(N)}=\sum_{k\geq0}\sum _{i=0}^{N}\sum_{m=0}^{k}(x)_{i} \binom{N}{i} \binom{k}{m}(-1)^{N-i+m}(i+m-1)_{m}a^{-i-m}C_{k-m}(x;a) \frac{t^{k}}{k!}.$$

By comparing coefficients of $$t^{k}$$ we complete the proof. □

## Actuarial polynomials

For brevity, we denote the generating functions $$F(x,t)=e^{\beta t+x(1-e^{t})}$$ and $$\frac{d^{j}}{dt^{j}}F(x;t)$$ by F and $$F^{(j)}$$ for $$j\geq0$$.

### Lemma 5

The generating function $$F^{(N)}$$ is given by $$(\sum_{i=0}^{N}b_{i}(N,x)e^{it} )F$$, where $$b_{0}(N,x)=\beta^{N}$$, $$b_{N}(N,x)=(-x)^{N}$$, and $$b_{i}(N,x)=-xb_{i-1}(N-1,x)+(\beta+i)b_{i}(N-1,x)$$ ($$1\leq i\leq N-1$$).

### Proof

Clearly, $$b_{0}(0,x)=1$$. For $$N=1$$, by (2) we have $$F^{(1)}=(\beta-xe^{t})F$$, which proves the lemma for $$N=1$$ (here $$b_{0}(1,x)=\beta$$ and $$b_{1}(1,x)=-x$$). Assume that $$F^{(N)}$$ is given by $$(\sum_{i=0}^{N} b_{i}(N,x)e^{it} )F$$. Then

\begin{aligned} F^{(N+1)}&= \Biggl(\sum_{i=0}^{N} b_{i}(N,x)ie^{it} \Biggr)F + \Biggl(\sum _{i=0}^{N} b_{i}(N,x)e^{it} \Biggr) \bigl(\beta-xe^{t}\bigr)F\\ &= \Biggl(\sum_{i=0}^{N}( \beta+i)a_{i}(N,x)e^{it} -x\sum_{i=1}^{N+1} b_{i-1}(N,x)e^{it} \Biggr)F, \end{aligned}

which shows that the generating function $$F^{(N+1)}$$ is given by

\begin{aligned} \Biggl(\beta b_{0}(N,x)+\sum_{i=1}^{N} \bigl(-xa_{i-1}(N,x) +(\beta+i)b_{i}(N,x) \bigr)e^{it}-xb_{N}(N,x)e^{(N+1)t} \Biggr)F. \end{aligned}

Comparing with $$F^{(N+1)}= (\sum_{i=0}^{N+1} b_{i}(N+1,x)e^{it} )C$$, we complete the proof. □

### Lemma 6

For all $$0\leq i\leq N$$, the coefficients $$b_{i}(N,x)$$ in Lemma 5 are given by

$$b_{i}(N,x)=(-x)^{i}\sum_{j=i}^{N} \binom{N}{j}\beta^{N-j}S(j,i),$$

where $$S(n,k)$$ are the Stirling numbers (for example, see [16]) of the second kind.

### Proof

By Lemma 5 we have that

$$b_{i}(N+1,x)=-xb_{i-1}(N,x)+(\beta+i)b_{i}(N,x),\quad 0 \leq i\leq N+1,$$

with $$b_{0}(0,x)=1$$ and $$b_{i}(N,x)=0$$ whenever $$i>N$$ or $$i<0$$. Define $$B_{i}(x;t)=\sum_{N\geq i}b_{i}(N,x)t^{N}$$. Then we have

$$B_{i}(x;t)=\frac{-xt}{1-(\beta+i)t}B_{i-1}(x)$$

with $$B_{0}(x;t)=\frac{1}{1-\beta t}$$. By induction on i we derive that

$$B_{i}(x,t)=\frac{(-xt)^{i}}{(1-\beta t)(1-(\beta+1)t)\cdots(1-(\beta +i)t)}=\frac{(-xt)^{i}}{(1-\beta t)^{i+1}}\prod _{j=0}^{i}\frac {1}{1-jt/(1-\beta t)}.$$

Hence, since $$\frac{x^{k}}{(1-x)(1-2x)\cdots(1-kx)}=\sum_{n\geq k}S(n,k)x^{n}$$ (for example, see [16]), where $$S(n,k)$$ are the Stirling numbers of the second kind, we obtain that

$$B_{i}(x,t)=(-x)^{i}\sum_{j\geq i}S(j,i) \frac{t^{j}}{(1-\beta t)^{j+1}}.$$

Since $$\frac{1}{(1+t)^{i+1}}=\sum_{j\geq0}\binom{i+j}{i}(-1)^{j}t^{j}$$, we obtain that

$$B_{i}(x,t)=(-x)^{i}\sum_{j\geq i} \sum_{\ell\geq0}\binom{j+\ell}{j}\beta ^{\ell}S(j,i)t^{J+\ell}.$$

Thus, by finding the coefficients of $$t^{N}$$ we complete the proof. □

Thus, by Lemmas 5 and 6 we can state the following result.

### Theorem 7

The linear differential equations

$$F^{(N)}=\sum_{i=0}^{N} \Biggl((-x)^{i}e^{it}\sum_{j=i}^{N} \binom{N-1}{j-1}\beta ^{N-j}S(j,i) \Biggr)F \quad(N=0,1,\ldots)$$

have a solution $$F(x,t)=e^{\beta t+x(1-e^{t})}$$.

Recall that $$F(x,t)=e^{\beta t+x(1-e^{t})}=\sum_{n\geq0}a_{n}^{(\beta )}(x)\frac{t^{n}}{n!}$$, which is the generating function for the actuarial polynomials $$a_{n}^{(\beta)}(x)$$ (see (2)). As an application of Theorem 7, we obtain the following corollary.

### Corollary 8

For all $$k,N\geq0$$,

$$a_{N+k}^{(\beta)}(x)=\sum_{i=0}^{N} \sum_{m=0}^{k}b_{i}(N;x) \binom {k}{m}i^{k-m}a_{m}^{(\beta)}(x),$$

where $$b_{i}(N,x)=(-x)^{i}\sum_{j=i}^{N}\binom{N-1}{j-1}\beta^{N-j}S(j,i)$$.

### Proof

By (2) and Theorem 7 we have $$F^{(N)}= (\sum_{i=0}^{N}b_{i}(N,x)e^{it} ) \sum_{\ell\geq0}a_{\ell}^{(\beta)}(x)\frac{t^{\ell}}{\ell!}$$. Thus,

$$F^{(N)}=\sum_{k\geq0}\sum _{i=0}^{N}\sum_{m=0}^{k}b_{i}(N,x) \binom {k}{m}i^{k-m}a_{m}^{(\beta)}(x) \frac{t^{k}}{k!}.$$

By comparing the coefficients of $$t^{N+k}$$ we complete the proof. □

## Meixner polynomials of the first kind

Recall that the rising polynomials $$\langle x\rangle_{N}$$ are defined by $$\langle x\rangle_{N}=x(x+1)\cdots(x+N-1)$$ with $$\langle x\rangle_{0}=1$$. For brevity, we denote the generating functions $$M(x,t)=(1-t/c)^{x}(1-x)^{-x-\beta}$$ and $$\frac{d^{j}}{dt^{j}}M(x;t)$$ by M and $$M^{(j)}$$ for $$j\geq0$$, respectively.

### Theorem 9

The linear differential equations

$$M^{(N)}= \Biggl(\sum_{i=0}^{N}(-1)^{i} \binom{N}{i}(x)_{N-i}\langle x+\beta\rangle_{i}(t-1)^{-i}(t-c)^{-(N-i)} \Biggr)M \quad(N=0,1,\ldots)$$

have a solution $$M=M(x,t)=(1-t/c)^{x}(1-x)^{-x-\beta}$$.

### Proof

We proceed the proof by induction on N. Clearly, the theorem holds for $$N=0$$. By (3) we have $$M^{(1)}=(x(t-c)^{-1}-(x+\beta)(t-1)^{-1})M$$, which proves the theorem for $$N=1$$. Assume that the theorem holds for $$N\geq1$$. Then by the induction hypothesis we have

\begin{aligned} &M^{(N+1)}\\ &=\frac{d}{dt} \Biggl(\sum_{i=0}^{N}(-1)^{i} \binom{N}{i}(x)_{N-i}\langle x+\beta\rangle_{i}(t-1)^{-i}(t-c)^{-(N-i)} \Biggr)M \\ &\quad= \Biggl\{ \Biggl(\sum_{i=0}^{N}(-1)^{i+1}i \binom{N}{i}(x)_{N-i}\langle x+\beta\rangle_{i}(t-1)^{-i-1}(t-c)^{-(N-i)} \Biggr)M \\ &\qquad{}+ \Biggl(\sum_{i=0}^{N}(-1)^{i+1}(N-i) \binom{N}{i}(x)_{N-i}\langle x+\beta\rangle_{i}(t-1)^{-i}(t-c)^{-(N+1-i)} \Biggr)M \\ &\qquad{}+ \Biggl(\sum_{i=0}^{N}(-1)^{i} \binom{N}{i}(x)_{N-i}\langle x+\beta\rangle_{i}(t-1)^{-i}(t-c)^{-(N-i)} \Biggr)\\ &\qquad{}\times \bigl(x(t-c)^{-1}-(x+\beta ) (t-1)^{-1}\bigr)M \Biggr\} . \end{aligned}

After rearranging the indices of the sums, we obtain

\begin{aligned} &M^{(N+1)}\\ &\quad= \Biggl(\sum_{i=1}^{N+1}(-1)^{i}(i-1) \binom {N}{i-1}(x)_{N+1-i}\langle x+\beta\rangle_{i-1}(t-1)^{-i}(t-c)^{-(N+1-i)} \Biggr)M\\ &\qquad{}+ \Biggl(\sum_{i=0}^{N}(-1)^{i+1}(N-i) \binom{N}{i}(x)_{N-i}\langle x+\beta\rangle_{i}(t-1)^{-i}(t-c)^{-(N+1-i)} \Biggr)M\\ &\qquad{}+ \Biggl(\sum_{i=0}^{N}(-1)^{i} \binom{N}{i}x(x)_{N-i}\langle x+\beta\rangle_{i}(t-1)^{-i}(t-c)^{-(N+1-i)} \Biggr)M\\ &\qquad{}+ \Biggl(\sum_{i=1}^{N+1}(-1)^{i} \binom{N}{i-1}(x)_{N+1-i}(x+\beta)\langle x +\beta\rangle_{i-1}(t-1)^{-i}(t-c)^{-(N+1-i)} \Biggr)M. \end{aligned}

This implies

\begin{aligned} M^{(N+1)}= \Biggl(\sum_{i=0}^{N+1}(-1)^{i} \binom{N+1}{i}(x)_{N+1-i}\langle x+\beta\rangle_{i}(t-1)^{-i}(t-c)^{-(N+1-i)} \Biggr)M, \end{aligned}

and the induction step is completed. □

From (3) we have $$M^{(N)}=\sum_{k\geq0}m_{k+N}(x;\beta,c)\frac {t^{k}}{k!}$$ for all $$N\geq0$$. Similarly to the previous section, we have a recurrence relation for the coefficients of $$m_{n}(x;\beta,c)$$.

### Corollary 10

For all $$k,N\geq0$$,

\begin{aligned} &m_{k+N}(x;\beta,c)=(-1)^{N}\sum _{i=0}^{N}(-1)^{i}\binom{N}{i}(x)_{N-i} \langle x+\beta\rangle_{i}\sum_{\ell+m+n=k} \frac{k!\binom{i+\ell-1}{\ell}\binom {N+m-i-1}{m}}{n!c^{N-i+m}} m_{n}(x;\beta,c). \end{aligned}

### Proof

By Theorem 9 we have

\begin{aligned} M^{(N)}= \Biggl(\sum_{i=0}^{N}(-1)^{i} \binom{N}{i}(x)_{N-i}\langle x+\beta\rangle_{i}(t-1)^{-i}(t-c)^{-(N-i)} \Biggr) \sum_{\ell\geq0}m_{\ell}(x;\beta,c) \frac{t^{\ell}}{\ell!}. \end{aligned}

Thus, since $$(t-c)^{-s}=(-1)^{s}\sum_{\ell\geq0}\binom{s+\ell-1}{\ell}c^{-s-\ell }t^{\ell}$$, we obtain

\begin{aligned} M^{(N)}={}&(-1)^{N}\sum_{i=0}^{N}(-1)^{i} \binom{N}{i}(x)_{N-i}\langle x+\beta\rangle_{i}\\ &{}\times\sum_{\ell\geq0}\sum_{m\geq0} \sum_{n\geq0} \binom{i+\ell-1}{\ell}\binom{N+m-i-1}{m}m_{n}(x; \beta,c)\frac {c^{-N-m+i}t^{\ell+m+n}}{n!}. \end{aligned}

Hence, by finding the coefficients of $$t^{k}$$ in the generating function $$M^{(N)}$$ we complete the proof. □

## Results and discussion

In this paper, the Poisson-Charlier polynomials, actuarial, and Meixner polynomial are introduced. We study linear differential equations arising from the Poisson-Charlier, actuarial, and Meixner polynomials and present some their recurrence relations. Linear differential equations for various families of polynomials are derived. Furthermore, some particular cases of the results are presented.

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## Acknowledgements

The present research has been conducted by the Research Grant of Kwangwoon University in 2016.

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Correspondence to Taekyun Kim.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally to the manuscript and typed, read, and approved the final manuscript.

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Kim, T., Kim, D.S., Mansour, T. et al. Linear differential equations for families of polynomials. J Inequal Appl 2016, 95 (2016). https://doi.org/10.1186/s13660-016-1038-8