Theory and Modern Applications

# Poly-Cauchy and Peters mixed-type polynomials

## Abstract

The Peters polynomials are a generalization of Boole polynomials. In this paper, we consider Peters and poly-Cauchy mixed-type polynomials and investigate the properties of those polynomials which are derived from umbral calculus. Finally, we give various identities of those polynomials associated with special polynomials.

## 1 Introduction

The Peters polynomials are defined by the generating function to be

$\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{S}_{n}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\frac{{t}^{n}}{n!}={\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{\left(1+t\right)}^{x}\phantom{\rule{1em}{0ex}}\left(\text{see [1]}\right).$
(1)

The first few of them are given by

${S}_{0}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)={2}^{âˆ’\mathrm{Î¼}},\phantom{\rule{2em}{0ex}}{S}_{1}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)={2}^{âˆ’\left(\mathrm{Î¼}+1\right)}\left(2xâˆ’\mathrm{Î»}\mathrm{Î¼}\right),\phantom{\rule{2em}{0ex}}â€¦.$

If $\mathrm{Î¼}=1$, then ${S}_{n}\left(x;\mathrm{Î»}\right)={S}_{n}\left(x:\mathrm{Î»},1\right)$ are called Boole polynomials.

In particular, for $\mathrm{Î¼}=1$, $\mathrm{Î»}=1$, ${S}_{n}\left(x;1,1\right)={Ch}_{n}\left(x\right)$ are Changhee polynomials which are defined by

$\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{Ch}_{n}\left(x\right)\frac{{t}^{n}}{n!}=\frac{1}{t+2}{\left(1+t\right)}^{x}\phantom{\rule{1em}{0ex}}\left(\text{see [2]}\right).$

The generating functions for the poly-Cauchy polynomials of the first kind ${C}_{n}^{\left(k\right)}\left(x\right)$ and of the second kind ${\stackrel{Ë†}{C}}_{n}^{\left(k\right)}\left(x\right)$ are, respectively, given by

${Lif}_{k}\left(log\left(1+t\right)\right){\left(1+t\right)}^{âˆ’x}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{C}_{n}^{\left(k\right)}\left(x\right)\frac{{t}^{n}}{n!}$
(2)

and

${Lif}_{k}\left(âˆ’log\left(1+t\right)\right){\left(1+t\right)}^{x}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{\stackrel{Ë†}{C}}_{n}^{\left(k\right)}\left(x\right)\frac{{t}^{n}}{n!},$
(3)

where ${Lif}_{k}\left(t\right)={âˆ‘}_{n=0}^{\mathrm{âˆž}}\frac{{t}^{n}}{n!{\left(n+1\right)}^{k}}$ ($kâˆˆ\mathbb{Z}$) (see [3, 4]).

In this paper, we consider the poly-Cauchy of the first kind and Peters (respectively the poly-Cauchy of the second kind and Peters) mixed-type polynomials as follows:

${\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(log\left(1+t\right)\right){\left(1+t\right)}^{âˆ’x}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\frac{{t}^{n}}{n!}$
(4)

and

${\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(âˆ’log\left(1+t\right)\right){\left(1+t\right)}^{x}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\stackrel{Ë†}{C}{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\frac{{t}^{n}}{n!}.$
(5)

For $\mathrm{Î±}âˆˆ{\mathbb{Z}}_{â‰¥0}$, the Bernoulli polynomials of order Î± are defined by the generating function to be

${\left(\frac{t}{{e}^{t}âˆ’1}\right)}^{\mathrm{Î±}}{e}^{xt}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{B}_{n}^{\left(\mathrm{Î±}\right)}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [2, 5â€“11]}\right).$
(6)

As is well known, the Frobenius-Euler polynomials of order Î± are given by

${\left(\frac{1âˆ’\mathrm{Î»}}{{e}^{t}âˆ’\mathrm{Î»}}\right)}^{\mathrm{Î±}}{e}^{xt}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{H}_{n}^{\left(\mathrm{Î±}\right)}\left(xâˆ£\mathrm{Î»}\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [2â€“14]}\right),$
(7)

where $\mathrm{Î»}âˆˆ\mathbb{C}$ with and $\mathrm{Î±}âˆˆ{\mathbb{Z}}_{â‰¥0}$.

When $x=0$, $C{P}_{n}^{\left(k\right)}\left(0;\mathrm{Î»},\mathrm{Î¼}\right)$ (or $\stackrel{Ë†}{C}{P}_{n}^{\left(k\right)}\left(0;\mathrm{Î»},\mathrm{Î¼}\right)$) are called the poly-Cauchy of the first kind and Peters (or the poly-Cauchy of the second kind and Peters) mixed-type numbers.

The higher-order Cauchy polynomials of the first kind are defined by the generating function to be

${\left(\frac{t}{log\left(1+t\right)}\right)}^{\mathrm{Î±}}{\left(1+t\right)}^{âˆ’x}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{\mathbb{C}}_{n}^{\left(\mathrm{Î±}\right)}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\mathrm{Î±}âˆˆ{\mathbb{Z}}_{â‰¥0}\right),$
(8)

and the higher-order Cauchy polynomials of the second kind are given by

${\left(\frac{t}{\left(1+t\right)log\left(1+t\right)}\right)}^{\mathrm{Î±}}{\left(1+t\right)}^{x}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{\stackrel{Ë†}{\mathbb{C}}}_{n}^{\left(\mathrm{Î±}\right)}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\mathrm{Î±}âˆˆ{\mathbb{Z}}_{â‰¥0}\right).$
(9)

The Stirling number of the first kind is given by

${\left(x\right)}_{n}=x\left(xâˆ’1\right)â‹¯\left(xâˆ’n+1\right)=\underset{l=0}{\overset{n}{âˆ‘}}{S}_{1}\left(n,l\right){x}^{l}.$
(10)

Thus, by (10), we get

(11)

It is easy to show that

${x}^{\left(n\right)}=x\left(x+1\right)â‹¯\left(x+nâˆ’1\right)={\left(âˆ’1\right)}^{n}{\left(âˆ’x\right)}_{n}=\underset{l=0}{\overset{n}{âˆ‘}}{S}_{1}\left(n,l\right){\left(âˆ’1\right)}^{nâˆ’l}{x}^{l}.$
(12)

Let â„‚ be the complex number field and let â„± be the algebra of all formal power series in the variable t over â„‚ as follows:

$\mathcal{F}=\left\{f\left(t\right)=\underset{k=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{a}_{k}\frac{{t}^{k}}{k!}|{a}_{k}âˆˆ\mathbb{C}\right\}.$
(13)

Let $\mathbb{P}=\mathbb{C}\left[x\right]$ and let ${\mathbb{P}}^{âˆ—}$ be the vector space of all linear functionals on â„™. $ã€ˆLâˆ£p\left(x\right)ã€‰$ denotes the action of the linear functional L on the polynomial $p\left(x\right)$, and we recall that the vector space operations on ${\mathbb{P}}^{âˆ—}$ are defined by $ã€ˆL+Mâˆ£p\left(x\right)ã€‰=ã€ˆLâˆ£p\left(x\right)ã€‰+ã€ˆMâˆ£p\left(x\right)ã€‰$, $ã€ˆcLâˆ£p\left(x\right)ã€‰=cã€ˆLâˆ£p\left(x\right)ã€‰$, where c is a complex constant in â„‚.

For $f\left(t\right)âˆˆ\mathcal{F}$, let us define the linear functional on â„™ by setting

$ã€ˆf\left(t\right)âˆ£{x}^{n}ã€‰={a}_{n}\phantom{\rule{1em}{0ex}}\left(nâ‰¥0\right).$
(14)

Then, by (13) and (14), we get

(15)

where ${\mathrm{Î´}}_{n,k}$ is the Kronecker symbol.

Let ${f}_{L}\left(t\right)={âˆ‘}_{k=0}^{\mathrm{âˆž}}\frac{ã€ˆLâˆ£{x}^{k}ã€‰}{k!}{t}^{k}.$ Then, by (14), we see that $ã€ˆ{f}_{L}\left(t\right)âˆ£{x}^{n}ã€‰=ã€ˆLâˆ£{x}^{n}ã€‰$. The map $Lâ†¦{f}_{L}\left(t\right)$ is a vector space isomorphism from ${\mathbb{P}}^{âˆ—}$ onto â„±. Henceforth, â„± denotes both the algebra of formal power series in t and the vector space of all linear functionals on â„™, and so an element $f\left(t\right)$ of â„± will be thought of as both a formal power series and a linear functional. We call â„± the umbral algebra, and the umbral calculus is the study of umbral algebra. The order $O\left(f\right)$ of the power series $f\left(t\right)$ (â‰ 0) is the smallest integer for which the coefficient of ${t}^{k}$ does not vanish. If $O\left(f\left(t\right)\right)=1$, then $f\left(t\right)$ is called a delta series; if $O\left(f\left(t\right)\right)=0$, then $f\left(t\right)$ is called an invertible series. For $f\left(t\right),g\left(t\right)âˆˆ\mathcal{F}$ with $O\left(f\left(t\right)\right)=1$ and $O\left(g\left(t\right)\right)=0$, there exists a unique sequence ${s}_{n}\left(x\right)$ of polynomials such that $ã€ˆg\left(t\right)f{\left(t\right)}^{k}âˆ£{s}_{n}\left(x\right)ã€‰=n!{\mathrm{Î´}}_{n,k}$ ($n,kâ‰¥0$).

The sequence ${s}_{n}\left(x\right)$ is called the Sheffer sequence for $\left(g\left(t\right),f\left(t\right)\right)$ which is denoted by ${s}_{n}\left(x\right)âˆ¼\left(g\left(t\right),f\left(t\right)\right)$.

For $f\left(t\right),g\left(t\right)âˆˆ\mathcal{F}$ and $p\left(x\right)âˆˆ\mathbb{P}$, we have

$ã€ˆf\left(t\right)g\left(t\right)âˆ£p\left(x\right)ã€‰=ã€ˆf\left(t\right)âˆ£g\left(t\right)p\left(x\right)ã€‰=ã€ˆg\left(t\right)âˆ£f\left(t\right)p\left(x\right)ã€‰$

and

$f\left(t\right)=\underset{k=0}{\overset{\mathrm{âˆž}}{âˆ‘}}ã€ˆf\left(t\right)âˆ£{x}^{k}ã€‰\frac{{t}^{k}}{k!},\phantom{\rule{2em}{0ex}}p\left(x\right)=\underset{k=0}{\overset{\mathrm{âˆž}}{âˆ‘}}ã€ˆ{t}^{k}âˆ£p\left(x\right)ã€‰\frac{{x}^{k}}{k!}.$
(16)

By (16), we get

${p}^{\left(k\right)}\left(0\right)=ã€ˆ{t}^{k}âˆ£p\left(x\right)ã€‰=ã€ˆ1âˆ£{p}^{\left(k\right)}\left(x\right)ã€‰\phantom{\rule{1em}{0ex}}\left(kâ‰¥0\right),$
(17)

where ${p}^{\left(k\right)}\left(0\right)=\frac{{d}^{k}p\left(x\right)}{d{x}^{k}}{|}_{x=0}$.

Thus, by (17), we have

${t}^{k}p\left(x\right)={p}^{\left(k\right)}\left(x\right)=\frac{{d}^{k}p\left(x\right)}{d{x}^{k}}\phantom{\rule{1em}{0ex}}\text{(see [1â€“3])}.$
(18)

Let ${s}_{n}\left(x\right)âˆ¼\left(g\left(t\right),f\left(t\right)\right)$. Then the following equations are known in [1]:

(19)

where $\stackrel{Â¯}{f}\left(t\right)$ is the compositional inverse for $f\left(t\right)$ with $f\left(\stackrel{Â¯}{f}\left(t\right)\right)=t$,

${s}_{n}\left(x\right)=\underset{j=0}{\overset{n}{âˆ‘}}\frac{1}{j!}ã€ˆ\frac{{\left(\stackrel{Â¯}{f}\left(t\right)\right)}^{j}}{g\left(\stackrel{Â¯}{f}\left(t\right)\right)}|{x}^{n}ã€‰{x}^{j},$
(20)
(21)

and

${s}_{n+1}\left(x\right)=\left(xâˆ’\frac{{g}^{â€²}\left(t\right)}{g\left(t\right)}\right)\frac{1}{{f}^{â€²}\left(t\right)}{s}_{n}\left(x\right),\phantom{\rule{1em}{0ex}}f\left(t\right){s}_{n}\left(x\right)=n{s}_{nâˆ’1}\left(x\right)\phantom{\rule{0.25em}{0ex}}\left(nâ‰¥0\right),$
(22)

and

$\frac{d}{dx}{s}_{n}\left(x\right)=\underset{l=0}{\overset{nâˆ’1}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right)ã€ˆ\stackrel{Â¯}{f}\left(t\right)âˆ£{x}^{nâˆ’l}ã€‰{s}_{l}\left(x\right).$
(23)

As is well known, the transfer formula for ${p}_{n}\left(x\right)âˆ¼\left(1,f\left(t\right)\right)$, ${q}_{n}\left(x\right)âˆ¼\left(1,g\left(t\right)\right)$ ($nâ‰¥1$) is given by

${q}_{n}\left(x\right)=x{\left(\frac{f\left(t\right)}{g\left(t\right)}\right)}^{n}{x}^{âˆ’1}{p}_{n}\left(x\right).$
(24)

For ${s}_{n}\left(x\right)âˆ¼\left(g\left(t\right),f\left(t\right)\right)$, ${r}_{n}\left(x\right)âˆ¼\left(h\left(t\right),l\left(t\right)\right)$, let

${s}_{n}\left(x\right)=\underset{m=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{C}_{n,m}{r}_{n}\left(x\right),$
(25)

where

${C}_{n,m}=\frac{1}{m!}ã€ˆ\frac{h\left(\stackrel{Â¯}{f}\left(t\right)\right)}{g\left(\stackrel{Â¯}{f}\left(t\right)\right)}{\left(l\left(\stackrel{Â¯}{f}\left(t\right)\right)\right)}^{m}|{x}^{n}ã€‰\phantom{\rule{1em}{0ex}}\text{(see [1])}.$
(26)

It is known that

$ã€ˆf\left(t\right)âˆ£xp\left(x\right)ã€‰=ã€ˆ{\mathrm{âˆ‚}}_{t}f\left(t\right)âˆ£p\left(x\right)ã€‰,\phantom{\rule{2em}{0ex}}{e}^{yt}p\left(x\right)=p\left(x+y\right),$
(27)

where $f\left(t\right)âˆˆ\mathcal{F}$ and $p\left(x\right)âˆˆ\mathbb{P}$ (see [1â€“3]).

In this paper, we consider Peters and poly-Cauchy mixed-type polynomials with umbral calculus viewpoint and investigate the properties of those polynomials which are derived from umbral calculus. Finally, we give some interesting identities of those polynomials associated with special polynomials.

## 2 Poly-Cauchy and Peters mixed-type polynomials

From (2), (3), and (19), we note that

$C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)âˆ¼\left({\left(1+{e}^{âˆ’\mathrm{Î»}t}\right)}^{\mathrm{Î¼}}\frac{1}{{Lif}_{k}\left(âˆ’t\right)},{e}^{âˆ’t}âˆ’1\right)$
(28)

and

$\stackrel{Ë†}{C}{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)âˆ¼\left({\left(1+{e}^{\mathrm{Î»}t}\right)}^{\mathrm{Î¼}}\frac{1}{{Lif}_{k}\left(âˆ’t\right)},{e}^{t}âˆ’1\right).$
(29)

It is not difficult to show that

$\begin{array}{rl}{\left(1+{e}^{âˆ’\mathrm{Î»}t}\right)}^{\mathrm{Î¼}}& ={2}^{\mathrm{Î¼}}{\left(1+\frac{1}{2}\underset{j=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\left(âˆ’\mathrm{Î»}t\right)}^{j}}{j!}\right)}^{\mathrm{Î¼}}\\ =\underset{i=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\underset{j=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\underset{{j}_{1}+â‹¯+{j}_{i}=j}{âˆ‘}{2}^{\mathrm{Î¼}âˆ’i}\left(\genfrac{}{}{0}{}{\mathrm{Î¼}}{i}\right)\left(\genfrac{}{}{0}{}{j+i}{{j}_{1}+1,â€¦,{j}_{i}+1}\right)\frac{{\left(âˆ’\mathrm{Î»}t\right)}^{j+i}}{\left(j+i\right)!}\end{array}$
(30)

and

$\begin{array}{rl}{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}& ={2}^{âˆ’\mathrm{Î¼}}{\left(1+\frac{1}{2}\underset{j=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\left(\genfrac{}{}{0}{}{\mathrm{Î»}}{j+1}\right){t}^{j+1}\right)}^{âˆ’\mathrm{Î¼}}\\ =\underset{i=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\underset{j=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\underset{{j}_{1}+â‹¯+{j}_{i}=j}{âˆ‘}{2}^{âˆ’\left(\mathrm{Î¼}+i\right)}\left(\genfrac{}{}{0}{}{âˆ’\mathrm{Î¼}}{i}\right)\left(\genfrac{}{}{0}{}{\mathrm{Î»}}{{j}_{1}+1}\right)â‹¯\left(\genfrac{}{}{0}{}{\mathrm{Î»}}{{j}_{i}+1}\right){t}^{j+i}.\end{array}$
(31)

From (14), we have

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(y;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=ã€ˆ\underset{l=0}{\overset{\mathrm{âˆž}}{âˆ‘}}C{P}_{l}^{\left(k\right)}\left(y;\mathrm{Î»},\mathrm{Î¼}\right)\frac{{t}^{l}}{l!}|{x}^{n}ã€‰\\ \phantom{\rule{1em}{0ex}}=ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(log\left(1+t\right)\right){\left(1+t\right)}^{âˆ’y}âˆ£{x}^{n}ã€‰\\ \phantom{\rule{1em}{0ex}}=ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}|\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){C}_{l}^{\left(k\right)}\left(y\right){x}^{nâˆ’l}ã€‰\\ \phantom{\rule{1em}{0ex}}=\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){C}_{l}^{\left(k\right)}\left(y\right)ã€ˆ\underset{m=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{S}_{m}\left(0;\mathrm{Î»},\mathrm{Î¼}\right)\frac{{t}^{m}}{m!}|{x}^{nâˆ’l}ã€‰\\ \phantom{\rule{1em}{0ex}}=\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{nâˆ’l}\left(0;\mathrm{Î»},\mathrm{Î¼}\right){C}_{l}^{\left(k\right)}\left(y\right).\end{array}$
(32)

Therefore, by (32), we obtain the following theorem.

Theorem 1 For $nâ‰¥0$, we have

$C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)=\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{nâˆ’l}\left(0;\mathrm{Î»},\mathrm{Î¼}\right){C}_{l}^{\left(k\right)}\left(x\right).$

Alternatively,

$\begin{array}{rcl}C{P}_{n}^{\left(k\right)}\left(y;\mathrm{Î»},\mathrm{Î¼}\right)& =& ã€ˆ\underset{l=0}{\overset{\mathrm{âˆž}}{âˆ‘}}C{P}_{l}^{\left(k\right)}\left(y;\mathrm{Î»},\mathrm{Î¼}\right)\frac{{t}^{l}}{l!}|{x}^{n}ã€‰\\ =& ã€ˆ{Lif}_{k}\left(log\left(1+t\right)\right)âˆ£{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{\left(1+t\right)}^{âˆ’y}{x}^{n}ã€‰\\ =& ã€ˆ{Lif}_{k}\left(log\left(1+t\right)\right)|\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{l}\left(âˆ’y;\mathrm{Î»},\mathrm{Î¼}\right){x}^{nâˆ’l}ã€‰\\ =& \underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{l}\left(âˆ’y;\mathrm{Î»},\mathrm{Î¼}\right)ã€ˆ{Lif}_{k}\left(log\left(1+t\right)\right)âˆ£{x}^{nâˆ’l}ã€‰\\ =& \underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{l}\left(âˆ’y;\mathrm{Î»},\mathrm{Î¼}\right){C}_{nâˆ’l}^{\left(k\right)}\left(0\right).\end{array}$
(33)

Therefore, by (33), we obtain the following theorem.

Theorem 2 For $nâ‰¥0$, let ${C}_{nâˆ’l}^{\left(k\right)}\left(0\right)={C}_{nâˆ’l}^{\left(k\right)}$. Then we have

$C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)=\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){C}_{nâˆ’l}^{\left(k\right)}{S}_{l}\left(âˆ’x;\mathrm{Î»},\mathrm{Î¼}\right).$

Remark By the same method, we get

$\stackrel{Ë†}{C}{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)=\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{nâˆ’l}\left(0;\mathrm{Î»},\mathrm{Î¼}\right){\stackrel{Ë†}{C}}_{l}^{\left(k\right)}\left(x\right)$
(34)

and

$\stackrel{Ë†}{C}{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)=\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){\stackrel{Ë†}{C}}_{nâˆ’l}^{\left(k\right)}{S}_{l}\left(x;\mathrm{Î»},\mathrm{Î¼}\right).$
(35)

From (20) and (28), we have

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=\underset{j=0}{\overset{n}{âˆ‘}}\frac{1}{j!}ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(log\left(1+t\right)\right){\left(âˆ’log\left(1+t\right)\right)}^{j}âˆ£{x}^{n}ã€‰{x}^{j}.\end{array}$
(36)

From (31), we note that

$\begin{array}{r}ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(log\left(1+t\right)\right){\left(âˆ’log\left(1+t\right)\right)}^{j}âˆ£{x}^{n}ã€‰\\ \phantom{\rule{1em}{0ex}}=\underset{m=0}{\overset{nâˆ’j}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{j}}{m!{\left(m+1\right)}^{k}}\underset{l=0}{\overset{nâˆ’jâˆ’m}{âˆ‘}}\frac{\left(m+j\right)!}{\left(l+m+j\right)!}{S}_{1}\left(l+m+j,m+j\right)\\ \phantom{\rule{2em}{0ex}}Ã—{\left(n\right)}_{l+m+j}ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}âˆ£{x}^{nâˆ’lâˆ’mâˆ’j}ã€‰\\ \phantom{\rule{1em}{0ex}}=\underset{m=0}{\overset{nâˆ’j}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{j}}{m!{\left(m+1\right)}^{k}}\underset{l=0}{\overset{nâˆ’jâˆ’m}{âˆ‘}}\frac{\left(m+j\right)!}{\left(l+m+j\right)!}\\ \phantom{\rule{2em}{0ex}}Ã—{S}_{1}\left(l+m+j,m+j\right){\left(n\right)}_{l+m+j}\underset{i=0}{\overset{nâˆ’jâˆ’mâˆ’l}{âˆ‘}}\underset{r=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\underset{{r}_{1}+â‹¯+{r}_{i}=r}{âˆ‘}{2}^{âˆ’\left(\mathrm{Î¼}+i\right)}\\ \phantom{\rule{2em}{0ex}}Ã—\left(\genfrac{}{}{0}{}{âˆ’\mathrm{Î¼}}{i}\right)\left(\genfrac{}{}{0}{}{\mathrm{Î»}}{{r}_{1}+1}\right)â‹¯\left(\genfrac{}{}{0}{}{\mathrm{Î»}}{{r}_{i}+1}\right)ã€ˆ{t}^{r+i}âˆ£{x}^{nâˆ’lâˆ’mâˆ’j}ã€‰\\ \phantom{\rule{1em}{0ex}}={2}^{âˆ’\mathrm{Î¼}}n!\underset{m=0}{\overset{nâˆ’j}{âˆ‘}}\underset{l=0}{\overset{nâˆ’jâˆ’m}{âˆ‘}}\underset{i=0}{\overset{nâˆ’jâˆ’mâˆ’l}{âˆ‘}}\underset{{r}_{1}+â‹¯+{r}_{i}=nâˆ’jâˆ’mâˆ’lâˆ’i}{âˆ‘}\frac{{2}^{âˆ’i}{\left(âˆ’1\right)}^{j}\left(m+j\right)!}{m!{\left(m+1\right)}^{k}\left(l+m+j\right)!}\\ \phantom{\rule{2em}{0ex}}Ã—\left(\genfrac{}{}{0}{}{âˆ’\mathrm{Î¼}}{i}\right)\left(\genfrac{}{}{0}{}{\mathrm{Î»}}{{r}_{1}+1}\right)â‹¯\left(\genfrac{}{}{0}{}{\mathrm{Î»}}{{r}_{i}+1}\right){S}_{1}\left(l+m+j,m+j\right).\end{array}$
(37)

Therefore, by (36) and (37), we obtain the following theorem.

Theorem 3 For $nâ‰¥0$, we have

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}={2}^{âˆ’\mathrm{Î¼}}n!\underset{j=0}{\overset{n}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{j}}{j!}\left\{\underset{m=0}{\overset{nâˆ’j}{âˆ‘}}\underset{l=0}{\overset{nâˆ’jâˆ’m}{âˆ‘}}\underset{i=0}{\overset{nâˆ’jâˆ’mâˆ’l}{âˆ‘}}\underset{{r}_{1}+â‹¯+{r}_{i}=nâˆ’jâˆ’mâˆ’lâˆ’i}{âˆ‘}\frac{{2}^{âˆ’i}}{m!{\left(m+1\right)}^{k}}\\ \phantom{\rule{2em}{0ex}}Ã—\frac{\left(m+j\right)!}{\left(l+m+j\right)!}\left(\genfrac{}{}{0}{}{âˆ’\mathrm{Î¼}}{i}\right)\left(\genfrac{}{}{0}{}{\mathrm{Î»}}{{r}_{1}+1}\right)â‹¯\left(\genfrac{}{}{0}{}{\mathrm{Î»}}{{r}_{i}+1}\right){S}_{1}\left(l+m+j,m+j\right)\right\}{x}^{j}.\end{array}$

Remark By the same method as Theorem 3, we get

$\begin{array}{r}\stackrel{Ë†}{C}{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}={2}^{âˆ’\mathrm{Î¼}}n!\underset{j=0}{\overset{n}{âˆ‘}}\frac{1}{j!}\left\{\underset{m=0}{\overset{nâˆ’j}{âˆ‘}}\underset{l=0}{\overset{nâˆ’jâˆ’m}{âˆ‘}}\underset{i=0}{\overset{nâˆ’jâˆ’mâˆ’l}{âˆ‘}}\underset{{r}_{1}+â‹¯+{r}_{i}=nâˆ’jâˆ’mâˆ’lâˆ’i}{âˆ‘}\frac{{2}^{âˆ’i}{\left(âˆ’1\right)}^{m}}{m!{\left(m+1\right)}^{k}}\\ \phantom{\rule{2em}{0ex}}Ã—\frac{\left(m+j\right)!}{\left(l+m+j\right)!}\left(\genfrac{}{}{0}{}{âˆ’\mathrm{Î¼}}{i}\right)\\ \phantom{\rule{2em}{0ex}}Ã—\left(\genfrac{}{}{0}{}{\mathrm{Î»}}{{r}_{1}+1}\right)â‹¯\left(\genfrac{}{}{0}{}{\mathrm{Î»}}{{r}_{i}+1}\right){S}_{1}\left(l+m+j,m+j\right)\right\}{x}^{j}.\end{array}$
(38)

From (28), we note that

${\left(1+{e}^{âˆ’\mathrm{Î»}t}\right)}^{\mathrm{Î¼}}\frac{1}{{Lif}_{k}\left(âˆ’t\right)}C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)âˆ¼\left(1,{e}^{âˆ’t}âˆ’1\right)$
(39)

and

${x}^{n}âˆ¼\left(1,t\right).$
(40)

By (24), (39), and (40), we get

$\begin{array}{r}{\left(1+{e}^{âˆ’\mathrm{Î»}t}\right)}^{\mathrm{Î¼}}\frac{1}{{Lif}_{k}\left(âˆ’t\right)}C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=x{\left(\frac{t}{{e}^{âˆ’t}âˆ’1}\right)}^{n}{x}^{nâˆ’1}\\ \phantom{\rule{1em}{0ex}}={\left(âˆ’1\right)}^{n}x{\left(\frac{âˆ’t}{{e}^{âˆ’t}âˆ’1}\right)}^{n}{x}^{nâˆ’1}\\ \phantom{\rule{1em}{0ex}}={\left(âˆ’1\right)}^{n}\underset{l=0}{\overset{nâˆ’1}{âˆ‘}}{\left(âˆ’1\right)}^{l}{B}_{l}^{\left(n\right)}\left(\genfrac{}{}{0}{}{nâˆ’1}{l}\right){x}^{nâˆ’l}.\end{array}$
(41)

Thus, by (41), we see that

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}={\left(âˆ’1\right)}^{n}\underset{l=0}{\overset{nâˆ’1}{âˆ‘}}{\left(âˆ’1\right)}^{l}\left(\genfrac{}{}{0}{}{nâˆ’1}{l}\right){B}_{l}^{\left(n\right)}{\left(1+{e}^{âˆ’\mathrm{Î»}t}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(âˆ’t\right){x}^{nâˆ’l}\\ \phantom{\rule{1em}{0ex}}={\left(âˆ’1\right)}^{n}\underset{l=0}{\overset{nâˆ’1}{âˆ‘}}{\left(âˆ’1\right)}^{l}\left(\genfrac{}{}{0}{}{nâˆ’1}{l}\right){B}_{l}^{\left(n\right)}\underset{m=0}{\overset{nâˆ’l}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{m}\left(\genfrac{}{}{0}{}{nâˆ’l}{m}\right)}{{\left(m+1\right)}^{k}}{\left(1+{e}^{âˆ’\mathrm{Î»}t}\right)}^{âˆ’\mathrm{Î¼}}{x}^{nâˆ’lâˆ’m}\\ \phantom{\rule{1em}{0ex}}={\left(âˆ’1\right)}^{n}\underset{l=0}{\overset{nâˆ’1}{âˆ‘}}{\left(âˆ’1\right)}^{l}\left(\genfrac{}{}{0}{}{nâˆ’1}{l}\right){B}_{l}^{\left(n\right)}\underset{m=0}{\overset{nâˆ’l}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{m}\left(\genfrac{}{}{0}{}{nâˆ’l}{m}\right)}{{\left(m+1\right)}^{k}}\underset{i=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\underset{j=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\underset{{j}_{1}+â‹¯+{j}_{n}=j}{âˆ‘}{2}^{âˆ’\mathrm{Î¼}âˆ’i}\left(\genfrac{}{}{0}{}{âˆ’\mathrm{Î¼}}{i}\right)\\ \phantom{\rule{2em}{0ex}}Ã—\left(\genfrac{}{}{0}{}{j+i}{{j}_{1}+1,â€¦,{j}_{i}+1}\right)\frac{{\left(âˆ’\mathrm{Î»}t\right)}^{j+i}}{\left(j+i\right)!}{x}^{nâˆ’lâˆ’m}\\ \phantom{\rule{1em}{0ex}}={\left(âˆ’1\right)}^{n}\underset{l=0}{\overset{n}{âˆ‘}}\underset{m=0}{\overset{nâˆ’l}{âˆ‘}}\underset{i=0}{\overset{nâˆ’lâˆ’m}{âˆ‘}}\underset{j=0}{\overset{nâˆ’lâˆ’mâˆ’i}{âˆ‘}}\underset{{j}_{1}+â‹¯+{j}_{i}=nâˆ’lâˆ’mâˆ’iâˆ’r}{âˆ‘}{\left(âˆ’1\right)}^{nâˆ’r}\frac{{2}^{âˆ’\mathrm{Î¼}âˆ’i}{\mathrm{Î»}}^{nâˆ’lâˆ’mâˆ’r}}{{\left(m+1\right)}^{k}}\left(\genfrac{}{}{0}{}{nâˆ’1}{l}\right)\\ \phantom{\rule{2em}{0ex}}Ã—\left(\genfrac{}{}{0}{}{nâˆ’l}{m}\right)\left(\genfrac{}{}{0}{}{âˆ’\mathrm{Î¼}}{i}\right)\left(\genfrac{}{}{0}{}{nâˆ’lâˆ’mâˆ’r}{{j}_{1}+1,â€¦,{j}_{i}+1}\right)\left(\genfrac{}{}{0}{}{nâˆ’lâˆ’m}{r}\right){B}_{l}^{\left(n\right)}{x}^{r}.\end{array}$
(42)

Therefore, by (42), we obtain the following theorem.

Theorem 4 For $nâ‰¥0$, we have

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=\frac{{\mathrm{Î»}}^{n}}{{2}^{\mathrm{Î¼}}}\underset{r=0}{\overset{n}{âˆ‘}}{\left(âˆ’{\mathrm{Î»}}^{âˆ’1}\right)}^{r}\left\{\underset{l=0}{\overset{nâˆ’r}{âˆ‘}}\underset{m=0}{\overset{nâˆ’râˆ’l}{âˆ‘}}\underset{i=0}{\overset{nâˆ’râˆ’lâˆ’m}{âˆ‘}}\underset{{j}_{1}+â‹¯+{j}_{i}=nâˆ’râˆ’lâˆ’mâˆ’i}{âˆ‘}\frac{{2}^{âˆ’i}{\mathrm{Î»}}^{âˆ’lâˆ’m}}{{\left(m+1\right)}^{k}}\left(\genfrac{}{}{0}{}{nâˆ’1}{l}\right)\\ \phantom{\rule{2em}{0ex}}Ã—\left(\genfrac{}{}{0}{}{nâˆ’l}{m}\right)\left(\genfrac{}{}{0}{}{âˆ’\mathrm{Î¼}}{i}\right)\left(\genfrac{}{}{0}{}{nâˆ’râˆ’lâˆ’m}{{j}_{1}+1,â€¦,{j}_{i}+1}\right)\left(\genfrac{}{}{0}{}{nâˆ’lâˆ’m}{r}\right){B}_{l}^{\left(n\right)}\right\}{x}^{r}.\end{array}$

Remark By the same method as Theorem 4, we get

$\begin{array}{r}\stackrel{Ë†}{C}{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=\frac{{\mathrm{Î»}}^{n}}{{2}^{\mathrm{Î¼}}}\underset{r=0}{\overset{n}{âˆ‘}}{\mathrm{Î»}}^{âˆ’r}\left\{\underset{l=0}{\overset{nâˆ’r}{âˆ‘}}\underset{m=0}{\overset{nâˆ’lâˆ’r}{âˆ‘}}\underset{i=0}{\overset{nâˆ’lâˆ’mâˆ’r}{âˆ‘}}\underset{{j}_{1}+â‹¯+{j}_{i}=nâˆ’râˆ’lâˆ’mâˆ’i}{âˆ‘}\frac{{\left(âˆ’1\right)}^{m}{2}^{âˆ’i}{\mathrm{Î»}}^{âˆ’lâˆ’m}}{{\left(m+1\right)}^{k}}\left(\genfrac{}{}{0}{}{nâˆ’1}{l}\right)\\ \phantom{\rule{2em}{0ex}}Ã—\left(\genfrac{}{}{0}{}{nâˆ’l}{m}\right)\left(\genfrac{}{}{0}{}{âˆ’\mathrm{Î¼}}{i}\right)\left(\genfrac{}{}{0}{}{nâˆ’râˆ’lâˆ’m}{{j}_{1}+1,â€¦,{j}_{i}+1}\right)\left(\genfrac{}{}{0}{}{nâˆ’lâˆ’m}{r}\right){B}_{l}^{\left(m\right)}\right\}{x}^{r}.\end{array}$
(43)

From (12), we note that

${x}^{\left(n\right)}=x\left(x+1\right)â‹¯\left(x+nâˆ’1\right)âˆ¼\left(1,1âˆ’{e}^{âˆ’t}\right).$
(44)

Thus, by (44), we see that

${\left(âˆ’1\right)}^{n}{x}^{\left(n\right)}={\left(âˆ’x\right)}_{n}=\underset{m=0}{\overset{n}{âˆ‘}}{S}_{1}\left(n,m\right){\left(âˆ’x\right)}^{m}âˆ¼\left(1,{e}^{âˆ’t}âˆ’1\right)$
(45)

and

${\left(1+{e}^{âˆ’\mathrm{Î»}t}\right)}^{\mathrm{Î¼}}\frac{1}{{Lif}_{k}\left(âˆ’t\right)}C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)âˆ¼\left(1,{e}^{âˆ’t}âˆ’1\right).$
(46)

From (45) and (46), we have

$\begin{array}{r}{\left(1+{e}^{âˆ’\mathrm{Î»}t}\right)}^{\mathrm{Î¼}}\frac{1}{{Lif}_{k}\left(âˆ’t\right)}C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}={\left(âˆ’1\right)}^{n}{x}^{\left(n\right)}\\ \phantom{\rule{1em}{0ex}}=\underset{l=0}{\overset{n}{âˆ‘}}{S}_{1}\left(n,l\right){\left(âˆ’x\right)}^{l}.\end{array}$
(47)

Thus, by (47), we get

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=\underset{l=0}{\overset{n}{âˆ‘}}{S}_{1}\left(n,l\right){\left(âˆ’1\right)}^{l}{\left(1+{e}^{âˆ’\mathrm{Î»}t}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(âˆ’t\right){x}^{l}\\ \phantom{\rule{1em}{0ex}}=\underset{l=0}{\overset{n}{âˆ‘}}{S}_{1}\left(n,l\right){\left(âˆ’1\right)}^{l}\underset{m=0}{\overset{l}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{m}\left(\genfrac{}{}{0}{}{l}{m}\right)}{{\left(m+1\right)}^{k}}{\left(1+{e}^{âˆ’\mathrm{Î»}t}\right)}^{âˆ’\mathrm{Î¼}}{x}^{lâˆ’m}\\ \phantom{\rule{1em}{0ex}}=\underset{l=0}{\overset{n}{âˆ‘}}{S}_{1}\left(n,l\right){\left(âˆ’1\right)}^{l}\underset{m=0}{\overset{l}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{m}\left(\genfrac{}{}{0}{}{l}{m}\right)}{{\left(m+1\right)}^{k}}\underset{i=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\underset{j=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\underset{{j}_{1}+â‹¯+{j}_{i}=j}{âˆ‘}{2}^{âˆ’\mathrm{Î¼}âˆ’i}\\ \phantom{\rule{2em}{0ex}}Ã—\left(\genfrac{}{}{0}{}{âˆ’\mathrm{Î¼}}{i}\right)\left(\genfrac{}{}{0}{}{j+i}{{j}_{1}+1,â€¦,{j}_{i}+1}\right)\frac{{\left(âˆ’\mathrm{Î»}\right)}^{j+i}}{\left(j+i\right)!}{t}^{j+i}{x}^{lâˆ’m}\\ \phantom{\rule{1em}{0ex}}=\underset{l=0}{\overset{n}{âˆ‘}}\underset{m=0}{\overset{l}{âˆ‘}}\underset{i=0}{\overset{lâˆ’m}{âˆ‘}}\underset{r=0}{\overset{lâˆ’mâˆ’i}{âˆ‘}}\underset{{j}_{1}+â‹¯+{j}_{i}=lâˆ’mâˆ’iâˆ’r}{âˆ‘}{\left(âˆ’1\right)}^{r}\frac{{2}^{âˆ’\mathrm{Î¼}âˆ’i}{\mathrm{Î»}}^{lâˆ’mâˆ’r}}{{\left(m+1\right)}^{k}}\\ \phantom{\rule{2em}{0ex}}Ã—\left(\genfrac{}{}{0}{}{l}{m}\right)\left(\genfrac{}{}{0}{}{âˆ’\mathrm{Î¼}}{i}\right)\left(\genfrac{}{}{0}{}{lâˆ’mâˆ’r}{{j}_{1}+1,â€¦,{j}_{i}+1}\right)\left(\genfrac{}{}{0}{}{lâˆ’m}{r}\right){S}_{1}\left(n,l\right){x}^{r}\\ \phantom{\rule{1em}{0ex}}={2}^{âˆ’\mathrm{Î¼}}\underset{r=0}{\overset{n}{âˆ‘}}{\left(âˆ’{\mathrm{Î»}}^{âˆ’1}\right)}^{r}\left\{\underset{l=r}{\overset{n}{âˆ‘}}\underset{m=0}{\overset{lâˆ’r}{âˆ‘}}\underset{i=0}{\overset{lâˆ’râˆ’m}{âˆ‘}}\underset{{j}_{1}+â‹¯+{j}_{i}=lâˆ’râˆ’mâˆ’i}{âˆ‘}\frac{{2}^{âˆ’i}{\mathrm{Î»}}^{lâˆ’m}}{{\left(m+1\right)}^{k}}\\ \phantom{\rule{2em}{0ex}}Ã—\left(\genfrac{}{}{0}{}{l}{m}\right)\left(\genfrac{}{}{0}{}{âˆ’\mathrm{Î¼}}{i}\right)\left(\genfrac{}{}{0}{}{lâˆ’mâˆ’r}{{j}_{1}+1,â€¦,{j}_{i}+1}\right)\left(\genfrac{}{}{0}{}{lâˆ’m}{r}\right){S}_{1}\left(n,l\right)\right\}{x}^{r}.\end{array}$
(48)

Therefore, by (48), we obtain the following theorem.

Theorem 5 For $nâ‰¥0$, we have

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}={2}^{âˆ’\mathrm{Î¼}}\underset{r=0}{\overset{n}{âˆ‘}}{\left(âˆ’{\mathrm{Î»}}^{âˆ’1}\right)}^{r}\left\{\underset{l=r}{\overset{n}{âˆ‘}}\underset{m=0}{\overset{lâˆ’r}{âˆ‘}}\underset{i=0}{\overset{lâˆ’râˆ’m}{âˆ‘}}\underset{{j}_{1}+â‹¯+{j}_{i}=lâˆ’râˆ’mâˆ’i}{âˆ‘}\frac{{2}^{âˆ’i}{\mathrm{Î»}}^{lâˆ’m}}{{\left(m+1\right)}^{k}}\\ \phantom{\rule{2em}{0ex}}Ã—\left(\genfrac{}{}{0}{}{l}{m}\right)\left(\genfrac{}{}{0}{}{âˆ’\mathrm{Î¼}}{i}\right)\left(\genfrac{}{}{0}{}{lâˆ’mâˆ’r}{{j}_{1}+1,â€¦,{j}_{i}+1}\right)\left(\genfrac{}{}{0}{}{lâˆ’m}{r}\right){S}_{1}\left(n,l\right)\right\}{x}^{r}.\end{array}$

It is easy to see that

${\left(1+{e}^{\mathrm{Î»}t}\right)}^{\mathrm{Î¼}}\frac{1}{{Lif}_{k}\left(âˆ’t\right)}\stackrel{Ë†}{C}{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)âˆ¼\left(1,{e}^{t}âˆ’1\right)$
(49)

and

${\left(x\right)}_{n}=x\left(xâˆ’1\right)â‹¯\left(xâˆ’n+1\right)=\underset{l=0}{\overset{n}{âˆ‘}}{S}_{1}\left(n,l\right){x}^{l}âˆ¼\left(1,{e}^{t}âˆ’1\right).$
(50)

By the same method as Theorem 5, we get

$\begin{array}{r}\stackrel{Ë†}{C}{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}={2}^{âˆ’\mathrm{Î¼}}\underset{r=0}{\overset{n}{âˆ‘}}{\mathrm{Î»}}^{âˆ’r}\left\{\underset{l=r}{\overset{n}{âˆ‘}}\underset{m=0}{\overset{lâˆ’r}{âˆ‘}}\underset{i=0}{\overset{lâˆ’râˆ’m}{âˆ‘}}\underset{{j}_{1}+â‹¯+{j}_{i}=lâˆ’râˆ’mâˆ’i}{âˆ‘}\frac{{\left(âˆ’1\right)}^{m}{2}^{âˆ’i}{\mathrm{Î»}}^{lâˆ’m}}{{\left(m+1\right)}^{k}}\\ \phantom{\rule{2em}{0ex}}Ã—\left(\genfrac{}{}{0}{}{l}{m}\right)\left(\genfrac{}{}{0}{}{âˆ’\mathrm{Î¼}}{i}\right)\left(\genfrac{}{}{0}{}{lâˆ’mâˆ’r}{{j}_{1}+1,â€¦,{j}_{i}+1}\right)\left(\genfrac{}{}{0}{}{lâˆ’m}{r}\right){S}_{1}\left(n,l\right)\right\}{x}^{r}.\end{array}$
(51)

From (20) and (28), we have

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=\underset{j=0}{\overset{n}{âˆ‘}}\frac{1}{j!}ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(log\left(1+t\right)\right){\left(âˆ’log\left(1+t\right)\right)}^{j}âˆ£{x}^{n}ã€‰{x}^{j}.\end{array}$
(52)

Now, we observe that

$\begin{array}{r}ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(log\left(1+t\right)\right){\left(âˆ’log\left(1+t\right)\right)}^{j}âˆ£{x}^{n}ã€‰\\ \phantom{\rule{1em}{0ex}}={\left(âˆ’1\right)}^{j}ã€ˆlog{\left(1+t\right)}^{j}|\underset{m=0}{\overset{\mathrm{âˆž}}{âˆ‘}}C{P}_{m}^{\left(k\right)}\left(0;\mathrm{Î»},\mathrm{Î¼}\right)\frac{{t}^{m}}{m!}{x}^{n}ã€‰\\ \phantom{\rule{1em}{0ex}}={\left(âˆ’1\right)}^{j}\underset{m=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{m}\right)C{P}_{m}^{\left(k\right)}\left(0;\mathrm{Î»},\mathrm{Î¼}\right)ã€ˆ{\left(log\left(1+t\right)\right)}^{j}âˆ£{x}^{nâˆ’m}ã€‰\\ \phantom{\rule{1em}{0ex}}={\left(âˆ’1\right)}^{j}\underset{m=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{m}\right)C{P}_{m}^{\left(k\right)}\left(0;\mathrm{Î»},\mathrm{Î¼}\right)j!{S}_{1}\left(nâˆ’m,j\right).\end{array}$
(53)

Therefore, by (52) and (53), we obtain the following theorem.

Theorem 6 For $nâ‰¥0$, we have

$C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)=\underset{j=0}{\overset{n}{âˆ‘}}{\left(âˆ’1\right)}^{j}\left\{\underset{m=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{m}\right){S}_{1}\left(nâˆ’m,j\right)C{P}_{m}^{\left(k\right)}\left(0;\mathrm{Î»},\mathrm{Î¼}\right)\right\}{x}^{j}.$

Remark By the same method as Theorem 6, we get

$\stackrel{Ë†}{C}{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)=\underset{j=0}{\overset{n}{âˆ‘}}\left\{\underset{m=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{m}\right){S}_{1}\left(nâˆ’m,j\right)\stackrel{Ë†}{C}{P}_{n}^{\left(k\right)}\left(0;\mathrm{Î»},\mathrm{Î¼}\right)\right\}{x}^{j}.$
(54)

From (21), we have

$C{P}_{n}^{\left(k\right)}\left(x+y;\mathrm{Î»},\mathrm{Î¼}\right)=\underset{j=0}{\overset{n}{âˆ‘}}{\left(âˆ’1\right)}^{j}\left(\genfrac{}{}{0}{}{n}{j}\right)C{P}_{nâˆ’j}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right){y}^{\left(j\right)}$
(55)

and

$\stackrel{Ë†}{C}{P}_{n}^{\left(k\right)}\left(x+y;\mathrm{Î»},\mathrm{Î¼}\right)=\underset{j=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{j}\right)\stackrel{Ë†}{C}{P}_{nâˆ’j}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right){\left(y\right)}_{j}.$
(56)

By (22) and (28), we get

$\left({e}^{âˆ’t}âˆ’1\right)C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)=nC{P}_{nâˆ’1}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)$
(57)

and

$\left({e}^{âˆ’t}âˆ’1\right)C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)=C{P}_{n}^{\left(k\right)}\left(xâˆ’1;\mathrm{Î»},\mathrm{Î¼}\right)âˆ’C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right).$
(58)

Therefore, by (57) and (58), we obtain the following theorem.

Theorem 7 For $nâ‰¥0$, we have

$C{P}_{n}^{\left(k\right)}\left(xâˆ’1;\mathrm{Î»},\mathrm{Î¼}\right)âˆ’C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)=nC{P}_{nâˆ’1}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right).$

Remark By the same method as Theorem 7, we get

$\stackrel{Ë†}{C}{P}_{n}^{\left(k\right)}\left(x+1;\mathrm{Î»},\mathrm{Î¼}\right)âˆ’\stackrel{Ë†}{C}{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)=n\stackrel{Ë†}{C}{P}_{nâˆ’1}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right).$
(59)

From (22), (28), and (29), we have

$\begin{array}{rl}C{P}_{n+1}^{\left(k\right)}\left(x;1,\mathrm{Î¼}\right)=& âˆ’xC{P}_{n}^{\left(k\right)}\left(x+1;1,\mathrm{Î¼}\right)+\mathrm{Î¼}\underset{m=0}{\overset{n}{âˆ‘}}{\left(âˆ’\frac{1}{2}\right)}^{m+1}{\left(n\right)}_{m}C{P}_{nâˆ’m}^{\left(k\right)}\left(x;1,\mathrm{Î¼}\right)\\ +{2}^{âˆ’\mathrm{Î¼}}\underset{r=0}{\overset{n}{âˆ‘}}{\left(âˆ’1\right)}^{r}\left\{\underset{m=r}{\overset{n}{âˆ‘}}\underset{l=r}{\overset{m}{âˆ‘}}\underset{i=0}{\overset{lâˆ’r}{âˆ‘}}\underset{{j}_{1}+â‹¯+{j}_{i}=lâˆ’iâˆ’r}{âˆ‘}\frac{{2}^{âˆ’i}}{{\left(mâˆ’l+2\right)}^{k}}\left(\genfrac{}{}{0}{}{m}{l}\right)\\ Ã—\left(\genfrac{}{}{0}{}{âˆ’\mathrm{Î¼}}{i}\right)\left(\genfrac{}{}{0}{}{lâˆ’r}{{j}_{1}+1,â€¦,{j}_{i}+1}\right)\left(\genfrac{}{}{0}{}{l}{r}\right){S}_{1}\left(n,m\right)\right\}{\left(x+1\right)}^{r}\end{array}$
(60)

and

$\begin{array}{r}\stackrel{Ë†}{C}{P}_{n+1}^{\left(k\right)}\left(x;1,\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=x\stackrel{Ë†}{C}{P}_{n}^{\left(k\right)}\left(xâˆ’1;1,\mathrm{Î¼}\right)+\mathrm{Î¼}\underset{m=0}{\overset{n}{âˆ‘}}{\left(âˆ’\frac{1}{2}\right)}^{m+1}{\left(n\right)}_{m}\stackrel{Ë†}{C}{P}_{nâˆ’m}^{\left(k\right)}\left(x;1,\mathrm{Î¼}\right)\\ \phantom{\rule{2em}{0ex}}âˆ’{2}^{âˆ’\mathrm{Î¼}}\underset{r=0}{\overset{n}{âˆ‘}}\left\{\underset{m=r}{\overset{n}{âˆ‘}}\underset{l=r}{\overset{m}{âˆ‘}}\underset{i=0}{\overset{lâˆ’r}{âˆ‘}}\underset{{j}_{1}+â‹¯+{j}_{i}=lâˆ’iâˆ’r}{âˆ‘}\frac{{\left(âˆ’1\right)}^{mâˆ’l}{2}^{âˆ’i}}{{\left(mâˆ’l+2\right)}^{k}}\left(\genfrac{}{}{0}{}{m}{l}\right)\left(\genfrac{}{}{0}{}{âˆ’\mathrm{Î¼}}{i}\right)\\ \phantom{\rule{2em}{0ex}}Ã—\left(\genfrac{}{}{0}{}{lâˆ’r}{{j}_{1}+1,â€¦,{j}_{i}+1}\right)\left(\genfrac{}{}{0}{}{l}{r}\right){S}_{1}\left(n,m\right)\right\}{\left(xâˆ’1\right)}^{r}.\end{array}$
(61)

By (14) and (27), we get

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(y;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=ã€ˆ\underset{l=0}{\overset{\mathrm{âˆž}}{âˆ‘}}C{P}_{l}^{\left(k\right)}\left(y;\mathrm{Î»},\mathrm{Î¼}\right)\frac{{t}^{l}}{l!}|{x}^{n}ã€‰\\ \phantom{\rule{1em}{0ex}}=ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(log\left(1+t\right)\right){\left(1+t\right)}^{âˆ’y}âˆ£xâ‹\dots {x}^{nâˆ’1}ã€‰\\ \phantom{\rule{1em}{0ex}}=ã€ˆ{\mathrm{âˆ‚}}_{t}\left({\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(log\left(1+t\right)\right){\left(1+t\right)}^{âˆ’y}\right)âˆ£{x}^{nâˆ’1}ã€‰\\ \phantom{\rule{1em}{0ex}}=ã€ˆ\left({\mathrm{âˆ‚}}_{t}{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}\right){Lif}_{k}\left(log\left(1+t\right)\right){\left(1+t\right)}^{âˆ’y}âˆ£{x}^{nâˆ’1}ã€‰\\ \phantom{\rule{2em}{0ex}}+ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}\left({\mathrm{âˆ‚}}_{t}{Lif}_{k}\left(log\left(1+t\right)\right)\right){\left(1+t\right)}^{âˆ’y}âˆ£{x}^{nâˆ’1}ã€‰\\ \phantom{\rule{2em}{0ex}}+ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(log\left(1+t\right)\right)\left({\mathrm{âˆ‚}}_{t}{\left(1+t\right)}^{âˆ’y}\right)âˆ£{x}^{nâˆ’1}ã€‰\\ \phantom{\rule{1em}{0ex}}=âˆ’\mathrm{Î¼}\mathrm{Î»}ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}âˆ’1}{Lif}_{k}\left(log\left(1+t\right)\right){\left(1+t\right)}^{âˆ’\left(yâˆ’\mathrm{Î»}+1\right)}âˆ£{x}^{nâˆ’1}ã€‰\\ \phantom{\rule{2em}{0ex}}âˆ’yã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}\left({Lif}_{k}\left(log\left(1+t\right)\right)\right){\left(1+t\right)}^{âˆ’yâˆ’1}âˆ£{x}^{nâˆ’1}ã€‰\\ \phantom{\rule{2em}{0ex}}+ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}\left({\mathrm{âˆ‚}}_{t}{Lif}_{k}\left(log\left(1+t\right)\right)\right){\left(1+t\right)}^{âˆ’y}âˆ£{x}^{nâˆ’1}ã€‰\\ \phantom{\rule{1em}{0ex}}=âˆ’\mathrm{Î¼}\mathrm{Î»}C{P}_{nâˆ’1}^{\left(k\right)}\left(yâˆ’\mathrm{Î»}+1;\mathrm{Î»},\mathrm{Î¼}+1\right)âˆ’yC{P}_{nâˆ’1}^{\left(k\right)}\left(y+1;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{2em}{0ex}}+ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}\frac{{Lif}_{kâˆ’1}\left(log\left(1+t\right)\right)âˆ’{Lif}_{k}\left(log\left(1+t\right)\right)}{\left(1+t\right)log\left(1+t\right)}{\left(1+t\right)}^{âˆ’y}âˆ£{x}^{nâˆ’1}ã€‰.\end{array}$
(62)

Now, we observe that

$\begin{array}{r}ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}\frac{{Lif}_{kâˆ’1}\left(log\left(1+t\right)\right)âˆ’{Lif}_{k}\left(log\left(1+t\right)\right)}{\left(1+t\right)log\left(1+t\right)}{\left(1+t\right)}^{âˆ’y}|{x}^{nâˆ’1}ã€‰\\ \phantom{\rule{1em}{0ex}}=ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}\frac{{Lif}_{kâˆ’1}\left(log\left(1+t\right)\right)âˆ’{Lif}_{k}\left(log\left(1+t\right)\right)}{t}{\left(1+t\right)}^{âˆ’y}|\\ \phantom{\rule{2em}{0ex}}\frac{t}{\left(1+t\right)log\left(1+t\right)}{x}^{nâˆ’1}ã€‰\\ \phantom{\rule{1em}{0ex}}=\underset{l=0}{\overset{nâˆ’1}{âˆ‘}}\left(\genfrac{}{}{0}{}{nâˆ’1}{l}\right){\stackrel{Ë†}{\mathbb{C}}}_{nâˆ’1âˆ’l}^{\left(1\right)}\left(0\right)\\ \phantom{\rule{2em}{0ex}}Ã—ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}\frac{{Lif}_{kâˆ’1}\left(log\left(1+t\right)\right)âˆ’{Lif}_{k}\left(log\left(1+t\right)\right)}{t}{\left(1+t\right)}^{âˆ’y}|{x}^{l}ã€‰\\ \phantom{\rule{1em}{0ex}}=\underset{l=0}{\overset{nâˆ’1}{âˆ‘}}\left(\genfrac{}{}{0}{}{nâˆ’1}{l}\right){\stackrel{Ë†}{\mathbb{C}}}_{nâˆ’1âˆ’l}^{\left(1\right)}\left(0\right)\\ \phantom{\rule{2em}{0ex}}Ã—ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}\frac{{Lif}_{kâˆ’1}\left(log\left(1+t\right)\right)âˆ’{Lif}_{k}\left(log\left(1+t\right)\right)}{t}{\left(1+t\right)}^{âˆ’y}|t\frac{{x}^{l+1}}{l+1}ã€‰\\ \phantom{\rule{1em}{0ex}}=\frac{1}{n}\underset{l=0}{\overset{nâˆ’1}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l+1}\right){\stackrel{Ë†}{\mathbb{C}}}_{nâˆ’1âˆ’l}^{\left(1\right)}\left(0\right)\left\{C{P}_{l+1}^{\left(kâˆ’1\right)}\left(y;\mathrm{Î»},\mathrm{Î¼}\right)âˆ’C{P}_{l+1}^{\left(k\right)}\left(y;\mathrm{Î»},\mathrm{Î¼}\right)\right\}.\end{array}$
(63)

Therefore, by (62) and (63), we obtain the following theorem.

Theorem 8 For $nâ‰¥0$, we have

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=âˆ’\mathrm{Î¼}\mathrm{Î»}C{P}_{nâˆ’1}^{\left(k\right)}\left(xâˆ’\mathrm{Î»}+1;\mathrm{Î»},\mathrm{Î¼}+1\right)âˆ’xC{P}_{nâˆ’1}^{\left(k\right)}\left(x+1;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{2em}{0ex}}+\frac{1}{n}\underset{l=0}{\overset{nâˆ’1}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l+1}\right){\stackrel{Ë†}{C}}_{nâˆ’1âˆ’l}\left\{C{P}_{l+1}^{\left(kâˆ’1\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)âˆ’C{P}_{l+1}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\right\},\end{array}$

where ${\stackrel{Ë†}{C}}_{nâˆ’1âˆ’l}={\stackrel{Ë†}{\mathbb{C}}}_{nâˆ’1âˆ’l}^{\left(1\right)}\left(0\right)$.

Remark By the same method as Theorem 8, we get

$\begin{array}{r}\stackrel{Ë†}{C}{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=âˆ’\mathrm{Î¼}\mathrm{Î»}\stackrel{Ë†}{C}{P}_{nâˆ’1}^{\left(k\right)}\left(x+\mathrm{Î»}âˆ’1;\mathrm{Î»},\mathrm{Î¼}+1\right)+x\stackrel{Ë†}{C}{P}_{nâˆ’1}^{\left(k\right)}\left(xâˆ’1;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{2em}{0ex}}+\frac{1}{n}\underset{l=0}{\overset{nâˆ’1}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l+1}\right){\stackrel{Ë†}{C}}_{nâˆ’1âˆ’l}\left(\stackrel{Ë†}{C}{P}_{l+1}^{\left(kâˆ’1\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)âˆ’\stackrel{Ë†}{C}{P}_{l+1}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\right).\end{array}$
(64)

By (23), we get

$\begin{array}{r}\frac{d}{dx}C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=\underset{l=0}{\overset{nâˆ’1}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right)ã€ˆâˆ’log\left(1+t\right)âˆ£{x}^{nâˆ’l}ã€‰C{P}_{l}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=\underset{l=0}{\overset{nâˆ’1}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right)ã€ˆ\underset{m=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{m}}{m}{t}^{m}|{x}^{nâˆ’l}ã€‰C{P}_{l}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=\underset{l=0}{\overset{nâˆ’1}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right)\underset{m=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{m}}{m}ã€ˆ{t}^{m}âˆ£{x}^{nâˆ’l}ã€‰C{P}_{l}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=\underset{l=0}{\overset{nâˆ’1}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){\left(âˆ’1\right)}^{nâˆ’l}C{P}_{l}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\left(nâˆ’lâˆ’1\right)!\\ \phantom{\rule{1em}{0ex}}=n!\underset{l=0}{\overset{nâˆ’1}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{nâˆ’l}}{\left(nâˆ’l\right)l!}C{P}_{l}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right).\end{array}$
(65)

By the same method as (65), we get

$\begin{array}{r}\frac{d}{dx}\stackrel{Ë†}{C}{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=n!\underset{l=0}{\overset{nâˆ’1}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{nâˆ’lâˆ’1}}{\left(nâˆ’l\right)l!}\stackrel{Ë†}{C}{P}_{l}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right).\end{array}$
(66)

Now, we compute the following equation in two different ways:

$ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(âˆ’log\left(1+t\right)\right){\left(log\left(1+t\right)\right)}^{m}âˆ£{x}^{n}ã€‰.$

On the one hand,

$\begin{array}{r}ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(âˆ’log\left(1+t\right)\right){\left(log\left(1+t\right)\right)}^{m}âˆ£{x}^{n}ã€‰\\ \phantom{\rule{1em}{0ex}}=ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(âˆ’log\left(1+t\right)\right)âˆ£{\left(log\left(1+t\right)\right)}^{m}{x}^{n}ã€‰\\ \phantom{\rule{1em}{0ex}}=\underset{l=0}{\overset{nâˆ’m}{âˆ‘}}m!\left(\genfrac{}{}{0}{}{n}{l+m}\right){S}_{1}\left(l+m,m\right)ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(âˆ’log\left(1+t\right)\right)âˆ£{x}^{nâˆ’lâˆ’m}ã€‰\\ \phantom{\rule{1em}{0ex}}=\underset{l=0}{\overset{nâˆ’m}{âˆ‘}}m!\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(nâˆ’l,m\right)\stackrel{Ë†}{C}{P}_{l}^{\left(k\right)}\left(0;\mathrm{Î»},\mathrm{Î¼}\right).\end{array}$
(67)

On the other hand,

$\begin{array}{r}ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(âˆ’log\left(1+t\right)\right){\left(log\left(1+t\right)\right)}^{m}âˆ£{x}^{n}ã€‰\\ \phantom{\rule{1em}{0ex}}=ã€ˆ{\mathrm{âˆ‚}}_{t}\left({\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(âˆ’log\left(1+t\right)\right){\left(log\left(1+t\right)\right)}^{m}\right)âˆ£{x}^{nâˆ’1}ã€‰\\ \phantom{\rule{1em}{0ex}}=ã€ˆ\left({\mathrm{âˆ‚}}_{t}{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}\right){Lif}_{k}\left(âˆ’log\left(1+t\right)\right){\left(log\left(1+t\right)\right)}^{m}âˆ£{x}^{nâˆ’1}ã€‰\\ \phantom{\rule{2em}{0ex}}+ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}\left({\mathrm{âˆ‚}}_{t}{Lif}_{k}\left(âˆ’log\left(1+t\right)\right)\right){\left(log\left(1+t\right)\right)}^{m}âˆ£{x}^{nâˆ’1}ã€‰\\ \phantom{\rule{2em}{0ex}}+ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(âˆ’log\left(1+t\right)\right)\left({\mathrm{âˆ‚}}_{t}{\left(log\left(1+t\right)\right)}^{m}\right)âˆ£{x}^{nâˆ’1}ã€‰.\end{array}$
(68)

Therefore, by (67) and (68), we obtain the following theorem.

Theorem 9 For $nâˆˆ\mathbb{N}$ with $nâ‰¥2$, let $nâˆ’1â‰¥mâ‰¥1$. Then we have

$\begin{array}{r}m\underset{l=0}{\overset{nâˆ’m}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(nâˆ’l,m\right)\stackrel{Ë†}{C}{P}_{l}^{\left(k\right)}\left(0;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=âˆ’\mathrm{Î¼}\mathrm{Î»}m\underset{l=0}{\overset{nâˆ’1âˆ’m}{âˆ‘}}\left(\genfrac{}{}{0}{}{nâˆ’1}{l}\right){S}_{1}\left(nâˆ’1âˆ’l,m\right)\stackrel{Ë†}{C}{P}_{l}^{\left(k\right)}\left(\mathrm{Î»}âˆ’1;\mathrm{Î»},\mathrm{Î¼}+1\right)\\ \phantom{\rule{2em}{0ex}}+\underset{l=0}{\overset{nâˆ’m}{âˆ‘}}\left(\genfrac{}{}{0}{}{nâˆ’1}{l}\right){S}_{1}\left(nâˆ’1âˆ’l,mâˆ’1\right)\stackrel{Ë†}{C}{P}_{l}^{\left(kâˆ’1\right)}\left(âˆ’1;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{2em}{0ex}}+\left(mâˆ’1\right)\underset{l=0}{\overset{nâˆ’m}{âˆ‘}}\left(\genfrac{}{}{0}{}{nâˆ’1}{l}\right){S}_{1}\left(nâˆ’1âˆ’l,mâˆ’1\right)\stackrel{Ë†}{C}{P}_{l}^{\left(k\right)}\left(âˆ’1;\mathrm{Î»},\mathrm{Î¼}\right).\end{array}$

Remark By the same method as Theorem 9, we get

$\begin{array}{r}m\underset{l=0}{\overset{nâˆ’m}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(nâˆ’l,m\right)C{P}_{l}^{\left(k\right)}\left(0;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=âˆ’\mathrm{Î¼}\mathrm{Î»}m\underset{l=0}{\overset{nâˆ’1âˆ’m}{âˆ‘}}\left(\genfrac{}{}{0}{}{nâˆ’1}{l}\right){S}_{1}\left(nâˆ’1âˆ’l,m\right)C{P}_{l}^{\left(k\right)}\left(1âˆ’\mathrm{Î»};\mathrm{Î»},\mathrm{Î¼}+1\right)\\ \phantom{\rule{2em}{0ex}}+\underset{l=0}{\overset{nâˆ’m}{âˆ‘}}\left(\genfrac{}{}{0}{}{nâˆ’1}{l}\right){S}_{1}\left(nâˆ’1âˆ’l,mâˆ’1\right)C{P}_{l}^{\left(kâˆ’1\right)}\left(1;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{2em}{0ex}}+\left(mâˆ’1\right)\underset{l=0}{\overset{nâˆ’m}{âˆ‘}}\left(\genfrac{}{}{0}{}{nâˆ’1}{l}\right){S}_{1}\left(nâˆ’1âˆ’l,mâˆ’1\right)C{P}_{l}^{\left(k\right)}\left(âˆ’1;\mathrm{Î»},\mathrm{Î¼}\right),\end{array}$

where $nâˆ’1â‰¥mâ‰¥1$.

Let us consider the following two Sheffer sequences:

$C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)âˆ¼\left({\left(1+{e}^{âˆ’\mathrm{Î»}t}\right)}^{\mathrm{Î¼}}\frac{1}{{Lif}_{k}\left(âˆ’t\right)},{e}^{âˆ’t}âˆ’1\right)$
(69)

and

${B}_{n}^{\left(s\right)}\left(x\right)âˆ¼\left({\left(\frac{{e}^{t}âˆ’1}{t}\right)}^{s},t\right)\phantom{\rule{1em}{0ex}}\left(sâˆˆ{\mathbb{Z}}_{â‰¥0}\right).$
(70)

Let

$C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)=\underset{m=0}{\overset{n}{âˆ‘}}{C}_{n,m}{B}_{m}^{\left(s\right)}\left(x\right).$
(71)

Then, by (26), we get

$\begin{array}{rl}{C}_{n,m}=& \frac{1}{m!}ã€ˆ\frac{{\left(\frac{{e}^{âˆ’log\left(1+t\right)}âˆ’1}{âˆ’log\left(1+t\right)}\right)}^{s}}{{\left(1+{e}^{\mathrm{Î»}log\left(1+t\right)}\right)}^{\mathrm{Î¼}}}{Lif}_{k}\left(log\left(1+t\right)\right){\left(âˆ’log\left(1+t\right)\right)}^{m}|{x}^{n}ã€‰\\ =& \frac{{\left(âˆ’1\right)}^{m}}{m!}ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(log\left(1+t\right)\right)\\ Ã—{\left(1+t\right)}^{âˆ’s}{\left(\frac{t}{log\left(1+t\right)}\right)}^{s}|{\left(log\left(1+t\right)\right)}^{m}{x}^{n}ã€‰\\ =& \frac{{\left(âˆ’1\right)}^{m}}{m!}\underset{l=0}{\overset{nâˆ’m}{âˆ‘}}m!\left(\genfrac{}{}{0}{}{n}{l+m}\right){S}_{1}\left(l+m,m\right)\underset{i=0}{\overset{nâˆ’lâˆ’m}{âˆ‘}}\left(\genfrac{}{}{0}{}{nâˆ’lâˆ’m}{i}\right){\mathbb{C}}_{i}^{\left(s\right)}\\ Ã—ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(log\left(1+t\right)\right){\left(1+t\right)}^{âˆ’s}âˆ£{x}^{nâˆ’lâˆ’mâˆ’i}ã€‰\\ =& {\left(âˆ’1\right)}^{m}\underset{l=0}{\overset{nâˆ’m}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(nâˆ’l,m\right)\underset{i=0}{\overset{l}{âˆ‘}}\left(\genfrac{}{}{0}{}{l}{i}\right){\mathbb{C}}_{i}^{\left(s\right)}C{P}_{lâˆ’i}^{\left(k\right)}\left(s;\mathrm{Î»},\mathrm{Î¼}\right).\end{array}$
(72)

Therefore, by (71) and (72), we obtain the following theorem.

Theorem 10 For $nâ‰¥0$, we have

$C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)=\underset{m=0}{\overset{n}{âˆ‘}}{\left(âˆ’1\right)}^{m}\left\{\underset{l=0}{\overset{nâˆ’m}{âˆ‘}}\underset{i=0}{\overset{l}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\genfrac{}{}{0}{}{l}{i}\right){S}_{1}\left(nâˆ’l,m\right){\mathbb{C}}_{i}^{\left(s\right)}C{P}_{lâˆ’i}^{\left(k\right)}\left(s;\mathrm{Î»},\mathrm{Î¼}\right)\right\}{B}_{n}^{\left(s\right)}\left(x\right).$

Remark By the same method as Theorem 10, we have

$\begin{array}{r}\stackrel{Ë†}{C}{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=\underset{m=0}{\overset{n}{âˆ‘}}\left\{\underset{l=0}{\overset{nâˆ’m}{âˆ‘}}\underset{i=0}{\overset{l}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\genfrac{}{}{0}{}{l}{i}\right){S}_{1}\left(nâˆ’l,m\right){\stackrel{Ë†}{\mathbb{C}}}_{i}^{\left(s\right)}\stackrel{Ë†}{C}{P}_{lâˆ’i}^{\left(k\right)}\left(s;\mathrm{Î»},\mathrm{Î¼}\right)\right\}{B}_{m}^{\left(s\right)}\left(x\right).\end{array}$
(73)

For $C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)âˆ¼\left({\left(1+{e}^{âˆ’\mathrm{Î»}t}\right)}^{\mathrm{Î¼}}\frac{1}{{Lif}_{k}\left(âˆ’t\right)},{e}^{âˆ’t}âˆ’1\right)$, ${H}_{n}^{\left(s\right)}\left(xâˆ£\mathrm{Î»}\right)âˆ¼\left({\left(\frac{{e}^{t}âˆ’\mathrm{Î»}}{1âˆ’\mathrm{Î»}}\right)}^{s},t\right)$, $sâˆˆ{\mathbb{Z}}_{â‰¥0}$, $\mathrm{Î»}âˆˆ\mathbb{C}$ with , let us assume that

$C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)=\underset{m=0}{\overset{n}{âˆ‘}}{C}_{n,m}{H}_{m}^{\left(s\right)}\left(x;\mathrm{Î»}\right).$
(74)

From (26), we have

$\begin{array}{rl}{C}_{n,m}=& \frac{{\left(âˆ’1\right)}^{m}}{m!}ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(log\left(1+t\right)\right)\\ Ã—{\left(1+t\right)}^{âˆ’s}{\left(1+\frac{\mathrm{Î»}}{\mathrm{Î»}âˆ’1}t\right)}^{s}|{\left(log\left(1+t\right)\right)}^{m}{x}^{n}ã€‰\\ =& \frac{{\left(âˆ’1\right)}^{m}}{m!}\underset{l=0}{\overset{nâˆ’m}{âˆ‘}}m!\left(\genfrac{}{}{0}{}{n}{l+m}\right){S}_{1}\left(l+m,m\right)\underset{i=0}{\overset{min\left\{s,nâˆ’lâˆ’m\right\}}{âˆ‘}}\left(\genfrac{}{}{0}{}{s}{i}\right){\left(\frac{\mathrm{Î»}}{\mathrm{Î»}âˆ’1}\right)}^{i}\\ Ã—ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(log\left(1+t\right)\right){\left(1+t\right)}^{âˆ’s}âˆ£{t}^{i}{x}^{nâˆ’lâˆ’m}ã€‰\\ =& {\left(âˆ’1\right)}^{m}\underset{l=0}{\overset{nâˆ’m}{âˆ‘}}\underset{i=0}{\overset{min\left\{s,nâˆ’lâˆ’m\right\}}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l+m}\right)\left(\genfrac{}{}{0}{}{s}{i}\right)\\ Ã—{\left(nâˆ’lâˆ’m\right)}_{i}{\left(\frac{\mathrm{Î»}}{\mathrm{Î»}âˆ’1}\right)}^{i}{S}_{1}\left(l+m,m\right)C{P}_{nâˆ’lâˆ’mâˆ’i}^{\left(k\right)}\left(s;\mathrm{Î»},\mathrm{Î¼}\right)\\ =& {\left(âˆ’1\right)}^{m}\underset{l=0}{\overset{nâˆ’m}{âˆ‘}}\underset{i=0}{\overset{min\left\{s,l\right\}}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\genfrac{}{}{0}{}{s}{i}\right){\left(l\right)}_{i}{\left(\frac{\mathrm{Î»}}{\mathrm{Î»}âˆ’1}\right)}^{i}{S}_{1}\left(nâˆ’l,m\right)C{P}_{lâˆ’i}^{\left(k\right)}\left(s;\mathrm{Î»},\mathrm{Î¼}\right).\end{array}$
(75)

Therefore, by (75) and (76), we obtain the following theorem.

Theorem 11 For $\mathrm{Î»}âˆˆ\mathbb{C}$ with , $nâ‰¥0$, we have

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=\underset{m=0}{\overset{n}{âˆ‘}}{\left(âˆ’1\right)}^{m}\left\{\underset{l=0}{\overset{nâˆ’m}{âˆ‘}}\underset{i=0}{\overset{min\left\{s,l\right\}}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\genfrac{}{}{0}{}{s}{i}\right){\left(l\right)}_{i}\\ \phantom{\rule{2em}{0ex}}â‹\dots {\left(\frac{\mathrm{Î»}}{\mathrm{Î»}âˆ’1}\right)}^{i}{S}_{1}\left(nâˆ’l,m\right)C{P}_{lâˆ’i}^{\left(k\right)}\left(s;\mathrm{Î»},\mathrm{Î¼}\right)\right\}{H}_{m}^{\left(s\right)}\left(x;\mathrm{Î»}\right).\end{array}$

Remark By the same method as Theorem 11, we get

$\begin{array}{r}\stackrel{Ë†}{C}{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)\\ \phantom{\rule{1em}{0ex}}=\underset{m=0}{\overset{n}{âˆ‘}}\left\{\underset{l=0}{\overset{nâˆ’m}{âˆ‘}}\underset{i=0}{\overset{min\left\{s,l\right\}}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\genfrac{}{}{0}{}{s}{i}\right){\left(l\right)}_{i}\\ \phantom{\rule{2em}{0ex}}â‹\dots {\left(\frac{1}{1âˆ’\mathrm{Î»}}\right)}^{i}{S}_{1}\left(nâˆ’l,m\right)\stackrel{Ë†}{C}{P}_{lâˆ’i}^{\left(k\right)}\left(0;\mathrm{Î»},\mathrm{Î¼}\right)\right\}{H}_{m}^{\left(s\right)}\left(x;\mathrm{Î»}\right).\end{array}$

For $C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)âˆ¼\left({\left(1+{e}^{âˆ’\mathrm{Î»}t}\right)}^{\mathrm{Î¼}}\frac{1}{{Lif}_{k}\left(âˆ’t\right)},{e}^{âˆ’t}âˆ’1\right)$ and ${x}^{\left(n\right)}âˆ¼\left(1,1âˆ’{e}^{âˆ’t}\right)$, let us assume that

$C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)=\underset{m=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{C}_{n,m}{x}^{\left(m\right)}.$
(76)

By (26), we get

$\begin{array}{rl}{C}_{n,m}=& \frac{1}{m!}ã€ˆ\frac{1}{{\left(1+{e}^{\mathrm{Î»}log\left(1+t\right)}\right)}^{\mathrm{Î¼}}}{Lif}_{k}\left(log\left(1+t\right)\right){\left(1âˆ’{e}^{log\left(1+t\right)}\right)}^{m}|{x}^{n}ã€‰\\ =& \frac{1}{m!}ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(log\left(1+t\right)\right){\left(âˆ’t\right)}^{m}âˆ£{x}^{n}ã€‰\\ =& \frac{{\left(âˆ’1\right)}^{m}}{m!}ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(log\left(1+t\right)\right)âˆ£{t}^{m}{x}^{n}ã€‰\\ =& {\left(âˆ’1\right)}^{m}\left(\genfrac{}{}{0}{}{n}{m}\right)ã€ˆ{\left(1+{\left(1+t\right)}^{\mathrm{Î»}}\right)}^{âˆ’\mathrm{Î¼}}{Lif}_{k}\left(log\left(1+t\right)\right)âˆ£{x}^{nâˆ’m}ã€‰\\ =& {\left(âˆ’1\right)}^{m}\left(\genfrac{}{}{0}{}{n}{m}\right)C{P}_{nâˆ’m}^{\left(k\right)}\left(0;\mathrm{Î»},\mathrm{Î¼}\right).\end{array}$
(77)

Therefore, by (77) and (78), we obtain the following theorem.

Theorem 12 For $nâ‰¥0$, we have

$C{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)=\underset{m=0}{\overset{n}{âˆ‘}}{\left(âˆ’1\right)}^{m}\left(\genfrac{}{}{0}{}{n}{m}\right)C{P}_{nâˆ’m}^{\left(k\right)}\left(0;\mathrm{Î»},\mathrm{Î¼}\right){x}^{\left(m\right)}.$

Remark By the same method as Theorem 12, we get

$\stackrel{Ë†}{C}{P}_{n}^{\left(k\right)}\left(x;\mathrm{Î»},\mathrm{Î¼}\right)=\underset{m=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{m}\right)\stackrel{Ë†}{C}{P}_{nâˆ’m}^{\left(k\right)}\left(0;\mathrm{Î»},\mathrm{Î¼}\right){\left(x\right)}_{m}.$
(78)

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## Acknowledgements

This work was supported by Kwangwoon University 2013.

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Correspondence to Taekyun Kim.

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

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Kim, D.S., Kim, T. Poly-Cauchy and Peters mixed-type polynomials. Adv Differ Equ 2014, 4 (2014). https://doi.org/10.1186/1687-1847-2014-4