1 Introduction
In this paper, we use umbral calculus techniques (see [1, 2]) to obtain several new and interesting identities of Barnes-type Daehee with λ-parameter and degenerate Euler mixed-type polynomials. To define the umbral calculus, let Π be the algebra of polynomials in a single variable x over \(\mathbb{C}\) and \(\Pi^{*}\) be the vector space of all linear functionals on Π. The action of a linear functional \(L\in\Pi^{*}\) on a polynomial \(p(x)\) is denoted by \(\langle L|p(x)\rangle \), and linearly extended as \(\langle cL+dL'|p(x)\rangle=c\langle L|p(x)\rangle+d\langle L'|p(x)\rangle\), where \(c,d\in\mathbb{C}\). Define \(\mathcal{H}=\{f(t)=\sum_{k\geq0} a_{k}\frac{t^{k}}{k!}\mid a_{k}\in\mathbb{C}\}\) to be the algebra of formal power series in a single variable t. The formal power series \(f(t)\in\mathcal{H}\) defines a linear functional on Π by setting \(\langle f(t)|x^{n}\rangle=a_{n}\) for all \(n\geq0\). Thus, we have (see [1, 2])
where \(\delta_{n,k}\) is the Kronecker symbol. Let \(f_{L}(t)=\sum_{n\geq0}\langle L|x^{n}\rangle\frac{t^{n}}{n!}\). By (1.1), we get that \(\langle f_{L}(t)|x^{n}\rangle=\langle L|x^{n}\rangle\). Thus, the map \(L\mapsto f_{L}(t)\) gives a vector space isomorphism from \(\Pi^{*}\) onto \(\mathcal{H}\). Therefore, \(\mathcal{H}\) is thought of as a set of both formal power series and linear functionals, which is called the umbral algebra. The umbral calculus is the study of umbral algebra.
$$ \bigl\langle t^{k}|x^{n}\bigr\rangle =n!\delta_{n,k} \quad \mbox{for all }n,k\geq 0, $$
(1.1)
The order
\(O(f(t))\) of the non-zero power series \(f(t)\) is defined to be k when \(f(t)=\sum_{n\geq k}a_{n}t^{n}\) and \(a_{k}\neq0\). Suppose that \(O(f(t))=1\) and \(O(g(t))=0\). Then there exists a unique sequence \(s_{n}(x)\) of polynomials such that \(\langle g(t)f(t)^{k}|s_{n}(x)\rangle=n!\delta_{n,k}\), where \(n,k\geq0\). The sequence \(s_{n}(x)\) is called the Sheffer sequence for \((g(t),f(t))\), and we write \(s_{n}(x)\sim(g(t),f(t))\) (see [1, 2]). For \(f(t)\in\mathcal{H}\) and \(p(x)\in\Pi\), we have that \(\langle e^{yt}|p(x)\rangle=p(y)\), \(\langle f(t)g(t)|p(x)\rangle=\langle g(t)|f(t)p(x)\rangle\), \(f(t)=\sum_{n\geq0}\langle f(t)|x^{n}\rangle \frac {t^{n}}{n!}\) and \(p(x)=\sum_{n\geq0}\langle t^{n}|p(x)\rangle\frac {x^{n}}{n!}\). Therefore, \(\langle t^{k}|p(x)\rangle=p^{(k)}(0)\), \(\langle1|p^{(k)}(x)\rangle=p^{(k)}(0)\), where \(p^{(k)}(0)\) denotes the kth derivative of \(p(x)\) with respect to x at \(x=0\). So, \(t^{k}p(x)=p^{(k)}(x)=\frac{d^{k}}{dx^{k}}p(x)\) for all \(k\geq0\) (see [1, 2]).
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Let \(s_{n}(x)\sim(g(t),f(t))\). Then we have
for all \(y\in\mathbb{C}\), where \(\bar{f}(t)\) is the compositional inverse of \(f(t)\) (see [1, 2]). For \(s_{n}(x)\sim(g(t),f(t))\) and \(r_{n}(x)\sim(h(t),\ell(t))\), let \(s_{n}(x)=\sum_{k=0}^{n} c_{n,k}r_{k}(x)\). Then we have
(see [1, 2]).
$$ \frac{1}{g(\bar{f}(t))}e^{y\bar{f}(t)}=\sum_{n\geq0}s_{n}(y) \frac {t^{n}}{n!} $$
(1.2)
$$ c_{n,k}=\frac{1}{k!} \biggl\langle \frac{h(\bar{f}(t))}{g(\bar {f}(t))}\bigl(\ell \bigl(\bar{f}(t)\bigr)\bigr)^{k}\Big|x^{n} \biggr\rangle $$
(1.3)
Throughout the paper, let \(r,s\in\mathbb{Z}_{>0}\), and let \(\mathbf{a}=(a_{1},a_{2},\ldots,a_{r})\), \(\mathbf{b}=(b_{1},b_{2},\ldots ,b_{s})\) with \(a_{j},b_{i}\neq0\) for all i, j. We define the Barnes-type Daehee with
λ-parameter and degenerate Euler mixed-type polynomials
\(D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})\) (for other Barnes-types, see [3‐5]) as
where we define
For \(x=0\), \(D\mathcal{E}_{n}(\lambda|\mathbf{a};\mathbf{b})=D\mathcal {E}_{n}(\lambda,0|\mathbf{a};\mathbf{b})\) are called the Barnes-type Daehee with
λ-parameter and degenerate Euler mixed-type numbers.
$$ P_{r,s}(t) (1+\lambda t)^{\frac{x}{\lambda}}=\sum _{n\geq0}D\mathcal{E}_{n}(\lambda ,x|\mathbf{a}; \mathbf{b})\frac{t^{n}}{n!}, $$
(1.4)
$$P_{r,s}(t)=\prod_{i=1}^{r} \biggl( \frac{\log(1+\lambda t)}{\lambda ((1+\lambda t)^{\frac{a_{i}}{\lambda}}-1)} \biggr)\prod_{i=1}^{s} \biggl(\frac{2}{(1+\lambda t)^{\frac{b_{i}}{\lambda}}+1} \biggr). $$
We recall here that the polynomials \(D_{n,\lambda}(x|\mathbf{a})\) given by
are called the Barnes-type Daehee polynomials with λ-parameter (see [6, 7]). Also, the polynomials \(\mathcal{E}_{n}(\lambda ,x|\mathbf{b})\) given by
are called the Barnes-type degenerate Euler polynomials which are studied in [8‐11]. In the case \(x=0\), we write \(\mathcal{E}_{n}(\lambda|\mathbf{b})=\mathcal{E}_{n}(\lambda,0|\mathbf {b})\), which are called the Barnes-type degenerate Euler numbers. Note that \(\lim_{\lambda\rightarrow0}\mathcal{E}_{n}(\lambda,x|\mathbf {b})=E_{n}(x|\mathbf{b})\) and \(\lim_{\lambda\rightarrow\infty}\lambda^{-n}\mathcal{E}_{n}(\lambda ,\lambda x|\mathbf{b})=(x)_{n}\), where \((x)_{n}=\prod_{i=0}^{n-1}(x-i)\) with \((x)_{0}=1\) and \(E_{n}(x|\mathbf{b})\) are the Barnes-type degenerate Euler polynomials given by
It is immediate from (1.2) and (1.4) to see that \(D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})\) is the Sheffer sequence for the pair \(g(t)=\prod_{i=1}^{r} (\frac {e^{a_{i}t}-1}{t} )\prod_{i=1}^{s} (\frac{e^{b_{i}t}+1}{2} )\) and \(f(t)=\frac{e^{\lambda t}-1}{\lambda}\). Thus,
$$P_{r,0}(t) (1+\lambda t)^{\frac{x}{\lambda}} =\sum _{n\geq0}D_{n,\lambda}(x|\mathbf{a})\frac{t^{n}}{n!} $$
$$ P_{0,s}(t) (1+\lambda t)^{\frac{x}{\lambda}}=\sum _{n\geq0}\mathcal{E}_{n}(\lambda ,x|\mathbf{b}) \frac{t^{n}}{n!} $$
(1.5)
$$\prod_{i=1}^{s} \biggl(\frac{2}{e^{b_{i}t}+1} \biggr)e^{xt}=\sum_{n\geq 0}E_{n}(x| \mathbf{b})\frac{t^{n}}{n!}. $$
$$ D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})\sim \Biggl(\prod_{i=1}^{r} \biggl( \frac{e^{a_{i}t}-1}{t} \biggr)\prod_{i=1}^{s} \biggl(\frac {e^{b_{i}t}+1}{2} \biggr),\frac{e^{\lambda t}-1}{\lambda} \Biggr). $$
(1.6)
The aim of the present paper is to present several new identities for Barnes-type Daehee with λ-parameter and degenerate Euler mixed-type polynomials by the use of umbral calculus. For some of the related works, one is referred to the papers [12‐20].
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2 Explicit formulas
In this section we suggest several explicit formulas for the Barnes-type Daehee with λ-parameter and degenerate Euler mixed-type polynomials. To do that, we recall that the Stirling numbers \(S_{1}(n,m)\) of the first kind are defined as \((x)_{n}=\sum_{m=0}^{n}S_{1}(n,m)x^{m}\sim(1,e^{t}-1)\) or \(\frac{1}{j!}(\log(1+t))^{j}=\sum_{\ell\geq j}S_{1}(\ell,j)\frac {t^{\ell}}{\ell!}\). Let \((x|\lambda)_{n}\) be the generalized falling factorials defined by \((x|\lambda)_{n}=\prod_{i=0}^{n-1}(x-i\lambda)\) with \((x|\lambda)_{0}=1\), namely \((x|\lambda)_{n}=\lambda^{n}(x/\lambda)_{n}\).
Let \(\mathit{BE}_{n}(x|\mathbf{a};\mathbf{b})\) be the Barnes-type Bernoulli and Euler mixed-type polynomials given by
Note that \(\mathit{BE}_{n}^{r,s}(x)\) denotes the special case \(\mathit{BE}_{n}(x|\underbrace{1,1,\ldots,1}_{r};\underbrace{1,1,\ldots,1}_{s})\) and was treated in [21, 22] by using p-adic integrals on \(\mathbb{Z}_{p}\).
$$ \prod_{i=1}^{r} \biggl( \frac{t}{e^{a_{i}t}-1} \biggr)\prod_{i=1}^{s} \biggl(\frac{2}{e^{b_{i}t}+1} \biggr)e^{xt}=\sum _{n\geq0}\mathit{BE}_{n}(x|\mathbf {a};\mathbf{b}) \frac{t^{n}}{n!}. $$
(2.1)
Theorem 2.1
For all
\(n\geq0\),
$$ D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})=\sum _{m=0}^{n}S_{1}(n,m)\lambda^{n-m} \mathit{BE}_{m}(x|\mathbf{a};\mathbf{b}). $$
Proof
By (1.6), we have that
Thus,
as claimed. □
$$ \prod_{i=1}^{r} \biggl( \frac{e^{a_{i}t}-1}{t} \biggr)\prod_{i=1}^{s} \biggl(\frac{e^{b_{i}t}+1}{2} \biggr)D\mathcal{E}_{n}(\lambda,x|\mathbf {a}; \mathbf{b})\sim \biggl(1,\frac{e^{\lambda t}-1}{\lambda} \biggr). $$
(2.2)
$$\begin{aligned} D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})&=\sum _{m=0}^{n}S_{1}(n,m)\lambda^{n-m} \prod_{i=1}^{r} \biggl(\frac {t}{e^{a_{i}t}-1} \biggr)\prod_{i=1}^{s} \biggl( \frac{2}{e^{b_{i}t}+1} \biggr)x^{m} \\ &=\sum_{m=0}^{n}S_{1}(n,m) \lambda^{n-m}\mathit{BE}_{m}(x|\mathbf{a};\mathbf{b}), \end{aligned}$$
Theorem 2.2
For all
\(n\geq0\),
$$D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})=\sum _{j=0}^{n} \Biggl(\sum_{\ell=j}^{n} \binom{n}{\ell}S_{1}(\ell,j)\lambda ^{\ell-j}D \mathcal{E}_{n-\ell}(\lambda|\mathbf{a};\mathbf {b}) \Biggr) x^{j}. $$
Proof
We proceed the proof by applying the conjugate representation: for \(s_{n}(x)\sim(g(t),f(t))\), we have \(S_{n}(x)=\sum_{j=0}^{n}\frac{1}{j!}\langle g(\bar{f}(t))^{-1}\bar{f}(t)^{j}|x^{n}\rangle x^{j}\). By (1.6), we obtain
Therefore, \(D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})=\sum_{j=0}^{n} (\sum_{\ell=j}^{n}\binom{n}{\ell}S_{1}(\ell,j)\lambda ^{\ell-j}D\mathcal{E}_{n-\ell}(\lambda|\mathbf{a};\mathbf {b}) ) x^{j}\), as claimed. □
$$\begin{aligned}& \bigl\langle g\bigl(\bar{f}(t)\bigr)^{-1}\bar{f}(t)^{j}|x^{n} \bigr\rangle \\& \quad = \biggl\langle P_{r,s}(t)\frac{\log^{j}(1+\lambda t)}{\lambda ^{j}}\Big|x^{n} \biggr\rangle =\lambda^{-j} \biggl\langle P_{r,s}(t)\Big|j!\sum _{\ell\geq j}S_{1}(\ell ,j)\frac{\lambda^{\ell}t^{\ell}}{\ell!}x^{n} \biggr\rangle \\& \quad =\lambda^{-j}j!\sum_{\ell=j}^{n} \binom{n}{\ell}S_{1}(\ell,j)\lambda ^{\ell} \bigl\langle P_{r,s}(t)|x^{n-\ell} \bigr\rangle =\lambda^{-j}j!\sum _{\ell =j}^{n}\binom{n}{\ell}S_{1}( \ell,j)\lambda^{\ell}D\mathcal{E}_{n-\ell }(\lambda|\mathbf{a}; \mathbf{b}). \end{aligned}$$
Theorem 2.3
For all
\(n\geq1\),
where
\(B_{\ell}^{(n)}\)
is the
ℓth Bernoulli number of order
n (see [23]).
$$ D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b}) =\sum _{\ell =0}^{n-1}\binom{n-1}{\ell}\lambda^{\ell}B_{\ell}^{(n)}\mathit{BE}_{n-\ell}(x|\mathbf{a}; \mathbf{b}), $$
Proof
We proceed the proof by using the following transfer formula: for \(p_{n}(x)\sim(1,f(t))\) and \(q_{n}(x)\sim(1,g(t))\), we have that \(q_{n}(x)=x (\frac{f(t)}{g(t)} )^{n}x^{-1}p_{n}(x)\) for all \(n\geq1\). So, by the fact that \(x^{n}\sim(1,t)\) and (2.2), we obtain
which, by (2.1), implies
as required. □
$$\begin{aligned}& \prod_{i=1}^{r} \biggl(\frac{e^{a_{i}t}-1}{t} \biggr)\prod_{i=1}^{s} \biggl( \frac{e^{b_{i}t}+1}{2} \biggr)D\mathcal{E}_{n}(\lambda,x|\mathbf {a}; \mathbf{b}) \\& \quad =x \biggl(\frac{\lambda t}{e^{\lambda t}-1} \biggr)^{n}x^{n-1}=x\sum _{\ell\geq0}B_{\ell}^{(n)} \frac{\lambda ^{\ell}t^{\ell}}{\ell!}x^{n-1}=\sum_{\ell=0}^{n-1} \binom{n-1}{\ell}\lambda ^{\ell}B_{\ell}^{(n)}x^{n-\ell}, \end{aligned}$$
$$\begin{aligned} D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})&=\sum _{\ell =0}^{n-1}\binom{n-1}{\ell}\lambda^{\ell}B_{\ell}^{(n)}\prod_{i=1}^{r} \biggl(\frac{t}{e^{a_{i}t}-1} \biggr)\prod_{i=1}^{s} \biggl(\frac{2}{e^{b_{i}t}+1} \biggr)x^{n-\ell} \\ &=\sum_{\ell=0}^{n-1}\binom{n-1}{\ell} \lambda^{\ell}B_{\ell}^{(n)}\mathit{BE}_{n-\ell}(x| \mathbf{a};\mathbf{b}), \end{aligned}$$
In order to state our next theorem, we recall the polynomials \(\beta\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})\), which are called the Barnes-type degenerate Bernoulli and Euler mixed-type polynomials. They are defined as
where \(Q_{r,s}(t)=\prod_{i=1}^{r} (\frac{t}{(1+\lambda t)^{\frac{a_{i}}{\lambda}}-1} )\prod_{i=1}^{s} (\frac {2}{(1+\lambda t)^{\frac{b_{i}}{\lambda}}+1} )\), for example, see [3].
$$ Q_{r,s}(t) (1+\lambda t)^{\frac{x}{\lambda}}=\sum _{n\geq0}\beta\mathcal{E}_{n}(\lambda ,x|\mathbf{a} ; \mathbf{b})\frac{t^{n}}{n!}, $$
(2.3)
Theorem 2.4
For all
\(n\geq0\),
$$ D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b}) =\sum _{\ell =0}^{n}\frac{\binom{n}{\ell}}{\binom{\ell+r}{r}}\lambda ^{\ell}S_{1}(\ell+r,r)\beta\mathcal{E}_{n-\ell}(\lambda,x|\mathbf {a}; \mathbf{b}). $$
Proof
By (1.4), we have
which, by (2.3), implies \(D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})=\sum_{\ell =0}^{n}\frac{\binom{n}{\ell}}{\binom{\ell+r}{r}}\lambda ^{\ell}S_{1}(\ell+r,r)\beta\mathcal{E}_{n-\ell}(\lambda,x|\mathbf {a};\mathbf{b})\), as required. □
$$\begin{aligned} D\mathcal{E}_{n}(\lambda,y|\mathbf{a};\mathbf{b})&= \biggl\langle \sum _{\ell\geq0}D\mathcal {E}_{\ell}(\lambda,y| \mathbf{a};\mathbf{b})\frac{t^{\ell}}{\ell !}\Big|x^{n} \biggr\rangle = \bigl\langle P_{r,s}(t) (1+\lambda t)^{\frac{y}{\lambda}}|x^{n} \bigr\rangle \\ &= \biggl\langle Q_{r,s}(t) (1+\lambda t)^{\frac{y}{\lambda}}\Big| \frac{\log^{r}(1+\lambda t)}{\lambda^{r} t^{r}}x^{n} \biggr\rangle \\ &= \biggl\langle Q_{r,s}(t) (1+\lambda t)^{\frac{y}{\lambda}}\Big| r!\sum _{\ell\geq0}\frac{S_{1}(\ell+r,r)\lambda^{\ell}t^{\ell}}{(\ell +r)!}x^{n} \biggr\rangle \\ &=\sum_{\ell=0}^{n}\frac{\binom{n}{\ell}}{\binom{\ell +r}{r}} \lambda^{\ell}S_{1}(\ell+r,r) \biggl\langle \sum _{m\geq0}\beta\mathcal{E}_{m}(\lambda ,y|\mathbf{a} ; \mathbf{b})\frac{t^{m}}{m!}\Big|x^{n-\ell} \biggr\rangle , \end{aligned}$$
In order to present our next theorem, we recall the polynomials \(\beta_{n}(\lambda,x|\mathbf{a})\), which are called the Barnes-type degenerate Bernoulli polynomials. They are given by
for example, see [8, 9, 23].
$$ Q_{r,0}(t) (1+\lambda t)^{\frac{x}{\lambda}}=\sum _{n\geq0}\beta_{n}(\lambda,x|\mathbf {a}) \frac{t^{n}}{n!}, $$
(2.4)
Theorem 2.5
For all
\(n\geq0\),
$$\begin{aligned} D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})&=\sum _{\ell =0}^{n}\sum_{m=0}^{n-\ell } \frac{\binom{n}{\ell}\binom{n-\ell}{m}}{\binom{\ell +r}{r}}\lambda^{\ell}S_{1}(\ell+r,r) \mathcal{E}_{n-\ell-m}(\lambda|\mathbf{b})\beta _{m}(\lambda,x| \mathbf{a}) \\ &=\sum_{\ell=0}^{n}\sum _{m=0}^{n-\ell}\frac{\binom{n}{\ell}\binom {n-\ell }{m}}{\binom{\ell+r}{r}}\lambda^{\ell}S_{1}(\ell+r,r)\beta_{n-\ell-m}(\lambda|\mathbf{a})\mathcal {E}_{m}(\lambda,x|\mathbf{b}). \end{aligned}$$
Proof
By the proof of Theorem 2.4, we have
Thus, by (1.5) and (2.4), we obtain
which completes the proof of the first formula.
$$\begin{aligned} D\mathcal{E}_{n}(\lambda,y|\mathbf{a};\mathbf{b})&=\sum _{\ell =0}^{n}\frac{\binom{n}{\ell }}{\binom{\ell+r}{r}}\lambda^{\ell}S_{1}(\ell+r,r) \bigl\langle Q_{r,s}(t) (1+\lambda t)^{\frac{y}{\lambda}}|x^{n-\ell} \bigr\rangle \\ &=\sum_{\ell=0}^{n}\frac{\binom{n}{\ell}}{\binom{\ell +r}{r}} \lambda^{\ell}S_{1}(\ell+r,r) \bigl\langle Q_{0,s}(t)|Q_{r,0}(t) (1+\lambda t)^{\frac{y}{\lambda}} x^{n-\ell} \bigr\rangle . \end{aligned}$$
$$\begin{aligned}& D\mathcal{E}_{n}(\lambda,y|\mathbf{a};\mathbf{b}) \\& \quad =\sum_{\ell=0}^{n}\frac{\binom{n}{\ell}}{\binom{\ell +r}{r}} \lambda^{\ell}S_{1}(\ell+r,r) \Biggl\langle Q_{0,s}(t)\bigg|\sum_{m=0}^{n-\ell} \binom {n-\ell }{m}\beta_{m}(\lambda,y|\mathbf{a})x^{n-\ell-m} \Biggr\rangle \\& \quad =\sum_{\ell=0}^{n}\frac{\binom{n}{\ell}}{\binom{\ell +r}{r}} \lambda^{\ell}S_{1}(\ell+r,r)\sum _{m=0}^{n-\ell}\binom{n-\ell}{m}\beta_{m}( \lambda ,y|\mathbf{a} ) \bigl\langle Q_{0,s}(t)|x^{n-\ell-m} \bigr\rangle \\& \quad =\sum_{\ell=0}^{n}\frac{\binom{n}{\ell}}{\binom{\ell +r}{r}} \lambda^{\ell}S_{1}(\ell+r,r)\sum _{m=0}^{n-\ell}\binom{n-\ell}{m}\beta_{m}( \lambda ,y|\mathbf{a} )\mathcal{E}_{n-\ell-m}(\lambda|\mathbf{b}), \end{aligned}$$
The second formula can be obtained by using very similar techniques. □
3 Recurrence relations
In this section, we present several recurrence relations for Barnes-type Daehee with λ-parameter and degenerate Euler mixed-type polynomials. Our first recurrence is based on the polynomials \((x|\lambda)_{n}\).
Theorem 3.1
For all
\(n\geq0\),
$$D\mathcal{E}_{n}(\lambda,x+y|\mathbf{a};\mathbf{b})=\sum _{j=0}^{n}\binom{n}{j}D\mathcal {E}_{j}(\lambda,x|\mathbf{a};\mathbf{b}) (y|\lambda)_{n-j}. $$
Proof
Let \(p_{n}(x)=\prod_{i=1}^{r} (\frac{e^{a_{i}t}-1}{t} )\prod_{i=1}^{s} (\frac{e^{b_{i}t}+1}{2} )D\mathcal{E}_{n}(\lambda ,x|\mathbf{a};\mathbf{b})\). By (2.2) we have that \(p_{n}(x)=(x|\lambda)_{n}\sim (1,\frac{e^{\lambda t}-1}{\lambda } )\), which leads to the required recurrence. □
The second recurrence is obtained from the fact that \(f(t)s_{n}(x)=ns_{n-1}(x)\) for all \(s_{n}(x)\sim(g(t),f(t))\) (see [1, 2]).
Theorem 3.2
For all
\(n\geq1\),
$$D\mathcal{E}_{n}(\lambda,x+\lambda|\mathbf{a};\mathbf{b})-D\mathcal {E}_{n}(\lambda,x|\mathbf{a} ;\mathbf{b})=n\lambda D \mathcal{E}_{n-1}(\lambda,x|\mathbf {a};\mathbf{b}). $$
Proof
By (1.6) and \(f(t)s_{n}(x)=ns_{n-1}(x)\) whenever \(s_{n}(x)\sim (g(t),f(t))\), we have
which implies \(D\mathcal{E}_{n}(\lambda,x+\lambda|\mathbf{a};\mathbf{b})-D\mathcal {E}_{n}(\lambda,x|\mathbf{a} ;\mathbf{b})=n\lambda D\mathcal{E}_{n-1}(\lambda,x|\mathbf {a};\mathbf{b})\), as required. □
$$\frac{e^{\lambda t}-1}{\lambda}D\mathcal{E}_{n}(\lambda,x|\mathbf {a};\mathbf{b} )=nD\mathcal{E}_{n-1}(\lambda,x|\mathbf{a};\mathbf{b}), $$
The next result gives an explicit formula for \(\frac{d}{dx}D\mathcal {E}_{n}(\lambda,x+\lambda|\mathbf{a};\mathbf{b})\).
Theorem 3.3
For all
\(n\geq1\),
$$\frac{d}{dx}D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})=n! \sum_{\ell=0}^{n-1}\frac {(-\lambda)^{n-\ell-1}}{\ell!(n-\ell)}D \mathcal{E}_{\ell}(\lambda ,x|\mathbf{a} ;\mathbf{b}). $$
Proof
It is well known that for \(s_{n}(x)\sim(g(t),f(t))\), \(\frac {d}{dx}s_{n}(x)=\sum_{\ell=0}^{n-1}\binom{n}{\ell}\langle\bar {f}(t)|x^{n-\ell}\rangle s_{\ell}(x)\) (see [1, 2]). In our case, by (1.6), we have
Thus \(\frac{d}{dx}D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf {b})=n!\sum_{\ell =0}^{n-1}\frac{(-\lambda)^{n-\ell-1}}{\ell!(n-\ell)}D\mathcal {E}_{\ell}(\lambda,x|\mathbf{a};\mathbf{b})\), as required. □
$$\begin{aligned} \bigl\langle \bar{f}(t)|x^{n-\ell}\bigr\rangle &= \biggl\langle \frac {1}{\lambda}\log (1+\lambda t)\Big|x^{n-\ell} \biggr\rangle \\ &= \lambda^{-1} \biggl\langle \sum_{m\geq1} \frac {(-1)^{m-1}(m-1)!\lambda ^{m}t^{m}}{m!}\Big|x^{n-\ell} \biggr\rangle \\ &=\lambda^{-1}(-1)^{n-\ell-1}\lambda^{n-\ell}(n-\ell-1)! \\ &=(- \lambda )^{n-\ell-1}(n-\ell-1)!. \end{aligned}$$
Another recurrence relation can be stated as follows.
Theorem 3.4
For all
\(n\geq1\),
where
\(\mathfrak{b}_{n}\)
is the
nth Bernoulli number of the second kind, which is defined by
\(\frac{t}{\log(1+t)}=\sum_{n\geq 0}\mathfrak {b}_{n}\frac{t^{n}}{n!}\).
$$\begin{aligned}& D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b}) \\& \quad = \Biggl(x-\sum_{i=1}^{r}a_{i}- \sum_{j=1}^{s}b_{j} \Biggr)D \mathcal {E}_{n-1}(\lambda,x-\lambda|\mathbf{a};\mathbf{b}) + \frac{r}{n}\sum_{\ell=0}^{n} \binom{n}{\ell}\lambda^{\ell}\mathfrak {b}_{\ell}D \mathcal{E}_{n-\ell}(\lambda,x-\lambda|\mathbf {a};\mathbf{b}) \\& \qquad {} -\frac{1}{n}\sum_{i=1}^{r}a_{i} \sum_{\ell=0}^{n}\binom{n}{\ell}\lambda ^{\ell}\mathfrak{b}_{\ell}D\mathcal{E}_{n-\ell}( \lambda,x-\lambda |a_{i},a_{1},\ldots ,a_{r}; \mathbf{b}) \\& \qquad {} +\frac{1}{2}\sum_{j=1}^{s} b_{j}D\mathcal{E}_{n-1}(\lambda,x-\lambda |\mathbf{a} ;b_{j},b_{1},\ldots,b_{s}), \end{aligned}$$
Proof
Let \(n\geq1\). Then
By (1.6), the term in (3.3) equals
For the term in (3.2), we observe that
So the term in (3.2) is
For the term in (3.1), we note that
where \(\frac{\lambda t}{\log(1+\lambda t)}-\frac{a_{i} t}{(1+\lambda t)^{a_{i}/\lambda}-1}\) has order at least 1. Thus, the term in (3.1) equals
which is equal to
By using (3.4), (3.5) and (3.6) instead of (3.3), (3.2) and (3.1), respectively, we complete the proof. □
$$\begin{aligned}& D\mathcal{E}_{n}(\lambda,y|\mathbf{a};\mathbf{b}) \\& \quad = \biggl\langle \sum _{\ell\geq0}D\mathcal {E}_{\ell}(\lambda,y| \mathbf{a};\mathbf{b})\frac{t^{\ell}}{\ell !}\Big|x^{n} \biggr\rangle \\& \quad = \bigl\langle P_{r,s}(t) (1+\lambda t)^{y/\lambda}|x^{n} \bigr\rangle = \biggl\langle \frac{d}{dt} \bigl(P_{r,s}(t) (1+ \lambda t)^{y/\lambda } \bigr)\Big|x^{n-1} \biggr\rangle \\& \quad = \Biggl\langle \frac{d}{dt}\prod_{i=1}^{r} \biggl(\frac{\log (1+\lambda t)}{\lambda((1+\lambda t)^{\frac{a_{i}}{\lambda}}-1)} \biggr)\prod_{i=1}^{s} \biggl(\frac{2}{(1+\lambda t)^{\frac{b_{i}}{\lambda }}+1} \biggr) (1+\lambda t)^{y/\lambda}\bigg|x^{n-1} \Biggr\rangle \end{aligned}$$
(3.1)
$$\begin{aligned}& \qquad {}+ \Biggl\langle \prod_{i=1}^{r} \biggl( \frac{\log(1+\lambda t)}{\lambda ((1+\lambda t)^{\frac{a_{i}}{\lambda}}-1)} \biggr)\frac{d}{dt}\prod_{i=1}^{s} \biggl(\frac{2}{(1+\lambda t)^{\frac{b_{i}}{\lambda}}+1} \biggr) (1+\lambda t)^{y/\lambda}\bigg|x^{n-1} \Biggr\rangle \end{aligned}$$
(3.2)
$$\begin{aligned}& \qquad {}+ \biggl\langle P_{r,s}(t)\frac{d}{dt}(1+\lambda t)^{y/\lambda }\Big|x^{n-1} \biggr\rangle . \end{aligned}$$
(3.3)
$$ y \bigl\langle P_{r,s}(t) (1+\lambda t)^{(y-\lambda)/\lambda }|x^{n-1} \bigr\rangle =yD\mathcal{E}_{n-1}(\lambda,y-\lambda |\mathbf{a}; \mathbf{b} ). $$
(3.4)
$$ \frac{d}{dt}\prod_{i=1}^{s} \biggl( \frac{2}{(1+\lambda t)^{\frac {b_{i}}{\lambda}}+1} \biggr) =\prod_{i=1}^{s} \biggl( \frac{2}{(1+\lambda t)^{\frac{b_{i}}{\lambda }}+1} \biggr)\sum_{i=1}^{s} \biggl(\frac{-b_{i}}{1+\lambda t}+\frac {b_{i}}{2(1+\lambda t)}\frac{2}{(1+\lambda t)^{b_{i}/\lambda}+1} \biggr). $$
$$\begin{aligned}& -\sum_{j=1}^{s}b_{j} \bigl\langle P_{r,s}(t) (1+\lambda t)^{(y-\lambda )/\lambda}|x^{n-1} \bigr\rangle +\frac{1}{2}\sum_{j=1}^{s}b_{j} \biggl\langle P_{r,s}(t)\frac{2(1+\lambda t)^{(y-\lambda)/\lambda}}{(1+\lambda t)^{b_{j}/\lambda}+1}\Big|x^{n-1} \biggr\rangle \\& \quad =-\sum_{j=1}^{s}b_{j}D \mathcal{E}_{n-1}(\lambda,y-\lambda|\mathbf {a};\mathbf{b}) + \frac{1}{2}\sum_{j=1}^{s}b_{j}D \mathcal{E}_{n-1}(\lambda,y-\lambda |\mathbf{a} ;b_{j},b_{1}, \ldots,b_{s}). \end{aligned}$$
(3.5)
$$\begin{aligned}& (1+\lambda t)\frac{d}{dt}\prod_{i=1}^{r} \biggl(\frac{\log(1+\lambda t)}{\lambda((1+\lambda t)^{\frac{a_{i}}{\lambda}}-1)} \biggr) \\& \quad =\prod_{i=1}^{r} \biggl( \frac{\log(1+\lambda t)}{\lambda((1+\lambda t)^{\frac{a_{i}}{\lambda}}-1)} \biggr) \Biggl(-\sum_{i=1}^{r}a_{i} +\frac {1}{t}\sum_{i=1}^{r} \biggl( \frac{\lambda t}{\log(1+\lambda t)}-\frac{a_{i} t}{(1+\lambda t)^{a_{i}/\lambda }-1} \biggr) \Biggr), \end{aligned}$$
$$\begin{aligned}& -\sum_{i=1}^{r}a_{i} \bigl\langle P_{r,s}(t) (1+\lambda t)^{(y-\lambda )/\lambda}|x^{n-1} \bigr\rangle \\& \qquad {} + \Biggl\langle P_{r,s}(t) (1+\lambda t)^{(y-\lambda)/\lambda}\bigg| \frac {1}{t}\sum_{i=1}^{r} \biggl( \frac{\lambda t}{\log(1+\lambda t)}-\frac{a_{i} t}{(1+\lambda t)^{a_{i}/\lambda}-1} \biggr)x^{n-1} \Biggr\rangle \\& \quad =-\sum_{i=1}^{r}a_{i}D \mathcal{E}_{n-1}(\lambda,y-\lambda|\mathbf {a};\mathbf{b}) \\& \qquad {} +\frac{1}{n} \Biggl\langle P_{r,s}(t) (1+\lambda t)^{(y-\lambda )/\lambda }\bigg| \sum_{i=1}^{r} \biggl( \frac{\lambda t}{\log(1+\lambda t)}-\frac{a_{i} t}{(1+\lambda t)^{a_{i}/\lambda}-1} \biggr)x^{n} \Biggr\rangle \\& \quad =-\sum_{i=1}^{r}a_{i}D \mathcal{E}_{n-1}(\lambda,y-\lambda|\mathbf {a};\mathbf{b}) \\& \qquad {}+ \frac {r}{n} \Biggl\langle P_{r,s}(t) (1+\lambda t)^{(y-\lambda)/\lambda}\bigg| \sum_{\ell\geq0}^{r}\mathfrak{b}_{\ell}\frac{\lambda^{\ell}t^{\ell}}{\ell !}x^{n} \Biggr\rangle \\& \qquad {} -\frac{1}{n}\sum_{i=1}^{r}a_{i} \Biggl\langle \frac{\log(1+\lambda t)}{\lambda((1+\lambda t)^{a_{i}/\lambda}-1)}P_{r,s}(t) (1+\lambda t)^{(y-\lambda)/\lambda}\Bigg| \sum_{\ell\geq0}^{r}\mathfrak{b}_{\ell}\frac {\lambda^{\ell}t^{\ell}}{\ell!}x^{n} \Biggr\rangle , \end{aligned}$$
$$\begin{aligned}& -\sum_{i=1}^{r}a_{i}D \mathcal{E}_{n-1}(\lambda,y-\lambda|\mathbf {a};\mathbf{b})+ \frac {r}{n}\sum_{\ell=0}^{n} \binom{n}{\ell}\lambda^{\ell}\mathfrak {b}_{\ell}D \mathcal{E}_{n-\ell}(\lambda,y-\lambda|\mathbf{a};\mathbf {b}) \\& \quad {} -\frac{1}{n}\sum_{i=1}^{r}a_{i} \sum_{\ell=0}^{n}\binom{n}{\ell}\lambda ^{\ell}\mathfrak{b}_{\ell}D\mathcal{E}_{n-\ell}( \lambda,y-\lambda |a_{i},a_{1},\ldots ,a_{r}; \mathbf{b}). \end{aligned}$$
(3.6)
Theorem 3.5
For all
\(n\geq0\),
where
\(B_{\ell}\)
is the
ℓth Bernoulli number and
\(E_{\ell}(1)\)
is the
ℓth Euler polynomial evaluated at 1.
$$\begin{aligned}& D\mathcal{E}_{n+1}(\lambda,x|\mathbf{a};\mathbf{b}) \\& \quad =xD\mathcal{E}_{n}(\lambda,x-\lambda|\mathbf{a};\mathbf{b})-\sum _{i=1}^{r}a_{i}\sum _{m=0}^{n}S_{1}(n,m)\lambda^{n-m} \mathit{BE}_{m}(x-\lambda|\mathbf{a};\mathbf{b}) \\& \qquad {} -\sum_{m=0}^{n}\sum _{\ell=0}^{m}S_{1}(n,m)\lambda^{n-m} \binom {n}{m} \Biggl(\frac{B_{\ell+1}}{\ell+1}\sum_{i=1}^{r}a_{i}^{\ell+1}+ \frac{E_{\ell}(1)}{2}\sum_{j=1}^{s}b_{j}^{\ell+1} \Biggr) \\& \qquad {}\times\mathit{BE}_{m-\ell}(x-\lambda |\mathbf{a};\mathbf{b}), \end{aligned}$$
Proof
It is well known that for \(s_{n}(x)\sim(g(t),f(t))\), \(s_{n+1}(x)=(x-g'(t)/g(t))\frac{1}{f'(t)} s_{n}(x)\) (see [1, 2]). In our case, by (1.6), we have
and by Theorem 2.1, we obtain
Note that
So
Hence, by substituting into (3.7), we complete the proof. □
$$ D\mathcal{E}_{n+1}(\lambda,x|\mathbf{a};\mathbf{b})=xD\mathcal {E}_{n}(\lambda,x-\lambda |\mathbf{a};\mathbf{b})-e^{-\lambda t} \frac{g'(t)}{g(t)}D\mathcal {E}_{n}(\lambda,x|\mathbf{a} ;\mathbf{b}), $$
$$\begin{aligned} D\mathcal{E}_{n+1}(\lambda,x|\mathbf{a};\mathbf{b}) =&xD \mathcal {E}_{n}(\lambda,x-\lambda |\mathbf{a};\mathbf{b}) \\ &{}-\sum_{m=0}^{n}S_{1}(n,m) \lambda^{n-m}e^{-\lambda t}\frac {g'(t)}{g(t)}\mathit{BE}_{m}(x| \mathbf{a};\mathbf{b}). \end{aligned}$$
(3.7)
$$\begin{aligned} \frac{g'(t)}{g(t)} &=\bigl(\log g(t)\bigr)'=\sum _{i=1}^{r} \frac{a_{i}e^{a_{i}t}}{e^{a_{i}t}-1}-\frac {r}{t}+\sum _{j=1}^{s}\frac{b_{j}e^{b_{j}t}}{e^{b_{j}t}+1} \\ &=\sum_{i=1}^{r}a_{i}+ \frac{1}{t}\sum_{i=1}^{r} \biggl( \frac {a_{i}t}{e^{a_{i}t}-1}-1 \biggr)+\frac{1}{2}\sum_{j=1}^{s} \frac {2b_{j}e^{b_{j}t}}{e^{b_{j}t}+1} \\ &=\sum_{i=1}^{r}a_{i}+ \frac{1}{t}\sum_{i=1}^{r}\sum _{\ell\geq0}\beta _{\ell}a_{i}^{\ell}\frac{t^{\ell}}{\ell!}+\frac{1}{2}\sum_{j=1}^{s} \sum_{\ell\geq 0}E_{\ell}(1)b_{j}^{\ell+1} \frac{t^{\ell}}{\ell!} \\ &=\sum_{i=1}^{r}a_{i}+\sum _{\ell\geq0}\frac{\beta_{\ell+1}}{(\ell +1)!}\sum _{i=1}^{r}a_{i}^{\ell+1}t^{\ell}+\frac{1}{2}\sum_{\ell\geq0}\frac{E_{\ell}(1)}{\ell!}\sum _{j=1}^{s}b_{j}^{\ell +1}t^{\ell}. \end{aligned}$$
$$\begin{aligned} \frac{g'(t)}{g(t)}\mathit{BE}_{m}(x|\mathbf{a};\mathbf{b}) =&\sum _{i=1}^{r}a_{i}BE_{m}(x| \mathbf{a};\mathbf{b})+\sum_{\ell=0}^{m} \binom{m}{\ell}\frac{\beta_{\ell+1}}{\ell+1}\sum_{i=1}^{r}a_{i}^{\ell+1} \mathit{BE}_{m-\ell}(x|\mathbf{a};\mathbf{b}) \\ &{}+\frac{1}{2}\sum_{\ell=0}^{m} \binom{m}{\ell}E_{\ell}(1)\sum_{j=1}^{s}b_{j}^{\ell +1} \mathit{BE}_{m-\ell}(x|\mathbf{a};\mathbf{b}). \end{aligned}$$
4 Relations with other families of polynomials
In this section, we establish a connection between Barnes-type Daehee with λ-parameter and degenerate Euler mixed-type polynomials and several known families of polynomials.
Theorem 4.1
For all
\(n\geq0\),
$$D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})=\sum _{m=0}^{n}\binom{n}{m}D\mathcal {E}_{n-m}(\lambda|\mathbf{a};\mathbf{b}) (x|\lambda)_{m}. $$
Proof
Note that \((x|\lambda)_{n}\sim(1,\frac{e^{\lambda t}-1}{\lambda})\). Let \(D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})=\sum_{m=0}^{n}c_{n,m}(x|\lambda)_{m}\). By (1.3) and (1.6), we have
which completes the proof. □
$$\begin{aligned} c_{n,m} &=\frac{1}{m!} \bigl\langle P_{r,s}(t)|t^{m}x^{n} \bigr\rangle =\binom {n}{m} \bigl\langle P_{r,s}(t)|x^{n-m} \bigr\rangle \\ &=\binom {n}{m}D\mathcal {E}_{n-m}(\lambda|\mathbf{a}; \mathbf{b}), \end{aligned}$$
For the following, we note that \(B_{n}^{(\alpha)}(x)\sim(\frac {(e^{t}-1)^{\alpha}}{t^{\alpha}},t)\).
Theorem 4.2
For all
\(n\geq0\), the polynomial
\(D\mathcal{E}_{n}(\lambda,x|\mathbf {a};\mathbf{b})\)
is given by
where
\(a_{\ell,k,q,p}=S_{1}(\ell,m)S_{1}(q+\alpha,q-p+\alpha)S_{2}(q-p+\alpha ,\alpha )\lambda^{k+\ell+p-m}b_{\ell}^{(\alpha)}\)
and
\(b_{\ell}^{(\alpha)}\)
is the
ℓth Bernoulli number of the second kind of order
α
given by
\((\frac{t}{\log(1+t)})^{\alpha}=\sum_{\ell\geq0}b_{\ell }^{(\alpha)}\frac {t^{\ell}}{k!}\).
$$\sum_{m=0}^{n} \Biggl(\sum _{\ell=m}^{n}\sum_{k=0}^{n-\ell} \sum_{q=0}^{n-\ell -k}\sum _{p=0}^{q} \frac{\binom{n}{\ell}\binom{n-\ell}{k}\binom{n-\ell -k}{q}}{\binom {q+\alpha}{\alpha}}a_{\ell,k,q,p}D \mathcal{E}_{n-\ell-k-q}(\lambda |\mathbf{a} ;\mathbf{b}) \Biggr)B^{(\alpha)}_{m}(x), $$
Proof
Let \(D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})=\sum_{m=0}^{n}c_{n,m}B_{m}^{(\alpha )}(x)\). By (1.3) and (1.6), we have
One can show that
where \(S_{2}(n,m)\) is the Stirling number of the second kind. Thus,
where \(\langle P_{r,s}(t)|x^{n-\ell-k-q}\rangle=D\mathcal{E}_{n-\ell -k-q}(\lambda|\mathbf{a};\mathbf{b})\). Hence,
which completes the proof. □
$$\begin{aligned} c_{n,m}&=\frac{1}{m!\lambda^{m}} \biggl\langle P_{r,s}(t) \biggl( \frac {(1+\lambda t)^{1/\lambda}-1}{t} \biggr)^{\alpha}\biggl(\frac{\lambda t}{\log(1+\lambda t)} \biggr)^{\alpha}\Big|\bigl(\log(1+\lambda t)\bigr)^{m}x^{n} \biggr\rangle \\ &=\frac{1}{\lambda^{m}}\sum_{\ell=m}^{n} \binom{n}{\ell}\lambda^{\ell }S_{1}(\ell,m) \biggl\langle P_{r,s}(t) \biggl(\frac{(1+\lambda t)^{1/\lambda }-1}{t} \biggr)^{\alpha}\bigg| \biggl( \frac{\lambda t}{\log(1+\lambda t)} \biggr)^{\alpha}x^{n-\ell} \biggr\rangle \\ &=\frac{1}{\lambda^{m}}\sum_{\ell=m}^{n}\sum _{k=0}^{n-\ell}\binom {n}{\ell } \binom{n-\ell}{k}S_{1}(\ell,m)\lambda^{\ell+k}b_{k}^{(\alpha)} \biggl\langle P_{r,s}(t) \biggl(\frac{(1+\lambda t)^{1/\lambda}-1}{t} \biggr)^{\alpha}\Big|x^{n-\ell-k} \biggr\rangle . \end{aligned}$$
$$\begin{aligned} \biggl(\frac{(1+\lambda t)^{1/\lambda}-1}{t} \biggr)^{\alpha}&= \biggl(\frac {e^{\frac{1}{\lambda}\log(1+\lambda t)}-1}{t} \biggr)^{\alpha}\\ &=\sum_{q\geq0}\sum_{p=0}^{q} \binom{q+\alpha}{\alpha }^{-1}S_{1}(q+\alpha ,q-p+ \alpha)S_{2}(q-p+\alpha,\alpha)\lambda^{p} \frac{t^{q}}{q!}, \end{aligned}$$
$$\begin{aligned}& \biggl\langle P_{r,s}(t) \biggl(\frac{(1+\lambda t)^{1/\lambda }-1}{t} \biggr)^{\alpha}\Big|x^{n-\ell-k} \biggr\rangle \\& \quad =\sum_{q=0}^{n-\ell-k}\sum _{p=0}^{q}\frac{\binom{n-\ell -k}{q}}{\binom {q+\alpha}{\alpha}}S_{1}(q+ \alpha,q-p+\alpha)S_{2}(q-p+\alpha,\alpha )\lambda ^{p}\bigl\langle P_{r,s}(t)|x^{n-\ell-k-q}\bigr\rangle , \end{aligned}$$
$$ c_{n,m} =\sum_{\ell=m}^{n}\sum _{k=0}^{n-\ell}\sum_{q=0}^{n-\ell -k} \sum_{p=0}^{q} \frac{\binom{n}{\ell}\binom{n-\ell}{k}\binom{n-\ell -k}{q}}{\binom {q+\alpha}{\alpha}}a_{\ell,k,q,p}D \mathcal{E}_{n-\ell-k-q}(\lambda |\mathbf{a} ;\mathbf{b}), $$
By similar techniques as in the proof of the last theorem, we can express our polynomials \(D\mathcal{E}_{n}(\lambda,x|\mathbf {a};\mathbf{b})\) in terms of the degenerate Bernoulli polynomials \(\beta_{n}^{(\alpha)}(\lambda,x)\) of order α. These polynomials are the Sheffer sequence which is given by \(\beta_{n}^{(\alpha)}(\lambda,x)\sim((\frac{\lambda (e^{t}-1)}{e^{\lambda t}-1})^{\alpha},\frac{e^{\lambda t}-1}{\lambda})\).
Theorem 4.3
For all
\(n\geq0\), the polynomial
\(D\mathcal{E}_{n}(\lambda,x|\mathbf {a};\mathbf{b})\)
is given by
where
\(c_{n,m}=\sum_{q=0}^{n-m}\sum_{p=0}^{q} \frac{\binom{n-m}{q}}{\binom{q+\alpha}{\alpha}}S_{1}(q+\alpha ,q-p+\alpha )S_{2}(q-p+\alpha,\alpha)\lambda^{p}D\mathcal{E}_{n-m-q}(\lambda |\mathbf{a};\mathbf{b})\).
$$\sum_{m=0}^{n}\binom{n}{m}c_{n,m} \beta^{(\alpha)}_{m}(\lambda,x), $$
Now we are interested in expressing our polynomials in terms of \(H_{n}^{(\alpha)}(x|\mu)\) which are called the Frobenius-Euler polynomials of order α. Note that \(H_{n}^{(\alpha)}(x|\mu)\sim ( (\frac{e^{t}-\mu}{1-\mu} )^{\alpha},t )\) (see [10, 24]).
Theorem 4.4
For all
\(n\geq0\),
where
$$D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})=\sum _{m=0}^{n} \biggl(\frac {a_{n,m}}{(1-\mu)^{\alpha}\lambda^{m}} \biggr)H^{(\alpha)}_{m}(x|\mu), $$
$$ a_{n,m} =\sum_{\ell=m}^{n}\sum _{k=0}^{n-\ell}\sum_{p=0}^{\alpha}\binom {n}{\ell }\binom{n-\ell}{k}\binom{\alpha}{p}S_{1}(\ell,m) \lambda^{\ell}(-\mu )^{\alpha-p}D\mathcal{E}_{k}(\lambda| \mathbf{a};\mathbf{b}) (p|\lambda)_{n-\ell-k}. $$
Proof
Let \(D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})=\sum_{m=0}^{n}c_{n,m}H^{(\alpha )}_{m}(x|\mu)\). By (1.3) and (1.6), we have
where
Thus, the constants \(c_{n,m}\) are given by
which completes the proof. □
$$\begin{aligned} c_{n,m}&=\frac{1}{m!(1-\mu)^{\alpha}\lambda^{m}} \bigl\langle P_{r,s}(t) \bigl((1+\lambda t)^{1/\lambda}-\mu \bigr)^{\alpha}|\bigl(\log(1+\lambda t)\bigr)^{m}x^{n} \bigr\rangle \\ &=\frac{1}{m!(1-\mu)^{\alpha}\lambda^{m}} \biggl\langle P_{r,s}(t) \bigl((1+\lambda t)^{1/\lambda}-\mu \bigr)^{\alpha}\Big|m!\sum_{\ell\geq m}S_{1}( \ell ,m)\frac{\lambda^{\ell}}{\ell!}t^{\ell}x^{n} \biggr\rangle \\ &=\frac{1}{(1-\mu)^{\alpha}\lambda^{m}}\sum_{\ell=m}^{n} \binom{n}{\ell }S_{1}(\ell,m)\lambda^{\ell}\bigl\langle \bigl((1+\lambda t)^{1/\lambda}-\mu \bigr)^{\alpha}|P_{r,s}(t)x^{n-\ell} \bigr\rangle \\ &=\frac{1}{(1-\mu)^{\alpha}\lambda^{m}}\sum_{\ell=m}^{n}\sum _{k=0}^{n-\ell }\binom{n}{\ell} \binom{n-\ell}{k}S_{1}(\ell,m)\lambda^{\ell}D\mathcal {E}_{k}(\lambda|\mathbf{a};\mathbf{b})w_{n,\ell,k}, \end{aligned}$$
$$\begin{aligned} w_{n,\ell,k}&= \bigl\langle \bigl((1+\lambda t)^{1/\lambda}-\mu \bigr)^{\alpha}|x^{n-\ell-k} \bigr\rangle \\ &= \Biggl\langle \sum_{p=0}^{\alpha}\binom{\alpha}{p}(-\mu)^{\alpha -p}(1+\lambda t)^{p/\lambda}\bigg|x^{n-\ell-k} \Biggr\rangle \\ &=\sum_{p=0}^{\alpha}\binom{\alpha}{p}(- \mu)^{\alpha-p} \biggl\langle \sum_{q\geq0}(p| \lambda)_{q}\frac{t^{q}}{q!}\Big|x^{n-\ell-k} \biggr\rangle \\ &=\sum_{p=0}^{\alpha}\binom{\alpha}{p}(- \mu)^{\alpha-p} (p|\lambda )_{n-\ell-k}. \end{aligned}$$
$$ \frac{1}{(1-\mu)^{\alpha}\lambda^{m}}\sum_{\ell=m}^{n}\sum _{k=0}^{n-\ell}\sum_{p=0}^{\alpha}\binom{n}{\ell}\binom{n-\ell}{k}\binom{\alpha }{p}S_{1}(\ell ,m) \lambda^{\ell}(-\mu)^{\alpha-p}D\mathcal{E}_{k}(\lambda| \mathbf {a};\mathbf{b}) (p|\lambda)_{n-\ell-k}, $$
Now we are interested in expressing our polynomials in terms of \(\mathcal{E}_{n}^{(\alpha)}(\lambda,x)\) which are called the degenerate Euler polynomials of order α. Note that
(see [10]). Using similar techniques as in the proof of the above theorem, we obtain the following relation.
$$\mathcal{E}_{n}^{(\alpha)}(\lambda,x)\sim \biggl( \biggl( \frac {e^{t}+1}{2} \biggr)^{\alpha},\frac{e^{\lambda t}-1}{\lambda} \biggr) $$
Theorem 4.5
For all
\(n\geq0\), the polynomial
\(D\mathcal{E}_{n}(\lambda,x|\mathbf {a};\mathbf{b})\)
is given by
$$\frac{1}{2^{\alpha}}\sum_{m=0}^{n} \binom{n}{m} \Biggl(\sum_{q=0}^{n-m} \sum_{p=0}^{\alpha}\binom{n-m}{q} \binom{\alpha}{p}(p|\lambda)_{q}D\mathcal {E}_{n-m-q}(\lambda| \mathbf{a};\mathbf{b}) \Biggr)\mathcal {E}^{(\alpha)}_{m}( \lambda,x). $$
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Authors’ contributions
All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.