Open Access

Differential equations arising from the generating function of general modified degenerate Euler numbers

Advances in Difference Equations20162016:129

https://doi.org/10.1186/s13662-016-0858-7

Received: 11 April 2016

Accepted: 9 May 2016

Published: 11 May 2016

Abstract

In this paper, we introduce the general modified degenerate Euler numbers and study ordinary differential equations arising from the generating function of these numbers. In addition, we give some new explicit identities for the general modified degenerate Euler numbers arising from our differential equations.

Keywords

general modified degenerate Euler numbersdifferential equations

MSC

05A1911B3711B8334A34

1 Introduction

As is known, the Euler numbers are defined by the generating function
$$ \frac{2}{e^{t}+1}=\sum_{n=0}^{\infty}E_{n} \frac{t^{n}}{n!}\quad (\text{see [1--9]} ). $$
(1.1)
Carlitz [2] considered the degenerate Euler numbers defined by the generating function
$$ \frac{2}{ (1+\lambda t )^{\frac{1}{\lambda}}+1}=\sum_{n=0}^{\infty} \mathcal{E}_{n,\lambda}\frac{t^{n}}{n!}. $$
(1.2)
In [7], the modified degenerate Euler numbers, which are slightly different from Carlitz’s degenerate Euler numbers, are defined by
$$ \frac{2}{ (1+\lambda )^{\frac{t}{\lambda}}+1}=\sum_{n=0}^{\infty}\tilde{ \mathcal{E}}_{n,\lambda}\frac{t^{n}}{n!}. $$
(1.3)
Note that \(\lim_{\lambda\rightarrow0}\tilde{\mathcal{E}}_{n,\lambda }=\lim_{\lambda\rightarrow0}\mathcal{E}_{n,\lambda}=E_{n}\) (\(n\ge0\)). Recently, Kim and Kim [6] studied nonlinear differential equations given by
$$ \biggl(\frac{d}{dt} \biggr)^{N} \biggl(\frac{1}{ (1+\lambda t )^{\frac{1}{\lambda}}+1} \biggr) =\frac{ (-1 )^{N}}{ (1+\lambda t )^{N}}\sum_{i=1}^{N+1}a_{i} (N,\lambda )F^{i}, $$
(1.4)
where \(F=\frac{1}{ (1+\lambda t )^{\frac{1}{\lambda}}+1}\).
Let α, a, b be nonzero real numbers. Then we consider the general modified degenerate Euler numbers as follows:
$$ \frac{2}{\alpha (1+\lambda )^{\frac{at}{\lambda}}+b}=\sum_{n=0}^{\infty}\tilde{ \mathcal{E}}_{n,\lambda} (\alpha\mid a,b )\frac{t^{n}}{n!}. $$
(1.5)
From (1.5) we note that
$$\begin{aligned} \lim_{\lambda\rightarrow0}\frac{2}{\alpha (1+\lambda )^{\frac{at}{\lambda}}+b} =&\frac{2}{\alpha e^{at}+b} \\ =&\frac{1}{b}\frac{2}{\frac{\alpha}{b}e^{at}+1} \\ =&\frac{1}{b}\sum_{n=0}^{\infty}E_{n,\frac{\alpha}{b}}a^{n} \frac {t^{n}}{n!}, \end{aligned}$$
(1.6)
where \(E_{n,q}\) (\(n\ge0\)) are the Apostol-Euler numbers given by the generating function
$$ \frac{2}{qe^{t}+1}=\sum_{n=0}^{\infty}E_{n,q} \frac{t^{n}}{n!}\quad (\text{see [1, 3]} ). $$
(1.7)
Thus, by (1.5) and (1.6) we get
$$\frac{a^{n}}{b}E_{n,\frac{\alpha}{b}}=\lim_{\lambda\rightarrow0}\tilde { \mathcal{E}}_{n,\lambda} (\alpha\mid a,b )\quad (n\ge 0 ). $$
Bayad and Kim [1] studied the following nonlinear differential equations related to Apostol-Euler numbers:
$$ F_{q}^{N}=\frac{1}{ (N-1 )!}\sum _{k=0}^{N}a_{k} (N )F_{q}^{ (k-1 )} \quad (N\in\mathbb{N} ), $$
(1.8)
where \(F_{q}^{ (k )}= (\frac{d}{dt} )^{k}F_{q} (t )\), \(F_{q} (t )=\frac{1}{qe^{t}+1}\).

In this paper, we study the ordinary differential equations associated with the generating function of general modified degenerate Euler numbers. In addition, we give some new and explicit formulas and identities for those numbers arising from our differential equations.

2 Generalized modified degenerate Euler numbers

For nonzero real numbers α, a, b, let
$$ F=F (t )=\frac{1}{\alpha (1+\lambda )^{\frac {at}{\lambda}}+b}. $$
(2.1)
Then by (2.1) we get
$$\begin{aligned} F^{ (1 )} & =\frac{dF}{dt} (t ) \\ & =\frac{ (-1 )\frac{a}{\lambda}\log (1+\lambda )}{ (\alpha (1+\lambda )^{\frac{at}{\lambda}}+b )^{2}} \bigl(\alpha (1+\lambda )^{\frac{\alpha t}{\lambda }} \bigr) \\ & =\frac{ (-1 )\frac{a}{\lambda}\log (1+\lambda )}{ (\alpha (1+\lambda )^{\frac{at}{\lambda}}+b )^{2}} \bigl\{ \alpha (1+\lambda )^{\frac{at}{\lambda }}+b-b \bigr\} \\ & = (-1 )\frac{a}{\lambda}\log (1+\lambda ) \bigl(F-bF^{2} \bigr). \end{aligned}$$
(2.2)
Thus, from (2.2) we have
$$ F^{ (1 )}=\frac{a}{\lambda}\log (1+\lambda ) \bigl(bF^{2}-F \bigr). $$
(2.3)
From (2.3) we derive the following equation:
$$\begin{aligned} F^{ (2 )} & =\frac{d}{dt}F^{ (1 )} \\ & =\frac{a}{\lambda}\log (1+\lambda ) \bigl\{ 2bFF^{ (1 )}-F^{ (1 )} \bigr\} \\ & =\frac{a}{\lambda}\log (1+\lambda ) (2bF-1 )F^{ (1 )} \\ & = \biggl(\frac{a}{\lambda}\log (1+\lambda ) \biggr)^{2} (2bF-1 ) \bigl(bF^{2}-F \bigr) \\ & = \biggl(\frac{a}{\lambda}\log (1+\lambda ) \biggr)^{2} \bigl(2b^{2}F^{3}-3bF^{2}+F \bigr). \end{aligned}$$
(2.4)
Continuing this process, we set
$$\begin{aligned} F^{ (N )} & = \biggl(\frac{d}{dt} \biggr)^{N}F (t ) \\ & = \biggl(\frac{a}{\lambda}\log (1+\lambda ) \biggr)^{N}\sum _{k=1}^{N+1}a_{k} (N )b^{k-1}F^{k}. \end{aligned}$$
(2.5)
By taking the derivative of (2.5) with respect to t we have
$$\begin{aligned} F^{ (N+1 )} & =\frac{dF^{ (N )}}{dt} \\ & = \biggl(\frac{a}{\lambda}\log (1+\lambda ) \biggr)^{N}\sum _{k=1}^{N+1}a_{k} (N )b^{k-1}kF^{k-1}F^{ (1 )} \\ & = \biggl(\frac{a}{\lambda}\log (1+\lambda ) \biggr)^{N+1}\sum _{k=1}^{N+1} \bigl(ka_{k} (N )b^{k}F^{k+1}-a_{k} (N )b^{k-1}kF^{k} \bigr) \\ & = \biggl(\frac{a}{\lambda}\log (1+\lambda ) \biggr)^{N+1} \Biggl\{ \sum_{k=2}^{N+2} (k-1 )a_{k-1} (N ) b^{k-1}F^{k}-\sum_{k=1}^{N+1}ka_{k} (N )b^{k-1}F^{k} \Biggr\} . \end{aligned}$$
(2.6)
Replacing N by \(N+1\) in (2.5), we get
$$ F^{ (N+1 )}= \biggl(\frac{a}{\lambda}\log (1+\lambda ) \biggr)^{N+1}\sum_{k=1}^{N+2}a_{k} (N+1 )b^{k-1}F^{k}. $$
(2.7)
Comparing the coefficients on both sides of (2.6) and (2.7), we obtain
$$ a_{1} (N+1 )=-a_{1} (N ). $$
(2.8)
Thus, by (2.8) we get
$$ a_{1} (N+1 )=-a_{1} (N )= (-1 )^{2}a_{1} (N-1 )=\cdots= (-1 )^{N}a_{1} (1 ). $$
(2.9)
From (2.5) we have
$$ \frac{a}{\lambda}\log (1+\lambda ) \bigl\{ bF^{2}-F \bigr\} = \frac{a}{\lambda}\log (1+\lambda ) \bigl\{ a_{1} (1 )F+a_{2} (1 )bF^{2} \bigr\} . $$
(2.10)
By (2.10) we get
$$ a_{1} (1 )=-1\quad \text{and}\quad a_{2} (1 )=1. $$
(2.11)
Thus, from (2.9) and (2.11) we have
$$ a_{1} (N+1 )= (-1 )^{N}a_{1} (1 )= (-1 )^{N+1}. $$
(2.12)
By (2.6) and (2.7) we see that
$$\begin{aligned} a_{N+2} (N+1 ) =& (N+1 )a_{N+1} (N ) \\ =& (N+1 )Na_{N} (N-1 ) \\ =& (N+1 )N (N-1 )a_{N-1} (N-2 ) \\ & \vdots \\ =& (N+1 )N (N-1 )\cdots2a_{2} (1 ) \\ =& (N+1 )!. \end{aligned}$$
(2.13)
Thus, by (2.13) we have
$$ a_{N+2} (N+1 )= (N+1 )!. $$
(2.14)
For \(2\le k\le N+1\), by comparing the coefficients on both sides of (2.6) and (2.7) we have
$$ a_{k} (N+1 )= (k-1 )a_{k-1} (N )-ka_{k} (N ). $$
(2.15)
Let \(k=2\) in (2.15). Then we have
$$\begin{aligned} a_{2} (N+1 ) =&a_{1} (N )-2a_{2} (N ) \\ =&a_{1} (N )-2 \bigl(a_{1} (N-1 )-2a_{2} (N-1 ) \bigr) \\ =&a_{1} (N )-2a_{1} (N-1 )+ (-1 )^{2}2^{2}a_{2} (N-1 ) \\ =&a_{1} (N )-2a_{1} (N-1 )+ (-1 )^{2}2^{2} \bigl\{ a_{1} (N-2 )-2a_{2} (N-2 ) \bigr\} \\ =&a_{1} (N )-2a_{1} (N-1 )+ (-1 )^{2}2^{2}a_{1} (N-2 )+ (-1 )^{3}2^{3}a_{2} (N-2 ) \\ & \vdots \\ =&\sum_{k=0}^{N-1} (-1 )^{k}a_{1} (N-k )2^{k}+ (-1 )^{N}2^{N}a_{2} (1 ) \\ =&\sum_{k=0}^{N} (-1 )^{k}a_{1} (N-k )2^{k}. \end{aligned}$$
(2.16)
For \(k=3\) in (2.15), we have
$$\begin{aligned} a_{3} (N+1 ) =&2a_{2} (N )-3a_{3} (N ) \\ =&2a_{2} (N )-3 \bigl\{ 2a_{2} (N-1 )-3a_{3} (N-1 ) \bigr\} \\ =&2a_{2} (N )-3\cdot2a_{2} (N-1 )+ (-1 )^{2}3^{2}a_{3} (N-1 ) \\ =&2a_{2} (N )-3\cdot2a_{2} (N-1 )+ (-1 )^{2}3^{2} \bigl\{ 2a_{2} (N-2 )-3a_{3} (N-2 ) \bigr\} \\ =&2a_{2} (N )-3\cdot2a_{2} (N-1 )+ (-1 )^{2}3^{2}2a_{2} (N-2 )+ (-1 )^{3}3^{3}a_{3} (N-2 ) \\ & \vdots \\ =&2\sum_{k=0}^{N-2}a_{2} (N-k ) (-1 )^{k}3^{k}+ (-1 )^{N-1}3^{N-1}a_{3} (2 ) \\ =&2\sum_{k=0}^{N-1}a_{2} (N-k ) (-1 )^{k}3^{k}. \end{aligned}$$
(2.17)
Continuing this process, we deduce
$$ a_{j} (N+1 )= (j-1 )\sum_{k=0}^{N-j+2}a_{j-1} (N-k ) (-1 )^{k}j^{k}, $$
(2.18)
where \(2\le j\le N+1\).
Now we give an explicit expression for \(a_{j} (N+1 )\) in (2.18). From (2.12) and (2.16) we can derive the following equation:
$$\begin{aligned} a_{2} (N+1 ) & =\sum_{k=0}^{N} (-1 )^{k}a_{1} (N-k )2^{k} \\ & =\sum_{k=0}^{N} (-1 )^{k} (-1 )^{N-k}2^{k} = (-1 )^{N}\sum_{k=0}^{N}2^{k}. \end{aligned}$$
(2.19)
By (2.17) we get
$$\begin{aligned} a_{3} (N+1 ) & =2\sum_{k_{2}=0}^{N-1}a_{2} (N-k_{2} ) (-1 )^{k_{2}}3^{k_{2}} \\ & =2\sum_{k_{2}=0}^{N-1} (-1 )^{N-k_{2}-1} \sum_{k_{1}=0}^{N-k_{2}-1}2^{k_{1}} (-1 )^{k_{2}}3^{k_{2}} \\ & =2 (-1 )^{N-1}\sum_{k_{2}=0}^{N-1} \sum_{k_{1}=0}^{N-1-k_{2}}2^{k_{1}}3^{k_{2}}. \end{aligned}$$
(2.20)
Continuing this process, we deduce that, for \(2\le j\le N+1\),
$$\begin{aligned}& a_{j} (N+1 ) \\& \quad = (j-1 )! (-1 )^{N-j+2} \\& \qquad {} \times\sum _{k_{j-1}=0}^{N-j+2}\sum_{k_{j-2}=0}^{N-j+2-k_{j-1}}\sum_{k_{j-3}=0}^{N-j+2-k_{j-1}-k_{j-2}}\cdots\sum _{k_{1}=0}^{N-j+2-k_{j-1}-\cdots-k_{2}}j^{k_{j-1}} (j-1 )^{k_{j-2}}\cdots3^{k_{2}}2^{k_{1}}. \end{aligned}$$
(2.21)

Therefore, by (2.5) and (2.21) we obtain the following theorem.

Theorem 1

Let α, a, b be nonzero real numbers. The family of nonlinear differential equations
$$F^{ (N )}= \biggl(\frac{a}{\lambda}\log (1+\lambda ) \biggr)^{N}\sum_{k=1}^{N+1}a_{k} (N )b^{k-1}F^{k} $$
has a solution \(F=F (t )=\frac{1}{\alpha (1+\lambda )^{\frac{at}{\lambda}}+b}\), where \(a_{1} (N )= (-1 )^{N}\), and
$$\begin{aligned} a_{j} (N ) =& (j-1 )! (-1 )^{N-j+1} \\ &{}\times\sum _{k_{j-1}=0}^{N-j+1}\sum_{k_{j-2}=0}^{N-j+1-k_{j-1}}\cdots\sum _{k_{1}=0}^{N-j+1-k_{j-1}-\cdots-k_{2}}j^{k_{j-1}} (j-1 )^{k_{j-2}}\cdots3^{k_{2}}2^{k_{1}} \end{aligned}$$
for \(2\le j\le N+1\).
Now we define the general modified degenerate Euler numbers given by the generating function
$$ \frac{2}{\alpha (1+\lambda )^{\frac{at}{\lambda}}+b}=\sum_{n=0}^{\infty}\tilde{ \mathcal{E}}_{n,\lambda} (\alpha;a,b )\frac{t^{n}}{n!}. $$
(2.22)
Note that \(\tilde{\mathcal{E}}_{n,\lambda} (1;1,1 )\) are the modified degenerate Euler numbers given by
$$\frac{2}{ (1+\lambda )^{\frac{t}{\lambda}}+1}=\sum_{n=0}^{\infty}\tilde{ \mathcal{E}}_{n,\lambda}\frac{t^{n}}{n!}. $$
Now we observe that
$$\begin{aligned} F^{ (N )} & =\frac{1}{2} \biggl(\frac{d}{dt} \biggr)^{N} \biggl(\frac{2}{\alpha (1+\lambda )^{\frac{at}{\lambda}}+b} \biggr) \\ & =\frac{1}{2}\sum_{n=0}^{\infty} \frac{\tilde{\mathcal{E}}_{n,\lambda } (\alpha;a,b )}{n!} \biggl(\frac{d}{dt} \biggr)^{N}t^{n} \\ & =\frac{1}{2}\sum_{n=0}^{\infty}\tilde{ \mathcal{E}}_{n+N,\lambda} (\alpha;a,b )\frac{t^{n}}{n!}. \end{aligned}$$
(2.23)
For \(r\in\mathbb{N}\), the higher-order general modified degenerate Euler numbers are defined by the generating function
$$ \biggl(\frac{2}{\alpha (1+\lambda )^{\frac{at}{\lambda }}+b} \biggr)^{r}=\sum _{n=0}^{\infty}\tilde{\mathcal{E}}_{n,\lambda }^{ (r )} (\alpha;a,b )\frac{t^{n}}{n!}. $$
(2.24)

Therefore, by Theorem 1, (2.23), and (2.21) we obtain the following theorem.

Theorem 2

Let α, a, b be nonzero real numbers. For \(n\ge0\), we have
$$\tilde{\mathcal{E}}_{n+N} (\alpha;a,b )= \biggl(\frac{a}{\lambda }\log (1+\lambda ) \biggr)^{N}\sum_{k=1}^{N+1}a_{k} (N )b^{k-1}2^{1-k}\tilde{\mathcal{E}}_{n,\lambda}^{ (k )} (\alpha;a,b ), $$
where \(a_{1} (N )=(-1)^{N}\), and, for \(2\le j\le N+1\),
$$\begin{aligned} a_{j} (N ) =& (j-1 )! (-1 )^{N-j+1} \\ &{}\times\sum _{k_{j-1}=0}^{N-j+1}\sum_{k_{j-2}=0}^{N-j+1-k_{j-1}}\cdots\sum _{k_{1}=0}^{N-j+1-k_{j-1}-\cdots-k_{2}}j^{k_{j-1}} (j-1 )^{k_{j-2}}\cdots3^{k_{2}}2^{k_{1}}. \end{aligned}$$

Declarations

Acknowledgements

The first author is appointed as a chair professor at Tianjin Polytechnic University, Tianjin City, China, from August 2015 to August 2019. We would like to thank the referee for his detailed suggestions that helped to improve the original manuscript.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics, College of Science, Tianjin Polytechnic University
(2)
Department of Mathematics, Kwangwoon University
(3)
Department of Mathematics, Sogang University
(4)
Department of Applied Mathematics, Pukyong National University

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