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

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.

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 $2\le j\le N+1$,

$$\begin{aligned} a_{j} (N ) =& (j-1 )! (-1 )^{N-j+1} \\ &{}\times\sum _{k_{j-1}=0}^{N-j+1}\sum_{k_{j-2}=0}^{N-j+1-k_{j-1}}\cdots\sum _{k_{1}=0}^{N-j+1-k_{j-1}-\cdots-k_{2}}j^{k_{j-1}} (j-1 )^{k_{j-2}}\cdots3^{k_{2}}2^{k_{1}}. \end{aligned}$$