Some identities of Frobenius-Euler polynomials arising from umbral calculus

Abstract

In this paper, we study some interesting identities of Frobenius-Euler polynomials arising from umbral calculus.

1 Introduction

Let C be the complex number field, and let F be the set of all formal power series in the variable t over C with

$$\mathbf{F}=\{f(t)=\sum _{k=0}^{\mathrm{\infty }}\frac{a_k}{k!}{t}^k|a_k\in \mathbf{C}\}.$$

We use notation $\mathbb{P}=\mathbf{C}[x]$ and ${\mathbb{P}}^{\ast }$ denotes the vector space of all linear functional on ℙ.

Also, $〈L|p(x)〉$ denotes the action of the linear functional L on the polynomial $p(x)$, and we remind that the vector space operations on ${\mathbb{P}}^{\ast }$ is defined by

$$\begin{array}{c}〈L+M|p(x)〉=〈L|p(x)〉+〈M|p(x)〉,\hfill \\ 〈cL|p(x)〉=c〈L|p(x)〉(\text{see [1]}),\hfill \end{array}$$

where c is any constant in C.

The formal power series

$$f(t)=\sum _{k=0}^{\mathrm{\infty }}\frac{a_k}{k!}{t}^k\in \mathbf{F}(\text{see [1, 2]}),$$

(1)

defines a linear functional on ℙ by setting

$$〈f(t)|x^n〉=a_n,\text{for all }n\ge 0.$$

(2)

In particular,

$$〈t^k|x^n〉=n!{\delta }_{n,k},$$

(3)

where ${\delta }_{n,k}$ is the Kronecker symbol. If $f_L(t)={\sum }_{k=0}^{\mathrm{\infty }}\frac{〈L|x^k〉}{k!}{t}^k$, then we get $〈f_L(t)|x^n〉=〈L|x^n〉$ and so as linear functionals $L=f_L(t)$ (see [1, 2]).

In addition, the map $L\mapsto f_L(t)$ is a vector space isomorphism from ${\mathbb{P}}^{\ast }$ onto F (see [1, 2]). Henceforth, F will denote both the algebra of formal power series in t and the vector space of all linear functionals on ℙ, and so an element $f(t)$ of F will be thought of as both a formal power series and a linear functional. We shall call F the umbral algebra (see [1, 2]).

Let us give an example. For y in C the evaluation functional is defined to be the power series $e^{yt}$. From (2), we have $〈e^{yt}|x^n〉=y^n$ and so $〈e^{yt}|p(x)〉=p(y)$ (see [1, 2]). Notice that for all $f(t)$ in F,

$$f(t)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈f(t)|x^t〉}{k!}{t}^k$$

(4)

and for all polynomial $p(x)$

$$p(x)=\sum _{k\ge 0}\frac{〈t^k|p(x)〉}{k!}{x}^k(\text{see [1, 2]}).$$

(5)

For $f_1(t),f_2(t),\dots ,f_m(t)\in \mathbf{F}$, we have

$$\begin{array}{c}〈f_1(t)f_2(t)\cdots f_m(t)|x^n〉\hfill \\ =\sum \left(\genfrac{}{}{0ex}{}{n}{i_1,\dots ,i_m}\right)〈f_1(t)|x^{i_1}〉\cdots 〈f_n(t)|x^{i_m}〉,\hfill \end{array}$$

where the sum is over all nonnegative integers $i_1,i_2,\dots ,i_m$ such that $i_1+\cdots +i_m=n$ (see [1, 2]). The order $o(f(t))$ of the power series $f(t)\ne 0$ is the smallest integer k for which $a_k$ does not vanish. We define $o(f(t))=\mathrm{\infty }$ if $f(t)=0$. We see that $o(f(t)g(t))=o(f(t))+o(g(t))$ and $o(f(t)+g(t))\ge min\{o(f(t)),o(g(t))\}$. The series $f(t)$ has a multiplicative inverse, denoted by $f{(t)}^{-1}$ or $\frac{1}{f(t)}$, if and only if $o(f(t))=0$. Such series is called an invertible series. A series $f(t)$ for which $o(f(t))=1$ is called a delta series (see [1, 2]). For $f(t),g(t)\in \mathbf{F}$, we have $〈f(t)g(t)|p(x)〉=〈f(t)|g(t)p(x)〉$.

A delta series $f(t)$ has a compositional inverse $\overline{f}(t)$ such that $f(\overline{f}(t))=\overline{f}(f(t))=t$.

For $f(t),g(t)\in \mathbf{F}$, we have $〈f(t)g(t)|p(x)〉=〈f(t)|g(t)p(x)〉$.

From (5), we have

$$p^{(k)}(x)=\frac{d^{k}p(x)}{d{x}^k}=\sum _{l=k}^{\mathrm{\infty }}\frac{〈t^l|p(x)〉}{l!}l(l-1)\cdots (l-k+1)x^{l-k}.$$

Thus, we see that

$$p^{(k)}(0)=〈t^k|p(x)〉=〈1|p^{(k)}(x)〉.$$

(6)

By (6), we get

$$t^{k}p(x)=p^{(k)}(x)=\frac{d^k(p(x))}{d{x}^k}(\text{see [1, 2]}).$$

(7)

By (7), we have

$$e^{yt}p(x)=p(x+y)(\text{see [1, 2]}).$$

(8)

Let $S_n(x)$ be a polynomial with $deg{S}_n(x)=n$.

Let $f(t)$ be a delta series, and let $g(t)$ be an invertible series. Then there exists a unique sequence $S_n(x)$ of polynomials such that $〈g(t)f{(t)}^k|S_n(x)〉=n!{\delta }_{n,k}$ for all $n,k\ge 0$. The sequence $S_n(x)$ is called the Sheffer sequence for $(g(t),f(t))$ or that $S_n(t)$ is Sheffer for $(g(t),f(t))$.

The Sheffer sequence for $(1,f(t))$ is called the associated sequence for $f(t)$ or $S_n(x)$ is associated to $f(t)$. The Sheffer sequence for $(g(t),t)$ is called the Appell sequence for $g(t)$ or $S_n(x)$ is Appell for $g(t)$ (see [1, 2]). The umbral calculus is the study of umbral algebra and the modern classical umbral calculus can be described as a systemic study of the class of Sheffer sequences. Let $p(x)\in \mathbb{P}$. Then we have

(9)

(10)

and

$$〈e^{yt}-1|p(x)〉=p(y)-p(0)(\text{see [1, 2]}).$$

(11)

Let $S_n(x)$ be Sheffer for $(g(t),f(t))$. Then

(12)

(13)

(14)

(15)

For $\lambda (\ne 1)\in \mathbf{C}$, we recall that the Frobenius-Euler polynomials are defined by the generating function to be

$$\frac{1-\lambda }{e^t-\lambda }{e}^{xt}=e^{H(x|\lambda )t}=\sum _{n=0}^{\mathrm{\infty }}{H}_n(x|\lambda )\frac{t^n}{n!},$$

(16)

with the usual convention about replacing $H^n(x|\lambda )$ by $H_n(x|\lambda )$ (see [3]). In the special case, $x=0$, $H_n(0|\lambda )=H_n(\lambda )$ are called the n th Frobenius-Euler numbers. By (16), we get

$$H_n(x|\lambda )={(H(\lambda )+x)}^n=\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)H_{n-l}^{(\lambda )}{x}^l,$$

(17)

and

$${(H(\lambda )+1)}^n-\lambda H_n(\lambda )=(1-\lambda ){\delta }_{0,n}(\text{see [1, 4–13]}).$$

(18)

From (17), we note that the leading coefficient of $H_n(x|\lambda )$ is $H_0(\lambda )=1$. So, $H_n(x|\lambda )$ is a monic polynomial of degree n with coefficients in $\mathbf{Q}(\lambda )$.

In this paper, we derive some new identities of Frobenius-Euler polynomials arising from umbral calculus.

2 Applications of umbral calculus to Frobenius-Euler polynomials

Let $S_n(x)$ be an Appell sequence for $g(t)$. From (14), we have

$$\frac{1}{g(t)}{x}^n=S_n(x)\text{if and only if}{x}^n=g(t)S_n(x)(n\ge 0).$$

(19)

For $\lambda (\ne 1)\in \mathbf{C}$, let us take $g_{\lambda }(t)=\frac{e^t-\lambda }{1-\lambda }\in \mathbf{F}$.

Then we see that $g_{\lambda }(t)$ is an invertible series.

From (16), we have

$$\sum _{k=0}^{\mathrm{\infty }}\frac{H_k(x|\lambda )}{k!}{t}^k=\frac{1}{g_{\lambda }(t)}{e}^{xt}.$$

(20)

By (20), we get

$$\frac{1}{g_{\lambda }(t)}{x}^n=H_n(x|\lambda )(\lambda (\ne 1)\in \mathbf{C},n\ge 0),$$

(21)

and by (17), we get

$$t{H}_n(x|\lambda )=H_n^{\mathrm{\prime }}(x|\lambda )=n{H}_{n-1}(x|\lambda ).$$

(22)

Therefore, by (21) and (22), we obtain the following proposition.

Proposition 1 For $\lambda (\ne 1)\in \mathbf{C}$, $n\ge 0$, we see that $H_n(x|\lambda )$ is the Appell sequence for $g_{\lambda }(t)=\frac{e^t-\lambda }{1-\lambda }$.

From (20), we have

$$\begin{array}{rcl}\sum _{k=1}^{\mathrm{\infty }}\frac{H_k(x|\lambda )}{k!}k{t}^{k-1}& =& \frac{x{g}_{\lambda }(t)e^{xt}-g_{\lambda }^{\mathrm{\prime }}(t)e^{xt}}{g_{\lambda }{(t)}^2}\\ =& \sum _{k=0}^{\mathrm{\infty }}\{x\frac{1}{g_{\lambda }(t)}{x}^k-\frac{g_{\lambda }^{\mathrm{\prime }}(t)}{g_{\lambda }(t)}\frac{1}{g_{\lambda }(t)}{x}^k\}\frac{t^k}{k!}.\end{array}$$

(23)

By (21) and (23), we get

$$H_{k+1}(x|\lambda )=x{H}_k(x|\lambda )-\frac{g_{\lambda }^{\mathrm{\prime }}(t)}{g_{\lambda }(t)}{H}_k(x|\lambda ).$$

(24)

Therefore, by (24) we obtain the following theorem.

Theorem 2 Let $g_{\lambda }(t)=\frac{e^t-\lambda }{1-\lambda }\in \mathbf{F}$. Then we have

$$H_{k+1}(x|\lambda )=(x-\frac{g_{\lambda }^{\mathrm{\prime }}(t)}{g_{\lambda }(t)})H_k(x|\lambda )(k\ge 0).$$

From (16), we have

$$\sum _{n=0}^{\mathrm{\infty }}(H_n(x+1|\lambda )-\lambda H_n(x|\lambda ))\frac{t^n}{n!}=\frac{1-\lambda }{e^t-\lambda }{e}^{(x+1)t}-\lambda \frac{1-\lambda }{e^t-\lambda }{e}^{xt}=(1-\lambda )e^{xt}.$$

(25)

By (25), we get

$$H_n(x+1|\lambda )-\lambda H_n(x|\lambda )=(1-\lambda )x^n.$$

(26)

From Theorem 2, we can derive the following equation (27):

$$g_{\lambda }(t)H_{k+1}(x|\lambda )=(g_{\lambda }(t)x-g_{\lambda }^{\mathrm{\prime }}(t))H_k(x|\lambda ).$$

(27)

By (27), we get

$$\left(\frac{e^t-\lambda }{1-\lambda }\right)H_{k+1}(x|\lambda )=\frac{e^t-\lambda }{1-\lambda }x{H}_k(x|\lambda )-\frac{e^t}{1-\lambda }{H}_k(x|\lambda ).$$

(28)

From (8) and (28), we have

$$\begin{array}{rcl}{H}_{k+1}(x+1|\lambda )-\lambda H_{k+1}(x|\lambda )& =& (x+1)H_k(x+1|\lambda )-\lambda x{H}_k(x|\lambda )-H_k(x+1|\lambda )\\ =& x{H}_k(x+1|\lambda )-\lambda x{H}_k(x|\lambda ).\end{array}$$

Therefore, by (26), we obtain the following theorem.

Theorem 3 For $k\ge 0$, we have

$$H_{k+1}(x+1|\lambda )=\lambda H_{k+1}(x|\lambda )+(1-\lambda )x^{k+1}.$$

From (16), (17), and (18), we note that

$$\begin{array}{rcl}{\int }_x^{x+y}{H}_n(u|\lambda )du& =& \frac{1}{n+1}\{H_{n+1}(x+y|\lambda )-H_{n+1}(x|\lambda )\}\\ =& \frac{1}{n+1}\sum _{k=1}^{\mathrm{\infty }}\left(\genfrac{}{}{0ex}{}{n+1}{k}\right)H_{n+1-k}(x|\lambda )y^k\\ =& \sum _{k=1}^{\mathrm{\infty }}\frac{n(n-1)\cdots (n-k+2)}{k!}{H}_{n+1-k}(x|\lambda )y^k\\ =& \sum _{k=1}^{\mathrm{\infty }}\frac{y^k}{k!}{t}^{k-1}{H}_n(x|\lambda )\\ =& \frac{1}{t}(\sum _{k=0}^{\mathrm{\infty }}\frac{y^k}{k!}{t}^k-1)H_n(x|\lambda )\\ =& \frac{e^{yt}-1}{t}{H}_n(x|\lambda ).\end{array}$$

(29)

Therefore, by (29), we obtain the following theorem.

Theorem 4 For $\lambda (\ne 1)\in \mathbf{C}$, $n\ge 0$, we have

$${\int }_x^{x+y}{H}_n(u|\lambda )du=\frac{e^{yt}-1}{t}{H}_n(x|\lambda ).$$

By (15) and Proposition 1, we get

$$t\{\frac{1}{n+1}{H}_{n+1}(x|\lambda )\}=H_n(x|\lambda ).$$

(30)

From (30), we can derive equation (31):

$$\begin{array}{rcl}〈e^{yt}-1|\frac{H_{n+1}(x|\lambda )}{n+1}〉& =& 〈\frac{e^{yt}-1}{t}|t\left\{\frac{H_{n+1}(x|\lambda )}{n+1}\right\}〉\\ =& 〈\frac{e^{yt}-1}{t}|H_n(x|\lambda )〉.\end{array}$$

(31)

By (11) and (31), we get

$$\begin{array}{rcl}〈\frac{e^{yt}-1}{t}|H_n(x|\lambda )〉& =& 〈e^{yt}-1|\frac{H_{n+1}(x|\lambda )}{n+1}〉\\ =& \frac{1}{n+1}\{H_{n+1}(y|\lambda )-H_{n+1}(\lambda )\}={\int }_0^y{H}_n(u|\lambda )du.\end{array}$$

(32)

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

Corollary 5 For $n\ge 0$, we have

$$〈\frac{e^{yt}-1}{t}|H_n(x|\lambda )〉={\int }_0^y{H}_n(u|\lambda )du.$$

Let $\mathbb{P}(\lambda )=\{p(x)\in \mathbf{Q}(\lambda )[x]|degp(x)\le n\}$ be a vector space over $\mathbf{Q}(\lambda )$.

For $p(x)\in {\mathbb{P}}_n(\lambda )$, let us take

$$p(x)=\sum _{k=0}^n{b}_k{H}_k(x|\lambda ).$$

(33)

By Proposition 1, $H_n(x|\lambda )$ is an Appell sequence for $g_{\lambda }(t)=\frac{e^t-\lambda }{1-\lambda }$ where $\lambda (\ne 1)\in \mathbf{C}$. Thus, we have

$$〈\frac{e^t-\lambda }{1-\lambda }{t}^k|H_n(x|\lambda )〉=n!{\delta }_{n,k}.$$

(34)

From (33) and (34), we can derive

$$\begin{array}{rcl}〈\frac{e^t-\lambda }{1-\lambda }{t}^k|p(x)〉& =& \sum _{l=0}^n{b}_l〈\frac{e^t-\lambda }{1-\lambda }{t}^k|H_l(x|\lambda )〉\\ =& \sum _{l=0}^n{b}_{l}l!{\delta }_{l,k}=k!b_k.\end{array}$$

(35)

Thus, by (35), we get

$$\begin{array}{rcl}{b}_k& =& \frac{1}{k!}〈\frac{e^t-\lambda }{1-\lambda }{t}^k|p(x)〉\\ =& \frac{1}{k!(1-\lambda )}〈(e^t-\lambda )t^k|p(x)〉\\ =& \frac{1}{k!(1-\lambda )}〈e^t-\lambda |p^{(k)}(x)〉.\end{array}$$

(36)

From (11) and (36), we have

$$b_k=\frac{1}{k!(1-\lambda )}\{p^{(k)}(1)-\lambda p^{(k)}(0)\},$$

(37)

where $p^{(k)}(x)=\frac{d^{k}p(x)}{d{x}^k}$.

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

Theorem 6 For $p(x)\in {\mathbb{P}}_n(\lambda )$, let us assume that $p(x)={\sum }_{k=0}^n{b}_k{H}_k(x|\lambda )$. Then we have

$$b_k=\frac{1}{k!(1-\lambda )}\{p^{(k)}(1)-\lambda p^{(k)}(0)\},$$

where $p^{(k)}(1)=\frac{d^{k}p(x)}{d{x}^k}{|}_{x=1}$.

The higher-order Frobenius-Euler polynomials are defined by

$${\left(\frac{1-\lambda }{e^t-\lambda }\right)}^r{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{H}_n^{(r)}(x|\lambda )\frac{t^n}{n!},$$

(38)

where $\lambda (\ne 1)\in \mathbf{C}$ and $r\in \mathbf{N}$ (see [4, 11]).

In the special case, $x=0$, $H_n^{(r)}(0|\lambda )=H_n^{(r)}(\lambda )$ are called the n th Frobenius-Euler numbers of order r. From (38), we have

$$\begin{array}{rcl}{H}_n^{(r)}(x)& =& \sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)H_{n-l}^{(r)}(\lambda )x^l\\ =& \sum _{n_1+\cdots +n_r=n}\left(\genfrac{}{}{0ex}{}{n}{n_1,\dots ,n_r}\right)H_{n_1}(x|\lambda )\cdots H_{n_r}(x|\lambda ).\end{array}$$

(39)

Note that $H_n^{(r)}(x|\lambda )$ is a monic polynomial of degree n with coefficients in $\mathbf{Q}(\lambda )$.

For $r\in \mathbf{N}$, $\lambda (\ne 1)\in \mathbf{C}$, let $g_{\lambda }^r(t)={(\frac{e^t-\lambda }{1-\lambda })}^r$. Then we easily see that $g_{\lambda }^r(t)$ is an invertible series.

From (38) and (39), we have

$$\frac{1}{g_{\lambda }^r(t)}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{H}_n^{(r)}(x|\lambda )\frac{t^n}{n!},$$

(40)

and

$$t{H}_n^{(r)}(x|\lambda )=n{H}_{n-1}^{(r)}(x|\lambda ).$$

(41)

By (40), we get

$$\frac{1}{g_{\lambda }^r(t)}{x}^n=H_n^{(r)}(x|\lambda )(n\in {\mathbf{Z}}_{+},r\in \mathbf{N}).$$

(42)

Therefore, by (41) and (42), we obtain the following proposition.

Proposition 7 For $n\in {\mathbf{Z}}_{+}$, $H_n^{(r)}(x|\lambda )$ is an Appell sequence for

$$g_{\lambda }^r(t)={\left(\frac{e^t-\lambda }{1-\lambda }\right)}^r.$$

Moreover,

$$\frac{1}{g_{\lambda }^r(t)}{x}^n=H_n^{(r)}(x|\lambda )\mathit{\text{and}}t{H}_n^{(r)}(x|\lambda )=n{H}_{n-1}^{(r)}(x|\lambda ).$$

Remark Note that

$$〈\frac{1-\lambda }{e^t-\lambda }|x^n〉=H_n(\lambda ).$$

(43)

From (43), we have

(44)

(45)

By (43), (44), and (45), we get

$$\sum _{n=i_1+\cdots +i_r}\left(\genfrac{}{}{0ex}{}{n}{i_1,\dots ,i_r}\right)H_{i_1}(\lambda )\cdots H_{i_r}(\lambda )=H_n^{(r)}(\lambda ).$$

Let us take $p(x)\in {\mathbb{P}}_n(\lambda )$ with

$$p(x)=\sum _{k=0}^n{C}_k^{(r)}{H}_k^{(r)}(x|\lambda ).$$

(46)

From the definition of Appell sequences, we have

$$〈{\left(\frac{e^t-\lambda }{1-\lambda }\right)}^r|H_n^{(r)}(x|\lambda )〉=n!{\delta }_{n,k}.$$

(47)

By (46) and (47), we get

$$\begin{array}{rcl}〈{\left(\frac{e^t-\lambda }{1-\lambda }\right)}^r{t}^k|p(x)〉& =& \sum _{l=0}^n{C}_l^{(r)}〈{\left(\frac{e^t-\lambda }{1-\lambda }\right)}^r{t}^k|H_l(x|\lambda )〉\\ =& \sum _{l=0}^n{C}_l^{(r)}l!{\delta }_{l,k}=k!C_k^{(r)}.\end{array}$$

(48)

Thus, from (48), we have

$$\begin{array}{rcl}{C}_k^{(r)}& =& \frac{1}{k!}〈{\left(\frac{e^t-\lambda }{1-\lambda }\right)}^r{t}^k|p(x)〉\\ =& \frac{1}{k!{(1-\lambda )}^r}〈{(e^t-\lambda )}^r{t}^k|p(x)〉\\ =& \frac{1}{k!{(1-\lambda )}^r}\sum _{l=0}^r\left(\genfrac{}{}{0ex}{}{r}{l}\right){(-\lambda )}^{r-l}〈e^{lt}|p^{(k)}(x)〉\\ =& \frac{1}{k!{(1-\lambda )}^r}\sum _{l=0}^r\left(\genfrac{}{}{0ex}{}{r}{l}\right){(-\lambda )}^{r-l}{p}^{(k)}(l).\end{array}$$

(49)

Therefore, by (46) and (49), we obtain the following theorem.

Theorem 8 For $p(x)\in {\mathbb{P}}_n(\lambda )$, let

$$p(x)=\sum _{k=0}^n{C}_k^{(r)}{H}_k^{(r)}(x|\lambda ).$$

Then we have

$$C_k^{(r)}=\frac{1}{k!{(1-\lambda )}^r}\sum _{l=0}^r\left(\genfrac{}{}{0ex}{}{r}{l}\right){(-\lambda )}^{r-l}{p}^{(k)}(l),$$

where $r\in \mathbf{N}$ and $p^{(k)}(l)=\frac{d^{k}p(x)}{d{x}^k}{|}_{x=l}$.

Remark Let $S_n(x)$ be a Sheffer sequence for $(g(t),f(t))$. Then Sheffer identity is given by

$$S_n(x+y)=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)P_k(y)S_{n-k}(x)=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)P_k(x)S_{n-k}(y),$$

(50)

where $P_k(y)=g(t)S_k(y)$ is associated to $f(t)$ (see [1, 2]).

From (21), Proposition 1, and (50), we have

$$\begin{array}{rcl}{H}_n(x+y|\lambda )& =& \sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)P_k(y)S_{n-k}(x)\\ =& \sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)H_{n-k}(y|\lambda )x^k.\end{array}$$

By Proposition 7 and (50), we get

$$H_n^{(r)}(x+y|\lambda )=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)H_{n-k}^{(r)}(y|\lambda )x^k.$$

Let $\alpha (\ne 0)\in \mathbf{C}$. Then we have

$$H_n(\alpha x|\lambda )={\alpha }^n\frac{g_{\lambda }(t)}{g_{\lambda }(\frac{t}{\alpha })}{H}_n(x|\lambda ).$$