Remarks on some properties of special polynomials with exponential distribution

TL;DR

This paper revisits recent work on special polynomials associated with exponential distribution, providing improvements and new proofs to deepen understanding of their properties.

Contribution

It offers enhancements and alternative proofs for existing results on special polynomials linked to exponential distribution.

Findings

Improved proofs of properties of special polynomials

New insights into their structural characteristics

Refined results compared to previous work

Abstract

In this notice, we revisit the recent work [1] of Jung Yoog Kang and Tai Sup about special polynomials with exponential distribution in order to state some improvements and get new proofs for results therein.

Equations

$$\frac{\lambda }{e^{\lambda t}}{e}^{xt}=\sum _{n⩾0}{\mathfrak{E}}_n\left(\lambda :x\right)\frac{t^n}{n!}$$

$$\frac{\lambda }{e^{\lambda t}}=\sum _{n⩾0}{\mathfrak{E}}_n\left(\lambda \right)\frac{t^n}{n!}$$

$${\mathfrak{E}}_n\left(\lambda :x\right)=\lambda {\left(x-\lambda \right)}^n$$

$${\mathfrak{E}}_n\left(\lambda \right)={\left(-1\right)}^n{\lambda }^{n+1}$$

$$D{\mathfrak{E}}_n\left(\lambda :x\right)=n{\mathfrak{E}}_{n-1}\left(\lambda :x\right)$$

$$D_y{\mathfrak{E}}_n\left(\lambda :x+y\right)=D_x{\mathfrak{E}}_n\left(\lambda :x+y\right)=n{\mathfrak{E}}_{n-1}\left(\lambda :x+y\right).$$

$${\int }_0^1{\mathfrak{E}}_n\left(\lambda :x+y\right)dy=\frac{{\mathfrak{E}}_{n+1}\left(\lambda :x+1\right)-{\mathfrak{E}}_{n+1}\left(\lambda :y\right)}{n+1}$$

$${\int }_0^1{\mathfrak{E}}_n\left(\lambda :x\right)dx=\frac{{\mathfrak{E}}_{n+1}\left(\lambda :1\right)-{\mathfrak{E}}_{n+1}\left(\lambda \right)}{n+1}.$$

$${\mathfrak{E}}_n\left(\lambda :x\right)=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){\mathfrak{E}}_n\left(\lambda \right)x^{n-k}$$

$${\mathfrak{E}}_n\left(\lambda :x\right)=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){\left(-1\right)}^k{\lambda }^{k+1}{x}^{n-k}.$$

$${\mathfrak{E}}_n\left(\lambda \right)={\left(-1\right)}^n{\lambda }^{n+1}$$

$$\frac{\lambda }{e^{\lambda t}}{e}^{xt}=\lambda e^{\left(x-\lambda \right)t}=\sum _{n⩾0}\lambda {\left(x-\lambda \right)}^n\frac{t^n}{n!}$$

$$\sum _{n⩾0}{\mathfrak{E}}_n\left(\lambda :x\right)\frac{t^n}{n!}=\sum _{n⩾0}\lambda {\left(x-\lambda \right)}^n\frac{t^n}{n!}.$$

$${\mathfrak{E}}_n\left(\lambda :x\right)=\lambda {\left(x-\lambda \right)}^n=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){\left(-1\right)}^k{\lambda }^{k+1}{x}^{n-k}.$$

$${\mathfrak{E}}_n\left(\lambda :x\right)=\lambda {\left(x-\lambda \right)}^n={\left(-1\right)}^n{\lambda }^{n+1}+\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{k}\right){\left(-1\right)}^k{\lambda }^{k+1}{x}^{n-k}$$

$${\mathfrak{E}}_n\left(\lambda \right)={\left(-1\right)}^n{\lambda }^{n+1}$$

$${\mathfrak{E}}_n\left(\lambda :x+y\right)=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){\mathfrak{E}}_n\left(\lambda :x\right)y^{n-k}$$

$${\mathfrak{E}}_n\left(\lambda :x+y\right)=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){\left(-1\right)}^k{\lambda }^{k+1}{\left(x+y\right)}^{n-k}$$

$${\mathfrak{E}}_n\left(\lambda :x+y\right)=\sum _{i=0}^n\sum _{k=0}^i\left(\genfrac{}{}{0ex}{}{n}{k}\right)\left(\genfrac{}{}{0ex}{}{n-k}{i-k}\right){\left(-1\right)}^k{\lambda }^{k+1}{x}^{i-k}{y}^{n-i}$$

$$\left(\genfrac{}{}{0ex}{}{n}{k}\right)\left(\genfrac{}{}{0ex}{}{n-k}{i-k}\right)=\frac{n!i!}{k!\left(i-k\right)!\left(n-i\right)!i!}=\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{i}{k}\right)$$

$${\mathfrak{E}}_n\left(\lambda :x+y\right)=\sum _{i=0}^n\sum _{k=0}^i\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{i}{k}\right){\left(-1\right)}^k{\lambda }^{k+1}{x}^{i-k}{y}^{n-i}$$

$$x^n=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){\lambda }^{n-k-1}{\mathfrak{E}}_k\left(\lambda :x\right)$$

$$\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){\lambda }^{n-k-1}{\mathfrak{E}}_k\left(\lambda :x\right)=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){\lambda }^{n-k}{\left(x-\lambda \right)}^k={\left(x-\lambda +\lambda \right)}^n=x^n.$$

$${\mathfrak{E}}_n\left(\lambda :x\right)={\left(-1\right)}^{n+1}{\mathfrak{E}}_n\left(-\lambda :-x\right)$$

$${\mathfrak{E}}_n\left(\lambda :x\right)=2{\mathfrak{E}}_n\left(\frac{\lambda }{2}:-\frac{\lambda }{2}+x\right)$$

$${\mathfrak{E}}_n\left(-\lambda :-x\right)=-\lambda {\left(-x+\lambda \right)}^n=-{\left(-1\right)}^n\lambda {\left(x-\lambda \right)}^n={\left(-1\right)}^{n+1}{\mathfrak{E}}_n\left(\lambda :x\right)$$

$${\mathfrak{E}}_n\left(\lambda :x\right)={\left(-1\right)}^{n+1}{\mathfrak{E}}_n\left(-\lambda :-x\right).$$

$${\mathfrak{E}}_n\left(\frac{\lambda }{2}:-\frac{\lambda }{2}+x\right)=\frac{\lambda }{2}{\left(-\frac{\lambda }{2}+x-\frac{\lambda }{2}\right)}^n=\frac{\lambda }{2}{\left(x-\lambda \right)}^n=\frac{1}{2}{\mathfrak{E}}_n\left(\lambda :x\right)$$

$${\mathfrak{E}}_n\left(\lambda :x\right)=2{\mathfrak{E}}_n\left(\frac{\lambda }{2}:-\frac{\lambda }{2}+x\right).$$

$$\displaystyle\sum_{k=0}^{n}{n\choose k}\left(\lambda-x\right)^{k}\mathfrak{E}_{n-k}\left(\lambda:x\right)=\left\{\begin{array}[]{ccc}\lambda&\quad\textrm{ if }\ n=0,\\ 0&\quad\textrm{ otherwise}\end{array}\right.$$

Full text

1. Introduction

In this notice we revisit the recent work [1] on some properties of special polynomials with exponential distribution of Jung Yoog Kang and Tai Sup Lee published in Commun. Korean Math. Soc. The object of this study is the family of polynomials $\mathfrak{E}_{n}\left(\lambda:x\right)$ generated by the generating function

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and associated numbers $\mathfrak{E}_{n}\left(\lambda\right)=\mathfrak{E}_{n}\left(\lambda:0\right)$ generated by

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First we prove that

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and then

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These results go alone to show that some results in this paper are trivial. For example the result in Theorem 2.10 p.387

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and the result in Theorem 3.1 p.387

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Furthermore the Theorem 3.2 p.388

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and then the Corollary 3.3 p.388

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2. Some basic properties

For any positive integer $n$, the polynomial $\mathfrak{E}_{n}\left(\lambda:x\right)$ is a binomial polynomial with weight $\lambda$, the following theorem states an improvement of the expression (i) Theorem 2.2 [1] p.384

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Theorem 2.1**.**

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Furthermore

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Proof.

Since

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then

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After comparison we deduce that

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To get the second formula just remark that

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and then for $x=0$ we conclude that

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∎

The identity (ii) Theorem 2.2 [1] p.384 is a consequence of the Theorem 2.1

Corollary 2.1**.**

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Proof.

The identity (2.3) Corollary 2.1 follows from the identity (2.1) Theorem 2.1 as follows.

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but

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then

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and the result (2.3) Corollary 2.1 follows. ∎

We attract attention that the identiy (2.4) is an improvement of the Theorem 3.4 [1] p.389. Only in means of the identity (2.1) Theorem 2.1 a sample proof of the identity in Theorem 2.4 [1] p.385

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is just to write

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Another proof of the identities (i) and (ii) in Theorem 2.3 [1] is explained in the following theorem.

Theorem 2.2**.**

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Proof.

Since we have

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then

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and

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and then

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∎

A sample proof of the identity in Theorem 2.5 [1] p.285

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is given as follows. It is trivial to see that for $n=0$ the sum is $\lambda$ and if $n\geqslant 1$ we have

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because

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Remark 2.1*.*

For any cupel $\left(a,b\right)$ of numbers, the formulae in Theorem 2.8 [1] p.386

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results from the identities

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and

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and the fact that

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In the case $x=y$ a new identity without the sum is obtained in the following corollary.

Corollary 2.2**.**

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Proof.

We have

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then

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with

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and the result follows. ∎

Finally for the cupel $\left(1,b\right)$, the identity (2.8) Corollary 2.2 becomes

Corollary 2.3**.**

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