# Representations by degenerate Daehee polynomials

Representations by degenerate Daehee polynomials 1Introduction and preliminariesThe aim of this paper is to derive formulas (see Theorem 3.1) expressing any polynomial in terms of the degenerate Daehee polynomials (see (1.12)) with the help of umbral calculus and to illustrate our results with some examples (see Chapter 6). This can be generalized to the higher-order degenerate Bernoulli polynomials (see (1.13)). Indeed, we deduce formulas (see Theorems 4.1) for representing any polynomial in terms of the higher-order degenerate Daehee polynomials again by using umbral calculus. Letting λ→0\lambda \to 0, we obtain formulas (see Remarks 3.2 and 4.2) for expressing any polynomial in terms of the Daehee polynomials (see (1.10)) and of the higher-order Daehee polynomials (see (1.11)). These formulas are also illustrated in Chapter 5. The contribution of this paper is the derivation of such formulas that, we think, have many potential applications.Let p(x)∈C[x]p\left(x)\in {\mathbb{C}}\left[x], with degp(x)=n{\rm{\deg }}\hspace{0.33em}p\left(x)=n. Write p(x)=∑k=0nakBk(x)p\left(x)={\sum }_{k=0}^{n}{a}_{k}{B}_{k}\left(x), where Bk(x){B}_{k}\left(x)are the Bernoulli polynomials (see (1.3)). Then, it is known (see [1]) that (1.1)ak=1k!∫01p(k)(x)dx,fork=0,1,…,n.{a}_{k}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\underset{0}{\overset{1}{\int }}{p}^{\left(k)}\left(x){\rm{d}}x,\hspace{1em}{\rm{for}}\hspace{1em}k=0,1,\ldots ,n.The following identity (see [1,2]) is obtained by applying the formula in (1.1) to the polynomial p(x)=∑k=1n−11k(n−k)Bk(x)Bn−k(x)p\left(x)={\sum }_{k=1}^{n-1}\frac{1}{k\left(n-k)}{B}_{k}\left(x){B}_{n-k}\left(x)and after slight modification: (1.2)∑k=1n−112k(2n−2k)B2k(x)B2n−2k(x)+22n−1B1(x)B2n−1(x)\begin{array}{l}\mathop{\displaystyle \sum }\limits_{k=1}^{n-1}\frac{1}{2k(2n-2k)}{B}_{2k}(x){B}_{2n-2k}(x)+\frac{2}{2n-1}{B}_{1}(x){B}_{2n-1}(x)\end{array}=1n∑k=1n12k2n2kB2kB2n−2k(x)+1nH2n−1B2n(x)+22n−1B1(x)B2n−1,\begin{array}{l}\hspace{1.0em}=\frac{1}{n}\mathop{\displaystyle \sum }\limits_{k=1}^{n}\frac{1}{2k}\left(\genfrac{}{}{0.0pt}{}{2n}{2k}\right){B}_{2k}{B}_{2n-2k}(x)+\frac{1}{n}{H}_{2n-1}{B}_{2n}(x)+\frac{2}{2n-1}{B}_{1}(x){B}_{2n-1},\end{array}where n≥2n\ge 2and Hn=1+12+⋯+1n{H}_{n}=1+\frac{1}{2}+\cdots +\frac{1}{n}.Letting x=0x=0and x=12x=\frac{1}{2}in (1.2), respectively, give a slight variant of Miki’s identity and the Faber-Pandharipande-Zagier (FPZ) identity. Here, it should be emphasized that the other proofs of Miki’s (see [3,4,5]) and FPZ identities (see [6,7]) are quite involved, while our proofs of Miki’s and FPZ identities follow from the simple formula in (1.1) involving only derivatives and integrals of the given polynomials.Analogous formulas to (1.1) can be obtained for the representations by Euler, Frobenius-Euler, ordered Bell and Genocchi polynomials. Many interesting identities have been derived by using these formulas (see [1,8,9, 10,11,12, 13,14] and references therein). The list in the references is far from being exhaustive. However, the interested reader can easily find more related papers in the literature. Also, we should mention here that there are other ways of obtaining the same result as the one in (1.2). One of them is to use Fourier series expansion of the function obtained by extending by periodicity 1 of the polynomial function restricted to the interval [0,1){[}0,1)(see [2,15,16]).The outline of this paper is as follows. In Section 1, we recall some necessary facts that are needed throughout this paper. In Section 2, we go over umbral calculus briefly. In Section 3, we derive formulas expressing any polynomial in terms of the degenerate Daehee polynomials. In Section 4, we derive formulas representing any polynomial in terms of the higher-order degenerate Daehee polynomials. In Section 5, we illustrate our results with examples of representation by the Daehee polynomials. In Section 6, we illustrate our results with examples of representation by the degenerate Daehee polynomials. Finally, we conclude our paper in Section 7.The Bernoulli polynomials Bn(x){B}_{n}\left(x)are defined by (1.3)tet−1ext=∑n=0∞Bn(x)tnn!.\frac{t}{{e}^{t}-1}{e}^{xt}=\mathop{\sum }\limits_{n=0}^{\infty }{B}_{n}\left(x)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}.When x=0x=0, Bn=Bn(0){B}_{n}={B}_{n}\left(0)are called the Bernoulli numbers. We observe that Bn(x)=∑j=0nnjBn−jxj{B}_{n}\left(x)={\sum }_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){B}_{n-j}{x}^{j}, ddxBn(x)=nBn−1(x)\frac{{\rm{d}}}{{\rm{d}}x}{B}_{n}\left(x)=n{B}_{n-1}\left(x), and Bn(x+1)−Bn(x)=nxn−1{B}_{n}\left(x+1)-{B}_{n}\left(x)=n{x}^{n-1}. The first few terms of Bn{B}_{n}are given by: B0=1,B1=−12,B2=16,B4=−130,B6=142,B8=−130,B10=566,B12=−6912730,…;B2k+1=0,(k≥1).\begin{array}{r}{B}_{0}=1,\hspace{0.5em}{B}_{1}=-\frac{1}{2},\hspace{0.5em}{B}_{2}=\frac{1}{6},\hspace{0.5em}{B}_{4}=-\frac{1}{30},\hspace{0.5em}{B}_{6}=\frac{1}{42},\hspace{0.5em}{B}_{8}=-\frac{1}{30},\hspace{0.5em}{B}_{10}=\frac{5}{66},\hspace{0.5em}{B}_{12}=-\frac{691}{\hspace{0.1em}\text{2730}\hspace{0.1em}},\ldots ;\\ {B}_{2k+1}=0,\hspace{0.5em}\left(k\ge 1).\end{array}More generally, for any nonnegative integer rr, the Bernoulli polynomials Bn(r)(x){B}_{n}^{\left(r)}\left(x)of order rrare given by (1.4)tet−1rext=∑n=0∞Bn(r)(x)tnn!.{\left(\frac{t}{{e}^{t}-1}\right)}^{r}{e}^{xt}=\mathop{\sum }\limits_{n=0}^{\infty }{B}_{n}^{\left(r)}\left(x)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}.When x=0x=0, Bn(r)=Bn(r)(0){B}_{n}^{\left(r)}={B}_{n}^{\left(r)}\left(0)are called the Bernoulli numbers of order rr. We observe that Bn(r)(x)=∑j=0nnjBn−j(r)xj{B}_{n}^{\left(r)}\left(x)={\sum }_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){B}_{n-j}^{\left(r)}{x}^{j}, ddxBn(r)(x)=nBn−1(r)(x)\frac{{\rm{d}}}{{\rm{d}}x}{B}_{n}^{\left(r)}\left(x)=n{B}_{n-1}^{\left(r)}\left(x), Bn(r)(x+1)−Bn(r)(x)=nBn−1(r−1)(x){B}_{n}^{\left(r)}\left(x+1)-{B}_{n}^{\left(r)}\left(x)=n{B}_{n-1}^{\left(r-1)}\left(x).The Euler polynomials En(x){E}_{n}\left(x)are defined by (1.5)2et+1ext=∑n=0∞En(x)tnn!.\frac{2}{{e}^{t}+1}{e}^{xt}=\mathop{\sum }\limits_{n=0}^{\infty }{E}_{n}\left(x)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}.When x=0x=0, En=En(0){E}_{n}={E}_{n}\left(0)are called the Euler numbers. We observe that En(x)=∑j=0nnjEn−jxj{E}_{n}\left(x)={\sum }_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){E}_{n-j}{x}^{j}, ddxEn(x)=nEn−1(x)\frac{{\rm{d}}}{{\rm{d}}x}{E}_{n}\left(x)=n{E}_{n-1}\left(x), En(x+1)+En(x)=2xn{E}_{n}\left(x+1)+{E}_{n}\left(x)=2{x}^{n}. The first few terms of En{E}_{n}are given by: E0=1,E1=−12,E3=14,E5=−12,E7=178,E9=−312,…;E2k=0,(k≥1).\hspace{-40.25em}\begin{array}{rcll}{E}_{0}& =& 1,\hspace{0.75em}{E}_{1}=-\frac{1}{2},\hspace{0.75em}{E}_{3}=\frac{1}{4},\hspace{0.75em}{E}_{5}=-\frac{1}{2},\hspace{0.75em}{E}_{7}=\frac{17}{8},\hspace{0.75em}{E}_{9}=-\frac{31}{2},\ldots ;\hspace{0.75em}{E}_{2k}=& 0,\hspace{0.75em}\left(k\ge 1).\end{array}The Genocchi polynomials Gn(x){G}_{n}\left(x)are defined by (1.6)2tet+1ext=∑n=0∞Gn(x)tnn!.\frac{2t}{{e}^{t}+1}{e}^{xt}=\mathop{\sum }\limits_{n=0}^{\infty }{G}_{n}\left(x)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}.\hspace{5.85em}When x=0x=0, Gn=Gn(0){G}_{n}={G}_{n}\left(0)are called the Genocchi numbers. We observe that Gn(x)=∑j=0nnjGn−jxj{G}_{n}\left(x)={\sum }_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){G}_{n-j}{x}^{j}, ddxGn(x)=nGn−1(x)\frac{{\rm{d}}}{{\rm{d}}x}{G}_{n}\left(x)=n{G}_{n-1}\left(x), Gn(x+1)+Gn(x)=2nxn−1{G}_{n}\left(x+1)+{G}_{n}\left(x)=2n{x}^{n-1}, and degGn(x)=n−1{\rm{\deg }}\hspace{0.33em}{G}_{n}\left(x)=n-1, for n≥1n\ge 1. The first few terms of Gn{G}_{n}are given by: G0=0,G1=1,G2=−1,G4=1,G6=−3,G8=17,G10=−155G12=2073,…;G2k+1=0,(k≥1).\begin{array}{l}{G}_{0}=0,\hspace{1em}{G}_{1}=1,\hspace{1em}{G}_{2}=-1,\hspace{1em}{G}_{4}=1,\hspace{1em}{G}_{6}=-3,\hspace{1em}{G}_{8}=17,\hspace{1em}{G}_{10}=-155\\ {G}_{12}=\hspace{0.1em}\text{2073}\hspace{0.1em},\ldots ;\hspace{1em}{G}_{2k+1}=0,\hspace{1em}\left(k\ge 1).\end{array}For any nonzero real number λ\lambda , the degenerate exponentials are given by (1.7)eλx(t)=(1+λt)xλ=∑n=0∞(x)n,λtnn!,eλ(t)=eλ1(t)=(1+λt)1λ=∑n=0∞(1)n,λtnn!.\begin{array}{rcl}{e}_{\lambda }^{x}\left(t)& =& {\left(1+\lambda t)}^{\tfrac{x}{\lambda }}=\mathop{\displaystyle \sum }\limits_{n=0}^{\infty }{\left(x)}_{n,\lambda }\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}},\\ {e}_{\lambda }\left(t)& =& {e}_{\lambda }^{1}\left(t)={\left(1+\lambda t)}^{\tfrac{1}{\lambda }}=\mathop{\displaystyle \sum }\limits_{n=0}^{\infty }{\left(1)}_{n,\lambda }\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}.\end{array}Here, we recall that the λ\lambda -falling factorials are given by (1.8)(x)0,λ=1,(x)n,λ=x(x−λ)⋯(x−(n−1)λ),(n≥1).{\left(x)}_{0,\lambda }=1,\hspace{1.0em}{\left(x)}_{n,\lambda }=x\left(x-\lambda )\cdots \left(x-\left(n-1)\lambda ),\hspace{1em}\left(n\ge 1).Especially, (x)n=(x)n,1{\left(x)}_{n}={\left(x)}_{n,1}are called the falling factorials and hence given by (1.9)(x)0=1,(x)n=x(x−1)⋯(x−n+1),(n≥1).{\left(x)}_{0}=1,\hspace{1.0em}{\left(x)}_{n}=x\left(x-1)\cdots \left(x-n+1),\hspace{1em}\left(n\ge 1).The compositional inverse of eλ(t){e}_{\lambda }\left(t)is called the degenerate logarithm and given by logλ(t)=1λ(tλ−1),{\log }_{\lambda }\left(t)=\frac{1}{\lambda }\left({t}^{\lambda }-1),\hspace{5.5em}which satisfies eλ(logλ(t))=logλ(eλ(t))=t{e}_{\lambda }\left({\log }_{\lambda }\left(t))={\log }_{\lambda }\left({e}_{\lambda }\left(t))=t.Note here that limλ→0eλx(t)=ext{\mathrm{lim}}_{\lambda \to 0}{e}_{\lambda }^{x}\left(t)={e}^{xt}, limλ→0logλ(t)=log(t){\mathrm{lim}}_{\lambda \to 0}{\log }_{\lambda }\left(t)=\log \left(t).Recall that the Daehee polynomials Dn(x){D}_{n}\left(x)are given by (1.10)log(1+t)t(1+t)x=∑n=0∞Dn(x)tnn!.\frac{\log \left(1+t)}{t}{\left(1+t)}^{x}=\mathop{\sum }\limits_{n=0}^{\infty }{D}_{n}\left(x)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}.When x=0x=0, Dn=Dn(0){D}_{n}={D}_{n}\left(0)are the Daehee numbers.More generally, for any nonnegative integer rr, the Daehee polynomials Dn(r)(x){D}_{n}^{\left(r)}\left(x)of order rrare given by (1.11)log(1+t)tr(1+t)x=∑n=0∞Dn(r)(x)tnn!.{\left(\frac{\log \left(1+t)}{t}\right)}^{r}{\left(1+t)}^{x}=\mathop{\sum }\limits_{n=0}^{\infty }{D}_{n}^{\left(r)}\left(x)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}.When x=0x=0, Dn(r)=Dn(r)(0){D}_{n}^{\left(r)}={D}_{n}^{\left(r)}\left(0)are the Daehee numbers of order rr.The degenerate Daehee polynomials Dn,λ(x){D}_{n,\lambda }\left(x)are defined by (1.12)logλ(1+t)t(1+t)x=∑n=0∞Dn,λ(x)tnn!,\frac{{\log }_{\lambda }\left(1+t)}{t}{\left(1+t)}^{x}=\mathop{\sum }\limits_{n=0}^{\infty }{D}_{n,\lambda }\left(x)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}},which are degenerate versions of the Daehee polynomials in (1.10). For x=0x=0, Dn,λ=Dn,λ(0){D}_{n,\lambda }={D}_{n,\lambda }\left(0)are called the degenerate Daehee numbers and introduced in [7] (see also [14]).More generally, for any nonnegative integer rr, the degenerate Daehee polynomials Dn,λ(r)(x){D}_{n,\lambda }^{\left(r)}\left(x)of order rrare defined by (1.13)logλ(1+t)tr(1+t)x=∑n=0∞Dn,λ(r)(x)tnn!,{\left(\frac{{\log }_{\lambda }\left(1+t)}{t}\right)}^{r}{\left(1+t)}^{x}=\mathop{\sum }\limits_{n=0}^{\infty }{D}_{n,\lambda }^{\left(r)}\left(x)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}},which are degenerate versions of the Daehe polynomials of order rrin (1.11). We remark that Dn,λ(x)→Dn(x){D}_{n,\lambda }\left(x)\to {D}_{n}\left(x), and Dn,λ(r)(x)→Dn(r)(x){D}_{n,\lambda }^{\left(r)}\left(x)\to {D}_{n}^{\left(r)}\left(x), as λ\lambda tends to 0.We recall some notations and facts about forward differences. Let ffbe any complex-valued function of the real variable xx. Then, for any real number aa, the forward difference Δa{\Delta }_{a}is given by (1.14)Δaf(x)=f(x+a)−f(x).{\Delta }_{a}f\left(x)=f\left(x+a)-f\left(x).\hspace{3em}If a=1a=1, then we let (1.15)Δf(x)=Δ1f(x)=f(x+1)−f(x).\Delta f\left(x)={\Delta }_{1}f\left(x)=f\left(x+1)-f\left(x).In general, the nnth oder forward differences are given by (1.16)Δanf(x)=∑i=0nni(−1)n−if(x+ia).{\Delta }_{a}^{n}\hspace{0.16em}f\left(x)=\mathop{\sum }\limits_{i=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{i}\right){\left(-1)}^{n-i}f\left(x+ia).For a=1a=1, we have (1.17)Δnf(x)=∑i=0nni(−1)n−if(x+i).{\Delta }^{n}f\left(x)=\mathop{\sum }\limits_{i=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{i}\right){\left(-1)}^{n-i}f\left(x+i).\hspace{.75em}Finally, we recall that the Stirling numbers of the second kind S2(n,k){S}_{2}\left(n,k)can be given by means of (1.18)1k!(et−1)k=∑n=k∞S2(n,k)tnn!.\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left({e}^{t}-1)}^{k}=\mathop{\sum }\limits_{n=k}^{\infty }{S}_{2}\left(n,k)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}.2Review of umbral calculusHere, we will briefly go over very basic facts about umbral calculus. For more details on this, we recommend the reader to refer to [3, 20, 22]. Let C{\mathbb{C}}be the field of complex numbers. Then, ℱ{\mathcal{ {\mathcal F} }}denotes the algebra of formal power series in ttover C{\mathbb{C}}, given by ℱ=f(t)=∑k=0∞aktkk!ak∈C,{\mathcal{ {\mathcal F} }}=\left\{f\left(t)=\mathop{\sum }\limits_{k=0}^{\infty }{a}_{k}\left.\frac{{t}^{k}}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\right|\hspace{0.33em}{a}_{k}\in {\mathbb{C}}\right\},and P=C[x]{\mathbb{P}}={\mathbb{C}}\left[x]indicates the algebra of polynomials in xxwith coefficients in C{\mathbb{C}}.Let P∗{{\mathbb{P}}}^{\ast }be the vector space of all linear functionals on P{\mathbb{P}}. If ⟨L∣p(x)⟩\langle L| p\left(x)\rangle denotes the action of the linear functional LLon the polynomial p(x)p\left(x), then the vector space operations on P∗{{\mathbb{P}}}^{\ast }are defined by ⟨L+M∣p(x)⟩=⟨L∣p(x)⟩+⟨M∣p(x)⟩,⟨cL∣p(x)⟩=c⟨L∣p(x)⟩,\langle L+M| p\left(x)\rangle =\langle L| p\left(x)\rangle +\langle M| p\left(x)\rangle ,\hspace{1.0em}\langle cL| p\left(x)\rangle =c\langle L| p\left(x)\rangle ,where ccis a complex number.For f(t)∈ℱf\left(t)\in {\mathcal{ {\mathcal F} }}with f(t)=∑k=0∞aktkk!f\left(t)={\sum }_{k=0}^{\infty }{a}_{k}\frac{{t}^{k}}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}, we define the linear functional on P{\mathbb{P}}by (2.1)⟨f(t)∣xk⟩=ak.\langle f\left(t)| {x}^{k}\rangle ={a}_{k}.From (2.1), we note that ⟨tk∣xn⟩=n!δn,k,(n,k≥0),\langle {t}^{k}| {x}^{n}\rangle =n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}{\delta }_{n,k},\hspace{1.0em}\left(n,k\ge 0),where δn,k{\delta }_{n,k}is the Kronecker’s symbol.Some remarkable linear functionals are as follows: (2.2)⟨eyt∣p(x)⟩=p(y),⟨eyt−1∣p(x)⟩=p(y)−p(0),eyt−1tp(x)=∫0yp(u)du.\begin{array}{rcl}\langle {e}^{yt}| p\left(x)\rangle & =& p(y),\\ \langle {e}^{yt}-1| p\left(x)\rangle & =& p(y)-p\left(0),\\ \left\langle \left.\frac{{e}^{yt}-1}{t}\right|p\left(x)\right\rangle & =& \underset{0}{\overset{y}{\displaystyle \int }}p\left(u){\rm{d}}u.\end{array}Let (2.3)fL(t)=∑k=0∞⟨L∣xk⟩tkk!.{f}_{L}\left(t)=\mathop{\sum }\limits_{k=0}^{\infty }\langle L| {x}^{k}\rangle \frac{{t}^{k}}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}.Then, by (2.1) and (2.3), we obtain ⟨fL(t)∣xn⟩=⟨L∣xn⟩.\hspace{.08em}\langle \hspace{.08em}{f}_{L}\left(t)| {x}^{n}\rangle =\langle L| {x}^{n}\rangle .\hspace{.8em}That is, fL(t)=L{f}_{L}\left(t)=L. In addition, the map L⟼fL(t)L\hspace{0.33em}\longmapsto \hspace{0.33em}{f}_{L}\left(t)is a vector space isomorphism from P∗{{\mathbb{P}}}^{\ast }onto ℱ{\mathcal{ {\mathcal F} }}.Henceforth, ℱ{\mathcal{ {\mathcal F} }}denotes both the algebra of formal power series in ttand the vector space of all linear functionals on P{\mathbb{P}}. ℱ{\mathcal{ {\mathcal F} }}is called the umbral algebra and the umbral calculus is the study of umbral algebra. For each nonnegative integer kk, the differential operator tk{t}^{k}on P{\mathbb{P}}is defined by (2.4)tkxn=(n)kxn−k,ifk≤n,0,ifk>n.{t}^{k}{x}^{n}=\left\{\begin{array}{ll}{\left(n)}_{k}{x}^{n-k},& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}k\le n,\\ 0,& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}k\gt n.\end{array}\right.\hspace{5.7em}Extending (2.4) linearly, any power series f(t)=∑k=0∞akk!tk∈ℱf\left(t)=\mathop{\sum }\limits_{k=0}^{\infty }\frac{{a}_{k}}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{t}^{k}\in {\mathcal{ {\mathcal F} }}gives the differential operator on P{\mathbb{P}}defined by (2.5)f(t)xn=∑k=0nnkakxn−k,(n≥0).f\left(t){x}^{n}=\mathop{\sum }\limits_{k=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{k}\right){a}_{k}{x}^{n-k},\hspace{1.0em}\left(n\ge 0).\hspace{2.75em}It should be observed that, for any formal power series f(t)f\left(t)and any polynomial p(x)p\left(x), we have (2.6)⟨f(t)∣p(x)⟩=⟨1∣f(t)p(x)⟩=f(t)p(x)∣x=0.\langle f\left(t)| p\left(x)\rangle =\langle 1| f\left(t)p\left(x)\rangle =f\left(t)p\left(x){| }_{x=0}.\hspace{0.275em}Here, we note that an element f(t)f\left(t)of ℱ{\mathcal{ {\mathcal F} }}is a formal power series, a linear functional, and a differential operator. Some notable differential operators are as follows: (2.7)eytp(x)=p(x+y),(eyt−1)p(x)=p(x+y)−p(x)=Δyp(x),eyt−1tp(x)=∫xx+yp(u)du.\begin{array}{l}{e}^{yt}p\left(x)=p\left(x+y),\\ \left({e}^{yt}-1)p\left(x)=p\left(x+y)-p\left(x)={\Delta }_{y}p\left(x),\\ \frac{{e}^{yt}-1}{t}p\left(x)=\underset{x}{\overset{x+y}{\displaystyle \int }}p\left(u){\rm{d}}u.\end{array}The order o(f(t))o(f\left(t))of the power series f(t)(≠0)f\left(t)\left(\ne 0)is the smallest integer for which ak{a}_{k}does not vanish. If o(f(t))=0o(f\left(t))=0, then f(t)f\left(t)is called an invertible series. If o(f(t))=1o(f\left(t))=1, then f(t)f\left(t)is called a delta series.For f(t),g(t)∈ℱf\left(t),g\left(t)\in {\mathcal{ {\mathcal F} }}with o(f(t))=1o(f\left(t))=1and o(g(t))=0o\left(g\left(t))=0, there exists a unique sequence sn(x){s}_{n}\left(x)(deg sn(x)=n{s}_{n}\left(x)=n) of polynomials such that (2.8)⟨g(t)f(t)k∣sn(x)⟩=n!δn,k,(n,k≥0).\langle g\left(t)f{\left(t)}^{k}| {s}_{n}\left(x)\rangle =n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}{\delta }_{n,k},\hspace{1.0em}\left(n,k\ge 0).The sequence sn(x){s}_{n}\left(x)is said to be the Sheffer sequence for (g(t),f(t))\left(g\left(t),f\left(t)), which is denoted by sn(x)∼(g(t),f(t)){s}_{n}\left(x)\hspace{0.33em} \sim \hspace{0.33em}\left(g\left(t),f\left(t)). We observe from (2.8) that (2.9)sn(x)=1g(t)pn(x),{s}_{n}\left(x)=\frac{1}{g\left(t)}{p}_{n}\left(x),where pn(x)=g(t)sn(x)∼(1,f(t)){p}_{n}\left(x)=g\left(t){s}_{n}\left(x)\hspace{0.33em} \sim \hspace{0.33em}\left(1,f\left(t)).In particular, if sn(x)∼(g(t),t){s}_{n}\left(x)\hspace{0.33em} \sim \hspace{0.33em}\left(g\left(t),t), then pn(x)=xn{p}_{n}\left(x)={x}^{n}, and hence, (2.10)sn(x)=1g(t)xn.{s}_{n}\left(x)=\frac{1}{g\left(t)}{x}^{n}.\hspace{1.25em}It is well known that sn(x)∼(g(t),f(t)){s}_{n}\left(x)\hspace{0.33em} \sim \hspace{0.33em}\left(g\left(t),f\left(t))if and only if (2.11)1g(f¯(t))exf¯(t)=∑k=0∞sk(x)k!tk,\frac{1}{g(\overline{f}\left(t))}{e}^{x\overline{f}\left(t)}=\mathop{\sum }\limits_{k=0}^{\infty }\frac{{s}_{k}\left(x)}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{t}^{k},for all x∈Cx\in {\mathbb{C}}, where f¯(t)\overline{f}\left(t)is the compositional inverse of f(t)f\left(t)such that f¯(f(t))=f(f¯(t))=t\overline{f}(f\left(t))=f(\overline{f}\left(t))=t.Equations (2.12)–(2.14) are equivalent to the fact that sn(x){s}_{n}(x)is Sheffer for (g(t),f(t))(g(t),f(t)), for some invertible g(t)g\left(t): (2.12)f(t)sn(x)=nsn−1(x),(n≥0),f(t){s}_{n}(x)=n{s}_{n-1}(x),\hspace{1.0em}(n\ge 0),(2.13)sn(x+y)=∑j=0nnjsj(x)pn−j(y),{s}_{n}(x+y)=\mathop{\sum }\limits_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){s}_{j}(x){p}_{n-j}(y),\hspace{0.25em}with pn(x)=g(t)sn(x){p}_{n}(x)=g(t){s}_{n}(x), (2.14)sn(x)=∑j=0n1j!⟨g(f¯(t))−1f¯(t)j∣xn⟩xj.\hspace{1.3em}{s}_{n}(x)=\mathop{\sum }\limits_{j=0}^{n}\frac{1}{j\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\langle g{(\overline{f}(t))}^{-1}\overline{f}{(t)}^{j}| {x}^{n}\rangle {x}^{j}.\hspace{0.25em}Let pn(x){p}_{n}\left(x), qn(x)=∑k=0nqn,kxk{q}_{n}\left(x)={\sum }_{k=0}^{n}{q}_{n,k}{x}^{k}be sequences of polynomials. Then, the umbral composition of qn(x){q}_{n}\left(x)with pn(x){p}_{n}\left(x)is defined to be the sequence (2.15)qn(p(x))=∑k=0nqn,kpk(x).{q}_{n}\left({\bf{p}}\left(x))=\mathop{\sum }\limits_{k=0}^{n}{q}_{n,k}{p}_{k}\left(x).\hspace{3em}3Representations by degenerate Daehee polynomialsOur interest here is to derive formulas expressing any polynomial in terms of the degenerate Daehee polynomials.From (1.7), (1.9), and (1.11), we first observe that (3.1)Dn,λ(x)∼g(t)=λf(t)eλt−1=λ(et−1)eλt−1,f(t)=et−1,{D}_{n,\lambda }\left(x)\hspace{0.33em} \sim \hspace{0.33em}\left(g\left(t)=\frac{\lambda f\left(t)}{{e}^{\lambda t}-1}=\frac{\lambda \left({e}^{t}-1)}{{e}^{\lambda t}-1},f\left(t)={e}^{t}-1\right),(3.2)(x)n∼(1,f(t)=et−1).{\left(x)}_{n}\hspace{0.33em} \sim \hspace{0.33em}\left(1,f\left(t)={e}^{t}-1).\hspace{0.5em}From (1.15), (2.7), (2.8), (2.12), (3.1), and (3.2), we note that (3.3)f(t)Dn,λ(x)=nDn−1,λ(x)=(et−1)Dn,λ(x)=ΔDn,λ(x),f\left(t){D}_{n,\lambda }\left(x)=n{D}_{n-1,\lambda }\left(x)=\left({e}^{t}-1){D}_{n,\lambda }\left(x)=\Delta {D}_{n,\lambda }\left(x),(3.4)f(t)(x)n=n(x)n−1,f\left(t){\left(x)}_{n}=n{\left(x)}_{n-1},\hspace{2.7em}(3.5)g(t)Dn,λ(x)=(x)n.g\left(t){D}_{n,\lambda }\left(x)={\left(x)}_{n}.\hspace{2.75em}Now, we assume that p(x)∈C[x]p\left(x)\in {\mathbb{C}}\left[x]has degree nn, and write p(x)=∑k=0nakDk,λ(x)p\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k,\lambda }\left(x). Then, from (3.5), we have (3.6)g(t)p(x)=∑k=0nakg(t)Dk,λ(x)=∑k=0nak(x)k.g\left(t)p\left(x)=\mathop{\sum }\limits_{k=0}^{n}{a}_{k}g\left(t){D}_{k,\lambda }\left(x)=\mathop{\sum }\limits_{k=0}^{n}{a}_{k}{\left(x)}_{k}.\hspace{5.35em}For k≥0k\ge 0, from (3.4) and (3.6), we obtain (3.7)f(t)kg(t)p(x)=f(t)k∑l=0nal(x)l,λ=∑l=knl(l−1)⋯(l−k+1)al(x)l−k,λ.f{\left(t)}^{k}g\left(t)p\left(x)=f{\left(t)}^{k}\mathop{\sum }\limits_{l=0}^{n}{a}_{l}{\left(x)}_{l,\lambda }=\mathop{\sum }\limits_{l=k}^{n}l\left(l-1)\cdots \left(l-k+1){a}_{l}{\left(x)}_{l-k,\lambda }.Letting x=0x=0in (3.7), we finally obtain (3.8)ak=1k!f(t)kg(t)p(x)∣x=0=1k!⟨g(t)f(t)k∣p(x)⟩,(k≥0).{a}_{k}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}f{\left(t)}^{k}g\left(t)p\left(x){| }_{x=0}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\langle g\left(t)f{\left(t)}^{k}| p\left(x)\rangle ,\hspace{1em}\left(k\ge 0).Now, we want to find more explicit expressions for (3.8). As λteλt−1ext=∑n=0∞λnBnxλtnn!\frac{\lambda t}{{e}^{\lambda t}-1}{e}^{xt}={\sum }_{n=0}^{\infty }{\lambda }^{n}{B}_{n}\left(\frac{x}{\lambda }\right)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}, we see from (2.10) that λnBnxλ=λteλt−1xn{\lambda }^{n}{B}_{n}\left(\frac{x}{\lambda }\right)=\frac{\lambda t}{{e}^{\lambda t}-1}{x}^{n}. To proceed further, we let p(x)=∑i=0nbixip\left(x)={\sum }_{i=0}^{n}{b}_{i}{x}^{i}.From (2.7), (2.15), and (3.1), noting that g(t)=et−1tλteλt−1g\left(t)=\frac{{e}^{t}-1}{t}\frac{\lambda t}{{e}^{\lambda t}-1}, we have (3.9)g(t)p(x)=et−1tλteλt−1p(x)=et−1t∑i=0nbiλteλt−1xi=et−1t∑i=0nbiλiBixλ=et−1tpλBxλ=∫xx+1pλBuλdu,\hspace{-19.65em}\begin{array}{rcl}g\left(t)p\left(x)& =& \frac{{e}^{t}-1}{t}\frac{\lambda t}{{e}^{\lambda t}-1}p\left(x)\\ & =& \frac{{e}^{t}-1}{t}\mathop{\displaystyle \sum }\limits_{i=0}^{n}{b}_{i}\frac{\lambda t}{{e}^{\lambda t}-1}{x}^{i}\\ & =& \frac{{e}^{t}-1}{t}\mathop{\displaystyle \sum }\limits_{i=0}^{n}{b}_{i}{\lambda }^{i}{B}_{i}\left(\frac{x}{\lambda }\right)\\ & =& \frac{{e}^{t}-1}{t}p\left(\lambda {\bf{B}}\left(\frac{x}{\lambda }\right)\right)\\ & =& \underset{x}{\overset{x+1}{\displaystyle \int }}p\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u,\end{array}where pλBxλp\left(\lambda {\bf{B}}\left(\frac{x}{\lambda }\right)\right)denotes the umbral composition of p(x)p\left(x)with λiBixλ{\lambda }^{i}{B}_{i}\left(\frac{x}{\lambda }\right), that is, it is given by pλBxλ=p\left(\lambda {\bf{B}}\left(\frac{x}{\lambda }\right)\right)=∑i=0nbiλiBixλ{\sum }_{i=0}^{n}{b}_{i}{\lambda }^{i}{B}_{i}\left(\frac{x}{\lambda }\right).We note from (3.5) and (3.9), in passing, that the following holds: (x)n=g(t)Dn,λ(x)=∫xx+1Dn,λλBuλdu.{\left(x)}_{n}=g\left(t){D}_{n,\lambda }\left(x)=\underset{x}{\overset{x+1}{\int }}{D}_{n,\lambda }\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u.\hspace{2.8em}From (2.7) and (3.9), we deduce (3.10)ak=1k!f(t)kg(t)p(x)∣x=0=1k!Δk∫xx+1pλBuλdux=0.\hspace{4.35em}{a}_{k}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}f{\left(t)}^{k}g\left(t)p\left(x){| }_{x=0}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\Delta }^{k}{\left.\left(\underset{x}{\overset{x+1}{\int }}p\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u\right),\right|}_{x=0}.By making use of (1.17) and (3.10), an alternative expression of (3.10) is given by (3.11)ak=1k!∑i=0kki(−1)k−i∫ii+1pλBuλdu.{a}_{k}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\mathop{\sum }\limits_{i=0}^{k}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right){\left(-1)}^{k-i}\underset{i}{\overset{i+1}{\int }}p\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u.\hspace{3.35em}We obtain yet another expression from (1.18), (3.8), and (3.9), which is given by (3.12)ak=1k!(et−1)k∫xx+1pλBuλdux=0=∑l=k∞S2(l,k)tll!∫xx+1pλBuλdux=0=∑l=knS2(l,k)1l!ddxl∫xx+1pλBuλdux=0,\begin{array}{rcl}{a}_{k}& =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left({e}^{t}-1)}^{k}{\left.\left(\underset{x}{\overset{x+1}{\displaystyle \int }}p\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u\right),\right|}_{x=0}\\ & =& \mathop{\displaystyle \sum }\limits_{l=k}^{\infty }{S}_{2}\left(l,k)\frac{{t}^{l}}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left.\left(\underset{x}{\overset{x+1}{\displaystyle \int }}p\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u\right),\right|}_{x=0}\\ & =& \mathop{\displaystyle \sum }\limits_{l=k}^{n}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left(\frac{{\rm{d}}}{{\rm{d}}x}\right)}^{l}{\left.\left(\underset{x}{\overset{x+1}{\displaystyle \int }}p\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u\right),\right|}_{x=0},\end{array}where we need to note that ∫xx+1pλBuλdu{\int }_{x}^{x+1}p\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}uhas degree nn.Finally, from (3.10)–(3.12), and (3.8), we obtain the following theorem.Theorem 3.1Let p(x)∈C[x]p\left(x)\in {\mathbb{C}}\left[x], with degp(x)=n{\rm{\deg }}\hspace{0.33em}p\left(x)=n. Then, we have p(x)=∑k=0nakDk,λ(x)p\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k,\lambda }\left(x), whereak=1k!f(t)kg(t)p(x)∣x=0=1k!Δk∫xx+1pλBuλdux=0=1k!∑i=0kki(−1)k−i∫ii+1pλBuλdu=∑l=knS2(l,k)1l!ddxl∫xx+1pλBuλdux=0,fork=0,1,…,n,\begin{array}{rcl}{a}_{k}& =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}f{\left(t)}^{k}g\left(t)p\left(x){| }_{x=0}\\ & =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\Delta }^{k}{\left.\left(\underset{x}{\overset{x+1}{\displaystyle \int }}p\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u\right)\right|}_{x=0}\\ & =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\mathop{\displaystyle \sum }\limits_{i=0}^{k}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right){\left(-1)}^{k-i}\underset{i}{\overset{i+1}{\displaystyle \int }}p\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u\\ & =& \mathop{\displaystyle \sum }\limits_{l=k}^{n}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left(\frac{{\rm{d}}}{{\rm{d}}x}\right)}^{l}{\left.\left(\underset{x}{\overset{x+1}{\displaystyle \int }}p\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u\right)\right|}_{x=0},\hspace{1em}{for}\hspace{1em}k=0,1,\ldots ,n,\end{array}where g(t)=λ(et−1)eλt−1g\left(t)=\frac{\lambda \left({e}^{t}-1)}{{e}^{\lambda t}-1}, f(t)=et−1f\left(t)={e}^{t}-1, and pλBxλp\left(\lambda {\bf{B}}\left(\frac{x}{\lambda }\right)\right)denotes the umbral composition of p(x)p\left(x)with λiBixλ{\lambda }^{i}{B}_{i}\left(\frac{x}{\lambda }\right).Remark 3.2Let p(x)∈C[x]p\left(x)\in {\mathbb{C}}\left[x], with degp(x)=n{\rm{\deg }}\hspace{0.33em}p\left(x)=n. Write p(x)=∑k=0nakDk(x)p\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k}\left(x). As λ\lambda tends to 0, g(t)→et−1tg\left(t)\to \frac{{e}^{t}-1}{t}, and pλBxλ→p(x)p\left(\lambda {\bf{B}}\left(\frac{x}{\lambda }\right)\right)\to p\left(x). Thus, we obtain the following result.ak=1k!Δk∫xx+1p(u)dux=0=1k!∑i=0kki(−1)k−i∫ii+1p(u)du=∑l=knS2(l,k)1l!ddxl∫xx+1p(u)dux=0,fork=0,1,…,n.\hspace{-30.1em}\begin{array}{rcl}{a}_{k}& =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\Delta }^{k}{\left.\left(\underset{x}{\overset{x+1}{\displaystyle \int }}p\left(u){\rm{d}}u\right)\right|}_{x=0}\\ & =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\mathop{\displaystyle \sum }\limits_{i=0}^{k}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right){\left(-1)}^{k-i}\underset{i}{\overset{i+1}{\displaystyle \int }}p\left(u){\rm{d}}u\\ & =& \mathop{\displaystyle \sum }\limits_{l=k}^{n}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left(\frac{{\rm{d}}}{{\rm{d}}x}\right)}^{l}{\left.\left(\underset{x}{\overset{x+1}{\displaystyle \int }}p\left(u){\rm{d}}u\right)\right|}_{x=0},\hspace{1em}{\rm{for}}\hspace{1em}k=0,1,\ldots ,n.\end{array}4Representations by higher-order degenerate Daehee polynomialsOur interest here is to derive formulas expressing any polynomial in terms of the higher-order degenerate Daehee polynomials.With g(t)=λf(t)eλt−1=λ(et−1)eλt−1g\left(t)=\frac{\lambda f\left(t)}{{e}^{\lambda t}-1}=\frac{\lambda \left({e}^{t}-1)}{{e}^{\lambda t}-1}, f(t)=et−1f\left(t)={e}^{t}-1, from (1.11), we note that (4.1)Dn,λ(r)(x)∼(g(t)r,f(t)),{D}_{n,\lambda }^{\left(r)}\left(x)\hspace{0.33em} \sim \hspace{0.33em}\left(g{\left(t)}^{r},f\left(t)),(4.2)(x)n∼(1,f(t)).{\left(x)}_{n}\hspace{0.33em} \sim \hspace{0.33em}\left(1,f\left(t)).\hspace{0.25em}From (1.15), (2.7), (2.8), (2.12), (4.1), and (4.2), we note that (4.3)f(t)Dn,λ(r)(x)=nDn−1,λ(r)(x)=(et−1)Dn,λ(r)(x)=ΔDn,λ(r)(x),f\left(t){D}_{n,\lambda }^{\left(r)}\left(x)=n{D}_{n-1,\lambda }^{\left(r)}\left(x)=\left({e}^{t}-1){D}_{n,\lambda }^{\left(r)}\left(x)=\Delta {D}_{n,\lambda }^{\left(r)}\left(x),(4.4)f(t)(x)n=n(x)n−1,f\left(t){\left(x)}_{n}=n{\left(x)}_{n-1},\hspace{1.01em}(4.5)g(t)rDn,λ(r)(x)=(x)n.\hspace{3.15em}g{\left(t)}^{r}{D}_{n,\lambda }^{\left(r)}\left(x)={\left(x)}_{n}.\hspace{3.9em}Now, we assume that p(x)∈C[x]p\left(x)\in {\mathbb{C}}\left[x]has degree nn, and write p(x)=∑k=0nakDk,λ(r)(x)p\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k,\lambda }^{\left(r)}\left(x). Then, from (4.5), we have (4.6)g(t)rp(x)=∑k=0nakg(t)rDk,λ(r)(x)=∑k=0nak(x)k.g{\left(t)}^{r}p\left(x)=\mathop{\sum }\limits_{k=0}^{n}{a}_{k}g{\left(t)}^{r}{D}_{k,\lambda }^{\left(r)}\left(x)=\mathop{\sum }\limits_{k=0}^{n}{a}_{k}{\left(x)}_{k}.For k≥0k\ge 0, from (4.4), we obtain (4.7)f(t)kg(t)rp(x)=f(t)k∑l=0nal(x)l=∑l=knl(l−1)⋯(l−k+1)al(x)l−k.f{\left(t)}^{k}g{\left(t)}^{r}p\left(x)=f{\left(t)}^{k}\mathop{\sum }\limits_{l=0}^{n}{a}_{l}{\left(x)}_{l}=\mathop{\sum }\limits_{l=k}^{n}l\left(l-1)\cdots \left(l-k+1){a}_{l}{\left(x)}_{l-k}.Letting x=0x=0in (4.7), we finally obtain (4.8)ak=1k!f(t)kg(t)rp(x)∣x=0=1k!⟨g(t)rf(t)k∣p(x)⟩,(k≥0).{a}_{k}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}f{\left(t)}^{k}g{\left(t)}^{r}p\left(x){| }_{x=0}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\langle g{\left(t)}^{r}f{\left(t)}^{k}| p\left(x)\rangle ,\hspace{1em}\left(k\ge 0).This also follows from the observation ⟨g(t)rf(t)k∣Dl,λ(r)(x)⟩=l!δl,k\langle g{\left(t)}^{r}f{\left(t)}^{k}| {D}_{l,\lambda }^{\left(r)}\left(x)\rangle =l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}\hspace{0.33em}{\delta }_{l,k}.Now, we want to find more explicit expressions for (4.8). As λteλt−1rext=∑n=0∞λnBn(r)xλtnn!{\left(\frac{\lambda t}{{e}^{\lambda t}-1}\right)}^{r}{e}^{xt}={\sum }_{n=0}^{\infty }{\lambda }^{n}{B}_{n}^{\left(r)}\left(\frac{x}{\lambda }\right)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}, we see from (2.10) that λnBn(r)xλ=λteλt−1rxn{\lambda }^{n}{B}_{n}^{\left(r)}\left(\frac{x}{\lambda }\right)={\left(\frac{\lambda t}{{e}^{\lambda t}-1}\right)}^{r}{x}^{n}. To proceed further, we let p(x)=∑i=0nbixip\left(x)={\sum }_{i=0}^{n}{b}_{i}{x}^{i}.From (2.7), (2.15), and (4.1), noting that g(t)=et−1tλteλt−1g\left(t)=\frac{{e}^{t}-1}{t}\frac{\lambda t}{{e}^{\lambda t}-1}, we have (4.9)g(t)rp(x)=et−1trλteλt−1rp(x)=et−1tr∑i=0nbiλteλt−1rxi=et−1tr∑i=0nbiλiBi(r)xλ=et−1trpλB(r)xλ=IrpλB(r)xλ,\hspace{-17.7em}\begin{array}{rcl}g{\left(t)}^{r}p\left(x)& =& {\left(\frac{{e}^{t}-1}{t}\right)}^{r}{\left(\frac{\lambda t}{{e}^{\lambda t}-1}\right)}^{r}p\left(x)\\ & =& {\left(\frac{{e}^{t}-1}{t}\right)}^{r}\mathop{\displaystyle \sum }\limits_{i=0}^{n}{b}_{i}{\left(\frac{\lambda t}{{e}^{\lambda t}-1}\right)}^{r}{x}^{i}\\ & =& {\left(\frac{{e}^{t}-1}{t}\right)}^{r}\mathop{\displaystyle \sum }\limits_{i=0}^{n}{b}_{i}{\lambda }^{i}{B}_{i}^{\left(r)}\left(\frac{x}{\lambda }\right)\\ & =& {\left(\frac{{e}^{t}-1}{t}\right)}^{r}p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)\\ & =& {I}^{r}p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right),\end{array}where pλB(r)xλp\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)denotes the umbral composition of p(x)p\left(x)with λiBi(r)xλ{\lambda }^{i}{B}_{i}^{\left(r)}\left(\frac{x}{\lambda }\right), that is, it is given by pλB(r)xλ=p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)=∑i=0nbiλiBi(r)xλ{\sum }_{i=0}^{n}{b}_{i}{\lambda }^{i}{B}_{i}^{\left(r)}\left(\frac{x}{\lambda }\right), and IIdenotes the linear integral operator given by q(x)→∫xx+1q(x)dxq\left(x)\to {\int }_{x}^{x+1}q\left(x){\rm{d}}x.We note from (4.5) and (4.9), in passing, that the following holds: (x)n=g(t)rDn,λ(x)=IrDn,λλB(r)xλ.{\left(x)}_{n}=g{\left(t)}^{r}{D}_{n,\lambda }\left(x)={I}^{r}{D}_{n,\lambda }\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right).From (2.7) and (4.9), we deduce (4.10)ak=1k!f(t)kg(t)rp(x)∣x=0=1k!ΔkIrpλB(r)xλx=0.{a}_{k}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}f{\left(t)}^{k}g{\left(t)}^{r}p\left(x){| }_{x=0}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\Delta }^{k}{\left.\left({I}^{r}p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)\right),\right|}_{x=0}.By making use of (1.17) and (4.10), an alternative expression of (3.10) is given by (4.11)ak=1k!∑i=0kki(−1)k−iIrpλB(r)xλx=i.{a}_{k}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\mathop{\sum }\limits_{i=0}^{k}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right){\left(-1)}^{k-i}{I}^{r}p{\left.\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right),\right|}_{x=i}.We obtain yet another expression from (1.18), (4.8), and (4.9), which is given by (4.12)ak=1k!(et−1)kIrpλB(r)xλx=0\begin{array}{rcl}{a}_{k}& =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left({e}^{t}-1)}^{k}{\left.\left({I}^{r}p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)\right),\right|}_{x=0}\end{array}\hspace{1.8em}=∑l=k∞S2(l,k)tll!IrpλB(r)xλx=0=∑l=knS2(l,k)1l!ddxlIrpλB(r)xλx=0,\hspace{-23.4em}\begin{array}{rcl}& =& \mathop{\displaystyle \sum }\limits_{l=k}^{\infty }{S}_{2}\left(l,k)\frac{{t}^{l}}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left.\left({I}^{r}p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)\right),\right|}_{x=0}\\ & =& \mathop{\displaystyle \sum }\limits_{l=k}^{n}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left(\frac{{\rm{d}}}{{\rm{d}}x}\right)}^{l}{\left.\left({I}^{r}p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)\right),\right|}_{x=0},\end{array}where we need to observe that IrpλB(r)xλ{I}^{r}p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)has degree nn.Finally, from (4.10)–(4.12) and (4.8), we obtain the following theorem.Theorem 4.1Let p(x)∈C[x]p\left(x)\in {\mathbb{C}}\left[x], with degp(x)=n{\rm{\deg }}\hspace{0.33em}p\left(x)=n. Then, we have p(x)=∑k=0nakDk,λ(r)(x)p\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k,\lambda }^{\left(r)}\left(x), whereak=1k!f(t)kg(t)rp(x)∣x=0=1k!ΔkIrpλB(r)xλx=0=1k!∑i=0kki(−1)k−iIrpλB(r)xλx=i=∑l=knS2(l,k)1l!ddxlIrpλB(r)xλx=0,fork=0,1,…,n,\hspace{3.55em}\begin{array}{rcl}{a}_{k}& =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}f{\left(t)}^{k}g{\left(t)}^{r}p\left(x){| }_{x=0}\\ & =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\Delta }^{k}{\left.\left({I}^{r}p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)\right),\right|}_{x=0}\\ & =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\mathop{\displaystyle \sum }\limits_{i=0}^{k}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right){\left(-1)}^{k-i}{I}^{r}p{\left.\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right),\right|}_{x=i}\\ & =& \mathop{\displaystyle \sum }\limits_{l=k}^{n}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left(\frac{{\rm{d}}}{{\rm{d}}x}\right)}^{l}{\left.\left({I}^{r}p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)\right),\right|}_{x=0},\hspace{1em}{for}\hspace{1em}k=0,1,\ldots ,n,\end{array}where g(t)=λ(et−1)eλt−1g\left(t)=\frac{\lambda \left({e}^{t}-1)}{{e}^{\lambda t}-1}, f(t)=et−1f\left(t)={e}^{t}-1, pλB(r)xλp\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)indicates the umbral composition of p(x)p\left(x)with λiBi(r)xλ{\lambda }^{i}{B}_{i}^{\left(r)}\left(\frac{x}{\lambda }\right), and IIdenotes the linear integral operator given by q(x)→∫xx+1q(x)dxq\left(x)\to {\int }_{x}^{x+1}q\left(x){\rm{d}}x.We observe that IrpλB(r)xλx=i=∫ii+1Ir−1pλB(r)xλdx{I}^{r}p{\left.\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)\right|}_{x=i}={\int }_{i}^{i+1}{I}^{r-1}p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right){\rm{d}}x.Remark 4.2Let p(x)∈C[x]p\left(x)\in {\mathbb{C}}\left[x], with degp(x)=n{\rm{\deg }}\hspace{0.33em}p\left(x)=n. Write p(x)=∑k=0nakDk(r)(x)p\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k}^{\left(r)}\left(x). As λ\lambda tends to 0, g(t)→et−1tg\left(t)\to \frac{{e}^{t}-1}{t}, and pλB(r)xλ→p(x)p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)\to p\left(x). Thus, we obtain the following result.ak=1k!Δk(Irp(x))x=0=1k!∑i=0kki(−1)k−iIrp(x)x=i=∑l=knS2(l,k)1l!ddxl(Irp(x))x=0,fork=0,1,…,n.\begin{array}{rcl}{a}_{k}& =& {\left.\frac{1}{k\text{&#x0021;}}{\Delta }^{k}({I}^{r}p\left(x))\right|}_{x=0}\\ & =& {\left.\frac{1}{k\text{&#x0021;}}\mathop{\displaystyle \sum }\limits_{i=0}^{k}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right){\left(-1)}^{k-i}{I}^{r}p\left(x)\right|}_{x=i}\\ & =& {\left.\mathop{\displaystyle \sum }\limits_{l=k}^{n}{S}_{2}\left(l,k)\frac{1}{l\text{&#x0021;}}{\left(\frac{{\rm{d}}}{{\rm{d}}x}\right)}^{l}({I}^{r}p\left(x))\right|}_{x=0},\hspace{1em}{\rm{for}}\hspace{1em}k=0,1,\ldots ,n.\end{array}We note that Irp(x)∣x=i=∫ii+1Ir−1p(x)dx{{I}^{r}p\left(x)| }_{x=i}={\int }_{i}^{i+1}{I}^{r-1}p\left(x){\rm{d}}x.5Examples on representation by Daehee polynomialsHere, we illustrate our formulas in Remarks 3.2 and 4.2 with some examples.(a) Let p(x)=Bn(x)=∑k=0nakDk(x)p\left(x)={B}_{n}\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k}\left(x). Then, as Bn(x+1)−Bn(x)=nxn−1{B}_{n}\left(x+1)-{B}_{n}\left(x)=n{x}^{n-1}, ∫xx+1Bn(u)du=xn{\int }_{x}^{x+1}{B}_{n}\left(u){\rm{d}}u={x}^{n}, from Remark 3.2, we have (5.1)ak=1k!Δkxn∣x=0=1k!∑i=0kki(−1)k−iin=S2(n,k),\hspace{-25.2em}{a}_{k}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\Delta }^{k}{x}^{n}{| }_{x=0}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\mathop{\sum }\limits_{i=0}^{k}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right){\left(-1)}^{k-i}{i}^{n}={S}_{2}\left(n,k),which are well known.Thus, we obtain the following identity: Bn(x)=∑k=0nS2(n,k)Dk(x).{B}_{n}\left(x)=\mathop{\sum }\limits_{k=0}^{n}{S}_{2}\left(n,k){D}_{k}\left(x).\hspace{7.25em}Next, we let p(x)=Bn(x)=∑k=0nakDk(r)(x)p\left(x)={B}_{n}\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k}^{\left(r)}\left(x). Then, we first observe that (5.2)IrBn(x)=1(n+r)r∑i=0rri(−1)r−iBn+r(x+i).{I}^{r}{B}_{n}\left(x)=\frac{1}{{\left(n+r)}_{r}}\mathop{\sum }\limits_{i=0}^{r}\left(\genfrac{}{}{0.0pt}{}{r}{i}\right){\left(-1)}^{r-i}{B}_{n+r}\left(x+i).Now, by making use of Remark 4.2, we obtain (5.3)ak=1k!(n+r)r∑i=0rri(−1)r−iΔkBn+r(x+i)∣x=0=1(n+r)r∑l=kn∑i=0r(−1)r−irin+rlS2(l,k)Bn+r−l(i).\begin{array}{rcl}{a}_{k}& =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}{\left(n+r)}_{r}}\mathop{\displaystyle \sum }\limits_{i=0}^{r}\left(\genfrac{}{}{0.0pt}{}{r}{i}\right){\left(-1)}^{r-i}{\Delta }^{k}{B}_{n+r}\left(x+i){| }_{x=0}\\ & =& \frac{1}{{\left(n+r)}_{r}}\mathop{\displaystyle \sum }\limits_{l=k}^{n}\mathop{\displaystyle \sum }\limits_{i=0}^{r}{\left(-1)}^{r-i}\left(\genfrac{}{}{0.0pt}{}{r}{i}\right)\left(\genfrac{}{}{0.0pt}{}{n+r}{l}\right){S}_{2}\left(l,k){B}_{n+r-l}\left(i).\end{array}Thus, we have the following: Bn(x)=1(n+r)r∑k=0n1k!∑i=0rri(−1)r−iΔkBn+r(x+i)∣x=0Dk(r)(x)=1(n+r)r∑k=0n∑l=kn∑i=0r(−1)r−irin+rlS2(l,k)Bn+r−l(i)Dk(r)(x).\begin{array}{rcl}{B}_{n}\left(x)& =& \frac{1}{{\left(n+r)}_{r}}\mathop{\displaystyle \sum }\limits_{k=0}^{n}\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left\{\mathop{\displaystyle \sum }\limits_{i=0}^{r}\left(\genfrac{}{}{0.0pt}{}{r}{i}\right){\left(-1)}^{r-i}{\Delta }^{k}{B}_{n+r}\left(x+i){| }_{x=0}\right\}{D}_{k}^{\left(r)}\left(x)\\ & =& \frac{1}{{\left(n+r)}_{r}}\mathop{\displaystyle \sum }\limits_{k=0}^{n}\left\{\mathop{\displaystyle \sum }\limits_{l=k}^{n}\mathop{\displaystyle \sum }\limits_{i=0}^{r}{\left(-1)}^{r-i}\left(\genfrac{}{}{0.0pt}{}{r}{i}\right)\left(\genfrac{}{}{0.0pt}{}{n+r}{l}\right){S}_{2}\left(l,k){B}_{n+r-l}\left(i)\right\}{D}_{k}^{\left(r)}\left(x).\end{array}(b) Here, we consider p(x)=∑k=1n−11k(n−k)Bk(x)Bn−k(x),(n≥2)p\left(x)={\sum }_{k=1}^{n-1}\frac{1}{k\left(n-k)}{B}_{k}\left(x){B}_{n-k}\left(x),\left(n\ge 2). For this, we first recall from [12] that (5.4)p(x)=2n∑m=0n−21n−mnmBn−mBm(x)+2nHn−1Bn(x),p\left(x)=\frac{2}{n}\mathop{\sum }\limits_{m=0}^{n-2}\frac{1}{n-m}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right){B}_{n-m}{B}_{m}\left(x)+\frac{2}{n}{H}_{n-1}{B}_{n}\left(x),where Hn=1+12+⋯+1n{H}_{n}=1+\frac{1}{2}+\cdots +\frac{1}{n}is the harmonic number and a slight modification of (5.4) gives the identity in (1.2). Let p(x)=∑k=0nakDk(x)p\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k}\left(x). Then, we have (5.5)ak=2n∑m=0n−21n−mnmBn−m∑l=knS2(l,k)mlδm,l+2nHn−1∑l=knS2(l,k)nlδn,l=2n∑m=kn−21n−mnmBn−mS2(m,k)+2nHn−1S2(n,k),\begin{array}{rcl}{a}_{k}& =& \frac{2}{n}\mathop{\displaystyle \sum }\limits_{m=0}^{n-2}\frac{1}{n-m}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right){B}_{n-m}\mathop{\displaystyle \sum }\limits_{l=k}^{n}{S}_{2}\left(l,k)\left(\genfrac{}{}{0.0pt}{}{m}{l}\right){\delta }_{m,l}+\frac{2}{n}{H}_{n-1}\mathop{\displaystyle \sum }\limits_{l=k}^{n}{S}_{2}\left(l,k)\left(\genfrac{}{}{0.0pt}{}{n}{l}\right){\delta }_{n,l}\\ & =& \frac{2}{n}\mathop{\displaystyle \sum }\limits_{m=k}^{n-2}\frac{1}{n-m}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right){B}_{n-m}{S}_{2}\left(m,k)+\frac{2}{n}{H}_{n-1}{S}_{2}\left(n,k),\end{array}where we understand that the sum in (5.5) is zero for k=n−1k=n-1or nn. Thus, we obtain the following identity: ∑k=1n−11k(n−k)Bk(x)Bn−k(x)=2n∑k=0n∑m=kn−21n−mnmBn−mS2(m,k)+Hn−1S2(n,k)Dk(x).\mathop{\sum }\limits_{k=1}^{n-1}\frac{1}{k\left(n-k)}{B}_{k}\left(x){B}_{n-k}\left(x)=\frac{2}{n}\mathop{\sum }\limits_{k=0}^{n}\left\{\mathop{\sum }\limits_{m=k}^{n-2}\frac{1}{n-m}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right){B}_{n-m}{S}_{2}\left(m,k)+{H}_{n-1}{S}_{2}\left(n,k)\right\}{D}_{k}\left(x).(c) In [12], it is shown that the following identity holds for n≥2n\ge 2: (5.6)∑k=1n−11k(n−k)Ek(x)En−k(x)=−4n∑m=0nnm(Hn−1−Hn−m)n−m+1En−m+1Bm(x),\mathop{\sum }\limits_{k=1}^{n-1}\frac{1}{k\left(n-k)}{E}_{k}\left(x){E}_{n-k}\left(x)=-\frac{4}{n}\mathop{\sum }\limits_{m=0}^{n}\frac{\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left({H}_{n-1}-{H}_{n-m})}{n-m+1}{E}_{n-m+1}{B}_{m}\left(x),where Hn=1+12+⋯+1n{H}_{n}=1+\frac{1}{2}+\cdots +\frac{1}{n}is the harmonic number.Write ∑k=1n−11k(n−k)Ek(x)En−k(x)=∑k=0nakDk(x){\sum }_{k=1}^{n-1}\frac{1}{k\left(n-k)}{E}_{k}\left(x){E}_{n-k}\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k}\left(x).By proceeding similarly to (b), we see that (5.7)ak=−4n∑m=0nnm(Hn−1−Hn−m)n−m+1En−m+1∑l=knS2(l,k)mlδm,l=−4n∑m=knnm(Hn−1−Hn−m)n−m+1En−m+1S2(m,k).\begin{array}{rcl}{a}_{k}& =& -\frac{4}{n}\mathop{\displaystyle \sum }\limits_{m=0}^{n}\frac{\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left({H}_{n-1}-{H}_{n-m})}{n-m+1}{E}_{n-m+1}\mathop{\displaystyle \sum }\limits_{l=k}^{n}{S}_{2}\left(l,k)\left(\genfrac{}{}{0.0pt}{}{m}{l}\right){\delta }_{m,l}\\ & =& -\frac{4}{n}\mathop{\displaystyle \sum }\limits_{m=k}^{n}\frac{\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left({H}_{n-1}-{H}_{n-m})}{n-m+1}{E}_{n-m+1}{S}_{2}\left(m,k).\end{array}Thus, (5.7) implies the next identity: ∑k=1n−11k(n−k)Ek(x)En−k(x)=−4n∑k=0n∑m=knnm(Hn−1−Hn−m)n−m+1En−m+1S2(m,k)Dk(x).\mathop{\sum }\limits_{k=1}^{n-1}\frac{1}{k\left(n-k)}{E}_{k}\left(x){E}_{n-k}\left(x)=-\frac{4}{n}\mathop{\sum }\limits_{k=0}^{n}\left\{\mathop{\sum }\limits_{m=k}^{n}\frac{\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left({H}_{n-1}-{H}_{n-m})}{n-m+1}{E}_{n-m+1}{S}_{2}\left(m,k)\right\}{D}_{k}\left(x).(d) In [16], it is proved that the following identity is valid for n≥2n\ge 2: (5.8)∑k=1n−11k(n−k)Gk(x)Gn−k(x)=−4n∑m=0n−2nmGn−mn−mBm(x).\mathop{\sum }\limits_{k=1}^{n-1}\frac{1}{k\left(n-k)}{G}_{k}\left(x){G}_{n-k}\left(x)=-\frac{4}{n}\mathop{\sum }\limits_{m=0}^{n-2}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\frac{{G}_{n-m}}{n-m}{B}_{m}\left(x).\hspace{5.8em}Again, by proceeding analogously to (b), we can show that (5.9)ak=−4n∑l=kn−2S2(l,k)ml∑m=0n−2nmGn−mn−mδm,l=−4n∑m=kn−2nmS2(m,k)Gn−mn−m.{a}_{k}=-\frac{4}{n}\mathop{\sum }\limits_{l=k}^{n-2}{S}_{2}\left(l,k)\left(\genfrac{}{}{0.0pt}{}{m}{l}\right)\mathop{\sum }\limits_{m=0}^{n-2}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\frac{{G}_{n-m}}{n-m}{\delta }_{m,l}=-\frac{4}{n}\mathop{\sum }\limits_{m=k}^{n-2}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right){S}_{2}\left(m,k)\frac{{G}_{n-m}}{n-m}.\hspace{4.1em}Therefore, we obtain the following identity: ∑k=1n−11k(n−k)Gk(x)Gn−k(x)=−4n∑k=0n−2∑m=kn−2nmS2(m,k)Gn−mn−mDk(x).\mathop{\sum }\limits_{k=1}^{n-1}\frac{1}{k\left(n-k)}{G}_{k}\left(x){G}_{n-k}\left(x)=-\frac{4}{n}\mathop{\sum }\limits_{k=0}^{n-2}\left\{\mathop{\sum }\limits_{m=k}^{n-2}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right){S}_{2}\left(m,k)\frac{{G}_{n-m}}{n-m}\right\}{D}_{k}\left(x).(e) Nielsen [2,19] also represented products of two Euler polynomials in terms of Bernoulli polynomials as follows: (5.10)Em(x)En(x)=−2∑r=1mmrErBm+n−r+1(x)m+n−r+1−2∑s=1nnsEsBm+n−s+1(x)m+n−s+1+2(−1)n+1m!n!(m+n+1)!Em+n+1.{E}_{m}\left(x){E}_{n}\left(x)=-2\mathop{\sum }\limits_{r=1}^{m}\left(\genfrac{}{}{0.0pt}{}{m}{r}\right){E}_{r}\frac{{B}_{m+n-r+1}\left(x)}{m+n-r+1}-2\mathop{\sum }\limits_{s=1}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{s}\right){E}_{s}\frac{{B}_{m+n-s+1}\left(x)}{m+n-s+1}+2{\left(-1)}^{n+1}\frac{m\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}\hspace{0.33em}n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left(m+n+1)\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{E}_{m+n+1}.In the same way as (b), we can show that (5.11)ak=2(−1)n+1m!n!(m+n+1)!Em+n+1δk,0−2∑r=1mmrErm+n−r+1S2(m+n−r+1,k)−2∑s=1nnsEsm+n−s+1S2(m+n−s+1,k).{a}_{k}=2{\left(-1)}^{n+1}\frac{m\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}\hspace{0.33em}n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left(m+n+1)\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{E}_{m+n+1}{\delta }_{k,0}-2\mathop{\sum }\limits_{r=1}^{m}\left(\genfrac{}{}{0.0pt}{}{m}{r}\right)\frac{{E}_{r}}{m+n-r+1}{S}_{2}\left(m+n-r+1,k)-2\mathop{\sum }\limits_{s=1}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{s}\right)\frac{{E}_{s}}{m+n-s+1}{S}_{2}\left(m+n-s+1,k).Thus, we arrive at the next identity: Em(x)En(x)=2(−1)n+1m!n!(m+n+1)!Em+n+1−2∑k=1m+n∑r=1mmrErm+n−r+1S2(m+n−r+1,k)Dk(x)−2∑k=1m+n∑s=1nnsEsm+n−s+1S2(m+n−s+1,k)Dk(x).\begin{array}{rcl}{E}_{m}\left(x){E}_{n}\left(x)& =& 2{\left(-1)}^{n+1}\frac{m\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}\hspace{0.33em}n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left(m+n+1)\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{E}_{m+n+1}-2\mathop{\displaystyle \sum }\limits_{k=1}^{m+n}\mathop{\displaystyle \sum }\limits_{r=1}^{m}\left(\genfrac{}{}{0.0pt}{}{m}{r}\right)\frac{{E}_{r}}{m+n-r+1}{S}_{2}\left(m+n-r+1,k){D}_{k}\left(x)\\ & & -2\mathop{\displaystyle \sum }\limits_{k=1}^{m+n}\mathop{\displaystyle \sum }\limits_{s=1}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{s}\right)\frac{{E}_{s}}{m+n-s+1}{S}_{2}\left(m+n-s+1,k){D}_{k}\left(x).\end{array}6Examples on representation by degenerate Daehee polynomialsHere, we illustrate our formulas in Theorems 3.1 and 4.1.(a) Let p(x)=Bn(x)=∑k=0nakDk,λ(x)p\left(x)={B}_{n}\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k,\lambda }\left(x). Then, as Bn(x)=∑j=0nnjBn−jxj{B}_{n}\left(x)={\sum }_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){B}_{n-j}{x}^{j}, we have (6.1)∫xx+1BnλBuλdu=∑j=0nnjBn−jλj+11j+1Bj+1(x+1λ)−Bj+1xλ.\underset{x}{\overset{x+1}{\int }}{B}_{n}\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u=\mathop{\sum }\limits_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){B}_{n-j}{\lambda }^{j+1}\frac{1}{j+1}\left({B}_{j+1}\left(\frac{x+1}{\lambda })-{B}_{j+1}\left(\frac{x}{\lambda }\right)\right).Thus, for 1≤l≤n1\le l\le n, (6.2)ddxl∫xx+1BnλBuλdu=∑j=0nnjBn−jλjddxl−1Bjx+1λ−Bjxλ=∑j=lnnjBn−jλj−l+1(j)l−1Bj−l+1x+1λ−Bj−l+1xλ.\begin{array}{rcl}{\left(\frac{{\rm{d}}}{{\rm{d}}x}\right)}^{l}\underset{x}{\overset{x+1}{\displaystyle \int }}{B}_{n}\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u& =& \mathop{\displaystyle \sum }\limits_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){B}_{n-j}{\lambda }^{j}{\left(\frac{{\rm{d}}}{{\rm{d}}x}\right)}^{l-1}\left({B}_{j}\left(\frac{x+1}{\lambda }\right)-{B}_{j}\left(\frac{x}{\lambda }\right)\right)\\ & =& \mathop{\displaystyle \sum }\limits_{j=l}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){B}_{n-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{x+1}{\lambda }\right)-{B}_{j-l+1}\left(\frac{x}{\lambda }\right)\right).\end{array}Now, from Theorem 3.1, (6.1), and (6.2), we obtain (6.3)ak=1k!∑j=0nnjλj+1j+1Bn−jΔkBj+1x+1λ−Bj+1xλx=0=1k!∑i=0k∑j=0n(−1)k−ikinjλj+1j+1Bn−jBj+1i+1λ−Bj+1iλ=∑l=kn∑j=lnS2(l,k)1l!njλj−l+1(j)l−1Bn−jBj−l+11λ−Bj−l+1,\hspace{-32.2em}\begin{array}{rcl}{a}_{k}& =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\mathop{\displaystyle \sum }\limits_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right)\frac{{\lambda }^{j+1}}{j+1}{B}_{n-j}{\Delta }^{k}{\left.\left({B}_{j+1}\left(\frac{x+1}{\lambda }\right)-{B}_{j+1}\left(\frac{x}{\lambda }\right)\right)\right|}_{x=0}\\ & =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\mathop{\displaystyle \sum }\limits_{i=0}^{k}\mathop{\displaystyle \sum }\limits_{j=0}^{n}{\left(-1)}^{k-i}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right)\left(\genfrac{}{}{0.0pt}{}{n}{j}\right)\frac{{\lambda }^{j+1}}{j+1}{B}_{n-j}\left({B}_{j+1}\left(\frac{i+1}{\lambda }\right)-{B}_{j+1}\left(\frac{i}{\lambda }\right)\right)\\ & =& \mathop{\displaystyle \sum }\limits_{l=k}^{n}\mathop{\displaystyle \sum }\limits_{j=l}^{n}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){\lambda }^{j-l+1}{\left(j)}_{l-1}{B}_{n-j}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right),\end{array}where we understand that (j)−1=1j+1{\left(j)}_{-1}=\frac{1}{j+1}.Hence, from (6.3), we obtain the following identity: Bn(x)=∑k=0n1k!∑j=0nnjλj+1j+1Bn−jΔkBj+1x+1λ−Bj+1xλx=0Dk,λ(x)=∑k=0n1k!∑i=0n∑j=0n(−1)k−ikinjλj+1j+1Bn−jBj+1i+1λ−Bj+1iλDk,λ(x)=∑k=0n∑l=kn∑j=lnS2(l,k)1l!njλj−l+1(j)l−1Bn−jBj−l+11λ−Bj−l+1Dk,λ(x).\hspace{-35.8em}\begin{array}{rcl}{B}_{n}\left(x)& =& \mathop{\displaystyle \sum }\limits_{k=0}^{n}\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left\{\mathop{\displaystyle \sum }\limits_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right)\frac{{\lambda }^{j+1}}{j+1}{B}_{n-j}{\Delta }^{k}{\left.\left({B}_{j+1}\left(\frac{x+1}{\lambda }\right)-{B}_{j+1}\left(\frac{x}{\lambda }\right)\right)\right|}_{x=0}\right\}{D}_{k,\lambda }\left(x)\\ & =& \mathop{\displaystyle \sum }\limits_{k=0}^{n}\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left\{\mathop{\displaystyle \sum }\limits_{i=0}^{n}\mathop{\displaystyle \sum }\limits_{j=0}^{n}{\left(-1)}^{k-i}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right)\left(\genfrac{}{}{0.0pt}{}{n}{j}\right)\frac{{\lambda }^{j+1}}{j+1}{B}_{n-j}\left({B}_{j+1}\left(\frac{i+1}{\lambda }\right)-{B}_{j+1}\left(\frac{i}{\lambda }\right)\right)\right\}{D}_{k,\lambda }\left(x)\\ & =& \mathop{\displaystyle \sum }\limits_{k=0}^{n}\left\{\mathop{\displaystyle \sum }\limits_{l=k}^{n}\mathop{\displaystyle \sum }\limits_{j=l}^{n}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){\lambda }^{j-l+1}{\left(j)}_{l-1}{B}_{n-j}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right)\right\}{D}_{k,\lambda }\left(x).\end{array}Next, we let p(x)=Bn(x)=∑k=0nakDk,λ(r)(x)p\left(x)={B}_{n}\left(x)=\mathop{\sum }\limits_{k=0}^{n}{a}_{k}{D}_{k,\lambda }^{\left(r)}\left(x). Then, we first observe that (6.4)IrBnλB(r)xλ=Ir∑j=0nnjBn−jλjBj(r)xλ=∑j=0n∑i=0r(−1)r−injriλj+r(j+r)rBn−jBj+r(r)x+iλ.{I}^{r}{B}_{n}\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)={I}^{r}\mathop{\sum }\limits_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){B}_{n-j}{\lambda }^{j}{B}_{j}^{\left(r)}\left(\frac{x}{\lambda }\right)=\mathop{\sum }\limits_{j=0}^{n}\mathop{\sum }\limits_{i=0}^{r}{\left(-1)}^{r-i}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right)\left(\genfrac{}{}{0.0pt}{}{r}{i}\right)\frac{{\lambda }^{j+r}}{{\left(j+r)}_{r}}{B}_{n-j}{B}_{j+r}^{\left(r)}\left(\frac{x+i}{\lambda }\right).So, for llwith j+r≥lj+r\ge l, we obtain (6.5)ddxlIrBnλB(r)xλ=∑j=0n∑i=0r(−1)r−injriλj+r(j+r)rBn−jddxlBj+r(r)x+iλ=∑j=0n∑i=0r(−1)r−injriλj+r−l(j+r)rBn−j(j+r)lBj+r−l(r)x+iλ.\begin{array}{rcl}{\left(\frac{{\rm{d}}}{{\rm{d}}x}\right)}^{l}{I}^{r}{B}_{n}\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)& =& \mathop{\displaystyle \sum }\limits_{j=0}^{n}\mathop{\displaystyle \sum }\limits_{i=0}^{r}{\left(-1)}^{r-i}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right)\left(\genfrac{}{}{0.0pt}{}{r}{i}\right)\frac{{\lambda }^{j+r}}{{\left(j+r)}_{r}}{B}_{n-j}{\left(\frac{{\rm{d}}}{{\rm{d}}x}\right)}^{l}{B}_{j+r}^{\left(r)}\left(\frac{x+i}{\lambda }\right)\\ & =& \mathop{\displaystyle \sum }\limits_{j=0}^{n}\mathop{\displaystyle \sum }\limits_{i=0}^{r}{\left(-1)}^{r-i}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right)\left(\genfrac{}{}{0.0pt}{}{r}{i}\right)\frac{{\lambda }^{j+r-l}}{{\left(j+r)}_{r}}{B}_{n-j}{\left(j+r)}_{l}{B}_{j+r-l}^{\left(r)}\left(\frac{x+i}{\lambda }\right).\end{array}Thus, from Theorem 4.1, (6.4), and (6.5), we have Bn(x)=∑k=0n1k!∑j=0n∑i=0r(−1)r−injriλj+r(j+r)rBn−jΔkBj+r(r)x+iλx=0Dk,λ(r)(x)=∑k=0n∑l=knS2(l,k)1l!∑j=max{0,l−r}n∑i=0r(−1)r−injriλj+r−l(j+r)rBn−j(j+r)lBj+r−l(r)iλDk,λ(r)(x).\hspace{2.2em}\begin{array}{rcl}{B}_{n}\left(x)& =& \mathop{\displaystyle \sum }\limits_{k=0}^{n}\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left\{\mathop{\displaystyle \sum }\limits_{j=0}^{n}\mathop{\displaystyle \sum }\limits_{i=0}^{r}{\left(-1)}^{r-i}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right)\left(\genfrac{}{}{0.0pt}{}{r}{i}\right)\frac{{\lambda }^{j+r}}{{\left(j+r)}_{r}}{B}_{n-j}{\Delta }^{k}{B}_{j+r}^{\left(r)}{\left.\left(\frac{x+i}{\lambda }\right)\right|}_{x=0}\right\}{D}_{k,\lambda }^{\left(r)}\left(x)\\ & =& \mathop{\displaystyle \sum }\limits_{k=0}^{n}\left\{\mathop{\displaystyle \sum }\limits_{l=k}^{n}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\mathop{\displaystyle \sum }\limits_{j={\rm{\max }}\{0,l-r\}}^{n}\mathop{\displaystyle \sum }\limits_{i=0}^{r}{\left(-1)}^{r-i}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right)\left(\genfrac{}{}{0.0pt}{}{r}{i}\right)\frac{{\lambda }^{j+r-l}}{{\left(j+r)}_{r}}{B}_{n-j}{\left(j+r)}_{l}{B}_{j+r-l}^{\left(r)}\left(\frac{i}{\lambda }\right)\right\}{D}_{k,\lambda }^{\left(r)}\left(x).\end{array}(b) Let p(x)=∑k=1n−11k(n−k)Bk(x)Bn−k(x)p\left(x)={\sum }_{k=1}^{n-1}\frac{1}{k\left(n-k)}{B}_{k}\left(x){B}_{n-k}\left(x), for n≥2n\ge 2. As we stated earlier, it was shown in [12] that (6.6)p(x)=2n∑m=0n−21n−mnmBn−mBm(x)+2nHn−1Bn(x),p\left(x)=\frac{2}{n}\mathop{\sum }\limits_{m=0}^{n-2}\frac{1}{n-m}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right){B}_{n-m}{B}_{m}\left(x)+\frac{2}{n}{H}_{n-1}{B}_{n}\left(x),where Hn=1+12+⋯+1n{H}_{n}=1+\frac{1}{2}+\cdots +\frac{1}{n}is the Harmonic number.Write p(x)=∑k=0nakDk(x)p\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k}\left(x). Then, from Theorem 4.1 and (6.6), we have (6.7)ak=2n∑l=kn−2∑m=ln−2∑j=lmS2(l,k)1l!1n−mnmmjBn−mBm−jλj−l+1(j)l−1Bj−l+11λ−Bj−l+1+2nHn−1∑l=kn∑j=lnS2(l,k)1l!njBn−jλj−l+1(j)l−1Bj−l+11λ−Bj−l+1.\begin{array}{rcl}{a}_{k}& =& \frac{2}{n}\left\{\mathop{\displaystyle \sum }\limits_{l=k}^{n-2}\mathop{\displaystyle \sum }\limits_{m=l}^{n-2}\mathop{\displaystyle \sum }\limits_{j=l}^{m}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\frac{1}{n-m}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left(\genfrac{}{}{0.0pt}{}{m}{j}\right){B}_{n-m}{B}_{m-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right)\right\}\\ & & +\frac{2}{n}{H}_{n-1}\mathop{\displaystyle \sum }\limits_{l=k}^{n}\mathop{\displaystyle \sum }\limits_{j=l}^{n}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){B}_{n-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right).\end{array}Thus, from (6.7), we obtain ∑k=1n−11k(n−k)Bk(x)Bn−k(x)=2n∑k=0n∑l=kn−2∑m=ln−2∑j=lmS2(l,k)1l!1n−mnmmjBn−mBm−jλj−l+1(j)l−1Bj−l+11λ−Bj−l+1+Hn−1∑l=kn∑j=lnS2(l,k)1l!njBn−jλj−l+1(j)l−1Bj−l+11λ−Bj−l+1Dk,λ(x),\hspace{-44.95em}\begin{array}{l}\mathop{\displaystyle \sum }\limits_{k=1}^{n-1}\frac{1}{k\left(n-k)}{B}_{k}\left(x){B}_{n-k}\left(x)\\ \hspace{1.0em}=\frac{2}{n}\mathop{\displaystyle \sum }\limits_{k=0}^{n}\left\{\mathop{\displaystyle \sum }\limits_{l=k}^{n-2}\mathop{\displaystyle \sum }\limits_{m=l}^{n-2}\mathop{\displaystyle \sum }\limits_{j=l}^{m}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\frac{1}{n-m}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left(\genfrac{}{}{0.0pt}{}{m}{j}\right){B}_{n-m}{B}_{m-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right)\right.\\ \hspace{1.0em}\hspace{1.0em}\left.+{H}_{n-1}\mathop{\displaystyle \sum }\limits_{l=k}^{n}\mathop{\displaystyle \sum }\limits_{j=l}^{n}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){B}_{n-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right)\right\}{D}_{k,\lambda }\left(x),\end{array}where we understand that the triple sum in the parentheses is zero for k=n−1k=n-1or k=nk=n, and (j)−1=1j+1{\left(j)}_{-1}=\frac{1}{j+1}.(c) Let p(x)=∑k=1n−11k(n−k)Ek(x)En−k(x)p\left(x)={\sum }_{k=1}^{n-1}\frac{1}{k\left(n-k)}{E}_{k}\left(x){E}_{n-k}\left(x), for n≥2n\ge 2. Then, as we saw earlier, it was proved in [12] that (6.8)p(x)=−4n∑m=0nnm(Hn−1−Hn−m)n−m+1En−m+1Bm(x).p\left(x)=-\frac{4}{n}\mathop{\sum }\limits_{m=0}^{n}\frac{\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left({H}_{n-1}-{H}_{n-m})}{n-m+1}{E}_{n-m+1}{B}_{m}\left(x).Write p(x)=∑k=0n−2akDk,λ(x)p\left(x)={\sum }_{k=0}^{n-2}{a}_{k}{D}_{k,\lambda }\left(x). Then, from Theorem 4.1 and (6.8), we can show that (6.9)ak=−4n∑l=kn∑m=ln∑j=lmS2(l,k)1l!nm(Hn−1−Hn−m)n−m+1En−m+1mjBm−jλj−l+1(j)l−1Bj−l+11λ−Bj−l+1,{a}_{k}=-\frac{4}{n}\left\{\mathop{\sum }\limits_{l=k}^{n}\mathop{\sum }\limits_{m=l}^{n}\mathop{\sum }\limits_{j=l}^{m}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\frac{\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left({H}_{n-1}-{H}_{n-m})}{n-m+1}{E}_{n-m+1}\left(\genfrac{}{}{0.0pt}{}{m}{j}\right){B}_{m-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right)\right\},where we understand that (j)−1=1j+1{\left(j)}_{-1}=\frac{1}{j+1}.Hence, from (6.9), we have ∑k=1n−11k(n−k)Ek(x)En−k(x)=−4n∑k=0n∑l=kn∑m=ln∑j=lmS2(l,k)1l!nm(Hn−1−Hn−m)n−m+1En−m+1mjBm−jλj−l+1(j)l−1Bj−l+11λ−Bj−l+1Dk,λ(x).\begin{array}{l}\mathop{\displaystyle \sum }\limits_{k=1}^{n-1}\frac{1}{k\left(n-k)}{E}_{k}\left(x){E}_{n-k}\left(x)\\ \hspace{1.0em}=-\frac{4}{n}\mathop{\displaystyle \sum }\limits_{k=0}^{n}\left\{\mathop{\displaystyle \sum }\limits_{l=k}^{n}\mathop{\displaystyle \sum }\limits_{m=l}^{n}\mathop{\displaystyle \sum }\limits_{j=l}^{m}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\frac{\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left({H}_{n-1}-{H}_{n-m})}{n-m+1}{E}_{n-m+1}\left(\genfrac{}{}{0.0pt}{}{m}{j}\right){B}_{m-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right)\right\}{D}_{k,\lambda }\left(x).\end{array}(d) Here, we consider p(x)=∑k=1n−11k(n−k)Gk(x)Gn−k(x)p\left(x)={\sum }_{k=1}^{n-1}\frac{1}{k\left(n-k)}{G}_{k}\left(x){G}_{n-k}\left(x), for n≥2n\ge 2. As we mentioned earlier, it was shown in [16] that (6.10)p(x)=−4n∑m=0n−2nmGn−mn−mBm(x).p\left(x)=-\frac{4}{n}\mathop{\sum }\limits_{m=0}^{n-2}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\frac{{G}_{n-m}}{n-m}{B}_{m}\left(x).Write p(x)=∑k=0n−2akDk,λ(x)p\left(x)={\sum }_{k=0}^{n-2}{a}_{k}{D}_{k,\lambda }\left(x). Then, from Theorem 4.1 and (6.10), we obtain that (6.11)ak=−4n∑l=kn−2∑m=ln−2∑j=lmS2(l,k)1l!nmGn−mn−mmjBm−jλj−l+1(j)l−1Bj−l+11λ−Bj−l+1,{a}_{k}=-\frac{4}{n}\left\{\mathop{\sum }\limits_{l=k}^{n-2}\mathop{\sum }\limits_{m=l}^{n-2}\mathop{\sum }\limits_{j=l}^{m}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\frac{{G}_{n-m}}{n-m}\left(\genfrac{}{}{0.0pt}{}{m}{j}\right){B}_{m-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right)\right\},where we understand that (j)−1=1j+1{\left(j)}_{-1}=\frac{1}{j+1}.Thus, from (6.11), we obtain ∑k=1n−11k(n−k)Gk(x)Gn−k(x)=−4n∑k=0n−2∑l=kn−2∑m=ln−2∑j=lmS2(l,k)1l!nmGn−mn−mmjBm−jλj−l+1(j)l−1Bj−l+11λ−Bj−l+1Dk,λ(x).\begin{array}{l}\mathop{\displaystyle \sum }\limits_{k=1}^{n-1}\frac{1}{k\left(n-k)}{G}_{k}\left(x){G}_{n-k}\left(x)\\ \hspace{1.0em}=-\hspace{-0.25em}\frac{4}{n}\mathop{\displaystyle \sum }\limits_{k=0}^{n-2}\left\{\mathop{\displaystyle \sum }\limits_{l=k}^{n-2}\mathop{\displaystyle \sum }\limits_{m=l}^{n-2}\mathop{\displaystyle \sum }\limits_{j=l}^{m}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\frac{{G}_{n-m}}{n-m}\left(\genfrac{}{}{0.0pt}{}{m}{j}\right){B}_{m-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right)\right\}{D}_{k,\lambda }\left(x).\end{array}(e) As we mentioned earlier, it was shown (see [17,18]) that, for positive integers mmand nn, with m+n≥2m+n\ge 2, we have (6.12)Bm(x)Bn(x)=∑rm2rn+n2rmB2rBm+n−2r(x)m+n−2r+(−1)m+1Bm+nm+nm.{B}_{m}\left(x){B}_{n}\left(x)=\sum _{r}\left\{\phantom{\rule[-1.25em]{}{0ex}},\left(\genfrac{}{}{0.0pt}{}{m}{2r}\right)n+\left(\genfrac{}{}{0.0pt}{}{n}{2r}\right)m\right\}\frac{{B}_{2r}{B}_{m+n-2r}\left(x)}{m+n-2r}+{\left(-1)}^{m+1}\frac{{B}_{m+n}}{\left(\genfrac{}{}{0.0pt}{}{m+n}{m}\right)}.\hspace{1.85em}Then, from Theorem 4.1 and (6.12), we can show that (6.13)ak=(−1)m+1Bm+nm+nmδk,0+∑r∑l=km+n∑j=lm+n−2rS2(l,k)1l!m2rn+n2rm×B2rm+n−2rm+n−2rjBm+n−2r−jλj−l+1(j)l−1Bj−l+11λ−Bj−l+1.\begin{array}{rcl}{a}_{k}& =& {\left(-1)}^{m+1}\frac{{B}_{m+n}}{\left(\genfrac{}{}{0.0pt}{}{m+n}{m}\right)}{\delta }_{k,0}+\displaystyle \sum _{r}\mathop{\displaystyle \sum }\limits_{l=k}^{m+n}\mathop{\displaystyle \sum }\limits_{j=l}^{m+n-2r}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left\{\phantom{\rule[-1.25em]{}{0ex}},\left(\genfrac{}{}{0.0pt}{}{m}{2r}\right)n+\left(\genfrac{}{}{0.0pt}{}{n}{2r}\right)m\right\}\\ & & \times \frac{{B}_{2r}}{m+n-2r}\left(\genfrac{}{}{0.0pt}{}{m+n-2r}{j}\right){B}_{m+n-2r-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right).\end{array}Thus, form (6.13), we obtain Bm(x)Bn(x)=(−1)m+1Bm+nm+nm+∑k=0m+n∑r∑l=km+n−2r∑j=lm+n−2rS2(l,k)1l!m2rn+n2rm×B2rm+n−2rm+n−2rjBm+n−2r−jλj−l+1(j)l−1Bj−l+11λ−Bj−l+1Dk,λ(x),\begin{array}{rcl}{B}_{m}\left(x){B}_{n}\left(x)& =& {\left(-1)}^{m+1}\frac{{B}_{m+n}}{\left(\genfrac{}{}{0.0pt}{}{m+n}{m}\right)}+\mathop{\displaystyle \sum }\limits_{k=0}^{m+n}\left\{\phantom{\rule[-1.25em]{}{0ex}},\displaystyle \sum _{r}\mathop{\displaystyle \sum }\limits_{l=k}^{m+n-2r}\mathop{\displaystyle \sum }\limits_{j=l}^{m+n-2r}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left\{\phantom{\rule[-1.25em]{}{0ex}},\left(\genfrac{}{}{0.0pt}{}{m}{2r}\right)n+\left(\genfrac{}{}{0.0pt}{}{n}{2r}\right)m\right\}\right.\\ & & \left.\phantom{\rule[-1.35em]{}{0ex}},\times \frac{{B}_{2r}}{m+n-2r}\left(\genfrac{}{}{0.0pt}{}{m+n-2r}{j}\right){B}_{m+n-2r-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right)\right\}{D}_{k,\lambda }\left(x),\end{array}where (j)−1=1j+1{\left(j)}_{-1}=\frac{1}{j+1}.7ConclusionIn this paper, we were interested in representing any polynomial in terms of the degenerate Daehee polynomials and of the higher-order degenerate Daehee polynomials. We were able to derive formulas for such representations with the help of umbral calculus. We showed that, by letting λ\lambda tends to zero, they give formulas for representations by the Daehee polynomials and by the higher-order Daehee polynomials. Further, we illustrated the formulas with some examples.As we mentioned in Section 1, both Faber-Pandharipande-Zagier (FPZ) identity and a variant of Miki’s identity follow from the one identity (see (1.2)) that can be derived from the formula (see (1.1)) involving only derivatives and integrals of the given polynomial, while all the other proofs are quite involved. We recall here that the FPZ identity was a conjectural relation between Hodge integrals in Gromov-Witten’s theory. It should be stressed that our method is very useful and powerful, even though it is elementary.It is one of our future research projects to continue to find formulas representing polynomials in terms of some specific special polynomials and to apply those in discovering some interesting identities. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Open Mathematics de Gruyter

# Representations by degenerate Daehee polynomials

, Volume 20 (1): 16 – Jan 1, 2022
16 pages

/lp/de-gruyter/representations-by-degenerate-daehee-polynomials-stbdD2DOJS
Publisher
de Gruyter
ISSN
2391-5455
eISSN
2391-5455
DOI
10.1515/math-2022-0013
Publisher site
See Article on Publisher Site

### Abstract

1Introduction and preliminariesThe aim of this paper is to derive formulas (see Theorem 3.1) expressing any polynomial in terms of the degenerate Daehee polynomials (see (1.12)) with the help of umbral calculus and to illustrate our results with some examples (see Chapter 6). This can be generalized to the higher-order degenerate Bernoulli polynomials (see (1.13)). Indeed, we deduce formulas (see Theorems 4.1) for representing any polynomial in terms of the higher-order degenerate Daehee polynomials again by using umbral calculus. Letting λ→0\lambda \to 0, we obtain formulas (see Remarks 3.2 and 4.2) for expressing any polynomial in terms of the Daehee polynomials (see (1.10)) and of the higher-order Daehee polynomials (see (1.11)). These formulas are also illustrated in Chapter 5. The contribution of this paper is the derivation of such formulas that, we think, have many potential applications.Let p(x)∈C[x]p\left(x)\in {\mathbb{C}}\left[x], with degp(x)=n{\rm{\deg }}\hspace{0.33em}p\left(x)=n. Write p(x)=∑k=0nakBk(x)p\left(x)={\sum }_{k=0}^{n}{a}_{k}{B}_{k}\left(x), where Bk(x){B}_{k}\left(x)are the Bernoulli polynomials (see (1.3)). Then, it is known (see [1]) that (1.1)ak=1k!∫01p(k)(x)dx,fork=0,1,…,n.{a}_{k}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\underset{0}{\overset{1}{\int }}{p}^{\left(k)}\left(x){\rm{d}}x,\hspace{1em}{\rm{for}}\hspace{1em}k=0,1,\ldots ,n.The following identity (see [1,2]) is obtained by applying the formula in (1.1) to the polynomial p(x)=∑k=1n−11k(n−k)Bk(x)Bn−k(x)p\left(x)={\sum }_{k=1}^{n-1}\frac{1}{k\left(n-k)}{B}_{k}\left(x){B}_{n-k}\left(x)and after slight modification: (1.2)∑k=1n−112k(2n−2k)B2k(x)B2n−2k(x)+22n−1B1(x)B2n−1(x)\begin{array}{l}\mathop{\displaystyle \sum }\limits_{k=1}^{n-1}\frac{1}{2k(2n-2k)}{B}_{2k}(x){B}_{2n-2k}(x)+\frac{2}{2n-1}{B}_{1}(x){B}_{2n-1}(x)\end{array}=1n∑k=1n12k2n2kB2kB2n−2k(x)+1nH2n−1B2n(x)+22n−1B1(x)B2n−1,\begin{array}{l}\hspace{1.0em}=\frac{1}{n}\mathop{\displaystyle \sum }\limits_{k=1}^{n}\frac{1}{2k}\left(\genfrac{}{}{0.0pt}{}{2n}{2k}\right){B}_{2k}{B}_{2n-2k}(x)+\frac{1}{n}{H}_{2n-1}{B}_{2n}(x)+\frac{2}{2n-1}{B}_{1}(x){B}_{2n-1},\end{array}where n≥2n\ge 2and Hn=1+12+⋯+1n{H}_{n}=1+\frac{1}{2}+\cdots +\frac{1}{n}.Letting x=0x=0and x=12x=\frac{1}{2}in (1.2), respectively, give a slight variant of Miki’s identity and the Faber-Pandharipande-Zagier (FPZ) identity. Here, it should be emphasized that the other proofs of Miki’s (see [3,4,5]) and FPZ identities (see [6,7]) are quite involved, while our proofs of Miki’s and FPZ identities follow from the simple formula in (1.1) involving only derivatives and integrals of the given polynomials.Analogous formulas to (1.1) can be obtained for the representations by Euler, Frobenius-Euler, ordered Bell and Genocchi polynomials. Many interesting identities have been derived by using these formulas (see [1,8,9, 10,11,12, 13,14] and references therein). The list in the references is far from being exhaustive. However, the interested reader can easily find more related papers in the literature. Also, we should mention here that there are other ways of obtaining the same result as the one in (1.2). One of them is to use Fourier series expansion of the function obtained by extending by periodicity 1 of the polynomial function restricted to the interval [0,1){[}0,1)(see [2,15,16]).The outline of this paper is as follows. In Section 1, we recall some necessary facts that are needed throughout this paper. In Section 2, we go over umbral calculus briefly. In Section 3, we derive formulas expressing any polynomial in terms of the degenerate Daehee polynomials. In Section 4, we derive formulas representing any polynomial in terms of the higher-order degenerate Daehee polynomials. In Section 5, we illustrate our results with examples of representation by the Daehee polynomials. In Section 6, we illustrate our results with examples of representation by the degenerate Daehee polynomials. Finally, we conclude our paper in Section 7.The Bernoulli polynomials Bn(x){B}_{n}\left(x)are defined by (1.3)tet−1ext=∑n=0∞Bn(x)tnn!.\frac{t}{{e}^{t}-1}{e}^{xt}=\mathop{\sum }\limits_{n=0}^{\infty }{B}_{n}\left(x)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}.When x=0x=0, Bn=Bn(0){B}_{n}={B}_{n}\left(0)are called the Bernoulli numbers. We observe that Bn(x)=∑j=0nnjBn−jxj{B}_{n}\left(x)={\sum }_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){B}_{n-j}{x}^{j}, ddxBn(x)=nBn−1(x)\frac{{\rm{d}}}{{\rm{d}}x}{B}_{n}\left(x)=n{B}_{n-1}\left(x), and Bn(x+1)−Bn(x)=nxn−1{B}_{n}\left(x+1)-{B}_{n}\left(x)=n{x}^{n-1}. The first few terms of Bn{B}_{n}are given by: B0=1,B1=−12,B2=16,B4=−130,B6=142,B8=−130,B10=566,B12=−6912730,…;B2k+1=0,(k≥1).\begin{array}{r}{B}_{0}=1,\hspace{0.5em}{B}_{1}=-\frac{1}{2},\hspace{0.5em}{B}_{2}=\frac{1}{6},\hspace{0.5em}{B}_{4}=-\frac{1}{30},\hspace{0.5em}{B}_{6}=\frac{1}{42},\hspace{0.5em}{B}_{8}=-\frac{1}{30},\hspace{0.5em}{B}_{10}=\frac{5}{66},\hspace{0.5em}{B}_{12}=-\frac{691}{\hspace{0.1em}\text{2730}\hspace{0.1em}},\ldots ;\\ {B}_{2k+1}=0,\hspace{0.5em}\left(k\ge 1).\end{array}More generally, for any nonnegative integer rr, the Bernoulli polynomials Bn(r)(x){B}_{n}^{\left(r)}\left(x)of order rrare given by (1.4)tet−1rext=∑n=0∞Bn(r)(x)tnn!.{\left(\frac{t}{{e}^{t}-1}\right)}^{r}{e}^{xt}=\mathop{\sum }\limits_{n=0}^{\infty }{B}_{n}^{\left(r)}\left(x)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}.When x=0x=0, Bn(r)=Bn(r)(0){B}_{n}^{\left(r)}={B}_{n}^{\left(r)}\left(0)are called the Bernoulli numbers of order rr. We observe that Bn(r)(x)=∑j=0nnjBn−j(r)xj{B}_{n}^{\left(r)}\left(x)={\sum }_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){B}_{n-j}^{\left(r)}{x}^{j}, ddxBn(r)(x)=nBn−1(r)(x)\frac{{\rm{d}}}{{\rm{d}}x}{B}_{n}^{\left(r)}\left(x)=n{B}_{n-1}^{\left(r)}\left(x), Bn(r)(x+1)−Bn(r)(x)=nBn−1(r−1)(x){B}_{n}^{\left(r)}\left(x+1)-{B}_{n}^{\left(r)}\left(x)=n{B}_{n-1}^{\left(r-1)}\left(x).The Euler polynomials En(x){E}_{n}\left(x)are defined by (1.5)2et+1ext=∑n=0∞En(x)tnn!.\frac{2}{{e}^{t}+1}{e}^{xt}=\mathop{\sum }\limits_{n=0}^{\infty }{E}_{n}\left(x)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}.When x=0x=0, En=En(0){E}_{n}={E}_{n}\left(0)are called the Euler numbers. We observe that En(x)=∑j=0nnjEn−jxj{E}_{n}\left(x)={\sum }_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){E}_{n-j}{x}^{j}, ddxEn(x)=nEn−1(x)\frac{{\rm{d}}}{{\rm{d}}x}{E}_{n}\left(x)=n{E}_{n-1}\left(x), En(x+1)+En(x)=2xn{E}_{n}\left(x+1)+{E}_{n}\left(x)=2{x}^{n}. The first few terms of En{E}_{n}are given by: E0=1,E1=−12,E3=14,E5=−12,E7=178,E9=−312,…;E2k=0,(k≥1).\hspace{-40.25em}\begin{array}{rcll}{E}_{0}& =& 1,\hspace{0.75em}{E}_{1}=-\frac{1}{2},\hspace{0.75em}{E}_{3}=\frac{1}{4},\hspace{0.75em}{E}_{5}=-\frac{1}{2},\hspace{0.75em}{E}_{7}=\frac{17}{8},\hspace{0.75em}{E}_{9}=-\frac{31}{2},\ldots ;\hspace{0.75em}{E}_{2k}=& 0,\hspace{0.75em}\left(k\ge 1).\end{array}The Genocchi polynomials Gn(x){G}_{n}\left(x)are defined by (1.6)2tet+1ext=∑n=0∞Gn(x)tnn!.\frac{2t}{{e}^{t}+1}{e}^{xt}=\mathop{\sum }\limits_{n=0}^{\infty }{G}_{n}\left(x)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}.\hspace{5.85em}When x=0x=0, Gn=Gn(0){G}_{n}={G}_{n}\left(0)are called the Genocchi numbers. We observe that Gn(x)=∑j=0nnjGn−jxj{G}_{n}\left(x)={\sum }_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){G}_{n-j}{x}^{j}, ddxGn(x)=nGn−1(x)\frac{{\rm{d}}}{{\rm{d}}x}{G}_{n}\left(x)=n{G}_{n-1}\left(x), Gn(x+1)+Gn(x)=2nxn−1{G}_{n}\left(x+1)+{G}_{n}\left(x)=2n{x}^{n-1}, and degGn(x)=n−1{\rm{\deg }}\hspace{0.33em}{G}_{n}\left(x)=n-1, for n≥1n\ge 1. The first few terms of Gn{G}_{n}are given by: G0=0,G1=1,G2=−1,G4=1,G6=−3,G8=17,G10=−155G12=2073,…;G2k+1=0,(k≥1).\begin{array}{l}{G}_{0}=0,\hspace{1em}{G}_{1}=1,\hspace{1em}{G}_{2}=-1,\hspace{1em}{G}_{4}=1,\hspace{1em}{G}_{6}=-3,\hspace{1em}{G}_{8}=17,\hspace{1em}{G}_{10}=-155\\ {G}_{12}=\hspace{0.1em}\text{2073}\hspace{0.1em},\ldots ;\hspace{1em}{G}_{2k+1}=0,\hspace{1em}\left(k\ge 1).\end{array}For any nonzero real number λ\lambda , the degenerate exponentials are given by (1.7)eλx(t)=(1+λt)xλ=∑n=0∞(x)n,λtnn!,eλ(t)=eλ1(t)=(1+λt)1λ=∑n=0∞(1)n,λtnn!.\begin{array}{rcl}{e}_{\lambda }^{x}\left(t)& =& {\left(1+\lambda t)}^{\tfrac{x}{\lambda }}=\mathop{\displaystyle \sum }\limits_{n=0}^{\infty }{\left(x)}_{n,\lambda }\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}},\\ {e}_{\lambda }\left(t)& =& {e}_{\lambda }^{1}\left(t)={\left(1+\lambda t)}^{\tfrac{1}{\lambda }}=\mathop{\displaystyle \sum }\limits_{n=0}^{\infty }{\left(1)}_{n,\lambda }\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}.\end{array}Here, we recall that the λ\lambda -falling factorials are given by (1.8)(x)0,λ=1,(x)n,λ=x(x−λ)⋯(x−(n−1)λ),(n≥1).{\left(x)}_{0,\lambda }=1,\hspace{1.0em}{\left(x)}_{n,\lambda }=x\left(x-\lambda )\cdots \left(x-\left(n-1)\lambda ),\hspace{1em}\left(n\ge 1).Especially, (x)n=(x)n,1{\left(x)}_{n}={\left(x)}_{n,1}are called the falling factorials and hence given by (1.9)(x)0=1,(x)n=x(x−1)⋯(x−n+1),(n≥1).{\left(x)}_{0}=1,\hspace{1.0em}{\left(x)}_{n}=x\left(x-1)\cdots \left(x-n+1),\hspace{1em}\left(n\ge 1).The compositional inverse of eλ(t){e}_{\lambda }\left(t)is called the degenerate logarithm and given by logλ(t)=1λ(tλ−1),{\log }_{\lambda }\left(t)=\frac{1}{\lambda }\left({t}^{\lambda }-1),\hspace{5.5em}which satisfies eλ(logλ(t))=logλ(eλ(t))=t{e}_{\lambda }\left({\log }_{\lambda }\left(t))={\log }_{\lambda }\left({e}_{\lambda }\left(t))=t.Note here that limλ→0eλx(t)=ext{\mathrm{lim}}_{\lambda \to 0}{e}_{\lambda }^{x}\left(t)={e}^{xt}, limλ→0logλ(t)=log(t){\mathrm{lim}}_{\lambda \to 0}{\log }_{\lambda }\left(t)=\log \left(t).Recall that the Daehee polynomials Dn(x){D}_{n}\left(x)are given by (1.10)log(1+t)t(1+t)x=∑n=0∞Dn(x)tnn!.\frac{\log \left(1+t)}{t}{\left(1+t)}^{x}=\mathop{\sum }\limits_{n=0}^{\infty }{D}_{n}\left(x)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}.When x=0x=0, Dn=Dn(0){D}_{n}={D}_{n}\left(0)are the Daehee numbers.More generally, for any nonnegative integer rr, the Daehee polynomials Dn(r)(x){D}_{n}^{\left(r)}\left(x)of order rrare given by (1.11)log(1+t)tr(1+t)x=∑n=0∞Dn(r)(x)tnn!.{\left(\frac{\log \left(1+t)}{t}\right)}^{r}{\left(1+t)}^{x}=\mathop{\sum }\limits_{n=0}^{\infty }{D}_{n}^{\left(r)}\left(x)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}.When x=0x=0, Dn(r)=Dn(r)(0){D}_{n}^{\left(r)}={D}_{n}^{\left(r)}\left(0)are the Daehee numbers of order rr.The degenerate Daehee polynomials Dn,λ(x){D}_{n,\lambda }\left(x)are defined by (1.12)logλ(1+t)t(1+t)x=∑n=0∞Dn,λ(x)tnn!,\frac{{\log }_{\lambda }\left(1+t)}{t}{\left(1+t)}^{x}=\mathop{\sum }\limits_{n=0}^{\infty }{D}_{n,\lambda }\left(x)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}},which are degenerate versions of the Daehee polynomials in (1.10). For x=0x=0, Dn,λ=Dn,λ(0){D}_{n,\lambda }={D}_{n,\lambda }\left(0)are called the degenerate Daehee numbers and introduced in [7] (see also [14]).More generally, for any nonnegative integer rr, the degenerate Daehee polynomials Dn,λ(r)(x){D}_{n,\lambda }^{\left(r)}\left(x)of order rrare defined by (1.13)logλ(1+t)tr(1+t)x=∑n=0∞Dn,λ(r)(x)tnn!,{\left(\frac{{\log }_{\lambda }\left(1+t)}{t}\right)}^{r}{\left(1+t)}^{x}=\mathop{\sum }\limits_{n=0}^{\infty }{D}_{n,\lambda }^{\left(r)}\left(x)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}},which are degenerate versions of the Daehe polynomials of order rrin (1.11). We remark that Dn,λ(x)→Dn(x){D}_{n,\lambda }\left(x)\to {D}_{n}\left(x), and Dn,λ(r)(x)→Dn(r)(x){D}_{n,\lambda }^{\left(r)}\left(x)\to {D}_{n}^{\left(r)}\left(x), as λ\lambda tends to 0.We recall some notations and facts about forward differences. Let ffbe any complex-valued function of the real variable xx. Then, for any real number aa, the forward difference Δa{\Delta }_{a}is given by (1.14)Δaf(x)=f(x+a)−f(x).{\Delta }_{a}f\left(x)=f\left(x+a)-f\left(x).\hspace{3em}If a=1a=1, then we let (1.15)Δf(x)=Δ1f(x)=f(x+1)−f(x).\Delta f\left(x)={\Delta }_{1}f\left(x)=f\left(x+1)-f\left(x).In general, the nnth oder forward differences are given by (1.16)Δanf(x)=∑i=0nni(−1)n−if(x+ia).{\Delta }_{a}^{n}\hspace{0.16em}f\left(x)=\mathop{\sum }\limits_{i=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{i}\right){\left(-1)}^{n-i}f\left(x+ia).For a=1a=1, we have (1.17)Δnf(x)=∑i=0nni(−1)n−if(x+i).{\Delta }^{n}f\left(x)=\mathop{\sum }\limits_{i=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{i}\right){\left(-1)}^{n-i}f\left(x+i).\hspace{.75em}Finally, we recall that the Stirling numbers of the second kind S2(n,k){S}_{2}\left(n,k)can be given by means of (1.18)1k!(et−1)k=∑n=k∞S2(n,k)tnn!.\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left({e}^{t}-1)}^{k}=\mathop{\sum }\limits_{n=k}^{\infty }{S}_{2}\left(n,k)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}.2Review of umbral calculusHere, we will briefly go over very basic facts about umbral calculus. For more details on this, we recommend the reader to refer to [3, 20, 22]. Let C{\mathbb{C}}be the field of complex numbers. Then, ℱ{\mathcal{ {\mathcal F} }}denotes the algebra of formal power series in ttover C{\mathbb{C}}, given by ℱ=f(t)=∑k=0∞aktkk!ak∈C,{\mathcal{ {\mathcal F} }}=\left\{f\left(t)=\mathop{\sum }\limits_{k=0}^{\infty }{a}_{k}\left.\frac{{t}^{k}}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\right|\hspace{0.33em}{a}_{k}\in {\mathbb{C}}\right\},and P=C[x]{\mathbb{P}}={\mathbb{C}}\left[x]indicates the algebra of polynomials in xxwith coefficients in C{\mathbb{C}}.Let P∗{{\mathbb{P}}}^{\ast }be the vector space of all linear functionals on P{\mathbb{P}}. If ⟨L∣p(x)⟩\langle L| p\left(x)\rangle denotes the action of the linear functional LLon the polynomial p(x)p\left(x), then the vector space operations on P∗{{\mathbb{P}}}^{\ast }are defined by ⟨L+M∣p(x)⟩=⟨L∣p(x)⟩+⟨M∣p(x)⟩,⟨cL∣p(x)⟩=c⟨L∣p(x)⟩,\langle L+M| p\left(x)\rangle =\langle L| p\left(x)\rangle +\langle M| p\left(x)\rangle ,\hspace{1.0em}\langle cL| p\left(x)\rangle =c\langle L| p\left(x)\rangle ,where ccis a complex number.For f(t)∈ℱf\left(t)\in {\mathcal{ {\mathcal F} }}with f(t)=∑k=0∞aktkk!f\left(t)={\sum }_{k=0}^{\infty }{a}_{k}\frac{{t}^{k}}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}, we define the linear functional on P{\mathbb{P}}by (2.1)⟨f(t)∣xk⟩=ak.\langle f\left(t)| {x}^{k}\rangle ={a}_{k}.From (2.1), we note that ⟨tk∣xn⟩=n!δn,k,(n,k≥0),\langle {t}^{k}| {x}^{n}\rangle =n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}{\delta }_{n,k},\hspace{1.0em}\left(n,k\ge 0),where δn,k{\delta }_{n,k}is the Kronecker’s symbol.Some remarkable linear functionals are as follows: (2.2)⟨eyt∣p(x)⟩=p(y),⟨eyt−1∣p(x)⟩=p(y)−p(0),eyt−1tp(x)=∫0yp(u)du.\begin{array}{rcl}\langle {e}^{yt}| p\left(x)\rangle & =& p(y),\\ \langle {e}^{yt}-1| p\left(x)\rangle & =& p(y)-p\left(0),\\ \left\langle \left.\frac{{e}^{yt}-1}{t}\right|p\left(x)\right\rangle & =& \underset{0}{\overset{y}{\displaystyle \int }}p\left(u){\rm{d}}u.\end{array}Let (2.3)fL(t)=∑k=0∞⟨L∣xk⟩tkk!.{f}_{L}\left(t)=\mathop{\sum }\limits_{k=0}^{\infty }\langle L| {x}^{k}\rangle \frac{{t}^{k}}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}.Then, by (2.1) and (2.3), we obtain ⟨fL(t)∣xn⟩=⟨L∣xn⟩.\hspace{.08em}\langle \hspace{.08em}{f}_{L}\left(t)| {x}^{n}\rangle =\langle L| {x}^{n}\rangle .\hspace{.8em}That is, fL(t)=L{f}_{L}\left(t)=L. In addition, the map L⟼fL(t)L\hspace{0.33em}\longmapsto \hspace{0.33em}{f}_{L}\left(t)is a vector space isomorphism from P∗{{\mathbb{P}}}^{\ast }onto ℱ{\mathcal{ {\mathcal F} }}.Henceforth, ℱ{\mathcal{ {\mathcal F} }}denotes both the algebra of formal power series in ttand the vector space of all linear functionals on P{\mathbb{P}}. ℱ{\mathcal{ {\mathcal F} }}is called the umbral algebra and the umbral calculus is the study of umbral algebra. For each nonnegative integer kk, the differential operator tk{t}^{k}on P{\mathbb{P}}is defined by (2.4)tkxn=(n)kxn−k,ifk≤n,0,ifk>n.{t}^{k}{x}^{n}=\left\{\begin{array}{ll}{\left(n)}_{k}{x}^{n-k},& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}k\le n,\\ 0,& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}k\gt n.\end{array}\right.\hspace{5.7em}Extending (2.4) linearly, any power series f(t)=∑k=0∞akk!tk∈ℱf\left(t)=\mathop{\sum }\limits_{k=0}^{\infty }\frac{{a}_{k}}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{t}^{k}\in {\mathcal{ {\mathcal F} }}gives the differential operator on P{\mathbb{P}}defined by (2.5)f(t)xn=∑k=0nnkakxn−k,(n≥0).f\left(t){x}^{n}=\mathop{\sum }\limits_{k=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{k}\right){a}_{k}{x}^{n-k},\hspace{1.0em}\left(n\ge 0).\hspace{2.75em}It should be observed that, for any formal power series f(t)f\left(t)and any polynomial p(x)p\left(x), we have (2.6)⟨f(t)∣p(x)⟩=⟨1∣f(t)p(x)⟩=f(t)p(x)∣x=0.\langle f\left(t)| p\left(x)\rangle =\langle 1| f\left(t)p\left(x)\rangle =f\left(t)p\left(x){| }_{x=0}.\hspace{0.275em}Here, we note that an element f(t)f\left(t)of ℱ{\mathcal{ {\mathcal F} }}is a formal power series, a linear functional, and a differential operator. Some notable differential operators are as follows: (2.7)eytp(x)=p(x+y),(eyt−1)p(x)=p(x+y)−p(x)=Δyp(x),eyt−1tp(x)=∫xx+yp(u)du.\begin{array}{l}{e}^{yt}p\left(x)=p\left(x+y),\\ \left({e}^{yt}-1)p\left(x)=p\left(x+y)-p\left(x)={\Delta }_{y}p\left(x),\\ \frac{{e}^{yt}-1}{t}p\left(x)=\underset{x}{\overset{x+y}{\displaystyle \int }}p\left(u){\rm{d}}u.\end{array}The order o(f(t))o(f\left(t))of the power series f(t)(≠0)f\left(t)\left(\ne 0)is the smallest integer for which ak{a}_{k}does not vanish. If o(f(t))=0o(f\left(t))=0, then f(t)f\left(t)is called an invertible series. If o(f(t))=1o(f\left(t))=1, then f(t)f\left(t)is called a delta series.For f(t),g(t)∈ℱf\left(t),g\left(t)\in {\mathcal{ {\mathcal F} }}with o(f(t))=1o(f\left(t))=1and o(g(t))=0o\left(g\left(t))=0, there exists a unique sequence sn(x){s}_{n}\left(x)(deg sn(x)=n{s}_{n}\left(x)=n) of polynomials such that (2.8)⟨g(t)f(t)k∣sn(x)⟩=n!δn,k,(n,k≥0).\langle g\left(t)f{\left(t)}^{k}| {s}_{n}\left(x)\rangle =n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}{\delta }_{n,k},\hspace{1.0em}\left(n,k\ge 0).The sequence sn(x){s}_{n}\left(x)is said to be the Sheffer sequence for (g(t),f(t))\left(g\left(t),f\left(t)), which is denoted by sn(x)∼(g(t),f(t)){s}_{n}\left(x)\hspace{0.33em} \sim \hspace{0.33em}\left(g\left(t),f\left(t)). We observe from (2.8) that (2.9)sn(x)=1g(t)pn(x),{s}_{n}\left(x)=\frac{1}{g\left(t)}{p}_{n}\left(x),where pn(x)=g(t)sn(x)∼(1,f(t)){p}_{n}\left(x)=g\left(t){s}_{n}\left(x)\hspace{0.33em} \sim \hspace{0.33em}\left(1,f\left(t)).In particular, if sn(x)∼(g(t),t){s}_{n}\left(x)\hspace{0.33em} \sim \hspace{0.33em}\left(g\left(t),t), then pn(x)=xn{p}_{n}\left(x)={x}^{n}, and hence, (2.10)sn(x)=1g(t)xn.{s}_{n}\left(x)=\frac{1}{g\left(t)}{x}^{n}.\hspace{1.25em}It is well known that sn(x)∼(g(t),f(t)){s}_{n}\left(x)\hspace{0.33em} \sim \hspace{0.33em}\left(g\left(t),f\left(t))if and only if (2.11)1g(f¯(t))exf¯(t)=∑k=0∞sk(x)k!tk,\frac{1}{g(\overline{f}\left(t))}{e}^{x\overline{f}\left(t)}=\mathop{\sum }\limits_{k=0}^{\infty }\frac{{s}_{k}\left(x)}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{t}^{k},for all x∈Cx\in {\mathbb{C}}, where f¯(t)\overline{f}\left(t)is the compositional inverse of f(t)f\left(t)such that f¯(f(t))=f(f¯(t))=t\overline{f}(f\left(t))=f(\overline{f}\left(t))=t.Equations (2.12)–(2.14) are equivalent to the fact that sn(x){s}_{n}(x)is Sheffer for (g(t),f(t))(g(t),f(t)), for some invertible g(t)g\left(t): (2.12)f(t)sn(x)=nsn−1(x),(n≥0),f(t){s}_{n}(x)=n{s}_{n-1}(x),\hspace{1.0em}(n\ge 0),(2.13)sn(x+y)=∑j=0nnjsj(x)pn−j(y),{s}_{n}(x+y)=\mathop{\sum }\limits_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){s}_{j}(x){p}_{n-j}(y),\hspace{0.25em}with pn(x)=g(t)sn(x){p}_{n}(x)=g(t){s}_{n}(x), (2.14)sn(x)=∑j=0n1j!⟨g(f¯(t))−1f¯(t)j∣xn⟩xj.\hspace{1.3em}{s}_{n}(x)=\mathop{\sum }\limits_{j=0}^{n}\frac{1}{j\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\langle g{(\overline{f}(t))}^{-1}\overline{f}{(t)}^{j}| {x}^{n}\rangle {x}^{j}.\hspace{0.25em}Let pn(x){p}_{n}\left(x), qn(x)=∑k=0nqn,kxk{q}_{n}\left(x)={\sum }_{k=0}^{n}{q}_{n,k}{x}^{k}be sequences of polynomials. Then, the umbral composition of qn(x){q}_{n}\left(x)with pn(x){p}_{n}\left(x)is defined to be the sequence (2.15)qn(p(x))=∑k=0nqn,kpk(x).{q}_{n}\left({\bf{p}}\left(x))=\mathop{\sum }\limits_{k=0}^{n}{q}_{n,k}{p}_{k}\left(x).\hspace{3em}3Representations by degenerate Daehee polynomialsOur interest here is to derive formulas expressing any polynomial in terms of the degenerate Daehee polynomials.From (1.7), (1.9), and (1.11), we first observe that (3.1)Dn,λ(x)∼g(t)=λf(t)eλt−1=λ(et−1)eλt−1,f(t)=et−1,{D}_{n,\lambda }\left(x)\hspace{0.33em} \sim \hspace{0.33em}\left(g\left(t)=\frac{\lambda f\left(t)}{{e}^{\lambda t}-1}=\frac{\lambda \left({e}^{t}-1)}{{e}^{\lambda t}-1},f\left(t)={e}^{t}-1\right),(3.2)(x)n∼(1,f(t)=et−1).{\left(x)}_{n}\hspace{0.33em} \sim \hspace{0.33em}\left(1,f\left(t)={e}^{t}-1).\hspace{0.5em}From (1.15), (2.7), (2.8), (2.12), (3.1), and (3.2), we note that (3.3)f(t)Dn,λ(x)=nDn−1,λ(x)=(et−1)Dn,λ(x)=ΔDn,λ(x),f\left(t){D}_{n,\lambda }\left(x)=n{D}_{n-1,\lambda }\left(x)=\left({e}^{t}-1){D}_{n,\lambda }\left(x)=\Delta {D}_{n,\lambda }\left(x),(3.4)f(t)(x)n=n(x)n−1,f\left(t){\left(x)}_{n}=n{\left(x)}_{n-1},\hspace{2.7em}(3.5)g(t)Dn,λ(x)=(x)n.g\left(t){D}_{n,\lambda }\left(x)={\left(x)}_{n}.\hspace{2.75em}Now, we assume that p(x)∈C[x]p\left(x)\in {\mathbb{C}}\left[x]has degree nn, and write p(x)=∑k=0nakDk,λ(x)p\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k,\lambda }\left(x). Then, from (3.5), we have (3.6)g(t)p(x)=∑k=0nakg(t)Dk,λ(x)=∑k=0nak(x)k.g\left(t)p\left(x)=\mathop{\sum }\limits_{k=0}^{n}{a}_{k}g\left(t){D}_{k,\lambda }\left(x)=\mathop{\sum }\limits_{k=0}^{n}{a}_{k}{\left(x)}_{k}.\hspace{5.35em}For k≥0k\ge 0, from (3.4) and (3.6), we obtain (3.7)f(t)kg(t)p(x)=f(t)k∑l=0nal(x)l,λ=∑l=knl(l−1)⋯(l−k+1)al(x)l−k,λ.f{\left(t)}^{k}g\left(t)p\left(x)=f{\left(t)}^{k}\mathop{\sum }\limits_{l=0}^{n}{a}_{l}{\left(x)}_{l,\lambda }=\mathop{\sum }\limits_{l=k}^{n}l\left(l-1)\cdots \left(l-k+1){a}_{l}{\left(x)}_{l-k,\lambda }.Letting x=0x=0in (3.7), we finally obtain (3.8)ak=1k!f(t)kg(t)p(x)∣x=0=1k!⟨g(t)f(t)k∣p(x)⟩,(k≥0).{a}_{k}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}f{\left(t)}^{k}g\left(t)p\left(x){| }_{x=0}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\langle g\left(t)f{\left(t)}^{k}| p\left(x)\rangle ,\hspace{1em}\left(k\ge 0).Now, we want to find more explicit expressions for (3.8). As λteλt−1ext=∑n=0∞λnBnxλtnn!\frac{\lambda t}{{e}^{\lambda t}-1}{e}^{xt}={\sum }_{n=0}^{\infty }{\lambda }^{n}{B}_{n}\left(\frac{x}{\lambda }\right)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}, we see from (2.10) that λnBnxλ=λteλt−1xn{\lambda }^{n}{B}_{n}\left(\frac{x}{\lambda }\right)=\frac{\lambda t}{{e}^{\lambda t}-1}{x}^{n}. To proceed further, we let p(x)=∑i=0nbixip\left(x)={\sum }_{i=0}^{n}{b}_{i}{x}^{i}.From (2.7), (2.15), and (3.1), noting that g(t)=et−1tλteλt−1g\left(t)=\frac{{e}^{t}-1}{t}\frac{\lambda t}{{e}^{\lambda t}-1}, we have (3.9)g(t)p(x)=et−1tλteλt−1p(x)=et−1t∑i=0nbiλteλt−1xi=et−1t∑i=0nbiλiBixλ=et−1tpλBxλ=∫xx+1pλBuλdu,\hspace{-19.65em}\begin{array}{rcl}g\left(t)p\left(x)& =& \frac{{e}^{t}-1}{t}\frac{\lambda t}{{e}^{\lambda t}-1}p\left(x)\\ & =& \frac{{e}^{t}-1}{t}\mathop{\displaystyle \sum }\limits_{i=0}^{n}{b}_{i}\frac{\lambda t}{{e}^{\lambda t}-1}{x}^{i}\\ & =& \frac{{e}^{t}-1}{t}\mathop{\displaystyle \sum }\limits_{i=0}^{n}{b}_{i}{\lambda }^{i}{B}_{i}\left(\frac{x}{\lambda }\right)\\ & =& \frac{{e}^{t}-1}{t}p\left(\lambda {\bf{B}}\left(\frac{x}{\lambda }\right)\right)\\ & =& \underset{x}{\overset{x+1}{\displaystyle \int }}p\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u,\end{array}where pλBxλp\left(\lambda {\bf{B}}\left(\frac{x}{\lambda }\right)\right)denotes the umbral composition of p(x)p\left(x)with λiBixλ{\lambda }^{i}{B}_{i}\left(\frac{x}{\lambda }\right), that is, it is given by pλBxλ=p\left(\lambda {\bf{B}}\left(\frac{x}{\lambda }\right)\right)=∑i=0nbiλiBixλ{\sum }_{i=0}^{n}{b}_{i}{\lambda }^{i}{B}_{i}\left(\frac{x}{\lambda }\right).We note from (3.5) and (3.9), in passing, that the following holds: (x)n=g(t)Dn,λ(x)=∫xx+1Dn,λλBuλdu.{\left(x)}_{n}=g\left(t){D}_{n,\lambda }\left(x)=\underset{x}{\overset{x+1}{\int }}{D}_{n,\lambda }\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u.\hspace{2.8em}From (2.7) and (3.9), we deduce (3.10)ak=1k!f(t)kg(t)p(x)∣x=0=1k!Δk∫xx+1pλBuλdux=0.\hspace{4.35em}{a}_{k}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}f{\left(t)}^{k}g\left(t)p\left(x){| }_{x=0}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\Delta }^{k}{\left.\left(\underset{x}{\overset{x+1}{\int }}p\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u\right),\right|}_{x=0}.By making use of (1.17) and (3.10), an alternative expression of (3.10) is given by (3.11)ak=1k!∑i=0kki(−1)k−i∫ii+1pλBuλdu.{a}_{k}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\mathop{\sum }\limits_{i=0}^{k}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right){\left(-1)}^{k-i}\underset{i}{\overset{i+1}{\int }}p\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u.\hspace{3.35em}We obtain yet another expression from (1.18), (3.8), and (3.9), which is given by (3.12)ak=1k!(et−1)k∫xx+1pλBuλdux=0=∑l=k∞S2(l,k)tll!∫xx+1pλBuλdux=0=∑l=knS2(l,k)1l!ddxl∫xx+1pλBuλdux=0,\begin{array}{rcl}{a}_{k}& =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left({e}^{t}-1)}^{k}{\left.\left(\underset{x}{\overset{x+1}{\displaystyle \int }}p\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u\right),\right|}_{x=0}\\ & =& \mathop{\displaystyle \sum }\limits_{l=k}^{\infty }{S}_{2}\left(l,k)\frac{{t}^{l}}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left.\left(\underset{x}{\overset{x+1}{\displaystyle \int }}p\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u\right),\right|}_{x=0}\\ & =& \mathop{\displaystyle \sum }\limits_{l=k}^{n}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left(\frac{{\rm{d}}}{{\rm{d}}x}\right)}^{l}{\left.\left(\underset{x}{\overset{x+1}{\displaystyle \int }}p\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u\right),\right|}_{x=0},\end{array}where we need to note that ∫xx+1pλBuλdu{\int }_{x}^{x+1}p\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}uhas degree nn.Finally, from (3.10)–(3.12), and (3.8), we obtain the following theorem.Theorem 3.1Let p(x)∈C[x]p\left(x)\in {\mathbb{C}}\left[x], with degp(x)=n{\rm{\deg }}\hspace{0.33em}p\left(x)=n. Then, we have p(x)=∑k=0nakDk,λ(x)p\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k,\lambda }\left(x), whereak=1k!f(t)kg(t)p(x)∣x=0=1k!Δk∫xx+1pλBuλdux=0=1k!∑i=0kki(−1)k−i∫ii+1pλBuλdu=∑l=knS2(l,k)1l!ddxl∫xx+1pλBuλdux=0,fork=0,1,…,n,\begin{array}{rcl}{a}_{k}& =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}f{\left(t)}^{k}g\left(t)p\left(x){| }_{x=0}\\ & =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\Delta }^{k}{\left.\left(\underset{x}{\overset{x+1}{\displaystyle \int }}p\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u\right)\right|}_{x=0}\\ & =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\mathop{\displaystyle \sum }\limits_{i=0}^{k}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right){\left(-1)}^{k-i}\underset{i}{\overset{i+1}{\displaystyle \int }}p\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u\\ & =& \mathop{\displaystyle \sum }\limits_{l=k}^{n}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left(\frac{{\rm{d}}}{{\rm{d}}x}\right)}^{l}{\left.\left(\underset{x}{\overset{x+1}{\displaystyle \int }}p\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u\right)\right|}_{x=0},\hspace{1em}{for}\hspace{1em}k=0,1,\ldots ,n,\end{array}where g(t)=λ(et−1)eλt−1g\left(t)=\frac{\lambda \left({e}^{t}-1)}{{e}^{\lambda t}-1}, f(t)=et−1f\left(t)={e}^{t}-1, and pλBxλp\left(\lambda {\bf{B}}\left(\frac{x}{\lambda }\right)\right)denotes the umbral composition of p(x)p\left(x)with λiBixλ{\lambda }^{i}{B}_{i}\left(\frac{x}{\lambda }\right).Remark 3.2Let p(x)∈C[x]p\left(x)\in {\mathbb{C}}\left[x], with degp(x)=n{\rm{\deg }}\hspace{0.33em}p\left(x)=n. Write p(x)=∑k=0nakDk(x)p\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k}\left(x). As λ\lambda tends to 0, g(t)→et−1tg\left(t)\to \frac{{e}^{t}-1}{t}, and pλBxλ→p(x)p\left(\lambda {\bf{B}}\left(\frac{x}{\lambda }\right)\right)\to p\left(x). Thus, we obtain the following result.ak=1k!Δk∫xx+1p(u)dux=0=1k!∑i=0kki(−1)k−i∫ii+1p(u)du=∑l=knS2(l,k)1l!ddxl∫xx+1p(u)dux=0,fork=0,1,…,n.\hspace{-30.1em}\begin{array}{rcl}{a}_{k}& =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\Delta }^{k}{\left.\left(\underset{x}{\overset{x+1}{\displaystyle \int }}p\left(u){\rm{d}}u\right)\right|}_{x=0}\\ & =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\mathop{\displaystyle \sum }\limits_{i=0}^{k}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right){\left(-1)}^{k-i}\underset{i}{\overset{i+1}{\displaystyle \int }}p\left(u){\rm{d}}u\\ & =& \mathop{\displaystyle \sum }\limits_{l=k}^{n}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left(\frac{{\rm{d}}}{{\rm{d}}x}\right)}^{l}{\left.\left(\underset{x}{\overset{x+1}{\displaystyle \int }}p\left(u){\rm{d}}u\right)\right|}_{x=0},\hspace{1em}{\rm{for}}\hspace{1em}k=0,1,\ldots ,n.\end{array}4Representations by higher-order degenerate Daehee polynomialsOur interest here is to derive formulas expressing any polynomial in terms of the higher-order degenerate Daehee polynomials.With g(t)=λf(t)eλt−1=λ(et−1)eλt−1g\left(t)=\frac{\lambda f\left(t)}{{e}^{\lambda t}-1}=\frac{\lambda \left({e}^{t}-1)}{{e}^{\lambda t}-1}, f(t)=et−1f\left(t)={e}^{t}-1, from (1.11), we note that (4.1)Dn,λ(r)(x)∼(g(t)r,f(t)),{D}_{n,\lambda }^{\left(r)}\left(x)\hspace{0.33em} \sim \hspace{0.33em}\left(g{\left(t)}^{r},f\left(t)),(4.2)(x)n∼(1,f(t)).{\left(x)}_{n}\hspace{0.33em} \sim \hspace{0.33em}\left(1,f\left(t)).\hspace{0.25em}From (1.15), (2.7), (2.8), (2.12), (4.1), and (4.2), we note that (4.3)f(t)Dn,λ(r)(x)=nDn−1,λ(r)(x)=(et−1)Dn,λ(r)(x)=ΔDn,λ(r)(x),f\left(t){D}_{n,\lambda }^{\left(r)}\left(x)=n{D}_{n-1,\lambda }^{\left(r)}\left(x)=\left({e}^{t}-1){D}_{n,\lambda }^{\left(r)}\left(x)=\Delta {D}_{n,\lambda }^{\left(r)}\left(x),(4.4)f(t)(x)n=n(x)n−1,f\left(t){\left(x)}_{n}=n{\left(x)}_{n-1},\hspace{1.01em}(4.5)g(t)rDn,λ(r)(x)=(x)n.\hspace{3.15em}g{\left(t)}^{r}{D}_{n,\lambda }^{\left(r)}\left(x)={\left(x)}_{n}.\hspace{3.9em}Now, we assume that p(x)∈C[x]p\left(x)\in {\mathbb{C}}\left[x]has degree nn, and write p(x)=∑k=0nakDk,λ(r)(x)p\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k,\lambda }^{\left(r)}\left(x). Then, from (4.5), we have (4.6)g(t)rp(x)=∑k=0nakg(t)rDk,λ(r)(x)=∑k=0nak(x)k.g{\left(t)}^{r}p\left(x)=\mathop{\sum }\limits_{k=0}^{n}{a}_{k}g{\left(t)}^{r}{D}_{k,\lambda }^{\left(r)}\left(x)=\mathop{\sum }\limits_{k=0}^{n}{a}_{k}{\left(x)}_{k}.For k≥0k\ge 0, from (4.4), we obtain (4.7)f(t)kg(t)rp(x)=f(t)k∑l=0nal(x)l=∑l=knl(l−1)⋯(l−k+1)al(x)l−k.f{\left(t)}^{k}g{\left(t)}^{r}p\left(x)=f{\left(t)}^{k}\mathop{\sum }\limits_{l=0}^{n}{a}_{l}{\left(x)}_{l}=\mathop{\sum }\limits_{l=k}^{n}l\left(l-1)\cdots \left(l-k+1){a}_{l}{\left(x)}_{l-k}.Letting x=0x=0in (4.7), we finally obtain (4.8)ak=1k!f(t)kg(t)rp(x)∣x=0=1k!⟨g(t)rf(t)k∣p(x)⟩,(k≥0).{a}_{k}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}f{\left(t)}^{k}g{\left(t)}^{r}p\left(x){| }_{x=0}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\langle g{\left(t)}^{r}f{\left(t)}^{k}| p\left(x)\rangle ,\hspace{1em}\left(k\ge 0).This also follows from the observation ⟨g(t)rf(t)k∣Dl,λ(r)(x)⟩=l!δl,k\langle g{\left(t)}^{r}f{\left(t)}^{k}| {D}_{l,\lambda }^{\left(r)}\left(x)\rangle =l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}\hspace{0.33em}{\delta }_{l,k}.Now, we want to find more explicit expressions for (4.8). As λteλt−1rext=∑n=0∞λnBn(r)xλtnn!{\left(\frac{\lambda t}{{e}^{\lambda t}-1}\right)}^{r}{e}^{xt}={\sum }_{n=0}^{\infty }{\lambda }^{n}{B}_{n}^{\left(r)}\left(\frac{x}{\lambda }\right)\frac{{t}^{n}}{n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}, we see from (2.10) that λnBn(r)xλ=λteλt−1rxn{\lambda }^{n}{B}_{n}^{\left(r)}\left(\frac{x}{\lambda }\right)={\left(\frac{\lambda t}{{e}^{\lambda t}-1}\right)}^{r}{x}^{n}. To proceed further, we let p(x)=∑i=0nbixip\left(x)={\sum }_{i=0}^{n}{b}_{i}{x}^{i}.From (2.7), (2.15), and (4.1), noting that g(t)=et−1tλteλt−1g\left(t)=\frac{{e}^{t}-1}{t}\frac{\lambda t}{{e}^{\lambda t}-1}, we have (4.9)g(t)rp(x)=et−1trλteλt−1rp(x)=et−1tr∑i=0nbiλteλt−1rxi=et−1tr∑i=0nbiλiBi(r)xλ=et−1trpλB(r)xλ=IrpλB(r)xλ,\hspace{-17.7em}\begin{array}{rcl}g{\left(t)}^{r}p\left(x)& =& {\left(\frac{{e}^{t}-1}{t}\right)}^{r}{\left(\frac{\lambda t}{{e}^{\lambda t}-1}\right)}^{r}p\left(x)\\ & =& {\left(\frac{{e}^{t}-1}{t}\right)}^{r}\mathop{\displaystyle \sum }\limits_{i=0}^{n}{b}_{i}{\left(\frac{\lambda t}{{e}^{\lambda t}-1}\right)}^{r}{x}^{i}\\ & =& {\left(\frac{{e}^{t}-1}{t}\right)}^{r}\mathop{\displaystyle \sum }\limits_{i=0}^{n}{b}_{i}{\lambda }^{i}{B}_{i}^{\left(r)}\left(\frac{x}{\lambda }\right)\\ & =& {\left(\frac{{e}^{t}-1}{t}\right)}^{r}p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)\\ & =& {I}^{r}p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right),\end{array}where pλB(r)xλp\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)denotes the umbral composition of p(x)p\left(x)with λiBi(r)xλ{\lambda }^{i}{B}_{i}^{\left(r)}\left(\frac{x}{\lambda }\right), that is, it is given by pλB(r)xλ=p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)=∑i=0nbiλiBi(r)xλ{\sum }_{i=0}^{n}{b}_{i}{\lambda }^{i}{B}_{i}^{\left(r)}\left(\frac{x}{\lambda }\right), and IIdenotes the linear integral operator given by q(x)→∫xx+1q(x)dxq\left(x)\to {\int }_{x}^{x+1}q\left(x){\rm{d}}x.We note from (4.5) and (4.9), in passing, that the following holds: (x)n=g(t)rDn,λ(x)=IrDn,λλB(r)xλ.{\left(x)}_{n}=g{\left(t)}^{r}{D}_{n,\lambda }\left(x)={I}^{r}{D}_{n,\lambda }\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right).From (2.7) and (4.9), we deduce (4.10)ak=1k!f(t)kg(t)rp(x)∣x=0=1k!ΔkIrpλB(r)xλx=0.{a}_{k}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}f{\left(t)}^{k}g{\left(t)}^{r}p\left(x){| }_{x=0}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\Delta }^{k}{\left.\left({I}^{r}p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)\right),\right|}_{x=0}.By making use of (1.17) and (4.10), an alternative expression of (3.10) is given by (4.11)ak=1k!∑i=0kki(−1)k−iIrpλB(r)xλx=i.{a}_{k}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\mathop{\sum }\limits_{i=0}^{k}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right){\left(-1)}^{k-i}{I}^{r}p{\left.\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right),\right|}_{x=i}.We obtain yet another expression from (1.18), (4.8), and (4.9), which is given by (4.12)ak=1k!(et−1)kIrpλB(r)xλx=0\begin{array}{rcl}{a}_{k}& =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left({e}^{t}-1)}^{k}{\left.\left({I}^{r}p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)\right),\right|}_{x=0}\end{array}\hspace{1.8em}=∑l=k∞S2(l,k)tll!IrpλB(r)xλx=0=∑l=knS2(l,k)1l!ddxlIrpλB(r)xλx=0,\hspace{-23.4em}\begin{array}{rcl}& =& \mathop{\displaystyle \sum }\limits_{l=k}^{\infty }{S}_{2}\left(l,k)\frac{{t}^{l}}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left.\left({I}^{r}p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)\right),\right|}_{x=0}\\ & =& \mathop{\displaystyle \sum }\limits_{l=k}^{n}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left(\frac{{\rm{d}}}{{\rm{d}}x}\right)}^{l}{\left.\left({I}^{r}p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)\right),\right|}_{x=0},\end{array}where we need to observe that IrpλB(r)xλ{I}^{r}p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)has degree nn.Finally, from (4.10)–(4.12) and (4.8), we obtain the following theorem.Theorem 4.1Let p(x)∈C[x]p\left(x)\in {\mathbb{C}}\left[x], with degp(x)=n{\rm{\deg }}\hspace{0.33em}p\left(x)=n. Then, we have p(x)=∑k=0nakDk,λ(r)(x)p\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k,\lambda }^{\left(r)}\left(x), whereak=1k!f(t)kg(t)rp(x)∣x=0=1k!ΔkIrpλB(r)xλx=0=1k!∑i=0kki(−1)k−iIrpλB(r)xλx=i=∑l=knS2(l,k)1l!ddxlIrpλB(r)xλx=0,fork=0,1,…,n,\hspace{3.55em}\begin{array}{rcl}{a}_{k}& =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}f{\left(t)}^{k}g{\left(t)}^{r}p\left(x){| }_{x=0}\\ & =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\Delta }^{k}{\left.\left({I}^{r}p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)\right),\right|}_{x=0}\\ & =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\mathop{\displaystyle \sum }\limits_{i=0}^{k}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right){\left(-1)}^{k-i}{I}^{r}p{\left.\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right),\right|}_{x=i}\\ & =& \mathop{\displaystyle \sum }\limits_{l=k}^{n}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left(\frac{{\rm{d}}}{{\rm{d}}x}\right)}^{l}{\left.\left({I}^{r}p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)\right),\right|}_{x=0},\hspace{1em}{for}\hspace{1em}k=0,1,\ldots ,n,\end{array}where g(t)=λ(et−1)eλt−1g\left(t)=\frac{\lambda \left({e}^{t}-1)}{{e}^{\lambda t}-1}, f(t)=et−1f\left(t)={e}^{t}-1, pλB(r)xλp\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)indicates the umbral composition of p(x)p\left(x)with λiBi(r)xλ{\lambda }^{i}{B}_{i}^{\left(r)}\left(\frac{x}{\lambda }\right), and IIdenotes the linear integral operator given by q(x)→∫xx+1q(x)dxq\left(x)\to {\int }_{x}^{x+1}q\left(x){\rm{d}}x.We observe that IrpλB(r)xλx=i=∫ii+1Ir−1pλB(r)xλdx{I}^{r}p{\left.\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)\right|}_{x=i}={\int }_{i}^{i+1}{I}^{r-1}p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right){\rm{d}}x.Remark 4.2Let p(x)∈C[x]p\left(x)\in {\mathbb{C}}\left[x], with degp(x)=n{\rm{\deg }}\hspace{0.33em}p\left(x)=n. Write p(x)=∑k=0nakDk(r)(x)p\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k}^{\left(r)}\left(x). As λ\lambda tends to 0, g(t)→et−1tg\left(t)\to \frac{{e}^{t}-1}{t}, and pλB(r)xλ→p(x)p\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)\to p\left(x). Thus, we obtain the following result.ak=1k!Δk(Irp(x))x=0=1k!∑i=0kki(−1)k−iIrp(x)x=i=∑l=knS2(l,k)1l!ddxl(Irp(x))x=0,fork=0,1,…,n.\begin{array}{rcl}{a}_{k}& =& {\left.\frac{1}{k\text{&#x0021;}}{\Delta }^{k}({I}^{r}p\left(x))\right|}_{x=0}\\ & =& {\left.\frac{1}{k\text{&#x0021;}}\mathop{\displaystyle \sum }\limits_{i=0}^{k}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right){\left(-1)}^{k-i}{I}^{r}p\left(x)\right|}_{x=i}\\ & =& {\left.\mathop{\displaystyle \sum }\limits_{l=k}^{n}{S}_{2}\left(l,k)\frac{1}{l\text{&#x0021;}}{\left(\frac{{\rm{d}}}{{\rm{d}}x}\right)}^{l}({I}^{r}p\left(x))\right|}_{x=0},\hspace{1em}{\rm{for}}\hspace{1em}k=0,1,\ldots ,n.\end{array}We note that Irp(x)∣x=i=∫ii+1Ir−1p(x)dx{{I}^{r}p\left(x)| }_{x=i}={\int }_{i}^{i+1}{I}^{r-1}p\left(x){\rm{d}}x.5Examples on representation by Daehee polynomialsHere, we illustrate our formulas in Remarks 3.2 and 4.2 with some examples.(a) Let p(x)=Bn(x)=∑k=0nakDk(x)p\left(x)={B}_{n}\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k}\left(x). Then, as Bn(x+1)−Bn(x)=nxn−1{B}_{n}\left(x+1)-{B}_{n}\left(x)=n{x}^{n-1}, ∫xx+1Bn(u)du=xn{\int }_{x}^{x+1}{B}_{n}\left(u){\rm{d}}u={x}^{n}, from Remark 3.2, we have (5.1)ak=1k!Δkxn∣x=0=1k!∑i=0kki(−1)k−iin=S2(n,k),\hspace{-25.2em}{a}_{k}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\Delta }^{k}{x}^{n}{| }_{x=0}=\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\mathop{\sum }\limits_{i=0}^{k}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right){\left(-1)}^{k-i}{i}^{n}={S}_{2}\left(n,k),which are well known.Thus, we obtain the following identity: Bn(x)=∑k=0nS2(n,k)Dk(x).{B}_{n}\left(x)=\mathop{\sum }\limits_{k=0}^{n}{S}_{2}\left(n,k){D}_{k}\left(x).\hspace{7.25em}Next, we let p(x)=Bn(x)=∑k=0nakDk(r)(x)p\left(x)={B}_{n}\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k}^{\left(r)}\left(x). Then, we first observe that (5.2)IrBn(x)=1(n+r)r∑i=0rri(−1)r−iBn+r(x+i).{I}^{r}{B}_{n}\left(x)=\frac{1}{{\left(n+r)}_{r}}\mathop{\sum }\limits_{i=0}^{r}\left(\genfrac{}{}{0.0pt}{}{r}{i}\right){\left(-1)}^{r-i}{B}_{n+r}\left(x+i).Now, by making use of Remark 4.2, we obtain (5.3)ak=1k!(n+r)r∑i=0rri(−1)r−iΔkBn+r(x+i)∣x=0=1(n+r)r∑l=kn∑i=0r(−1)r−irin+rlS2(l,k)Bn+r−l(i).\begin{array}{rcl}{a}_{k}& =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}{\left(n+r)}_{r}}\mathop{\displaystyle \sum }\limits_{i=0}^{r}\left(\genfrac{}{}{0.0pt}{}{r}{i}\right){\left(-1)}^{r-i}{\Delta }^{k}{B}_{n+r}\left(x+i){| }_{x=0}\\ & =& \frac{1}{{\left(n+r)}_{r}}\mathop{\displaystyle \sum }\limits_{l=k}^{n}\mathop{\displaystyle \sum }\limits_{i=0}^{r}{\left(-1)}^{r-i}\left(\genfrac{}{}{0.0pt}{}{r}{i}\right)\left(\genfrac{}{}{0.0pt}{}{n+r}{l}\right){S}_{2}\left(l,k){B}_{n+r-l}\left(i).\end{array}Thus, we have the following: Bn(x)=1(n+r)r∑k=0n1k!∑i=0rri(−1)r−iΔkBn+r(x+i)∣x=0Dk(r)(x)=1(n+r)r∑k=0n∑l=kn∑i=0r(−1)r−irin+rlS2(l,k)Bn+r−l(i)Dk(r)(x).\begin{array}{rcl}{B}_{n}\left(x)& =& \frac{1}{{\left(n+r)}_{r}}\mathop{\displaystyle \sum }\limits_{k=0}^{n}\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left\{\mathop{\displaystyle \sum }\limits_{i=0}^{r}\left(\genfrac{}{}{0.0pt}{}{r}{i}\right){\left(-1)}^{r-i}{\Delta }^{k}{B}_{n+r}\left(x+i){| }_{x=0}\right\}{D}_{k}^{\left(r)}\left(x)\\ & =& \frac{1}{{\left(n+r)}_{r}}\mathop{\displaystyle \sum }\limits_{k=0}^{n}\left\{\mathop{\displaystyle \sum }\limits_{l=k}^{n}\mathop{\displaystyle \sum }\limits_{i=0}^{r}{\left(-1)}^{r-i}\left(\genfrac{}{}{0.0pt}{}{r}{i}\right)\left(\genfrac{}{}{0.0pt}{}{n+r}{l}\right){S}_{2}\left(l,k){B}_{n+r-l}\left(i)\right\}{D}_{k}^{\left(r)}\left(x).\end{array}(b) Here, we consider p(x)=∑k=1n−11k(n−k)Bk(x)Bn−k(x),(n≥2)p\left(x)={\sum }_{k=1}^{n-1}\frac{1}{k\left(n-k)}{B}_{k}\left(x){B}_{n-k}\left(x),\left(n\ge 2). For this, we first recall from [12] that (5.4)p(x)=2n∑m=0n−21n−mnmBn−mBm(x)+2nHn−1Bn(x),p\left(x)=\frac{2}{n}\mathop{\sum }\limits_{m=0}^{n-2}\frac{1}{n-m}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right){B}_{n-m}{B}_{m}\left(x)+\frac{2}{n}{H}_{n-1}{B}_{n}\left(x),where Hn=1+12+⋯+1n{H}_{n}=1+\frac{1}{2}+\cdots +\frac{1}{n}is the harmonic number and a slight modification of (5.4) gives the identity in (1.2). Let p(x)=∑k=0nakDk(x)p\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k}\left(x). Then, we have (5.5)ak=2n∑m=0n−21n−mnmBn−m∑l=knS2(l,k)mlδm,l+2nHn−1∑l=knS2(l,k)nlδn,l=2n∑m=kn−21n−mnmBn−mS2(m,k)+2nHn−1S2(n,k),\begin{array}{rcl}{a}_{k}& =& \frac{2}{n}\mathop{\displaystyle \sum }\limits_{m=0}^{n-2}\frac{1}{n-m}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right){B}_{n-m}\mathop{\displaystyle \sum }\limits_{l=k}^{n}{S}_{2}\left(l,k)\left(\genfrac{}{}{0.0pt}{}{m}{l}\right){\delta }_{m,l}+\frac{2}{n}{H}_{n-1}\mathop{\displaystyle \sum }\limits_{l=k}^{n}{S}_{2}\left(l,k)\left(\genfrac{}{}{0.0pt}{}{n}{l}\right){\delta }_{n,l}\\ & =& \frac{2}{n}\mathop{\displaystyle \sum }\limits_{m=k}^{n-2}\frac{1}{n-m}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right){B}_{n-m}{S}_{2}\left(m,k)+\frac{2}{n}{H}_{n-1}{S}_{2}\left(n,k),\end{array}where we understand that the sum in (5.5) is zero for k=n−1k=n-1or nn. Thus, we obtain the following identity: ∑k=1n−11k(n−k)Bk(x)Bn−k(x)=2n∑k=0n∑m=kn−21n−mnmBn−mS2(m,k)+Hn−1S2(n,k)Dk(x).\mathop{\sum }\limits_{k=1}^{n-1}\frac{1}{k\left(n-k)}{B}_{k}\left(x){B}_{n-k}\left(x)=\frac{2}{n}\mathop{\sum }\limits_{k=0}^{n}\left\{\mathop{\sum }\limits_{m=k}^{n-2}\frac{1}{n-m}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right){B}_{n-m}{S}_{2}\left(m,k)+{H}_{n-1}{S}_{2}\left(n,k)\right\}{D}_{k}\left(x).(c) In [12], it is shown that the following identity holds for n≥2n\ge 2: (5.6)∑k=1n−11k(n−k)Ek(x)En−k(x)=−4n∑m=0nnm(Hn−1−Hn−m)n−m+1En−m+1Bm(x),\mathop{\sum }\limits_{k=1}^{n-1}\frac{1}{k\left(n-k)}{E}_{k}\left(x){E}_{n-k}\left(x)=-\frac{4}{n}\mathop{\sum }\limits_{m=0}^{n}\frac{\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left({H}_{n-1}-{H}_{n-m})}{n-m+1}{E}_{n-m+1}{B}_{m}\left(x),where Hn=1+12+⋯+1n{H}_{n}=1+\frac{1}{2}+\cdots +\frac{1}{n}is the harmonic number.Write ∑k=1n−11k(n−k)Ek(x)En−k(x)=∑k=0nakDk(x){\sum }_{k=1}^{n-1}\frac{1}{k\left(n-k)}{E}_{k}\left(x){E}_{n-k}\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k}\left(x).By proceeding similarly to (b), we see that (5.7)ak=−4n∑m=0nnm(Hn−1−Hn−m)n−m+1En−m+1∑l=knS2(l,k)mlδm,l=−4n∑m=knnm(Hn−1−Hn−m)n−m+1En−m+1S2(m,k).\begin{array}{rcl}{a}_{k}& =& -\frac{4}{n}\mathop{\displaystyle \sum }\limits_{m=0}^{n}\frac{\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left({H}_{n-1}-{H}_{n-m})}{n-m+1}{E}_{n-m+1}\mathop{\displaystyle \sum }\limits_{l=k}^{n}{S}_{2}\left(l,k)\left(\genfrac{}{}{0.0pt}{}{m}{l}\right){\delta }_{m,l}\\ & =& -\frac{4}{n}\mathop{\displaystyle \sum }\limits_{m=k}^{n}\frac{\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left({H}_{n-1}-{H}_{n-m})}{n-m+1}{E}_{n-m+1}{S}_{2}\left(m,k).\end{array}Thus, (5.7) implies the next identity: ∑k=1n−11k(n−k)Ek(x)En−k(x)=−4n∑k=0n∑m=knnm(Hn−1−Hn−m)n−m+1En−m+1S2(m,k)Dk(x).\mathop{\sum }\limits_{k=1}^{n-1}\frac{1}{k\left(n-k)}{E}_{k}\left(x){E}_{n-k}\left(x)=-\frac{4}{n}\mathop{\sum }\limits_{k=0}^{n}\left\{\mathop{\sum }\limits_{m=k}^{n}\frac{\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left({H}_{n-1}-{H}_{n-m})}{n-m+1}{E}_{n-m+1}{S}_{2}\left(m,k)\right\}{D}_{k}\left(x).(d) In [16], it is proved that the following identity is valid for n≥2n\ge 2: (5.8)∑k=1n−11k(n−k)Gk(x)Gn−k(x)=−4n∑m=0n−2nmGn−mn−mBm(x).\mathop{\sum }\limits_{k=1}^{n-1}\frac{1}{k\left(n-k)}{G}_{k}\left(x){G}_{n-k}\left(x)=-\frac{4}{n}\mathop{\sum }\limits_{m=0}^{n-2}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\frac{{G}_{n-m}}{n-m}{B}_{m}\left(x).\hspace{5.8em}Again, by proceeding analogously to (b), we can show that (5.9)ak=−4n∑l=kn−2S2(l,k)ml∑m=0n−2nmGn−mn−mδm,l=−4n∑m=kn−2nmS2(m,k)Gn−mn−m.{a}_{k}=-\frac{4}{n}\mathop{\sum }\limits_{l=k}^{n-2}{S}_{2}\left(l,k)\left(\genfrac{}{}{0.0pt}{}{m}{l}\right)\mathop{\sum }\limits_{m=0}^{n-2}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\frac{{G}_{n-m}}{n-m}{\delta }_{m,l}=-\frac{4}{n}\mathop{\sum }\limits_{m=k}^{n-2}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right){S}_{2}\left(m,k)\frac{{G}_{n-m}}{n-m}.\hspace{4.1em}Therefore, we obtain the following identity: ∑k=1n−11k(n−k)Gk(x)Gn−k(x)=−4n∑k=0n−2∑m=kn−2nmS2(m,k)Gn−mn−mDk(x).\mathop{\sum }\limits_{k=1}^{n-1}\frac{1}{k\left(n-k)}{G}_{k}\left(x){G}_{n-k}\left(x)=-\frac{4}{n}\mathop{\sum }\limits_{k=0}^{n-2}\left\{\mathop{\sum }\limits_{m=k}^{n-2}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right){S}_{2}\left(m,k)\frac{{G}_{n-m}}{n-m}\right\}{D}_{k}\left(x).(e) Nielsen [2,19] also represented products of two Euler polynomials in terms of Bernoulli polynomials as follows: (5.10)Em(x)En(x)=−2∑r=1mmrErBm+n−r+1(x)m+n−r+1−2∑s=1nnsEsBm+n−s+1(x)m+n−s+1+2(−1)n+1m!n!(m+n+1)!Em+n+1.{E}_{m}\left(x){E}_{n}\left(x)=-2\mathop{\sum }\limits_{r=1}^{m}\left(\genfrac{}{}{0.0pt}{}{m}{r}\right){E}_{r}\frac{{B}_{m+n-r+1}\left(x)}{m+n-r+1}-2\mathop{\sum }\limits_{s=1}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{s}\right){E}_{s}\frac{{B}_{m+n-s+1}\left(x)}{m+n-s+1}+2{\left(-1)}^{n+1}\frac{m\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}\hspace{0.33em}n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left(m+n+1)\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{E}_{m+n+1}.In the same way as (b), we can show that (5.11)ak=2(−1)n+1m!n!(m+n+1)!Em+n+1δk,0−2∑r=1mmrErm+n−r+1S2(m+n−r+1,k)−2∑s=1nnsEsm+n−s+1S2(m+n−s+1,k).{a}_{k}=2{\left(-1)}^{n+1}\frac{m\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}\hspace{0.33em}n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left(m+n+1)\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{E}_{m+n+1}{\delta }_{k,0}-2\mathop{\sum }\limits_{r=1}^{m}\left(\genfrac{}{}{0.0pt}{}{m}{r}\right)\frac{{E}_{r}}{m+n-r+1}{S}_{2}\left(m+n-r+1,k)-2\mathop{\sum }\limits_{s=1}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{s}\right)\frac{{E}_{s}}{m+n-s+1}{S}_{2}\left(m+n-s+1,k).Thus, we arrive at the next identity: Em(x)En(x)=2(−1)n+1m!n!(m+n+1)!Em+n+1−2∑k=1m+n∑r=1mmrErm+n−r+1S2(m+n−r+1,k)Dk(x)−2∑k=1m+n∑s=1nnsEsm+n−s+1S2(m+n−s+1,k)Dk(x).\begin{array}{rcl}{E}_{m}\left(x){E}_{n}\left(x)& =& 2{\left(-1)}^{n+1}\frac{m\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}\hspace{0.33em}n\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{\left(m+n+1)\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}{E}_{m+n+1}-2\mathop{\displaystyle \sum }\limits_{k=1}^{m+n}\mathop{\displaystyle \sum }\limits_{r=1}^{m}\left(\genfrac{}{}{0.0pt}{}{m}{r}\right)\frac{{E}_{r}}{m+n-r+1}{S}_{2}\left(m+n-r+1,k){D}_{k}\left(x)\\ & & -2\mathop{\displaystyle \sum }\limits_{k=1}^{m+n}\mathop{\displaystyle \sum }\limits_{s=1}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{s}\right)\frac{{E}_{s}}{m+n-s+1}{S}_{2}\left(m+n-s+1,k){D}_{k}\left(x).\end{array}6Examples on representation by degenerate Daehee polynomialsHere, we illustrate our formulas in Theorems 3.1 and 4.1.(a) Let p(x)=Bn(x)=∑k=0nakDk,λ(x)p\left(x)={B}_{n}\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k,\lambda }\left(x). Then, as Bn(x)=∑j=0nnjBn−jxj{B}_{n}\left(x)={\sum }_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){B}_{n-j}{x}^{j}, we have (6.1)∫xx+1BnλBuλdu=∑j=0nnjBn−jλj+11j+1Bj+1(x+1λ)−Bj+1xλ.\underset{x}{\overset{x+1}{\int }}{B}_{n}\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u=\mathop{\sum }\limits_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){B}_{n-j}{\lambda }^{j+1}\frac{1}{j+1}\left({B}_{j+1}\left(\frac{x+1}{\lambda })-{B}_{j+1}\left(\frac{x}{\lambda }\right)\right).Thus, for 1≤l≤n1\le l\le n, (6.2)ddxl∫xx+1BnλBuλdu=∑j=0nnjBn−jλjddxl−1Bjx+1λ−Bjxλ=∑j=lnnjBn−jλj−l+1(j)l−1Bj−l+1x+1λ−Bj−l+1xλ.\begin{array}{rcl}{\left(\frac{{\rm{d}}}{{\rm{d}}x}\right)}^{l}\underset{x}{\overset{x+1}{\displaystyle \int }}{B}_{n}\left(\lambda {\bf{B}}\left(\frac{u}{\lambda }\right)\right){\rm{d}}u& =& \mathop{\displaystyle \sum }\limits_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){B}_{n-j}{\lambda }^{j}{\left(\frac{{\rm{d}}}{{\rm{d}}x}\right)}^{l-1}\left({B}_{j}\left(\frac{x+1}{\lambda }\right)-{B}_{j}\left(\frac{x}{\lambda }\right)\right)\\ & =& \mathop{\displaystyle \sum }\limits_{j=l}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){B}_{n-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{x+1}{\lambda }\right)-{B}_{j-l+1}\left(\frac{x}{\lambda }\right)\right).\end{array}Now, from Theorem 3.1, (6.1), and (6.2), we obtain (6.3)ak=1k!∑j=0nnjλj+1j+1Bn−jΔkBj+1x+1λ−Bj+1xλx=0=1k!∑i=0k∑j=0n(−1)k−ikinjλj+1j+1Bn−jBj+1i+1λ−Bj+1iλ=∑l=kn∑j=lnS2(l,k)1l!njλj−l+1(j)l−1Bn−jBj−l+11λ−Bj−l+1,\hspace{-32.2em}\begin{array}{rcl}{a}_{k}& =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\mathop{\displaystyle \sum }\limits_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right)\frac{{\lambda }^{j+1}}{j+1}{B}_{n-j}{\Delta }^{k}{\left.\left({B}_{j+1}\left(\frac{x+1}{\lambda }\right)-{B}_{j+1}\left(\frac{x}{\lambda }\right)\right)\right|}_{x=0}\\ & =& \frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\mathop{\displaystyle \sum }\limits_{i=0}^{k}\mathop{\displaystyle \sum }\limits_{j=0}^{n}{\left(-1)}^{k-i}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right)\left(\genfrac{}{}{0.0pt}{}{n}{j}\right)\frac{{\lambda }^{j+1}}{j+1}{B}_{n-j}\left({B}_{j+1}\left(\frac{i+1}{\lambda }\right)-{B}_{j+1}\left(\frac{i}{\lambda }\right)\right)\\ & =& \mathop{\displaystyle \sum }\limits_{l=k}^{n}\mathop{\displaystyle \sum }\limits_{j=l}^{n}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){\lambda }^{j-l+1}{\left(j)}_{l-1}{B}_{n-j}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right),\end{array}where we understand that (j)−1=1j+1{\left(j)}_{-1}=\frac{1}{j+1}.Hence, from (6.3), we obtain the following identity: Bn(x)=∑k=0n1k!∑j=0nnjλj+1j+1Bn−jΔkBj+1x+1λ−Bj+1xλx=0Dk,λ(x)=∑k=0n1k!∑i=0n∑j=0n(−1)k−ikinjλj+1j+1Bn−jBj+1i+1λ−Bj+1iλDk,λ(x)=∑k=0n∑l=kn∑j=lnS2(l,k)1l!njλj−l+1(j)l−1Bn−jBj−l+11λ−Bj−l+1Dk,λ(x).\hspace{-35.8em}\begin{array}{rcl}{B}_{n}\left(x)& =& \mathop{\displaystyle \sum }\limits_{k=0}^{n}\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left\{\mathop{\displaystyle \sum }\limits_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right)\frac{{\lambda }^{j+1}}{j+1}{B}_{n-j}{\Delta }^{k}{\left.\left({B}_{j+1}\left(\frac{x+1}{\lambda }\right)-{B}_{j+1}\left(\frac{x}{\lambda }\right)\right)\right|}_{x=0}\right\}{D}_{k,\lambda }\left(x)\\ & =& \mathop{\displaystyle \sum }\limits_{k=0}^{n}\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left\{\mathop{\displaystyle \sum }\limits_{i=0}^{n}\mathop{\displaystyle \sum }\limits_{j=0}^{n}{\left(-1)}^{k-i}\left(\genfrac{}{}{0.0pt}{}{k}{i}\right)\left(\genfrac{}{}{0.0pt}{}{n}{j}\right)\frac{{\lambda }^{j+1}}{j+1}{B}_{n-j}\left({B}_{j+1}\left(\frac{i+1}{\lambda }\right)-{B}_{j+1}\left(\frac{i}{\lambda }\right)\right)\right\}{D}_{k,\lambda }\left(x)\\ & =& \mathop{\displaystyle \sum }\limits_{k=0}^{n}\left\{\mathop{\displaystyle \sum }\limits_{l=k}^{n}\mathop{\displaystyle \sum }\limits_{j=l}^{n}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){\lambda }^{j-l+1}{\left(j)}_{l-1}{B}_{n-j}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right)\right\}{D}_{k,\lambda }\left(x).\end{array}Next, we let p(x)=Bn(x)=∑k=0nakDk,λ(r)(x)p\left(x)={B}_{n}\left(x)=\mathop{\sum }\limits_{k=0}^{n}{a}_{k}{D}_{k,\lambda }^{\left(r)}\left(x). Then, we first observe that (6.4)IrBnλB(r)xλ=Ir∑j=0nnjBn−jλjBj(r)xλ=∑j=0n∑i=0r(−1)r−injriλj+r(j+r)rBn−jBj+r(r)x+iλ.{I}^{r}{B}_{n}\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)={I}^{r}\mathop{\sum }\limits_{j=0}^{n}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){B}_{n-j}{\lambda }^{j}{B}_{j}^{\left(r)}\left(\frac{x}{\lambda }\right)=\mathop{\sum }\limits_{j=0}^{n}\mathop{\sum }\limits_{i=0}^{r}{\left(-1)}^{r-i}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right)\left(\genfrac{}{}{0.0pt}{}{r}{i}\right)\frac{{\lambda }^{j+r}}{{\left(j+r)}_{r}}{B}_{n-j}{B}_{j+r}^{\left(r)}\left(\frac{x+i}{\lambda }\right).So, for llwith j+r≥lj+r\ge l, we obtain (6.5)ddxlIrBnλB(r)xλ=∑j=0n∑i=0r(−1)r−injriλj+r(j+r)rBn−jddxlBj+r(r)x+iλ=∑j=0n∑i=0r(−1)r−injriλj+r−l(j+r)rBn−j(j+r)lBj+r−l(r)x+iλ.\begin{array}{rcl}{\left(\frac{{\rm{d}}}{{\rm{d}}x}\right)}^{l}{I}^{r}{B}_{n}\left(\lambda {{\bf{B}}}^{\left(r)}\left(\frac{x}{\lambda }\right)\right)& =& \mathop{\displaystyle \sum }\limits_{j=0}^{n}\mathop{\displaystyle \sum }\limits_{i=0}^{r}{\left(-1)}^{r-i}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right)\left(\genfrac{}{}{0.0pt}{}{r}{i}\right)\frac{{\lambda }^{j+r}}{{\left(j+r)}_{r}}{B}_{n-j}{\left(\frac{{\rm{d}}}{{\rm{d}}x}\right)}^{l}{B}_{j+r}^{\left(r)}\left(\frac{x+i}{\lambda }\right)\\ & =& \mathop{\displaystyle \sum }\limits_{j=0}^{n}\mathop{\displaystyle \sum }\limits_{i=0}^{r}{\left(-1)}^{r-i}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right)\left(\genfrac{}{}{0.0pt}{}{r}{i}\right)\frac{{\lambda }^{j+r-l}}{{\left(j+r)}_{r}}{B}_{n-j}{\left(j+r)}_{l}{B}_{j+r-l}^{\left(r)}\left(\frac{x+i}{\lambda }\right).\end{array}Thus, from Theorem 4.1, (6.4), and (6.5), we have Bn(x)=∑k=0n1k!∑j=0n∑i=0r(−1)r−injriλj+r(j+r)rBn−jΔkBj+r(r)x+iλx=0Dk,λ(r)(x)=∑k=0n∑l=knS2(l,k)1l!∑j=max{0,l−r}n∑i=0r(−1)r−injriλj+r−l(j+r)rBn−j(j+r)lBj+r−l(r)iλDk,λ(r)(x).\hspace{2.2em}\begin{array}{rcl}{B}_{n}\left(x)& =& \mathop{\displaystyle \sum }\limits_{k=0}^{n}\frac{1}{k\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left\{\mathop{\displaystyle \sum }\limits_{j=0}^{n}\mathop{\displaystyle \sum }\limits_{i=0}^{r}{\left(-1)}^{r-i}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right)\left(\genfrac{}{}{0.0pt}{}{r}{i}\right)\frac{{\lambda }^{j+r}}{{\left(j+r)}_{r}}{B}_{n-j}{\Delta }^{k}{B}_{j+r}^{\left(r)}{\left.\left(\frac{x+i}{\lambda }\right)\right|}_{x=0}\right\}{D}_{k,\lambda }^{\left(r)}\left(x)\\ & =& \mathop{\displaystyle \sum }\limits_{k=0}^{n}\left\{\mathop{\displaystyle \sum }\limits_{l=k}^{n}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\mathop{\displaystyle \sum }\limits_{j={\rm{\max }}\{0,l-r\}}^{n}\mathop{\displaystyle \sum }\limits_{i=0}^{r}{\left(-1)}^{r-i}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right)\left(\genfrac{}{}{0.0pt}{}{r}{i}\right)\frac{{\lambda }^{j+r-l}}{{\left(j+r)}_{r}}{B}_{n-j}{\left(j+r)}_{l}{B}_{j+r-l}^{\left(r)}\left(\frac{i}{\lambda }\right)\right\}{D}_{k,\lambda }^{\left(r)}\left(x).\end{array}(b) Let p(x)=∑k=1n−11k(n−k)Bk(x)Bn−k(x)p\left(x)={\sum }_{k=1}^{n-1}\frac{1}{k\left(n-k)}{B}_{k}\left(x){B}_{n-k}\left(x), for n≥2n\ge 2. As we stated earlier, it was shown in [12] that (6.6)p(x)=2n∑m=0n−21n−mnmBn−mBm(x)+2nHn−1Bn(x),p\left(x)=\frac{2}{n}\mathop{\sum }\limits_{m=0}^{n-2}\frac{1}{n-m}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right){B}_{n-m}{B}_{m}\left(x)+\frac{2}{n}{H}_{n-1}{B}_{n}\left(x),where Hn=1+12+⋯+1n{H}_{n}=1+\frac{1}{2}+\cdots +\frac{1}{n}is the Harmonic number.Write p(x)=∑k=0nakDk(x)p\left(x)={\sum }_{k=0}^{n}{a}_{k}{D}_{k}\left(x). Then, from Theorem 4.1 and (6.6), we have (6.7)ak=2n∑l=kn−2∑m=ln−2∑j=lmS2(l,k)1l!1n−mnmmjBn−mBm−jλj−l+1(j)l−1Bj−l+11λ−Bj−l+1+2nHn−1∑l=kn∑j=lnS2(l,k)1l!njBn−jλj−l+1(j)l−1Bj−l+11λ−Bj−l+1.\begin{array}{rcl}{a}_{k}& =& \frac{2}{n}\left\{\mathop{\displaystyle \sum }\limits_{l=k}^{n-2}\mathop{\displaystyle \sum }\limits_{m=l}^{n-2}\mathop{\displaystyle \sum }\limits_{j=l}^{m}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\frac{1}{n-m}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left(\genfrac{}{}{0.0pt}{}{m}{j}\right){B}_{n-m}{B}_{m-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right)\right\}\\ & & +\frac{2}{n}{H}_{n-1}\mathop{\displaystyle \sum }\limits_{l=k}^{n}\mathop{\displaystyle \sum }\limits_{j=l}^{n}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){B}_{n-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right).\end{array}Thus, from (6.7), we obtain ∑k=1n−11k(n−k)Bk(x)Bn−k(x)=2n∑k=0n∑l=kn−2∑m=ln−2∑j=lmS2(l,k)1l!1n−mnmmjBn−mBm−jλj−l+1(j)l−1Bj−l+11λ−Bj−l+1+Hn−1∑l=kn∑j=lnS2(l,k)1l!njBn−jλj−l+1(j)l−1Bj−l+11λ−Bj−l+1Dk,λ(x),\hspace{-44.95em}\begin{array}{l}\mathop{\displaystyle \sum }\limits_{k=1}^{n-1}\frac{1}{k\left(n-k)}{B}_{k}\left(x){B}_{n-k}\left(x)\\ \hspace{1.0em}=\frac{2}{n}\mathop{\displaystyle \sum }\limits_{k=0}^{n}\left\{\mathop{\displaystyle \sum }\limits_{l=k}^{n-2}\mathop{\displaystyle \sum }\limits_{m=l}^{n-2}\mathop{\displaystyle \sum }\limits_{j=l}^{m}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\frac{1}{n-m}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left(\genfrac{}{}{0.0pt}{}{m}{j}\right){B}_{n-m}{B}_{m-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right)\right.\\ \hspace{1.0em}\hspace{1.0em}\left.+{H}_{n-1}\mathop{\displaystyle \sum }\limits_{l=k}^{n}\mathop{\displaystyle \sum }\limits_{j=l}^{n}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left(\genfrac{}{}{0.0pt}{}{n}{j}\right){B}_{n-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right)\right\}{D}_{k,\lambda }\left(x),\end{array}where we understand that the triple sum in the parentheses is zero for k=n−1k=n-1or k=nk=n, and (j)−1=1j+1{\left(j)}_{-1}=\frac{1}{j+1}.(c) Let p(x)=∑k=1n−11k(n−k)Ek(x)En−k(x)p\left(x)={\sum }_{k=1}^{n-1}\frac{1}{k\left(n-k)}{E}_{k}\left(x){E}_{n-k}\left(x), for n≥2n\ge 2. Then, as we saw earlier, it was proved in [12] that (6.8)p(x)=−4n∑m=0nnm(Hn−1−Hn−m)n−m+1En−m+1Bm(x).p\left(x)=-\frac{4}{n}\mathop{\sum }\limits_{m=0}^{n}\frac{\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left({H}_{n-1}-{H}_{n-m})}{n-m+1}{E}_{n-m+1}{B}_{m}\left(x).Write p(x)=∑k=0n−2akDk,λ(x)p\left(x)={\sum }_{k=0}^{n-2}{a}_{k}{D}_{k,\lambda }\left(x). Then, from Theorem 4.1 and (6.8), we can show that (6.9)ak=−4n∑l=kn∑m=ln∑j=lmS2(l,k)1l!nm(Hn−1−Hn−m)n−m+1En−m+1mjBm−jλj−l+1(j)l−1Bj−l+11λ−Bj−l+1,{a}_{k}=-\frac{4}{n}\left\{\mathop{\sum }\limits_{l=k}^{n}\mathop{\sum }\limits_{m=l}^{n}\mathop{\sum }\limits_{j=l}^{m}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\frac{\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left({H}_{n-1}-{H}_{n-m})}{n-m+1}{E}_{n-m+1}\left(\genfrac{}{}{0.0pt}{}{m}{j}\right){B}_{m-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right)\right\},where we understand that (j)−1=1j+1{\left(j)}_{-1}=\frac{1}{j+1}.Hence, from (6.9), we have ∑k=1n−11k(n−k)Ek(x)En−k(x)=−4n∑k=0n∑l=kn∑m=ln∑j=lmS2(l,k)1l!nm(Hn−1−Hn−m)n−m+1En−m+1mjBm−jλj−l+1(j)l−1Bj−l+11λ−Bj−l+1Dk,λ(x).\begin{array}{l}\mathop{\displaystyle \sum }\limits_{k=1}^{n-1}\frac{1}{k\left(n-k)}{E}_{k}\left(x){E}_{n-k}\left(x)\\ \hspace{1.0em}=-\frac{4}{n}\mathop{\displaystyle \sum }\limits_{k=0}^{n}\left\{\mathop{\displaystyle \sum }\limits_{l=k}^{n}\mathop{\displaystyle \sum }\limits_{m=l}^{n}\mathop{\displaystyle \sum }\limits_{j=l}^{m}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\frac{\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left({H}_{n-1}-{H}_{n-m})}{n-m+1}{E}_{n-m+1}\left(\genfrac{}{}{0.0pt}{}{m}{j}\right){B}_{m-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right)\right\}{D}_{k,\lambda }\left(x).\end{array}(d) Here, we consider p(x)=∑k=1n−11k(n−k)Gk(x)Gn−k(x)p\left(x)={\sum }_{k=1}^{n-1}\frac{1}{k\left(n-k)}{G}_{k}\left(x){G}_{n-k}\left(x), for n≥2n\ge 2. As we mentioned earlier, it was shown in [16] that (6.10)p(x)=−4n∑m=0n−2nmGn−mn−mBm(x).p\left(x)=-\frac{4}{n}\mathop{\sum }\limits_{m=0}^{n-2}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\frac{{G}_{n-m}}{n-m}{B}_{m}\left(x).Write p(x)=∑k=0n−2akDk,λ(x)p\left(x)={\sum }_{k=0}^{n-2}{a}_{k}{D}_{k,\lambda }\left(x). Then, from Theorem 4.1 and (6.10), we obtain that (6.11)ak=−4n∑l=kn−2∑m=ln−2∑j=lmS2(l,k)1l!nmGn−mn−mmjBm−jλj−l+1(j)l−1Bj−l+11λ−Bj−l+1,{a}_{k}=-\frac{4}{n}\left\{\mathop{\sum }\limits_{l=k}^{n-2}\mathop{\sum }\limits_{m=l}^{n-2}\mathop{\sum }\limits_{j=l}^{m}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\frac{{G}_{n-m}}{n-m}\left(\genfrac{}{}{0.0pt}{}{m}{j}\right){B}_{m-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right)\right\},where we understand that (j)−1=1j+1{\left(j)}_{-1}=\frac{1}{j+1}.Thus, from (6.11), we obtain ∑k=1n−11k(n−k)Gk(x)Gn−k(x)=−4n∑k=0n−2∑l=kn−2∑m=ln−2∑j=lmS2(l,k)1l!nmGn−mn−mmjBm−jλj−l+1(j)l−1Bj−l+11λ−Bj−l+1Dk,λ(x).\begin{array}{l}\mathop{\displaystyle \sum }\limits_{k=1}^{n-1}\frac{1}{k\left(n-k)}{G}_{k}\left(x){G}_{n-k}\left(x)\\ \hspace{1.0em}=-\hspace{-0.25em}\frac{4}{n}\mathop{\displaystyle \sum }\limits_{k=0}^{n-2}\left\{\mathop{\displaystyle \sum }\limits_{l=k}^{n-2}\mathop{\displaystyle \sum }\limits_{m=l}^{n-2}\mathop{\displaystyle \sum }\limits_{j=l}^{m}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\frac{{G}_{n-m}}{n-m}\left(\genfrac{}{}{0.0pt}{}{m}{j}\right){B}_{m-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right)\right\}{D}_{k,\lambda }\left(x).\end{array}(e) As we mentioned earlier, it was shown (see [17,18]) that, for positive integers mmand nn, with m+n≥2m+n\ge 2, we have (6.12)Bm(x)Bn(x)=∑rm2rn+n2rmB2rBm+n−2r(x)m+n−2r+(−1)m+1Bm+nm+nm.{B}_{m}\left(x){B}_{n}\left(x)=\sum _{r}\left\{\phantom{\rule[-1.25em]{}{0ex}},\left(\genfrac{}{}{0.0pt}{}{m}{2r}\right)n+\left(\genfrac{}{}{0.0pt}{}{n}{2r}\right)m\right\}\frac{{B}_{2r}{B}_{m+n-2r}\left(x)}{m+n-2r}+{\left(-1)}^{m+1}\frac{{B}_{m+n}}{\left(\genfrac{}{}{0.0pt}{}{m+n}{m}\right)}.\hspace{1.85em}Then, from Theorem 4.1 and (6.12), we can show that (6.13)ak=(−1)m+1Bm+nm+nmδk,0+∑r∑l=km+n∑j=lm+n−2rS2(l,k)1l!m2rn+n2rm×B2rm+n−2rm+n−2rjBm+n−2r−jλj−l+1(j)l−1Bj−l+11λ−Bj−l+1.\begin{array}{rcl}{a}_{k}& =& {\left(-1)}^{m+1}\frac{{B}_{m+n}}{\left(\genfrac{}{}{0.0pt}{}{m+n}{m}\right)}{\delta }_{k,0}+\displaystyle \sum _{r}\mathop{\displaystyle \sum }\limits_{l=k}^{m+n}\mathop{\displaystyle \sum }\limits_{j=l}^{m+n-2r}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left\{\phantom{\rule[-1.25em]{}{0ex}},\left(\genfrac{}{}{0.0pt}{}{m}{2r}\right)n+\left(\genfrac{}{}{0.0pt}{}{n}{2r}\right)m\right\}\\ & & \times \frac{{B}_{2r}}{m+n-2r}\left(\genfrac{}{}{0.0pt}{}{m+n-2r}{j}\right){B}_{m+n-2r-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right).\end{array}Thus, form (6.13), we obtain Bm(x)Bn(x)=(−1)m+1Bm+nm+nm+∑k=0m+n∑r∑l=km+n−2r∑j=lm+n−2rS2(l,k)1l!m2rn+n2rm×B2rm+n−2rm+n−2rjBm+n−2r−jλj−l+1(j)l−1Bj−l+11λ−Bj−l+1Dk,λ(x),\begin{array}{rcl}{B}_{m}\left(x){B}_{n}\left(x)& =& {\left(-1)}^{m+1}\frac{{B}_{m+n}}{\left(\genfrac{}{}{0.0pt}{}{m+n}{m}\right)}+\mathop{\displaystyle \sum }\limits_{k=0}^{m+n}\left\{\phantom{\rule[-1.25em]{}{0ex}},\displaystyle \sum _{r}\mathop{\displaystyle \sum }\limits_{l=k}^{m+n-2r}\mathop{\displaystyle \sum }\limits_{j=l}^{m+n-2r}{S}_{2}\left(l,k)\frac{1}{l\hspace{0.1em}\text{&#x0021;}\hspace{0.1em}}\left\{\phantom{\rule[-1.25em]{}{0ex}},\left(\genfrac{}{}{0.0pt}{}{m}{2r}\right)n+\left(\genfrac{}{}{0.0pt}{}{n}{2r}\right)m\right\}\right.\\ & & \left.\phantom{\rule[-1.35em]{}{0ex}},\times \frac{{B}_{2r}}{m+n-2r}\left(\genfrac{}{}{0.0pt}{}{m+n-2r}{j}\right){B}_{m+n-2r-j}{\lambda }^{j-l+1}{\left(j)}_{l-1}\left({B}_{j-l+1}\left(\frac{1}{\lambda }\right)-{B}_{j-l+1}\right)\right\}{D}_{k,\lambda }\left(x),\end{array}where (j)−1=1j+1{\left(j)}_{-1}=\frac{1}{j+1}.7ConclusionIn this paper, we were interested in representing any polynomial in terms of the degenerate Daehee polynomials and of the higher-order degenerate Daehee polynomials. We were able to derive formulas for such representations with the help of umbral calculus. We showed that, by letting λ\lambda tends to zero, they give formulas for representations by the Daehee polynomials and by the higher-order Daehee polynomials. Further, we illustrated the formulas with some examples.As we mentioned in Section 1, both Faber-Pandharipande-Zagier (FPZ) identity and a variant of Miki’s identity follow from the one identity (see (1.2)) that can be derived from the formula (see (1.1)) involving only derivatives and integrals of the given polynomial, while all the other proofs are quite involved. We recall here that the FPZ identity was a conjectural relation between Hodge integrals in Gromov-Witten’s theory. It should be stressed that our method is very useful and powerful, even though it is elementary.It is one of our future research projects to continue to find formulas representing polynomials in terms of some specific special polynomials and to apply those in discovering some interesting identities.

### Journal

Open Mathematicsde Gruyter

Published: Jan 1, 2022

Keywords: degenerate Daehee polynomial; higher-order degenerate Daehee polynomial; umbral calculus; 05A19; 05A40; 11B68; 11B83