Some Formulae for the Product of Two Bernoulli and Euler Polynomials

Abstract

We investigate some formulae for the product of two Bernoulli and Euler polynomials arising from the Euler and Bernoulli basis polynomials.

1. Introduction

As is well known, the Bernoulli polynomials are defined by the generating function as follows:𝑡𝑒^𝑡𝑒−1^𝑥𝑡=𝑒^{𝐵(𝑥)𝑡}=^∞_𝑛=0𝐵_𝑛(𝑥)𝑡^𝑛𝑛! (1.1) (see [1–21]), with the usual convention about replacing 𝐵^𝑛(𝑥) by 𝐵_𝑛(𝑥). In the special case, 𝑥=0, 𝐵_𝑛(0)=𝐵_𝑛 are called the 𝑛th Bernoulli numbers. The Euler polynomials are also defined by the generating function as follows:2𝑒^𝑡𝑒+1^𝑥𝑡=𝑒^{𝐸(𝑥)𝑡}=^∞_𝑛=0𝐸_𝑛(𝑥)𝑡^𝑛𝑛! (1.2) (see [6–11]), with the usual convention about replacing 𝐸^𝑛(𝑥) by 𝐸_𝑛(𝑥). In the special case, 𝑥=0, 𝐸_𝑛(0)=𝐸_𝑛 are called the 𝑛th Euler numbers. From (1.1) and (1.2), we can derive the following recurrence relations for the Bernoulli and Euler numbers:𝐵₀=1,𝐵_𝑛(1)−𝐵_𝑛=𝛿_1,𝑛,𝐸₀=1,𝐸_𝑛(1)+𝐸_𝑛=2𝛿_0,𝑛, (1.3) where 𝛿_𝑘,𝑛 is the Kronecker symbol. By (1.1) and (1.2), we get𝐵_𝑛(𝑥)=^𝑛_ℓ=0𝑥^𝑛−ℓ𝑛ℓ𝐵_ℓ,𝐸_𝑛(𝑥)=^𝑛_ℓ=0𝑥^𝑛−ℓ𝑛ℓ𝐸_ℓ. (1.4) From (1.4), we can derive𝑑𝐵𝑑𝑥^𝑛(𝑥)=𝑛𝐵_𝑛−1(𝑥),𝑑𝐸𝑑𝑥^𝑛(𝑥)=𝑛𝐸_𝑛−1(𝑥). (1.5) By (1.4) and (1.5), we get¹₀𝐵_𝑛(𝑥)𝛿𝑑𝑥=^0,𝑛𝑛+1=𝛿_0,𝑛,¹₀𝐸_𝑛(𝑥)𝐸𝑑𝑥=−2^𝑛+1.𝑛+1 (1.6) It is easy to show that𝑒^𝑡𝑥=1𝑡𝑡𝑒^(𝑥+1)𝑡𝑒^𝑡−−1𝑡𝑒^𝑥𝑡𝑒^𝑡=1−1𝑡^∞_𝑛=0𝐵_𝑛(𝑥+1)−𝐵_𝑛(𝑥)𝑡^𝑛.𝑛! (1.7) Thus, we have𝑥^𝑛=1𝑛+1^𝑛_ℓ=0ℓ𝐵𝑛+1_ℓ(𝑥) (1.8) (see [11–18]). By the definition of the Euler polynomials, we get𝑒^𝑡𝑥=122𝑒^(𝑥+1)𝑡𝑒^𝑡++12𝑒^𝑥𝑡𝑒^𝑡=1+12^∞_𝑛=0𝐸_𝑛(𝑥+1)+𝐸_𝑛(𝑥)𝑡^𝑛.𝑛! (1.9) From (1.9), we have𝑥^𝑛=𝐸_𝑛(𝑥)+12^𝑛−1_ℓ=0𝑛ℓ𝐸_ℓ(𝑥) (1.10) (see [1–18]). By (1.8) and (1.10), we see that the set {𝐸₀(𝑥),…,𝐸_𝑛(𝑥)} and {𝐵₀(𝑥),…,𝐵_𝑛(𝑥)} are the basis for the space of polynomials of degree less than or equal to 𝑛 with coefficients in ℚ (see [1–21]).

From 𝑚,𝑛∈ℤ₊=ℕ⋃{0}, let 𝐼_𝑚,𝑛=∫¹₀𝐵_𝑚(𝑥)𝑥^𝑛𝑑𝑥. Then, we note that𝐼_0,𝑛=¹₀𝑥^𝑛1𝑑𝑥=𝑛+1,𝐼_𝑚,0=¹₀𝐵_𝑚(𝑥)𝑑𝑥=𝛿_0,𝑚. (1.11) Let us assume that 𝑚,𝑛≥1. Then, we have𝐼_𝑚,𝑛=𝐵_𝑚+1−𝑛𝑚+1𝐼𝑚+1^{𝑚+1,𝑛−1}=𝐵_𝑚+1+()𝑛𝑚+1−1()𝐵𝑚+1)(𝑚+2^𝑚+2+(−1)²𝑛(𝑛−1)()𝐼𝑚+1)(𝑚+2^{𝑚+2,𝑛−2}. (1.12)

Continuing this process, we get𝐼_𝑚,𝑛=^𝑛−1_𝑗=1(−1)^𝑗−1𝑛−𝑗+1^𝑗_ℓ=1𝑛−ℓ+1𝐵𝑚+ℓ^𝑚+𝑗+(−1)^𝑛−1𝑛(𝑛−1)⋯2(𝑚+1)⋯(𝑚+𝑛−1)¹₀𝐵_{𝑚+𝑛−1}(𝑥)=𝑥𝑑𝑥_𝑛𝑗=1(−1)^𝑗−1_𝑗^𝑛+1(𝑛+1)_𝑚^𝑚+𝑗𝐵_𝑚+𝑗. (1.13) Let 𝐽_𝑚,𝑛=∫¹₀𝐸_𝑚(𝑥)𝑥^𝑛𝑑𝑥 for 𝑚,𝑛∈ℤ₊. Then, we have𝐽_0,𝑛=¹₀𝑥^𝑛1𝑑𝑥=𝑛+1,𝐽_𝑚,0=¹₀𝐸_𝑚(𝑥)2𝑑𝑥=−𝐸𝑚+1^𝑚+1. (1.14) Assume that 𝑚,𝑛≥1. Then, we get𝐽_𝑚,𝑛=¹₀𝐸_𝑚(𝑥)𝑥^𝑛𝐸𝑑𝑥=−^𝑚+1−𝑛𝑚+1𝑚+1¹₀𝐸_𝑚+1(𝑥)𝑥^𝑛−11𝑑𝑥=−𝐸𝑚+1^𝑚+1−𝑛𝐽𝑚+1^{𝑚+1,𝑛−1}𝐸=−^𝑚+1+()𝑚+1−1²𝑛()𝐸𝑚+1)(𝑚+2^𝑚+2+(−1)²𝑛(𝑛−1)()𝐽𝑚+1)(𝑚+2^{𝑚+2,𝑛−2}. (1.15)

Continuing this process, we get𝐽_𝑚,𝑛=^𝑛−1_𝑗=1(−1)^𝑗1𝑛−𝑗+1^𝑗_ℓ=1𝑛−ℓ+1𝐸𝑚+ℓ^𝑚+𝑗+(−1)^𝑛−1𝑛(𝑛−1)⋯2()⋯()𝐽𝑚+1𝑚+𝑛−1^{𝑚+𝑛−1,1},𝐽_{𝑚+𝑛−1,1}=¹₀𝐸_{𝑚+𝑛−1}(𝑥)𝐸𝑥𝑑𝑥=−^𝑚+𝑛+()𝑚+𝑛−1²2𝐸_𝑚+𝑛+1().𝑚+𝑛)(𝑚+𝑛+1 (1.16) By (1.16), we get𝐽_𝑚,𝑛=1𝑛+1^𝑛_𝑗=1(−1)^{𝑗𝑗𝑛+1}_{𝑚𝑚+𝑗}𝐸_𝑚+𝑗()+2−1^𝑛+1𝐸_𝑛+𝑚+1(𝑛+𝑚+1)(_𝑚^𝑛+𝑚). (1.17) From the properties of the Bernoulli and Euler basis for the space of the polynomials of degree less than or equal to 𝑛 with coefficients in ℚ, we derive some identities for the product of two Bernoulli and Euler polynomials.

2. Some Identities for the Bernoulli and Euler Numbers

Let us consider the polynomial 𝑝(𝑥)=∑^𝑛_𝑘=0𝐵_𝑘(𝑥)𝑥^𝑛−𝑘, with 𝑛∈ℕ. Then, we have𝑝^(𝑘)(𝑥)=(𝑛+1)!(𝑛−𝑘+1)!^𝑛_ℓ=𝑘𝐵_ℓ−𝑘(𝑥)𝑥^𝑛−ℓ(𝑘=0,1,2,…,𝑛). (2.1) From the properties of the Bernoulli basis for the space of polynomials of degree less than or equal to 𝑛 with coefficients in ℚ, 𝑝(𝑥) is given by𝑝(𝑥)=^𝑛_𝑘=0𝑎_𝑘𝐵_𝑘(𝑥). (2.2) Thus, by (2.2), we get𝑎₀=¹₀𝑝(𝑡)𝑑𝑡=^𝑛_𝑘=0¹₀𝐵_𝑘(𝑡)𝑥^𝑛−𝑘𝑑𝑡=^𝑛_𝑘=0𝐼_{𝑘,𝑛−𝑘}=𝐼_0,𝑛+^𝑛−1_𝑘=1𝐼_{𝑘,𝑛−𝑘}+𝐼_𝑛,0=1+𝑛+1^𝑛−1_𝑘=1^𝑛−𝑘_𝑗=1(−1)^𝑗−1_𝑗^{𝑛−𝑘+1}(𝑛−𝑘+1)_𝑘^𝑘+𝑗𝐵_𝑘+𝑗+𝛿_0,𝑛. (2.3) By (2.1) and (2.2), we get𝑎_𝑘=𝑝^(𝑘−1)(1)−𝑝^(𝑘−1)(0=()!)𝑛+1𝑘!(𝑛−𝑘+2)!^𝑛_ℓ=𝑘−1𝐵_ℓ−𝑘+1(1)−𝐵_ℓ−𝑘+10^𝑛−ℓ=^𝑘𝑛+2𝑛+2^𝑛−1_ℓ=𝑘−1𝐵_ℓ−𝑘+1+1. (2.4) Therefore, by (2.3) and (2.4), we obtain the following theorem.

Theorem 2.1. For 𝑛∈ℕ, one has ^𝑛_𝑘=0𝐵_𝑘(𝑥)𝑥^𝑛−𝑘=1+𝑛+1^𝑛−1_ℓ=1^𝑛−ℓ_𝑗=1(−1)^𝑗−1_𝑗^𝑛−ℓ+1(𝑛−ℓ+1)_ℓ^ℓ+𝑗𝐵_ℓ+𝑗+1𝑛+2^𝑛_𝑘=1𝑘𝑛+2^𝑛−1_ℓ=𝑘−1𝐵_ℓ−𝑘+1+1𝐵_𝑘(𝑥). (2.5)

From the properties of the Euler basis for the space of polynomials of degree less than or equal to 𝑛 with coefficients in ℚ, 𝑝(𝑥) is given by𝑝(𝑥)=^𝑛_𝑘=0𝑏_𝑘𝐸_𝑘(𝑥). (2.6) By (2.1) and (2.6), we get𝑏_𝑘=1𝑝2𝑘!^(𝑘)(1)+𝑝^(𝑘)(0=1)()!2𝑘!𝑛+1(𝑛−𝑘+1)!^𝑛_ℓ=𝑘𝐵_ℓ−𝑘(1)+𝐵_ℓ−𝑘0^𝑛−ℓ=^𝑘𝑛+12^𝑛_ℓ=𝑘𝐵_ℓ−𝑘+1−𝛿_𝑘,𝑛+𝐵_𝑛−𝑘=2^{𝑘𝑛+1𝑛}_ℓ=𝑘𝐵_ℓ−𝑘+1+𝐵_𝑛−𝑘if𝑘≠𝑛,𝑛+1if𝑘=𝑛. (2.7) Therefore, by (2.6) and (2.7), we obtain the following theorem.

Theorem 2.2. For 𝑛∈ℤ₊, one has ^𝑛_𝑘=0𝐵_𝑘(𝑥)𝑥^𝑛−𝑘=12^𝑛−1_𝑘=0𝑘𝑛+1^𝑛_ℓ=𝑘B_ℓ−𝑘+1+𝐵_𝑛−𝑘𝐸_𝑘(𝑥)+(𝑛+1)𝐸_𝑛(𝑥). (2.8)

Let us take polynomial 𝑝(𝑥) with 𝑝(𝑥)=∑^𝑛_𝑘=0𝐸_𝑘(𝑥)𝑥^𝑛−𝑘. Then, we have𝑝^(𝑘)(𝑥)=(𝑛+1)!(𝑛−𝑘+1)!^𝑛_ℓ=𝑘𝐸_ℓ−𝑘(𝑥)𝑥^𝑛−ℓ(𝑘=0,1,2,…,𝑛). (2.9) By the basis set {𝐵₀(𝑥),…,𝐵_𝑛(𝑥)} for the space of polynomials of degree less than or equal to 𝑛 with coefficients in ℚ, we see that 𝑝(𝑥) is given by𝑝(𝑥)=^𝑛_𝑘=0𝑎_𝑘𝐵_𝑘(𝑥). (2.10) From (2.10), we note that𝑎₀=¹₀𝑝(𝑡)𝑑𝑡=^𝑛_ℓ=0¹₀𝐸_ℓ(𝑡)𝑥^𝑛−ℓ𝑑𝑡=^𝑛_ℓ=0𝐽_{ℓ,𝑛−ℓ}=𝐽_0,𝑛+^𝑛−1_ℓ=1𝐽_{ℓ,𝑛−ℓ}+𝐽_𝑛,0=1+𝑛+1^𝑛−1_ℓ=1𝐽_{ℓ,𝑛−ℓ}−2𝐸𝑛+1^𝑛+1=^𝑛_𝑗=1(−1)^{𝑗𝑗𝑛+1}𝐸𝑛+1^𝑗+2(−1)^𝑛+1𝐸𝑛+1^𝑛+1+^𝑛−1_ℓ=1^𝑛−ℓ_𝑗=1(−1)^{𝑗𝑗𝑛−ℓ+1}(𝑛−ℓ+1)_ℓ^ℓ+𝑗𝐸_ℓ+𝑗+2(−1)^𝑛−ℓ+1(𝑛+1)(^𝑛_ℓ)𝐸_𝑛+1−2𝐸𝑛+1^𝑛+1=^𝑛_ℓ=0^𝑛−ℓ_𝑗=1(−1)^{𝑗𝑗𝑛−ℓ+1}(𝑛−ℓ+1)_ℓ^ℓ+𝑗𝐸_ℓ+𝑗+2(−1)^𝑛−ℓ+1(𝑛+1)(^𝑛_ℓ)𝐸_𝑛+1. (2.11) Note that^𝑛_ℓ=0(−1)^ℓ−1(𝑛+1)(^𝑛_ℓ)=−^𝑛_ℓ=0(𝑛−ℓ)!ℓ!()!()𝑛+1−1^ℓ=−^𝑛_ℓ=0𝐵(𝑛−ℓ+1,ℓ+1)(−1)^ℓ()=−1+−1^𝑛,𝑛+2 (2.12) where 𝐵(𝛼,𝛽) is the beta function.

From (2.11) and (2.12), we have𝑎₀=^𝑛_ℓ=0^𝑛−ℓ_𝑗=1(−1)^𝑗_𝑗^𝑛−ℓ+1(𝑛−ℓ+1)_ℓ^ℓ+𝑗𝐸_ℓ+𝑗+4(−1)^𝑛+1𝐸𝑛+2^𝑛+1. (2.13) For 𝑘=1,2,…,𝑛, by (2.9) and (2.10), we get𝑎_𝑘=1𝑝𝑘!^(𝑘−1)(1)−𝑝^(𝑘−1)(0=()!)𝑛+1𝑘!(𝑛−𝑘+2)!^𝑛_ℓ=𝑘−1𝐸_ℓ−𝑘+1(1)−𝐸_ℓ−𝑘+10^𝑛−ℓ=1𝑘−𝑛+2𝑛+2^𝑛_ℓ=𝑘−1𝐸_ℓ−𝑘+1+2−𝐸_{𝑛−𝑘+1}. (2.14) Therefore, by (2.13) and (2.14), we obtain the following theorem.

Theorem 2.3. For 𝑛∈ℤ₊, one has ^𝑛_𝑘=0𝐸_𝑘(𝑥)𝑥^𝑛−𝑘=^𝑛−1_𝑘=0^𝑛−ℓ_𝑗=1(−1)^𝑗_𝑘^𝑛−ℓ+1(𝑛−ℓ+1)_ℓ^ℓ+𝑗𝐸_ℓ+𝑗+4(−1)^𝑛+1𝐸𝑛+2^𝑛+1+1𝑛+2^𝑛_𝑘=1𝑘−𝑛+2^𝑛_ℓ=𝑘−1𝐸_ℓ−𝑘+1+2−𝐸_{𝑛−𝑘+1}𝐵_𝑘(𝑥). (2.15)

From the Euler basis {𝐸₀(𝑥),𝐸₁(𝑥),…,𝐸_𝑛(𝑥)} for the space of polynomials of degree less than or equal to 𝑛 with coefficients in ℚ, we note that 𝑝(𝑥) can be written as follows:𝑝(𝑥)=^𝑛_𝑘=0𝑏_𝑘𝐸_𝑘(𝑥). (2.16) Thus, we have𝑏_𝑘=1𝑝2𝑘!^(𝑘)(1)+𝑝^(𝑘)(0=1)()!2𝑘!𝑛+1(𝑛−𝑘+1)!^𝑛_ℓ=𝑘𝐸_ℓ−𝑘(1)+𝐸_ℓ−𝑘0^𝑛−ℓ=12𝑘−𝑛+1^𝑛−1_ℓ=𝑘𝐸_ℓ−𝑘+2. (2.17) Therefore, by (2.16) and (2.17), we obtain the following theorem.

Theorem 2.4. For 𝑛∈ℤ₊, one has ^𝑛_𝑘=0𝐸_𝑘(𝑥)𝑥^𝑛−𝑘=12^𝑛−1_𝑘=0𝑘−𝑛+1^𝑛−1_ℓ=𝑘𝐸_ℓ−𝑘+2𝐸_𝑘(𝑥)+(𝑛+1)𝐸_𝑛(𝑥). (2.18)

Let us consider the polynomial 𝑝(𝑥)=∑^𝑛_𝑘=0(𝐵_𝑘(𝑥)𝑥^𝑛−𝑘)/(𝑘!(𝑛−𝑘)!)=∑^𝑛_𝑘=0𝑎_𝑘𝐵_𝑘(𝑥). Then, we have𝑎₀=¹₀𝑝(𝑡)𝑑𝑡=^𝑛_ℓ=01()!𝐼ℓ!𝑛−ℓ^{ℓ,𝑛−ℓ}=1𝐼𝑛!^0,𝑛+^𝑛−1_ℓ=1𝐼_{ℓ,𝑛−ℓ}()!+𝐼𝑙!𝑛−ℓ^𝑛,0=1𝑛!()!+1𝑛+1(𝑛+1)!^𝑛−1_ℓ=1^𝑛−ℓ_𝑗=1(−1)^𝑗−1𝐵𝑛+1ℓ+𝑗^ℓ+𝑗. (2.19) It is easy to show that𝑝^(𝑘)(𝑥)=2^𝑘𝑛_ℓ=𝑘1()!()!𝐵ℓ−𝑘𝑛−ℓ^ℓ−𝑘(𝑥)𝑥^𝑛−ℓ. (2.20) For 𝑘=1,2,…,𝑛, we have𝑎_𝑘=1𝑝𝑘!^(𝑘−1)(1)−𝑝^(𝑘−1)(0=2_𝑘−1)𝑘!^𝑛_ℓ=𝑘−11()!()!𝐵ℓ−𝑘+1𝑛−ℓ^ℓ−𝑘+1(1)−𝐵_ℓ−𝑘+10^𝑛−ℓ=2^𝑘−1𝑘!^𝑛_ℓ=𝑘−1𝐵_ℓ−𝑘+1+𝛿_{1,ℓ−𝑘+1}()!()!−𝐵ℓ−𝑘+1𝑛−ℓ(^{𝑛−𝑘+1})!=2𝑛−𝑘+1_𝑘−1𝑘!^𝑛−1_ℓ=𝑘−1𝐵_ℓ−𝑘+1()!()!+1ℓ−𝑘+1𝑛−ℓ()!.𝑛−𝑘 (2.21) Therefore, by (2.19) and (2.21), we obtain the following theorem.

Theorem 2.5. For 𝑛∈ℤ₊, one has ^𝑛_𝑘=0𝐵_𝑘(𝑥)()!𝑥𝑘!𝑛−𝑘^𝑛−𝑘=1()!+1𝑛+1(𝑛+1)!^𝑛−1_ℓ=1^𝑛−ℓ_𝑗=1(−1)^𝑗−1𝐵𝑛+1ℓ+𝑗^ℓ+𝑗+^𝑛_𝑘=12^𝑘−1𝑘!^𝑛−1_ℓ=𝑘−1𝐵_ℓ−𝑘+1()!()!+1ℓ−𝑘+1𝑛−ℓ()!𝐵𝑛−𝑘^𝑘(𝑥). (2.22)

Remark 2.6. If 𝑝(𝑥)=∑^𝑛_𝑘=0𝑏_𝑘𝐸_𝑘(𝑥), by the same method, we get ^𝑛_𝑘=0𝐵_𝑘(𝑥)𝑥^𝑛−𝑘()!=𝑘!𝑛−𝑘^𝑛−1_𝑘=02^𝑘−1𝑘!^𝑛_ℓ=𝑘𝐵_ℓ−𝑘()!()!+1ℓ−𝑘𝑛−ℓ()!+𝐵𝑛−𝑘−1(^𝑛−𝑘)!𝐸𝑛−𝑘^𝑘(𝑥)+2^𝑛𝐸𝑛!^𝑛(𝑥). (2.23)

Let us consider the polynomial 𝑝(𝑥)=∑^𝑛_𝑘=0(𝐸_𝑘(𝑥)𝑥^𝑛−𝑘)/(𝑘!(𝑛−𝑘)!)=∑^𝑛_𝑘=0𝑎_𝑘𝐵_𝑘(𝑥). Then, we have𝑎₀=¹₀𝑝(𝑡)𝑑𝑡=^𝑛_ℓ=01()!𝐽ℓ!𝑛−ℓ^{ℓ,𝑛−ℓ}=^𝑛_ℓ=01ℓ!(𝑛−ℓ)!^𝑛−ℓ_𝑗=1(−1)^{𝑗𝑗𝑛−ℓ+1}(𝑛−ℓ+1)_ℓ^ℓ+𝑗𝐸_ℓ+𝑗+2(−1)^𝑛−ℓ+1(𝑛+1)(^𝑛_ℓ)𝐸_𝑛+1=^𝑛_ℓ=0^𝑛−ℓ_𝑗=1(−1)^𝑗𝐸_ℓ+𝑗!!+ℓ+𝑗𝑛+1−ℓ−𝑗2𝐸_𝑛+1(−1)^𝑛+1(𝑛+1)!^𝑛_ℓ=0(−1)^ℓ=^𝑛_ℓ=0^𝑛−ℓ_𝑗=1(−1)^𝑗𝐸_ℓ+𝑗!!+ℓ+𝑗𝑛+1−ℓ−𝑗2𝐸_𝑛+1(−1)^𝑛+1()!.𝑛+1 (2.24) It is easy to show that𝑝^(𝑘)(𝑥)=2^𝑘𝑛_ℓ=𝑘1()!()!𝐸ℓ−𝑘𝑛−ℓ^ℓ−𝑘(𝑥)𝑥^𝑛−ℓ. (2.25)

For 𝑘=1,2,…,𝑛, we have𝑎_𝑘=1𝑝𝑘!^(𝑘−1)(1)−𝑝^(𝑘−1)(0=2_𝑘−1)𝑘!^𝑛_ℓ=𝑘−11()!()!𝐸ℓ−𝑘+1𝑛−ℓ^ℓ−𝑘+1(1)−𝐸_ℓ−𝑘+10^𝑛−ℓ=2^𝑘−1−𝑘!^𝑛_ℓ=𝑘−1𝐸_ℓ−𝑘+1()!()!+2ℓ−𝑘+1𝑛−ℓ()!−𝐸𝑛−𝑘+1(^{𝑛−𝑘+1})!.𝑛−𝑘+1 (2.26) Therefore, by (2.24) and (2.26), we obtain the following theorem.

Theorem 2.7. For 𝑛∈ℤ₊, one has ^𝑛_𝑘=0𝑛𝑘𝐸_𝑘(𝑥)𝑥^𝑛−𝑘=^𝑛_ℓ=0^𝑛−ℓ_𝑗=1(−1)^𝑗𝑛+1_ℓ+𝑗𝐸𝑛+1^ℓ+𝑗+2(−1)^𝑛+1()𝐸𝑛+1^𝑛+1+^𝑛_𝑘=1−1𝑛+1^𝑛_ℓ=𝑘−12^𝑘−1𝑘𝐸𝑛+1𝑛−𝑘+1ℓ−𝑘+1^ℓ−𝑘+1+2()𝑘−2𝑛+1𝑛+1^𝑘−1𝑘𝐸𝑛+1𝑛+1_{𝑛−𝑘+1}𝐵_𝑘(𝑥). (2.27)

Let us consider the polynomial 𝑝(𝑥)=∑^𝑛_𝑘=0(𝐸_𝑘(𝑥)𝑥^𝑛−𝑘)/(𝑘!(𝑛−𝑘)!)=∑^𝑛_𝑘=0𝑏_𝑘𝐸_𝑘(𝑥). By the same method, we obtain the following identity:^𝑛_𝑘=0𝐸_𝑘(𝑥)𝑥^𝑛−𝑘()!=𝑘!𝑛−𝑘^𝑛_𝑘=02^𝑘−1−𝑘!^𝑛_ℓ=𝑘𝐸_ℓ−𝑘()!()!+2ℓ−𝑘𝑛−ℓ()!+𝐸𝑛−𝑘(^𝑛−𝑘)!𝐸𝑛−𝑘^𝑘(𝑥). (2.28) Let us take 𝑝(𝑥)=∑^𝑛−1_𝑘=1(1/(𝑘(𝑛−𝑘)))𝐵_𝑘(𝑥)𝑥^𝑛−𝑘. Then, the 𝑘th derivative of 𝑝(𝑥) is given by𝑝^(𝑘)(𝑥)=𝐶_𝑘𝑥^𝑛−𝑘+𝐵_𝑛−𝑘(𝑥)+(𝑛−1)⋯(𝑛−𝑘)^𝑛−1_ℓ=𝑘+1𝐵_ℓ−𝑘(𝑥)𝑥^𝑛−ℓ(),𝑛−ℓ)(ℓ−𝑘 (2.29) where𝐶_𝑘=∑^𝑘_𝑗=1(𝑛−1)⋯𝑛−𝑗+1𝑛−𝑗−1⋯(𝑛−𝑘)()𝑛−𝑘𝑘=1,2,…,𝑛−1,𝐶₀=0. (2.30) Note that𝑝^(𝑛)(𝑥)=𝑝^(𝑛−1)(𝑥)^′=𝐶_𝑛−1𝑥+𝐵₁(𝑥)^′=2𝐶_𝑛−1=2(𝑛−1)!𝐻_𝑛−1, (2.31) where 𝐻_𝑛−1=∑^𝑛−1_𝑗=1(1/𝑗).

By the properties of the Bernoulli basis for the space of polynomials of degree less than or equal to 𝑛 with coefficients in ℚ, 𝑝(𝑥) is given by𝑝(𝑥)=^𝑛_𝑘=0𝑎_𝑘𝐵_𝑘(𝑥). (2.32) Thus, by (2.32), we get𝑎₀=^𝑛−1_ℓ=11ℓ(𝑛−ℓ)¹₀𝐵_ℓ(𝑡)𝑥^𝑛−ℓ𝑑𝑡=^𝑛−1_ℓ=01ℓ()𝐼𝑛−ℓ^{ℓ,𝑛−ℓ}=^𝑛−1_ℓ=01ℓ(𝑛−ℓ)^𝑛−ℓ_𝑗=1(−1)^𝑗−1_𝑗^𝑛−ℓ+1𝐵_ℓ+𝑗(𝑛−ℓ+1)_ℓ^ℓ+𝑗=1𝑛𝑛²−1^𝑛−1_ℓ=1^𝑛−ℓ_𝑗=1(−1)^{𝑗−1𝑛+1}_{𝑗+ℓ𝑛−2ℓ−1}𝐵_ℓ+𝑗. (2.33)

From (2.29), (2.31), and (2.32), we note that𝑎_𝑘=1𝑝𝑘!^(𝑘−1)(1)−𝑝^(𝑘−1)(0=𝐶_𝑘−1)𝐵𝑘!^{𝑛−𝑘+1}(1)−𝐵_{𝑛−𝑘+1}+1−0^{𝑛−𝑘+1}+(^𝑛_𝑘)𝑛^𝑛−1_ℓ=𝑘1()𝐵𝑛−ℓ)(ℓ−𝑘+1^ℓ−𝑘+1(1)−𝐵_ℓ−𝑘+10^𝑛−ℓ=𝐶_𝑘−1𝛿𝑘!^{1,𝑛−𝑘+1}+(+1^𝑛𝑘)𝑛^𝑛−1_ℓ=𝑘𝐵_ℓ−𝑘+1+𝛿_{1,ℓ−𝑘+1}()=𝐶ℓ−𝑘+1)(𝑛−ℓ_𝑘−1𝛿𝑘!^{1,𝑛−𝑘+1}+(+1^𝑛𝑘)𝑛^𝑛−1_ℓ=𝑘𝐵_ℓ−𝑘+1()+1ℓ−𝑘+1)(𝑛−ℓ=𝐶𝑛−𝑘_𝑘−1+(𝑘!^𝑛𝑘)𝑛^𝑛−1_ℓ=𝑘𝐵_ℓ−𝑘+1()+1ℓ−𝑘+1)(𝑛−ℓ𝑛−𝑘if1≤𝑘≤𝑛−1,2𝐻_𝑛−1𝑛if𝑘=𝑛. (2.34) From (2.30), we have𝐶_𝑘−1=1𝑘!𝑘!^𝑘−1_𝑗=1(𝑛−1)!()!=𝑛−𝑘+1)(𝑛−𝑘𝑛−𝑗𝑛!()!1𝑘!𝑛−𝑘𝑛(𝑛−𝑘+1)^𝑘−1_𝑗=11=𝑛𝑘1𝑛−𝑗𝑛(𝑛−𝑘+1)^𝑛−1_𝑗=11𝑗−^𝑛−𝑘_𝑗=11𝑗=(^𝑛_𝑘)𝑛()𝐻𝑛−𝑘+1^𝑛−1−𝐻_𝑛−𝑘. (2.35) Therefore, by (2.32), (2.34), and (2.35), we obtain the following theorem.

Theorem 2.8. For 𝑛∈ℕ, one has ^𝑛−1_𝑘=11𝑘()𝐵𝑛−𝑘^𝑘(𝑥)𝑥^𝑛−𝑘=1𝑛𝑛²−1^𝑛−1_ℓ=1^𝑛−ℓ_𝑗=1(−1)^{𝑗−1𝑛+1}_ℓ+𝑗𝐵_{ℓ+𝑗𝑛−2ℓ−1}+1𝑛^𝑛−1_𝑘=1𝑛𝑘×𝐻_𝑛−1−𝐻_𝑛−𝑘+𝑛−𝑘+1^𝑛−1_ℓ=𝑘𝐵_ℓ−𝑘+1()+1ℓ−𝑘+1)(𝑛−ℓ𝑛−𝑘×𝐵_𝑘(𝑥)+2𝐻_𝑛−1𝐵_𝑛(𝑥)𝑛. (2.36)

We assume that 𝑝(𝑥)=∑^𝑛−1_𝑘=1(1/𝑘(𝑛−𝑘))𝐵_𝑘(𝑥)𝑥^𝑛−𝑘=∑^𝑛_𝑘=0𝑏_𝑘𝐸_𝑘(𝑥). For 𝑘=0,1,2,…,𝑛−1, we have𝑏_𝑘=1𝑝2𝑘!^(𝑘)(1)+𝑝^(𝑘)(0=𝐶_𝑘)𝐵2𝑘!^𝑛−𝑘(1)+1+𝐵_𝑛−𝑘0^𝑛−𝑘+(𝑛−1)⋯(𝑛−𝑘)2𝑘!^𝑛−1_ℓ=𝑘+11()𝐵ℓ−𝑘)(𝑛−ℓ^ℓ−𝑘(1)+𝐵_ℓ−𝑘0^𝑛−ℓ=𝐶_𝑘2𝑘!2𝐵_𝑛−𝑘+1+^𝑘𝑛−12^𝑛−1_ℓ=𝑘+1𝐵_ℓ−𝑘()+1ℓ−𝑘)(𝑛−ℓ𝐶𝑛−𝑘−1if0≤𝑘≤𝑛−2,_𝑛−12()!=1𝑛−12𝐻_𝑛−1if𝑘=𝑛−1. (2.37) Finally,𝑏_𝑛=1𝑝2𝑛!^(𝑛)(1)+𝑝^(𝑛)(0)=4𝐶_𝑛−1=2𝑛!2𝐶_𝑛−1=𝑛!2𝐻_𝑛−1𝑛. (2.38)

By the same method, we obtain the following identity:^𝑛−1_𝑘=11𝑘()𝐵𝑛−𝑘^𝑘(𝑥)𝑥^𝑛−𝑘=12^𝑛−2_𝑘=0(^𝑛_𝑘)𝑛𝐻_𝑛−1−𝐻_{𝑛−𝑘−1}2𝐵_𝑛−𝑘+𝑘+1𝑛−1^𝑛−1_ℓ=𝑘+1𝐵_ℓ−𝑘()+1ℓ−𝑘)(𝑛−ℓ𝑛−𝑘−1×𝐸_𝑘(𝑥)+12𝐻_𝑛−1𝐸_𝑛−1(𝑥)+2𝐻_𝑛−1𝑛𝐸_𝑛(𝑥). (2.39) Let us take 𝑝(𝑥)=∑^𝑛−1_𝑘=1(1/𝑘(𝑛−𝑘))𝐸_𝑘(𝑥)𝑥^𝑛−𝑘. Then, for 𝑘=0,1,2,…,𝑛−1, we have𝑝^(𝑘)(𝑥)=𝐶_𝑘𝐸_𝑛−𝑘(𝑥)+𝑥^𝑛−𝑘+(𝑛−1)⋯(𝑛−𝑘)^𝑛−1_ℓ=𝑘+1𝐸_ℓ−𝑘(𝑥)𝑥^𝑛−ℓ(),ℓ−𝑘)(𝑛−ℓ (2.40) where 𝐶_𝑘=(∑^𝑘_𝑗=1(𝑛−1)⋯(𝑛−𝑗+1)(𝑛−𝑗−1)⋯(𝑛−𝑘))/(𝑛−𝑘).

Note that𝑝^(𝑛)(𝑥)=𝑝^(𝑛−1)(𝑥)^′=𝐶_𝑛−1𝐸₁(𝑥)+𝑥^′=2𝐶_𝑛−1=2(𝑛−1)!𝐻_𝑛−1. (2.41) By the properties of the Bernoulli basis for the space of polynomials of degree less than or equal to 𝑛 with coefficients in ℚ, 𝑝(𝑥) can be written as𝑝(𝑥)=^𝑛_𝑘=0𝑎_𝑘𝐵_𝑘(𝑥). (2.42) Thus, by (2.42), we get𝑎₀=¹₀𝑝(𝑡)𝑑𝑡=^𝑛−1_ℓ=11ℓ(𝑛−ℓ)¹₀𝐸_ℓ(𝑡)𝑡^𝑛−ℓ𝑑𝑡=^𝑛−1_ℓ=11ℓ()𝐽𝑛−ℓ^{ℓ,𝑛−ℓ}=1𝑛𝑛²−1^𝑛−1_ℓ=1^𝑛−ℓ_𝑗=1(−1)^𝑗𝑛+1_{𝑗+ℓ𝑛−2ℓ−1}𝐸_ℓ+𝑗+2(−1)^𝑛+1𝑛𝑛²𝐸−1^{𝑛+1𝑛−1}_ℓ=1(−1^ℓ_{𝑛−2ℓ−1}). (2.43)

It is easy to show that^𝑛−1_ℓ=1(−1)^ℓ(𝑛−1)^𝑛−2_ℓ−1=^𝑛−1_ℓ=1(−1)^ℓ𝐵()=4ℓ,𝑛−ℓ𝑛()−2()𝑛+1)(𝑛+2𝑛−1𝑛²(𝑛+1)²𝐸_𝑛+1. (2.44) For 𝑘=1,2,…,𝑛, one has𝑎_𝑘=1𝑝𝑘!^(𝑘−1)(1)−𝑝^(𝑘−1)(0=𝐶_𝑘−1)𝐸𝑘!^{𝑛−𝑘+1}(1)+1−𝐸_{𝑛−𝑘+1}−0^{𝑛−𝑘+1}+(𝑛−1)⋯(𝑛−𝑘+1)𝑘!^𝑛−1_ℓ=𝑘1()𝐸ℓ−𝑘+1)(𝑛−ℓ^ℓ−𝑘+1(1)−𝐸_ℓ−𝑘+10^𝑛−ℓ=1𝐶𝑘!^𝑘−1−2𝐸_{𝑛−𝑘+1}−()⋯()+1𝑛−1𝑛−𝑘+1𝑘!^𝑛−1_ℓ=𝑘𝐸_ℓ−𝑘+1()=(ℓ−𝑘+1)(𝑛−ℓ_𝑛𝑘)𝐻_𝑛−1−𝐻_𝑛−𝑘𝑛(𝑛−𝑘+1)−2𝐸_{𝑛−𝑘+1}−(+1^𝑛𝑘)𝑛^𝑛−1_ℓ=𝑘𝐸_ℓ−𝑘+1().ℓ−𝑘+1)(𝑛−ℓ (2.45) Therefore, by (2.42), (2.44), and (2.45), we obtain the following theorem.

Theorem 2.9. For 𝑛∈ℕ, one has ^𝑛_𝑘=1𝐸_𝑘(𝑥)𝑥^𝑛−𝑘𝑘()=1𝑛−𝑘𝑛𝑛²−1^𝑛−1_ℓ=1^𝑛−ℓ_𝑗=1(−1)^𝑗𝑛+1_ℓ+𝑗𝐸_{ℓ+𝑗𝑛−2ℓ−1}+2(−1)^𝑛+1𝑛()4𝑛+1𝑛()−2()𝑛+1)(𝑛+2𝑛−1𝑛²(𝑛+1)²𝐸_𝑛+1+1𝑛^𝑛_𝑘=1𝑛𝑘𝐻_𝑛−1−𝐻_𝑛−𝑘𝑛−𝑘+11−2𝐸_{𝑛−𝑘+1}−𝑛𝑘^{𝑛−1ℓ=𝑘}𝐸_ℓ−𝑘+1()𝐵ℓ−𝑘+1)(𝑛−ℓ^𝑘(𝑥). (2.46)

We may assume that 𝑝(𝑥)=∑^𝑛_𝑘=0(1/(𝑘(𝑛−𝑘)))𝐸_𝑘(𝑥)𝑥^𝑛−𝑘=∑^𝑛_𝑘=0𝑏_𝑘𝐸_𝑘(𝑥). Then, we note that𝑏_𝑘=1𝑝2𝑘!^(𝑘)(1)+𝑝^(𝑘)(0)(𝑘=0,1,2,…,𝑛−1). (2.47)

Thus, we have𝑏_𝑘=𝐶_𝑘𝐸2𝑘!^𝑛−𝑘(1)+1+𝐸_𝑛−𝑘+0^𝑛−𝑘+^𝑘𝑛−12^𝑛−1_ℓ=𝑘+11()𝐸ℓ−𝑘)(𝑛−ℓ^ℓ−𝑘(1)+𝐸_ℓ−𝑘0^𝑛−ℓ=𝐶_𝑘+2𝑘!2^{𝑘𝑛−1𝑛−1}_ℓ=𝑘+1−𝐸_ℓ−𝑘()=𝐶ℓ−𝑘)(𝑛−ℓ_𝑘−2𝑘!2^{𝑘𝑛−1𝑛−1}_ℓ=𝑘+1𝐸_ℓ−𝑘(),𝑏ℓ−𝑘)(𝑛−ℓ_𝑛=1𝑝2𝑛!^(𝑛)(1)+𝑝^(𝑛)(0)=12𝑛!4𝐶_𝑛−1=2𝐶_𝑛−1=𝑛!2𝐻_𝑛−1𝑛. (2.48)

By the same method, we obtain the following identity:^𝑛−1_𝑘=11𝑘()𝐸𝑛−𝑘^𝑘(𝑥)𝑥^𝑛−𝑘=^𝑛−1_𝑘=01𝑛𝑘𝐻2𝑛^𝑛−1−𝐻_{𝑛−𝑘−1}−^𝑘𝑛−12^𝑛−1_ℓ=𝑘+1𝐸_ℓ−𝑘(ℓ−𝑘)(𝑛−ℓ)×𝐸_𝑘(𝑥)+2𝑛𝐻_𝑛−1𝐸_𝑛(𝑥). (2.49)