Hermite Polynomials and their Applications Associated with Bernoulli and Euler Numbers

Abstract

We derive some interesting identities and arithmetic properties of Bernoulli and Euler polynomials from the orthogonality of Hermite polynomials. Let 𝐏_𝑛={𝑝(𝑥)∈ℚ[𝑥]∣deg𝑝(𝑥)≤𝑛} be the (𝑛+1)-dimensional vector space over ℚ. Then we show that {𝐻₀(𝑥),𝐻₁(𝑥),…,𝐻_𝑛(𝑥)} is a good basis for the space 𝐏_𝑛 for our purpose of arithmetical and combinatorial applications.

1. Introduction

As is well known, the Euler polynomials, 𝐸_𝑛(𝑥), are defined by the generating function as follows: 2𝑒^𝑡𝑒+1^𝑥𝑡=𝑒^{𝐸(𝑥)𝑡}=^∞_𝑛=0𝐸_𝑛(𝑥)𝑡^𝑛𝑛! (1.1) (see [1–8]), with the usual convention about replacing 𝐸^𝑛(𝑥) by 𝐸_𝑛(𝑥).

In the special case, 𝑥=0, 𝐸_𝑛(0)=𝐸_𝑛 is called the nth Euler number. From (1.1) and definition of Euler numbers, we note that 𝐸_𝑛(𝑥)=(𝐸+𝑥)^𝑛=^𝑛_𝑙=0𝑛𝑙𝐸_𝑙𝑥^𝑛−𝑙=^𝑛_𝑙=0𝑛𝑙𝐸_𝑛−𝑙𝑥^𝑙 (1.2) with the usual convention about replacing 𝐸^𝑛 by 𝐸_𝑛.

The Bernoulli numbers are defined as 𝐵₀=1,(𝐵+1)^𝑛−𝐵_𝑛=𝛿_1,𝑛 (1.3) (see [9–14]), where 𝛿_𝑘,𝑛 is a Kronecker symbol.

As is well known, Bernoulli polynomials are also defined by 𝐵_𝑛(𝑥)=(𝐵+𝑥)^𝑛=^𝑛_𝑙=0𝑛𝑙𝐵_𝑙𝑥^𝑛−𝑙=^𝑛_𝑙=0𝑛𝑙𝐵_𝑛−𝑙𝑥^𝑙 (1.4) with the usual convention about replacing 𝐵^𝑛 by 𝐵_𝑛 (see [1, 15–18]).

The Hermite polynomials are defined by the generating function as follows: 𝑒^{2𝑥𝑡−𝑡2}=𝑒^{𝐻(𝑥)𝑡}=^∞_𝑛=0𝐻_𝑛(𝑥)𝑡^𝑛𝑛! (1.5) (see [5, 19]), with the usual convention about replacing 𝐻^𝑛(𝑥) by 𝐻_𝑛(𝑥).

From (1.5), we can derive the following identities: 𝐻_𝑛(𝑥)=𝜕𝜕𝑡^𝑛𝑒^{2𝑥𝑡−𝑡2}_𝑡=0=𝑒^𝑥2𝜕𝜕𝑡^𝑛𝑒^{−(𝑥−𝑡)2}_𝑡=0=(−1)^𝑛𝑒^𝑥2𝜕𝜕𝑥^𝑛𝑒^{−(𝑥−𝑡)2}_𝑡=0=(−1)^𝑛𝑒^𝑥2𝑑^𝑛𝑑𝑥^𝑛𝑒^−𝑥2. (1.6)

Let us consider two operators as follows: 𝑓⟼𝑂₁𝑓=−𝑒^𝑥2𝑑𝑒𝑑𝑥^−𝑥2𝑓=2𝑥𝑓−𝑑𝑓,𝑑𝑥𝑓⟼𝑂₂𝑓=𝑒^𝑥2/2𝑑𝑥−𝑒𝑑𝑥^−𝑥2/2𝑓=2𝑥𝑓−𝑑𝑓.𝑑𝑥 (1.7)

By (1.7), we get 𝑂₁=𝑂₂. In particular, if we take 𝑓=1, then we have −𝑒^𝑥2𝑑𝑒𝑑𝑥^−𝑥2=𝑒^𝑥2/2𝑑𝑥−𝑒𝑑𝑥^−𝑥2/2. (1.8)

We note that (−1)^𝑛𝑒^𝑥2𝑑^𝑛𝑑𝑥^𝑛𝑒^−𝑥2=−𝑒^𝑥2𝑑𝑒𝑑𝑥^−𝑥2𝑛. (1.9)

From (1.8), we note that (−1)^𝑛𝑒^𝑥2𝑑^𝑛𝑒^−𝑥2𝑑𝑥^𝑛=−𝑒^𝑥2𝑑𝑒^−𝑥2𝑑𝑥^𝑛=𝑒^𝑥2/2𝑑𝑥−𝑒𝑑𝑥^{−𝑥2/2𝑛}=𝑒^𝑥2/2𝑑𝑥−𝑑𝑥^𝑛𝑒^−𝑥2/2. (1.10)

Thus, by (1.10), we get 𝐻_𝑛(𝑥)=𝑒^𝑥2/2𝑑𝑥−𝑑𝑥^𝑛𝑒^−𝑥2/2 (1.11) (see [5, 19–23]). In the special case, 𝑥=0, 𝐻_𝑛(0)=𝐻_𝑛 are called the Hermite numbers.

From (1.5), we can derive the following identities: 𝐻_𝑛(𝑥)=(𝐻+2𝑥)^𝑛=^𝑛_𝑙=0𝑛𝑙𝐻_𝑛−𝑙2^𝑙𝑥^𝑙 (1.12) (cf. [5, 19]), with the usual convention about replacing 𝐻^𝑛 by 𝐻_𝑛. It is easy to show that ^∞_𝑛=0𝐻_𝑛𝑡^𝑛𝑛!=𝑒^−𝑡2=^∞_𝑙=0(−1)^𝑛𝑡𝑛!^2𝑛. (1.13)

By comparing coefficients on the both sides of (1.13), we get 𝐻_2𝑛=(−1)^𝑛()⋯()=()2𝑛2𝑛−1𝑛+1−1^𝑛(2𝑛)!𝑛!,𝐻_2𝑛−1=0, (1.14) where 𝑛∈ℕ. From (1.12), we have 𝑑𝐻_𝑛(𝑥)𝑑𝑥=2𝑛𝐻_𝑛−1(𝑥)(𝑛∈ℕ). (1.15)

Let 𝐏_𝑛={𝑝∈ℚ[𝑥]∣deg𝑝(𝑥)≤𝑛} be the (𝑛+1)-dimensional vector space over ℚ. Probably, {1,𝑥,𝑥²,…,𝑥^𝑛} is the most natural basis for this space. But {𝐻₀(𝑥),𝐻₁(𝑥),𝐻₂(𝑥),…,𝐻_𝑛(𝑥)} is also a good basis for the space 𝐏_𝑛, for our purpose of arithmetical and combinatorial applications.

For 𝑝(𝑥)∈𝐏_𝑛, 𝑝(𝑥)=^𝑛_𝑘=0𝐶_𝑘𝐻_𝑘(𝑥), (1.16) for some uniquely determined 𝑏_𝑙∈ℚ.

The purpose of this paper is to develop methods for computing 𝐶_𝑘 from the information of 𝑝(𝑥). By using these methods, we define some interesting identities.

2. Properties of Hermite Polynomials

From (1.5) and (1.13), we note that 1=^∞_𝑚=0𝐻_𝑚𝑡^𝑚𝑚!^∞_𝑙=0𝑡^2𝑙=𝑙!_∞𝑚=0𝐻_2𝑚𝑡^2𝑚(2𝑚)!^∞_𝑙=0(2𝑙)(2𝑙−1)⋯(𝑙+1)()!𝑡2𝑙^2𝑙=^∞_𝑛=0^𝑛_𝑙=0(2𝑙)(2𝑙−1)⋯(𝑙+1)()!()!𝐻2𝑙2𝑛−2𝑙^{2𝑛−2𝑙}()!𝑡2𝑛^2𝑛()!=2𝑛_{∞𝑛=0𝑛𝑙=0}𝑙𝑙!𝐻2𝑙2𝑛2𝑙^{2𝑛−2𝑙}𝑡^2𝑛()!.2𝑛 (2.1)

Thus, by (2.1), we obtain the following recurrence formula.

Proposition 2.1. For 𝑛∈ℤ₊=ℕ∪{0}, one has ^𝑛𝑙=0𝑙𝑙!𝐻2𝑙2𝑛2𝑙^{2𝑛−2𝑙}=1,𝑖𝑓𝑛=00,𝑖𝑓𝑛≠0. (2.2)

By, (1.5), we get ^∞_𝑛=0𝐻_𝑛()𝑡−𝑥^𝑛𝑛!=𝑒^{2𝑡(−𝑥)−𝑡2}=𝑒^{2𝑥(−𝑡)−(−𝑡)2}=^∞_𝑛=0𝐻_𝑛(𝑥)(−1)^𝑛𝑡^𝑛.𝑛! (2.3)

From (2.3), we can derive the following reflection symmetric identity of 𝐻_𝑛(𝑥): 𝐻_𝑛(−𝑥)=(−1)^𝑛𝐻_𝑛(𝑥). (2.4)

By (1.5), we easily see that 𝜕𝑒𝜕𝑡^{2𝑥𝑡−𝑡2}=(2𝑥−2𝑡)𝑒^{2𝑥𝑡−𝑡2}. (2.5)

Thus, by (1.5) and (2.5), we get 𝜕𝜕𝑡^∞_𝑛=0𝐻_𝑛(𝑥)𝑡^𝑛=(𝑛!2𝑥−2𝑡)^∞_𝑛=0𝐻_𝑛(𝑥)𝑡^𝑛.()=𝑛!LHSof2.5^∞_𝑛=1𝐻_𝑛(𝑥)𝑡^𝑛−1()!=𝑛−1^∞_𝑛=0𝐻_𝑛+1(𝑥)𝑡^𝑛,()=𝑛!RHSof2.5^∞_𝑛=02𝑥𝐻_𝑛(𝑥)𝑡^𝑛−𝑛!^∞_𝑛=02𝐻_𝑛(𝑥)𝑡^𝑛+1=𝑛!_∞𝑛=02𝑥𝐻_𝑛(𝑥)𝑡^𝑛−𝑛!^∞_𝑛=12𝐻_𝑛−1(𝑥)𝑡^𝑛()!=𝑛−1_∞𝑛=02𝑥𝐻_𝑛(𝑥)𝑡^𝑛−𝑛!^∞_𝑛=12𝑛𝐻_𝑛−1(𝑥)𝑡^𝑛.𝑛! (2.6)(2.7)(2.8)

Thus, by (2.6) and (2.7), we get 𝐻_𝑛+1(𝑥)=2𝑥𝐻_𝑛(𝑥)−2𝑛𝐻_𝑛−1(𝑥),(𝑛∈ℕ). (2.9)

From (1.15) and (2.9), we note that 𝐻_𝑛+1(𝑥)−2𝑥𝐻_𝑛(𝑥)+𝐻^′_𝑛(𝑥)=0. (2.10)

Differentiating on both sides, we have 2(𝑛+1)𝐻_𝑛(𝑥)−2𝐻_𝑛(𝑥)−2𝑥𝐻^′_𝑛(𝑥)+𝐻^′_𝑛(𝑥)=0. (2.11)

Thus, we have 𝐻_𝑛^′′(𝑥)−2𝑥𝐻^′_𝑛(𝑥)+2𝑛𝐻_𝑛(𝑥)=0. (2.12)

From (2.12), we note that 𝐻_𝑛(𝑥) is a solution of the following second-order linear differential equation: 𝑢^′′−2𝑥𝑢^′+2𝑛𝑢=0. (2.13)

From (1.5), we note that ^∞_𝑚=0𝐻_𝑛(𝑥)𝑡^𝑛𝑛!=𝑒^{2𝑡𝑥−𝑡2}=^∞_𝑙=0(2𝑥)^𝑙𝑡𝑙!^𝑙∞_𝑘=0(−1)^𝑘𝑡𝑘!^2𝑘=^∞_𝑛=0^[𝑛/2]_𝑘=0(−1)^𝑘𝑛!(2𝑥)^𝑛−2𝑘()!𝑡𝑘!𝑛−2𝑘^𝑛.𝑛! (2.14)

Thus, by (2.14), we get 𝐻_𝑛(𝑥)=^[𝑛/2]_𝑘=0(−1)^𝑘𝑛!()!()𝑘!𝑛−2𝑘2𝑥^𝑛−2𝑘=^𝑛/2_𝑙∑₌₀(−1)^{𝑛/2−𝑙}𝑛!2^2𝑙()!()!𝑥𝑛/2−𝑙2𝑙^2𝑙,if𝑛≡0(mod2_(𝑛−1)/2),_𝑙∑₌₀(−1)^{(𝑛−1)/2−𝑙}𝑛!2^2𝑙+1))!()!𝑥((𝑛−1/2−𝑙2𝑙+1^2𝑙+1,if𝑛≡1(mod2). (2.15)

3. Main Results

By (1.6), we easily get ^∞_−∞𝑒^−𝑥2𝐻_𝑛(𝑥)𝐻_𝑚(𝑥)𝑑𝑥=(−1)^𝑛∞_−∞𝑑^𝑛𝑑𝑥^𝑛𝑒^−𝑥2𝐻_𝑚(𝑥)𝑑𝑥. (3.1)

From (3.1), we note that ^∞_−∞𝑒^−𝑥2𝐻_𝑛(𝑥)𝐻_𝑚(𝑥)𝑑𝑥=2^𝑛𝑛!√𝜋𝛿_𝑚,𝑛. (3.2)

It is easy to show that ^∞_−∞𝑒^−𝑥2𝑥^𝑙(),𝑑𝑥=√0if𝑙≡1mod2𝑙!𝜋2^𝑙()!(),𝑙/2if𝑙≡0mod2 (3.3) where 𝑙∈ℤ₊=ℕ∪{0}. By (3.3), we get ^∞_−∞𝑑^𝑛𝑒^−𝑥2𝑑𝑥^𝑛𝑥^𝑚(),𝑑𝑥=()0if𝑛>𝑚or𝑛≤𝑚with𝑛−𝑚≡1mod2𝑚!−1^𝑛√𝜋2^𝑚−𝑛))!().((𝑚−𝑛/2if𝑛≤𝑚with𝑛−𝑚≡0mod2 (3.4)

From (3.2), we note that 𝐻₀(𝑥),𝐻₁(𝑥),…,𝐻_𝑛(𝑥) are orthogonal basis for the space ℙ_𝑛={𝑝(𝑥)∈ℚ[𝑥]∣deg𝑝(𝑥)≤𝑛} with respect to the inner product 𝑝(𝑥),𝑞(𝑥)=^∞_−∞𝑒^−𝑥2𝑝(𝑥)𝑞(𝑥)𝑑𝑥. (3.5)

For 𝑝(𝑥)∈ℙ_𝑛, the polynomial 𝑝(𝑥) is given by 𝑝(𝑥)=^∞_𝑘=0𝐶_𝑘𝐻_𝑘(𝑥), (3.6) where 𝐶_𝑘=12^𝑘𝑘!√𝜋𝑝(𝑥),𝐻_𝑘(𝑥=())−1^𝑘2^𝑘𝑘!√𝜋^∞_−∞𝑑^𝑘𝑒^−𝑥2𝑑𝑥^𝑘𝑝(𝑥)𝑑𝑥. (3.7)

Let us take 𝑝(𝑥)=𝑥^𝑛∈ℙ_𝑛. For 𝑛≡0(mod2), we compute 𝐶_𝑘 in (3.6) as follows 𝐶_𝑘=(−1)^𝑘2^𝑘𝑘!√𝜋^∞_−∞𝑑^𝑘𝑒^−𝑥2𝑑𝑥^𝑘𝑥^𝑛=()𝑑𝑥−1^𝑘2^𝑘𝑘!√𝜋×(−1)^𝑘𝑛!√𝜋2^𝑛−𝑘))!(),().((𝑛−𝑘/2if𝑘≡0mod20if𝑘≡1mod2 (3.8)

Let 𝑛≡1(mod2). Then we have 𝐶_𝑘=(−1)^𝑘2^𝑘𝑘!√𝜋^∞_−∞𝑑^𝑘𝑒^−𝑥2𝑑𝑥^𝑘𝑥^𝑛=𝑑𝑥𝑛!2^𝑛))!(),().𝑘!((𝑛−𝑘/2if𝑘≡1mod20if𝑘≡0mod2 (3.9)

Therefore, by (3.6), (3.8), and (3.9), we obtain the following proposition.

Proposition 3.1. One has 𝑥^2𝑛=(2𝑛)!2^𝑛2𝑛_𝑘=01()!()!𝐻2𝑘𝑛−𝑘^2𝑘(𝑥),𝑥^2𝑛+1=(2𝑛+1)!2^𝑛2𝑛+1_𝑘=01()!()!𝐻2𝑘+1𝑛−𝑘^2𝑘+1(𝑥). (3.10)

Let us take 𝑝(𝑥)=𝐵_𝑛(𝑥). From (3.4), 𝑃(𝑥) can be rewritten by 𝐵_𝑛(𝑥)=^𝑛_𝑘=0𝐶_𝑘𝐻_𝑘(𝑥), (3.11) where 𝐶_𝑘=(−1)^𝑘2^𝑘𝑘!√𝜋^∞_−∞𝑑^𝑘𝑒^−𝑥2𝑑𝑥^𝑘𝐵_𝑛(𝑥)𝑑𝑥. (3.12)

By integrating by parts, we get ^∞_−∞𝑑^𝑘𝑒^−𝑥2𝑑𝑥^𝑘𝐵_𝑛(𝑥)=(−𝑛)(−(𝑛−1))⋯(−(𝑛−𝑘+1))^∞_−∞𝑒^−𝑥2𝐵_𝑛−𝑘(𝑥)=()𝑑𝑥−1^𝑘𝑛!(𝑛−𝑘)!^𝑛−𝑘_𝑙=0𝑙𝐵𝑛−𝑘_{𝑛−𝑘−𝑙}^∞_−∞𝑒^−𝑥2𝑥^𝑙=()𝑑𝑥−1^𝑘𝑛!(𝑛−𝑘)!_{(0≤𝑙≤𝑛−𝑘𝑙≡0mod2)}(𝑛−𝑘)!𝐵_{𝑛−𝑘−𝑙}()!×√𝑙!𝑛−𝑘−𝑙𝑙!𝜋2^𝑙()!=()𝑙/2−1^𝑘𝑛!√𝜋_{(0≤𝑙≤𝑛−𝑘𝑙≡0mod2)}𝐵_{𝑛−𝑘−𝑙}(𝑛−𝑘−𝑙)!2^𝑙()!.𝑙/2 (3.13)

Thus, from (3.11) and (3.13), we have 𝐶_𝑘=𝑛!2^𝑘𝑘!_{(0≤𝑙≤𝑛−𝑘𝑙≡0mod2)}𝐵_{𝑛−𝑘−𝑙}(𝑛−𝑘−𝑙)!2^𝑙()!.𝑙/2 (3.14)

Therefore, by (3.11) and (3.14), we obtain the following theorem.

Theorem 3.2. For 𝑛∈ℤ₊, one has 𝐵_𝑛(𝑥)=𝑛!^𝑛_{𝑘=00≤𝑙≤𝑛−𝑘𝑙≡0(mod2)}𝐵_{𝑛−𝑘−𝑙}2^𝑘+𝑙()!()!𝐻𝑘!𝑛−𝑘−𝑙𝑙/2^𝑘(𝑥). (3.15)

Remark 3.3. Let us take 𝑝(𝑥)=𝐸_𝑛(𝑥). Then, by the same method, we obtain the following identity: 𝐸_𝑛(𝑥)=𝑛!^𝑛_{𝑘=0(0≤𝑙≤𝑛−𝑘𝑙≡0mod2)}𝐸_{𝑛−𝑘−𝑙}2^𝑘+𝑙()!()!𝐻𝑘!𝑛−𝑘−𝑙𝑙/2^𝑘(𝑥). (3.16)

Now, we consider 𝑝(𝑥)=𝐻_𝑛(𝑥). From (3.6), we note that 𝑝(𝑥) can be rewritten as 𝐻_𝑛(𝑥)=^𝑛_𝑘=0𝐶_𝑘𝐻_𝑘(𝑥), (3.17) where 𝐶_𝑘=(−1)^𝑘2^𝑘𝑘!√𝜋^∞_−∞𝑑^𝑘𝑒^−𝑥2𝑑𝑥^𝑘𝐻_𝑛(𝑥)𝑑𝑥. (3.18)

By integrating by parts, we get ^∞_−∞𝑑^𝑘𝑒^−𝑥2𝑑𝑥^𝑘𝐻_𝑛(𝑥)𝑑𝑥=(−2𝑛)⋯(−2(𝑛−𝑘+1))^∞_−∞𝑒^−𝑥2𝐻_𝑛−𝑘(𝑥)=()𝑑𝑥−1^𝑘2^𝑘𝑛!(𝑛−𝑘)!^𝑛−𝑘_𝑙=0𝑙2𝑛−𝑘_𝑙𝐻_{𝑛−𝑘−𝑙}^∞_−∞𝑒^−𝑥2𝑥^𝑙=()𝑑𝑥−1^𝑘2^𝑘𝑛!(𝑛−𝑘)!^𝑛−𝑘_{(𝑙=0𝑙≡0mod2)}2^𝑙(𝑛−𝑘)!()!𝐻𝑙!𝑛−𝑘−𝑙^{𝑛−𝑘−𝑙}𝑙!√𝜋2^𝑙(𝑙/2)!=(−1)^𝑘2^𝑘𝑛!√𝜋^𝑛−𝑘_{(𝑙=0𝑙≡0mod2)}𝐻_{𝑛−𝑘−𝑙}()!()!.𝑛−𝑘−𝑙𝑙/2 (3.19)

From (3.17) and (3.19), we note that 𝐶_𝑘=(−1)^𝑘2^𝑘𝑘!√𝜋×(−1)^𝑘2^𝑘𝑛!√𝜋_{(0≤𝑙≤𝑛−𝑘𝑙≡0mod2)}𝐻_{𝑛−𝑘−𝑙}()!()!=𝑛−𝑘−𝑙𝑙/2𝑛!𝑘!_{(0≤𝑙≤𝑛−𝑘𝑙≡0mod2)}𝐻_{𝑛−𝑘−𝑙}()!()!.𝑛−𝑘−𝑙𝑙/2 (3.20)

Therefore, by (3.17) and (3.20), we obtain the following theorem.

Theorem 3.4. For 𝑛∈ℤ₊, one has 𝐻_𝑛(𝑥)=𝑛!^𝑛_{𝑘=0(0≤𝑙≤𝑛−𝑘𝑙≡0mod2)}𝐻_{𝑛−𝑘−𝑙}()!()!𝐻𝑘!𝑛−𝑘−𝑙𝑙/2^𝑘(𝑥). (3.21)

From Theorem 3.4, we note that 𝐻_𝑛(𝑥)=𝑛!^𝑛−1_{𝑘=0(0≤𝑙≤𝑛−𝑘𝑙≡0mod2)}𝐻_{𝑛−𝑘−𝑙}()!()!𝐻𝑘!𝑛−𝑘−𝑙𝑙/2^𝑘(𝑥)+𝑛!𝐻_𝑛(𝑥).𝑛! (3.22)

Thus, we have, for 0≤𝑘≤𝑛−𝑘, _{(0≤𝑙≤𝑛−𝑘𝑙≡0mod2)}𝐻_{𝑛−𝑘−𝑙}(𝑛−𝑘−𝑙)!(𝑙/2)!=0. (3.23)

Let 𝑙,𝑘∈ℤ₊ with 𝑘≤𝑙. Then we easily see that ^∞_−∞𝑑^𝑘𝑒^−𝑥2𝑑𝑥^𝑘𝐵_𝑙(𝑥)𝑑𝑥=(−1)^𝑘𝑙!√𝜋_{0≤𝑗≤𝑙−𝑘𝑗≡0(mod2)}𝐵_{𝑙−𝑘−𝑗}𝑙−𝑘−𝑗!2^𝑗!,𝑗/2_∞−∞𝑑^𝑘𝑒^−𝑥2𝑑𝑥^𝑘𝐸_𝑙(𝑥)𝑑𝑥=(−1)^𝑘𝑙!√𝜋_{(0≤𝑗≤𝑙−𝑘𝑗≡0mod2)}𝐸_{𝑙−𝑘−𝑗}𝑙−𝑘−𝑗!2^𝑗!.𝑗/2 (3.24)(3.25)

Let us consider the following polynomial of degree 𝑛 in ℙ_𝑛: 𝑝(𝑥)=^𝑛_𝑘=0𝐵_𝑘(𝑥)𝐵_𝑛−𝑘(𝑥). (3.26)

From (3.6), we note that 𝑝(𝑥) can be rewritten as 𝑝(𝑥)=^𝑛_𝑘=0𝐶_𝑘𝐻_𝑘(𝑥), (3.27) where 𝐶_𝑘=(−1)^𝑘2^𝑘𝑘!√𝜋^∞_−∞𝑑^𝑘𝑒^−𝑥2𝑑𝑥^𝑘𝑝(𝑥)𝑑𝑥. (3.28)

In [15], it is known that 𝑝(𝑥)=^𝑛_𝑘=0𝐵_𝑘(𝑥)𝐵_𝑛−𝑘(𝑥)=2𝑛+2^𝑛−2_𝑙=0𝑙𝐵𝑛+2_𝑛−𝑙𝐵_𝑙(𝑥)+(𝑛+1)𝐵_𝑛(𝑥). (3.29)

From (3.23) and (3.29), we have the following: 𝐶_𝑘=(−1)^𝑘2^𝑘𝑘!√𝜋2𝑛+2^𝑛−2_𝑙=0𝑙𝑛+2^∞_−∞𝑑^𝑘𝑒^−𝑥2𝑑𝑥^𝑘𝐵_𝑙(𝑥)𝑑𝑥+(𝑛+1)^∞_−∞𝑑^𝑘𝑒^−𝑥2𝑑𝑥^𝑘𝐵_𝑛(𝑥)𝑑𝑥, (3.30)

By (3.24) and (3.30), we get 𝐶_𝑛=(−1)^𝑛2^𝑛𝑛!√𝜋×(𝑛+1)^∞_−∞𝑑^𝑛𝑒^−𝑥2𝑑𝑥^𝑛𝐵_𝑛(𝑥)=()𝑑𝑥−1^𝑛2^𝑛𝑛!√𝜋×()()𝑛+1−1^𝑛𝑛!√𝜋𝐵₀0!2⁰=0!𝑛+12^𝑛,𝐶_𝑛−1=(−1)^𝑛−12^𝑛−1(𝑛−1)!√𝜋×(𝑛+1)^∞_−∞𝑑^𝑛−1𝑒^−𝑥2𝑑𝑥^𝑛−1𝐵_𝑛(𝑥)=()𝑑𝑥−1^𝑛−12^𝑛−1(𝑛−1)!√𝜋×(𝑛+1)(−1)^𝑛−1𝑛!√𝜋¹_{(𝑗=0𝑗≡0mod2)}𝐵_1−𝑗1−𝑗!2^𝑗!=()𝑗/2−1^𝑛−12^𝑛−1(𝑛−1)!√𝜋×(𝑛+1)(−1)^𝑛−1𝑛!√𝜋𝐵₁=−𝑛(𝑛+1)2^𝑛. (3.31)

For 0≤𝑘≤𝑛−2, we have 𝐶_𝑘=(−1)^𝑘2^𝑘𝑘!√𝜋2𝑛+2^𝑛−2_𝑙=𝑘𝑙𝐵𝑛+2_𝑛−𝑙^∞_−∞𝑑^𝑘𝑒^−𝑥2𝑑𝑥^𝑘𝐵_𝑙(𝑥)𝑑𝑥+(𝑛+1)^∞_−∞𝑑^𝑘𝑒^−𝑥2𝑑𝑥^𝑘𝐵_𝑛(𝑥)=()𝑑𝑥−1^𝑘2^𝑘𝑘!√𝜋2𝑛+2^𝑛−2_𝑙=𝑘𝑙𝐵𝑛+2_𝑛−𝑙(−1)^𝑘𝑙!√𝜋×_{(0≤𝑗≤𝑙−𝑘𝑗≡0mod2)}𝐵_{𝑙−𝑘−𝑗}𝑙−𝑘−𝑗!2^𝑗!+()𝑗/2𝑛+1)(−1^𝑘𝑛!√𝜋_{(0≤𝑗≤𝑛−𝑘𝑗≡0mod2)}𝐵_{𝑛−𝑘−𝑗}𝑛−𝑘−𝑗!2^𝑗!=2𝑗/2𝑛+2^𝑛−2_{𝑙=𝑘(0≤𝑗≤𝑙−𝑘𝑗≡0mod2)}𝑙𝑛+2𝐵_𝑛−𝑙𝐵_{𝑙−𝑘−𝑗}𝑙!2^𝑘+𝑗!!+()!𝑘!𝑙−𝑘−𝑗𝑗/2𝑛+1_{(0≤𝑗≤𝑛−𝑘𝑗≡0mod2)}𝐵_{𝑛−𝑘−𝑗}𝑘!𝑛−𝑘−𝑗!𝑗/2!2^𝑘+𝑗. (3.32)

Therefore, by (3.27) and (3.32), we obtain the following theorem.

Theorem 3.5. For 𝑛∈ℤ₊, one has ^𝑛_𝑘=0𝐵_𝑘(𝑥)𝐵_𝑛−𝑘(𝑥)=^𝑛−2_𝑘=02𝑛+2^𝑛−2_{𝑙=𝑘(0≤𝑗≤𝑛−𝑘𝑗≡0mod2)}𝑙𝑛+2𝑙!𝐵_𝑛−𝑙𝐵_{𝑙−𝑘−𝑗}2^𝑘+𝑗!!+()𝑘!𝑙−𝑘−𝑗𝑗/2𝑛+1!_{(0≤𝑗≤𝑛−𝑘𝑗≡0mod2)}𝐵_{𝑛−𝑘−𝑗}2^𝑘+𝑗!!𝐻𝑘!𝑛−𝑘−𝑗𝑗/2^𝑘(𝑥)−𝑛(𝑛+1)2^𝑛𝐻_𝑛−1(𝑥)+𝑛+12^𝑛𝐻_𝑛(𝑥). (3.33)

Acknowledgment

This research was supported by Basic Science Research Program through the National Research Foundation of Korea NRF funded by the Ministry of Education, Science and Technology 2012R1A1A2003786.