Old and New Identities for Bernoulli Polynomials via Fourier Series

Abstract

The Bernoulli polynomials 𝐵_𝑘 restricted to [0,1) and extended by periodicity have nth sine and cosine Fourier coefficients of the form 𝐶_𝑘/𝑛^𝑘. In general, the Fourier coefficients of any polynomial restricted to [0,1) are linear combinations of terms of the form 1/𝑛^𝑘. If we can make this linear combination explicit for a specific family of polynomials, then by uniqueness of Fourier series, we get a relation between the given family and the Bernoulli polynomials. Using this idea, we give new and simpler proofs of some known identities involving Bernoulli, Euler, and Legendre polynomials. The method can also be applied to certain families of Gegenbauer polynomials. As a result, we obtain new identities for Bernoulli polynomials and Bernoulli numbers.

1. Introduction

The Bernoulli polynomials 𝐵_𝑘(𝑥) (with 𝑘∈ℕ∪{0}) are defined via the generating function 𝑡𝑒^𝑡𝑥𝑒^𝑡=−1^∞_𝑘=0𝐵_𝑘(𝑥)𝑡^𝑘,𝑘! (1.1) from which one also obtains the Bernoulli numbers as the values 𝐵_𝑘=𝐵_𝑘(0). Although it does not immediately yield their explicit form, the manipulation of (1.1), along with the uniqueness theorem for power series expansions, leads to many properties of these polynomials. For instance, the following properties concerning derivation and symmetry are easily obtained in this way 𝐵^′_𝑘(𝑥)=𝑘𝐵_𝑘−1(𝑥)𝐵,𝑘≥1,_𝑘(1−𝑥)=(−1)^𝑘𝐵_𝑘(𝑥),𝑘≥0. (1.2)

These two properties are all that is needed to obtain the Fourier series of the 1-periodic functions which coincide with 𝐵_𝑘(𝑥) on the interval [0,1)𝐵_2𝑘(𝑥)=2(−1)^𝑘−1(2𝑘)!(2𝜋)^∞2𝑘_𝑛=1cos(2𝜋𝑛𝑥)𝑛^2𝑘𝐵,𝑘≥1,_2𝑘+1(𝑥)=2(−1)^𝑘−1(2𝑘+1)!(2𝜋)^∞2𝑘+1_𝑛=1sin(2𝜋𝑛𝑥)𝑛^2𝑘+1,𝑘≥0. (1.3) In fact, (1.3) and (1.5) are equivalent to the statements that the sine and cosine Fourier coefficients of the polynomials 𝐵^∘_2𝑘(𝑥)=(−1)^𝑘−1(2𝜋)^2𝑘()!𝐵2𝑘^2𝑘(𝑥),𝑘≥1,_∘𝐵_2𝑘+1(𝑥)=(−1)^𝑘−1(2𝜋)^2𝑘+1()!𝐵2𝑘+1^2𝑘+1(𝑥),𝑘≥0, (1.4) are ¹₀𝐵^∘_2𝑘(𝑥)()1cos2𝜋𝑚𝑥𝑑𝑥=𝑚^2𝑘,𝑚>0,0,𝑚=0,_10∘𝐵_2𝑘(𝑥)sin(2𝜋𝑚𝑥)𝑑𝑥=0,𝑚>0,_10∘𝐵_2𝑘+1(𝑥)()1sin2𝜋𝑚𝑥𝑑𝑥=𝑚^2𝑘+1,𝑚>0,_10∘𝐵_2𝑘+1(𝑥)cos(2𝜋𝑚𝑥)𝑑𝑥=0,𝑚≥0, (1.5)(1.6)(1.7)(1.8) and the properties (1.5)–(1.8) can be deduced easily using only induction and (1.2) and (1.3).

On the other hand, as we show in the following lemma, the sine and cosine coefficients on [0,1) of any polynomial of degree 𝑛 are a linear combination of the terms 1/𝑚^𝑘, 𝑘≤𝑛. Therefore, every polynomial of degree 𝑛 is a linear combination of the normalized polynomials 𝐵^∘_𝑘, 𝑘≤𝑛. Of course, this last assertion is also obvious without the use of Fourier series. The point here is that in many cases we can obtain the coefficients of the linear combination more quickly and easily by computing the Fourier coefficients of the polynomial.

We will illustrate the method with a very simple example: the monomial 𝑥^𝑛. First, we compute its Fourier coefficients.

Lemma 1.1. Let 𝑛,𝑚∈ℕ. Then ¹₀𝑥^𝑛cos(2𝜋𝑚𝑥)𝑑𝑥=^[𝑛/2]_𝑘=1(−1)^𝑘−1𝑛!()!1𝑛−2𝑘+1(2𝜋𝑚)^2𝑘₁₀,𝑥^𝑛sin(2𝜋𝑚𝑥)𝑑𝑥=^[𝑛/2]_𝑘=1(−1)^𝑘−1𝑛!()!1𝑛−2𝑘+2(2𝜋𝑚)^2𝑘−1. (1.9)

Proof. This is a straightforward computation using integration by parts and induction.

Now, as an application of the method, we give a proof of a well-known theorem.

Theorem 1.2. For 𝑛∈ℕ, one has ^𝑛𝑗=0𝑗𝐵𝑛+1_𝑗(𝑥)=(𝑛+1)𝑥^𝑛. (1.10)

Proof. Let us suppose that 𝑛 is even (the odd case is entirely similar) and use 2𝑛 instead. We define 𝑉_𝑛(𝑥)=^𝑛_𝑘=1(−1)^𝑘−1(2𝑛)!()!12𝑛−2𝑘+1(2𝜋)^∘2𝑘𝐵_2𝑘(𝑥)=^𝑛_𝑘=1(2𝑛)!()!()!𝐵2𝑛−2𝑘+12𝑘^2𝑘(𝑥),𝑊_𝑛(𝑥)=^𝑛_𝑘=1(−1)^𝑘(2𝑛)!()!12𝑛−2𝑘+2(2𝜋)^∘2𝑘−1𝐵_2𝑘−1(𝑥)=^𝑛_𝑘=1(2𝑛)!()!()!𝐵2𝑛−2𝑘+22𝑘−1^2𝑘−1(𝑥). (1.11) By the lemma and formulas (1.5)–(1.8), it is immediate that 𝑉_𝑛+𝑊_𝑛 and 𝑥^2𝑛 have the same sine and cosine Fourier coefficients, except for the constant term, which is 0 for the first polynomial and 1/(2𝑛+1) for the second one. Therefore, by uniqueness of Fourier series 𝑉_𝑛(𝑥)+𝑊_𝑛(𝑥)=𝑥^2𝑛−1.2𝑛+1 (1.12) Finally, observe that 𝑉_𝑛(𝑥)+𝑊_𝑛(𝑥)=^2𝑛_𝑗=1(2𝑛)!𝐵2𝑛−𝑗+1!𝑗!^𝑗(𝑥). (1.13) hence, the result follows now from a couple of simple manipulations.

Remark 1.3. The identities extend from [0,1) to ℂ by analytic continuation.

The classical proof of this theorem uses the formulas 𝐵_𝑛(1+𝑥)=𝑛𝑥^𝑛−1+𝐵_𝑛(𝑥),𝐵_𝑛(1+𝑥)=^𝑛_𝑘=0𝑛𝑘𝐵_𝑘(𝑥), (1.14) which can be easily deduced from the defining relation (1.1).

In this specific case, the difficulties of the above proof and the classical one are comparable. However, in more complicated situations, we believe that our method is simpler. It is the purpose of this paper to provide evidence for this claim, by obtaining, via the use of Fourier series, identities between polynomial families, some of which are new while others are well known, but are given simpler proofs.

In addition, properties of one family are carried over to another, and hence, we obtain new recurrence relations for Bernoulli polynomials and Bernoulli numbers. It should be mentioned that new relations between classical and generalized Bernoulli polynomials and numbers are still obtainable through the use of different types of expansions (see [1–4] and the references cited within). These identities are similar in appearance to ours but not identical to them. Power series expansions are the most often used, due to the natural starting point given by the generating series (1.1), but it is not the only one, for instance, the related power series for the cosecant is used in [5]. We believe that, for the purpose of proving these types of identities, Fourier expansion is also a useful tool, and one that is not typically used in the mainstream. The authors have previously employed this method to prove some facts about Bernoulli and Euler polynomials in [6] and for Apostol-Bernoulli polynomials in [7].

The organization of the paper is as follows. In Section 2, we obtain a relationship between Legendre and Bernoulli polynomials, which was already proved in [8, 9], but with longer and more complicated proofs. In Section 3, which we have separated from the previous one for the sake of clarity, we extend the results to Gegenbauer polynomials with polynomial weight, the results here are new and lead us also to some new identities for Bernoulli polynomials which are developed in Section 4. In Section 5, we show how to use the Fourier coefficients to obtain a known formula relating Euler and Bernoulli polynomials in a very simple way.

Remark 1.4. The formulas appearing in this paper were checked with Maple to avoid possible mistakes in their transcription, especially the longer ones.

2. Bernoulli and Legendre Polynomials

We denote the classical Legendre polynomials by 𝑃_𝑛(𝑥). Recall that they are orthogonal on the interval [−1,1] and have the same parity as 𝑛. It is straightforward to prove, by substituting 2𝑥−1=𝑦, that ¹₀𝑃_2𝑛(2𝑥−1)sin(2𝜋𝑚𝑥)𝑑𝑥=0,𝑛,𝑚∈ℕ,₁₀𝑃_2𝑛(2𝑥−1)𝑑𝑥=0,𝑛∈ℕ. (2.1) The key point here is that it is also possible to explicitly find the cosine coefficients by applying classical integral formulas.

Lemma 2.1. Let 𝑛,𝑚∈ℕ. Then ¹₀𝑃_2𝑛(2𝑥−1)cos(2𝜋𝑚𝑥)𝑑𝑥=^𝑛_𝑘=1(−1)^𝑘−12^2𝑘−1(2𝑛+2𝑘−1)!()!()!12𝑘−12𝑛−2𝑘+1(𝜋𝑚)^2𝑘. (2.2)

Proof. By substituting 2𝑥−1=𝑦, the proof is immediate if we take into account the two following classical formulas relating Bessel functions: ¹₀𝑃_2𝑛𝑦cos𝜋𝑚𝑦𝑑𝑦=(−1)^𝑛1𝐽2𝑚^2𝑛+1/2(𝜋𝑚) (2.3) (see [10, formula 2.17.7-1, page 433]) and 1𝐽2𝑚^2𝑛+(1/2)()=1𝜋𝑚(𝜋(𝜋𝑚cos𝑚−𝑛))^𝑛−1_𝑘=0(−1)^𝑘(2𝑛+2𝑘+1)!()!()!12𝑘+12𝑛−2𝑘−1(2𝜋𝑚)^2𝑘+1 (2.4) (see [11, formula 10.1.8, page 437]).

This lemma and the uniqueness of Fourier series are all that we need to state the following theorem.

Theorem 2.2. Let 𝑛∈ℕ. Then 12𝑃_2𝑛(2𝑥−1)=^𝑛_𝑘=1(2𝑛+2𝑘−1)!()!()!()!𝐵2𝑘−12𝑘2𝑛−2𝑘+1^2𝑘(𝑥). (2.5)

Proof. The previous lemma and (1.5), (1.6) show that the polynomials on both sides of (2.5) have the same sine and cosine Fourier coefficients on [0,1]. Therefore, the formula is true on [0,1) and, by analytic continuation, on ℝ.

In the odd case the same references, [10, 11], serve to show the following result.

Lemma 2.3. Let 𝑛,𝑚∈ℕ. Then (i)∫¹₀𝑃_2𝑛+1(2𝑥−1)sin(2𝜋𝑚𝑥)𝑑𝑥=−∑^𝑛_𝑘=0((−1)^𝑘/2^2𝑘)((2𝑛+2𝑘+1)!/(2𝑘)!(2𝑛−2𝑘+1)!)(1/(𝜋𝑚)^2𝑘+1),(ii)∫¹₀𝑃_2𝑛+1(2𝑥−1)cos(2𝜋𝑚𝑥)𝑑𝑥=∫¹₀𝑃_2𝑛+1(2𝑥−1)𝑑𝑥=0.

We then have the corresponding consequence.

Theorem 2.4. Let 𝑛∈ℕ. Then 12𝑃_2𝑛+1(2𝑥−1)=^𝑛_𝑘=0(2𝑛+2𝑘+1)!()!()!()!𝐵2𝑘2𝑘+12𝑛−2𝑘+1^2𝑘+1(𝑥). (2.6)

As we said in Section 1, the same formulas for Legendre polynomials were obtained in [8, 9] by much more complicated techniques.

3. Bernoulli and Gegenbauer Polynomials

We now consider the classical Gegenbauer polynomials 𝐺^𝜆_𝑛(𝑥), which are orthogonal on [−1,1] with respect to the weight 𝜔_𝜆(𝑥)=(1−𝑥²)^𝜆−1/2, when 𝜆−1/2∈ℕ∪{0}, with 𝐺_𝑛^1/2(𝑥)=𝑃_𝑛(𝑥) being a particular case. We recall that 𝐺^𝜆_𝑛(𝑥) also has the same parity as 𝑛. The following identities are straightforward to prove by substituting 2𝑥−1=𝑦. For 𝑛,𝑚∈ℕ, ¹₀(𝑥(1−𝑥))^𝜆−1/2𝐺^𝜆_𝑛(2𝑥−1)𝑑𝑥=0,₁₀(𝑥(1−𝑥))^𝜆−1/2𝐺^𝜆_2𝑛(2𝑥−1)sin(2𝜋𝑚𝑥)𝑑𝑥=0,₁₀(𝑥(1−𝑥))^𝜆−1/2𝐺^𝜆_2𝑛+1(2𝑥−1)cos(2𝜋𝑚𝑥)𝑑𝑥=0. (3.1)

We can extend the results of the Section 2 since there are also classical formulas which permit us to compute the remaining Fourier coefficients of such polynomials. The formulas are slightly different according to whether 𝜆 is of the form 2𝑗+1/2 or 2𝑗−1/2, with 𝑗∈ℕ, and whether the degree of the polynomial is odd or even. Thus we, present them separately.

3.1. The Case 2𝑗+1/2 and Even Degree

The cosine Fourier coefficients of the polynomial (𝑥(1−𝑥))^2𝑗𝐺^𝜆_2𝑛(2𝑥−1) are given by the following lemma.

Lemma 3.1. Let 𝜆=2𝑗+1/2, 𝑗∈ℕ∪{0}. Then for 𝑛,𝑚∈ℕ, one has¹₀(𝑥(1−𝑥))^2𝑗𝐺^𝜆_2𝑛()()=()2𝑥−1cos2𝜋𝑚𝑥𝑑𝑥−1^𝑗2^6𝑗√𝜋Γ(!2𝑗+1/2)4𝑗+2𝑛()!×2𝑛_{𝑛+𝑗𝑘=1}(−1)^𝑘−12^2𝑘−12𝑛+2𝑗+2𝑘−1!()!!12𝑘−12𝑛+2𝑗−2𝑘+1(𝜋𝑚)^2𝑘+2𝑗. (3.2)

Proof. We apply [10, formula 2.21.7-1, page 534] and [11, formula 10.1.8, page 437] in the same way as in Lemma 2.1.
This is enough to prove the next relation.

Theorem 3.2. Let 𝜆=2𝑗+(1/2), 𝑗∈ℕ∪{0}. For 𝑛∈ℕ, one has (𝑥(1−𝑥))^2𝑗𝐺^𝜆_2𝑛()=12𝑥−12^4𝑗−1√𝜋Γ()!2𝑗+1/24𝑗+2𝑛()!×2𝑛_{𝑛+𝑗𝑘=1}2𝑛+2𝑗+2𝑘−1!()!!!𝐵2𝑘−12𝑘+2𝑗2𝑛+2𝑗−2𝑘+1^2𝑘+2𝑗(𝑥). (3.3)

The same references cited in the last lemma contain the formulas necessary for the remaining cases, which we present without further comments.

3.2. The Case 2𝑗−1/2 and Even Degree

Lemma 3.3. Let 𝜆=2𝑗−1/2, 𝑗∈ℕ. Then for 𝑛,𝑚∈ℕ, one has ¹₀(𝑥(1−𝑥))^2𝑗−1𝐺^𝜆_2𝑛()()()2𝑥−1cos2𝜋𝑚𝑥𝑑𝑥=−1^𝑗2^6𝑗−3√𝜋Γ(!2𝑗−1/2)4𝑗+2𝑛−2()!×2𝑛_{𝑛+𝑗−1𝑘=0}(−1)^𝑘2^2𝑘2𝑛+2𝑗+2𝑘−1!()!!12𝑘2𝑛+2𝑗−2𝑘−1(𝜋𝑚)^2𝑘+2𝑗. (3.4)

Theorem 3.4. Let 𝜆=2𝑗−1/2, 𝑗∈ℕ. For 𝑛∈ℕ, one has (𝑥(1−𝑥))^2𝑗−1𝐺^𝜆_2𝑛()12𝑥−1=−2^4𝑗−3√𝜋Γ!2𝑗−1/24𝑗+2𝑛−2()!×2𝑛_{𝑛+𝑗−1𝑘=0}2𝑛+2𝑗+2𝑘−1!()!!!𝐵2𝑘2𝑘+2𝑗2𝑛+2𝑗−2𝑘−1^2𝑘+2𝑗(𝑥). (3.5)

3.3. The Case 2𝑗+1/2 and Odd Degree

Lemma 3.5. Let 𝜆=2𝑗+1/2, 𝑗∈ℕ∪{0}. Then for 𝑛,𝑚∈ℕ, one has ¹₀(𝑥(1−𝑥))^2𝑗𝐺^𝜆_2𝑛+1()()()2𝑥−1sin2𝜋𝑚𝑥𝑑𝑥=−1^𝑗+12^6𝑗√𝜋Γ!2𝑗+1/24𝑗+2𝑛+1()!×2𝑛+1_{𝑛+𝑗𝑘=0}(−1)^𝑘2^2𝑘2𝑛+2𝑗+2𝑘+1!()!!12𝑘2𝑛+2𝑗−2𝑘+1(𝜋𝑚)^{2𝑘+2𝑗+1}. (3.6)

Theorem 3.6. Let 𝜆=2𝑗+1/2, 𝑗∈ℕ∪{0}. For 𝑛∈ℕ, one has (𝑥(1−𝑥))^2𝑗𝐺^𝜆_2𝑛+1()=12𝑥−12^4𝑗−1√𝜋Γ!2𝑗+1/24𝑗+2𝑛+1()!×2𝑛+1_{𝑛+𝑗𝑘=0}2𝑛+2𝑗+2𝑘+1!()!!!𝐵2𝑘2𝑘+2𝑗+12𝑛+2𝑗−2𝑘+1^{2𝑘+2𝑗+1}(𝑥). (3.7)

3.4. The Case 2𝑗−1/2 and Odd Degree

Lemma 3.7. Let 𝜆=2𝑗−1/2, 𝑗∈ℕ. Then, for 𝑛,𝑚∈ℕ, one has ¹₀(𝑥(1−𝑥))^2𝑗−1𝐺^𝜆_2𝑛+1()()=()2𝑥−1sin2𝜋𝑚𝑥𝑑𝑥−1^𝑗2^6𝑗−3√𝜋Γ!2𝑗−1/24𝑗+2𝑛−1()!×2𝑛+1_{𝑛+𝑗𝑘=1}(−1)^𝑘−12^2𝑘−12𝑛+2𝑗+2𝑘−1!()!!12𝑘−12𝑛+2𝑗−2𝑘+1(𝜋𝑚)^{2𝑘+2𝑗−1}. (3.8)

Theorem 3.8. Let 𝜆=2𝑗−1/2, 𝑗∈ℕ. For 𝑛∈ℕ, one has (𝑥(1−𝑥))^2𝑗−1𝐺^𝜆_2𝑛+1(2𝑥−1)=−12^4𝑗−3√𝜋Γ!2𝑗−1/24𝑗+2𝑛−1()!×2𝑛+1_{𝑛+𝑗𝑘=1}2𝑛+2𝑗+2𝑘−1!()!!!𝐵2𝑘−12𝑘+2𝑗−12𝑛+2𝑗−2𝑘+1^{2𝑘+2𝑗−1}(𝑥). (3.9)

4. New Identities for Bernoulli Polynomials and Numbers

Using the results of the previous sections, we can transfer formulas for various orthogonal polynomials to the Bernoulli polynomials and vice versa. We are mainly interested in the first direction. For example, a simple (and very easy to prove) property which relates Legendre and Gegenbauer polynomials is 𝑑^𝑚𝑑𝑥^𝑚𝑃_𝑛(𝑥)=1⋅3⋯(2𝑚−1)𝐺^𝑚+1/2_𝑛−𝑚(𝑥),𝑚≤𝑛 (4.1) (see, for instance, [11, formula 22.5.37, page 779]). Now, if we differentiate in formulas (2.5) or (2.6) and use (4.1) and Theorems 3.2–3.8, we obtain some identities between Bernoulli polynomials. Let us carry out the details in the case of even Legendre polynomials and an even number of derivatives.

From (4.1) we have, for 𝑛, 𝑗∈ℕ, 𝑑^2𝑗𝑑𝑥^2𝑗𝑃_2𝑛+2𝑗(𝑥)=1⋅3⋯4𝑗−1𝐺^2𝑗+1/2_2𝑛(𝑥). (4.2) Differentiating (2.5) 2𝑗 times and applying (4.2), we get 2^2𝑗−1⋅1⋅3⋯4𝑗−1𝐺^2𝑗+1/2_2𝑛()𝑑2𝑥−1=2^2𝑗𝑑𝑥^2𝑗𝑃_2𝑛+2𝑗()=2𝑥−1_{𝑛+𝑗𝑘=𝑗}2𝑛+2𝑗+2𝑘−1!()!()!!𝑑2𝑘−12𝑘2𝑛+2𝑗−2𝑘+1𝑑𝑥^2𝑗2𝑗𝐵_2𝑘(𝑥). (4.3) We multiply both sides of (4.3) by 𝑥^2𝑗(1−𝑥)^2𝑗 and use (3.3) along with the (easily proved) formula 2^2𝑗−1√⋅1⋅3⋯4𝑗−1𝜋2^4𝑗−1Γ2𝑗+1/2=1,𝑗∈ℕ. (4.4) Next, we recall the expression for the derivative of a Bernoulli polynomial 𝑑^2𝑗𝑑𝑥^2𝑗𝐵_2𝑘(𝑥)=(2𝑘)!!𝐵2𝑘−2𝑗^{2𝑘−2𝑗}(𝑥),𝑘≥𝑗. (4.5) Denoting 𝛼=!𝑘,𝑗,𝑛2𝑛+2𝑗+2𝑘−1(2𝑘−1)!2𝑛+2𝑗−2𝑘+1!, (4.6) we have then proved the identity.

Proposition 4.1. Let 𝑛,𝑗∈ℕ. Then 2𝑛+4𝑗!(2𝑛)!^𝑛+𝑗_𝑘=1𝛼𝑘,𝑗,𝑛!𝐵2𝑘+2𝑗^2𝑘+2𝑗(𝑥)=^𝑛+𝑗_𝑘=𝑗𝛼𝑘,𝑗,𝑛!𝑥2𝑘−2𝑗^2𝑗(1−𝑥)^2𝑗𝐵_{2𝑘−2𝑗}(𝑥). (4.7)

In the same way, if we repeat the above argument with even polynomials and an odd number of derivatives (using now (2.6) and (3.5)), and we denote 𝛽=!𝑘,𝑗,𝑛2𝑛+2𝑗+2𝑘−1(2𝑘)!2𝑛+2𝑗−2𝑘−1!, (4.8) then we obtain.

Proposition 4.2. Let 𝑛,𝑗∈ℕ. Then 2𝑛+4𝑗−2!(2𝑛)!^{𝑛+𝑗−1}_𝑘=0𝛽𝑘,𝑗,𝑛!𝐵2𝑘+2𝑗^2𝑘+2𝑗(𝑥)=−^{𝑛+𝑗−1}_{𝑘=𝑗−1}𝛽𝑘,𝑗,𝑛!𝑥2𝑘−2𝑗+2^2𝑗−1(1−𝑥)^2𝑗−1𝐵_{2𝑘−2𝑗+2}(𝑥). (4.9)

By differentiating odd Legendre polynomials an even number of times and using the notation 𝛾=!𝑘,𝑗,𝑛2𝑛+2𝑗+2𝑘+1(2𝑘)!2𝑛+2𝑗−2𝑘+1!, (4.10) we get the following.

Proposition 4.3. Let 𝑛,𝑗∈ℕ. Then 2𝑛+4𝑗+1!(2𝑛+1)!^𝑛+𝑗_𝑘=0𝛾𝑘,𝑗,𝑛!𝐵2𝑘+2𝑗+1^{2𝑘+2𝑗+1}(𝑥)=^𝑛+𝑗_𝑘=𝑗𝛾𝑘,𝑗,𝑛!𝑥2𝑘−2𝑗+1^2𝑗(1−𝑥)^2𝑗𝐵_{2𝑘−2𝑗+1}(𝑥). (4.11)

Finally, with odd Legendre polynomials and an odd number of derivatives, we have the following proposition:

Proposition 4.4. Let 𝑛,𝑗∈ℕ. Then 2𝑛+4𝑗−1!(2𝑛+1)!^𝑛+𝑗_𝑘=1𝛼𝑘,𝑗,𝑛!𝐵2𝑘+2𝑗−1^{2𝑘+2𝑗−1}(𝑥)=−^𝑛+𝑗_𝑘=𝑗𝛼𝑘,𝑗,𝑛!𝑥2𝑘−2𝑗+1^2𝑗−1(1−𝑥)^2𝑗−1𝐵_{2𝑘−2𝑗+1}(𝑥). (4.12)

As a consequence of these results, we also obtain some identities for Bernoulli numbers, for instance.

Corollary 4.5. Let 𝑗,𝑛∈ℕ. Then ^𝑛+𝑗_𝑘=1𝐵2𝑛+4𝑗+12𝑘+2𝑗2𝑛+2𝑗+2𝑘−12𝑘−1^2𝑘+2𝑗=0,_{𝑛+𝑗𝑘=1}𝐵2𝑛+2𝑗2𝑘−12𝑛+2𝑗+2𝑘−12𝑘+2𝑗^2𝑘+2𝑗=0,_{𝑛+𝑗−1𝑘=0}𝐵2𝑛+4𝑗−12𝑘+2𝑗2𝑛+2𝑗+2𝑘−12𝑘^2𝑘+2𝑗=0,_{𝑛+𝑗−1𝑘=0}𝐵2𝑛+2𝑗−12𝑘2𝑛+2𝑗+2𝑘−12𝑘+2𝑗^2𝑘+2𝑗=0. (4.13)

Proof. Taking 𝑥=0 in Proposition 4.1, we obtain ^𝑛+𝑗_𝑘=1𝛼𝑘,𝑗,𝑛!𝐵2𝑘+2𝑗^2𝑘+2𝑗=0. (4.14) If we multiply both sides by the term (2𝑛+4𝑗+1)!/(2𝑛+2𝑗)!, which is independent of 𝑘, we obtain the first identity. If we multiply by (2𝑛+2𝑗)!/(2𝑛−1)!, we obtain the second one. The two remaining identities are obtained using Proposition 4.2 and a similar argument.

Remarks. We do not think that Propositions 4.1–4.4 can be easily obtained directly from formulas for the Bernoulli polynomials. For instance, there can be no direct relation between 𝑥^2𝑗(1−𝑥)^2𝑗𝐵_{2𝑘−2𝑗}(𝑥) and 𝐵_2𝑘+2𝑗(𝑥) because the even Bernoulli polynomials have no roots at neither 0 nor 1. For the same reasons, we do not see that formula (4.3) together with (2.5) directly imply (3.3). For this reason, we believe that our results above for Bernoulli and Gegenbauer polynomials are new. They are neither an easy consequence nor a generalization of the results in [8, 9].
Similarly, observe that Propositions 4.3 and 4.4 cannot be obtained directly from Propositions 4.1 and 4.2 by differentiation in a simple way, because of the presence of the factors 𝑥^𝑟(1−𝑥)^𝑟. Thus, we believe that these propositions give new identities for Bernoulli polynomials, which are essentially different for each 𝑗.
We have seen in the proof that the first two identities in the corollary are equivalent, in fact proportional. However, we do not see whether the last two properties are equivalent to the first ones. In any case, they remind us of other known formulas for Bernoulli numbers which can be obtained by methods involving hypergeometric functions. For instance, the Gessel-Viennot identity (see [2]), of which the following is a special case: 4𝑚−2𝑘+2𝐵^𝑚_𝑘=02𝑚+22𝑛+12𝑘+1^{2𝑛−2𝑘}=2𝑚−𝑘+12𝑛+14𝑚−2𝑛+34𝑚−2𝑛+32𝑚+2,𝑚≤𝑛<2𝑚+1 (4.15) (see also [3] and the references there). We have not gotten around to deducing some of our formulas from others such as these. Anyway, our proofs via Fourier series are still much easier than those based on hypergeometric functions.

5. A Remark on Euler Polynomials

The Euler polynomials 𝐸_𝑛(𝑥), which are defined by means of the generating function 2𝑒^𝑡𝑥𝑒^𝑡=+1^∞_𝑛=0𝐸_𝑛(𝑥)𝑡^𝑛,𝑛! (5.1) also have readily handled expressions for their Fourier coefficients. Indeed, from (5.1) we can easily get the properties 𝐸_𝑛(1−𝑥)=(−1)^𝑛𝐸_𝑛(𝑥),𝐸^′_𝑛(𝑥)=𝑛𝐸_𝑛−1(𝑥). (5.2) Now, from (5.2) the proof of the following lemma is straightforward.

Lemma 5.1. Let 𝑛,𝑚∈ℕ. Then one has the following:(i)∫¹₀𝐸_2𝑛(𝑥)sin(2𝜋𝑚𝑥)𝑑𝑥=0, (ii)∫¹₀𝐸_2𝑛(𝑥)𝑑𝑥=−(2/2𝑛+1)𝐸_2𝑛+1(0), (iii)∫¹₀𝐸_2𝑛(𝑥)cos(2𝜋𝑚𝑥)𝑑𝑥=2∑^𝑛_𝑘=1(−1)^𝑘𝐸_{2𝑛−2𝑘+1}(0)((2𝑛)!/(2𝑛−2𝑘+1)!)(1/(2𝜋𝑚)^2𝑘).

Consequently, by uniqueness of the Fourier series, we obtain the following theorem.

Theorem 5.2. Let 𝑛∈ℕ. Then 𝐸_2𝑛(𝑥)+2𝐸2𝑛+1^2𝑛+1(0)=−2^𝑛_𝑘=1𝐸_{2𝑛−2𝑘+1}(0)(2𝑛)!()!()!𝐵2𝑛−2𝑘+12𝑘^2𝑘(𝑥) (5.3) or, equivalently, 𝐸_2𝑛(𝑥)=−2^𝑛_𝑘=0𝐸_{2𝑛−2𝑘+1}(0)(2𝑛)!()!()!𝐵2𝑛−2𝑘+12𝑘^2𝑘(𝑥). (5.4)

We proceed in the same way for the odd Euler polynomials.

Lemma 5.3. Let 𝑛,𝑚∈ℕ. Then one has the following: (i)∫¹₀𝐸_2𝑛−1(𝑥)cos(2𝜋𝑚𝑥)𝑑𝑥=0, (ii)∫¹₀𝐸_2𝑛−1(𝑥)𝑑𝑥=0, (iii)∫¹₀𝐸_2𝑛(𝑥)cos(2𝜋𝑚𝑥)𝑑𝑥=2∑^𝑛−1_𝑘=0(−1)^𝑘𝐸_{2𝑛−2𝑘−1}(0)((2𝑛−1)!/(2𝑛−2𝑘−1)!)(1/(2𝜋𝑚)^2𝑘+1).

Theorem 5.4. For 𝑛∈ℕ, one has 𝐸_2𝑛−1(𝑥)=−2^𝑛−1_𝑘=0𝐸_{2𝑛−2𝑘−1}(0)(2𝑛−1)!()!()!𝐵2𝑛−2𝑘−12𝑘+1^2𝑘+1(𝑥). (5.5)

If we use the identities 𝐸_𝑛(0)2=−2𝑛+1^𝑛+1−1𝐵_𝑛+1,𝑛∈ℕ, (5.6) (which have immediate proofs) and we put together the even and odd cases, we obtain the following result.

Corollary 5.5. Let 𝑛∈ℕ and 𝜎∈{0,1}. Then 𝐸_2𝑛−𝜎(𝑥)=4^𝑛−𝜎_𝑘=0(2𝑛−𝜎)!()!()!22𝑛−2𝑘+2−2𝜎2𝑘+𝜎^{2𝑛−2𝑘+2−2𝜎}−1𝐵_{2𝑛−2𝑘+2−2𝜎}𝐵_2𝑘+𝜎(𝑥). (5.7)

This formula is also well known (see, e.g., [11, formula 23.1.28, page 806]). However, we have obtained it in a very simple and natural way.

Acknowledgment

Research of the second and third authors supported by grant MTM2009-12740-C03-03 of the DGI.