Lecture Notes, Bernoulli Polynomials and Applications

Home , Bernoulli polynomials

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1. Introduction

There are many ways to introduce Bernoulli polynomials and numbers. We opted for the algebraic approach relying on the difference operator. But first, let us introduce some notation. Let the real vector space of polynomials with real coefficients be denoted by R[X]. For a nonnegative integer n, let Rn[X] be the subspace of R[X] consisting of polynomials of degree smaller or equal to n. If P is a polynomial from R[X], we define ∆P def= P (X + 1) P (X), and we denote by ∆ the linear operator, defined on R[X], by P ∆P . − 7→ Lemma 1.1. The linear operator Φ defined by 1 Φ: R[X] R[X] R, P ∆P, P (t) dt (1.1) −→ × 7→ Z0 is bijective. 1 Proof. Consider P ker Φ, then P ker ∆ and 0 P (t) dt = 0. Now, if we consider Q(X)= P (X) P (0), then clearly we have∈ ∈ − R Q(X +1)= P (X + 1) P (0) = P (X) P (0) = Q(X) − − This implies by induction that Q(n) = 0 for every nonnegative integer n, so Q = 0, since it has infinitely 1 many zeros. Thus, P (X)= P (0), but we have also 0 P (t) dt = 0, so P (0) = 0, and consequently P = 0. This proves that Φ is injective. R Clearly, for a nonnegative integer n we have deg ∆(Xn+1)= n. Thus P R [X] = ∆P R [X] ∈ n+1 ⇒ ∈ n Therefore, n N, Φ(R [X]) R [X] R. ∀ ∈ n+1 ⊂ n × But the fact that Φ is injective implies that dim Φ(R [X]) = dim R [X] = 1 + dim R [X] = dim(R [X] R), n+1 n+1 n n × and consequently n N, Φ(R [X]) = R [X] R ∀ ∈ n+1 n × This, proves that Φ is surjective, and the lemma follows.

n 1 Let us consider the basis = (en)n N of R[X] R deﬁned by e0 = (0, 1) and en = (nX − , 0) for E ∈ × n N∗. We can deﬁne the Bernoulli polynomials, In terms of this basis and of the isomorphism Φ of Lemma∈ 1.1 as follows:

Definition 1.2. The sequence of Bernoulli polynomials (Bn)n N is defined by ∈ 1 B =Φ− (e ) for n 0. n n ≥ According to Lemma 1.1, this definition takes a more practical form as follows :

Corollary 1.3. The sequence of Bernoulli polynomials (Bn)n N is uniquely deﬁned by the conditions: ∈ 1 B (X)=1. 0 n 1 2 n N∗, B (X + 1) B (X)= nX − . (1.2) ∀ ∈ n − n 1 3 n N∗, B (t) dt =0. ∀ ∈ n Z0 For instance, it is straightforward to see that 1 1 B (X)= X , and B (X)= X2 X + . 1 − 2 2 − 6 BERNOULLI POLYNOMIALS AND APPLICATIONS 3

2. Properties of Bernoulli polynomials

In the next proposition, we summarize some simple properties of Bernoulli polynomials :

Proposition 2.1. The sequence of Bernoulli polynomials (Bn)n N satisfies the following properties: ∈ i. For every positive integer n we have Bn′ (X)= nBn 1(X). − n ii. For every positive integer n we have Bn(1 X) = ( 1) Bn(X). iii. For every nonnegative integer n and every− positive integer− p we have p 1 1 − X + k 1 B = B (X). p n p pn n Xk=0 1 Proof. (i ) Consider the sequence of polynomials (Qn)n N defined by Qn = Bn′ +1. It is straightforward ∈ n+1 to see that Q (X)=1 and for n 1 : 0 ≥ 1 1 n n 1 ∆Q = (∆B )′ = ((n + 1)X )′ = nX − n n +1 n+1 n +1 and 1 1 1 ∆Bn+1(0) Q (t) dt = B′ (t) dt = =0. n n +1 n+1 n +1 Z0 Z0 This proves that the sequence of (Qn)n N satisfies the conditions 1 , 2 and 3 of Corollary 1.3, and (i ) ∈ follows because of the unicity assertion. n (ii) Consider again the sequence (Qn)n N defined by Qn(X) = ( 1) Bn(1 X). Clearly Q0(X) = 1 and for n 1 : ∈ − − ≥ n n 1 ∆Q = ( 1) (B ( X) B (1 X)) = ( 1) − ∆B ( X) n − n − − n − − n − n 1 n 1 n 1 = ( 1) − n( X) − = nX − . − − 1 n 1 Moreover, for n 1, Qn(t) dt = ( 1) Bn(1 t) dt = 0. This proves that the sequence of (Qn)n N ≥ 0 − 0 − ∈ satisfies the conditions 1 , 2 and 3 of Corollary 1.3, and (ii ) follows from the unicity assertion. R R (iii) Similarly, consider the sequence of polynomials (Qn)n N defined by ∈ p 1 − n 1 X + k Q (X)= p − B n n p Xk=0 Clearly Q (X)=1 and for n 1 : 0 ≥ p 1 p 1 − − n 1 X + k +1 n 1 X + k ∆Q = p − B p − B n n p − n p Xk=0 Xk=0 p p 1 − n 1 X + k X + k = p − Bn Bn p − p ! Xk=1 Xk=0 n 1 X + p X n 1 1 n 1 = p − B B = p − ∆B X = nX − . n p − n p n p Moreover, for n 1, ≥ p 1 1 − 1 n 1 t + k Qn(t)dt = p − Bn dt 0 0 p Z kX=0 Z p 1 − (k+1)/p 1 n 1 n 1 = p − Bn(t) dt = p − Bn(t) dt =0. k/p 0 kX=0 Z Z This proves that the sequence of (Qn)n N satisfies the conditions 1 , 2 and 3 of Corollary 1.3, and (iii ) ∈ follows by unicity. The proof of Proposition 2.1 is complete. 4 OMRAN KOUBA

Deﬁnition 2.2. The sequence of Bernoulli Numbers (bn)n N is deﬁned by ∈ b = B (0) for n 0. n n ≥ The following proposition summarizes some properties of Bernoulli numbers.

Proposition 2.3. The sequence of Bernoulli numbers (bn)n N satisﬁes the following properties: ∈ i. For every positive integer n we have b2n+1 =0 and b2n = B2n(1). 1 1 n ii. For every nonnegative integer n we have Bn 2 = (2 − 1)bn. iii. For every nonnegative integer n we have − n n k Bn(X)= bn kX . k − kX=0 iv. For every positive integer n we have

n 1 1 − n +1 b = b . n −n +1 k k Xk=0 Proof. (i) Using Proposition 2.1(i ) we have, for n 2 : ≥ 1 1 Bn(1) Bn(0) = Bn′ (t) dt = n Bn 1(t)dt =0 − − Z0 Z0 n and according to Proposition 2.1(ii ) we have Bn(1) = ( 1) Bn(0) for every n 1. This proves that b = 0 if n is an odd integer greater than 2, and that B −(1) = b for n 1. This≥ is (i ). n 2n 2n ≥ (ii) According to Proposition 2.1(iii ) with p = 2 we see that, for every nonnegative integer n we have

X X +1 1 n B + B =2 − B (X) n 2 n 2 n Substituting X = 0 we obtain (ii ). (iii) Consider n N, according Now, we use again Proposition 2.1(i ) to conclude that ∈ (k) Bn = n(n 1) ... (n k + 1)Bn k, for 0 < k n. − − − ≤ It follows that (k) Bn (Y ) n = Bn k(Y ), for 0 < k n k! k − ≤ and using Taylor’s formula for polynomials we conclude that n n k Bn(X + Y )= Bn k(Y )X k − Xk=0 Finally, substituting Y = 0, we obtain (iii ).

(iv) Let n be an integer such that n 2. We have shown that Bn(1) = bn, and using the preceding point n n ≥ n n n 1 n we conclude that Bn(1) = bn k, that is bn = bn k or − bk =0. So k=0 k − k=0 k − k=0 k P n nP+1 P n 1, b =0 ∀ ≥ k k Xk=0 which is equivalent to (iv ). BERNOULLI POLYNOMIALS AND APPLICATIONS 5

Application 1. In Proposition 2.3 (iii ), Bernoulli polynomials are expressed in terms of the canonical k basis (X )k N of R[X]. In fact, we have proved that ∈ n n k Bn(X + Y )= Bn k(X)Y (2.1) k − Xk=0 k and this can be used to, conversely, express the canonical basis (X )k N of R[X] in terms of Bernoulli polynomials. ∈ Indeed, substituting Y = 1 in (2.1) we obtain

n+1 n +1 Bn+1(X +1)= Bn+1 k(X), k − Xk=0 n but according to Corollary 1.3 we have also Bn+1(X +1)= Bn+1(X) + (n + 1)X . Thus n+1 n n n +1 n +1 (n + 1)X = Bn+1 k(X)= Bk(X) k − k Xk=1 Xk=0 Finally, 1 n n +1 Xn = B (X) n +1 k k Xk=0

Remark 2.4. Using the recurrence relation of Proposition 2.3 (iv ) we can determine the sequence of 1 Bernoulli Numbers. In particular, b1 = 2 . We find in Table 1 the the values of the first six Bernoulli numbers with even indices. Also we find− in Table 2 the list of the first six Bernoulli Polynomials.

n 0 1 2 3 4 5 b 1 1 1 1 1 5 2n 6 − 30 42 − 30 66

Table 1. The ﬁrst Bernoulli Numbers with even indces.

B0(X) =1 B (X) = X 1 1 − 2 B (X) = X2 X + 1 2 − 6 B (X) = X3 3 X2 + 1 X = X X 1 (X 1) 3 − 2 2 − 2 − 4 3 2 1 B4(X) = X 2X + X − − 30 B (X) = X5 5 X4 + 5 X3 1 X = B (X) X2 X 1 5 − 2 3 − 6 3 − − 3 6 5 5 4 1 2 1 B6(X) = X 3X + X X + − 2 − 2 42

Table 2. The ﬁrst Bernoulli Polynomials. 6 OMRAN KOUBA

Application 2. For a nonnegative integer n and a positive integer m, we deﬁne Sn(m) to be the sum m S (m)=1n +2n + + mn = kn. n ··· kX=1 Noting that 1 kn = (B (k + 1) B (k)) n +1 n+1 − n+1 we see that 1 S (m)= (B (m + 1) B (0)) . n n +1 n+1 − n+1 In Table 3 we have listed the ﬁrst values of these sums using the results from Table 2.

m(m + 1) 1 +2 + + m = ··· 2 m(m + 1)(2m + 1) 12 +22 + + m2 = ··· 6 m2(m + 1)2 13 +23 + + m3 = ··· 4 m(m + 1)(2m + 1)(3m2 + m 1) 14 +24 + + m4 = − ··· 30 Table 3. The sum of consecutive powers.

It was on studying these sums that Jacob Bernoulli introduced the numbers named after him.

. / ?.

. ./ +?/ . . . / +3

. + /

Figure 1. The graphs of B and B on [0, 1]. 1 − 2

It is clear that x B (x) = x 1 is negative on 0, 1 and positive on 1 , 1 . It follows that 7→ 1 − 2 2 2 x B (x)= x2 + x 1 is increasing on the interval 0, 1 , and decreasing on the interval 1 , 1 , and 7→ − 2 − − 6 2 2 since it has opposit signs at 0 and 1 we conclude that B vanishes exactly once on the interval 0, 1 and 2 2 2 exactly once on the interval 1 , 1 . 2 These results can be generalized as follows: BERNOULLI POLYNOMIALS AND APPLICATIONS 7

Proposition 2.5. For every positive integer n we have : The function x ( 1)nB (x) is increasing on 0, 1 , and decreasing on 1 , 1 , and conse- Pn 7→ − 2n 2 2 quently, it vanishes exactly once on each of the intervals 0, 1 and 1 , 1 . 2 2 : The function x ( 1)nB (x) is negative on 0, 1 , and positive on 1 , 1 . Moreover, 0, Qn 7→ − 2n+1 2 2 1 and 1 are simple zeros of B in the interval [0, 1]. 2 2n+1 Proof. We have already proved . P1 n n n n. Let f(x) = ( 1) B2n+1(x), then we have f ′(x)=(2n + 1)( 1) B2n(x), and according to n, P ⇒Q −1 1 − P there exists an α in 0, 2 and a β in 2 , 1 such that f ′ is negative on each of the intervals (0, α) and (β, 1), and positive on the interval (α, β). Therefore, f has the following table of variations: 1 x 0 α 2 β 1

f ′(x) 0 + 0 − − f(x) 0 ⌣ 0 ⌢ 0 ց ր ր ց where we used Proposition 2.1 (ii ) and Proposition 2.3 (i ) and (ii ) to conclude that 1 B (0) = B = B (1) = 0. 2n+1 2n+1 2 2n+1 1 1 Moreover, n implies that f ′ does not vanish at any of the points 0, 2 and 1. This proves that 0, 2 and 1 are the onlyP zeros of f in the interval [0, 1] and that they are simple. follows immediately. Qn n+1 n n n+1. Let f(x) = ( 1) B2n+2(x), then we have f ′(x)= (2n+2)( 1) B2n+1(x), and according Q ⇒P − 1 1 − − to n, the derivative f ′ is positive on 0, 2 and negative on 2 , 1 . Therefore, f has the following table of variations:Q 1 x 0 2 1

f ′(x) 0+ 0 0 − 1 2n f(x) A (1 2− − )A A ր − − ց with A = ( 1)n+1b . Clearly A = 0 because f is increasing on 0, 1 . Consequently, f(0)f 1 < 0 − 2n+2 6 2 2 and f(1)f 1 < 0. Thus, f vanishes exactly once on each of intervals 0, 1 and 1 , 1 , and that, it is 2 2 2 increasing on 0, 1 and decreasing 1 , 1 . This proves , and achieves the proof of the proposition. 2 2 Pn+1

. +?/ ./ +?0 . . . +0- / ?1

. +3

Figure 2. Illustration of Proposition 2.5

Remark 2.6. It follows from the preceding proof that ( 1)n+1b > 0 for every positive integer n. − 2n 8 OMRAN KOUBA

Corollary 2.7. For every positive integer n we have 2n +1 sup B2n(x) = b2n and sup B2n+1(x) b2n x [0,1] | | | | x [0,1] | |≤ 4 | | ∈ ∈ Proof. In fact, we conclude from Proposition 2.5 that

1 sup B2n(x) = max B2n(0) , B2n x [0,1] | | | | 2 ∈

1 = max b , 1 b = b | 2n| − 22n 1 | 2n| | 2n| −

In order to show the second inequality we consider several cases: If x 0, 1 , then we have • ∈ 4 x B (x)= B (x) B (0) = B (t) dt 2n+1 2n+1 − 2n+1 2n Z0 So x B (x) (2n + 1) B (t) dt | 2n+1 |≤ | 2n | Z0 2n +1 (2n + 1) b x b ≤ | 2n| ≤ 4 | 2n| If x 1 , 1 , then we have • ∈ 4 2 1 x B (x)= B (x) B = (2n + 1)B (t) dt 2n+1 2n+1 − 2n+1 2 2n Z1/2 Thus 1/2 B (x) (2n + 1) B (t) dt | 2n+1 |≤ | 2n | Zx 1 2n +1 (2n + 1) b x b ≤ | 2n| 2 − ≤ 4 | 2n| 1 Finally, when x 2 , 1 , we recall that B2n+1(x)= B2n+1(1 x) according to Proposition 2.1 • (ii ). Thus, in this∈ case we have also − − 2n +1 B2n+1(x) sup B2n+1(t) b2n | |≤ 0 t 1 | |≤ 4 | | ≤ ≤ 2 and the second part of the proposition follows.

Remark 2.8. We will show later in these notes that 2n +1 sup B2n+1(x) b2n x [0,1] | |≤ 2π | | ∈ which is, asymptotically, the best possible result. That is, we have also 2π lim sup B2n+1(x) =1 n (2n + 1) b2n · x [0,1] | | →∞ | | ∈ BERNOULLI POLYNOMIALS AND APPLICATIONS 9

3. Fourier series and Bernoulli polynomials

Extending periodically the restriction B of the Bernoulli polynomial B to the interval [0, 1), n|[0,1) n we obtain a 1-periodic piecewise continuous function denoted by Bn. In fact, for every real x we have B (x) = B ( x ) where t = t t is the fractional part of the real t. In Figure 3 the graphs of the n n { } { } −⌊ ⌋ functions B1, B2 and B3 are depicted. e e e e e v . / á ?. á ?0 u á . ?/ */

Figure 3. The graphs of B1, B2 and B3

e e e In this section we consider the Fourier series expansion of these periodic functions.

Proposition 3.1. i. For every x (0, 1), we have ∈ 1 ∞ sin(2πkx) B (x)= 1 −π k kX=1 ii. For every positive integer n, and every x [0, 1], we have ∈ 2(2n)! ∞ cos(2πkx) B (x) = ( 1)n+1 2n − (2π)2n k2n kX=1 2(2n + 1)! ∞ sin(2πkx) B (x) = ( 1)n+1 2n+1 − (2π)2n+1 k2n+1 kX=1

Proof. First, let us consider the case of B1. It is clear that 1 Ce0(B1)= B1(t)dt =0. Z0 and for k = 0 we have e 6 1 1 2iπkt 1 2iπkt C (B )= B (t)e− dt = t e− dt k 1 1 − 2 Z0 Z0 2iπkt 1 1 e 1 e− 1 2iπkt i = t + e− dt = − − 2 2iπk 2iπk 2πk 0 Z0 Thus, according to Dirichlet’s theorem [14, Corollary 3.3.9] we conclude that for x R Z we have ∈ \ i 1 ∞ sin(2πkx) B (x)= e2iπkx = 1 2πk −π k k Z 0 k=1 ∈X\{ } X e and (i ) follows. 10 OMRAN KOUBA

Let us now consider the case of B , for n 2. According to Corollary 1.3 we have n ≥ 1 e C0(Bn)= Bn(t)dt =0. Z0 and for k = 0 we ﬁnd that e 6 1 2iπkt Ck(Bn)= Bn(t)e− dt Z0 2iπkt 1 1 e e− 1 2iπkt = B (t) + B′ (t)e− dt − n 2iπk 2iπk n 0 Z0 1 Bn(0) Bn(1) n 2iπkt = − + Bn 1(t)e− dt 2iπk 2iπk 0 − n Z = Ck(Bn 1) 2iπk − where we used Corollary 1.3 and Propositione 2.1 (i ). This allows us to prove by induction on n that, n! n N∗, k Z 0 , C (B )= ∀ ∈ ∀ ∈ \{ } k n −(2iπk)n Thus, because of the continuity of B for n 2, and of thee uniform convergence of the Fourier series of n ≥ B in this case, we conclude that [20, Ch. I, Sec. 3], for x R we have n ∈ e 2iπkx n 2iπkx n! n! ∞ e + ( 1) e− e B (x)= e2iπkx = − n − (2iπk)n −(2iπ)n kn k Z 0 k=1 ∈X\{ } X e and (ii ) follows by considering separately the cases of n even and n odd.

In particular, we have the following well-known result: Corollary 3.2. For n 1 we have ≥ ∞ 2n def 1 n 1 (2π) ζ(2n) = = ( 1) − b k2n − 2 (2n)! 2n kX=1 · k 1 2n 2n 2n def ∞ ( 1) − n (2π) 1 n 1 (2 2)π η(2n) = − = ( 1) B = ( 1) − − b k2n − 2 (2n)! 2n 2 − 2 (2n)! 2n kX=1 · · (Here ζ( ) is the well-know “Riemann Zeta function,”.) · Using Bessel-Parseval’s identity [20, Ch. I, Sec. 5] we obtain the following corollary :

Corollary 3.3. If n and m are positive integers then 1 m 1 ( 1) − Bn(x)Bm(x) dx = −n+m bn+m Z0 n In particular, for n N∗ we have ∈ 1 2 Bn(x) n 1 b2n b2n dx = ( 1) − = | | n! − (2n)! (2n)! Z0 Proof. Indeed, if n + m 1 mod 2 then the change of variables x 1 x proves, using Proposi- ≡ 1 ← − tion 2.1 (ii ), that the considered integral 0 Bn(x)Bm(x) dx equals 0, and the conclusion follows from Proposition 2.3 (i ). R BERNOULLI POLYNOMIALS AND APPLICATIONS 11

So, let us suppose that n m mod 2. In this case, by Bessel-Parseval’s identity, we have ≡ 1 Bn(x)Bm(x) dx = Ck(Bn)Ck(Bm) 0 k Z Z X∈ en! m! e 1 = ( 1)m − (2iπ)n+m kn+m k Z 0 ∈X\{ } and the desired conclusion follows by Corollary 3.2.

Application 3. One more formula for Bernoulli polynomials. Let us consider the polynomial Tn deﬁned by 1 n Tn(X)= Bk(X)Bn k(X). n +1 − kX=0 Clearly, for n 1, we have ≥ n n 1 − (n + 1)Tn′ (X)= kBk 1(X)Bn k(X)+ (n k)Bk(X)Bn k 1(X) − − − − − Xk=1 Xk=0 n 1 n 1 − − = (k + 1)Bk(X)Bn k 1(X)+ (n k)Bk(X)Bn k 1(X) − − − − − Xk=0 Xk=0 n 1 − n +1 = (n + 1) Bk(X)Bn 1 k(X)= Tn 1 − − n − Xk=0 (n) That is Tn′ = nTn 1. Now, since Tn is a polynomial of degree n there are (λk )0 k n such that n (n)− ≤ ≤ Tn(X)= λ Bk, and from Tn′ = nTn 1 we conclude that k=0 k − n 1 n 1 P − (n) − (n 1) (k + 1)λk+1Bk = nλk − Bk. kX=0 kX=0 (n) (n 1) (n) n (n k) − − Thus (k + 1)λk+1 = nλk for 0 k n 1. It follows that λk = k λ0 for 0 k n. So, we have proved that ≤ ≤ − ≤ ≤ n n 1 n (k) Bk(X)Bn k(X)= λ0 Bn k n +1 − k − Xk=0 Xk=0 Integrating on [0, 1], and using Corollaries 1.3 and 3.3 , we obtain λ(0) = 1, λ(1) = 0, and for n 2: 0 0 ≥ n 1 n 1 k 1 (n) 1 bn − ( 1) − λ0 = Bk(t)Bn k(t) dt = − − n n +1 0 n +1 k kX=0 Z Xk=1 n 1 1 n 1 − k 1 k!(n k)! − n k k = bn ( 1) − − = bn t − (t 1) dt − (n + 1)! − 0 − ! kX=1 Z Xk=1 1 1 + ( 1)n = b tn+1 (t 1)n+1 tn (t 1)n dt = − b . − n − − − − − (n + 2)(n + 1) n Z0 (2m+1) (2m) 1 That is λ0 = 0 and λ0 = (m+1)(2m+1) b2m. Therefore, we have proved that

n n/2 1 ⌊ ⌋ n b2k Bk(X)Bn k(X)= Bn 2k(X). n +1 − 2k (k + 1)(2k + 1) − Xk=0 kX=0 12 OMRAN KOUBA

In particular, taking n =2m and x = 0 we ﬁnd, for m = 1, that 6 m m 1 2m +2 b2kb2m 2k = b2kb2m 2k. − m +1 2k +2 − Xk=0 Xk=0 or equivalently, for m = 2 : 6 m m 1 2m b2(k 1)b2(m k) = b2(k 1)b2(m k). − − m 2k − − Xk=1 Xk=1 This is an unusual formula since it combines both convolution and binomial convolution.

4. Bernoulli polynomials on the unit interval [0, 1]

Our ﬁrst result concerns the sequence of Bernoulli numbers and it follows immediately from Corollary 3.2:

Proposition 4.1. The sequence of Bernoulli numbers (b2n)n 1 satisﬁes the following: ≥ 3 (2n)! (2n)! i. For every positive integer n, we have b < 2 1+ < 4 . | 2n| 22n (2π)2n (2π)2n (2n)! ii. Asymptotically, for n in the neighborhood of + , we have b2n + 2 . ∞ | |∼ ∞ (2π)2n

2n 2n Proof. Noting that, for t [k 1, k], we have k− t− we conclude that ∈ − ≤ k ∞ 1 ∞ dt ∞ dt 2 1 2n 2n = 2n = 2n k ≤ k 1 t 2 t 2n 1 · 2 kX=3 Xk=3 Z − Z − Thus, for every n 1 we have ≥ ∞ 1 1 2 1 1 < ζ(2n)= 1+ + k2n ≤ 22n 2n 1 · 22n Xk=1 − or, equivalently 2n +1 1 3 3 1 < ζ(2n) 1+ 1+ 1+ < 2 ≤ 2n 1 · 22n ≤ 22n ≤ 4 − Hence, we have proved that 1 < ζ(2n) < 2 for every n 1 and that lim ζ(2n) = 1. This implies the ≥ n desired conclusion using Corollary 3.2. →∞

Remark 4.2. Using Stirling’s Formula [1, pp. 257] we see that for large n we have n 2n b2n + 4√πn | |∼ ∞ eπ

It is clear, according to Proposition 2.5 that the function x B2n(x) attains its maximum on the interval [0, 1] at x = 0. Thus, for n 1 we have 7→ | | ≥ sup B2n(x) = B2n(0) = b2n x [0,1] | | | | | | ∈

Determining the maximum of x B2n+1(x) on the interval [0, 1] is more diﬃcult. In this regard, we have the following result. 7→ | | BERNOULLI POLYNOMIALS AND APPLICATIONS 13

Proposition 4.3. For every positive integer n we have 2n +1 i. sup B2n+1(x) < b2n . x [0,1] | | 2π | | ∈ 1 4 2n +1 ii. B 1 b . 2n+1 4 ≥ − 22n 2π | 2n|

Proof. In fact, using Proposition 3.1 (ii ) we see that

2(2n + 1)! ∞ 1 2(2n + 1)! ∞ 1 sup B2n+1(x) 2n+1 2n+1 < 2n+1 2n . x [0,1] | |≤ (2π) · k (2π) · k ∈ kX=1 Xk=1 Thus, according to Corollary 3.2 we conclude that 2(2n + 1)! (2π)2n 2n +1 sup B2n+1(x) < 2n+1 b2n = b2n . x [0,1] | | (2π) · 2 (2n)! | | 2π | | ∈ · which is (i ). On the other hand, for n N, using Proposition 3.1 (ii ) once more, we obtain ∈ 1 2(2n + 1)! ∞ ( 1)k B = ( 1)n+1 − , 2n+1 4 − (2π)2n+1 (2k + 1)2n+1 Xk=0 but the series above is alternating, so ∞ ( 1)k 1 − > 1 (2k + 1)2n+1 − 32n+1 kX=0 Thus 1 1 2(2n + 1)! ( 1)n+1B > 1 . − 2n+1 4 − 32n+1 (2π)2n+1 Now, using Proposition 4.1 (i ) we see that 2n 1 1 1 3− − 2n +1 ( 1)n+1B > − b , − 2n+1 4 1+3 2 2n 2π | 2n| · − and since 2n 1 1 3− − 4 n N∗, − 2n 1 2n ∀ ∈ 1+3 2− ≥ − 2 we obtain (ii ). ·

Remark 4.4. Combining Corollary 2.7 and Propositions 4.1 and 4.3 we see that, for large n we have 2 n! sup Bn(x) + · n . x [0,1] | |∼ ∞ (2π) ∈

Next, we will study the behavior of the unique zero of B2n in the interval (0, 1/2).

Proposition 4.5. For a positive integer n, let αn be the unique zero of B2n that belongs to the interval (0, 1/2). Then the sequence (αn)n 1 satisﬁes the following inequality: ≥ 1 1 1 < α < α < . 4 − π 22n n n+1 4 · Proof. First, note that α = 1 1 so 1 1 1 < α < 1 . On the other hand, since α (1 α )= 1 1 2 − 2√3 4 − π 1 4 1 − 1 6 and B (X)= X2(1 X)2 1 we conclude that 4 − − 30 1 1 269 B (α )= < 0 and B = > 0. 4 1 −180 4 4 7680 This proves the desired inequality for n = 1. 14 OMRAN KOUBA

Now, let us suppose that n 2. Using Proposition 2.1 (iii ) with p = 4 and X = 0 we obtain ≥ 1 1 3 1 B (0) + B + B + B = B (0). 2n 2n 4 2n 2 2n 4 42n 1 2n − 1 3 But, according to Proposition 2.1 (ii ), B2n 4 = B2n 4 , and using Proposition 2.3 (ii ), we have also 1 1 2n B = (2 − 1)b . Hence 2n 2 − 2n 1 1 1 2n 1 1 B = 2 − 1 b = B . 2n 4 22n − 2n 22n 2n 2 1 1 1 1 This proves that B2n 4 B2n 2 > 0 and B2n 4 B2n(0) < 0. It follows that 0 < αn < 4 . Now, let us deﬁne the function hn by ∞ cos(2πkx) h (x)= . n k2n kX=1 1 1 Also, let xn = 4 π 22n . Clearly we have − · π 2 ∞ cos(2πkx ) h (x ) = cos + n n n 2 − 22n k2n Xk=2 2 ∞ 1 2 2n +1 1 sin > sin ≥ 22n − k2n 22n − 2n 1 · 22n Xk=2 − 2n+1 2n where we used the inequality ζ(2n) 1 < 2n 1 2− from Proposition 4.1. − 3 x − Finally, using the inequality sin x x 6 which is valid for x 0, and recalling that n 2 we conclude that ≥ − ≥ ≥ 1 2n +1 4 h (x ) > 2 n n 22n − 2n 1 − 3 24n − · 1 2n 3 4 1 1 h (x ) > − 1 > 0 n n 22n 2n 1 − 3 24n ≥ 3 22n − 26 − · · This proves, according to Proposition 3.1 (ii ), that B2n(xn)B2n(0) > 0 and consequently xn < αn.

Next, let us show that hn+1(αn) > 0, because this implies, according to Proposition 3.1 (ii ), that B2n+2(αn)B2n+2(0) > 0 and consequently αn < αn+1. First, on one hand, we have

∞ cos(2πkα ) ∞ cos(2πkα ) 0= h (α )= n = cos(2πα )+ n n n k2n n k2n Xk=1 kX=2 and on the other ∞ cos(2πkα ) ∞ cos(2πkα ) h (α )= n = cos(2πα )+ n . n+1 n k2n+2 n k2n+2 kX=1 Xk=2 Hence ∞ 1 cos(2πkα ) h (α )= 1 n n+1 n − − k2 k2n Xk=2 3 cos(4πα ) ∞ 1 cos(2πkα ) = n 1 n . −4 · 22n − − k2 k2n Xk=3 1 1 1 4 But, from 4 π 22n < αn < 4 we conclude that π 22n < 4παn < π, and consequently − · − 8 4 4 1 1 2 sin2 = cos < cos(4πα ), − 24n ≤ − 22n 22n − n BERNOULLI POLYNOMIALS AND APPLICATIONS 15

1 2 ∞ Thus, using the estimate k=3 k2n < 3 22n obtained on the occasion of proving Proposition 4.1, we get · P 3 6 ∞ 1 1 h (α ) 1 n+1 n ≥ 22n+2 − 26n − − k2 k2n Xk=3 3 6 ∞ 1 3 8 2 > > 22n+2 − 26n − k2n 22n+2 − 26n − 3 22n Xk=3 · 1 1 8 1 1 1 > 0 ≥ 22n 12 − 24n ≥ 22n 12 − 32 This concludes the proof of the desired inequality.

1 1 2n Remark 4.6. The better inequality: αn > 4 2π 2− , is proved in [26], but the increasing behaviour of the sequence is not discussed there. Concerning− the rational zeros of Bernoulli polynomials, it was 1 proved in [18] that the only possible rational zeros for a Bernoulli polynomial are 0, 2 and 1. A detailed account of the complex zeros of Bernoulli polynomials can be found in [10].

1 1 In the next proposition, we will show how to estimate the “L -norm” 0 Bn in terms of the “L∞- norm” sup B . | | [0,1] | n+1| R Proposition 4.7. The following two properties hold: i. For every positive integer n, we have 1 n! B (x) dx < 16 . | n | (2π)n+1 Z0 ii. Asymptotically, for large n, we have 1 n! Bn(x) dx + 8 | | ∼ ∞ (2π)n+1 Z0

Proof. According to Proposition 2.1 (ii ) we have B (x) = B (1 x) . Thus | n | | n − | 1 1/2 B (x) =2 B (x) dx. | n | | n | Z0 Z0 So, we consider two cases:

(a) The case n = 2m. We have seen (Proposition 2.5) that x ( 1)mB (x) is increasing on [0, 1/2] 7→ − 2m and has a unique zero αm in this interval. Hence

1/2 αm 1/2 B (x) dx = ( 1)m B (x) dx + B (x) dx | n | − − n n Z0 Z0 Zαm ! 1 B (x) αm B (x) 2 = ( 1)m n+1 + n+1 − − n +1 n +1 0 αm ! m+1 B2m+1(αm) 2 = 2( 1) = sup Bn+1(x) . − n +1 n +1 x [0,1] | | ∈ Thus, if n is even, we have 1 4 Bn(x) dx = sup Bn+1(x) ( ) 0 | | n +1 x [0,1] | | † Z ∈ 16 OMRAN KOUBA

m+1 (b) The case n =2m + 1. Using again Proposition 2.5, we see that the function x ( 1) B2m+1(x) is nonnegative on [0, 1/2], thus 7→ −

1/2 1/2 B (x) 1/2 B (x) dx = ( 1)m+1 B (x) dx = ( 1)m+1 n+1 | n | − n − n +1 Z0 Z0 0 ( 1)m+1 1 = − B B (0) n +1 2m+2 2 − 2m+2 ( 1)m+1 1 = − 1 b b n +1 22m+1 − 2m+2 − 2m+2 1 b = 2 | n+1|. − 2n n +1 So, according to Corollary 2.7, if n is odd, we have

1 1 n 4 2 − Bn(x) dx = − sup Bn+1(x) ( ) 0 | | n +1 x [0,1] | | ‡ Z ∈ Thus, combinning ( ), ( ) and Remark 4.4 we obtain † ‡ 1 4 8 n! Bn(x) dx + sup Bn+1(x) + · n+1 0 | | ∼ ∞ n +1 x [0,1] | |∼ ∞ (2π) Z ∈ Also, using Propositions 4.1 and 4.3 we obtain

1 16 n! B (x) dx < · | n | (2π)n+1 Z0 which is the desired conclusion.

5. Asymptotic behavior of Bernoulli polynomials

Proposition 3.1 shows that, for x [0, 1], we have ∈ (2π)2n ∞ 1 3 ( 1)n+1 B (x) cos(2πx) = ζ(2n) 1 < − 2 (2n)! 2n − ≤ k2n − 22n k=2 · X and

(2π)2n+1 ∞ 1 3 ( 1)n+1 B (x) sin(2πx) = ζ(2n + 1) 1 < − 2 (2n + 1)! 2n+1 − ≤ k2n+1 − 22n+1 k=2 · X where we used the following simple inequality, valid for m 2: ≥ 1 ∞ dt m +1 1 3 ζ(m) 1 < + = < − 2m tm m 1 2m 2m Z2 − 2n n+1 (2π) Thus, the sequence ( 1) 2(2n)! B2n converges uniformly on the interval [0, 1] to cos(2π ), and − n 1 · 2n+1 ≥ n+1 (2π) similarly, the sequence ( 1) B2n+1 converges uniformly on the interval [0, 1] to sin(2π ). − 2(2n+1)! n 1 · In fact, this conclusion is a particular case of a more≥ general result proved by K. Dilcher [9]. BERNOULLI POLYNOMIALS AND APPLICATIONS 17

Let us ﬁrst introduce some notation. Let (Tn)n N be the sequence of polynomials deﬁned by the formula : ∈ n/2 ⌊ ⌋ n 2k n/2 k (2πz) − T (z) = ( 1)⌊ ⌋ ( 1) . n − − (n 2k)! Xk=0 − So that n (2πz)2k T (z)= ( 1)k , 2n − (2k)! Xk=0 n (2πz)2k+1 T (z)= ( 1)k . 2n+1 − (2k + 1)! Xk=0 With this notation we have:

Proposition 5.1. For every integer n, with n 2, and every complex number z we have ≥ n 4π z n/2 (2π) 1 e | | ( 1)⌊ ⌋ B z + T (z) < − 2 n! n 2 − n 2n · Proof. Note that B 1 = 0 for every k 0, and according to Corollary 3.2 we have 2k+1 2 ≥ 2 (2k)! B 1 = ( 1)k · η(2k), for k 1 2k 2 − (2π)2k ≥ m 1 2k where η(2k)= ∞ ( 1) − /m . Thus, using Taylor’s expansion we have m=1 − n P 1 n n 1 n k B z + = z + B z − n 2 k k 2 kX=1 n n 1 n 2k = z + B z − 2k 2k 2 1 k n/2 ≤X≤ n 2k 2 n! (2πz) − = zn + · ( 1)kη(2k) (2π)n − (n 2k)! 1 k n/2 ≤X≤ − Thus n n n 2k (2π) 1 n/2 (2πz) k (2πz) − B z + ( 1)⌊ ⌋T (z)= + ( 1) (η(2k) 1) 2 n! n 2 − − n − 2 n! − − (n 2k)! 1 k n/2 · · ≤X≤ − 2k and consequently, since 0 < 1 η(2k) < 2− , we get − n n n 2k n/2 (2π) 1 (2π z ) (2π z ) − ( 1)⌊ ⌋ B z + T (z) | | + (1 η(2k)) | | − 2 n! n 2 − n ≤ 2 n! − (n 2k)! · · 1 k n/2 − ≤X≤

n n 2k 1 (4π z ) (4π z ) − | | + | | ≤ 2n  2 n! (n 2k)!  1 k n/2 · ≤X≤ −   n 2k 4π z 1 (4π z ) − e | | | | ≤ 2n  (n 2k)!  ≤ 2n 0 k n/2 ≤X≤ − and the desired inequality follows.  

Clearly, the sequences of polynomial functions (T2n)n N and (T2n+1)n N converge uniformly on every compact subset of C to z cos(2πz) and z sin(2πz)∈ respectively. So,∈ the next corollary is obtained on replacing z by z 1/27→ in Proposition 5.1.7→ − 18 OMRAN KOUBA

Corollary 5.2. The following two properties hold: 2n n+1 (2π) i. The sequence ( 1) B2n converges uniformly on every compact subset of C to the − 2(2n)! n 1 function cos(2π ). ≥ 2n+1 · n+1 (2π) ii. The sequence ( 1) B2n+1 converges uniformly on every compact subset of C to the − 2(2n+1)! n 1 function sin(2π ). ≥ ·

6. The generating function of Bernoulli polynomials

In what follows, we will write D(a, r) to denote the open disk of center a and radius r in the complex plane C: D(a, r)= z C : z a < r . ∈ | − | The next result gives the generating function of the sequence of Bernoulli polynomials.

∞ B (z) Proposition 6.1. For every (z, w) C D(0, 2π), the series n wn is convergent and ∈ × n! n=0 X ∞ B (z) wezw (z, w) C D(0, 2π), n wn = ∀ ∈ × n! ew 1 n=0 X − Proof. Using Proposition 4.1 (i ), and the facts that b = 1, b = 1/2 we see that 0 1 − n! n N, b 4 . ∀ ∈ | n|≤ (2π)n So, using Proposition 2.3 (iii ), we see that for every nonnegative integer n and complex number z we have: n n k Bn(z) bn k z | |≤ k | − | | | k=0 Xn n (n k)! k 4 −n k z ≤ k (2π) − | | Xk=0 n k n! (2π z ) n! 2π z =4 | | 4 e | |. (2π)n k! ≤ (2π)n Xk=0 Hence, for every (z, w) C D(0, 2π) and every nonnegative integer n we have ∈ × n Bn(z) n 2π z w | | w 4e | | | | n! | | ≤ 2π Bn(z) ∞ n This implies the convergence of the series n=0 n! w . Therefore, we can deﬁne

P ∞ Bn(z) F : C D(0, 2π) C, F (z, w)= wn × −→ n! n=0 X Bn( ) n Moreover, for w D(0, 2π), the normal convergence of the series ∞ · w on every compact subset n=0 n! ′ ∈ B ( ) ∞ n · n of C, implies, using Proposition 2.1 (i ), the normal convergenceP of the series n=0 n! w on every compact subset of C. So, the function F ( , w) has a derivative on C and · P ∂F ∞ B (z) (z, w)= n′ wn ∂z n! n=0 X ∞ nBn 1(z) n = − w = wF (z, w) n! n=1 X BERNOULLI POLYNOMIALS AND APPLICATIONS 19

Thus, there exists a function f : D(0, 2π) C such that F (z, w) = ezwf(w) for every (z, w) in B−→n( ) n ∞ C D(0, 2π). Now, since the series n=0 n!· w is normally convergent on the compact set [0, 1], and× using Corollary 1.3, we obtain P 1 1 ∞ wn F (t, w) dt = B (t) dt =1 n! n 0 n=0 0 Z X Z But, on the other hand, we have 1 1 ew 1 F (t, w) dt = f(w) etw dt = f(w) − w Z0 Z0 Hence, f(w)= w/(ew 1), and consequently F (z, w)= wezw/(ew 1), which is the desired conclusion. − −

The above result allows us to ﬁnd the power series expansion of some well-known functions.

Proposition 6.2. The functions z z cot z, z tan z and z z/ sin z have the following power series expansions in the neighbourhood of zero:7→ 7→ 7→ ∞ 22n( 1)nb i. z D(0, π), z cot z = − 2n z2n. ∀ ∈ (2n)! n=0 X 2n 2n n+1 π ∞ 2 (2 1)( 1) b2n 2n 1 ii. z D 0, , tan z = − − z − . ∀ ∈ 2 (2n)! n=1 X z ∞ (22n 2)( 1)n+1b iii. z D(0, π), = − − 2n z2n. ∀ ∈ sin z (2n)! n=0 X Proof. Indeed, choosing z = 0 in Proposition 6.1 and using Proposition 2.3 (i ) we obtain

1 ∞ b w w D(0, 2π), w + 2n w2n = . ∀ ∈ −2 (2n)! ew 1 n=0 X − Thus ∞ 2b ew +1 w D(0, 2π), 2n w2n = w . ∀ ∈ (2n)! ew 1 n=0 X − Substituting w =2iz we obtain

∞ 22n( 1)nb e2iz +1 z D(0, π), − 2n z2n = iz = z cot z. ∀ ∈ (2n)! e2iz 1 n=0 X − This proves (i ). On the other hand. Noting that tan z = cot z 2 cot(2z) − 1 z = cot cot z sin z 2 − we obtain (ii ) and (iii ).

Remark 6.3. Recalling Corollary 3.2 we see that for z D(0, 1) we have ∈ ∞ πz cot(πz)=1 2 ζ(2n)z2n (6.1) − n=1 X 20 OMRAN KOUBA

Interchanging the signs of summation we find that ∞ ∞ z 2n ∞ z2 πz cot(πz)=1 2 =1 2 − k − k2 z2 n=1 X kX=1 Xk=1 − ∞ 1 1 =1+ z + z k z + k Xk=1 − This yields the following simple fraction expansion of the cotangent function: n 1 ∞ 1 1 1 π cot(πz)= + + = lim (6.2) z z k z + k n z k k=1 →∞ k= n X − X− − Note that we have proved this for z D(0, 1) but the result is valid for every z C Z using analytic continuation [2, Chap. 8, 1]. Similarly,∈ for z D(0, 1) we have ∈ \ § ∈ πz ∞ =1 2 η(2n)z2n (6.3) sin(πz) − n=1 X Interchanging the signs of summation we find that 2n 2 πz ∞ ∞ k 1 z ∞ k 1 z =1 2 ( 1) − =1 2 ( 1) − sin(πz) − − k − − k2 z2 n=1 X Xk=1 Xk=1 − ∞ 1 1 =1+ z ( 1)k + − z k z + k Xk=1 − This yields the following simple fraction expansion of the cosecant function: n π 1 ∞ 1 1 ( 1)k = + ( 1)k + = lim − (6.4) sin(πz) z − z k z + k n z k k=1 →∞ k= n X − X− − This is also valid for every z C Z using analytic continuation. ∈ \ Application 4. Using the power series expansion of z tan z obtained in the previous result, we see that for every positive integer n we have 7→ 2n 2n (2n 1) 2 (2 1) n+1 tan − (0) = − ( 1) b 2n − 2n (n) So, let us define an = tan (0). We note that a2n = 0 for every n 0 since “tan” is an odd function. If 2 ≥ we use tan′ = 1 + tan and the Leibniz formula, we obtain, 2n (2n+1) 2 (2n) 2n (k) (2n k) tan (z)= 1 + tan z = tan (z) tan − (z) k k=0 X for every positive integer n. Thus n 1 − 2n n 1, a2n+1 = a2k+1a2(n k) 1 ∀ ≥ 2k +1 − − kX=0 But, a1 = 1 and the above formula shows inductively that a2n+1 is an integer for every n. This proves that 22n(22n 1) n 1, − b Z ∀ ≥ 2n 2n ∈ and considering the separate case of b1 we see that 2n(2n 1) n 1, − b Z. ∀ ≥ n n ∈ This result is to be compared with Corollary 7.5. BERNOULLI POLYNOMIALS AND APPLICATIONS 21

2 2 Application 5. Let f(z)= z cot z. Since f(z) zf ′(z)= z + f (z) we conclude from Proposition 6.2 (i ) that − 2n n 2n n n ∞ 2 (1 2n)( 1) b2n 2n 2 ∞ 2 ( 1) 2n 2n − − z = z + − b2kb2n 2k z (2n)! (2n)! 2k − n=0 n=0 ! X X Xk=0 2n Comparing the coeﬃcients of z we see that the sequence (b2n)n 1 can be deﬁned recursively by the formula ≥ n 1 1 1 − 2n b2 = , n 2, b2n = b2kb2n 2k. 6 ∀ ≥ −2n +1 2k − Xk=1 Application 6. A multiplication formula for Bernoulli polynomials. Consider an integer q, with q 2, clearly we have ≥ q 1 q 1 eqw 1 − ∞ − wn − = ekw =1+ kn , ew 1 n! n=0 ! − kX=0 X kX=1 ∞ wn ∞ B (q) B (0) wn =1+ S (q 1) =1+ n+1 − n+1 , n − n! n +1 · n! n=0 n=0 X X ∞ B (q) b wn = q + n+1 − n+1 . n +1 · n! n=1 X Where we used the notation of Application 2. Noting the identity we(qz)w (qw)ez(qw) eqw 1 q = − · ew 1 eqw 1 · ew 1 − − − we conclude that

∞ wn ∞ wn ∞ B (q) b wn qB (qz) = qnB (z) q + n+1 − n+1 n n! n n! n +1 · n! n=0 n=0 ! n=1 ! X X X ∞ wn = G (q,z) n n! n=0 X where n 1 n+1 1 − j n +1 Gn(q,z)= q Bn(z)+ q Bj (z)(Bn+1 j (q) bn+1 j ). n +1 j − − − j=0 X But, because a power series expansion is unique, we have qBn(qz)= Gn(q,z) for every n. Now, ﬁx z in C and consider the polynomial

n 1 − n 1 n +1 j 1 Q(X)= Bn(zX) X Bn(z)+ Bj (z)(Bn+1 j (X) bn+1 j )X − −  n +1 j − − −  j=0 X   (Note that X (Bn+1 j (X) bn+1 j).) Clearly deg Q n, and Q has inﬁnitely many zeros, (namely, every integer q greater| than− 1.)− Thus Q− (X) = 0 and we have≤ proved the following “Multiplication Formula”, valid for every complex numbers z and w: n 1 − n 1 n +1 j 1 Bn(zw)= w Bn(z)+ Bj (z)w − (Bn+1 j (w) bn+1 j) (6.5) n +1 j − − − j=0 X For example, taking w = 2 and z = 0 we obtain, the following recurrence n 1 1 − n b = 2j b n 2(1 2n) j j j=0 − X 22 OMRAN KOUBA

since Bn+1 j(2) Bn+1 j (0) = n +1 j for 0 j

ew 1 Application 7. More formulæ for Bernoulli numbers. The function w w− is entire, and has a power series expansion, that converges in the whole complex plane. Let ρ be7→ deﬁned by 1 ew 1 = sup − > 1. ρ w w D(0,1) ∈ w Now, for the disk w D(0,ρ) we have e 1 < 1 and consequently ∈ | − | ∞ 1 w = Log(1 (1 ew)) = (1 ew)n. − − − n − n=1 X Thus, for every m 1 we have ≥ m w (1 ew)n = − + g (w) ew 1 n +1 m n=0 − X 1 where g (w) = ∞ (1 ew)n. Clearly, w = 0 is a zero of g of order greater than m. Thus m n=m+1 n+1 − m (m) th w gm (0) = 0. But,P using Proposition 6.1 the Bernoulli number bm is the m derivative of w ew 1 at 0 so 7→ − m 1 w n (m) bm = (1 e ) n +1 − w=0 n=0 X i But (1 ew)n = n n ( 1)kekw. Thus, − k=0 k − P m 1 n n b = ( 1)kkm , for m 1. (6.6) m n +1 k − ≥ n=0 ! X kX=0 This is quite an old formula for Bernoulli numbers (see [12] and the references therein.) Noting that

m n 1 x m 1 x m (1 x) − n 1 − t 1 − = t − dt = − dt n t 1 n=1 0 n=1 ! 0 X Z X Z − x m x m (1 u) 1 m n n 1 = − − du = ( 1) u − du u n − Z1 Z1 n=1 ! m X m xn 1 = ( 1)n − , n − n n=1 X we can rearrange our previous calculation, as follows

m w n 1 m n nw m n n 1 (1 e ) − m ( 1) e 1 m ( 1) − − = − − = − ekw . n n n ew 1 n n n=1 n=1 n=1 ! X X − X Xk=0 Hence, taking as before the mth derivative at 0, another formula is obtained [5]:

m n 1 m ( 1)n − b = − km , for m 1. (6.7) m n n ≥ n=1 ! X Xk=1 BERNOULLI POLYNOMIALS AND APPLICATIONS 23

7. The von Staudt-Clausen theorem

In this section we give the proof of a famous theorem that determines the fractional part of a Bernoulli number. First, let us introduce some notation, the reader is invited to take a look at [19, Chapter 15], and the references therein, for a deeper insight on the role played by Bernoulli numbers in Number Theory. Let us denote by A the set of functions f that are analytic in the neighborhood of 0 and such that f (n)(0) is an integer for every nonnegative integer n. Let f and g be two members of A, and let m be a positive integer. We will write f g (mod m) if f (n)(0) g(n)(0) (mod m) for every nonnegative integer n. Finally, for two functions≡f and g that are analytic≡ functions in the neighborhood of 0, we write f g (mod A) if f g A. ≡ − ∈ Lemma 7.1. The following properties hold: z i. If f belongs to A, then both f ′ and z 0 f(t) dt belong to A. ii. If f and g belong to A, then fg belongs7→ also to A. 1 Rm iii. If f belongs to A and f(0) = 0, then m! f belongs also to A for every positive integer m. a N ∞ n n Proof. Consider f A. There is a sequence of integers (an)n such that f(z) = n=0 n! z in a neighbourhood of 0.∈ But then ∈ P z ∞ an+1 n ∞ an 1 n f ′(z)= z , and f(t)dt = − z . n! n! n=0 0 n=1 X Z X z and consequently both f ′ and z 0 f(t) dt belong to A. This proves (i ). Property (ii ) follows from Leibniz7→ formula. R Property (iii ) is proved by mathematical induction. It is true for m = 1, and if we suppose that 1 m 1 1 m 1 (m 1)! f − belongs to A, then according to (i ) and (ii ) the function (m 1)! f − f ′ belongs also to A, and− using (i ) once more we conclude that − z 1 m 1 m 1 z f (z)= f − (t)f ′(t)dt 7→ m! (m 1)! − Z0 also belongs A. This achieves the proof of the lemma.

Lemma 7.2. If m is a composite positive number such that m> 4 then (m 1)! 0 (mod m). − ≡ Proof. Let p be the smallest prime that divides m, and let q = m/p. Since m is composite we conclude that q p so, there are two cases: ≥ q>p. In this case 1 4, implies that p > 2 and consequently− 1

Proposition 7.3. The function g deﬁned by g(z) = ez 1 belongs to A and it satisﬁes the following properties: − m 1 i. If m is composite and greater than 4 then g − 0 (mod m). 2k+1 ≡ m 1 ∞ z ii. If m =4 then g − (z) 2 (mod m). ≡ (2k + 1)! Xk=0 ∞ k(m 1) m 1 z − iii. If m is prime then g − (z) (mod m). ≡− (km k)! Xk=1 − 24 OMRAN KOUBA

Proof. The fact that g A is immediate. ∈ m−1 Suppose that m is a composite integer greater than 4. Using Lemma 7.1 (iii ) we see that g A, (m 1)! ∈ and (i ) follows from Lemma 7.2. − Consider the case m = 4. Noting that g3(z) = e3z 3e2z +3ez 1 we conclude that g3(z) = an n − − n=3 n! z with a =3n 3 2n +3. P n − · n But, for n 3 we have an ( 1) 1 (mod 4) so a2k 0 (mod 4) and a2k+1 2 (mod 4). This proves (ii ) ≥ ≡ − − ≡ ≡ Finally, suppose that m is a prime. Here

m 1 − m 1 m 1 m k 1 kz g − (z)= − ( 1) − − e , k − Xk=0 m 1 bn n and consequently g − (z)= n=m 1 n! z with − Pb0 = b1 = ... = bm 2 =0, bm 1 = (m 1)!, − − − and for n m 1 ≥ − m 1 − m 1 m k 1 n b = − ( 1) − − k n k − Xk=1 m 1 But according to Fermat’s Little Theorem [16, Theorem 71] we have k − 1 (mod m)for1 k m 1, ≡ ≤ ≤ − and consequently bn+m 1 bn (mod m) for every n. Thus, bn 0 (mod m) if n is not a multiple of m 1, and if n is a multiple− ≡ of m 1 then ≡ − − b (m 1)! 1 (mod m) n ≡ − ≡− where the last congruence follows from Wilson’s Theorem [16, Theorem 80], and (iii ) follows.

Theorem 7.4 (von Staudt-Clausen Theorem). For a given positive integer n, let the set of primes p such that p 1 divides 2n be denoted by p . Then − n 1 b + Z. 2n p ∈ p pn X∈ Proof. Indeed, consider the function g of Proposition 7.3. Note that

m 1 ∞ ( 1) − z = Log(1 + g(z)) = − gm(z) m m=1 X Thus

∞ m 1 3 p 1 z ( 1) − m 1 g(z) g (z) g − (z) = − g − (z)=1 + (mod A) ez 1 m − 2 − 4 p − m=1 p>2 X p Xprime 1 ∞ z2k+1 1 ∞ zk(p 1) =1 − (mod A) − 2 (2k + 1)! − p (kp k)! k=1 p 2 k=1 − X p Xprime≥ X

b But, since z/(ez 1) = ∞ n zn according to Proposition 6.1, the desired conclusion follows from the − n=0 n! above equality, on comparing the coeﬃcients of z2n. P For instance, p = 2, 3 and b + 1 + 1 = 1. Also, p = 2, 3, 5 , and b + 1 + 1 + 1 = 1. Generally, 1 { } 2 2 3 2 { } 4 2 3 5 for a positive integer n, we have 2, 3 pn and consequently the denominator of b2n is always a multiple of 6. { }⊂ BERNOULLI POLYNOMIALS AND APPLICATIONS 25

k Corollary 7.5. For every positive integer m and every nonnegative integer k the quantity m(m 1)bk is an integer. − Proof. We only need to consider the case k =2n for some positive integer n because the other cases are trivial. Consider a prime p such that p 1 divides 2n, (i.e. p p .) Consider also a positive integer m. − ∈ n If p m then clearly p m(mk 1). • | | − p 1 If p ∤ m then, according to Fermat’s Little Theorem [16, Theorem 71] we have m − 1 (mod p), • and since (p 1) k we conclude that mk 1 (mod p). Thus p m(mk 1) also in≡ this case. −k | ≡ | − It follows that m(m 1) is a multiple of every prime p pn, and the result follows according to Proposition 7.4. − ∈

Corollary 7.6. For every positive integer n there are inﬁnitely many integers m such that B2m B2n is an integer. − Proof. Consider τ = lcm(d + 1 : d (2n)); (the least common multiple of the numbers d + 1 where d is a divisor of 2n,) and let q be a prime| number such that q 1 (mod τ). ≡ Now, if m = nq then p = p . Indeed, if p′ p then p′ 1 divides 2nq, so, there are two cases: m n ∈ m − If p′ 1 divides 2n then clearly p′ p since p′ is prime. • − ∈ n If p′ 1= dq for some d 2n, then • − | p′ =1+ dq = (d + 1)q +1 q = (d + 1)q λτ, for some integer λ, − − so, p′ is a multiple of d + 1, and since it is a prime, we conclude that p′ = d + 1 which is absurd since q = 1. 6 Thus, we have proved that p p . But, the inverse inclusion is trivially true, and p = p , or m ⊂ n m n equivalently B2m B2n is an integer. Finally, using− Dirichlet’s Theorem [19, Chapter 16], we know that there are inﬁnitely many primes q such that q 1 (mod τ), and the corollary follows. ≡

8. The Euler-Maclaurin’s formula

For a function g deﬁned on the interval [0, 1] we introduce the notation δg to denote the diﬀerence g(1) g(0). Also we recall the notation Bn for the 1-periodic function that coincides with x Bn(x) on the interval− [0, 1], or equivalently, 7→ e x R, B (x)= B ( x ) ∀ ∈ n n { } where t = t t is the fractional part of t. { } −⌊ ⌋ e Proposition 8.1. Consider a positive integer m and a function f : [0, 1] C having a continuous mth derivative. For every x in [0, 1] we have −→

1 m 1 1 − Bk+1(x) (k) 1 (m) f(t) dt f(x)+ δf = Bm(x t)f (t) dt 0 − (k + 1)! m! 0 − Z kX=0 Z Proof. For an integer k with 0 k m we deﬁne F (x) by the formulae ≤ ≤ k 1 1 F (x)= B (x t)f (k)(t) dt. k k! k − Z0 Clearly we have e 1 1 F (x)= B (x t)f(t) du = f(t) dt. 0 0 − Z0 Z0 e 26 OMRAN KOUBA

Also, for 0 k

Corollary 8.2. Consider a positive integer m, and a function f that has a continuous (2m 1)st derivative (2m 1) − on [0, 1]. If f − is decreasing, then 1 m 1 f(1) + f(0) − b2k (2k 1) m+1 f(t) dt = δf − + ( 1) Rm 0 2 − (2k)! − Z kX=1 with 1/2 B2m 1(t) (2m 1) (2m 1) R = | − | f − (t) f − (1 t) dt m (2m 1)! − − Z0 − and 6 (2m 1) (2m 1) 0 R f − (0) f − (1) . ≤ m ≤ (2π)2m − Proof. Indeed, choosing x = 1 in Proposition 8.1 with 2m 1 for m, we obtain − 1 2m 2 1 − Bk+1(1) (k) 1 (2m 1) f(t) dt f(1) + δf = B2m 1(1 t)f − (t) dt 0 − (k + 1)! (2m 1)! 0 − − Z Xk=0 − Z Now, using Proposition 2.1 (ii ), Proposition 2.3 (i ) and the fact that B1(1) = 1/2, we see that 1 m 1 f(1) + f(0) − b2k (2k 1) rm f(t) dt + δf − = 0 − 2 (2k)! −(2m 1)! Z kX=1 − with 1 (2m 1) rm = B2m 1(t)f − (t) dt − Z0 BERNOULLI POLYNOMIALS AND APPLICATIONS 27

But, 1/2 1 (2m 1) (2m 1) rm = B2m 1(t) f − (t) dt + B2m 1(t) f − (t) dt − − Z0 Z1/2 1/2 1/2 (2m 1) (2m 1) = B2m 1(t) f − (t) dt + B2m 1(1 t) f − (1 t) dt − − − − Z0 Z0 1/2 (2m 1) (2m 1) = B2m 1(t) f − (t) f − (1 t) dt − − − Z0 m Now, according to Proposition 2.5, we know that ( 1) B2m 1 is positive on (0, 1/2). Thus − − 1/2 m (2m 1) (2m 1) rm = ( 1) B2m 1(t) f − (t) f − (1 t) dt, − 0 | − | − − Z and the expression of Rm follows. (2m 1) In particular, when f − is decreasing, the maximum on the interval [0, 1/2] of the quantity (2m 1) (2m 1) (2m 1) f − (t) f − (1 t) is δf − , attained at t = 0, and its minimum on the same interval is 0 and it is attained− at t =1− /2. Consequently,− using Proposition 4.1, we have 1/2 m+1 B2m 1(t) (2m 1) 0 R ( 1) − dt δf − ≤ m ≤ − (2m 1)! Z0 − ! m+1 B2m(1/2) B2m(0) (2m 1) ( 1) − δf − ≤ − (2m)!

1 2m b2m (2m 1) (2 2 − ) | | δf − ≤ − (2m)!

1 3 1 (2m 1) 4 1 1+ δf − ≤ − 22m 22m (2π)2n 1 2m 1+2 − (2m 1) 6 (2m 1) 4 δf − δf − ≤ (2π)2m ≤ (2π)2m

and the desired conclusion follows.

Before proceeding to the next result, we will prove the following property that generalises the well- known “Riemann Lebesgue’s lemma”.

Lemma 8.3. Consider an integrable function h : [0, 1] C, and a piecewise continuous 1-periodic function g : R C. Then −→ −→ 1 1 1 lim g(pt)h(t) dt = g(t) dt h(t) dt p · →∞ Z0 Z0 Z0 Proof. First, suppose that 1 g = 0. This implies that x G(x) = x g(t)dt is a continuous 1-periodic 0 7→ 0 function. Particularly G is bounded, and we can deﬁne M = supR G . R | R| Now, assume that h = χ[α,β); the characteristic function of an interval [α, β). In this case 1 β G(pβ) G(pα) g(pt)h(t) dt = g(pt) dt = − p Z0 Zα So 1 2M g(pt)h(t) dt ≤ p Z0 and consequently lim 1 g(pt)h(t) dt = 0. p 0 →∞ R 28 OMRAN KOUBA

Using linearity, we see that the same conclusion holds if h is a step function. Finally, the density of the space of step functions in L1([0, 1]) implies that lim 1 g(pt)h(t) dt = 0 for every integrable function p 0 →∞ h : [0, 1] C. −→ R 1 1 Applying the preceding case to the functiong ˜ = g 0 g, that satisﬁes 0 g˜(t)dt = 0, we conclude that − 1 R R lim g˜(pt)h(t) dt =0 p →∞ Z0 for every h L1([0, 1]). Which is the desired conclusion. ∈ The next theorem is the main result of this section.

Theorem 8.4. For a positive integer p, and a function f having at least a continuous mth derivative on the interval [0, 1], we deﬁne the quantities (f; x) and (p,m,f; x) for x [0, 1] by Hp E ∈ p 1 1 − k + x (f; x)= f Hp p p Xk=0 and

1 m (k 1) Bk(x) δf − (p,m,f; x)= f(t) dt p(f; x)+ k . E 0 − H k! · p Z kX=1 Then, i. The quantity (p,m,f; x) has the following expression in terms of B : E m 1 1 B (x pt) (p,m,f; x)= m − f (m)(t) dt.e E pm m! Z0 ii. It satisﬁes also the following inequality: e

8 1 (m) (p,m,f; x) m sup f . |E |≤ π · (2πp) · [0,1]

iii. Moreover, lim pm (p,m,f; x)=0. p →∞ ·E Proof. Applying Proposition 8.1 to the function x (f; x) we obtain 7→ Hp m 1 1 − 1 Bk+1(x) (k) 1 (m) p(f; t) dt p(f; x)+ δ p (f; )= Bm(x t) p (f; t) dt. ( ) 0 H − H (k + 1)! H · m! 0 − H ∗ Z Xk=0 Z But e p 1 p 1 1 1 − 1 k + t − (k+1)/p 1 p(f; t) dt = f dt = f(u)du = f(u) du. 0 H p 0 p k/p 0 Z kX=0 Z kX=0 Z Z Also, p 1 1 − k + x 1 (k)(f; x)= f (k) = (f (k); x). Hp pk+1 p pk Hp Xk=0 Thus p 1 p 1 − − (k) 1 (k) k +1 (k) k δ p (f; )= k+1 f f , H · p p − p ! Xk=0 Xk=0 f (k)(1) f (k)(0) 1 = − = δf (k). pk+1 pk+1 BERNOULLI POLYNOMIALS AND APPLICATIONS 29

Replacing the above results in ( ) we conclude that ∗ 1 1 (p,m,f; x)= B (x t) (f (m); t) dt, E m! pm m − Hp · Z0 and (i ) follows, since e p 1 1 − 1 (m) 1 (m) k + t Bm(x t) p(f ; t) dt = Bm(x t)f dt 0 − H p 0 − p Z Xk=0 Z p 1 e − (k+1)e/p (m) = Bm(x + k pt)f (t) dt k/p − Xk=0 Z p 1 − (k+1)/p e (m) = Bm(x pt)f (t) dt k/p − Xk=0 Z 1 e = B (x pt)f (m)(t) dt m − Z0 Using (i ), and recalling that Bm is 1-periodic, wee see that 1 1 (m) (p,m,f; x) e m Bm(x pt) f (t) dt |E |≤ m! p 0 − · · Z 1 1 e (m) m sup f (t) Bm(x pt) dt ≤ m! p t [0,1] · 0 − · ∈ Z p 1 (m) 1 e m sup f (t) Bm(x u) du ≤ m! p t [0,1] · p 0 − · ∈ Z 1 1 (m) e = m sup f (t) Bm(t) dt, m! p t [0,1] · 0 · ∈ Z and (ii ) follows using Proposition 4.7 (i ). e

Finally, applying Lemma 8.3 to the 1-periodic function u g(t) = Bm(x t) and the integrable (m) 1 7→ − function t h(t)= f (t) we obtain (iii ) because 0 g = 0 in this case. 7→ e R 9. Asymptotic expansions for numerical quadrature formulæ

In this section we only consider functions deﬁned on the intervall [0, 1]. The more general case of a functions deﬁned on [a,b] can be obtained by applying the results after using the change of variable t a + t(b a). 7→ − We consider a function f : [0, 1] C, having a continuous mth derivative, with m 2, and we will use freely the notation of the previous−→ section. ≥

9.1. Riemann sums. The Riemann sum of f obtained by taking the values of the function f at the lower bound of each subdivision interval, is given by p 1 1 − k L(f)= f . (9.1) Rp p p Xk=0 According to Theorem 8.4 we have L(f)= (f, 0), so Rp Hp 1 L δf b2k (2k 1) f(t)dt = (f)+ δf − + (p,m,f; 0) (9.2) Rp 2p − (2k)! p2k · E 0 1 k m Z ≤X≤ 2 · 30 OMRAN KOUBA

Similarly, the Riemann sum of f obtained by taking the values of the function f at the upper bound of each subdivision interval, is given by

p 1 k R(f)= f (9.3) Rp p p kX=1 And again using Theorem 8.4 we have R(f)= (f, 1), so Rp Hp 1 R δf b2k (2k 1) f(t)dt = (f) δf − + (p,m,f; 0) (9.4) Rp − 2p − (2k)! p2k · E 0 1 k m Z ≤X≤ 2 · where we noted that (p,m,f;1) = (p,m,f; 0), since B is 1-periodic. E E m Also, the Riemann sum of f obtained by taking thee values of the function f at the midpoint of each subdivision interval, is given by p 1 1 − 2k +1 M (f)= f (9.5) Rp p 2p Xk=0 This is the “Midpoint Quadrature Rule”. By Theorem 8.4 we have M (f)= f, 1 , so Rp Hp 2 1 1 2k M (2 − 1)b2k (2k 1) 1 f(t)dt = p (f) − 2k δf − + p,m,f; (9.6) 0 R − m (2k)! p · E 2 1 k 2 Z ≤X≤ · For example, taking m = 2, we obtain from Theorem 8.4 (iii):

1 f ′(1) f ′(0) lim p2 f(t)dt M (f) = − . (9.7) p − Rp 24 →∞ Z0 Thus, the midpoint quadrature rule is a second order rule.

L R 9.2. The trapezoidal rule. Taking the half sum of p (f) and p (f) we obtain the trapesoidal rule that corresponds to approximating f linearly on each intervalR of theR subdivision.

p 1 1 f(0) + f(1) 1 − k (f)= L(f)+ R(f) = + f (9.8) Tp 2 Rp Rp 2p p p k=1 X Using (9.2) and (9.4) we see that

1 b2k (2k 1) f(t)dt = (f) δf − + (p,m,f; 0) (9.9) Tp − (2k)! p2k · E 0 1 k m Z ≤X≤ 2 · In particular, choosing m = 2, we obtain from Theorem 8.4 (iii):

1 2 f ′(1) f ′(0) lim p f(t)dt p(f) = − . (9.10) p − T − 12 →∞ Z0 Thus, the trapezoidal quadrature rule is a second order rule.

In fact, for the case of the trapezoidal rule we have a more reﬁned result in some cases. This is the object of the following proposition. BERNOULLI POLYNOMIALS AND APPLICATIONS 31

Proposition 9.1. Consider a positive integer m, and a function f that has a continuous (2m 1)st (2m 1) − derivative on [0, 1]. If f − is decreasing, then, for every positive integer p we have

1 m 1 − b2k (2k 1) m+1 f(t) dt = p(f) 2k δf − + ( 1) Rm,p 0 T − (2k)! p · − Z Xk=1 · with 6 (2m 1) (2m 1) 0 R f − (0) f − (1) . ≤ m,p ≤ (2πp)2m − Proof. Our starting point will be Corollary 8.2 applied to the function x f j+x , with 0 j

P 6 2m 1 (2m 1 0 R f − (0) f − (1) ≤ m,p ≤ (2πp)2m − which is the desired conclusion.

Application 8. An asymptotic expansion for a trigonometric sum. For a positive integer p, we consider the trigonometric sum p 1 − jπ J = j cot . p p j=1 X This sum will be studied in detail later, but we want here to illustrate the use of the Proposition 9.1.

Proposition 9.2. For every positive integers p and m, there is a real number θp,m such that

m 1 m 1 2 ln(2π) 2 p − b2k(1+2ζ(2k)) ( 1) Jp = p Hp + p 2k 2 + −2m 2 θp,m −π π − 2π − 2πk p − p − Xk=1 · and b (1+2ζ(2m)) 0 <θ < | 2m| , p,m 2πm p th where Hp = j=1 1/j is the p harmonic number. Proof. Indeed,P let ϕ be the function deﬁned by 1 ϕ(x)= πx cot(πx)+ . 1 x − 32 OMRAN KOUBA

According to formula (6.2) we know that

x ∞ x x ϕ(x)=2+ + + . x +1 x n x + n n=2 X − Thus, ϕ is deﬁned and analytic on the interval ( 1, 2). Let us show that, for every positive integer k, − the derivative ϕ(2k) is negative on the interval [0, 1]. To this end, we note, using (6.1), that 2 1 ϕ(x)= πx cot(πx)+ 1 x2 − 1+ x − 1 ∞ =3 2 (ζ(2n) 1)x2n. − 1+ x − − n=1 X Hence, (2k) ∞ ϕ (x) 1 2n 2n 2k = 2 (ζ(2n) 1)x − , (2k)! −(1 + x)2k+1 − 2k − nX=k which is clearly negative on [0, 1].

(2k 1) Now, we can apply Proposition 9.1 to ϕ. We only need to calculate δϕ − for every k. Note that 1 π cot(πx) 1 ϕ(x)+ ϕ(1 x)= − +2πx cot(πx)+ − x 1 x − Thus, using (6.1) again we obtain, for x < 1, | |

∞ 2n 1 ∞ 2n ϕ(x)+ ϕ(1 x)=3+ (2ζ(2n)+1)x − + (1 4ζ(2n))x − − n=1 n=1 X X Taking the (2k 1)st derivative at x = 0, we get − δϕ2k 1 − = 1 2ζ(2k). (2k 1)! − − − So, applying Proposition 9.1, we obtain

1 m 1 − b2k(1+2ζ(2k)) m+1 ϕ(t) dt = p(ϕ)+ 2k + ( 1) Rm,p (*) 0 T 2k p − Z Xk=1 · with 6(2m 1)! (1+2ζ(2m)) 0 R − · ≤ m,p ≤ (2πp)2m But p 1 ϕ(0) + ϕ(1) 1 − j (ϕ)= + ϕ Tp 2p p p j=1 X p 1 p 1 3 π − πj − 1 1 π = + j cot + = H + + J 2p p2 p p j p 2p p2 p j=1 j=1 X X − Also, for x [0, 1), we have ∈ x x ϕ(t) dt = ln(1 x)+ x ln sin(πx) ln sin(πt) dt − − − Z0 Z0 and, letting x tend to 1 we obtain 1 1 ϕ(t) dt = ln(π) ln sin(πt) dt = ln(2π) − Z0 Z0 BERNOULLI POLYNOMIALS AND APPLICATIONS 33

where we used the fact 1 ln sin(πt) dt = ln 2, (see [13, 4.224 Formula 3.]. Thus (*) is equivalent to 0 − m 1 R 1 π − b (1+2ζ(2k)) ln(2π)= H + + J + 2k + ( 1)m+1R p 2p p2 p 2k p2k − m,p Xk=1 · or m 1 π 1 − b (1+2ζ(2k)) J = ln(2π) H 2k + ( 1)mR p2 p − p − 2p − p2k − m,p Xk=1 Thus, we have shown that for every nonnegative integer m we have 2m π 1 b (1+2ζ(2k)) J < ln(2π) H 2k p2 p − p − 2p − 2k p2k Xk=1 · and 2m+1 π 1 b (1+2ζ(2k)) J > ln(2π) H 2k p2 p − p − 2p − 2k p2k kX=1 · So, for every positive integer m we have, m 1 2 − m p p b2k(1+2ζ(2k)) b2m (1+2ζ(2m)) 0 < ( 1) Jp (ln(2π) Hp)+ + 2k 2 < | | 2m 2 . − − π − 2π 2πk p − ! 2πm p − kX=1 · · Which is the desired conclusion.

The result of this proposition is not completely satisfactory, because of the sum Hp. That is why it is just the beginning of the story! It will be pursued in a later section.

9.3. Simpson’s rule. Comparing (9.10) and (9.7) we see that 1 M p(f)+2 (f) lim p2 f T Rp =0, p − 3 →∞ Z0 ! 1 M so the quantity 3 p(f)+2 p (f) is a better quarature rule than the second order ones. Hence, let us deﬁne the “SimpsonT quadratureR rule” by 1 (f)= L(f)+2 M (f) Sp 3 Tp Rp p 1 1 − 2k 2k +1 2k +2 = f + f + f (9.11) 6p 2p 2p 2p Xk=1 Using (9.9) and (9.6) we see that 1 1 k (1 4 − )b2k (2k 1) S f(t)dt = (f)+ − δf − + (p,m,f) (9.12) Sp 3(2k)! p2k · E 0 2 k m Z ≤X≤ 2 · with 1 1 S (p,m,f)= (p,m,f;0)+2 p,m,f; E 3 E E 2 Using Theorem 8.4 we see that

S 8 1 (2m) 2m S (p,m,f) 2m sup f and lim p (p,m,f)=0 E ≤ π · (2p) · [0,1] p ·E →∞

In particular, choosing m = 4, we obtain : 1 (3) (3) 4 f (1) f (0) lim p f(t)dt p(f) = − . (9.13) p − S − 2880 →∞ Z0 Thus, the Simpson quadrature rule is a forth order rule. 34 OMRAN KOUBA

9.4. The two point Gauss rule. Applying Theorem 8.4 at x and 1 x and using Proposition 2.1 (ii ), we obtain after taking the half sum: − p 1 1 − 1 k+x k+1 x B2k(x) (2k 1) f(t)dt = f + f − δf − + (p,m,f; x) 2p p p − (2k)! p2k E 0 k=0 1 k m Z X ≤X≤ 2 with e 1 (p,m,f; x)= ( (p,m,f; x)+ (p,m,f;1 x)) . E 2 E E − Here satisﬁes the same propertiese as in Theorem 8.4. The case x = 0 corresponds to the trapezoidal E 1 E rule, and the case x = 2 corresponds the midpoint point rule. But the best choice for x is when x = αe= 1 1 which is a zero of B . Then, we obtain “the two point Gauss quadrature rule”: 2 − √12 2 p 1 k + 1 1 k + 1 + 1 1 − 2 √12 2 √12 p(f)= f − + f (9.14) G 2p p ! p !! Xk=0 with 1 B2k(α) (2k 1) f(t)dt = (f) δf − + (p,m,f; α) (9.15) Gp − (2k)! p2k E 0 2 k m Z ≤X≤ 2 For example, with m = 4 we ﬁnd that e 1 (3) (3) 4 f (1) f (0) lim p f(t)dt p(f) = − (9.16) p − G 4320 →∞ Z0 Thus, the two point Gauss quadrature rule is a forth order rule.

9.5. Romberg’s rule. Let us consider again the case of the trapezoidal rule (9.8), and the error asymptotic expansion: 1 b2k (2k 1) f(t)dt = (f) δf − + (p,m,f; 0) (9.17) Tp − (2k)! p2k · E 0 1 k m Z ≤X≤ 2 · (0) (0) We deﬁne p (f)= (f) and for simplicity we write m (p) for (p,m,f; 0). Next, we deﬁne inductively T Tp E E ℓ (ℓ 1) (ℓ 1) 4 − (f) p − (f) (ℓ)(f)= T2p − T Tp 4ℓ 1 − ℓ (ℓ 1) (ℓ 1) 4 m− (2p) m− (p) (ℓ)(p)= E − E Em 4ℓ 1 − for ℓ =1, 2,.... It is easy to prove by induction, starting from (9.17) that

1 1 k (1) 4 − 1 b2k (2k 1) (1) f(t)dt = p (f) − 2k δf − + m (p) 0 T − m 4 1 (2k)! p E 1 k 2 Z ≤X≤ − · 1 1 k 2 k (2) 4 − 1 4 − 1 b2k (2k 1) (2) f(t)dt = p (f) − 2 − 2k δf − + m (p) 0 T − m 4 1 4 1 (2k)! p E 1 k 2 Z ≤X≤ − − · . . . . = . .

1 ℓ j k (ℓ) 4 − 1 b2k (2k 1) (ℓ) f(t)dt = p (f) j − 2k δf − + m (p) 0 T − m 4 1 (2k)! p E 1 k 2 j=1 Z ≤X≤ Y − · BERNOULLI POLYNOMIALS AND APPLICATIONS 35

Note that ℓ j k 4 − 1 − =0 for k =1, 2,...,ℓ, 4j 1 j=1 Y − so, in fact, we have

1 ℓ j k (ℓ) 4 − 1 b2k (2k 1) (ℓ) f(t)dt = p (f) j − 2k δf − + m (p) (9.18) 0 T − m 4 1 (2k)! p E ℓ

In order to simplfy a little bit the notation we recall that the finite q-Pochhammer (z; q)n symbol is defined as the product n k 1 (z; q) = 1 zq − . n − k=1 Y The limit as n tend to + defines the q-Pochhammer symbol (z; q) when q < 1. Also, we define the n ∞ ∞ | | q-binomial coefficient m q by the formula

n (q; q) = n , for 0 m n. (9.19) m (q; q)n m(q; q)m ≤ ≤ q − With this notation we see that for k>ℓ and q =1/4, we have

ℓ j k ℓ ℓ k j 4 − 1 ( 1) 1 q − − = − − 4j 1 2ℓ(1+ℓ) 1 qj j=1 j=1 Y − Y − ℓ ℓ ( 1) (q; q)k 1 ( 1) k 1 − = ℓ−(1+ℓ) = ℓ−(1+ℓ) − 2 (q; q)k 1 ℓ(q; q)ℓ 2 ℓ − − q Thus, we can write (9.18) as follows

1 ℓ (ℓ) ( 1) k 1 b2k (2k 1) (ℓ) f(t)dt = p (f) ℓ−(1+ℓ) − 2k δf − + m (p) (9.20) 0 T − 2 m ℓ 1 (2k)! p E ℓ

m (0) Also, since according to Theorem 8.4 (iii ), we have lim p m (p) = 0, we conclude by induction on ℓ p E that we have →∞ m (ℓ) lim p m (p)=0, for ℓ =0, 1, 2,.... p →∞ E In particular, when m 2ℓ + 2 we have ≥ 1 b lim p2ℓ+2 f(t)dt (ℓ)(f) = | 2ℓ+2| δf (2ℓ+1). p − Tp −2ℓ(1+ℓ) (2ℓ + 2)! →∞ Z0 On the other hand, using Theorem 8.4 (ii ) we have

(0) 8 (m) m (p) m sup f E ≤ (2πp) · [0,1]

and, for ℓ =1, 2,... we have

ℓ (ℓ 1) (ℓ 1) 4 m− (2p) + m− (p) (ℓ)(p) E E Em ≤ 4ℓ 1 −

36 OMRAN KOUBA

So m (1) 8 1+4/2 (m) m (p) m sup f E ≤ (2πp) 4 1 · [0,1] − m 2 m (2) 8 1+4/2 1+4 /2 (m) m (p) m 2 sup f E ≤ (2πp) 4 1 4 1 · [0,1] − − . . . .

ℓ j m (ℓ) 8 1+4 /2 (m) m (p) m j sup f (9.21) E ≤ (2πp) 4 1 · [0,1] j=1 − Y But, for m 2ℓ + 2 we have ≥ ℓ ℓ ℓ j m (ℓ j+1) j 1+4 /2 1+4− − = 1+4− ≤ j=1 j=1 j=1 Y Y Y So ℓ ℓ 1+4j /2m 1 1+4 j − (9.22) 4j 1 ≤ 2ℓ(1+ℓ) 1 4 j j=1 j=1 − Y − Y − 1+x Now, since x ln 1 x is convex on the interval [0, 1/4] we conclude that for x [0, 1/4] we have 7→ − ∈ 1+ x 1+4 1 5 ln 4 ln − x = 4 ln x 1 x ≤ 1 4 1 3 − − − Thus ℓ 1+4 j 5 ∞ 1 4 5 ln − < 4 ln = ln 1 4 j 3 4j 3 3 j=1 − j=1 X − X Finally, since (5/3)4/3 < 2 we obtain from (9.22) that

ℓ 1+4j /2m 2 m 2ℓ +2, ∀ ≥ 4j 1 ≤ 2ℓ(1+ℓ) j=1 Y − Thus, (9.21) implies the following more appealing form

(ℓ) 16 1 (m) m (p) ℓ(1+ℓ) m sup f (9.23) E ≤ π · 2 (2πp) [0,1]

We have proved the following result: Proposition 9.3. Let m be a positive integer, and let f be a function having a continuous mth derivative (0) on [0, 1]. For a positive p, let p (f) be the trapezoidal quadrature rule applied to f deﬁned by (9.8). Next, for ℓ 1, deﬁne inductivelyT the Romberg’s rule of order ℓ by ≥ ℓ (ℓ 1) (ℓ 1) 4 − (f) p − (f) (ℓ)(f)= T2p − T Tp 4ℓ 1 − Then 1 ℓ (ℓ) ( 1) k 1 b2k (2k 1) (ℓ) f(t)dt = p (f) ℓ−(1+ℓ) − 2k δf − + m (p) 0 T − 2 m ℓ 1 (2k)! p E ℓ

BERNOULLI POLYNOMIALS AND APPLICATIONS 37

Consider the particular case where m =2ℓ + 2. In this case, Proposition 9.3 implies 1 1 (2π)2ℓ+2 b 16 f(t)dt (ℓ)(f) | 2ℓ+2| δf (2ℓ+1) + sup f (2ℓ+2) − Tp ≤ 2ℓ(1+ℓ)(2πp)2ℓ+2 (2ℓ + 2)! π Z0 [0,1] !

1 (2ℓ+1) 16 (2 ℓ+2) ℓ(ℓ+1) 2ℓ+2 4 δf + sup f ≤ 2 (2πp) π [0,1] !

4+16/π (2ℓ+2) ℓ(ℓ+1) 2ℓ+2 sup f ≤ 2 (2πp) [0,1]

10 (2ℓ+2) (ℓ+2)(ℓ+1) 2ℓ+2 sup f ≤ 2 (πp) [0,1]

where we used Proposition 4.1 (i ), and the fact that δf (2ℓ+1) sup f (2ℓ+2) . ≤ [0,1] (ℓ) ℓ Note that the Romberg’s rule p subdivides the interval [0 , 1] into 2 p equal subintervals. In the particular case p = 1, we obtainT 1 10 f(t)dt (ℓ)(f) sup f (2ℓ+2) . − T1 ≤ 2(ℓ+2)(ℓ+1) π2ℓ+2 Z0 [0,1]

10. Asymptotic expansions for the sum of certain series related to harmonic numbers

n Recall that the sequence of harmonic numbers (Hn)n N is deﬁned by Hn = 1/k (with the ∈ k=1 convention H0 = 0). It is well-known that lim (Hn ln n)= γ, where γ 0.57721 56649 is the so called n − ≈ P Euler-Mascheroni Constant. →∞

In the next proposition, the asymptotic expansion of (Hn)n N is presented. ∈ Proposition 10.1. For every positive integer n and nonnegative integer m, we have m 1 1 − b 1 H = ln n + γ + 2k + ( 1)mR , n 2n − 2k · n2k − n,m Xk=1 with 1/2 ∞ 1 1 Rn,m = B2m 1(t) dt | − | (j + t)2m − (j +1 t)2m 0 j=n Z X − b Moreover, 0 < R < | 2m| . n,m 2m n2m · Proof. Note that for j 1 we have ≥ 1 1 1 1 1 1 t ln 1+ = dt = dt j − j j − j + t j(j + t) Z0 Z0 Adding these equalities as j varies from 1 to n 1 we conclude that − 1 n 1 1 − t H ln n = dt. n − − n  j(j + t) 0 j=1 Z X Thus, letting n tend to , and using the monotone convergence theorem [4, Corollary 2.3.5], we conclude ∞ 1 ∞ t γ = dt.  j(j + t) 0 j=1 Z X   38 OMRAN KOUBA

It follows that 1 1 ∞ t γ + ln n H + = dt. − n n  j(j + t) 0 j=n Z X So, let us consider the function f : [0, 1] R deﬁned by  n −→ ∞ t f (t)= n j(j + t) j=n X Note that fn(0) = 0, fn(1) = 1/n, and that fn is inﬁnitely continuously derivable with (k) f (t) ∞ 1 n = ( 1)k+1 , for k 1. k! − (j + t)k+1 ≥ j=n X In particular, (2k 1) f − (t) ∞ 1 n = , for k 1. (2k 1)! (j + t)2k ≥ j=n − X (2m 1) So, fn − is decreasing on the interval [0, 1], and (2k 1) δf − ∞ 1 ∞ 1 1 n = = (2k 1)! (j + 1)2k − j2k −n2k j=n j=n − X X Applying Corollary 8.2 to fn, and using the above data, we get m 1 1 − b γ + ln n H + = 2k + ( 1)m+1R − n 2n 2kn2k − n,m kX=1 with 1/2 ∞ 1 1 Rn,m = B2m 1(t) dt | − | (j + t)2m − (j +1 t)2m 0 j=n Z X − and 6 (2m 1)! 0 < R < · − . n,m (2π)2mn2m

What is important in this estimate is the lower bound, i.e. Rn,m > 0. In fact, considering separately the cases m odd and m even, we obtain, for every nonnegative integer m′:

′ 1 2m b 1 H < ln n + γ + 2k , n 2n − 2k · n2k Xk=1 and ′ 2m +1 1 b 1 H > ln n + γ + 2k . n 2n − 2k · n2k Xk=1

This yields the following more precise estimate for the error term:

m 1 − m 1 b2k b2m 0 < ( 1) Hn ln n γ + 2k < | 2| m (10.1) − − − − 2n 2k n ! 2m n kX=1 · · which is valid for every positive integer m. (see [15, Chapter 9].) BERNOULLI POLYNOMIALS AND APPLICATIONS 39

For example, for every positive integer n, we have 1 1 1 1 1 ln n + γ +

+ n γn− γn 1 0.5750000000 0.5833333333 2 0.5771653194 0.5776861528 4 0.5772147535 0.5772473055 8 0.5772156500 0.5772176845 16 0.5772156647 0.5772157918 32 0.5772156649 0.5772156728 64 0.5772156649 0.5772156654 128 0.5772156649 0.5772156649

+ Table 4. Euler’s γ belongs to the interval (γn−,γn ) for each n.

Now, consider the two sequences (cn)n 1 and (dn)n 1 defined by ≥ ≥ 1 c = H ln n γ and d = H ln n γ n n − − − 2n n n − − 1 For a positive integer p, we know according to Proposition 10.1 that cpn = n2 , it follows that the series 1 O n 1 ∞ c is convergent. Similarly, since d = c + and the series ∞ ( 1) − /n is convergent, we n=1 pn pn pn 2pn n=1 − conclude that ∞ ( 1)n 1d is also convergent. In what follows we aim to find asymptotic expansions, P n=1 − pn P (for large p,) of the following− sums: P ∞ ∞ 1 C = c = H ln(pn) γ (10.2) p pn pn − − − 2pn n=1 n=1 X X ∞ n 1 ∞ n 1 D = ( 1) − d = ( 1) − (H ln(pn) γ) (10.3) p − pn − pn − − n=1 n=1 X X Proposition 10.2. If p and m are positive integers and Cp is defined by (10.2), then m 1 − b ζ(2k) ζ(2m) C = 2k + ( 1)m ε , with 0 <ε < b , p − 2k p2k − 2m p2m p,m p,m | 2m| Xk=1 · · where ζ is the well-known Riemann zeta function. Proof. Indeed, we conclude from Proposition 10.1 that m 1 1 − b 1 ( 1)m r H ln(pn) γ = 2k + − pn,m . pn − − − 2pn − 2k p2k · n2k 2m p2m · n2m Xk=1 · · with 0 < r b . It follows that pn,m ≤ | 2m| m 1 − b ∞ 1 ( 1)m C = 2k + − r˜ . p − 2kp2k · n2k 2m p2m · p,m n=1 ! Xk=1 X · 40 OMRAN KOUBA

rpn,m ∞ wherer ˜p,m = n=1 n2m . Hence, P ∞ r ∞ 1 0 < r˜ = pn,m < b = b ζ(2m) p,m n2m | 2m| n2m | 2m| n=1 n=1 X X and the desired conclusion follows with εp,m =r ˜p,m/ζ(2m).

For example, when m = 3, we obtain ∞ 1 π2 π4 1 H ln(pn) γ = + + . pn − − − 2pn −72p2 10800p4 O p6 n=1 X Similarly, in the next proposition we have the analogous result corresponding to Dp.

Proposition 10.3. If p and m are positive integers and Dp is deﬁned by (10.3), then m 1 ln 2 − b2kη(2k) m η(2m) D = + ( 1) ε′ , with 0 <ε′ < b , p 2p − 2k p2k − 2m p2m p,m p,m | 2m| Xk=1 · · where η(2k) is deﬁned in Corollary 3.2.

Proof. Indeed, let us deﬁne an,m by the formula m 1 1 − b a = H ln n γ + 2k n,m n − − − 2n 2k n2k Xk=1 · with empty sum equal to 0. We have shown in the proof of Proposition 10.1 that 1/2 m ( 1) an,m = B2m 1(t) gn,m(t) dt − | − | Z0 where gn,m is the positive decreasing function on [0, 1/2] deﬁned by

∞ 1 1 g (t)= . n,m (j + t)2m − (j +1 t)2m j=n X − Now, for every t [0, 1/2] the sequence (gnp,m(t))n 1 is positive and decreasing to 0. So, Using the alternating series criterion∈ [3, Theorem 7.8, and Corollary≥ 7.9] we see that, for every N 1 and t [0, 1/2], ≥ ∈ ∞ n 1 1 ( 1) − gnp,m(t) gNp,m(t) gNp,m(0) = . − ≤ ≤ (Np)2m n=N X This proves the uniform convergence on [0, 1/ 2] of the series

∞ n 1 G (t)= ( 1) − g (t). p,m − np,m n=1 X Consequently 1/2 m ∞ n 1 ( 1) ( 1) − apn,m = B2m 1(t) Gp,m(t) dt. − − | − | n=1 0 X Z Now using the properties of alternating series, we see that for t (0, 1/2) we have ∈ ∞ 1 1 1 0 < G (t)

1/2 with 0 <ρp,m < B2m 1(t) dt. 0 | − | R On the other hand we have

m 1 ∞ ∞ n 1 − ∞ n 1 n 1 1 ( 1) − b2k ( 1) − ( 1) − a = D − + − − pn,m p − 2p n 2kp2k n2k n=1 n=1 n=1 X X Xk=1 X m 1 ln 2 − b η(2k) = D + 2k . p − 2p 2k p2k Xk=1 ·

Thus

m 1 ln 2 − b η(2k) ( 1)m D = 2k + − ρ p 2p − 2k p2k p2m p,m Xk=1 ·

Now, the important estimate for ρp,m is the lower bound, i.e. ρp,m > 0. In fact, considering separately the cases m odd and m even, we obtain, for every nonnegative integer m′:

′ ln 2 2m b η(2k) D < 2k , p 2p − 2k p2k kX=1 · and ′ 2m +1 ln 2 b η(2k) D > 2k . p 2p − 2k p2k Xk=1 ·

This yields the following more precise estimate for the error term:

m 1 m ln 2 − b2kη(2k) b2m η(2m) 0 < ( 1) Dp + 2k < | | 2m , − − 2p 2kp ! 2m p kX=1 ·

and the desired conclusion follows.

In the next lemma, we will show that there are other alternating series that can be expressed in terms of D , This lemma will be helpful in 11. p §

Lemma 10.4. For a positive integer p, we have

∞ π E def= ( 1)n(H H ) = ln p + γ ln +2D , (10.4) p − p(n+1) − pn − 2 p n=0 X

where Dp is the sum deﬁned by (10.3). 42 OMRAN KOUBA

Proof. Indeed

∞ n 1 ∞ n 1 2D = d + ( 1) − d + ( 1) − d p p − pn − pn n=2 n=1 X X ∞ n ∞ n 1 = d + ( 1) d + ( 1) − d p − p(n+1) − pn n=1 n=1 X X ∞ n 1 = d + ( 1) − (d d ) p − pn − p(n+1) n=1 X ∞ n ∞ n 1 n +1 = d + ( 1) (H H )+ ( 1) − ln p − p(n+1) − pn − n n=1 n=1 X X ∞ ∞ n n 1 n +1 = ln p γ + ( 1) (H H )+ ( 1) − ln − − − p(n+1) − pn − n n=0 n=1 X X

Using Wallis formula for π [13, Formula 0.262], we have

∞ n 1 n +1 ∞ 2n 2n ( 1) − ln = ln − n 2n 1 · 2n +1 n=1 n=1 X X − ∞ 1 π = ln 1 = ln − − 4n2 2 n=1 Y

and the desired formula follows.

11. Asymptotic expansions for certain trigonometric sums

In this section we aim to exploit the results of the previous sections to study the following trigonometric sums deﬁned for a positive integer p by the formulæ:

p 1 p 1 − 1 − kπ I = = csc (11.1) p sin(kπ/p) p kX=1 kX=1 p 1 − kπ J = k cot (11.2) p p kX=1 with empty sums interpreted as 0.

p 2kπ While there is a of favourable result [24] concerning the sum k=1 sec 2p+1 , and many favourable p 1 2n results [6] concerning the power sums k−=1 csc (kπ/p), it seemsP that there is no known closed form for Ip, and the same can be said about the sum Jp. Therefore, we will look for asymptotic expansions for P these sums and will give some tight inequalities that bound Ip and Jp. This investigation complements the work of H. Chen in [7, Chapter 7.], and answers an open question raised there.

In the next lemma we give some equivalent forms for the trigonometric sums under consideration. BERNOULLI POLYNOMIALS AND APPLICATIONS 43

Lemma 11.1. For a positive integer p let p 1 p 1 − kπ − kπ K = tan , K = cot , p 2p p 2p kX=1 kX=1 p 1 p 1 − k e − (2k + 1)π L = ,M = (2k + 1)cot p sin(kπ/p) p 2p kX=1 kX=0 Then, i. Kp = Kp = Ip. ii. Lp = (p/2) Ip. iii. M = (ep/2) J 2J = p I . p 2p − p − p Proof. First, note that the change of summation variable j p j proves that Kp = Kp. So, using the trigonometric identity tan θ + cot θ = 2csc(2θ) we obtain ← − p 1 e − kπ kπ 2K = K + K = tan + cot p p p 2p 2p Xk=1 p 1 − e kπ =2 csc =2I p p Xk=1 This proves (i). Similarly, (ii) follows from the change of summation variable j p j in L : ← − p p 1 − p k L = − = pI L p sin(kπ/p) p − p j=1 X Also, 2p 1 kπ − kπ kπ M = k cot = k cot k cot p 2p 2p − 2p 1 k<2p k=1 1 k<2p ≤kXodd X k≤Xeven 2p 1 p 1 − kπ − kπ = k cot 2 k cot = J 2J . 2p − p 2p − p Xk=1 Xk=1 But p 1 2p 1 − kπ − kπ J = k cot + k cot 2p 2p 2p kX=1 k=Xp+1 p 1 p 1 − kπ − kπ = k cot (2p k)cot 2p − − 2p kX=1 Xk=1 p 1 − kπ =2 k cot 2pK 2p − p Xk=1 Thus, using (i) and the trigonometric identity cot(θ/2) ecot θ = csc θ we obtain − p 1 − kπ kπ M = J 2J =2 k cot cot 2pI p 2p − p 2p − p − p Xk=1 p 1 − kπ =2 k csc 2pI =2L 2pI = pI p − p p − p − p Xk=1 This concludes the proof of (iii). 44 OMRAN KOUBA

Proposition 11.2. For p 2, let I be the sum of cosecants deﬁned by the (11.1). Then ≥ p 2 ln 2 2p I = + E , p − π π p 2 ln 2 2p 4p = + (ln p + γ ln(π/2)) + D , − π π − π p where Dp and Ep are deﬁned by formulæ (10.3) and (10.4) respectively. Proof. Indeed, our starting point will be the “simple fractions” expansion (6.4) of the cosecant function:

π ( 1)n 1 ∞ 1 1 = − = + ( 1)n + sin(πα) α n α − α n α + n n Z n=1 X∈ − X − which is valid for α C Z. Using this formula with α = k/p for k = 1, 2,...,p 1 and adding, we conclude that ∈ \ − p 1 p 1 π − 1 ∞ − 1 1 I = + ( 1)n + p p k − k np k + np n=1 Xk=1 X Xk=1 − p 1 pn 1 p(n+1) 1 − 1 ∞ − 1 − 1 = + ( 1)n + , k − − j j  k=1 n=1 j=p(n 1)+1 j=pn+1 X X X− X   and this result can be expressed in terms of the Harmonic numbers as follows

∞ π n Ip = Hp 1 + ( 1) Hpn 1 + Hp(n 1) + Hp(n+1) 1 Hpn p − − − − − − − n=1 X ∞ n 1 ∞ n 1 1 = Hp 1 + ( 1) Hp(n+1) 2Hpn + Hp(n 1) + ( 1) − − − − p − n − n +1 n=1 n=1 X X n n ∞ n 1 ∞ ( 1) ∞ ( 1) = Hp 1 + ( 1) Hp(n+1) 2Hpn + Hp(n 1) + − + − − − − − p n n n=1 n=1 n=2 ! X X X ∞ ∞ n 2 n 1 1 = Hp + ( 1) Hp(n+1) 2Hpn + Hp(n 1) ( 1) − − − − − p − n n=1 n=1 X X 2 ln 2 ∞ n = Hp + ( 1) Hp(n+1) 2Hpn + Hp(n 1) . − p − − − n=1 X Thus

π 2 ln 2 ∞ n ∞ n Ip + = Hp + ( 1) Hp(n+1) Hpn + ( 1) Hp(n 1) Hpn p p − − − − − n=1 n=1 X X ∞ ∞ n n = ( 1) Hp(n+1) Hpn + ( 1) Hp(n 1) Hpn − − − − − n=0 n=1 X X = Ep + Ep =2Ep, and the desired formula follows according to Lemma 10.4. Combining Proposition 11.2 and Proposition 10.3, we obtain: Proposition 11.3. For p 2 and m 1, we have ≥ ≥ m 1 − 2b2kη(2k) m 2η(2m) πIp =2p ln p + 2(γ ln(π/2))p 2k 1 + ( 1) 2m 1 εp,m′ − − k p − − m p − Xk=1 · · with 0 <ε′ < b . p,m | 2m| BERNOULLI POLYNOMIALS AND APPLICATIONS 45

Using the values of the η(2k)’s from Corollary 3.2, and considering separately the cases m even and m odd we obtain the following corollary.

Corollary 11.4. For every positive integer p and every nonnegative integer n, the sum of cosecants Ip deﬁned by (11.1) satisﬁes the following inequalities: 2n 2k 1 2p (22k 2)b2 π − I < (ln p + γ ln(π/2)) + ( 1)k − 2k , p π − − k (2k)! p kX=1 · and 2n+1 2k 1 2p (22k 2)b2 π − I > (ln p + γ ln(π/2)) + ( 1)k − 2k . p π − − k (2k)! p Xk=1 · As an example, for n = 0 we obtain the following inequality, valid for every p 1 : ≥ 2p π 2p (ln p + γ ln(π/2)) < I < (ln p + γ ln(π/2)). π − − 36p p π − This answers positively the open problem proposed in [7, Section 7.4].

Remark 11.5. The asymptotic expansion of Ip was proposed as an exercise in [17, Exercise 13, p. 460], and it was attributed to P. Waldvogel, but the result there is less precise than Corollary 11.4

Now we turn our attention to the other trigonometric sum Jp. Next we ﬁnd an analogous result to Proposition 11.2.

Proposition 11.6. For a positive integer p, let Jp be the sum of cotangents defined by the (11.2). Then πJ = p2 ln p + (ln(2π) γ)p2 p +2p2C p − − − p where Cp is defined by the formula (10.2). Proof. Recall that c = H ln n γ 1 satisfies c = (1/n2). Thus, both series n n − − − 2n n O ∞ ∞ n 1 C = c and C = ( 1) − c p pn p − pn n=1 n=1 X X are convergent. Further, we note that C = D ln 2 ewhere D is defined by (10.3). p p − 2p p According to Remark 11.10 we have e ln(π/2) γ ln p π C = − − + I . (11.3) p 2 4p p Now, noting that e ∞ Cp = cpn + cpn = cpn + c2pn n 1 n 1 n 1 n=1 nXodd≥ nXeven≥ nXodd≥ X ∞ C = c c = c c p pn − pn pn − 2pn n 1 n 1 n 1 n=1 X≥ nXeven≥ X≥ X e n odd n odd we conclude that C C =2C , or equivalently p − p 2p C 2C = C (11.4) e p − 2p p On the other hand, for a positive integer p let us define Fp by e ln p + γ ln(2π) 1 π F = − + + J . (11.5) p 2 2p 2p2 p 46 OMRAN KOUBA

It is easy to check, using Lemma 11.1 (iii), that ln(π/2) ln p γ π F 2F = − − (J 2J ) p − 2p 2 − 4p2 2p − p ln(π/2) ln p γ π = − − + I (11.6) 2 4p p We conclude from (11.4) and (11.6) that C 2C = F 2F , or equivalently p − 2p p − 2p C F = 2(C F ). p − p 2p − 2p Hence, m m 1, C F =2 (C m F m ) (11.7) ∀ ≥ p − p 2 p − 2 p Now, using Proposition 10.1 to replace Hp in Proposition 9.2, we obtain π 1 1 J = ln(2π) H + p2 p − p − 2p O p2 1 1 = ln(2π) ln p γ + − − − p O p2 1 1 1 Thus F = 2 . Similarly, from the fact that c = 2 we conclude also that C = 2 . p O p n O n p O p 2 Consequently, there exists a constant κ such that, for large values of p we have Cp Fp κ/p . So, from (11.7), we see that for large values of m we have | − | ≤ κ C F | p − p|≤ 2mp2 and letting m tend to + we obtain C = F , which is equivalent to the announced result. ∞ p p

Combining Proposition 11.6 and Proposition 10.2, we obtain: Proposition 11.7. For p 2 and m 1, we have ≥ ≥ m 1 2 2 − b2kζ(2k) m ζ(2m) πJp = p ln p + (ln(2π) γ)p p 2k 2 + ( 1) 2m 2 εp,m, − − − − k p − − m p − kX=1 · · with 0 <ε < b , where ζ is the well-known Riemann zeta function. p,m | 2m| Using the values of the ζ(2k)’s from Corollary 3.2, and considering separately the cases m even and m odd we obtain the next corollary.

Corollary 11.8. For every positive integer p and every nonnegative integer n, the sum of cotangents Jp deﬁned by (11.2) satisﬁes the following inequalities:

2n 2k 2 1 b2 2π − J < p2 ln p + (ln(2π) γ)p2 p +2π ( 1)k 2k , p π − − − − k (2k)! p Xk=1 · and 2n+1 2k 2 1 b2 2π − J > p2 ln p + (ln(2π) γ)p2 p +2π ( 1)k 2k . p π − − − − k (2k)! p k=1 · X As an example, for n = 0 we obtain the following double inequality, which is valid for p 1 : ≥ 1 π 0 < p2 ln p + (ln(2π) γ)p2 p J < π − − − − p 36 BERNOULLI POLYNOMIALS AND APPLICATIONS 47

Remark 11.9. It is not clear that one can prove Proposition 11.7 by combining Propositions 9.2 and 10.1 directly. Remark 11.10. Note that we have proved the following results. For a postive integer p: p 1 ∞ − n 1 ln(π/2) γ ln p ln 2 π kπ ( 1) − (H ln(pn) γ)= − − + + csc . − pn − − 2 2p 4p p n=1 X kX=1 p 1 ∞ ln 2 π − kπ ( 1)n(H H )= + csc . − p(n+1) − pn p 2p p n=0 X kX=1 p 1 ∞ 1 ln p + γ ln(2π) 1 π − kπ H ln(pn) γ = − + + k cot . pn − − − 2pn 2 2p 2p2 p n=1 X Xk=1 These results are to be compared with those in [22], see also [23].

12. Endnotes

Bernoulli numbers appeared in the work Jacob Bernoulli (1654-1705, Basel, Switzerland) in connec- n k tion with evaluating the sums of the form i=1 i . They appear in Bernoulli’s most original work Ars Conjectandi, (“The Art of Conjecturing”,) published by his nephew in Basel in 1713, eight years after his death. They also appear independantly in theP work of the Japanese Mathematician Seki Takakazu (Seki K¯owa) (1642-1708). Bernoulli numbers are related to Fermat’s Last Theorem by Kummer’s theorem [25], this opened a very large and furctuous ﬁeld of investigation (see [19, Chapter 15].) The Euler-Maclaurin’s Summation Formula was developed independantly by Leonhard Euler (1736) and Colin MacLaurin (1742). The aim was to provide a powerful tool for the computation of several sums like the harmonic numbers. Finally, a thorough bibliography on Bernoulli numbers and their applications that contains more than 3000 entries can be found on the web [11]. References [1] Abramowitz, M. and Stegan, I. A., Handbook of Mathematical Functions, with Formulas, Graphs, and Mathematical Tables, Dover Books on Mathematics, Dover Publication, Inc., New York, (1972). [2] Ahlfors, L. V., Complex Analysis, McGraw-Hill, Inc. (1979). [3] Amann, H. and Escher, J., Analysis I, Birkh¨auser Verlag, Basel -Boston-Berlin. (2005). [4] Athreya, K. B. and Lahiri, S. N., Measure Theory and Probability Theory, Springer Science+Business Media, LLC. (2006). [5] Bergmann, H., Eine explizite Darstellung der Bernoullischen Zahlen, Math. Nachr., 34, (1967), pp.377-378. [6] Chen, H., On some trigonometric power sums, Internat. J. Math. Math. Sci, 30, (2002), 185–191. [7] , Excursions in Classical Analysis, Mathematical Association of America, Inc. (2010). [8] Deeba, E. Y. and Rodrigues, D. M., Stirling’s Series and Bernoulli Numbers, The American Mathematical Monthly, 98, 5 (1991), pp.423–426. [9] Dilcher, K., Asymptotic Behaviour of Bernoulli, Euler, and Generalized Bernoulli Polynomials, Journal of Approxima- tion Theory, 49, (1987), 321–330. [10] , Zeros of Bernoulli, generalized Bernoulli and Euler polynomials, Mem. Amer. Math. Soc., Number 386, (1988). [11] Dilcher, K. and Slavutskii, I. Sh., A Bibliography of Bernoulli Numbers (1713-2007). [ONLINE : http://http://www.mathstat.dal.ca/∼dilcher/bernoulli.html]. [12] Gould. H. W., Explicit formulas for Bernoulli numbers, The American Mathematical Monthly, 79, 1 (1972), pp.44–51. [13] Gradshteyn, I. and Ryzhik, I., Tables of Integrals, Series and Products, 7th ed., Academic Press, (2007). [14] Grafakos, L., Classical Fourier Analysis, Second edition., Graduate Texts in Mathemetics, Springer Science+Business Media, LLC. (2008). 48 OMRAN KOUBA

[15] Graham, R. L., Knuth, D. E., and Patashnik, O. Concrete Mathematics : a foundation for computer science, 2nd ed. Addison-Wesley Publishing Company, Inc, (1994). [16] Hardy, G. H. and Wright, E. M., An introduction to the theory of numbers, 6th ed., Oxford University Press, (2007). [17] Henrici, P., Applied and Computational Complex Analysis, Vol. 2, John Wiley & Sons, New York, (1977). [18] Inkeri, K., The real roots of Bernoulli polynomials, Ann. Univ. Turku. Ser. A I37 (1959), 20pp. [19] Ireland, K. and Rosen, M., A Classical Introduction to Modern Number Theory, 2nd ed., Springer-Verlag, New York, Inc, (1990). [20] Katznelson, Y., Introduction To Harmonic Analysis, 3rd ed., Cambridge University Press, (2004). [21] Knuth, D. E., Eulers constant to 1271 places, Math. Comput., 16, (1962), pp. 275–281. [22] Kouba, O., The sum of certain series related to harmonic numbers, Octogon Mathematical Magazine, 19, No. 1 (2011), pp.3–18. [ONLINE : http://www.uni-miskolc.hu/∼matsefi/Octogon/]. [23] , Proposed Problem 11499, The American Mathematical Monthly, 117, 7 (2010), p.371. [24] Kouba, O., and Andreescu, T., Solution to Problem U207, Mathematical Reflections, Issue 5, (2011), p.16. [ONLINE : http://www.awesomemath/home]. [25] Kummer E. E.,Allgemeiner Beweis des Fermat’schen Satzes, dass die Gleichung xλ + yλ = zλ durch ganze Zahlen unlösbar ist, für alle diejenigen Potenz-Exponenten λ, welche ungerade Primzahlen sind und in den Zählern der ersten (λ − 3)/2 Bernoulli’schen Zahlen als Factoren nicht vorkommen, J. Reine Angew. Math. 40 (1850), pp.131–138. [26] Lehmer, D. H., On the maxima and minima of Bernoulli Polynomials, The American Mathematical Monthly, 47, 8 (1940). pp, 533–358. [27] Namias, V., A simple Derivation of Stirling’s Asymptotic Series, The American Mathematical Monthly, 93, 1 (1986). pp, 25–29.

Department of Mathematics, Higher Institute for Applied Sciences and Technology, P.O. Box 31983, Damascus, Syria. E-mail address: omran [email protected]