Notes on Number Theory and Discrete Mathematics Vol. 19, 2013, No. 1, 59–65

A note on Bernoulli numbers

Ramesh Kumar Muthumalai

Department of Mathematics, D.G. Vaishnav College Chennai-600106, Tamil Nadu, India e-mail: [email protected]

Abstract: Some explicit formulae for Bernoulli numbers and Bernoulli polynomials are derived. Some rational approimations to powers of π are given in terms of Bernoulli numbers. Keywords: Bernoulli numbers, Bernoulli polynomials, Stirling numbers, Identities, π. AMS Classification: 11B68.

1 Introduction

The Bernoulli numbers Bn, n ∈ W are a sequence of rational numbers with many interesting arithmetic properties. They are defined by the following generating function [5, p. 525].

∞ t X tn = B , |t| < 2π. (1.1) et − 1 n n! k=0

The appearances of Bernoulli numbers throughout mathematics are abundant and include find- ing a formula for the sum of powers of the first n positive integers, values of L-functions and Euler-Macluarin summation formulae [3]. There exists many recursion and explicit formulae for Bernoulli numbers. The explicit formula in terms of Stirling numbers of second kind given in [5, p. 461] is as follows

k X m! B = (−1)m S(k). (1.2) k m + 1 m m=1

(k) Where Sm striling number of second kind. Also, double series representaion given in [5, p. 537] is as follows

m k X X k vm B = (−1)v . (1.3) m v k + 1 k=0 v=0

59 In the present study, similar identities for Bernoulli numbers as shown in (1.2) and (1.3) are derived through Stirling numbers of second kind and double series in different form. Further, some rational approximations to powers of π are given in terms of Bernoulli numbers.

2 Some lemmas and remarks

th Lemma 2.1. Let n ∈ W and Bn denotes n Bernoulli number. If " #  x  1 − ex n+1 T (x) = + 1 − 1 , (2.1) 1 − ex x then

(n) T (0) = Bn. (2.2)

(n) dnT (x) Where T (x) = dxn . Proof. It is well known that

∞ 1 − ex X xk−1 + 1 = − . (2.3) x k! k=2

Differentiating (2.1) n times and using Leibniz rule, then

n " n+1 # X  n   x  1 − ex  DnT (x) = Dm Dn−m + 1 − 1 . (2.4) m 1 − ex x m=0

n dn Where D = dxn . Using (1.1) and (2.3) in (2.4), then setting x = 0 and after simplification, completes the Lemma.

th Lemma 2.2. Let n ∈ W and Bn(t) denotes n degree Bernoulli polynomial. If

U(t, x) = etxT (x), (2.5)

then

(n) U (t, 0) = Bn(t). (2.6)

(n) dnU(x,t) Where U (x, t) = dxn . Proof. Given that

U(t, x) = etxT (x).

60 Differentiating n times and using Leibnitz rule, then

n dn X n dn−k U(t, x) = tketx T (x). dxn k dxn−k k=0

Putting x = 0, gives

n X n U (n)(t, 0) = tkT (n−k)(0). k k=0

(n−k) Lemma (2.1) shows that T (0) = Bn−k, then gives

n X n U (n)(t, 0) = B tk. (2.7) k n−k k=0

It is well known that [3, p. 231]

n X n B (t) = B tk. (2.8) n k n−k k=0

Using (2.8) in (2.7) gives (2.6).

(m) Remark 2.3. Let m, r ∈ W and Sm+r is Stirling number of second kind [2, p. 1037]. Then

 x m r e − 1 m!r! (m) D = Sm+r. (2.9) x x=0 (m + r)!

(−n) Remark 2.4. Let m, n ∈ W and Bm (t) is Bernoulli polynomials of degree m of order −n [4]. Then

 x m r e − 1 xt (−n) D e = Br (t). (2.10) x x=0

r dr Where D = dxr .

3 Main Results

(n) Theorem 3.1. Let Sm be Stirling number of second kind. Then

n−1   X n + 1 (n − k)! (n−k) B = n! (−1)n−k S . (3.1) n k (2n − k)! 2n−k k=0

Proof. From equation (2.1), that " #  x  1 − ex n+1 T (x) = + 1 − 1 . 1 − ex x

61 After simplification, it gives

n n−k X n + 1 1 − ex  T (x) = (−1)n−k . (3.2) k x k=0

Differentiating (3.2) n times with respect to x and putting x = 0 and then using Remark 2.3, yields

n   X n + 1 (n − k)! (n−k) T (n)(0) = n! S . k (2n − k)! 2n−k k=0

Using Lemma 2.1 and after simplification, gives (3.1). This completes the theorem.

Example 3.2. Let n = 1, 2, 3, 4 in (3.1). Then the first four Bernoulli numbers can be expressed through Stirling numbers of second kind as follows

1 B = − S(1). 1 2 2 1 B = S(2) − S(1). 2 6 4 3 1 2 3 B = − S(3) + S(2) − S(1). 3 20 6 5 5 2 4 1 1 2 B = S(4) − S(3) + S(2) − 2S(1). 4 70 8 7 7 3 6 5 Theorem 3.3. Let n ∈ W . Then

n−1 n−k X X n + 1n − k r2n−k B = n! (−1)2n−2k−r . (3.3) n k r (2n − k)! k=0 r=0

Proof. Let us consider the identity given in Theorem 3.1

n−1   X n + 1 (n − k)! (n−k) B = n! (−1)n−k S . n k (2n − k)! 2n−k k=0

Using explicit formula for Stirling number of second kind [2, p.1031], then

n−1 n−k X n + 1 (n − k)! 1 X n − k B = n! (−1)n−k (−1)n−k−r r2n−k. n k (2n − k)! (n − k)! r k=0 r=0

After simplification, it gives

n−1 n−k X X n + 1n − k r2n−k B = n! (−1)2n−2k−r . n k r (2n − k)! k=0 r=0

This completes theorem.

62 (−n) th Theorem 3.4. Let Bm (t) be the m degree Bernoulli polynomial of order −n. Then

n   X n + 1 (k−n) B (t) = B (t). (3.4) n k k k=0

Proof. Let us consider the identity (2.5) " #  xext  1 − ex n+1 U(t, x) = + 1 − 1 . 1 − ex x

After simplification, it gives

n n−k X n + 1 1 − ex  U(t, x) = (−1)n−k ext. k x k=0

Differentiating above equation n times with respect to x

n n−k X n + 1 1 − ex  DnU(t, x) = (−1)n−k Dn ext. k x k=0

Putting x = 0 and then using Remark 2.4, yields

n   X n + 1 (k−n) B (t) = B (t). n k k k=0

This completes the theorem.

4 Some approximations to powers of π

22 355 The simplest rational appromiations to π are 7 and 113 correction up to two and six decimal places respectively. But, the powers of 22/7 and 355/113 does not converge to two and six decimals. For instance 234256 π4 − (22/7)4 = π4 − = 0.15692. 2401 and 15882300625 π8 − (355/113)8 = π8 − = 0.00645. 163047361 In this section, some rational approximations to powers of π are given in terms of Bernoulli numbers.

Theorem 4.1. Let p1, p2, . . . , pn are first m prime numbers and n ∈ N.

2.2n! 1 4n! |B | 1 2(2π)2n ≈ + 2n . (4.1) |B2n| P (2n) 2n! |B4n| Q(2n)

63 where

m Y  1  P (n) = 1 − (4.2) pn k=1 k m Y  1  Q(n) = 1 + (4.3) pn k=1 k

Proof. It is well known that from Ref [2, p. 1031]

Y  1  2.2n! 1 1 − = (−1)n−1 . (4.4) p2n (2π)2n B p 2n

It can be written as follows ∞   Y 1 n−1 2.2n! 1 1 1 − 2n = (−1) 2n . (4.5) p (2π) B2n P (2n) k=m+1 k

Similarly,

∞   Y 1 n−1 4n! 1 B2n 1 1 + 2n = (−1) 2n . (4.6) p 2n! (2π) B4n Q(2n) k=m+1 k

Using (4.5) and (4.6), gives

2.2n! 1 1 4n! 1 |B2n| 1 2 ≈ 2n + 2n . (4.7) (2π) |B2n| P (2n) 2n! (2π) |B4n| Q(2n)

After simplification, it completes the theorem. Example 4.2. The following are some rational approximations to powers of π. 1. Let n = 2 and m = 2 in (4.1). Then 1656 339471 π4 = and π4 = 17 3485 gives correction up to 2 and 4 decimals respectively.

2. Let n = 3 and m = 2 in (4.1). Then 664320 4413077316 π6 = and π6 = 691 4590313 gives correction up to 4 and 6 decimals respectively.

3. Similarly, for n = 4 and m = 1

149944152960 π8 = 15802673 gives correction up 5 decimals.

64 References

[1] Apostal, T. M., Introduction to Analytic Number Theory, Springer International Student Edition, New York, 1989.

[2] Gradshteyn, I. S., I. M. Ryzhik, Tables of Integrals, Series and Products, 6th Edition, Aca- demic Press, USA, 2000.

[3] Ireland, K., M. Rosen, A Classical Introduction to Modern Number Theory, 2nd Edition, Springer-verlag, New York, 1990.

[4] Muthumalai, R. K. Some properties and applications of generalized Bernoulli polynomials, Int. Jour. Appl. Math., Vol. 25, 2012, No. 1, 83–93.

[5] Sandor,´ J., B. Crstici, Handbook of Number Theory II, Springer, Kluwer Acedemic Publish- ers, The Netherlands, 2004.

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