Bernoulli Numbers and Polynomials

Transcription

1 Bernoull Numbers and Polynomals T. Muthukumar 17 Jun 2014 The sum of frst n natural numbers 1, 2, 3,..., n s n n(n + 1 S 1 (n := m = = n n 2. Ths formula can be derved by notng that S 1 (n = n Therefore, summng term-by-term, S 1 (n = n + (n S 1 (n = (n (n + 1 = n(n + 1. }{{} n tmes An alternate way of obtanng the above sum s by usng the followng two dentty: ( (m m 2 = 2m + 1 and, hence, n n [ 2 m = (m m 2] n. ( n [ (m m 2] = [ ] + [ ] [n 2 (n 1 2 ] + [(n n 2 ] = (n

2 Thus, S 1 (n = (n + 12 (n + 1 n(n + 1 =. 2 2 More generally, the sum of k-th power of frst n natural numbers s denoted as S k (n := 1 k + 2 k n k. Snce a 0 = 1, for any a, we have S 0 (n = n. For k N, one can compute S k (n usng the denttes: ( and, hence, ( n m k = (m + 1 k+1 m k+1 = ( m =0 n [ (m + 1 k+1 m k+1] k 1 =0 n ( m. ( n [ (m + 1 k+1 m k+1] = (n + 1 k+1 1. Thus, ( n k 1 ( n m k = (n + 1 k+1 1 m =0 (n + 1k+1 (n + 1 S k (n = 1 k 1 ( S (n. The formula obtaned n RHS s a (-degree polynomal of n. Usng the =1 2

3 above formula, one can compute S 2 (n = n3 3 + n2 2 + n 6, S 3 (n = n4 4 + n3 2 + n2 4, S 4 (n = n5 5 + n4 2 + n3 3 n 30, S 5 (n = n6 6 + n n4 12 n2 12, S 6 (n = n7 7 + n6 2 + n5 2 n3 6 + n 42, S 7 (n = n8 8 + n n6 12 7n n2 12,... James/Jacques/Jakob Bernoull observed that the sum of frst n whole numbers rased to the k-th power can be concsely wrtten as, S k (n = nk n 12 knk n k Note that the coeffcents 1, 1/2, 1/12, 0,... are ndependent of k. Bernoull rewrote the above expresson as Jakob and, hence, (S k (n = n k (n k( n k (k 2(k 1k( ! S k (n = 1 ( B n k+1, (1 =0 where B are the -th Bernoull numbers and the formula s called Bernoull formula. An easer way to grasp the above formula for S k (n s to rewrte 1 t as 1 Umbral Calculus S k (n = (n + Bk+1 B k+1 3

4 where B s a notaton used to dentfy the -th power of B wth the -th Bernoull number B and k+1 ( (n + B k+1 := B n k+1. =0 Ths notaton also motvates the defnton of Bernoull polynomal of degree k as ( k B k (t := B t k =0 where B are the Bernoull numbers. In terms of Bernoull polynomals, the k-th Bernoull number B k = B k (0. Two quck observaton can be made from (1. ( There s no constant term n S k (n because does not take. ( The k-th Bernoull number, B k, s the coeffcent of n n S k (n. For nstance, B 0 s coeffcent of n n S 0 (n = n and, hence, B 0 = 1. Smlarly, B 1 = 1/2, B 2 = 1/6, B 3 = 0, B 4 = 1/30, B 5 = 0, B 6 = 1/42, B 7 = 0,.... The beauty about the sequence of Bernoull numbers s that one can compute them a pror and use t to calculate S k (n,.e., gven n N and k N {0} t s enough to know B, for all 0 k, to compute S k (n. We already computed B 0 = 1. Usng n = 1 n the Bernoull formula (1, we get 1 = 1 ( =0 B (2 and, ths mples that the k-th Bernoull number, for k > 0, s defned as B k = 1 1 k 1 =0 ( B. Recall that B 3, B 5, B 7 vansh. In fact, t turns out that B k = 0 for all odd k > 1. The odd ndexed Bernoull numbers vansh because they have no n-term. Snce there are no constant terms n S k (n, the vanshng of Bernoull numbers s equvalent to the fact that n 2 s a factor of S k (n. Some propertes of Bernoull numbers: 4

5 ( For odd k > 1, B k = 0. ( For even k, B k 0. ( B k Q for all k N {0}. (v B 0 = 1 s the only non-zero nteger. (v B 4j s negatve ratonal and B 4j 2 s postve ratonal, for all j N. (v B 6 = 1/42 < B 2k, for all k N. L. Euler gave a nce generatng functon for the Bernoull numbers. He seeked a functon f(x such that f (k (0 = B k where f (k denotes the k-th dervatve of f wth the conventon that f (0 = f. If such an f exsts then t admts the Taylor seres expanson, around 0, Therefore, for such a functon Recall the Taylor seres of e x, f(x = f(x = e x = f (k (0 xk k!. =0 B k x k k!. defned for all x R. Consder the product (dscrete convoluton/cauchy product of the nfnte seres ( ( f(xe x x k x k = B k k! k! ( x x k = B! (k! =0 ( B = x k!(k! =0 ( ( k x = B k k!. 5 x k k!

6 Let Then f(xe x = c k := ( k k 1 ( k B = B + B k = k + B k. =0 =0 (k + B k xk k! = x k (k 1! + f(x = xex + f(x. k=1 Thus, the f we seek satsfes f(x = xex e x 1 and s called the generatng functon. Snce e x > 0 for all x R, we may rewrte f(x as x f(x =. (3 1 e x The entre exercse of seekng f can be generalsed to complex numbers and f(z = z 1 e z z C wth 0 I(z < 2π. A word of cauton that denttes (2 and (3 are dfferent from the standard formulae because we have derved them for second Bernoull numbers, vz., wth B 1 = 1/2. The standard conventon s to work wth frst Bernoull numbers, vz., wth B 1 = 1/2. The frst Bernoull numbers can be obtaned by followng the approach of summng the k-th powers of frst n 1 natural numbers, for any gven n. The Bernoull numbers wth appeared whle computng S k (n s appears n many crucal places. (a In the expanson of tan z. tan z = k=1 ( 1 k 1 22k (2 2k 1B 2k 2n! z 2k 1. for all z < π/2. 6

7 (b In computng the sum of Remann-zeta functon ζ(s = for postve even ntegers s. The case s = 2 s the famous Basel problem computed by Euler to be π 2 /6. Theorem 1 (Euler. For all k N, n=1 1 n s k+1 (2π2k ζ(2k = ( 1 2(2k! B 2k. Further, the relaton ζ( 2k = B 2k+1 2k+1 gves the trval zeroes of the Remann zeta functon. (c The Bernoull numbers were also used as an attempt to prove Fermat s last theorem (already dscussed n a prevous artcle/blog. Defnton 1. An odd prme number p s called regular f p does not dvde the numerator of B k, for all even k p 3. Any odd prme whch s not regular s called rregular. The odd prmes 3, 5, 7,..., 31 are all regular prmes. The frst rregular prme s 37. It s an open queston: are there nfntely many regular prmes? However, t s known that there are nfntely many rregular prmes. Theorem 2 (Kummer, If p s a regular odd prme then the equaton a p + b p = c p has no soluton n N. 7