THE BERNOULLI NUMBERS. t k. = lim. = lim = 1, d t B 1 = lim. 1+e t te t = lim t 0 (e t 1) 2. = lim = 1 2.
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1 THE BERNOULLI NUMBERS The Bernoulli numbers are defined here by he exponenial generaing funcion ( e The firs one is easy o compue: (2 and (3 B 0 lim 0 e lim, 0 e ( d B lim 0 d e +e e lim 0 (e 2 lim 0 2(e lim 0 2e 2. Ideally one would lie o obain a recurrence for hese numbers. The only ool we have is he ordinary generaing funcion, so we wor wih i. The relaion ( is wrien as (4 ( e ( ha can be wrien as (5! Dividing by we can wrie he series on he lef as (6! ( +! and (5 becomes (7 ( +!
2 2 THE BERNOULLI NUMBERS How does one muliply series. In order o simplify (7 we will obain an expression for he produc of wo power series: (8 a b a b +. The double sum on he righ corresponds o summing over all poin on he firs quadran N 0 N 0. The same se of indices can be covered by lines of slope r, ha is, summing over all indices (i, wih r i+ fixed and hen summing over all values of r N 0. This gives (9 a b + a b +. r0 +r In he inner sum he value of he index r is fixed and we now eliminae he index, o obain r wih he range 0 r. This gives (0 a b + a r b r. r0 i+r r0 The conclusion is ha ( ( a b a r b r. This can also be wrien in he following form: ( a b a r b (2! r (r r0 ( ( r a r r b r! r0 r0 Theorem. The coefficien of r in he produc (3 a b is (4 a r b a b r. The coefficien of r /r! in he produc (5 a is (6! ( r a r b b ( r a b r.
3 THE BERNOULLI NUMBERS 3 Now apply he rule in (6 o he ideniy (7 wrien in he form (7 +! o obain (8 ( r r + { 0 if r 0 if r 0 Theorem 2. The Bernoulli numbers saisfy he recurrence r ( r (9 B r, for r > 0. r + Proof. Solve he relaion (8 for B r. Corollary 3. The Bernoulli numbers are raional numbers. This recurrence can be used o generae he sequence of Bernoulli numbers. The firs few are { (20, 2, 6, 0, 30, 0, 42, 0, 30, 0, 5 69, 0, , 0, 7 } 6 and i can be seen from his able ha, aside from B 2 he Bernoulli numbers wih odd index vanish. This mus no be hard o prove. Consider he generaing funcion (2 G( e and modify i o eliminae he erm corresponding o B. Tha is, define (22 G ( e + 2. This can be reduced o (23 and he second facor is (24 [ G ( e + ] 2 2 e + e e + e e/2 +e /2 e /2 e /2 and is is clear ha his is an odd funcion. Therefore, becuase of he exra facor /2, he funcion G ( is an even funcion. As such i has only even erms in is generaing funcion expansion. Theorem 4. For n odd and n 3, he Bernoulli number B n vanishes. Tha is (25 B 2n+ 0, for n.
4 4 THE BERNOULLI NUMBERS Wih his resul, he generaing funcion ( for he Bernoulli numbers can be wrien as (26 e 2 + B 2 2 (2. Many properies of he Bernoulli numbers are esablished by clever manipulaions of he generaing funcion. The deails given nex appear in a paper by L. J. Mordell in he American Mahemaical Monhly, volume 80, 973, pages Sar wih he ideniy (27 e + e 2 e 2 and expand boh sides in series. The righ-hand side is easy: (28 e 2 e 2 ( 2 2 (2. The expansion of he lef-hand side is no so obvious, bu he clever idea is o muliply by a nice facor. Indeed, (29 e + e 2 e 2 and he righ-hand side can be wrien as (30 and his can be expanded as (3 2 e e 2 2 e e Now ae he ideniy (27 and muliply i by /(e o produce ( ( ( e 2 e e 2 e 2 Expanding in series gives ( [ ] B (2 i i B i! i! The produc on he righ hand he form in (6 wih i (34 a B and b (2.
5 THE BERNOULLI NUMBERS 5 Therefore, he coefficien of r /r! for he produc on he righ is given by ( r (35 (2 i B i B r i i i0 and on he lef-hand side his coefficien is (36 B r 2 r 2. Therefore, if r is even, say r 2s wih s > he lef-hand side is 0 and we have 2s ( 2s (37 (2 i B i B 2s i 0. i i0 The erm for i 0 vanishes, he erm for i also vansihes because of he facor B 2s (his is an odd index for Bernoulli number and 2s >. Therefore he sum sars a i 2 and i mus conain only even indices i, because B i 0 for i odd. Le i 2 and wrie (37 as (38 The summand for s is s ( 2s 2 (2 2 B 2 B 2s 2 0. (39 (2 2s B 2s and if we solve for i leads o s (40 B 2s s ( 2s B 2 B 2s 2. 2 This is a recurrence for he Bernoulli numbers ha involve only even indices. Theorem 5. The Bernoulli numbers of even index saisfy he recurrence (4 B 2s s ( 2s 2 2s (2 2 B 2 B 2s 2 2 wih iniial condiion B 0. Anoher observaion coming from he lis in (20 is ha he sign of he non-zero Bernoulli numbers alernae. This is easy o prove from he recurrence (4. Corollary 6. For n N (42 ( n B 2n > 0. Proof. Define (43 b n ( n B 2n and replace in (4 o obain (44 b n n n ( 2n b b n. 2 The iniial condiion b 6 shows ha b n > 0 for all n N.
6 6 THE BERNOULLI NUMBERS A second ideniy comes by manipulaions of he generaing funcion. The Bernoulli numbers have been defined here by he exponenial generaing in ( by (45 f( e (46 The ideniy comes by differeniaing he generaing funcion o obain o produce (47 f ( f( This is wrien as (48 d de e e (e 2 f( f2 (. r0 ( r B B r Now compare he coefficiens of r o produce B r+ (49 B r+ r! (r +! B r r+ ( r + B B r+ r! (r +! ha can be wrien as r+ ( r + (50 rb r+ (r +B r B B r+ and his leads o (5 (r +2B r+ (r +B r ( r + B B r+. Theorem 7. The Bernoulli numbers saisfy he recurrence ( r + (52 (r +2B r+ (r +B r B B r+ for r 0. Tae r 2u o be odd, hen (52 gives (53 (2u+B 2u 2u ( 2u B B 2u r r! and i follows ha in he sum we should only consider even indices, o produce Theorem 8. The Bernoulli numbers saisfy he recurrence u ( 2u (54 (2u+B 2u B 2r B 2u 2r. 2r r
7 THE BERNOULLI NUMBERS 7 Now wrie (55 b r ( r B r and replace in (52 o produce (56 (r +2b r+ (r +b r + ( r + b b r+. This recurrence gives anoher proof of Corollary 6. Now replace r + by n in (52 o obain n ( n (57 (n+b n nb n B B n ha holds for n. Wrie he sum as n ( n (58 B B n n ( n B B n +2nB B n n 2 ( n B B n nb n. Then (57 becomes n 2 ( n (59 (n+b n B B n. This relaion is valid only for n 3. Wha happens for n and n 2? 2
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