Bernoulli Numbers. Junior Number Theory Seminar University of Texas at Austin September 6th, 2005 Matilde N. Lalín. m 1 ( ) m + 1 k. B m.
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1 Bernoulli Nubers Junior Nuber Theory Seinar University of Texas at Austin Septeber 6th, 5 Matilde N. Lalín I will ostly follow []. Definition and soe identities Definition 1 Bernoulli nubers are defined as B = 1 and recursively as ( + 1B = B, so we find B 1 = 1, B = 1 6, B 3 =, B 4 = 1 3, B 5 =, B 6 = 1 4,...,B 1 = ,... Lea = t e t 1 = = B t!. Write t e t 1 = = a t! and ultiply by e t 1, t = n=1 t n n! t a!, = equate coefficients for t +1 gives a = 1 and + 1 a =. Theore 3 (J. Bernoulli Let be a positive integer and define then = S (n = (n 1, ( + 1S (n = + 1 B n +1. = In e t = = t! substitute =, 1,...,n 1 and add, S (n t! = 1 + et e (n 1t = ent 1 t t e t 1 = = =1 n t 1! j= B j t j j! now equate the coefficients of t and ultiply by ( + 1!. 1
2 Definition 4 are called Bernoulli polynoials. B (x = So B (x = 1, B 1 (x = x 1, etc. Then Theore 3 ay be stated as = B x Lea 5 First note that B (x = S (n = (B +1(n B = te xt e t 1 = = B (x t!. ( B x 1 = B 1 (x. Also B (xdx = B =, 1. = Now let F(x, t = = B (x t!, differentiating, x F(x, t = =1 t B 1 (x = tf(x, t. ( 1! Now we solve using separation of variables, F(x, t = T(te xt, then F(x, tdx = but F(x, tdx = = and this proves the stateent (Castellanos, [1]. Proposition 6. B (1 x = ( 1 B (x. 3. B = = T(te xt dx = T(t et 1 t t B (xdx = 1! 1. B (x + 1 B (x = = ( B (x = x 1 (Roan [5] = ( 1+( 5. B (x = q 1 1 j= B r= ( 1r( r r (Radeacher [4]. B ( =! (Ruiz [6]. x + j.
3 Euler MacLaurin su forula (Radeacher, [4]. Let f(x sooth. Since B 1 (x = 1, =... = Evaluating in x = 1, f(1 = f(xdx + f(xdx = B 1 (xf(x 1 q ( 1 1 B (x f ( 1 (x! =1 1 B 1 (xf (xdx + ( 1 q q ( 1 B! (f( 1 (1 f ( 1 ( + ( 1 q 1 =1 Changing f(x by f(n 1 + x and adding, we obtain the forula b n=a+1 f(n = b a f(xdx + B q (x f (q (xdx q! q ( 1 B! (f( 1 (b f ( 1 (a + R q =1 B q (x f (q (xdx. q! where R q = ( 1q 1 q! b a B q (x [x]f (q (xdx An integral and soe identities Proposition 7 We have: xlog xdx (x + a (x + b = ( π +1 P log a π P log b π a b. where P (x = i+1 ( ( x ( x B +1 B +1 + (+1 i i i + 1 (Idea We first prove that B +1 x α dx (x + a (x + b = π(aα 1 b α 1 cos πα (b a for < α < 1, a, b by first coputing x α dx x + a = πaα 1 cos πα. We note that the polynoials P ay be defined recursively as P (x = x j>1(odd ( 1 j+1 ( + 1 j P +1 j (x. 3
4 The idea, suggested by Rodriguez-Villegas, is to obtain the value of the integral in the stateent by differentiating ties the integral of with α and then evaluating at α = 1. Let f(α = π(aα 1 b α 1 cos πα (b a which is the value of the integral with α. In other words, we have f ( (1 = xlog xdx (x + a (x + b. By developing in power series around α = 1, we obtain f(αcos πα = By differentiating ties, j= ( f ( j (α cos πα j We evaluate in α = 1, j=(odd ( 1 j+1 As a consequence, we obtain f ( (1 = When =, +1 j>1(odd ( 1 j+1 f ( (1 = f(1 = π (b a n= (j = π (b a ( π f ( j (1 j ( + 1 j log n a log n b (α 1 n. n! n= f (+1 j (1 log a log b a b = π log n+ a log n+ b (α 1 n. n! j π(log a log b = (b a. ( π P log a π j 1 + log +1 a log +1 b ( + 1(a b. P log b π a b. The general result follows by induction on and the definition of P. Theore 8 We have the following identities: For 1 l n: s n l (1,...,(n 1 n l = n s n l s (,...,(n 1 (l + s l + s B s ( s ( 1 s+1. s s= For 1 n: (n! n = n n! n s n s (,...,(n 1 s B s( s 1( 1 s+1. s=1 4
5 For l n: (l + 1s n l (,...,(n n l (l + s = (n + 1 s n l s (1,...,(n 1 B s ( s ( 1 s+1. s where For 1 n: n s=1 s= s n s (,...,(n ( 1 s+1s ( s 1 B s = (n 1! s 1 if l = s l (a 1,...,a = i 1 <...<i l a i1...a il if < l if < l are the eleentary syetric polynoials, i.e., (x + a i = i=1 s l (a 1,...,a x l l= Soe big classic results Theore 9 (Euler ζ( = ( 1 +1(π (! B. We will need cot x = 1 x n=1 x n π x. This identity ay be deduced by applying the logarithic derivative to Then On the other hand, xcot x = 1 sinx = x n=1 x n π n=1 = (1 x n π. xcot x = x cos x sinx = xi(eix + e ix e ix e ix = ix + ix and copare coefficients of x. For instance, ζ( = π π4 6, ζ(4 = 9, etc. Corollary 1 1. ( 1 +1 B >. ( x = 1 ζ( x nπ π =1 5 e ix 1 = 1 + n= B n (ix n n!
6 . B or B ( 1 +1 (! as. (π The first assertion is consequence of the fact that ζ( is positive. The second is consequence of the fact that ζ( > 1 iplies B > (! (π. Theore 11 (Claussen, von Staudt For 1 We will need the following B (p 1, p prie 1 p (od 1 Definition 1 For every rational nuber r and p prie write r = p a b where a, b are integers such that p ab. Then ord p (r =. We say that r is p-integral if ord p (r. Lea 13 Let p be a prie nuber and a positive integer, then 1.. p +1 is p-integral. p +1 (od p if. 3. p +1 is p-integral if 3 and p 5. By induction, + 1 p. Let + 1 = p a q. Then p +1 = p a q 1 iplies a. For the second case use that + 1 < p for. The third case is consequence of + 1 < p for 3 and p 5. Proposition 14 Let p be a prie and a positive integer. Then pb is p-integral. Also, if is even pb S (p(od p For the first stateent we will use induction. It is clear for = 1. Now note that for we have + 1 = Then Theore 3 becoes S (n = = ( n B = ( n B = Now set n = p and since S (p is integer, it suffices to prove that ( p pb + 1 is p-integral for = 1,...,, but that is true by induction and Lea (1
7 For the congruence it suffices to see that ( p ord p (pb for 1. By Lea 13 this is true for. The case with = 1 corresponds to (pb 1p and it is true because is even and the only nontrivial case is with =. Lea 15 Let p be a prie. If p 1, then S (p (od p. If p 1 then S (p 1 (odp First suppose that p 1. Let g be a priitive root odulo p. Then and S (p = (p g g (p (odp (g 1S (p g (p 1 1 (od p the result follows. Now suppose that p 1, then S (p p 1 (od p (Theore 11 Assue is even. By Proposition 14, pb is p-integral and S (p(od p. By Lea 15, B is a p-integer if p 1 and pb 1 (od p if p 1. Then B + 1 p p 1 is a p-integer for all pries p, then it ust be integral. More Congruences Corollary 16 If p 1, then B is p-integral. If p 1 then pb +1 is p-integral and ( ord p (pb + 1 = ord p (p B p so pb 1 (od p. Also 6 always divides the denoinator of B. Fro now on write B = U V as a fraction in lowest ters with V >. Proposition 17 For even and > 1, V S (n U n (od n We will use equation (1, for 1 write n (B 1 n = A + 1 n. 7
8 If we show that for p n, p, 3, then ord p (A and if p n, p = or 3, ord p(a 1, then (A, n ust divide 6 and the sae is true for the greater coon divisor between the su of A and n. Then we ay write S (n = B n + An lb with (B, n = 1 and l 6. Multiplying by BV and using the fact that 6 V (by Corollary 16 the result is proved. Use Corollary 16 to see that ord p (B 1. Assue p n and p, 3. The cases = 1, are siple. If 3, ( n 1 ord p B 1 + ( 1ord p n ord p ( + 1 ord p ( by Lea 15. Now let p =. If = 1, then B 1 = for > and A 1 = B 1 1 = 1. For > 1 note that B = unless is even or = 1. even iplies ord ( + 1 = and = 1, A 1 = 1 n which has order greater or equal to 1. When p = 3, ord 3 (A 1 and ord 3(A 3 1. For 4, one shows that ord Corollary 18 Let be even and p prie with p 1. Then S (p B p (odp. By Theore 11, p V. Now put n = p in the above Proposition and divide by V which is perissible since p V. Proposition 19 (Voronoi s congruence Let even and > 1. Suppose that a and n are positive coprie integers. Then n 1 [ ] ja (a 1U a 1 V j 1 (od n. n j=1 Write ja = q j n + r j with r j < n. Then j a r j + q j nr 1 j (odn. But r j ja(od n, then Suing for j = 1,...,n 1, j a r j + a 1 q j nj 1 (od n. n 1 [ ja S (na S (n + a 1 n j 1 n Now ultiply by V and use Proposition j=1 ] (odn.
9 Proposition If p 1, then B is p-integral. By Theore 11, B is a p-integer. Let = p t with p. In Voronoi congruence put n = p t. Then (a 1U (odp t. Now let a be a priitive root odulo p. Since p 1, then p a 1. Then U (od p t. Then B = U V is p-integer. Theore 1 (Kuer congruences Suppose is even, p prie, and p 1. Let C = (1 p 1 B. If (od φ(p e, then C C (odp e. We will see the case e = 1. Let t = ord p (. By Proposition, p t U. In Voronoi s congruence, set n = p e+t. Since p t divides both U and, and V p is prie to t p, we obtain, (a pe+t 1 1B [ ] 1 ja a j 1 p e+t (od p e. The right-hand side is unchanged if we replace by (od p 1. Then (a 1B j=1 (a 1B (odp. Choose a to be a priitive root odulo p. Since p 1 we have a 1 a 1 (od p. Then B B (od p. Definition An odd prie nuber p is said to be regular if p does not divide the nuerator of any of the nubers B, B 4,...,B p 3. The prie 3 is regular. Equivalently, p is regular if it does not divide the class nuber of Q(ξ p The first irregular pries are 37 and 59. Theore 3 (Kuer Let p be a regular prie. Then x p + y p = z p has no solution in positive integers. Theore 4 (Jensen The set of irregular pries is infinite. Let {p 1,...,p s } be the set of irregular pries. Let be even and n = (p (p s 1. Choose large such that B n ( n > 1 and p prie such that ord Bn p n >. Then p 1 n and so p p i. We will prove that p is also irregular. Let n (od p 1 where < < p 1. Then is even and p 3. By the Kuer congruence, B n n B (od p. ( Since ord Bn ( p n > and Bn ordp n B >, then B ord p = ord p B > and p is irregular. 9
10 References [1] D. Castellanos, The Ubiquitous Pi. Part I.Math. Mag. 61, 67 98, [] K. Ireland, M. Rosen, A classical introduction to odern nuber theory. Second edition. Graduate Texts in Matheatics, 84. Springer-Verlag, New Yor, 199. xiv+389 pp. [3] M. N. Lalín, Mahler easure of soe n-variable polynoial failies, (preprint, Septeber 4, to appear in J. Nuber Theory [4] H. Radeacher, Topics in Analytic Nuber Theory, Die Grundlehren der Matheatischen, Wissenschaften. New Yor: Springer-Verlag, [5] S. Roan, The Bernoulli Polynoials. 4.. The Ubral Calculus. New Yor: Acadeic Press, pp. 93 1, [6] Eric Weisstein s World of Matheatics, 1
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