Bernoulli Polynomials Talks given at LSBU, October and November 2015 Tony Forbes

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1 Beroulli Polyoials Tals give at LSBU, October ad Noveber 5 Toy Forbes Beroulli Polyoials The Beroulli polyoials B (x) are defied by B (x), Thus B (x) B (x) ad B (x) x, B (x) x x + 6, B (x) dx,. () B 3 (x) x 3 3x + x, B 4(x) x 4 x 3 + x 3, B 5 (x) x 5 5x4 + 5x3 3 x 6, B 6(x) x 6 3x 5 + 5x4 x + 4, B 7 (x) x 7 7x6 + 7x5 7x3 6 + x 6, B 8(x) x 8 4x 7 + 4x6 7x x 3 3, B 9 (x) x 9 9x8 +6x7 x5 5 +x3 3x, B (x) x 5x 9 + 5x8 7x6 +5x 4 3x , B (x) x x + 55x9 6 B (x) x 6x + x 33x8 x 7 + x 5 x3 + x 6 33x4 The costat ter gives the th Beroulli uber, B B (), + 5x 6, + 5x 69 73,.... B, B, B 6, B 3, B 4 3, B 5, B 6 4, B 7, B 8 3, B 9, B 5 66, B, B 69 73, B 3, B 4 7 6, B 5, B , B 7, B , B 9, B ,.... There is soe erit i defiig what oe ight call the alterative Beroulli polyoials: B (x) B (x) B (x), ad B (x) dx,. The B (x) B (x) + x ( ) B ( x). The correspodig alterative Beroulli ubers are B B () B (). The oly differece occurs whe : B B, B B,,, 3, 4,.... We wo t ivestigate the alterative polyoials, but watch out for B i what follows.

2 The graphs show that B ( x) ( ) B (x). Fro the defiitio, the coefficiet of x is B () B. More geerally, the coefficiet of x i B (x) is ( ) B ; hece ( ) B (x) B x. Moreover, for x [, ] we have these approxiatios for large : B (x) B cos πx, eve, B (x) B (/4) si πx, odd, ad B (x)b (x) dx whe + is odd. Exercise for reader: copute B (x)b (x) dx whe + is eve Beroulli polyoials B (x) for, 3,..., 8 The geeratig fuctios for the Beroulli polyoials ad the Beroulli ubers are respectively te xt e t B (x) t! ad t e t B t!. () These could be used as defiitios; the the left-had equality i () is recovered by differetiatig the left-had equality i () with respect to x: t e xt e x B (x) t text! e x B (x) t B +(x)! + t!. Fro the right-had equality i () we obtai two facts: (i) B + for sice the fuctio t/(e t ) + t/ is eve, ad (ii) ultiplyig the series by the Taylor expasio of (e t )/t gives ( ) + B. This last equality ca be used to copute the B recursively; see Lovelace [3], for exaple.

3 Stirlig ubers of the secod id The Stirlig uber of the secod id S(, ) is the uber of ways to partitio a set of obects ito o-epty subsets: S(, )! ( ( ) ) ( )! or they ca be defied recursively ( ) ( ), S(, ), S(, ) for >, S(, ) for >, S( +, ) S(, ) + S(, ) for > It is clear fro the recursive defiitio that the S(, ) are itegers, that S(, ) ad that S(, ) if >. Iterestig proble: prove directly that ( )( ) wheever > > Plot of log S(, ) + for 5,,... ad,,..., 3 3

4 Euler Maclauri suatio Let f(x) be a sufficietly well-behaved fuctio. Suppose we wat to su f(x) over iteger values of x ad that we ow how to itegrate f(x). Let < be itegers. The + f(r) f(x) dx + r ( f ( ) () f ( ) () ) B! or, aig use of the fact that B for odd 3, + f() f(x) dx + f() f() f (r) (x) B r(x x ) (r)! + s f (r) (x) B r ( x x) r! ( f ( ) () f ( ) () ) B ()! dx, (r ). If f(x) is a polyoial, the last ter will vaish for sufficietly large r. For exaple, to get the forula for the su of the first squares put ad f(x) x : x dx + + B! ( + )( + ). 6 dx, Poly-Beroulli ubers If we defie the ore geeral poly-beroulli ubers B () Li (t) t, Li ( e t ) e t by B () t!, the gives Li (x) log( x) ad the geeratig fuctio t/( e t ) of B () B. Properties of the Beroulli ubers Theore (Explicit forula for the Beroulli ubers) We have B + ( ) ( ) ( )! S(, ). (3) + Proof We refer to the geeratig fuctio: x e x log( + (ex )) e x e x ( ) + (ex ) ( ) (ex ) +. Now expad the bioial (e x ) ad the expad e x : x e x ( ) + x! x! ( ) ( ) e x ( ) ( ) ( ) + ( )!S(, ). + 4 ( ) + ( ) ( ) x!

5 But S(, ) whe >. Hece x e x x! ( )!S(, ), + as required to get the right-had expressio i (3). substitutig for S(, ). The other expressio follows by Theore (vo Staudt Clause) If is a positive eve iteger, the B + p, p prie p (od ). (4) Proof Suppose is eve. We cosider each of the ters i the su over i either of the two expressios of (3). If + is ot prie ad > 3, the!/( + ) is a iteger, as also is S(, ); so the ter correspodig to this is a iteger. Also oe ca verify directly that the ters correspodig to ad 3 are itegers. Now suppose + is prie ad does ot divide. Write q + r with < r <. The for <, q (od + ). Hece odulo + the su over becoes ( )( ) r ±!S(r, ) sice > r. Fially, suppose + is prie ad divides. The for <, (od + ), ad. Therefore ( )( ) (od + ) ad hece odulo there is a cotributio of /( + ) to the su. Whe is odd the su is zero odulo ; the ter cotributes /, as before, but ow the 3 ter also cotributes /. All other ters are itegers. The Vo Staudt Clause theore allows you to copute the fractioal part of B for soe quite large. For exaple, if 8, the divisors of are powers of two ad the deoiators of the fractios i (4) are ust ad the Ferat pries, 3, 5, 7, 57 ad O the other had, the calculatio is curretly ipossible for, say, 9. It is coveiet to have a expressio for the absolute value of B. So we rearrage (4) to get, writig {x} for x x, { } p, p prie /p, >, (od 4), { } p, p prie B B + /p, (od 4), (5),, /,,, >, odd. Thus for, 4,..., we get 6, 3, 4, 3, 5 66, 69 73, 6, 47 5, , 4 33, 7 38, 69 73, 6, 59 87, , 47 5, 6, , 6, 353, 86, 53 69, 4 8, , 5 66, 83 59, , 59 87, , , 6, 47 5, , 3, , , 6, 3, , , , , 6, , , 77 4, 6, , 6,

6 Theore 3 The Beroulli ubers are related to the Riea zeta-fuctio by B ζ( ), ; B! ζ(), eve. (6) (πi) Proof (Titcharsh [6, sectio.4]) Let s be a coplex veriable with Re s >. Usig we get Γ(s)ζ(s) x s e s dx x s e x dx Γ(s) s x s e x dx x s e x dx. Let C ρ, ρ >, deote the cotour that goes fro + to ρ o the real axis, circles the origi oce aticlocwise ad the returs to +. Let ρ. Sice the itegral alog the circle teds to zero as ρ teds to zero, we have C z s e z dz z s e z dz + (e πi z) s e z dz (e πis )Γ(s)ζ(s). Usig the well-ow forula Γ( z)γ(z) π/(si πz), we obtai ζ(s) z (e πis )Γ(s) C s e z dz e πis Γ( s) z πi C s e z dz, which exteds the doai of ζ(s) to the whole coplex plae except for s. Now let be a iteger ad put s. The ζ( ) ( ) ( )! πi C z e z dz ( ) ( )! πi C B! z dz. But the value of itegral o the right is πi ties the residue of the pole at z, ad this is ust the coefficiet of /z i the su. Hece, recallig that B B ad B for odd 3, ζ( ) ( ) ( )! πi πi B! ( ) B B. This is the left-had equality i (6) for. For the reaiig case, recall that ζ(s) γ + O(s ) as s. s The right-had equality follows fro the fuctioal equatio of ζ(s), ( ( ) s s Γ π ) s/ ζ(s) Γ π ( s)/ ζ( s). (See Edwards [], for exaple.) Of course we ca discard the sig if we are iterested oly i the absolute value: B!! ζ() ( + (π) (π) + 3 ) +..., eve. (7) 6

7 We ca use (5) ad (7) to deterie B exactly. Write b() (!) / /(π). The split the su ito three parts, B I + E + R, b() ( ) I b(), E b() ( ) b() I, R b()+ ( ) b(). We copute I exactly ad E approxiately, we assue that R is sall eough to be igored, ad we get the exact value of F, the fractioal part of B, usig the vo Staudt Clause forula, (5). The for positive eve, we have B I + ɛ + F, where ɛ if E < F, ɛ otherwise. Exaple: b() 59, I , E ad F > E. There is a iterestig applicatio cocerig prie geeratig fuctios, [4], ad also a iterestig proble. Is there a that requires ɛ? By siple-ided direct calculatio there are o istaces for eve up to 3. But could it possibly happe that E is very early equal to but less tha, F is very sall, ad R, which is very sall ayway, is ust large eough so that E + R >? The we would have F E + R ad hece B I +. A siilar proble is preseted i M5 [, 5]. Refereces [] H. M. Edwards, Riea s Zeta-Fuctio. [] TF, Proble 67.4 Beroulli ubers, M5 67 (Deceber 5). [3] A. A. Lovelace, Note G i: L. F. Meabrea, Setch of The Aalytical Egie Iveted by Charles Babbage With otes upo the Meoir by the Traslator, Ada Augusta, Coutess of Lovelace, Bibliothèque Uiverselle de Geève 8 (October 84). [4] R. Thopso, Beroulli ubers ad prie geeratig fuctios, M5 67 (Deceber 5). [5] R. Thopso, Solutio 67.4 Beroulli ubers, M5, to appear. [6] E. C. Titcharsh, The Theory of the Riea Zeta-Fuctio. 7