The Bernoulli Numbers John C. Baez, December 23, x k. x e x 1 = n 0. B k n = n 2 (n + 1) 2

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1 The Bernoulli Numbers John C. Bez, December 23, 2003 The numbers re defined by the eqution e 1 n 0 k. They re clled the Bernoulli numbers becuse they were first studied by Johnn Fulhber in book published in 1631, nd mthemticl discoveries re never nmed fter the people who mde them. For emple, the Čech compctifiction ws invented by Tychonoff, while Tychonoff s Theorem is due to Čech. It hd been known since ntiquity tht n n 2 n(n + 1) 2 n(n + 1)(2n + 1) n 3 n 2 (n + 1) 2 but Fulhber, the Arithmeticin of Ulm, seems to hve been the first to seek generl formul for the sum of pth powers of the numbers from 1 to n. This generl formul requires wht re now clled the Bernoulli numbers. Fulhber s work ws cited nd further developed by Bernoulli in his fmous Ars Conjectndi, published posthumously in In this book, Bernoulli wrote tht he used the method described below to compute in less thn hlf qurter n hour tht Wht showoff! Let s see how he did it. I ll give modern eplntion of the trick. Sums re to integrls s the difference opertor ( f)(z) : f(z + 1) f(z) is to derivtives. It would be crzy to tckle fncy integrls before mstering derivtives. So, before trying to do fncy sums, we should mster the difference opertor! We will think of this s n opertor : E E where E is the spce of entire functions on the comple plne. Note tht E is contined in the spce of forml power series C[[z]], so it gives us slight vrition on theme we know nd love: the Fock representtion. In prticulr, we hve nnihiltion nd cretion opertors given by the usul formuls: : E E, : E E (f)(z) d dz f(z) ( f)(z) zf(z) The dvntge of E is tht we cn evlute its elements t ny point of the comple plne, while forml power series my not converge. We need this to mke well-defined. 4

2 Now: since the difference opertor is discretized version of the derivtive, let us find formul for in terms of! 1. For ny t C, define the opertor by Show tht 2. Show tht e t : E E e t (t) k. (e t f)(z) f(z + t). e 1. Net: just s integrtion is the inverse of differentition, summtion should be the inverse of the difference opertor. So, to do sums we just need n inverse for the difference opertor. To see wht I men, prove the following discrete version of the Fundmentl Theorem of Clculus: 3. Show tht if F E nd then F f n 1 f(i) F (n) F (0). i0 So, if we could compute F from f vi F 1 f we d hve nifty formul for summing the vlues of f. There s just one slight ctch: the opertor : E E isn t invertible! The reson is tht F 0 whenever F is constnt function. Since hs nontrivil kernel, it cn t hve n inverse. In fct, this problem is lredy fmilir from ordinry clculus. I ws lying slightly: integrtion isn t relly the inverse of differentition s n opertor on functions. The derivtive of ny constnt is zero, so the integrl of function is only well-defined up to constnt. However, the opertors nd do hve one-sided inverses, nd this ll we relly need! 4. Define the opertor 1 : E E by Show tht but for ll f E. ( 1 )f(z) z 0 f(u)du. 1 f f not necessrily 1 f f We now use 1 to concoct one-sided inverse for the difference opertor s follows. We sw how the opertor e 1 cn be written s power series in. We d like to do the sme for 1 1 e 1,

3 but we cn t, since the function 1 e 1 hs simple pole t 0. To get rid of this pole, we cn work insted with e 1. This is n entire function if we define its vlue t 0 to be 1, so we cn write it s power series: e 1 k where re the Bernoulli numbers. We cn then define n opertor by the sme power series: Net, we define the opertor 1 by e 1 : E E e 1 : 1 k e 1 1. where 1 is the one-sided inverse of. And believe it or not, 1 is one-sided inverse of : 5. Strting from the definition of the Bernoulli numbers, show tht j 1 j j! k. (1) Use this to show nd therefore j 1 j j! k e 1. (2) 6. Using prts 4 nd 5 together with the definition of 1, show tht for ll f E, but Wrning: here s how not to do the first prt: 1 f f not necessrily 1 f f. 1 (e 1) e You cn t simply cncel the fctors of e 1, becuse e is not defined s the product of nd the 1 (noneistent) opertor (e 1) 1 it s defined s power series involving the Bernoulli numbers! This is why we need the rigmrole in prt 5.

4 Now, here s how we use this bstrct nonsense to clculte sum! Let s do the emple Fulhber did, where the function to be summed is Then we hve but by definition, so ( 1 f)(z) 1 f ( 1 f)(z) f(z) z p. z e 1 1 f 0 e 1 p + 1 z p+1 f(u)du zp+1 p + 1 d k z p+1 dz k p + 1 p+1 (p + 1)p (p + 1 k)zp+1 k p p+1 z p+1 k nd since 1 is one-sided inverse of, prt 3 gives or if you prefer, n 1 i p ( 1 f)(n) ( 1 f)(0) i0 n i1 1 p+1 n p+1 k i p 1 p+1 (n + 1) p+1 k. Since the right-hnd sum hs just p + 1 terms, regrdless of n, it s ctully much esier to clculte thn the left-hnd sum when n gets big.... or t lest it would be esy if we knew the Bernoulli numbers! We could clculte these by grinding out the Tylor series of /(e 1), but here s much quicker wy: 7. By tking the coefficient of i+1 on both sides of eqution (1), show tht for i Using prt 7, show tht 0 B 0 0! 1 (i + 1)! + B 1 1 1! i! + + B i 1 i! 1! 0 1B 0 + 2B 1 0 1B 0 + 3B 1 + 3B 2 0 1B 0 + 4B 1 + 6B 2 + 4B 3 0 1B 0 + 5B B B 3 + 5B 4

5 nd so on, where the numbers in boldfce re tken from Pscl s tringle in n obvious wy. 9. Show tht B 0 1, nd by using the recurrence in prt 8 work out B 1, B 2, B 3 nd B Use everything we ve done to work out n eplicit formul for n Show tht 1 p + + n p (B + n + 1)p+1 B p+1 p + 1 where the rules of the gme re tht we epnd the right-hnd side using the binomil theorem nd then set B i equl to the Bernoulli number B i. Wht formul do you get if you set n 0 here? Where hve you seen tht formul before? 12. Etr Credit: ctegorify this homework ssignment s much s possible, replcing E by the ctegory of entire structure types: tht is, structure types whose generting functions re entire functions on the comple plne. Why is this hrd?

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