Power-sum problem, Bernoulli Numbers and Bernoulli Polynomials.

Transcription

1 Power-sum roblem, Bernoull Numbers and Bernoull Polynomals. Arady M. Alt Defnton 1 Power um Problem Fnd the sum n : 1... n where, n N or, usng sum notaton, n n n closed form. Recurrence for n Exercse Usng reresentatons 1 1, 1 1, fnd n for 1,, 3 and n N. Exercse 3 By summng dfferences 1 1, , for runnng from 1 to n fnd n for, 3, 4. General case Exercse 4 For any N by summng dfferences n rove that n n 1 1 n 1 1 n 1 1 for runnng from 1 to Exercse 5 For any N by summng dfferences that n 1 n n 1 Recurrences 1 and gve oortunty, startng from 0 n n 1 for runnng from 1 to n rove 0 n, constructvely fnd reresentaton of n as olynomal of n. nce any olynomal degree of m unquely defned by ther values n m 1 dstnct onts 1 or holds for any natural n then, by such, olynomals x are defned for any x R and N, more recsely, defned sequence of olynomals x by recurrence N x x or by recurrence x 1 x 1 wth ntal condton 0 x x. x x 1 Mathematcal Reflectons

2 1 Bernoull Numbers and Bernoull Polynomals Our goal s to solve ths recurrence n closed form, that s to fnd a regular olynomal reresentaton of x. nce 0 0 for any 0, 1,,... then we should fnd real numbers s 1,..., s 1 such that x s 1 x...s 1 x 1. Note that the roblem would smly be solved f we had nown for some olynomal H x of degree such that H x 1 H x cx where c s some constant. Then n n n 1 n H n 1 H 1 H 1 H. c c In a sense, we already have one such olynomal u to an arbtrary constant c, H x x 1 c snce H x 1 H x x x 1 x But our roblem s that x s not yet reresented n terms of owers of x. By dfferentaton of 1 x 1 x 1 x 1 we obtan 1 x 1 x 1 x ; then 1 x 1 can be co sdered as another canddate for the role of H x, whch does not loo better than x 1 for the same reason. x x 1 x x 1 x 1 We now that 0 x x, 1 x, x, 3 x x x Alyng the recurrences 1 or we obtan 4 x x x 1 x 1 3x 3x 1 and 5 x x x 1 x x Accordngly, we also have 0 x 1, 1 x x 1, x x x 1 6, 3 x x3 3x x, 4 x x4 x 3 x 1 30, 5 x x5 5 x4 5 3 x3 x 6 0 x 0, 1 x 1, x x x x, 3 x 3x 3x 1 3 x x x, 4 x 4x3 6x x 4 x 3 3x x 4 3 x, 5 x 5x4 10x 3 5x 1 x x 3 x x. The above equatons lead to the concluson that the correlaton x 1 x holds for any N. In fact, assumng x 1 x, 1,,..., 1,and by dfferentatng 1 twce, we obtan x 1 1 x 0 x x x and 1 x 1 1 x x x x x x x x x 0 1 x Exercse 6 Prove that x 1 x, for any N usng. 1 x. Thus, by nducton, x 1 x for any N. Comng bac to the olynomal x 1, we denote t by B x, and then by relacng x wth x 1 n the recurrence Mathematcal Reflectons 3 018

3 x 1 1 x 0 x we obtan the followng recurrence for olynomals B x, N : 1 1 B x B x x 1 1. B Proertes. P0. deg B x deg x 1 ; P1. B 0 x 1 x 1 1; P. B x x 1 x 1 1 x B 1 x ; P3. B x 1 B x x x 1 x 1, N. We call such olynomals Bernoull Polynomals. We already have the frst few olynomals B x, namely, B 1 x 1 x 1 x 1 1 x 1, B x x 1 x 1 x x x 1 6, B 3 x 3 x 1 x 13 3 x 1 x 1 x 3 3 x 1 x; B 4 x 4 x 1 x4 x 3 x 1 30, B 5 x 5 x 1 x5 5x4 5x3 3 x 6. We can see that B 1 0 1, B 1 1 1, but B 0 B 1 1 6, B 3 0 B 3 1 0, B 4 0 B , B 5 0 B and n general B 0 B 1 for any. Furthermore, B 1 0 B nce B x 1 B x x 1, then for x 0 we obtan B 1 B B 1 B 0, for all. Hyothess B 1 0 B 1 1 0, N s equvalent to dvdng B 1 x by x whch we wll rove later. Note that the recurson B x B 1 x, N wth ntal condton B 0 x 1 allows us to obtan olynomals B 1 x, B x, B 3 x,...and thus easer than by recurrence B1 or B. Indeed, assume that we already now olynomal B 1 x,then B x B 1 x 1 B t dt x 1 B 1 t dt B x B 0 x 1 B 1 t dt. Let B : B 0, N {0}. We call such numbers Bernoull Numbers. By relacng x n B1 or n B wth 0 we obtan or B 0 1 B B3 1 1 B B 1 1. B4 Any of these recurrences allows to get consstently numbers B 1, B, B 3,... Exercse 7 Fnd the frst 5 terms of sequence B 0. We show that by B, 1,,..., we can obtan olynomal B x. Let B x b x b 1 x 1... b 1 x b 0,where b should be determned. nce B 0 B then b 0 B. Also snce B x B 1 x then B x B x and B x b x b 1 x 1... b 1 x b 0 b x b 1 x 1... b 1 x b! yelds B! B 0 b! B 0 b! b Mathematcal Reflectons

4 b B, 1,,...,. Hence, B x B B 1 x B 1 x 1 B 0 x 0 B x. In artcular B 0 x x, B 1 x x 1, B x x x 1 6, B 3 x B 0 x 3 3B 1 x 3B x B 3 x x x x3 3 x 1 x. More roertes of Bernoull olynomals and numbers. 1 P4. 0 B x dx 0 for any N. Proof. Because of P. we have B 1 x B x then 1 0 B x dx 1 0 B 1 x dx B 1 x 1 B B B x dx 0. We wll rove that roertes P1.,P.,P3. determne olynomals B x unquely. Let Q x be a sequence of olynomals such that Q 0 0 x 1, Q n x nq n 1 x, n N and Q x 1 Q x 1 x 1, N. Frst note that Q 0 x 1 B 0 x.also, Q n 1 Q n 0 for n snce Q 1 Q ,. Ths yelds 1 0 Q x dx 0, N. Indeed, 1 0 Q x dx 1 0 Q 1 x dx Q n1 1 Q n nce Q 1 x 1 Q 0 x 1 then Q 1 x x c and, therefore, Q x Q 1 x yelds Q x x cx d. Then Q x 1 Q x x x 1 c x 1 x cx x c 1 0 c 1.Hence, Q 1 x x 1 B 1 x Assume that Q x B x then Q 1 x Q x B x B 1 x Q 1 x B 1 x c.therefore Q 1 x dx 1 B1 x c dx B 1 x dx c c. o, by nducton Q x B x for any N. P5. B x 1 B 1 x, 0.Comlement roerty Proof. Let Q x : 1 B 1 x, N {0} then: 1. By P1 Q 0 x B 0 1 x 1;. By P. Q x 1 B 1 x 1 B 1 x 1 B 1 x 1 1 B 1 1 x Q 1 x ; 3. By P3. Q x 1 Q x 1 B 1 x 1 1 B 1 x 1 B x B 1 x 1 1 B x 1 B x 1 1 x 1 x 1. Therefore, by roerty of unqueness we get 1 B 1 x B x. Corollary 8 For m 1, m N holds B 0 0. Indeed, f m 1 then B x B 1 x and, therefore, for x 0 we have B 0 B 1 B 0 B 0 0 B 0 0. Corollary 9 By relacng x n B x 1 B 1 x wth x 1 we obtan B x 1 1 B 1 x 1 1 B x 1 B x 0 1 n B x 1 B x. 0 0 Now, we wrte n n olynomal form by owers of n. nce B 1 x 1 B 1 x x then n n n B1 1 B 1 B 1 n 1 B 1 1 B 1 n 1 B 1 0 and, therefore, n B 1 n 1 B 1 Mathematcal Reflectons

5 n B 1 n 1 B B 1 n. n B 1 n B 1 B 1 n Faulhaber s Formula. Problem 1 Prove that B m1 x s dvsble by x 1 for any m N. Problem Prove that gn B m 1 m1 and max B 4m x B 4m, mn [0,1] Hnt use nducton. [0,1] B 4m x B 4m, m N. 1. A. M. Alt-Varatons on a theme-the sum of equal owers of natural numbers, art 1 Crux vol.40,n.8;. A. M. Alt-Varatons on a theme-the sum of equal owers of natural numbers, art Crux vol.40,n.10.. Mathematcal Reflectons