Evaluating ζ(2m) via Telescoping Sums.

Transcription

1 /25 Evaluating ζ(2m) via Telescoping Sums. Brian Sittinger CSU Channel Islands February 205

2 2/25 Outline

3 3/25 Basel Problem Find the exact value of n= n 2 =

4 3/25 Basel Problem Find the exact value of n= n 2 = First posed by Pietro Mengoli in 644.

5 3/25 Basel Problem Find the exact value of n= n 2 = First posed by Pietro Mengoli in 644. Gained notoriety when Jakob Bernoulli wrote about it in 689.

6 3/25 Basel Problem Find the exact value of n= n 2 = First posed by Pietro Mengoli in 644. Gained notoriety when Jakob Bernoulli wrote about it in 689. Remained unsolved into the 730 s.

7 4/25 Answer to the Basel Problem n= n 2 = = π2 6.

8 4/25 Answer to the Basel Problem n= n 2 = = π2 6. First established by Leonhard Euler in 734.

9 4/25 Answer to the Basel Problem n= n 2 = = π2 6. First established by Leonhard Euler in 734. Subsequently proved in many distinct ways (Multivariate change of variables from Calculus, Fourier Series, Complex Analysis, etc.)

10 4/25 Answer to the Basel Problem n= n 2 = = π2 6. First established by Leonhard Euler in 734. Subsequently proved in many distinct ways (Multivariate change of variables from Calculus, Fourier Series, Complex Analysis, etc.) Daniel Daners published a proof of this fact using basic Calculus in 202.

11 5/25 using telescoping sums Define the following families of integrals (with n being a nonnegative integer): A n = π 2 0 cos 2n x dx, B n = π 2 0 x 2 cos 2n x dx.

12 5/25 using telescoping sums Define the following families of integrals (with n being a nonnegative integer): A n = π 2 0 cos 2n x dx, B n = Initial values: A 0 = π 2, and B 0 = π3 24. π 2 0 x 2 cos 2n x dx.

13 5/25 using telescoping sums Define the following families of integrals (with n being a nonnegative integer): A n = π 2 0 cos 2n x dx, B n = Initial values: A 0 = π 2, and B 0 = π3 24. π 2 0 x 2 cos 2n x dx. Recurrence Relations for n > 0 (via integration by parts): A n = 2n 2n A n and A n = n(2n )B n 2n 2 B n.

14 6/25 using telescoping sums, continued Divide both sides of the second recurrence by n 2 A n and substitute the first recurrence: ( n 2 = 2 Bn B ) n. A n A n (Telescoping step) Sum both sides from n = to k: k n= n 2 = 2 ( B0 A 0 B k A k ) = π2 6 2B k A k.

15 7/25 using telescoping sums, concluded Since sin x 2x π on [0, π 2 ], observe that B k = π 2 0 x 2 cos 2k x dx π 2 0 ( 2 π sin x ) 2 cos 2k x dx = π2 A k 8(k + ).

16 7/25 using telescoping sums, concluded Since sin x 2x π on [0, π 2 ], observe that B k = π 2 0 Then 0 < B k A k x 2 cos 2k x dx π 2 0 ( 2 π sin x ) 2 cos 2k x dx = π2 A k 8(k + ). π 2 B k, and so lim = 0. 8(k + ) k A k

17 7/25 using telescoping sums, concluded Since sin x 2x π on [0, π 2 ], observe that B k = π 2 0 Then 0 < B k A k x 2 cos 2k x dx π 2 Finally, letting k yields 0 π 2 B k, and so lim 8(k + ) k ( 2 π sin x ) 2 cos 2k x dx = π2 A k 8(k + ). n= A k = 0. n 2 = π2 6.

18 8/25 Beyond the Basel Problem More generally, define ζ(s) = n= n s.

19 8/25 Beyond the Basel Problem More generally, define ζ(s) = n= n s. Euler also proved that for any m N ζ(2m) = k= k 2m = ( )m+ 2 2m B 2m (2m)! where B k denotes the k-th Bernoulli number. π 2m First few: B 0 =, B = 2, B 2 = 6, B 4 = 30, B 6 = Moreover, B 3 = B 5 = B 7 =... =

20 9/25 A few even zeta constants ζ(2) = = π2 6

21 9/25 A few even zeta constants ζ(2) = = π2 6 ζ(4) = = π4 90

22 9/25 A few even zeta constants ζ(2) = = π2 6 ζ(4) = = π4 90 ζ(6) = = π6 945

23 9/25 A few even zeta constants ζ(2) = = π2 6 ζ(4) = = π4 90 ζ(6) = = π6 945 ζ(8) = = π8 9450

24 9/25 A few even zeta constants ζ(2) = = π2 6 ζ(4) = = π4 90 ζ(6) = = π6 945 ζ(8) = = π ζ(0) = π0 + =

25 9/25 A few even zeta constants ζ(2) = = π2 6 ζ(4) = = π4 90 ζ(6) = = π6 945 ζ(8) = = π ζ(0) = π0 + = ζ(2) = π2 + =

26 0/25 Solving for the even zeta constants As with the Basel Problem, there are many distinct solutions (though not quite as many).

27 0/25 Solving for the even zeta constants As with the Basel Problem, there are many distinct solutions (though not quite as many). For the remainder of this discussion, I will demonstrate how to extend Daner s telescoping sum technique to compute ζ(2m) for any positive integer m.

28 0/25 Solving for the even zeta constants As with the Basel Problem, there are many distinct solutions (though not quite as many). For the remainder of this discussion, I will demonstrate how to extend Daner s telescoping sum technique to compute ζ(2m) for any positive integer m. As a warm-up for the general case, I will first use this technique to compute ζ(4).

29 /25 using telescoping sums Define A n = and C n = π 2 0 π 2 0 cos 2n x dx, B n = x 4 cos 2n x dx. π 2 0 x 2 cos 2n x dx,

30 /25 using telescoping sums Define A n = and C n = π 2 0 π 2 0 cos 2n x dx, B n = x 4 cos 2n x dx. π 2 Initial values: A 0 = π 2, B 0 = π3 24, C 0 = π x 2 cos 2n x dx,

31 /25 using telescoping sums Define A n = and C n = π 2 0 π 2 0 cos 2n x dx, B n = x 4 cos 2n x dx. π 2 Initial values: A 0 = π 2, B 0 = π3 24, C 0 = π5 60. Recurrence relations (for n > 0): A n = 2n 2n A n A n = n(2n )B n 2n 2 B n ) B n = 6 (n(2n )C n 2n 2 C n 0 x 2 cos 2n x dx,

32 2/25 using telescoping sums, continued Divide both sides of the last two recurrences by n 2 A n and substitute the first recurrence: ( n 2 = 2 Bn B ) n and B n A n A n n 2 = ( Cn C ) n A n 3 A n A n

33 2/25 using telescoping sums, continued Divide both sides of the last two recurrences by n 2 A n and substitute the first recurrence: ( n 2 = 2 Bn B ) n and B n A n A n n 2 = ( Cn C ) n A n 3 A n A n Telescoping step - Sum the first equation from n = to k and the second from k = to N (after changing its index): k n= ( π 2 n 2 = 2 2 B ) k and A k N k= B k k 2 = ( π 4 A k 3 80 C ) N A N

34 3/25 using telescoping sums, continued The first equation yields B k A k = π2 2 2 k n= n 2.

35 3/25 using telescoping sums, continued The first equation yields B k A k = π2 2 2 Substitute this into the second equation: π 2 2 N k= k 2 2 N k k= n= k n= n 2. k 2 n 2 = ( π C ) N. A N

36 3/25 using telescoping sums, continued The first equation yields B k A k = π2 2 2 Substitute this into the second equation: π 2 2 N k= k 2 2 N k= n= Rewrite this to expose k 4 : π 2 2 N k= k 2 ( N 2 k= k k 4 + k n= n 2. k 2 n 2 = ( π C ) N. A N N j<k j,k= ) j 2 k 2 = ( π C ) N. A N

37 4/25 using telescoping sums, continued However, ( N k= ) 2 N = k 2 k k= N j<k j,k= j 2 k 2.

38 4/25 using telescoping sums, continued However, ( N k= ) 2 N = k 2 k k= N j<k j,k= Use this relation to eliminate the nested sum: π 2 2 N k= k 2 [ N 4 k= ( N k 4 + k= j 2 k 2. ) 2 ] k 2 = ( π C ) N. A N

39 5/25 using telescoping sums, concluded Let N, noting that C N A N π 2 2 k= k 2 [ 4 k= 0 (as before): ( k 4 + k= ) 2 ] k 2 = π4 240.

40 5/25 using telescoping sums, concluded Let N, noting that C N A N π 2 2 k= k 2 [ 4 k= 0 (as before): ( k 4 + k= ) 2 ] k 2 = π Since k 2 = π2, substituting this into the previous relation 6 k= and solving for the remaining sum gives k 4 = π4 90. k=

41 6/25 Overview We now prove the general case: ζ(2m) = k= k 2m = ( )m+ 2 2m B 2m (2m)! π 2m This proof is similar to the previous two cases, but with a couple more difficulties: Evaluating the nested sum S(k) = i 2 i i k <... i k 2. Using the Bernoulli numbers.

42 7/25 Dealing with the nested sum Lemma : For any k N, we have S(k) = 2( 2 2k )ζ(2k).

43 7/25 Dealing with the nested sum Lemma : For any k N, we have S(k) = 2( 2 2k )ζ(2k). Proof (sketch): Define S(0) =, and consider the generating function G(x) = k=0 S(k)x 2k = + S()x 2 + S(2)x 4 +.

44 7/25 Dealing with the nested sum Lemma : For any k N, we have S(k) = 2( 2 2k )ζ(2k). Proof (sketch): Define S(0) =, and consider the generating function G(x) = k=0 S(k)x 2k = + S()x 2 + S(2)x 4 +. G(x) = j= = πx x2 sin πx, by the infinite product for sine. j 2

45 7/25 Dealing with the nested sum Lemma : For any k N, we have S(k) = 2( 2 2k )ζ(2k). Proof (sketch): Define S(0) =, and consider the generating function G(x) = k=0 S(k)x 2k = + S()x 2 + S(2)x 4 +. G(x) = j= = πx x2 sin πx, by the infinite product for sine. j 2 Next, G(x) = 2 n= fraction expansion of cosecant. ( ) n+ x 2 x 2 n 2, by the infinite partial

46 7/25 Dealing with the nested sum Lemma : For any k N, we have S(k) = 2( 2 2k )ζ(2k). Proof (sketch): Define S(0) =, and consider the generating function G(x) = k=0 S(k)x 2k = + S()x 2 + S(2)x 4 +. G(x) = j= = πx x2 sin πx, by the infinite product for sine. j 2 Next, G(x) = 2 n= fraction expansion of cosecant. ( ) n+ x 2 x 2 n 2, by the infinite partial x Since 2 = x 2 n 2 k= ( x n )2k by the geometric series, applying this yields G(x) = + 2 ( ) n+ k= n= x 2k. n 2k

47 8/25 Dealing with the nested sum, continued Proof (sketch), continued: However, n= ( ) n+ n 2k = n= n 2k n= 2 (2n) 2k = ( 2 2k )ζ(2k).

48 8/25 Dealing with the nested sum, continued Proof (sketch), continued: However, n= ( ) n+ n 2k = n= n 2k n= 2 (2n) 2k So, G(x) = + k= 2( 2 2k )ζ(2k)x 2k. = ( 2 2k )ζ(2k).

49 8/25 Dealing with the nested sum, continued Proof (sketch), continued: However, n= ( ) n+ n 2k = n= n 2k n= 2 (2n) 2k So, G(x) = + k= 2( 2 2k )ζ(2k)x 2k. = ( 2 2k )ζ(2k). Since G(x) = k=0 S(k)x 2k, equating like coefficients of x 2k yields the desired result.

50 9/25 A Bernoulli Identity Lemma 2: For any m N, m ( ) 2m + ( 2 2k )B 2k = 0. 2k k=0

51 9/25 A Bernoulli Identity Lemma 2: For any m N, m ( ) 2m + ( 2 2k )B 2k = 0. 2k k=0 Proof (sketch): Substitute the generating function power series for e t into t e t = +et 2 2t e 2t. t e t = n=0 B n tn n! This can be rewritten as r= B r ( 2 r ) tr r! = r ( r ) r=0 j=0 j 2 j t B r j r!. and the By equating the coefficients of t 2m+, using r j=0 ( r j) Bj = B r, and noting that B 2k+ = 0, the result follows.

52 20/25 using telescoping sums Define for non-negative integers k and n: I k,n = π 2 0 x 2k cos 2n x dx. Observe that A n = I 0,n, B n = I,n, C n = I 2,n. π 2k+ Initial values: I k,0 = 2 2k+ (2k + ) Recurrences: I 0,n = 2n I 0,n. I k,n = 2n (2k + 2)(2k + ) (2n(2n )I k+,n 4n 2 I k+,n ).

53 2/25 using telescoping sums, continued Divide both sides of the last recurrence by n 2 I 0,n and substitute the first recurrence: I k,n n 2 I 0,n = 4 (2k + 2)(2k + ) ( Ik+,n I 0,n I k+,n I 0,n ). Note that we have a family of recurrences as k varies.

54 22/25 using telescoping sums, continued Let S N (0) =, and for k > 0, S N (k) = i 2... i k 2, a truncated sum of S(k). i i k N

55 22/25 using telescoping sums, continued Let S N (0) =, and for k > 0, S N (k) = i 2... i k 2, a truncated sum of S(k). i i k N Telescoping step - Summing over n and back substituting these equations (as k varies) yields m k=0 ( ) k (2m)! 2 2k (2m 2k)! Im k,0 S N (k) = I m,n. I 0,0 I 0,N

56 23/25 using telescoping sums, continued Let N, noting that I m,n I 0,N 0: m k=0 ( ) k (2m)! 2 2k (2m 2k)! Im k,0 I 0,0 S(k) = 0. Apply Lemma and solve for ζ(2m) (taking ζ(0) = 2 as shorthand): ζ(2m) = m 2 2m ( ) m k π 2m 2k ( 2 2k )ζ(2k). (2m 2k + )! k=0 The desired result now follows from Lemma 2 and strong induction.

57 24/25 Recommended reading Euler s Solution of the Basel Problem - The Longer Story reznick/sandifer.pdf R. Chapman, Evaluating ζ(2): D. Daners, A short elementary proof of /k 2 = π 2 /6, Mathematics Magazine 85(202) T. Osler, Finding ζ(2p) from a Product of Sines, The American Mathematical Monthly (2004) F. Beukers, E. Calabi, and J. Kolk, Sums of generalized harmonic series and volumes, Nieuw Archief Wiskunde (4) (993)

58 25/25 Thank you!