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Revisiting the Riemann Zeta Function at Positive Even Integers 1. Introduction One of the Most Famous Mathematical Problems

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Revisiting the Riemann Zeta Function at Positive Even Integers 1. Introduction One of the Most Famous Mathematical Problems July 10, 2018 13:29 WSPC/S1793-0421 203-IJNT 1850110 International Journal of Number Theory Vol. 14, No. 7 (2018) 1849–1856 c World Scientific Publishing Company DOI: 10.1142/S1793042118501105 Revisiting the Riemann Zeta function at positive even integers Krishnaswami Alladi Department of Mathematics University of Florida, Gainesville, FL 32611, USA alladik@ufl.edu Colin Defant Department of Mathematics Princeton University, Princeton, NJ 08544, USA [email protected] Received 4 September 2017 Accepted 27 November 2017 Published 26 April 2018 Using Parseval’s identity for the Fourier coefficients of xk, we provide a new proof that k+1 2k ζ k (−1) B2k (2π) (2 )= 2(2k)! . Keywords: Riemann zeta function; even positive integers, Euler’s formula; Parseval iden- tity; Bernoulli numbers. Mathematics Subject Classification 2010: 11M06, 11B68, 11B83 1. Introduction One of the most famous mathematical problems was the evaluation of ∞ 1 . n2 n=1 The Italian mathematician Pietro Mengoli originally posed this problem in 1644. The problem was later popularized by the Bernoullis who lived in Basel, Switzer- land, so this became known as the Basel Problem [3]. Leonhard Euler solved this brilliantly when he was just 28 years old by showing that the sum in question was equal to π2/6. Indeed, this was Euler’s first mathematical work, and it brought him world fame. In solving the Basel Problem, Euler found closed-form evaluations more generally of ∞ ζ k 1 (2 )= n2k n=1 1849 July 10, 2018 13:29 WSPC/S1793-0421 203-IJNT 1850110 1850 K. Alladi & C. Defant for all even integers 2k ≥ 2. The value of ζ(2k) is given as a rational multiple of π2k. More precisely, Euler showed that [6, p. 16] k+1 2k (−1) B2k(2π) ζ(2k)= , (1) 2(2k)! where the Bernoulli numbers Bm are defined by the exponential generating function x ∞ xm B . ex − = m m 1 m=0 ! Since Euler’s time, numerous proofs of his formula for ζ(2k) have been given. Our goal here is to give yet another proof of (1) that involves a new identity for Bernoulli numbers and a different induction process than used previously. We consider the Fourier coefficients an(k)andbn(k) of the function that is periodic of period 2π and is given by f(x)=xk on the interval (−π, π], and we obtain a pair of intertwining recurrences for these coefficients (see (7) and (8) below). We then apply Parseval’s theorem to connect these Fourier coefficients with ζ(2k) and then establish (1) by solving this recurrence. The novelty in our approach is a new identity for Bernoulli numbers that we obtain and use. A very recent paper by Navas, Ruiz, and Varona [5] establishes new connections between the Fourier coefficients of many fundamental sequences of polynomials such as the Legendre polynomials and the Gegenbauer polynomials by making them periodic of period 1. In doing so, these authors consider the Fourier coefficients of xk, but they do not use their ideas to establish Euler’s formula (1). Another paper by Kuo [4] provides a recurrence for the values of ζ(2k) involving only the values ζ(2j)forj ≤ k/2 instead of requiring j ≤ k − 1aswedo.Ourmethodisvery different from Kuo’s. Other recent proofs of Euler’s formula appear in [1, 2]. 2. Warm Up For each positive integer k,letan(k)andbn(k)bethenth Fourier coefficients of x → xk a k 1 π xk dx a k 1 π xk nx dx the function .Thatis, 0( )= 2π −π , n( )= π −π cos( ) , 1 π k and bn(k)= π −π x sin(nx) dx for n, k ≥ 1. Parseval’s identity [7, p. 191] informs us that π ∞ 1 x2k dx a k 2 a k 2 b k 2 . π =2 0( ) + ( n( ) + n( ) ) (2) −π n=1 Let us apply (2) in the case k = 1. We easily calculate that a0(1) = an(1) = 0 (−1)n+1 and bn(1) = 2 n for all n ≥ 1. This implies that π2 π ∞ ∞ 2 1 x2 dx b 2 4 , = π = n(1) = n2 3 −π n=1 n=1 π2 so we obtain Euler’s classic result ζ(2) = 6 . In Sec. 4, we use Parseval’s identity to obtain a new inductive proof of Euler’s famous formula (1). July 10, 2018 13:29 WSPC/S1793-0421 203-IJNT 1850110 Revisiting the Riemann Zeta function at positive even integers 1851 3. An Identity for Bernoulli Numbers Before we proceed, let us recall some well-known properties of Bernoulli numbers (see [6, Chap. 1]). In Lemma 3.1, we also establish one new identity involving these numbers. 1 1 The first few Bernoulli numbers are B0 =1,B1 = − 2 ,andB2 = 6 .Ifm ≥ 3is odd, then Bm = 0. The equation n−1 n B m m =0 (3) m=0 holds for any integer n ≥ 2. If n is odd, then n n B m . m2 m =0 (4) m=0 n n n− One can prove (4) by evaluating the Bernoulli polynomial Bn(x)= =0 B x n at x =1/2 and using the known identity Bn(x)=(−1) Bn(1 − x). n In what follows, we write r1,r2,r3 to denote the trinomial coefficient given by n! r1!r2!r3! . Lemma 3.1. For any positive integer k, k k B 2t 2 +2 k 2k − k 2 , 2t2 t, i , k − t − i =( +1) 2 +( 1) k i+t≤k/2 2 2 +1 2 2 2 +1 where the sum ranges over all non-negative integers i and t satisfying i + t ≤k/2. Proof. The difficulty in evaluating the given sum spawns from its unusual limits of summation. Therefore, we will first evaluate the sum obtained by allowing i and t to k−t 2(k−t)+2 range over all non-negative integers satisfying i + t ≤ k. Because i=0 2i+1 = 22k−2t+1,wehave k B 2t 2 +2 2t2 t, i , k − t − i i+t≤k 2 2 +1 2 2 2 +1 k k k−t k − t B 2t 2 +2 2( )+2 = 2t2 t i t=0 2 i=0 2 +1 k k 2k+1 B 2 +2 . =2 2t t t=0 2 Since Bm =0foralloddm ≥ 3, k B 2t 2 +2 2t2 t, i , k − t − i i+t≤k 2 2 +1 2 2 2 +1 2k+1 k k 2k+1 B 2 +2 − B 2 +2 2k+1 k . =2 1 =2 ( +1) (5) =0 1 July 10, 2018 13:29 WSPC/S1793-0421 203-IJNT 1850110 1852 K. Alladi & C. Defant 1 Note that we used (3) along with the fact that B1 = − 2 to deduce the last equality above. We next compute k B 2t 2 +2 2t2 t, i , k − t − i k/2<i+t≤k 2 2 +1 2 2 2 +1 k k m m 2 +2 B 2t 2 +1 = m 2t2 t m=k/2+1 2 +1 t=0 2 k k 2m+1 m m 2 +2 B 2 +1 − B 2 +1 = m 2 2 1 m=k/2+1 2 +1 =0 1 k k 2 +2 m , = m (2 +1) m=k/2+1 2 +1 2m+1 2m+1 wherewehaveused(4)toseethat B2 = 0. Therefore, =0 k B 2t 2 +2 2t2 t, i , k − t − i k/2<i+t≤k 2 2 +1 2 2 2 +1 k k k k 2 2 = (2 +2) m + m − m=k/2+1 2 2 1 k k k k k k 2 2 2 2 =( +1) m + k − m + m − + k − m m=k/2+1 2 2 2 2 1 2 2 +1 k k 2k − − k 2 . =( +1) 2 ( 1) k (6) Lemma 3.1 now follows if we subtract (6) from (5). 4. The Formula for ζ(2k) To begin this section, let us use integration by parts to see that for any n ≥ 1and k ≥ 2, π π 1 k k 1 k−1 k an(k)= x cos(nx) dx = − x sin(nx) dx = − bn(k − 1). (7) π −π n π −π n Similarly, if k is odd, then π n+1 k−1 1 k (−1) π k bn(k)= x sin(nx) dx =2 + an(k − 1). (8) π −π n n Now, it is clear from the definitions of an(k)andbn(k)thatan(k) = 0 whenever k 2 2 is odd and bn(k) = 0 whenever k is even. Hence, an(k) + bn(k) is actually equal July 10, 2018 13:29 WSPC/S1793-0421 203-IJNT 1850110 Revisiting the Riemann Zeta function at positive even integers 1853 2 to (an(k)+bn(k)) . If we appeal to (7) and (8) recursively and use the fact that (−1)n+1 bn(1) = 2 n , then a simple inductive argument shows that when n, k ≥ 1, k−1 2 π2 a k b k c k, , n( )+ n( )= n( )nk−2 (9) =0 wherewedefine k k − 2 ! − k/2++n+1, ≤ ≤ 1 ( 1) if 0 ; cn(k, )= (2 +1)! 2 (10) 0, otherwise. Gathering (2), (9), and (10) together yields 2 k−1 π2k π ∞ 2 π2 − a k 2 1 x2k dx − a k 2 c k, 2 k 2 0( ) = π 2 0( ) = n( )nk−2 2 +1 −π n=1 =0 2 k−1 2 k−1 ∞ 2 π2j 2 r k, j r k, j π2j ζ k − j , = ( )n2k−2j = ( ) (2 2 ) (11) n=1 j=0 j=0 where min( k−1 ,j) j 2 r(k, j)= cn(k, i)cn(k, j − i)= cn(k, i)cn(k, j − i) i=0 k−1 i=max(0,j− 2 ) min( k−1 ,j) 2 k k 2 ! − k/2+i+n+1 2 ! − k/2+j−i+n+1 = i ( 1) j − i ( 1) k−1 (2 +1)! (2 2 +1)! i=max(0,j− 2 ) min( k−1 ,j) − jk 2 2 j 4( 1) ! 2 +2 .
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