A q -analogue for Euler's evaluations of the Riemann zeta function
aa r X i v : . [ m a t h . N T ] S e p A q -ANALOGUE FOR EULER’S EVALUATIONS OF THERIEMANN ZETA FUNCTION ANKUSH GOSWAMI
Abstract.
We provide a q -analogue of Euler’s formula for ζ (2 k ) for k ∈ Z + .Our main results are stated in Theorems 3.1 and 3.2 below. The result gen-eralizes a recent result of Z.W. Sun who obtained q -analogues of ζ (2) = π / ζ (4) = π / Introduction
Recently, Sun obtained a very nice q -analogue of Euler’s formula ζ (2) = π / q -analogue of ζ (4) and noted that it wassimultaneously and independently obtained by Sun [9]. The author then obtainedthe q -analogue of ζ (6) in [4] but realized that this transition to the q -analogue of ζ (6) is more difficult as compared to ζ (2) and ζ (4). This difficulty arises due to anextra term that shows up in the identity; however in the limit as q → − , this term →
0. Thus it is necessary to study the q -analogue of Euler’s celebrated formula ζ (2 k ) = ( − k +1 k B k π k k )!(1.1)for all k ∈ Z + . We will see shortly that this requires a consideration of two cases: k even and k odd separately (see Theorems 3.1, 3.2 below). We also mention herethat Zudilin [10] and Krattenthaler-Rivoal-Zudilin [8] have studied the Diophantineproperties of q -zeta values, the sums appearing in the left-hand side of Theorems3.1 and 3.2. 2. Notations
For a positive integer k we use the following standard notations. Let B k denotethe 2 k th Bernoulli number. Let (cid:8) nk (cid:9) denote a Stirling number of the second kind,which is the number of ways of partitioning a set of n objects into k non-emptysubsets. Let the complex upper half-plane be denoted by H = { τ ∈ C : Im( τ ) > } and let SL ( Z ) denote the full modular group which is defined to be the set of all2 × (4) denote thewell-known principal congruence subgroup of SL ( Z ) defined byΓ (4) = (cid:26)(cid:18) a bc d (cid:19) ∈ SL ( Z ) : c ≡ (cid:27) , Mathematics Subject Classification.
Key words and phrases.
Riemann zeta function, Stirling numbers of second kind, triangularnumbers, upper half plane.
Finally, let M k (Γ (4)) be the vector space of all weight 2 k modular forms overΓ (4) and S k (Γ (4)) denote the subspace of M k (Γ (4)) of all weight 2 k cusp formsover Γ (4). Let q = e πiτ where τ ∈ H . We define the Dedekind eta function, awell-known modular form of weight 1/2, by η ( τ ) = q / ∞ Y n =1 (1 − q n ) . (2.1)For a thorough treatment on modular forms, the interested readers should consult[5, 7]. Finally, we denote the n th triangular number T n by T n = n ( n + 1)2 , n = 1 , , .... (2.2)and the corresponding generating function by ψ ( q ) = ∞ X n =1 q T n . (2.3)Also let d k be given by d k = − ( − k B k (4 k − k ∈ Q . Main theorems
Theorem 3.1.
Let k ≥ be an even integer. For a complex number q with | q | < we have ∞ X n =0 k − q n +1 P e k − ( q n +1 )(1 − q n +1 ) k − T k ( τ /
2) = q k/ d k ∞ Y n =1 (1 − q n ) k (1 − q n − ) k (3.1) where P e k − ( z ) = k − X l =1 ( − l b k ( l ) z l − (3.2) is a polynomial of degree (2 k − with integer coefficients b k ( l ) = k − X m =0 ( − m a k ( m ) (cid:18) k − m − l (cid:19) (3.3) where the a k ( m ) are defined by a k ( m ) = k − X j =0 j !( − j (cid:26) k − j (cid:27)(cid:18) jm (cid:19) . (3.4) and T k ( τ ) ∈ S k (Γ (4)) , thus T k ( τ / → as q → , where the limit is takenfrom inside the unit disk. In other words, Theorem . gives a q -analogue of (1 . for ζ (2 k ) with k even. Theorem 3.2.
Let k ≥ be an odd integer. For a complex number q with | q | < we have ∞ X n =0 q n +1 P o k − ( q n +1 )(1 − q n +1) ) k − T k ( τ ) = q k d k ∞ Y n =1 (1 − q n ) k (1 − q n − ) k (3.5) q -ANALOGUE FOR EULER’S EVALUATIONS OF THE RIEMANN ZETA FUNCTION 3 where P o k − ( z ) = (1 + z ) k P e k − ( z ) − k − zP e k − ( z )(3.6) is a polynomial of degree (4 k − with integer coefficients and where P e k − ( z ) isthe polynomial defined in (3 . and T k ( τ ) ∈ S k (Γ (4)) , thus T k ( τ ) → as q → ,where the limit is taken from inside the unit disk. In other words, Theorem . gives a q -analogue of (1 . for ζ (2 k ) with k odd.Two remarks: (1) Note that we are using (4.5) of Theorem 4.2 (see below) to prove Theorems3.1 and 3.2. Clearly the left-hand of (4.5) is q k times a function of q . Ifthe right-hand side of (4.5) also turns out to be a function of q , we canreplace q → √ q without affecting our results. This happens to be the casein Theorem 3.1 where we obtain expressions in q for both the sum andproduct, thereby giving us T k ( τ /
2) in (3.1). However we do not obtainsuch expressions in q on both sides of (3.5) in Theorem 3.2 and thus we get T k ( τ ) instead of T k ( τ / k = 1 , , q for the sum in(3.5) so that we can replace q → √ q , thereby getting T k ( τ /
2) in (3.5).(2) The cusp form T k ( τ ) in Theorems 3.1, 3.2 is well-defined and uniquelydetermined by the difference of a q -series and a q -product as follows: T k ( τ ) = ∞ X n =0 k − q n +2 P e k − ( q n +2 )(1 − q n +2 ) k − q k d k ∞ Y n =1 (1 − q n ) k (1 − q n − ) k ( k even) ∞ X n =0 q n +1 P o k − ( q n +1 )(1 − q n +2 ) k − q k d k ∞ Y n =1 (1 − q n ) k (1 − q n − ) k ( k odd)4. Some useful lemmas
We next state an important theorem which follows from Jacobi triple productidentity, originally proved by Gauss (see [2], p.10, Cor. 1.3.4 and notes in p.23).
Lemma 4.1.
For | q | < we have ψ ( q ) = ∞ Y n =1 (1 − q n )(1 − q n − ) . (4.1)Thus Lemma 4 . ψ k ( q ) = ∞ Y n =1 (1 − q n ) k (1 − q n − ) k = ∞ X n =1 t k ( n ) q n (4.2)where t k ( n ) is the number of ways of representing a positive integer n as a sum of4 k triangular numbers.Next, the following well-known result in [1] due to Atanosov et al. gives us an exactformula for t k ( n ). Indeed, the authors show that t k ( n ) behaves when n becomes ANKUSH GOSWAMI large like σ k − (2 n + k ), the modified divisor function defined by σ k ( n ) := X d | nn/d odd d k = ( σ k ( n ) when n is odd , k σ k (cid:16) n (cid:17) when n is even(4.3)where σ k ( n ) is the k th divisor function defined as σ k ( n ) = X d | n d k . (4.4) Theorem 4.2.
Let k ∈ N . Then q k ψ k ( q ) = 1 d k ( H k ( τ ) − T k ( τ ))(4.5) where d k is defined as in Theorem . and . , T k ( τ ) ∈ S (Γ (4)) and H k ( τ ) is anEisenstein series of weight k on Γ (4) defined by H k ( τ ) = X n> n even σ k − ( n ) q n for k even , X n> n odd σ k − ( n ) q n for k odd . (4.6)By comparing cofficients in (4 . t k ( n ) in [1] (Cor. 2.6, p.119): t k ( n ) = 1 d k ( σ k − (2 n + k ) − a (2 n + k ))(4.7)where T k ( τ ) = ∞ X n =0 a ( n ) q n ∈ S k (Γ (4)). Indeed, Theorem 4.2 follows easily from[6] where the authors detail a closed formula for t k ( n ).We next state an important theorem for the generating function transformationinvolving Stirling numbers. Let F ( z ) denote the infinite geometric series F ( z ) := ∞ X n =0 z n = 11 − z (4.8)with | z | <
1. Then choosing f n = 1 in Prop. 3.1, p.135 of [3] we obtain Proposition 4.3.
Let l be a fixed positive integer. Then we have ∞ X n =0 n l z n = l X j =0 j ! (cid:26) lj (cid:27) z j (1 − z ) j +1 . (4.9)The proof follows by induction on l in conjuction with the recurrence relation ofStirling numbers (cid:26) nl (cid:27) = l (cid:26) n − l (cid:27) + (cid:26) n − l − (cid:27) . (4.10) q -ANALOGUE FOR EULER’S EVALUATIONS OF THE RIEMANN ZETA FUNCTION 5 Proofs of Theorems 3.1 and 3.2
Since ζ (2 k ) = ( − k +1 k B k k )! π k has the following equivalent form ∞ X n =0 n + 1) k = (cid:18) k − k (cid:19) ζ (2 k ) = ( − k +1 (4 k − B k k )! π k (5.1)it will be sufficient to get the q -analogue of (5.1). From the q -analogue of Euler’sGamma function we know thatlim q ↑ (1 − q ) ∞ Y n =1 (1 − q n ) (1 − q n − ) = π .
2) we havelim q ↑ (1 − q ) k ∞ Y n =1 (1 − q n ) k (1 − q n − ) k = π k k (5.3)where q ↑ q → Proof of Theorem 3.1.
Let k ≥ .
6) and(4 .
3) we have H k ( τ ) = ∞ X n =1 σ k − (2 n ) q n = 2 k − ∞ X n =1 σ k − ( n ) q n . Using the definition of σ k − ( n ) in the expression above we obtain H k ( τ ) = 2 k − ∞ X n =1 X d | nn/d odd d k − q n = 2 k − ∞ X i =0 ∞ X j =0 ( j + 1) k − q j +1)(2 i +1) = 2 k − ∞ X i =0 ∞ X j =0 ( j + 1) k − ( q i +1) ) j +1 . (5.4)We wish to find a polynomial Q e k − ( z ) such that the expression in parentheses inthe right-hand equation of (5 .
4) can be written as Q e k − ( q i +1) )(1 − q i +1) ) k = ∞ X j =0 ( j + 1) k − ( q i +1) ) j +1 . (5.5)For notational simplicity let us write z = q i +1) so that (5 .
5) can be rewritten as Q e k − ( z )(1 − z ) k = ∞ X j =0 ( j + 1) k − z j +1 . (5.6) Lemma 5.1. Q e k − ( z ) is a polynomial of degree (2 k − with integer coefficients. ANKUSH GOSWAMI
Proof.
The right hand side of (5 .
6) can be identified with the left-hand side of (4.9)so that using Proposition 4.3 we can rewrite the right-hand side of (5.6) as Q e k − ( z )(1 − z ) k = k − X j =0 j ! (cid:26) k − j (cid:27) z j (1 − z ) j +1 (5.7)Noting that z = 1 − (1 − z ) we use binomial expansion in z j = (1 − (1 − z )) j followedby rearrangements of the sums above to obtain Q e k − ( z )(1 − z ) k = k − X j =0 j ! (cid:26) k − j (cid:27) (1 − (1 − z )) j (1 − z ) j +1 = k − X j =0 j ! (cid:26) k − j (cid:27) j X m =0 ( − j − m (cid:18) jm (cid:19) (1 − z ) j − m − z ) j +1 = k − X m =0 ( − m k − X j =0 ( − j j ! (cid:26) k − j (cid:27)(cid:18) jm (cid:19) − z ) m +1 = k − X m =0 ( − m a k ( m )(1 − z ) m +1 , (5.8)where a k ( m ) is defined by a k ( m ) = k − X j =0 ( − j j ! (cid:26) k − j (cid:27)(cid:18) jm (cid:19) . (5.9)Note that in going from the second step to the third step in (5.8) we used the factthat (cid:0) jm (cid:1) = 0 if m > j and hence we are able to interchange the sums over m and j above. Thus multiplying both sides of (5.8) by (1 − z ) k and using the binomialexpansion yields Q e k − ( z ) = k − X m =0 ( − m a k ( m )(1 − z ) k − m − = k − X m =0 ( − m a k ( m ) k − m − X l =0 ( − l (cid:18) k − m − l (cid:19) z l = k − X l =0 ( − l k − X m =0 ( − m a k ( m ) (cid:18) k − m − l (cid:19)! z l = k − X l =0 ( − l b k ( l ) z l , (5.10)where the b k ( l ) are defined by b k ( l ) = k − X m =0 ( − m a k ( m ) (cid:18) k − m − l (cid:19) , (5.11)which establishes Lemma 5.1. (cid:3) q -ANALOGUE FOR EULER’S EVALUATIONS OF THE RIEMANN ZETA FUNCTION 7 We also note from (5.6) that Q e k − (0) = 0. Thus we define the polynomial P e k − ( z ) of degree (2 k −
2) by Q e k − ( z ) := zP e k − ( z )(5.12)where P e k − ( z ) = k − X l =1 ( − l b k ( l ) z l − . (5.13)We rewrite (5.4) using (5.12) as H k ( τ ) = 2 k − ∞ X i =0 q i +1) P e k − ( q i +1) )(1 − q i +1) ) k . (5.14)Now from (4.5) of Theorem 4.2 we have H k ( τ ) − T k ( τ ) = d k q k ψ k ( q ) . (5.15)Thus from (4.2), (5.14) and (5.15) we get ∞ X n =0 k − q n +1) P e k − ( q n +1) )(1 − q n +1) ) k − T k ( τ ) = d k q k ∞ Y n =1 (1 − q n ) k (1 − q n − ) k . (5.16)Making the change of variable q → √ q in (5.16) we obtain ∞ X n =0 k − q n +1 P e k − ( q n +1 )(1 − q n +1 ) k − T k ( τ /
2) = d k q k/ ∞ Y n =1 (1 − q n ) k (1 − q n − ) k . (5.17)On multiplying both sides of (5.7) by (1 − z ) k we obtain Q e k − ( z ) = k − X j =0 j ! (cid:26) k − j (cid:27) z j (1 − z ) k − j − . (5.18)As z → − , each summand in (5.18) vanishes except the term corresponding to j = 2 k −
1. Thus we get lim z → − Q e k − ( z ) = (2 k − . (5.19)In view of (5.19) and the fact that T k ( τ / → q → − , (5.17)gives 2 k − ∞ X n =0 (2 k − n + 1) k = d k π k k . (5.20)Using the definition of d k we obtain identity (5.1). Thus Theorem 3.1 follows fromall the above observations. ANKUSH GOSWAMI
Proof of Theorem 3.2.
Let k ≥ H k ( τ ) = X n> n odd σ k − ( n ) q n = ∞ X n =0 n odd σ k − (2 n + 1) q n +1 = ∞ X n =1 σ k − ( n ) q n − ∞ X n =1 σ k − (2 n ) q n = ∞ X n =1 X d | nn/d odd d k − q n − k − ∞ X n =1 X d | nn/d odd d k − q n = ∞ X i =0 ∞ X j =0 ( j + 1) k − q ( j +1)(2 i +1) − k − ∞ X i =0 ∞ X j =0 ( j + 1) k − q j +1)(2 i +1) . (5.21)In view of (5.6) and Lemma 5.1 we can rewrite (5.21) as H k ( τ ) = ∞ X i =0 Q e k − ( q i +1 )(1 − q i +1 ) k − k − ∞ X i =0 Q e k − ( q i +1) )(1 − q i +1) ) k = ∞ X i =0 Q o k − ( q i +1 )(1 − q i +1) ) k (5.22)where Q o k − ( q i +1 ) is the polynomial in w = q i +1 of degree 4 k − Q o k − ( w ) = (1 + w ) k Q e k − ( w ) − k − Q e k − ( w ) . (5.23)Since Q e k − (0) = 0, in view of (5.6) and (5.23) we also have Q o k − (0) = 0. There-fore we define the polynomial P o k − ( w ) of degree 4 k − Q o k − ( w ) = w (1 + w ) k P e k − ( w ) − k − w P e k − ( w ):= wP o k − ( w )(5.24)where P o k − ( w ) = (1 + w ) k P e k − ( w ) − k − wP e k − ( w ) . (5.25)Hence from (5.22), (5.24) and Theorem 4.2 we obtain H k ( τ ) − T k ( τ ) = d k q k ψ k ( q ) , ∞ X n =0 q n +1 P o k − ( q n +1 )(1 − q n +1) ) k − T k ( τ ) = d k q k ∞ Y n =1 (1 − q n ) k (1 − q n − ) k . (5.26)Also as w → − , (5.19) and (5.24) yieldlim w → − Q o k − ( w ) = 2 k (2 k − − k − (2 k − k − (2 k − . (5.27) q -ANALOGUE FOR EULER’S EVALUATIONS OF THE RIEMANN ZETA FUNCTION 9 Thus on multiplying both sides of (5.26) by (1 − q ) k and taking the limit as q → k − (2 k − ∞ X n =0 n + 1) k = d k π k k , (5.28)Using definition of d k we immediately obtain identity (5.1). Thus Theorem 3.2follows from all of the above observations.6. Explicit computations of P e k − ( z ) and P o k − ( z ) for different k We used the Python programming language to compute the co-efficients a k ( m )and b k ( l ) to determine the polynomials P e k − ( z ) and P o k − ( z ) for a few differentvalues of k . We will see that our results for k = 1 , , Case k=1 : Sun’s result.
Since k = 1 is odd we use (3.6) to get P o ( z ) = (1 + z ) P e ( z ) − zP e ( z ) , where we define P e ( z ) = 1. Therefore, P o ( z ) = (1 + z ) − z = 1 + z . (6.1)Thus, (6.1) and (3.5) yield ∞ X n =1 q n (1 + q n +1) )(1 − q n +1) ) = ∞ Y n =1 (1 − q n ) (1 − q n − ) , (6.2)where, d = 1 and T ( τ ) = 0, (Table I, p.120, [1]). This is Theorem 1.1, (1.1), of[9] with q → √ q in (6.2).6.2. Case k=2.
Here k = 2 is even, so we use (3.2) to get P e ( z ) = − b (1) + b (2) z − b (3) z where b ( l ), 1 ≤ l ≤ a (0) , a (1) , a (2)and a (3). Using (3.4) we obtain a (0) = − , a (1) = − , a (2) = − , a (3) = − . Thus (3.3) yields b (1) = − , b (2) = 4 , b (3) = − . Hence we obtain P e ( z ) = 1 + 4 z + z (6.3)which when used in (3.1) yields the results in [9] and [4]. Here again T ( τ /
2) = 0(Table I, p.120, [1]).
Case: k=3.
Here we need to evaluate the coefficients b (1) , b (2) , b (3) , b (4) ,b (5) and the corresponding a (0) , a (1) , a (2) , a (3) , a (4) , a (5). Using (3.4) weget a (0) = − , a (1) = − , a (2) = − , a (3) = − , a (4) = − ,a (5) = − b (1) = − , b (2) = 26 , b (3) = − , b (4) = 26 , b (5) = − . Using these values in (3.6) we obtain P o ( z ) = (1 + z ) P e ( z ) − zP e ( z )= (1 + z ) (1 + 26 z + 66 z + 26 z + z ) − z (1 + 26 z + 66 z + 26 z + z )= z + 237 z + 1682 z + 1682 z + 237 z + 1= ( z + 1)( z + 236 z + 1446 z + 236 z + 1)which when used in (3.5) along with the change of variable q → √ q gives us theresult in [4]. We note here that in [4] we obtained explicitly T ( τ /
2) = φ ( q ) where φ ( q ) = Q ∞ n =1 (1 − q n ) is the Euler’s function.6.4. Case: k=4.
Here we use (3.4) and (3.3) to obtain a (0) = − , a (1) = − , a (2) = − , a (3) = − , a (4) = − ,a (5) = − , a (6) = − , a (7) = − b (1) = − , b (2) = 120 , b (3) = − , b (4) = 2416 , b (5) = − ,b (6) = 120 , b (7) = − . Thus we have P e ( z ) = z + 120 z + 1191 z + 2416 z + 1191 z + 120 z + 1which when used in (3.1) gives us the following q -analogue of ζ (8) = π / ∞ X n =0 q n P e ( q n +1 )(1 − q n +1 ) − T ( τ /
2) = 136 q ∞ Y n =1 (1 − q n ) (1 − q n − ) . (6.4)6.5. Case: k=5.
Here we use (3.4) and (3.3) to obtain a (0) = − , a (1) = − , a (2) = − , a (3) = − ,a (4) = − , a (5) = − , a (6) = − , a (7) = − ,a (8) = − , a (9) = − b (1) = − , b (2) = 502 , b (3) = − , b (4) = 88234 , b (5) = − ,b (6) = 88234 , b (7) = − , b (8) = 502 , b (9) = − . q -ANALOGUE FOR EULER’S EVALUATIONS OF THE RIEMANN ZETA FUNCTION 11 Thus from (3.6) we obtain the polynomial P o ( z ) = (1 + z ) P e ( z ) − zP e ( z )= (1 + z ) (cid:0) z + 502 z + 14608 z + 88234 z + 156190 z + 88234 z +14608 z + 502 z + 1 (cid:1) − z (cid:0) z + 502 z + 14608 z + 88234 z +156190 z + 88234 z + 14608 z + 502 z + 1 (cid:1) = (1 + z )( z + 19672 z + 1736668 z + 19971304 z + 49441990 z +19971304 z + 1736668 z + 19672 z + 1) . Using this in (3.5) with q → √ q we obtain the following q -analogue of ζ (10) = π / ∞ X n =0 q n (1 + q n +1 ) S ( q n +1 )(1 − q n +1 ) − T ( τ /
2) = 2031616 q ∞ Y n =1 (1 − q n ) (1 − q n − ) where S ( z ) = z + 19672 z + 1736668 z + 19971304 z + 49441990 z +19971304 z + 1736668 z + 19672 z + 1 . Acknowledgement
The author is grateful to Krishnaswami Alladi for his constant support, en-couragement and stimulating discussions. He sincerely thanks Frank Garvan forseveral interesting discussions on the problem and providing him with some usefulreferences. He also expresses his appreciation to George Andrews for his support.Finally, he thanks the anonymous referees for their feedback on the manuscriptwhich improved exposition.
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