The summation of infinite partial fraction decomposition I: some formulae related to the Hurwitz zeta function
aa r X i v : . [ m a t h . C A ] F e b T HE SUMMATION OF INFINITE PARTIAL FR AC TIONDEC OMPOSITION I: SOME FOR MULAE R ELATED TO THE H URWITZ ZETA FUNC TION
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Xiaowei Wang ∗ February 9, 2021 A BSTRACT
In this paper we establish a new summation method by expanding Q k (1 − za k ) − with two ap-proaches: the Taylor expansion and the infinite partial fraction decomposition. Here we focus on thecase when a k is arithmetic sequence. By this summation we obtain many equalities involve Hurwitzzeta function and Gammma function. K eywords Partial fraction, Summation, Hurwitz zeta function
Let S be the series of rational functions S = P P ( n ) Q ( n ) with deg( Q ) ≥ deg( P ) + 2 . By decomposing PQ into partialfractions, S can be rewritten as finite linear combination of Hurwitz zeta function ζ ( m, a ) = P ∞ k =0 1( a + k ) m with posi-tive interger m . In this paper, we construct another series representation of ζ ( m, a ) by discuss two kinds of expansionsof the infinite product Q ∞ k =0 (1 − ( za + k ) m ) .In fact, the correspondence is natural. In general, there is some relation between P ∞ k =0 1 a mk and P ∞ k =0 1 F ′ ( a k ) 1 a m +1 k ,where F ( z ) is the product Q ∞ k =0 (1 − ( za k ) m ) . Basically, the relation is derived by the infinite partial fraction decompo-sition of F ( z ) . And the correspondence is not only valid for infinite series, but also valid for finite sum. Therefore theconvergence can be discussed easily. In this we paper we study this relation and focus on the case when a k = a + k ,namely, the problem involves the Hurwitz zeta function.Throughout the paper we denote the coefficient of z n in the Laurent expansion of f ( z ) around by [ f ( z )] n . Thenotation F ′ ( a n ) refers to lim z → a n F ′ ( z ) . Lemma 1. (Homogeneous partial fraction decomposition)Let a , ..., a n be distinct complex numbers, x ∈ C \{ a , ..., a n } , then there exist µ , ..., µ n ∈ C such that followingidentity is true, n Y i =1 x − a i = n X i =1 µ i x − a i (1)where µ i = Q nj =1 ,j = i a i − a j . Further, The relation P ni =1 µ i = 0 holds. Proof.
The proof is easy, one can consider the Lagrange’s interpolation formula. For more details, See author’s paper[1]. ∗ This paper is written in Dec 2020
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Lemma 2. (One point lifting)Suppose that
N, L ∈ Z ≥ and a n are complex numbers that differ from each other, then the following identity holds x L N Y n =1 x − a n = L X j =1 N X n =1 − µ n a L − j +1 n x j + N X n =1 µ n a Ln x − a n where µ n = N Y s =1 ,s = n a n − a s Proof.
Firstly one can see the case L = 0 is exactly Lemma 1, hence the the identity is true for ℓ = 0 . Fol-lowing we assume that L ∈ Z ≥ . Suppose that z , z , ..., z N be complex variables that take distinct values. Let H ( z , z , ..., z n ) = Q Nn =0 1 x − z n . Then by Lemma 1 we have the identity H ( z , z , ..., z n ) = N X n =0 λ n x − z n Where λ n = Q Ns =0 ,s = n z n − z s . Now we taking partial derivatives of H with respect to z . There are two ways to carryout. On the one hand, ∂ l H∂z l = l !( x − z ) l +1 ∞ Y n =1 x − z n (2)On the other hand, ∂ l H∂z l = ∂ l ∂z l ( λ x − z ) + N X n =1 x − z n ∂ l λ n ∂z l = l X j =0 (cid:18) lj (cid:19) ∂ l − j λ ∂z l − j j !( x − z ) j +1 + N X n =1 x − z n ∂ l λ n ∂z l Recall that λ = Q Ns =1 1 z − z s . reapplying Lemma 1 λ = N X n =1 M n z − z n where M n = Q Ns =1 ,s = n z n − z s . Therefore ∂ l − j λ ∂z l − j = N X n =1 M n ( − l − j ( l − j )!( z − z n ) l − j +1 Further, for n = 1 , ..., N ∂ l λ n ∂z l = l ! M n ( z n − z ) l +1 Now compare to (2) we obtain x − z ) l +1 ∞ Y n =1 x − z n = l X j =0 N X n =1 M n ( − l − j ( l − j )!( z − z n ) l − j +1 x − z ) j +1 + N X n =1 x − z n M n ( z n − z ) l +1 Lastly let L = l + 1 , z = 0 , and z i = a i , µ n = M n | z i = a i for i = 1 , ..., N respectively, we obtain what required.2 PREPRINT - F
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In the last section we introduce some lemmas about partial fraction decomposition. In follow we discuss the partialfraction expansion of the infinite product ( z L F ( z )) − = z − L Q ∞ n =1 (1 − za n ) − . We then show that the rearrangementto the power series of z remains invariant for different L . Theorem 1. ( Partial Fraction Summation )Let ( a n ) be a sequence in C \{ } and satisfy that: 1, a n differ from each other for all n ; 2, P ∞ n =1 1 a n absolutelyconverges. Define F ( z ) = ∞ Y n =1 (1 − za n ) then following identity holds for k = 0 , , , ... . [ 1 F ( z ) ] k = ∞ X n =1 − F ′ ( a n ) 1 a k +1 n On the other hand, for L = 0 , , , ... , F ( z ) has the following partial fraction decompositions F ( z ) = [ 1 F ( z ) ] + [ 1 F ( z ) ] z + ... + [ 1 F ( z ) ] L − z L − + ∞ X n =1 z L F ′ ( a n ) a Ln ( z − a n ) For given L , we call L the order of this expansion .Proof. Let M = inf( | a n | ) , in the proof we always assume that | z | < M . Consider the expansion of F ( z ) , namely, F ( z ) = exp( − ∞ X n =1 log(1 − za n )) = exp( ∞ X k =1 ∞ X n =1 a kn z k k ) Denote [ F ( z ) ] k by c k , it can be computed by expanding the right hand side in the above formula. Then z L F ( z ) = 1 z L ( c + c z + c z + ... ) (3)On the other hand, consider the another expansion via Lemma 2 and let N → ∞ , that is z L F ( z ) = 1 z L ( ∞ X n =1 − λ n a n ) + ... + 1 z ( ∞ X n =1 − λ n a L − n ) + 1 z ( ∞ X n =1 − λ n a Ln ) + ∞ X n =1 λ n a Ln ( z − a n ) (4)Where λ n = − a n Q ∞ s =1 ,s = n a s a s − a n . Further, we can reformulate λ n as following λ n = − a n lim z − a n → a n − za n ∞ Y s =1 a s a s − z = 1 F ′ ( a n ) Now consider the expansion of P ∞ n =1 λ n a Ln ( z − a n ) , due to the absolute convergence of P a n , for | z | < M we have ∞ X n =1 λ n a Ln ( z − a n ) = ∞ X n =1 − λ n a L +1 n + ∞ X n =1 − λ n a L +2 n z + ∞ X n =1 − λ n a L +3 n z + ... Then equation (4) becomes z L F ( z ) = 1 z L ( ∞ X n =1 − λ n a n + ∞ X n =1 − λ n a n z + ... + ∞ X n =1 − λ n a L +1 n z L + ... ) (5)Compare to (3) we have for k = 0 , , , , ... , the equalities hold c k = ∞ X n =1 − F ′ ( a n ) 1 a k +1 n PREPRINT - F
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Corollary 1.
As a direct corollary of theorem 1, ∞ X n =1 − F ′ ( a n ) 1 a n ≡ ∞ X n =1 − F ′ ( a n ) 1 a n = ∞ X n =1 a n ∞ X n =1 − F ′ ( a n ) 1 a n = 12 (( ∞ X n =1 a n ) + ∞ X n =1 a n ) ... Theorem 2. ( Partial fraction summation for symmetric products )Let ( a n ) be a sequence in C \{ } and satisfy that: 1, a n differ from each other for all n ; 2, P ∞ n =1 1 a n absolutelyconverges. Define F ( z ) = ∞ Y n =1 (1 − ( za n ) ) then following identity holds for k = 0 , , , ... . [ 1 F ( z ) ] k = ∞ X n =1 − F ′ ( a n ) 1 a k +1 n On the other hand, for L = 0 , , , ... , F ( z ) has the following partial fraction decompositions F ( z ) = [ 1 F ( z ) ] + [ 1 F ( z ) ] z + ... + [ 1 F ( z ) ] L − z L − + ∞ X n =1 z L F ′ ( a n ) a L − n ( z − a n ) (6) for L = 1 , , , ... , F ( z ) has the following partial fraction decompositions F ( z ) = [ 1 F ( z ) ] + [ 1 F ( z ) ] z + ... + [ 1 F ( z ) ] L − z L − + ∞ X n =1 z L +1 F ′ ( a n ) a Ln ( z − a n ) (7) Proof.
By Theorem 1, one can let F ( z ) = Q ∞ n =1 (1 − zb n ) where b n − = a n , b n = − a n . In order the find F ′ ( b n ) ,just need to note that F ( z ) = f ( z ) f ( − z ) , where f ( z ) = Q ∞ n =1 (1 − za n ) ( This infinite product may be divergent, onecan use finite product to approximate ), Then F ′ ( b n − ) = ( f ( z ) f ( − z )) ′ | z = a n = f ′ ( a n ) f ( − a n ) − f ( a n ) f ′ ( − a n ) = f ′ ( a n ) f ( − a n ) F ′ ( b n ) = ( f ( z ) f ( − z )) ′ | z = − a n = f ′ ( − a n ) f ( a n ) − f ( − a n ) f ′ ( a n ) = − f ′ ( a n ) f ( − a n ) That is F ′ ( b n − ) = − F ′ ( b n ) . By Theorem 1 we have [ 1 F ( z ) ] k − = ∞ X n =1 − F ′ ( b n ) 1 b kn = ∞ X n =1 − F ′ ( b n − ) 1 b k n − + ∞ X n =1 − F ′ ( b n ) 1 b k n = ∞ X n =1 − F ′ ( b n − ) 1 a kn + ∞ X n =1 − F ′ ( b n ) 1 a kn = 0 PREPRINT - F
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9, 2021This is trivial since F ( z ) is even function. [ 1 F ( z ) ] k = ∞ X n =1 − F ′ ( b n ) 1 b k +1 n = ∞ X n =1 − F ′ ( b n − ) 1 b k +12 n − + ∞ X n =1 − F ′ ( b n ) 1 b k +12 n = ∞ X n =1 − F ′ ( b n − ) 1 a k +1 n − ∞ X n =1 − F ′ ( b n ) 1 a k +1 n = ∞ X n =1 − F ′ ( a n ) 1 a k +1 n The remaining partial fraction decomposition formulae (6), (7) are straightforward, we omit the computation.Let a ∈ C \{ , − , − , ... } , define g a ( z ) := Γ( a + z )Γ( a − z )Γ( a ) Then g a ( z ) has the infinite product representation g a ( z ) = ∞ Y k =0 (1 − ( za + k ) ) − (8)It can be proved easily by using the formula Γ( a + z )Γ( a ) = ae − γz a + z ∞ Y n =1 (1 + za + n ) − e zn (9)Now by Theorem 2 we have the partial fraction decomposition of g a ( z ) of order a + z )Γ( a − z ) = Γ( a ) + ∞ X k =0 − k +1 Γ(2 a + k )( a + k ) k ! z z − ( a + k ) (10)In order to keep the convergence, we shall keep ℜ ( a ) < and | z | < | a | . Further, it’s easy to obtain [Γ( a + z )Γ( a − z )] by (8), namely [Γ( a + z )Γ( a − z )] = ζ (2 , a ) . On the other hand, by finding the Taylor expansion of (10), one has Corollary 2. ( The PFS representation of ζ (2 , a ) ) ζ (2 , a ) = ∞ X k =0 a + k ) = ∞ X k =0 − k Γ(2 a + k )Γ( a ) k !( a + k ) , ( ℜ ( a ) < , a = 0 , − , − , − , ... ) (11) Theorem 3. ( Partial fraction summation for cyclotomic product )Let ( a n ) be a sequence in C \{ } and satisfy that:I, a n differ from each other for all n ;II, P ∞ n =1 1 a mn absolutely converges.Define F ( z ) = ∞ Y n =1 (1 − ( za n ) m ) then following identity holds for J = 0 , , , ... . [ 1 F ( z ) ] J = ( , if mod ( J, m ) = 0 P ∞ n =1 − mF ′ ( a n ) 1 a J +1 n , if mod ( J, m ) = 0 (12)5
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Proof.
The proof is quite similar to Theorem 2. For k = 1 , , .. , let k = m ( n −
1) + r and b k = b m ( n − r = a n ω r − m , r = 1 , , ..., m where ω m = e iπm is the m − th root of unity. Then F ( z ) = ∞ Y k =1 (1 − zb k ) On the othe hand, let f r,N ( z ) = N Y n =1 (1 − za n ω r − m ) , r = 1 , , ..., m Then f r,N ( z ) = f ,N ( z/ω r − m ) , we denote f ,N ( . ) directly by f N ( . ) . Then f r,N ( a n ω r − m ) = f N ( a n ) = 0 for n ≤ N .To compute F ′ ( b k ) = F ′ ( b m ( n − r ) : F ′ ( b m ( n − r ) = lim N →∞ ( f ,N ( z ) f ,N ( z ) ...f m,N ( z )) ′ | z = a n ω r − m = lim N →∞ f N ( z ) f N ( zω m ) ...f N ( zω m − m )( f ′ N ( z ) f N ( z ) + 1 ω m f ′ N ( z/ω m ) f N ( z/ω m ) + ... + 1 ω m − m f ′ N ( z/ω m − m ) f N ( z/ω m − m ) ) | z = a n ω r − m = 1 ω r − m f ′ ( a n ) f ( a n ω m ) ...f ( a n ω m − m ) One can implies that for all n = 1 , , , ... and r = 1 , , ..., m , F ′ ( b m ( n − r ) = 1 ω r − m F ′ ( b m ( n − ) According to Theorem 1, we have [ 1 F ( z ) ] J = ∞ X k =1 − F ′ ( b k ) 1 b J +1 k = m X r =1 ∞ X n =1 − F ′ ( b m ( n − r ) 1( a n ω r − m ) J +1 = m X r =1 ω J ( r − m ∞ X n =1 − F ′ ( b m ( n − ) 1 a J +1 n Therefore [ 1 F ( z ) ] J = ( , if mod ( J, m ) = 0 P ∞ n =1 − mF ′ ( a n ) 1 a J +1 n , if mod ( J, m ) = 0
Corollary 3. ( The PFS representation of ζ ( m, a ) )For m = 2 , , , ... , following identity holds, ζ ( m, a ) = ∞ X k =0 m ( − k Q m − r =1 Γ( a − ω rm ( a + k ))Γ( a ) m k !( a + k ) m +1 (13) Proof.
In the Theorem 3 for convenience let k = n − , let a n = a − n = a + k , then F ( z ) = Q ∞ k =0 (1 − ( za + k ) m ) − ,it easy to see that [ F ( z ) ] m = ζ ( m, a ) , now according to Theorem 3, it remains to find F ′ ( a + k ) . Note that F ( z ) = Γ( a ) m Q m − r =0 Γ( a − ω rm z ) PREPRINT - F
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9, 2021this can be easily derived by the formula (9). After some computations F ′ ( z ) = Γ( a ) m Q m − r =0 Γ( a − ω rm z ) m − X r =0 ω rm ψ ( a − ω rm z ) Note that lim z →− k ψ ( z )Γ( z ) = ( − k − k ! , hence F ′ ( a + k ) = Γ( a ) m k !( − k − Q m − r =1 Γ( a − ω rm ( a + k )) Finally by (12) we obtain what required, namely ζ ( m, a ) = ∞ X k =0 − mF ′ ( a + k ) 1( a + k ) m +1 = ∞ X k =0 m ( − k Q m − r =1 Γ( a − ω rm ( a + k ))Γ( a ) m k !( a + k ) m +1 Corollary 4. ( The PFS representation of ζ ( m ) )Let a = 1 , we have then ζ ( m ) = ∞ X n =1 m ( − n − Γ(1 − ω m n ) ... Γ(1 − ω m − m n ) n ! n m As a special case, after some simplification, we have a series representation for Aprèy’s constant ζ (3) = ∞ X n =1 − n − Γ( √ n )Γ( −√ n ) n · n ! Theorem 4. ( The partial fraction summation for general product)
Let F ( z ) = ∞ Y k =0 M Y m =1 (1 − za m,k ) then formally we have [ 1 F ( z ) ] ℓ = ∞ X k =0 M X m =1 − a ℓ +1 m,k F ′ ( a m,k ) especially, if F ( z ) normally converges in | z | < M for some M > , then [ 1 F ( z ) ] = ∞ X k =0 M X m =1 − a m,k F ′ ( a m,k ) = 1[ 1 F ( z ) ] = ∞ X k =0 M X m =1 − a m,k F ′ ( a m,k ) = ∞ X k =0 M X m =1 a m,k Proof.
The proof is similar to the proof of Theorem 3. We left it to the reader.
Example 5.
Let F ( z ) = Γ( )Γ(1+ z )Γ( − z ) , then we can reformulate F ( z ) as F ( z ) = ∞ Y k =1 (1 − za k )(1 − zb k ) where a k = 2 k − , b k = − k . It’s not hard to obtain F ′ ( z ) = √ π z )Γ( − z ) ( ψ ( 12 − z − ψ (1 + z PREPRINT - F
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9, 2021It turns out that F ′ ( a k ) = − F ′ ( b k ) = √ π − k ( k − k + ) By the Theorem 4 ∞ X k =1 ( − k − Γ( k + )( k − k − k ) = √ π ∞ X k =1 ( − k − Γ( k + )( k − k − − k ) ) = √ π log 22 ∞ X k =1 ( − k − Γ( k + )( k − k − + 1(2 k ) ) = √ π (log ζ (2))4 Note that Γ( k + )( k − = √ π (2 k − k − , in fact ∞ X k =1 ( − k − (2 k − k − k − k ) = 12 ∞ X k =1 ( − k − (2 k − k − k − − k ) ) = log 22 ∞ X k =1 ( − k − (2 k − k − k − + 1(2 k ) ) = log ζ (2)2 More general, we have
Example 6.
Assume that ≤ ℜ ( a ) , ℜ ( b ) < and a, b = 0 . Let a k = a + k , b k = − b − k , k = 0 , , , ... . Consider F ( z ) = ∞ Y k =0 (1 − za k )(1 − zb k ) = Γ( a )Γ( b )Γ( a − z )Γ( b + z ) One can show that F ′ ( z ) = Γ( a )Γ( b )Γ( a − z )Γ( b + z ) ( ψ ( a − z ) − ψ ( b − z )) That is F ′ ( a k ) = − F ′ ( b k ) = ( − k − Γ( a + b + k ) k !Γ( a )Γ( b ) Therefore [ 1 F ( z ) ] J = ∞ X k =0 ( − k Γ( a + b + k ) k !Γ( a )Γ( b ) ( 1( a + k ) J +1 + ( − m ( b + k ) J +1 ) If we consider some special cases for instance [ F ( z ) ] and [ F ( z ) ] , then Γ( a )Γ( b ) = ∞ X k =0 ( − k Γ( a + b + k ) k ! ( 1 a + k + 1 b + k ) and Γ( a )Γ( b )( ∞ X k =0 a + k − b + k ) = ∞ X k =0 ( − k Γ( a + b + k ) k ! ( 1( a + k ) − b + k ) ) PREPRINT - F
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9, 2021Moreover, if we let a = , b = , then it becomes Γ( 14 )Γ( 34 )( ∞ X k =0 + k − + k ) = ∞ X k =0 ( − k ( 1( + k ) − + k ) ) or ∞ X k =0 ( − k ( k +1)2 k + 1) = √ π We can take derivative with respect to a in (13). As a simple application, following we discuss the case when m = 2 . Theorem 7. Γ( a + z )Γ( a − z )( ψ ( a + z ) + ψ ( a − z )) = 2 ψ ( a )Γ( a ) + S ( a, z ) − S ( a, z ) where S ( a, z ) = ∞ X k =0 ∞ X n =1 − k ψ (2 a + k )Γ(2 a + k ) k !( a + k ) n +1 z n S ( a, z ) = ∞ X k =0 ∞ X n =1 n + 1)( − k Γ(2 a + k ) k !( a + k ) n +2 z n Proof.
The proof is easy, recall (10) Γ( a + z )Γ( a − z ) = Γ( a ) + ∞ X k =0 − k +1 Γ(2 a + k )( a + k ) k ! z z − ( a + k ) where it converges in | z | < a . The right hand side z z − ( a + k ) can be represented as − P ∞ n =1 ( za + k ) n . It can bereformulated as Γ( a + z )Γ( a − z ) = Γ( a ) + ∞ X k =0 ∞ X n =1 − k Γ(2 a + k ) k !( a + k ) n +1 z n Take ∂∂a on both sides then obtain what required.
Corollary 5.
Denote ζ A ( s ) = ∞ X k =1 ( − k − k s ζ A,H ( s ) = ∞ X k =1 ( − k − H k k s where H k is the harmonic number H k = 1 + + ... + k . Then the following relation is valid for n ∈ Z + . ζ A,H (2 n ) = ( n − n n − n +1 ) ζ (2 n + 1) − n − X j =1 ζ (2 j + 1) ζ A (2 n − j ) Proof.
Let a = 1 then according to Theorem 7 πz sin( πz ) ( ψ (1 + z ) + ψ (1 − z )) = − γ + S (1 , z ) − S (1 , z ) PREPRINT - F
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9, 2021 S (1 , z ) = ∞ X k =0 ∞ X n =1 − k ψ (2 + k )(1 + k ) n z n = − γ ∞ X k =0 ∞ X n =1 − k (1 + k ) n z n + ∞ X k =0 ∞ X n =1 − k H k (1 + k ) n z n or rewritten as [ S (1 , z )] n = − γζ A (2 n ) + 4 ζ A,H (2 n ) , ( n ≥ On the other hand S ( a, z ) = ∞ X k =0 ∞ X n =1 n + 1)( − k Γ(2 a + k ) k !( a + k ) n +2 z n = ∞ X n =1 ∞ X k =0 n + 1)( − k ( k + 1) n +1 z n or rewritten as [ S (1 , z )] n = 2(2 n + 1) ζ A (2 n + 1) , ( n ≥ Therefore [ S (1 , z ) − S (1 , z )] n = − γζ A (2 n ) + 4 ζ A,H (2 n ) − n + 1) ζ A (2 n + 1) , ( n ≥ If we denote [ πz sin( πz ) ] n by τ n , and note that ( ψ (1 + z ) + ψ (1 − z )) = − γ + P ∞ n =1 ζ (2 n + 1) z n ) then [ πz sin( πz ) ( ψ (1 + z ) + ψ (1 − z ))] n = − γτ n + ζ (2 n + 1) + n − X j =1 ζ (2 j + 1) τ n − j ) Therefore γζ A (2 n ) − ζ A,H (2 n ) + (2 n + 1) ζ A (2 n + 1) = γτ n + ζ (2 n + 1) + n − X j =1 ζ (2 j + 1) τ n − j One can show that τ n = 2 ζ A (2 n ) and ζ A ( s ) = (1 − s − ) ζ ( s ) . It follows that ζ A,H (2 n ) = ( n − n n − n +1 ) ζ (2 n + 1) − n − X j =1 ζ (2 j + 1) ζ A (2 n − j ) Corollary 6.
Denote β ( s ) = ∞ X k =0 ( − k (2 k − s β H ( s ) = ∞ X k =1 ( − k − H k − (2 k − s Then for n ∈ Z + the following relation is valid. β H (2 n + 1) = (2 n + 1) β (2 n + 2) − β (2 n + 1) log 4 − n − X j =0 (1 − n − j +1 ) β (2 j + 1) ζ (2 n − j + 1) PREPRINT - F
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Proof.
Let u n = [ ψ ( + z ) + ψ ( − z )] n , one can show that u n = (cid:26) − n +1 ) ζ (2 n + 1) , if n ≥ − γ − log 4) , if n = 0 On the other hand, π cos( πz ) = ∞ X k =0 ( − k E k π k +1 (2 k )! z k where E k is Euler number E = 1 , E = − , E = 5 , E = − , ... . Recall the formula about the Dirichlet betafunction β (2 n + 1) = ( − n E n π n +1 n +2 (2 n )! we have [ π cos( πz ) ] n = 2 n +2 β (2 n + 1) One can show that S ( 12 , z ) = ∞ X k =0 ∞ X n =1 − k ψ (1 + k )( + k ) n +1 z n S ( 12 , z ) = ∞ X k =0 ∞ X n =1 n + 1)( − k ( + k ) n +2 z n or rewritten as [ S ( 12 , z )] n = 2 n +3 ( − γβ (2 n + 1) + β H (2 n + 1))[ S ( 12 , z )] n = 2 n +3 (2 n + 1) β (2 n + 2) According to Theorem 7, by some computation one can get β H (2 n + 1) = (2 n + 1) β (2 n + 2) − β (2 n + 1) log 4 − n − X j =0 (1 − n − j +1 ) β (2 j + 1) ζ (2 n − j + 1) Theorem 8. ( Differential Relation )Suppose that the sequence ( a k ) satisfies the condictions in Theorem 2. Let F ( z ) = z Q ∞ k =0 (1 − ( za k ) ) , H ( z ) = F ′ ( z ) F ( z ) ,then F ′′ ( z ) F ( z ) = H ′ ( z ) + H ( z ) = F ′′ ( ∞ ) F ( ∞ ) + ∞ X k =0 F ′′ ( a k ) F ′ ( a k ) 2 a k z − a k where F ′′ ( ∞ ) F ( ∞ ) = ∞ X k =0 F ′′ ( a k ) a k F ′ ( a k ) − ∞ X k =0 a k Equivalently, the relation can be reformulated as [ F ′′ ( z ) F ( z ) ] J = , if J odd − P ∞ k =0 1 a k , if J = 0 P ∞ k =0 − F ′′ ( a k ) a J +1 k F ′ ( a k ) , if J = 2 , , , ... PREPRINT - F
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Proof.
Note that H = (log F ( z )) ′ . It turns out that H ( z ) = 1 z + ∞ X k =0 zz − a k (14)For convenience define e H ( z ) := H ( z ) − z , by the convergence of P | a k | one can see that e H ( z ) converges normallyin | z | < M By some easy computation one has e H ′ ( z ) = 1 z e H ( z ) − ∞ X k =0 ( 2 zz − a k ) On the other hand, by (14) e H ( z ) = ∞ X k =0 ( 2 zz − a k ) + X i = j,i,j ≥ zz − a i zz − a j Therefore the following relation is valid e H ′ ( z ) + e H ( z ) − z e H ( z ) = ∆ where ∆ = X i = j,i,j ≥ zz − a i zz − a j Taking the partial fraction decomposition, then ∆4 z = ∞ X k =0 e δ k z − a k where a k e δ k = ∞ X s = k,s =0 a k a k − a s = lim z → a k ( e H ( z ) − zz − a k ) = F ′′ ( a k )2 F ′ ( a k ) − a k Hence e H ′ ( z ) + e H ( z ) − z e H ( z ) = ∞ X k =0 ( F ′′ ( a k )2 a k F ′ ( a k ) − a k ) 4 z z − a k or reformulated as F ′′ ( z ) F ( z ) − F ′ ( z ) zF ( z ) + 3 z = ∞ X k =0 ( F ′′ ( a k )2 a k F ′ ( a k ) − a k ) 4 z z − a k Moreover, after some simplification, one has a more beautiful relation F ′′ ( z ) F ( z ) = F ′′ ( ∞ ) F ( ∞ ) + ∞ X k =0 F ′′ ( a k ) F ′ ( a k ) 2 a k z − a k (15)where F ′′ ( ∞ ) F ( ∞ ) = ∞ X k =0 F ′′ ( a k ) a k F ′ ( a k ) − ∞ X k =0 a k If we expand (15) as power series around , we can find that [ F ′′ ( z ) F ( z ) ] J = , if J odd − P ∞ k =0 1 a k , if J = 0 P ∞ k =0 − F ′′ ( a k ) a J +1 k F ′ ( a k ) , if J = 2 , , , ... PREPRINT - F
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Corollary 7.
If we define c m = P ∞ k =0 1 a mk and δ k = F ′′ ( a k ) F ′ ( a k ) , then the following relation holds for m = 1 , , , ... (2 m + 3) c m +1 − m X i =1 c i c m +1 − i = ∞ X k =0 δ k a m +1 k Proof.
In Theorem 8 H ′ ( z ) + H ( z ) = F ′′ ( z ) F ( z ) For H ( z ) = z + P ∞ k =1 2 zz − a k , there is an expansion H ( z ) = 1 z − ∞ X n =1 c n z n − By the rearrangement we have H ′ ( z ) + H ( z ) = − c + ∞ X m =1 e C m z m where e C m = − m + 3) c m +1 + 4 m X i =1 c i c m +1 − i That is [ H ′ ( z ) + H ( z )] m = , if m odd − c , if m = 0 − m + 3) c m +1 + 4 P mi =1 c i c m +1 − i , if m = 2 , , , ... According to Theorem 8 [ F ′′ ( z ) F ( z ) ] J = , if J odd − P ∞ k =0 1 a k , if J = 0 P ∞ k =0 − δ k a J +1 k , if J = 2 , , , ... By comparing the coefficients one obtain the conclusion.
Example 9.
A special case is when F ( z ) = π sin( πz ) . It follows from F ( z ) = z Q ∞ k =0 (1 − ( z k ) ) = π sin( πz ) one has F ′′ ( z ) F ( z ) = − π where a k = 1 + k . Therefore δ k = F ′′ ( a k ) F ′ ( a k ) = 0 . If we denote ζ (2 n ) by c n , then according toCorollary 7 (2 m + 3) c m +1 − m X i =1 c i c m +1 − i = 0 or rewritten as (2 m + 3) ζ (2 m + 2) = 2 m X i =1 ζ (2 i ) ζ (2 m + 2 − i ) This is a well-known recursion relation for ζ (2 n ) .Following example provides a similar relation for ζ (2 n, a ) Example 10.
Let F ( z ) = z Q ∞ k =0 (1 − ( xa + k ) ) = z Γ( a ) Γ( a + z )Γ( a − z ) . It’s routine to to find F ′′ ( z ) F ′ ( z ) . Here a k = a + k . F ′′ ( z ) F ′ ( z ) = z ( ψ ( a − z ) − ψ ( a + z )) + 2( ψ ( a − z ) − ψ ( a + z )) − z ( ψ ′ ( a − z ) + ψ ′ ( a + z ))1 + z ( ψ ( a − z ) − ψ ( a + z )) PREPRINT - F
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9, 2021therefore F ′′ ( a k ) F ′ ( a k ) = ( a + k )( ψ ( − k ) − ψ (2 a + k )) + 2( ψ ( − k ) − ψ (2 a + k )) − ( a + k )( ψ ′ ( − k ) + ψ ′ (2 a + k ))1 + ( a + k )( ψ ( − k ) − ψ (2 a + k ))= ( a + k )( ψ ( − k ) − ψ ′ ( − k )) + (2 − a + k ) ψ (2 a + k )) ψ ( − k ) + M ( a + k ) ψ ( − k ) + M = lim z →− k ψ ( z ) − ψ ′ ( z ) ψ ( z ) + 2 a + k − ψ (2 a + k ) Note that lim z → ψ ( z ) − ψ ′ ( z ) ψ ( z ) = − γ and recall ψ ( z + k ) = ψ ( z ) + z + z +1 + ... + z + k − . Hence lim z →− k ψ ( z ) − ψ ′ ( z ) ψ ( z ) = 2( H k − γ ) Therefore δ k = F ′′ ( a k ) F ′ ( a k ) = 2( H k + 1 a + k − γ − ψ (2 a + k ))= 2( ψ (1 + k ) + 1 a + k − ψ (2 a + k )) This makes δ k = F ′′ ( a k ) F ′ ( a k ) = 2( H k + 1 a + k − γ − ψ (2 a + k )) = 2( ψ (1 + k ) + 1 a + k − ψ (2 a + k )) Therefore we have the conclusion according to Corollary 7:
Theorem 11. ( recursion formula for ζ (2 m, a ) ) For m = 1 , , , ... we have ( m + 12 ) ζ (2 m + 2 , a ) − m X i =1 ζ (2 i, a ) ζ (2 m − i + 2 , a ) = ∞ X k =0 ψ (1 + k ) − ψ (2 a + k )( a + k ) m +1 Observe the formula (15), we conjecture that there is a more general relation
Conjecture 1. F ( L ) ( z ) F ( z ) = F ( L ) ( ∞ ) F ( ∞ ) + ∞ X k =0 m − X r =0 F ( L ) ( a k ) F ′ ( a k ) ( ω rm z − a k ω rm ) References [1] Xiaowei Wang, A new series representation involving root of unity for the values of Riemann zeta function atinteger arguments, Arxiv preprint: https://arxiv.org/abs/2010.07112v2Xiaowei Wang( 王 骁 威 )Institut für Mathematik, Universität Potsdam, Potsdam OT Golm, GermanyEmail: [email protected]@gmx.de