Multiple zeta values for classical special functions Tanay Wakhare, - PowerPoint PPT Presentation

Multiple zeta values for classical special functions Tanay Wakhare, Christophe Vignat May 23, 2018

Tanay Wakhare

MZV Given some function G ( z ) , we denote by { a n } the set of its zeros (assumed to be non-zero complex numbers). Then define a zeta function associated with G as ∞ 1 � ζ G ( s ) = . a s n n = 1

MZV Given some function G ( z ) , we denote by { a n } the set of its zeros (assumed to be non-zero complex numbers). Then define a zeta function associated with G as ∞ 1 � ζ G ( s ) = . a s n n = 1 We also construct a multiple zeta function as 1 � ζ G ( s 1 , . . . , s r ) = a s 1 n 1 · · · a s r n r n 1 > n 2 > ··· > n r ≥ 1 and a multiple zeta-starred function as 1 ζ ∗ � G ( s 1 , . . . , s r ) = . a s 1 n 1 · · · a s r n r n 1 ≥ n 2 ≥···≥ n r ≥ 1

MZV Denote ζ G ( { s } n ) = ζ G ( { s , s , . . . } ) ζ ∗ G ( { s }} n ) = ζ ∗ G ( { s , s , . . . } )

MZV and ground states Given a quantum system with eigenvalues { E 1 , E 2 , . . . } , arranged in decreasing order of magnitude, the quantum zeta function Z ( s ) is defined as the series ∞ 1 � Z ( s ) = , E s n n = 1 and the ground state energy E 1 can be approximated as 1 E 1 ∼ Z ( s ) s

MZV and ground states Given a quantum system with eigenvalues { E 1 , E 2 , . . . } , arranged in decreasing order of magnitude, the quantum zeta function Z ( s ) is defined as the series ∞ 1 � Z ( s ) = , E s n n = 1 and the ground state energy E 1 can be approximated as 1 E 1 ∼ Z ( s ) s In the case of a quantum multiple zeta function Z ( s 1 , s 2 , . . . , s r ) , an equivalent is 1 1 Z ( { s } r ) s ∼ E 1 E 2 · · · E r

outline We address two questions in this talk: ◮ what does the knowledge of the Weierstrass product factorization of the function G tell us about the MZV built on its zeros ?

outline We address two questions in this talk: ◮ what does the knowledge of the Weierstrass product factorization of the function G tell us about the MZV built on its zeros ? ◮ what are the benefits of introducing generalized Bernoulli numbers in the computation of the MZV’s ?

Definitions and Methodology Symmetric functions - the elementary symmetric functions e r = � i 1 < i 2 <...< i r x i 1 x i 2 · · · x i r , - the complete symmetric functions h r = � i 1 ≤ i 2 ≤ ... ≤ i r x i 1 x i 2 · · · x i r i x r - the power sums p r = � i .

Definitions and Methodology Symmetric functions - the elementary symmetric functions e r = � i 1 < i 2 <...< i r x i 1 x i 2 · · · x i r , - the complete symmetric functions h r = � i 1 ≤ i 2 ≤ ... ≤ i r x i 1 x i 2 · · · x i r i x r - the power sums p r = � i . The symmetric functions have generating functions ∞ ∞ e k t k = � � E ( t ) = ( 1 + tx i ) , i = 1 k = 0 ∞ ∞ h k t k = ( 1 − tx i ) − 1 = E ( − t ) − 1 , � � H ( t ) = k = 0 i = 1 ∞ ∞ = H ′ ( t ) H ( t ) = E ′ ( − t ) x i p k t k − 1 = � � P ( t ) = E ( − t ) . 1 − tx i k = 1 i = 1

Definitions and Methodology Lemma When x i = 1 i , a s ∞ 1 � p k = ζ G ( ks ) = a ks n n = 1 1 � { s } k � � e k = ζ G = a s n 1 a s n 2 . . . a s n k n 1 > n 2 > ··· > n r ≥ 1 and 1 � { s } k � h k = ζ ∗ � = G a s n 1 a s n 2 . . . a s n k n 1 ≥ n 2 ≥···≥ n r ≥ 1

Definitions and Methodology As a consequence define the averaged zeta � S ( mn , k ) = ζ G ( ma 1 , . . . , ma k ) , | a | = n S ∗ � ζ ∗ G ( mn , k ) = G ( ma 1 , . . . , ma k ) | a | = n

Definitions and Methodology As a consequence define the averaged zeta � S ( mn , k ) = ζ G ( ma 1 , . . . , ma k ) , | a | = n S ∗ � ζ ∗ G ( mn , k ) = G ( ma 1 , . . . , ma k ) | a | = n The following generating products hold: � � 1 + ( y − 1 ) t ∞ ∞ n a s � � � S G ( sn , k ) y k t n , k = � � 1 − t k = 1 n = 0 k = 0 a s k � � 1 − t ∞ ∞ n a s � � � S ∗ G ( sn , k ) y k t n k � = � 1 − ( y + 1 ) t k = 1 n = 0 k = 0 a s k

Hypergeometric Zeta The hypergeometric zeta function is defined by 1 � ζ a , b ( s ) = , z s a , b ; k k ≥ 1 where z a , b ; k are the zeros of the classic Kummer (confluent hypergeometric) function � � a Φ a , b ( z ) := 1 F 1 a + b ; z . The zeros z a , b ; k are pair-wise complex conjugated.

Hypergeometric Zeta For a = 1 , b = 0, we have � 1 � = e z . Φ 1 , 0 ( z ) = 1 F 1 1 ; z For b = a , we have � a � I a − 1 2 ( z ) � � a + 1 z 2 2 2 a − 1 Γ Φ a , a ( z ) = 1 F 1 2 a ; z = e . 2 z a − 1 2

Hypergeometric Zeta This function has the Weierstrass factorization � � � z � z a a a + b z � za , b ; k . a + b ; z = e 1 − 1 F 1 e z a , b ; k k ≥ 1

Hypergeometric Zeta This function has the Weierstrass factorization � � � z � z a a a + b z � za , b ; k . a + b ; z = e 1 − 1 F 1 e z a , b ; k k ≥ 1 We deduce a (nontrivial) generating function for ζ a , b ( s ) as ∞ � Φ a , b + 1 ( z ) � b ζ a , b ( k + 1 ) z k = � − 1 . a + b Φ a , b ( z ) k = 1

Hypergeometric Zeta This function has the Weierstrass factorization � � � z � z a a a + b z � za , b ; k . a + b ; z = e 1 − 1 F 1 e z a , b ; k k ≥ 1 We deduce a (nontrivial) generating function for ζ a , b ( s ) as ∞ � Φ a , b + 1 ( z ) � b ζ a , b ( k + 1 ) z k = � − 1 . a + b Φ a , b ( z ) k = 1 Moreover, from the Weierstrass factorization we deduce � � z 2 � � � � a a � a + b ; ı z a + b ; − ı z = 1 + 1 F 1 1 F 1 . z 2 a , b ; k k ≥ 1 We need an identity originally due to Ramanujan [1, Entry 18, p.61] and later proved by Preece.

Hypergeometric Zeta Lemma We have the identity ; − z 2 � � � � � � a a a , b a + b ; ı z a + b ; − ı z = 2 F 3 1 F 1 1 F 1 . a + b , a + b 2 , a + b + 1 4 2 The multiple zeta value ζ a , b ( { 2 } n ) can be deduced as ζ a , b ( { 2 } n ) = ( − 1 ) n ( a ) n ( b ) n . n ! ( a + b ) n ( a + b ) 2 n For example, ab ζ a , b ( { 2 } ) = − ( a + b ) 2 ( a + b + 1 )

Hypergeometric Zeta The values of ζ a , b ( { 2 r } n ) for r = 1 , 2 , . . . can be recursively computed from ζ a , b ( { 2 } n ) using the following general result. Theorem Take m ∈ Z , m > 0 and ω a primitive m − th root of unity. Then m − 1 ζ G ( { ms } n ) = ( − 1 ) n ( m − 1 ) � � ζ G ( { s } l j ) ω jl j . l 0 + l 1 + ··· + l m − 1 = mn j = 0

Hypergeometric Zeta Proof. Start with the identity 1 − z m = � m − 1 j = 0 ( 1 − z ω j ) for m > 0 , so that ∞ ∞ ∞ m − 1 1 − t m 1 − t ω j � � � � ( − 1 ) n ζ G ( { ms } n ) t mn = � � � � = a ms a s k k n = 0 k = 1 k = 1 j = 0 � ∞ m − 1 ∞ m − 1 � 1 − t ω j � � � � � � ω jn ( − 1 ) n ζ G ( { s } n ) t n = = . a s k j = 0 k = 1 j = 0 n = 0

Hypergeometric Zeta Theorem The multiple zeta value ζ a , b ( { 4 } n ) is equal to ζ a , b ( { 4 } n ) = ( − 1 ) n ( a ) 2 n ( b ) 2 n ( 2 n )! ( a + b ) 2 n ( a + b ) 4 n � − 2 n , 1 − 2 n − a − b , 1 − 2 n − a + b 2 , 1 − 2 n − a + b + 1 � , a , b × 6 F 5 2 ; − 1 1 − 2 n − a , 1 − 2 n − b , a + b , a + b 2 , a + b + 1 2 For example, ζ a , b ( { 4 } ) = − ab ( a 3 + a 2 ( 1 − 2 b ) + b 2 ( 1 + b ) − 2 ab ( 2 + b ) ( a + b ) 4 ( 1 + a + b ) 2 ( 2 + a + b )( 3 + a + b ) .

Hypergeometric Zeta Proof. We want to compute ζ a , b ( { 4 } n ) from the generating function � � z 4 ζ a , b ( { 4 } n ) z 4 n = � � 1 + z 4 a , b ; k n ≥ 0 k ≥ 1 1 + ( √ ı z ) 2 1 + ( ı √ ı z ) 2 � � � � � � = . z 2 z 2 a , b ; k a , b ; k k ≥ 1 k ≥ 1 � � � � a , b z 2 ; − z 2 Since � 1 + = 2 F 3 , we k ≥ 1 a + b , a + b 2 , a + b + 1 z 2 4 a , b ; k 2 deduce � ζ a , b ( { 4 } n ) z 4 n n ≥ 0 ; − ı z 2 ; ı z 2 � � � � a , b a , b = 2 F 3 2 F 3 . a + b , a + b 2 , a + b + 1 a + b , a + b 2 , a + b + 1 4 4 2 2

Hypergeometric Zeta Proof. We have z k � � � � ( α 1 ) k ( α 2 ) k a 1 , a 2 α 1 , α 2 � k ! d k b 1 , b 2 , b 3 ; cz β 1 , β 2 , β 3 ; dz = 2 F 3 2 F 3 ( β 1 ) k ( β 2 ) k ( β 3 ) k k ≥ 0 � − k , 1 − k − β 1 , 1 − k − β 2 , 1 − k − β 3 , a 1 , a 2 ; c � × 6 F 5 . 1 − k − α 1 , 1 − k − α 2 , b 1 , b 2 , b 3 d

Hypergeometric Zeta Proof. We have z k � � � � ( α 1 ) k ( α 2 ) k a 1 , a 2 α 1 , α 2 � k ! d k b 1 , b 2 , b 3 ; cz β 1 , β 2 , β 3 ; dz = 2 F 3 2 F 3 ( β 1 ) k ( β 2 ) k ( β 3 ) k k ≥ 0 � − k , 1 − k − β 1 , 1 − k − β 2 , 1 − k − β 3 , a 1 , a 2 ; c � × 6 F 5 . 1 − k − α 1 , 1 − k − α 2 , b 1 , b 2 , b 3 d Therefore our desired hypergeometric product is � ı z 2 n ( a ) n ( b ) n � n � � a + b � a + b + 1 n ! 4 � � ( a + b ) n n ≥ 0 2 n 2 n � − n , 1 − n − a − b , 1 − n − a + b 2 , 1 − n − a + b + 1 � , a , b 2 × 6 F 5 ; − 1 1 − n − a , 1 − n − b , a + b , a + b 2 , a + b + 1 2

Hypergeometric Zeta * a , b ( { 2 } n ) with generating function We want to compute now ζ ∗ � − 1 � z 2 a , b ( { 2 } n ) z 2 n = � ζ ∗ � 1 − z 2 a , b ; k n ≥ 0 k ≥ 1 1 = � . � � � a a 1 F 1 a + b ; z 1 F 1 a + b ; − z