Identities involving harmonic numbers revisited Helmut Prodinger - PowerPoint PPT Presentation

Identities involving harmonic numbers revisited Helmut Prodinger Stellenbosch 10th April 2008

Knuth (TAOCP, Vol 3) has an exercise, attributed to S.O.Rice: Show that � n � 1 � ( − 1) k U n = 2 k − 1 − 1 k k ≥ 2 = ( − 1) n n ! � 1 dz 2 z − 1 − 1 , 2 π i z ( z − 1) . . . ( z − n ) C where C is a skinny closed curve encircling the points 2 , 3 , . . . , n . Changing C to an arbitrarily large circle centered at the origin, derive the convergent series U n = ( H n − 1 − 1) n + further terms . log 2

H n = 1 + 1 2 + 1 3 + · · · + 1 n (harmonic number) In Computer Science circles, this method is now called Rice’s method, although the integral representation of the alternating sum was known to N¨ orlund.

H n = 1 + 1 2 + 1 3 + · · · + 1 n (harmonic number) In Computer Science circles, this method is now called Rice’s method, although the integral representation of the alternating sum was known to N¨ orlund.

Solving an open problem by Knuth about digital search trees, Flajolet and Sedgewick had to compute n � n � � ( − 1) k R k − 2 , n − k k =2 with � 1 + · · · + 1 � R n = Q n Q 0 Q n and 1 − 1 1 − 1 1 − 1 � �� � � � Q n = . . . . 2 n 2 4

Solving an open problem by Knuth about digital search trees, Flajolet and Sedgewick had to compute n � n � � ( − 1) k R k − 2 , n − k k =2 with � 1 + · · · + 1 � R n = Q n Q 0 Q n and 1 − 1 1 − 1 1 − 1 � �� � � � Q n = . . . . 2 n 2 4

Solving an open problem by Knuth about digital search trees, Flajolet and Sedgewick had to compute n � n � � ( − 1) k R k − 2 , n − k k =2 with � 1 + · · · + 1 � R n = Q n Q 0 Q n and 1 − 1 1 − 1 1 − 1 � �� � � � Q n = . . . . 2 n 2 4

n � n � � ( − 1) k R k − 2 k k =2 n !( − 1) n 1 � = z ( z − 1) . . . ( z − n ) R ( z − 2) dz 2 π i The construction of the meromorphic function R ( z ) is a bit tricky here. But the asymptotic evaluation can now be done by computing residues.

n � n � � ( − 1) k R k − 2 k k =2 n !( − 1) n 1 � = z ( z − 1) . . . ( z − n ) R ( z − 2) dz 2 π i The construction of the meromorphic function R ( z ) is a bit tricky here. But the asymptotic evaluation can now be done by computing residues.

n � n � � ( − 1) k R k − 2 k k =2 n !( − 1) n 1 � = z ( z − 1) . . . ( z − n ) R ( z − 2) dz 2 π i The construction of the meromorphic function R ( z ) is a bit tricky here. But the asymptotic evaluation can now be done by computing residues.

Consider n � n � � ( − 1) k ϕ ( k ) . k k = n 0 Rice’s method leads to an identity in at least the following cases: ϕ ( z ) is rational, analytic on [ n 0 , ∞ ). ϕ ( z ) is meromorphic, of polynomial growth.

Consider n � n � � ( − 1) k ϕ ( k ) . k k = n 0 Rice’s method leads to an identity in at least the following cases: ϕ ( z ) is rational, analytic on [ n 0 , ∞ ). ϕ ( z ) is meromorphic, of polynomial growth.

Consider n � n � � ( − 1) k ϕ ( k ) . k k = n 0 Rice’s method leads to an identity in at least the following cases: ϕ ( z ) is rational, analytic on [ n 0 , ∞ ). ϕ ( z ) is meromorphic, of polynomial growth.

Consider n � n � � ( − 1) k ϕ ( k ) . k k = n 0 Rice’s method leads to an identity in at least the following cases: ϕ ( z ) is rational, analytic on [ n 0 , ∞ ). ϕ ( z ) is meromorphic, of polynomial growth.

q -version of Rice’s formula (HP): � n � ( q ; q ) n := , ( q ; q ) k ( q ; q ) n − k k q with ( z ; q ) n := (1 − z )(1 − zq ) . . . (1 − zq n − 1 ). n � n � 1 ( q ; q ) n � ( − 1) k − 1 q ( k 2 ) � f ( q − k ) = f ( z ) dz , 2 π i ( z ; q ) n +1 k C q k =1 where C encircles the poles q − 1 , . . . , q − n and no others.

q -version of Rice’s formula (HP): � n � ( q ; q ) n := , ( q ; q ) k ( q ; q ) n − k k q with ( z ; q ) n := (1 − z )(1 − zq ) . . . (1 − zq n − 1 ). n � n � 1 ( q ; q ) n � ( − 1) k − 1 q ( k 2 ) � f ( q − k ) = f ( z ) dz , 2 π i ( z ; q ) n +1 k C q k =1 where C encircles the poles q − 1 , . . . , q − n and no others.

For f ( z ) rational, we get identities:

Van Hamme: n ( − 1) k − 1 q ( k +1 2 ) n q k � n � � � S := = 1 − q k . 1 − q k k q k =1 k =1 1 For that one, f ( z ) = z − 1 .

Uchimura’s generalization for m ∈ N : ( − 1) k − 1 q ( k +1 n 2 ) n q k � n � �� k + m � � � S := = . 1 − q k + m 1 − q k k k q q k =1 k =1 1 Here, f ( z ) = z − q m . ( q ; q ) n 1 ( q ; q ) n 1 S = − Res z − q m − Res ( z ; q ) n +1 ( z ; q ) n +1 z − q m z = q m z =1 1 − q m − ( q ; q ) n ( q ; q ) m − 1 1 = , ( q ; q ) m + n which is better (closed form!) than Uchimura’s formula.

Dilcher’s sum: ( − 1) k − 1 q ( k 2 ) + mk � n � � (1 − q k ) m k q 1 ≤ k ≤ n q i 1 q i m � = 1 − q i 1 . . . 1 − q i m . 1 ≤ i 1 ≤ i 2 ≤···≤ i m ≤ n This time, 1 f ( z ) = ( z − 1) m .

As Dilcher noted, the limit for q → 1 is � n � ( − 1) k − 1 1 1 � � k m = . k i 1 . . . i m 1 ≤ k ≤ n 1 ≤ i 1 ≤ i 2 ≤···≤ i m ≤ n If n is replaced by infinity, we are in the realm of multiple ζ - values , and there is a big industry about finding identities for them. Hern´ andez proved the following identity: � n � 1 1 � ( − 1) k − 1 � � = k m . k i 1 i 2 . . . i m 1 ≤ k ≤ n 1 ≤ i 1 ≤ i 2 ≤···≤ i m = k 1 ≤ k ≤ n This identity does not really require a proof, since it is just an inverted form Dilcher’s identity!

As Dilcher noted, the limit for q → 1 is � n � ( − 1) k − 1 1 1 � � k m = . k i 1 . . . i m 1 ≤ k ≤ n 1 ≤ i 1 ≤ i 2 ≤···≤ i m ≤ n If n is replaced by infinity, we are in the realm of multiple ζ - values , and there is a big industry about finding identities for them. Hern´ andez proved the following identity: � n � 1 1 � ( − 1) k − 1 � � = k m . k i 1 i 2 . . . i m 1 ≤ k ≤ n 1 ≤ i 1 ≤ i 2 ≤···≤ i m = k 1 ≤ k ≤ n This identity does not really require a proof, since it is just an inverted form Dilcher’s identity!

As Dilcher noted, the limit for q → 1 is � n � ( − 1) k − 1 1 1 � � k m = . k i 1 . . . i m 1 ≤ k ≤ n 1 ≤ i 1 ≤ i 2 ≤···≤ i m ≤ n If n is replaced by infinity, we are in the realm of multiple ζ - values , and there is a big industry about finding identities for them. Hern´ andez proved the following identity: � n � 1 1 � ( − 1) k − 1 � � = k m . k i 1 i 2 . . . i m 1 ≤ k ≤ n 1 ≤ i 1 ≤ i 2 ≤···≤ i m = k 1 ≤ k ≤ n This identity does not really require a proof, since it is just an inverted form Dilcher’s identity!

As Dilcher noted, the limit for q → 1 is � n � ( − 1) k − 1 1 1 � � k m = . k i 1 . . . i m 1 ≤ k ≤ n 1 ≤ i 1 ≤ i 2 ≤···≤ i m ≤ n If n is replaced by infinity, we are in the realm of multiple ζ - values , and there is a big industry about finding identities for them. Hern´ andez proved the following identity: � n � 1 1 � ( − 1) k − 1 � � = k m . k i 1 i 2 . . . i m 1 ≤ k ≤ n 1 ≤ i 1 ≤ i 2 ≤···≤ i m = k 1 ≤ k ≤ n This identity does not really require a proof, since it is just an inverted form Dilcher’s identity!

Thus, inverting the q -version of Dilcher’s formula, I got a q -Hern´ andez formula: Assume that a 0 = b 0 = 0, then � n � ( − 1) k q ( k 2 ) a k , � � b k = k q 1 ≤ k ≤ n 1 ≤ k ≤ n � n � ( − 1) k q − kn + ( k 2 ) b k . � � q − k a k = k q 1 ≤ k ≤ n 1 ≤ k ≤ n I found these inversion formulæ myself, but it is due to Carlitz.

Thus, inverting the q -version of Dilcher’s formula, I got a q -Hern´ andez formula: Assume that a 0 = b 0 = 0, then � n � ( − 1) k q ( k 2 ) a k , � � b k = k q 1 ≤ k ≤ n 1 ≤ k ≤ n � n � ( − 1) k q − kn + ( k 2 ) b k . � � q − k a k = k q 1 ≤ k ≤ n 1 ≤ k ≤ n I found these inversion formulæ myself, but it is due to Carlitz.

Thus, inverting the q -version of Dilcher’s formula, I got a q -Hern´ andez formula: Assume that a 0 = b 0 = 0, then � n � ( − 1) k q ( k 2 ) a k , � � b k = k q 1 ≤ k ≤ n 1 ≤ k ≤ n � n � ( − 1) k q − kn + ( k 2 ) b k . � � q − k a k = k q 1 ≤ k ≤ n 1 ≤ k ≤ n I found these inversion formulæ myself, but it is due to Carlitz.

q -analogue of Hern´ andez’ formula q i 1 q i m � n � ( − 1) k − 1 q − kn + ( k 2 ) � � 1 − q i 1 . . . 1 − q i m k q 1 ≤ k ≤ n 1 ≤ i 1 ≤ i 2 ≤···≤ i m = k q k ( m − 1) � = (1 − q k ) m . 1 ≤ k ≤ n Always invert!

q -analogue of Hern´ andez’ formula q i 1 q i m � n � ( − 1) k − 1 q − kn + ( k 2 ) � � 1 − q i 1 . . . 1 − q i m k q 1 ≤ k ≤ n 1 ≤ i 1 ≤ i 2 ≤···≤ i m = k q k ( m − 1) � = (1 − q k ) m . 1 ≤ k ≤ n Always invert!