Method of summation of some slowly convergent series Paweł Woźny Rafał Nowak e-mail: rafal.nowak@cs.uni.wroc.pl Institute of Computer Science University of Wrocław, ul. Joliot-Curie 15, 50-383 Wrocław, Poland Approximation and extrapolation of convergent and divergent sequences and series CIRM Luminy, France September 28 – October 2, 2009 P. Woźny, R. Nowak (UWr) Method of summation Luminy ’09 1 / 29
Motivation Outline Motivation 1 Annihilation by Linear Difference Operators Approach Method 2 Recurrent construction Algorithm Results 3 Relation to ε -algorithm Generalized hypergeometric series Basic hypergeometric series Orthogonal polynomials P. Woźny, R. Nowak (UWr) Method of summation Luminy ’09 2 / 29
Motivation Annihilation by Linear Difference Operators Notations and definitions Series, partial sums, remainders Consider an infinite series ∞ � s = a n n = 0 with terms a n , partial sums n − 1 � s n = a n , n ∈ ◆ , j = 0 and remainders ∞ � r n = a n + j , n ∈ ◆ ∪ { 0 } . j = 0 Thus s = s n + r n P. Woźny, R. Nowak (UWr) Method of summation Luminy ’09 3 / 29
Motivation Annihilation by Linear Difference Operators Notations and definitions Linear Difference Operators Identity operator ■ x n = x n Shift operator ❊ k x n = x n + k , ❊ x n = x n + 1 , k ∈ ❩ Linear difference operator ▲ of order ord ▲ = ℓ k 0 + ℓ � λ k ( n ) · ❊ k , ▲ = λ k 0 ( n ) , λ k 0 + ℓ ( n ) � = 0 k = k 0 Example — forward difference operator: ∆ := ❊ − ■ , ∆ s n = s n + 1 − s n = a n (1) P. Woźny, R. Nowak (UWr) Method of summation Luminy ’09 4 / 29
Motivation Annihilation by Linear Difference Operators Notations and definitions Linear Difference Operators Identity operator ■ x n = x n Shift operator ❊ k x n = x n + k , ❊ x n = x n + 1 , k ∈ ❩ Linear difference operator ▲ of order ord ▲ = ℓ k 0 + ℓ � λ k ( n ) · ❊ k , ▲ = λ k 0 ( n ) , λ k 0 + ℓ ( n ) � = 0 k = k 0 Example — forward difference operator: ∆ := ❊ − ■ , ∆ s n = s n + 1 − s n = a n (1) P. Woźny, R. Nowak (UWr) Method of summation Luminy ’09 4 / 29
Motivation Annihilation by Linear Difference Operators Notations and definitions Multiplication P x n = y n , ◗ y n = z n ⇒ ( ◗ · P ) x n = z n = Powers ▲ 0 := ■ , ▲ k + 1 := ▲ · ▲ k Operator ▲ annihilates the sequence x n , if ▲ x n = 0 Example If x n is a polynomial in n of degree k , then ∆ k + 1 x n = 0. P. Woźny, R. Nowak (UWr) Method of summation Luminy ’09 5 / 29
Motivation Annihilation by Linear Difference Operators Notations and definitions Multiplication P x n = y n , ◗ y n = z n ⇒ ( ◗ · P ) x n = z n = Powers ▲ 0 := ■ , ▲ k + 1 := ▲ · ▲ k Operator ▲ annihilates the sequence x n , if ▲ x n = 0 Example If x n is a polynomial in n of degree k , then ∆ k + 1 x n = 0. P. Woźny, R. Nowak (UWr) Method of summation Luminy ’09 5 / 29
Motivation Annihilation by Linear Difference Operators Notations and definitions Multiplication P x n = y n , ◗ y n = z n ⇒ ( ◗ · P ) x n = z n = Powers ▲ 0 := ■ , ▲ k + 1 := ▲ · ▲ k Operator ▲ annihilates the sequence x n , if ▲ x n = 0 Example If x n is a polynomial in n of degree k , then ∆ k + 1 x n = 0. P. Woźny, R. Nowak (UWr) Method of summation Luminy ’09 5 / 29
Motivation Annihilation by Linear Difference Operators Motivation We have s = s n + r n . (2) Let the linear difference operator ▲ ( ∞ ) annihilate the remainder r n . Then ▲ ( ∞ ) ( s ) = ▲ ( ∞ ) ( s n ) + ▲ ( ∞ ) ( r n ) , s · ▲ ( ∞ ) ( 1 ) = ▲ ( ∞ ) ( s n ) , and consequently s = ▲ ( ∞ ) ( s n ) ▲ ( ∞ ) ( 1 ) , (3) if ▲ ( ∞ ) ( 1 ) � = 0 . Problems Does ▲ ( ∞ ) exist? How to find annihilator ▲ ( ∞ ) ? P. Woźny, R. Nowak (UWr) Method of summation Luminy ’09 6 / 29
Motivation Annihilation by Linear Difference Operators Motivation We have s = s n + r n . (2) Let the linear difference operator ▲ ( ∞ ) annihilate the remainder r n . Then ▲ ( ∞ ) ( s ) = ▲ ( ∞ ) ( s n ) + ▲ ( ∞ ) ( r n ) , s · ▲ ( ∞ ) ( 1 ) = ▲ ( ∞ ) ( s n ) , and consequently s = ▲ ( ∞ ) ( s n ) ▲ ( ∞ ) ( 1 ) , (3) if ▲ ( ∞ ) ( 1 ) � = 0 . Problems Does ▲ ( ∞ ) exist? How to find annihilator ▲ ( ∞ ) ? P. Woźny, R. Nowak (UWr) Method of summation Luminy ’09 6 / 29
Motivation Approach Levin-Type Sequence Transformation Let ▲ ( m ) , m ∈ ◆ , be an approximation of ▲ ( ∞ ) in the sense that � ▲ ( m ) ( r ( m ) ) = 0, n (4) ▲ ( m ) ( 1 ) � = 0, where r ( m ) = r n − r n + m = a n + a n + 1 + · · · + a n + m − 1 . n Since s ≈ s n + r ( m ) , n we can expect that := ▲ ( m ) ( s n ) Q ( m ) (5) n ▲ ( m ) ( 1 ) gives an approximation of s , of accuracy growing when m → ∞ . P. Woźny, R. Nowak (UWr) Method of summation Luminy ’09 7 / 29
Motivation Approach Levin-Type Sequence Transformation Let ▲ ( m ) , m ∈ ◆ , be an approximation of ▲ ( ∞ ) in the sense that � ▲ ( m ) ( r ( m ) ) = 0, n (4) ▲ ( m ) ( 1 ) � = 0, where r ( m ) = r n − r n + m = a n + a n + 1 + · · · + a n + m − 1 . n Since s ≈ s n + r ( m ) , n we can expect that := ▲ ( m ) ( s n ) Q ( m ) (5) n ▲ ( m ) ( 1 ) gives an approximation of s , of accuracy growing when m → ∞ . P. Woźny, R. Nowak (UWr) Method of summation Luminy ’09 7 / 29
Method Outline Motivation 1 Annihilation by Linear Difference Operators Approach Method 2 Recurrent construction Algorithm Results 3 Relation to ε -algorithm Generalized hypergeometric series Basic hypergeometric series Orthogonal polynomials P. Woźny, R. Nowak (UWr) Method of summation Luminy ’09 8 / 29
Method Recurrent construction Method Of Obtaining The Annihilators ▲ ( m ) Recurrent Construction P. Wozny, R. Nowak, Method of summation of some slowly convergent, Applied Mathematics and Computation (2009), accepted. According to ▲ ( m ) ( a n + a n + 1 + . . . + a n + m − 1 ) = 0, (6) � �� � r ( m ) n operator ▲ ( 1 ) should annihilate a n . A possible choice is the first-order operator � 1 � 1 ❊ − 1 ▲ ( 1 ) := ∆ · = (7) ■ ■ . a n a n + 1 a n The operators ▲ ( 2 ) , ▲ ( 3 ) , . . . are constructed recursively by ▲ ( m ) = P ( m ) ▲ ( m − 1 ) , m � 2. (8) P. Woźny, R. Nowak (UWr) Method of summation Luminy ’09 9 / 29
Method Recurrent construction Method Of Obtaining The Annihilators ▲ ( m ) Recurrent Construction P. Wozny, R. Nowak, Method of summation of some slowly convergent, Applied Mathematics and Computation (2009), accepted. According to ▲ ( m ) ( a n + a n + 1 + . . . + a n + m − 1 ) = 0, (6) � �� � r ( m ) n operator ▲ ( 1 ) should annihilate a n . A possible choice is the first-order operator � 1 � 1 ❊ − 1 ▲ ( 1 ) := ∆ · = (7) ■ ■ . a n a n + 1 a n The operators ▲ ( 2 ) , ▲ ( 3 ) , . . . are constructed recursively by ▲ ( m ) = P ( m ) ▲ ( m − 1 ) , m � 2. (8) P. Woźny, R. Nowak (UWr) Method of summation Luminy ’09 9 / 29
Method Recurrent construction Method Of Obtaining The Annihilators ▲ ( m ) Recurrent Construction P. Wozny, R. Nowak, Method of summation of some slowly convergent, Applied Mathematics and Computation (2009), accepted. According to ▲ ( m ) ( a n + a n + 1 + . . . + a n + m − 1 ) = 0, (6) � �� � r ( m ) n operator ▲ ( 1 ) should annihilate a n . A possible choice is the first-order operator � 1 � 1 ❊ − 1 ▲ ( 1 ) := ∆ · = (7) ■ ■ . a n a n + 1 a n The operators ▲ ( 2 ) , ▲ ( 3 ) , . . . are constructed recursively by ▲ ( m ) = P ( m ) ▲ ( m − 1 ) , m � 2. (8) P. Woźny, R. Nowak (UWr) Method of summation Luminy ’09 9 / 29
Method Recurrent construction Method Of Obtaining The Annihilators ▲ ( m ) Recurrent Construction P. Wozny, R. Nowak, Method of summation of some slowly convergent, Applied Mathematics and Computation (2009), accepted. According to ▲ ( m ) ( a n + a n + 1 + . . . + a n + m − 1 ) = 0, (6) � �� � r ( m ) n operator ▲ ( 1 ) should annihilate a n . A possible choice is the first-order operator � 1 � 1 ❊ − 1 ▲ ( 1 ) := ∆ · = (7) ■ ■ . a n a n + 1 a n The operators ▲ ( 2 ) , ▲ ( 3 ) , . . . are constructed recursively by ▲ ( m ) = P ( m ) ▲ ( m − 1 ) , m � 2. (8) P. Woźny, R. Nowak (UWr) Method of summation Luminy ’09 9 / 29
Method Recurrent construction Step Of Construction r ( m ) = r ( m − 1 ) + a n + m − 1 n n ▲ ( m − 1 ) ( r ( m − 1 ) ) = 0 n ▲ ( m − 1 ) := ❊ m − 1 ▲ ( 1 ) ❊ − m + 1 � � ▲ ( m − 1 ) ( a n + m − 1 ) = 0. ⇒ = Assume that operators P ( m ) and ❘ ( m ) are such that P ( m ) ▲ ( m − 1 ) = ❘ ( m ) � ▲ ( m − 1 ) . (9) Then ▲ ( m ) := P ( m ) ▲ ( m − 1 ) , ▲ ( m ) ( r ( m ) ) = 0 n Proof: ) + ▲ ( m ) ( a n + m − 1 ) ▲ ( m ) ( r ( m ) ) = ▲ ( m ) ( r ( m − 1 ) n n ▲ ( m − 1 ) ( a n + m − 1 ) = P ( m ) ( 0 ) + ❘ ( m ) ( 0 ) = 0, ) + ❘ ( m ) � = P ( m ) ▲ ( m − 1 ) ( r ( m − 1 ) n P. Woźny, R. Nowak (UWr) Method of summation Luminy ’09 10 / 29
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