A solution of Sun’s $520 challenge concerning 520 π SIAM Annual Meeting, San Diego Symbolic Computation and Special Functions Armin Straub July 10, 2013 University of Illinois & Max-Planck-Institut at Urbana–Champaign f¨ ur Mathematik, Bonn Based on joint work with : Mathew Rogers University of Montreal A solution of Sun’s $520 challenge concerning 520/ π Armin Straub 1 / 25
Sun’s challenge � 2 n � � n � 2 � 2 k � n CONJ � ∞ � 520 1054 n + 233 ( − 1) k 8 2 k − n = 480 n π n k n n =0 k =0 • roughly, each two terms of the outer sum give one correct digit “ I would like to offer $520 (520 US dollars) for the person who could give the first correct proof of (*) in 2012 because ” May 20 is the day for Nanjing University. Zhi-Wei Sun (2011) A solution of Sun’s $520 challenge concerning 520/ π Armin Straub 2 / 25
The earliest series for 1 /π ? Introduction � 1 � 3 � 1 . 3 � 3 � 1 . 3 . 5 � 3 2 π = 1 − 5 + 9 − 13 + . . . 2 2 . 4 2 . 4 . 6 � ∞ (1 / 2) 3 n ( − 1) n (4 n + 1) = n ! 3 n =0 • Included in first letter of Ramanujan to Hardy but already given by Bauer in 1859 and further studied by Glaisher A solution of Sun’s $520 challenge concerning 520/ π Armin Straub 3 / 25
The earliest series for 1 /π ? Introduction � 1 � 3 � 1 . 3 � 3 � 1 . 3 . 5 � 3 2 π = 1 − 5 + 9 − 13 + . . . 2 2 . 4 2 . 4 . 6 � ∞ (1 / 2) 3 n ( − 1) n (4 n + 1) = n ! 3 n =0 • Included in first letter of Ramanujan to Hardy but already given by Bauer in 1859 and further studied by Glaisher • Limiting case of the terminating (Zeilberger, 1994) � ∞ (1 / 2) 2 Γ(3 / 2 + m ) n ( − m ) n ( − 1) n (4 n + 1) Γ(3 / 2)Γ( m + 1) = n ! 2 (3 / 2 + m ) n n =0 which has a WZ proof Carlson’s theorem justifies setting m = − 1 / 2 . A solution of Sun’s $520 challenge concerning 520/ π Armin Straub 3 / 25
The first of Ramanujan’s series Introduction � 1 � 3 � 1 . 3 � 3 � 1 . 3 . 5 � 3 π = 1 + 7 4 + 13 + 19 + . . . 4 2 4 3 4 2 2 . 4 2 . 4 . 6 � ∞ (1 / 2) 3 (6 n + 1) 1 n = n ! 3 4 n n =0 � ∞ (1 / 2) 3 16 (42 n + 5) 1 n π = n ! 3 2 6 n n =0 Srinivasa Ramanujan Modular equations and approximations to π Quart. J. Math., Vol. 45, p. 350–372, 1914 A solution of Sun’s $520 challenge concerning 520/ π Armin Straub 4 / 25
The first of Ramanujan’s series Introduction � 1 � 3 � 1 . 3 � 3 � 1 . 3 . 5 � 3 π = 1 + 7 4 + 13 + 19 + . . . 4 2 4 3 4 2 2 . 4 2 . 4 . 6 � ∞ (1 / 2) 3 (6 n + 1) 1 n = n ! 3 4 n n =0 � ∞ (1 / 2) 3 16 (42 n + 5) 1 n π = n ! 3 2 6 n n =0 • Starred in High School Musical, a 2006 Disney production • Both series also have WZ proof (Guillera, 2006) but no such proof known for more general series Srinivasa Ramanujan Modular equations and approximations to π Quart. J. Math., Vol. 45, p. 350–372, 1914 A solution of Sun’s $520 challenge concerning 520/ π Armin Straub 4 / 25
Another one of Ramanujan’s series Introduction √ � ∞ 1 π = 2 2 (4 n )! 1103 + 26390 n n ! 4 396 4 n 9801 n =0 • Instead of proof, Ramanujan hints at “corresponding theories” which he unfortunately never developed • Used by R. W. Gosper in 1985 to compute 17 , 526 , 100 digits of π A solution of Sun’s $520 challenge concerning 520/ π Armin Straub 5 / 25
Another one of Ramanujan’s series Introduction √ � ∞ 1 π = 2 2 (4 n )! 1103 + 26390 n n ! 4 396 4 n 9801 n =0 • Instead of proof, Ramanujan hints at “corresponding theories” which he unfortunately never developed • Used by R. W. Gosper in 1985 to compute 17 , 526 , 100 digits of π Correctness of first 3 million digits showed that the series sums to 1 /π in the first place. • First proof of all of Ramanujan’s 17 series for 1 /π by Borwein brothers Jonathan M. Borwein and Peter B. Borwein Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity Wiley, 1987 A solution of Sun’s $520 challenge concerning 520/ π Armin Straub 5 / 25
The Chudnovsky series Introduction � ∞ ( − 1) n (6 n )! 1 13591409 + 545140134 n π = 12 (3 n )! n ! 3 640320 3 n +3 / 2 n =0 • Used by David and Gregory Chudnovsky in 1988 to compute 2 , 260 , 331 , 336 digits of π A solution of Sun’s $520 challenge concerning 520/ π Armin Straub 6 / 25
The Chudnovsky series Introduction � ∞ ( − 1) n (6 n )! 1 13591409 + 545140134 n π = 12 (3 n )! n ! 3 640320 3 n +3 / 2 n =0 • Used by David and Gregory Chudnovsky in 1988 to compute 2 , 260 , 331 , 336 digits of π • This is the m = 163 case of the following: For τ = (1 + √− m ) / 2 , THM Chud- � novskys (1993) � ∞ 1 m ( J ( τ ) − 1) (6 n )! (1 − s 2 ( τ )) / 6 + n π = , J ( τ ) (3 n )! n ! 3 (1728 J ( τ )) n n =0 where � � E 3 4 ( τ ) s 2 ( τ ) = E 4 ( τ ) 3 J ( τ ) = 6 ( τ ) , E 2 ( τ ) − . E 3 4 ( τ ) − E 2 E 6 ( τ ) π Im τ A solution of Sun’s $520 challenge concerning 520/ π Armin Straub 6 / 25
Review: Singular moduli Introduction f a modular function, τ 0 a quadratic irrationality FACT = f ( τ 0 ) is an algebraic number. ⇒ A solution of Sun’s $520 challenge concerning 520/ π Armin Straub 7 / 25
Review: Singular moduli Introduction f a modular function, τ 0 a quadratic irrationality FACT = f ( τ 0 ) is an algebraic number. ⇒ • Such τ 0 is fixed by some A ∈ GL 2 ( Z ) : A · τ 0 = aτ 0 + b cτ 0 + d = τ 0 • Two modular functions are related by a modular equation : P ( f ( A · τ ) , f ( τ )) = 0 • Hence: Q ( f ( τ 0 )) = 0 where Q ( x ) = P ( x, x ) A solution of Sun’s $520 challenge concerning 520/ π Armin Straub 7 / 25
Review: Singular moduli Introduction f a modular function, τ 0 a quadratic irrationality FACT = f ( τ 0 ) is an algebraic number. ⇒ • Such τ 0 is fixed by some A ∈ GL 2 ( Z ) : A · τ 0 = aτ 0 + b cτ 0 + d = τ 0 • Two modular functions are related by a modular equation : P ( f ( A · τ ) , f ( τ )) = 0 • Hence: Q ( f ( τ 0 )) = 0 where Q ( x ) = P ( x, x ) Trouble: Complexity of modular equation increases very quickly. A solution of Sun’s $520 challenge concerning 520/ π Armin Straub 7 / 25
Review: Values of the j -function Introduction • j ( τ ) = q − 1 + 744 + 196884 q + 21493760 q 2 + · · · • Modular polynomial Φ N ∈ Z [ x, y ] such that Φ N ( j ( Nτ ) , j ( τ )) = 0 . A solution of Sun’s $520 challenge concerning 520/ π Armin Straub 8 / 25
Review: Values of the j -function Introduction • j ( τ ) = q − 1 + 744 + 196884 q + 21493760 q 2 + · · · • Modular polynomial Φ N ∈ Z [ x, y ] such that Φ N ( j ( Nτ ) , j ( τ )) = 0 . Φ N is O ( N 3 log N ) bits. RK EG Φ 2 ( x, y ) = x 3 + y 3 − x 2 y 2 + 2 4 · 3 · 31( x 2 + xy 2 ) − 2 4 · 3 4 · 5 3 ( x 2 + y 2 ) + 3 4 · 5 3 · 4027 xy + 2 8 · 3 7 · 5 6 ( x + y ) − 2 12 · 3 9 · 5 9 A solution of Sun’s $520 challenge concerning 520/ π Armin Straub 8 / 25
Review: Values of the j -function Introduction • j ( τ ) = q − 1 + 744 + 196884 q + 21493760 q 2 + · · · • Modular polynomial Φ N ∈ Z [ x, y ] such that Φ N ( j ( Nτ ) , j ( τ )) = 0 . Φ N is O ( N 3 log N ) bits. RK EG Φ 2 ( x, y ) = x 3 + y 3 − x 2 y 2 + 2 4 · 3 · 31( x 2 + xy 2 ) − 2 4 · 3 4 · 5 3 ( x 2 + y 2 ) + 3 4 · 5 3 · 4027 xy + 2 8 · 3 7 · 5 6 ( x + y ) − 2 12 · 3 9 · 5 9 Φ 11 ( x, y ) = x 12 + y 12 x 11 y 11 + 8184 x 11 y 10 − 28278756 x 11 y 9 + . . . several pages . . . + + 392423345094527654908696 . . . 100 digits . . . 000 Φ 11 ( x, y ) due to Kaltofen–Yui, 1984. A solution of Sun’s $520 challenge concerning 520/ π Armin Straub 8 / 25
Review: Values of the j -function Introduction • j ( τ ) = q − 1 + 744 + 196884 q + 21493760 q 2 + · · · • Modular polynomial Φ N ∈ Z [ x, y ] such that Φ N ( j ( Nτ ) , j ( τ )) = 0 . Φ N is O ( N 3 log N ) bits. RK EG Φ 2 ( x, y ) = x 3 + y 3 − x 2 y 2 + 2 4 · 3 · 31( x 2 + xy 2 ) − 2 4 · 3 4 · 5 3 ( x 2 + y 2 ) + 3 4 · 5 3 · 4027 xy + 2 8 · 3 7 · 5 6 ( x + y ) − 2 12 · 3 9 · 5 9 Φ 11 ( x, y ) = x 12 + y 12 x 11 y 11 + 8184 x 11 y 10 − 28278756 x 11 y 9 + . . . several pages . . . + + 392423345094527654908696 . . . 100 digits . . . 000 Φ 11 ( x, y ) due to Kaltofen–Yui, 1984. Φ 10007 ( x, y ) would require an estimated 4.8TB. A solution of Sun’s $520 challenge concerning 520/ π Armin Straub 8 / 25
Review: Computation of singular moduli Introduction Options for computation of singular moduli: • via modular equations • via PSLQ/LLL and rigorous bounds • via class field theory (Shimura reciprocity) A solution of Sun’s $520 challenge concerning 520/ π Armin Straub 9 / 25
Review: Computation of singular moduli Introduction Options for computation of singular moduli: • via modular equations • via PSLQ/LLL and rigorous bounds • via class field theory (Shimura reciprocity) Let us evaluate j ( 1+ √− 23 EG ) . 2 A solution of Sun’s $520 challenge concerning 520/ π Armin Straub 9 / 25
Review: Computation of singular moduli Introduction Options for computation of singular moduli: • via modular equations • via PSLQ/LLL and rigorous bounds • via class field theory (Shimura reciprocity) Let us evaluate j ( 1+ √− 23 EG ) . 2 CFT: The Galois conjugates are j ( 1+ √− 23 ) , j ( − 1+ √− 23 ) . 4 4 A solution of Sun’s $520 challenge concerning 520/ π Armin Straub 9 / 25
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