In the Beginning ... RAMANUJANS THEORY OF THETA FUNCTIONS Bruce - PowerPoint PPT Presentation

In the Beginning ... RAMANUJAN’S THEORY OF THETA FUNCTIONS Bruce Berndt University of Illinois at Urbana-Champaign May 26; June 1, 2009

G. H. Hardy 12 January 1920 Ramanujan to Hardy I discovered very interesting functions recently which I call “Mock” ϑ -functions. Unlike the “False” ϑ -functions (studied partially by Prof. Rogers in his interesting paper) they enter into mathematics as beautifully as the ordinary ϑ -functions.

Definitions Definition Ramanujan’s general theta function f ( a , b ) is defined by ∞ � a n ( n +1) / 2 b n ( n − 1) / 2 , f ( a , b ) := | ab | < 1 . n = −∞

Definitions Definition Ramanujan’s general theta function f ( a , b ) is defined by ∞ � a n ( n +1) / 2 b n ( n − 1) / 2 , f ( a , b ) := | ab | < 1 . n = −∞ ∞ q n 2 , � ϕ ( q ) := f ( q , q ) = n = −∞ ∞ � ψ ( q ) := f ( q , q 3 ) = q n ( n +1) / 2 , n =0 ∞ f ( − q ) := f ( − q , − q 2 ) = � ( − 1) n q n (3 n − 1) / 2 . n = −∞

Elliptic Integrals Jacobi Triple Product Identity f ( a , b ) = ( − a ; ab ) ∞ ( − b ; ab ) ∞ ( ab ; ab ) ∞ .

Elliptic Integrals Jacobi Triple Product Identity f ( a , b ) = ( − a ; ab ) ∞ ( − b ; ab ) ∞ ( ab ; ab ) ∞ . Definition Complete Elliptic Integral of the First Kind � π/ 2 d θ K ( k ) := , 1 − k 2 sin 2 θ � 0 where k , 0 < k < 1, is the modulus .

Modular Equation Definition Let K , K ′ , L , and L ′ denote complete elliptic integrals of the first √ kind associated with the moduli k , k ′ := 1 − k 2 , ℓ, and √ ℓ ′ := 1 − ℓ 2 , respectively, where 0 < k , ℓ < 1 . Suppose that, for n ∈ Z + , nK ′ K = L ′ L . (1) A relation between k and ℓ induced by (1) is called a modular equation of degree n . Set α = k 2 β = ℓ 2 . and We often say that β has degree n over α.

Main Theorem in Elliptic Functions q = exp( − π K ′ / K ) , ( a ) 0 := 1 , ( a ) n := a ( a + 1) · · · ( a + n − 1) , n ≥ 1 . ∞ ( a ) n ( b ) n � ( c ) n n ! x n , 2 F 1 ( a , b ; c ; x ) := | x | < 1 . n =0 ϕ 2 ( q ) = 2 F 1 ( 1 2 , 1 2 ; 1; k 2 ) � π/ 2 = 2 d θ =: 2 π K ( k ) . 1 − k 2 sin 2 θ � π 0 Multiplier 2 F 1 ( 1 2 , 1 m = ϕ 2 ( q ) 2 ; 1; α ) ϕ 2 ( q n ) = 2 ; 1; β ) . 2 F 1 ( 1 2 , 1

Examples of Modular Equations z = ϕ 2 ( q ) , z n = ϕ 2 ( q n ) Third Order ( αβ ) 1 / 4 + { (1 − α )(1 − β ) } 1 / 4 = 1 , Seventh Order ( αβ ) 1 / 8 + { (1 − α )(1 − β ) } 1 / 8 = 1 .

Examples of Modular Equations z = ϕ 2 ( q ) , z n = ϕ 2 ( q n ) Third Order ( αβ ) 1 / 4 + { (1 − α )(1 − β ) } 1 / 4 = 1 , Seventh Order ( αβ ) 1 / 8 + { (1 − α )(1 − β ) } 1 / 8 = 1 . 1 17 3,9 3,13,39 5,27,135 11,13,143 3 19 5,25 3,21,63 7,9,63 11,21,231 7 23 3,5,15 3,29,87 7,17,119 13,19,247 11 31 3,7,21 5,7,35 7,25,175 15,17,255 13 47 3,9,27 5,11,55 9,15,135 15 51 3,11,33 5,19,95 9,23,207

Catalogue Theorem Let x = α = k 2 . ϕ ( q ) = √ z , (i) ϕ ( − q ) = √ z (1 − x ) 1 / 4 , (ii) ϕ ( − q 2 ) = √ z (1 − x ) 1 / 8 , (iii) � √ ϕ ( q 2 ) = √ z � � 1 (iv) 1 + 1 − x , 2 √ z � 1 + (1 − x ) 1 / 4 � ϕ ( q 4 ) = 1 (v) , 2 ϕ ( √ q ) = √ z 1 + √ x � 1 / 2 , � (vi) ϕ ( −√ q ) = √ z 1 − √ x � 1 / 2 . � (vii)

Catalogue Theorem � 1 2 z ( x / q ) 1 / 8 , (i) ψ ( q ) = � 2 z ( x (1 − x ) / q ) 1 / 8 , 1 (ii) ψ ( − q ) = √ z ( x / q ) 1 / 4 , ψ ( q 2 ) = 1 (iii) 2 √ � 1 / 2 � �� � ψ ( q 4 ) = 1 1 (iv) 2 z 1 − 1 − x / q , 2 √ z { 1 − (1 − x ) 1 / 4 } / q , ψ ( q 8 ) = 1 (v) 4 ψ ( √ q ) = √ z 2 (1 + √ x ) � 1 � 1 / 4 ( x / q ) 1 / 16 , (vi) ψ ( −√ q ) = √ z 2 (1 − √ x ) � 1 � 1 / 4 ( x / q ) 1 / 16 . (vii)

Catalogue Theorem f ( q ) = √ z 2 − 1 / 6 { x (1 − x ) / q } 1 / 24 , (i) f ( − q ) = √ z 2 − 1 / 6 (1 − x ) 1 / 6 ( x / q ) 1 / 24 , (ii) f ( − q 2 ) = √ z 2 − 1 / 3 { x (1 − x ) / q } 1 / 12 , (iii) f ( − q 4 ) = √ z 4 − 1 / 3 (1 − x ) 1 / 24 ( x / q ) 1 / 6 . (iv)

Examples of multipliers Entry (p. 351) If β and the multiplier m have degree 3 , then � � � β 1 − β β (1 − β ) m 2 = α + 1 − α − α (1 − α ) . Entry (p. 351) If β and the multiplier m have degree 5 , then � 1 / 4 � 1 / 4 � 1 / 4 � β � 1 − β � β (1 − β ) m = + − . α 1 − α α (1 − α )

Examples of multipliers Entry (p. 351) If β and the multiplier m have degree 7 , then � 1 / 2 � 1 / 2 � 1 / 2 � β � 1 − β � β (1 − β ) m 2 = + − α 1 − α α (1 − α ) � 1 / 3 � β (1 − β ) − 8 . α (1 − α ) Entry (p. 351) If β and the multiplier m have degree 9 , then � 1 / 8 � 1 / 8 � 1 / 8 √ m = � β � 1 − β � β (1 − β ) + − . α 1 − α α (1 − α )

Examples of multipliers Entry (p. 352) If β and the multiplier m have degree 13 , then � 1 / 4 � 1 / 4 � 1 / 4 � β � 1 − β � β (1 − β ) m = + − α 1 − α α (1 − α ) � 1 / 6 � β (1 − β ) − 4 . α (1 − α )

Examples of multipliers Entry (p. 352) If β and the multiplier m have degree 17 , then � 1 / 4 � 1 / 4 � 1 / 4 � β � 1 − β � β (1 − β ) m = + + α 1 − α α (1 − α ) � 1 / 8 � � 1 / 8 � β (1 − β ) � β − 2 1 + α (1 − α ) α � 1 / 8 � � 1 − β + . 1 − α

More General Transformation Formulas Recall that m can be represented as a quotient of hypergeometric functions. Thus, the formulas for m can be regarded as transformation formulas for hypergeometric functions. Can any of these transformation formulas be generalized by replacing the parameters 1 2 , 1 2 , 1 by functions of α and β ?

More General Transformation Formulas Recall that m can be represented as a quotient of hypergeometric functions. Thus, the formulas for m can be regarded as transformation formulas for hypergeometric functions. Can any of these transformation formulas be generalized by replacing the parameters 1 2 , 1 2 , 1 by functions of α and β ? � 4 x � a , b ; 2 b ; 2 F (1 + x ) 2 � a , a + 1 2 − b ; b + 1 � = (1 + x ) 2 a 2 F 1 2; x 2

More General Transformation Formulas Recall that m can be represented as a quotient of hypergeometric functions. Thus, the formulas for m can be regarded as transformation formulas for hypergeometric functions. Can any of these transformation formulas be generalized by replacing the parameters 1 2 , 1 2 , 1 by functions of α and β ? � 4 x � a , b ; 2 b ; 2 F (1 + x ) 2 � a , a + 1 2 − b ; b + 1 � = (1 + x ) 2 a 2 F 1 2; x 2 How did Ramanujan derive formulas for m ?

More General Transformation Formulas Recall that m can be represented as a quotient of hypergeometric functions. Thus, the formulas for m can be regarded as transformation formulas for hypergeometric functions. Can any of these transformation formulas be generalized by replacing the parameters 1 2 , 1 2 , 1 by functions of α and β ? � 4 x � a , b ; 2 b ; 2 F (1 + x ) 2 � a , a + 1 2 − b ; b + 1 � = (1 + x ) 2 a 2 F 1 2; x 2 How did Ramanujan derive formulas for m ? How did Ramanujan derive modular equations, in general.

Sums of Theta Functions Theorem (Entry 31, Chapter 16) Let, for each positive integer n, U n = a n ( n +1) / 2 b n ( n − 1) / 2 , V n = a n ( n − 1) / 2 b n ( n +1) / 2 . Then n − 1 � U n + r , V n − r � � f ( a , b ) = . U r f U r U r r =0

Sums of Theta Functions ∞ � x n q n 2 , T ( x , q ) := x � = 0 , | q | < 1 n = −∞ Theorem (Schr¨ oter’s Formula) For positive integers a, b, a + b − 1 y n q bn 2 T ( xyq 2 bn , q a + b ) T ( x , q a ) T ( y , q b ) = � n =0 × T ( x − b y a q 2 abn , q ab 2 + a 2 b ) .

Sums of Theta Functions Example If µ is an odd positive integer, ψ ( q µ + ν ) ψ ( q µ − ν ) = q µ 2 / 4 − µ/ 4 ψ ( q 2 µ ( µ 2 − ν 2 ) ) f ( q µ + µν , q µ − µν ) ( µ − 3) / 2 � q ( µ +2 m +1)( µ 2 − ν 2 ) , � q µ m ( m +1) f + m =0 q ( µ − 2 m − 1)( µ 2 − ν 2 ) � f ( q µ + ν +2 ν m , q µ − ν − 2 ν m ) .

Eta Function Identities ∞ � (1 − aq n ) , ( a ; q ) ∞ = | q | < 1 n =0 f ( − q ) = ( q ; q ) ∞ = e − 2 π i τ/ 24 η ( τ ) , q = e 2 π i τ . Theorem Let f ( − q 3 ) f ( − q ) P = and Q = q 3 / 4 f ( − q 21 ) . q 1 / 4 f ( − q 7 ) Then � P � 2 � 2 7 � Q PQ + PQ = − 3 + . P Q

Combinatorics of Theta Function Identities Theorem Let S be the set consisting of one copy of the positive integers and one additional copy of those positive integers that are multiples of 7 . If k ∈ Z + , the number of partitions of 2 k into even elements of S is equal to the number of partitions of 2 k + 1 into odd elements of S.

Combinatorics of Theta Function Identities Theorem Let S denote the set consisting of two copies, say in colors orange and blue, of the positive integers and one additional copy, say in color red, of those positive integers that are not multiples of 3 . Let A ( N ) and B ( N ) denote the number of partitions of 2 N into odd elements and even elements, respectively, of S. Then, for N ≥ 1 , A ( N ) = B ( N ) .