Srivastava, Bhaskar

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Ramanujan's Mock Theta Functions

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Mathematical Journal of OkayamaUniversityVolume 47, Issue 1 2005 Article 15JANUARY 2005Ramanujan’s Mock Theta FunctionsBhaskar Srivastava∗∗Lucknow UniversityCopyright c©2005 by the authors. Mathematical Journal of Okayama University is produced byThe Berkeley Electronic Press (bepress). http://escholarship.lib.okayama-u.ac.jp/mjouMath. J. Okayama Univ. 47 (2005), 163{174RAMANUJAN'S MOCK THETA FUNCTIONSBhaskar SRIVASTAVA1. IntroductionIn his last letter to G. H. Hardy, S. Ramanujan wrote, " I discovered veryinteresting functions recently which I call mock theta functions". He alsogave a long list of 'third order', '¯fth order' and 'seventh order' mock thetafunctions together with identities satis¯ed by them. All the results on thirdand ¯fth order mock theta functions were proved by G. N. Watson in histwo celebrated papers [8,9].Andrews rediscovered Ramanujan's notebook while visiting CambridgeUniversity and called it Ramanujan's 'lost' notebook. The 'lost' notebookcontains several further results on mock theta functions which were provedby Andrews [1,2], Andrews and Garvan [3], Andrews and Hickerson [4].Lastly the 'lost' notebook contains eight identities for the tenth order mocktheta function. Choi [5] has proved the ¯rst two of Ramanujan's tenth ordermock theta function identities and said that further identities will be provedin subsequent papers.In this paper we have given relations and expansions of partial mocktheta functions, mock theta functions of tenth, third, ¯fth and sixth order.We have also given two Continued Fractions for tenth order mock thetafunctions.In section 4, we give a proof of a simple identity. In section 5, usingthis identity we connect the tenth order mock theta functions, partial tenthorder mock theta functions with mock theta functions, partial mock thetafunctions of third, ¯fth, and sixth order. In section 6, we give relations be-tween ¯fth order mock theta functions and third order mock theta functionsand their partial sums. In section 7, we have given expansions of a tenthorder mock theta functions in terms of partial mock theta function of tenthorder.In section 8, we have proved two lemmas and with the help of these lem-mas expressed the tenth order mock theta functions as a Continued Fraction.2. NotationThe following q-notations have been used.Mathematics Subject Classi¯cation. 33D.Key words and phrases. Mock theta function, Partial mock theta functions, ContinuedFractions.1631Srivastava: Ramanujan's Mock Theta FunctionsProduced by The Berkeley Electronic Press, 2005164 B. SRIVASTAVAFor jqkj < 1,(a; qk)n =n¡1Yj=0(1¡ aqkj); n ¸ 1(a; qk)0 = 1;(a; qk)1 =1Yj=0(1¡ aqkj);(a)n = (a; q)n;(a1; a2; : : : ; am; qk)n = (a1; qk)n(a2; qk)n ¢ ¢ ¢ (am; qk)n:A generalized basic hypergeometric series with base q1 is de¯ned asAÁA¡1 [ a1; a2; : : : ; aA; b1; b2; : : : ; bA¡1; q1; z ]=1Xn=0(a1; q1)n ¢ ¢ ¢ (aA; q1)nzn(b1; q1)n ¢ ¢ ¢ (bA¡1; q1)n(q; q1)n ; jzj < 1:3. Definition of Partial Mock Theta FunctionsThird order mock theta functions, (Watson [8])f(q) :=1Xn=0qn2(¡q)2n;©(q) :=1Xn=0qn2(¡q2; q2)n ;ª(q) :=1Xn=0qn2(q; q2)n;°(q) :=1Xn=0qn(n+1)(¡q; q2)n+1 :Fifth order mock theta functions, (Watson [9])f0(q) :=1Xn=0qn2(¡q; q)n ;F0(q) :=1Xn=0q2n2(q; q2)n:2Mathematical Journal of Okayama University, Vol. 47 [2005], Iss. 1, Art. 15http://escholarship.lib.okayama-u.ac.jp/mjou/vol47/iss1/15RAMANUJAN'S MOCK THETA FUNCTIONS 165Sixth order mock theta functions, (Andrews and Hickerson [4])©L(q) :=1Xn=0(¡1)nqn2(q; q2)n(¡q)2n ;ªL(q) :=1Xn=0(¡1)nq(n+1)2(q; q2)n(¡q)2n+1 ;½L(q) :=1Xn=0qn(n+1)=2(¡q)n(q; q2)n+1;¾L(q) :=1Xn=0q(n+1)(n+2)=2(¡q)n(q; q2)n+1:To distinguish these functions from the others in notation, we have put asu±x L in their symbols.Tenth order mock theta functions, ( Choi [5] )©R(q) :=1Xn=0qn(n+1)=2(q; q2)n+1;ªR(q) :=1Xn=0q(n+1)(n+2)=2(q; q2)n+1;XR(q) :=1Xn=0(¡1)nqn2(¡q; q)2n ;ÂR(q) :=1Xn=0(¡1)nq(n+1)2(¡q; q)2n+1 :To distinguish these functions from the others in notation, we have put asu±x R in their symbols.By taking the partial sums of the series de¯ning the mock theta functionsfrom 0 to N we have partial mock theta functions. We put N in the su±xto denote the partial sum from 0 to N . Thus partial mock theta functionsof tenth order will be©RN (q) :=NXn=0qn(n+1)=2(q; q2)n+1:Similar notation will be used for the other tenth order mock theta functionsand mock theta functions of third order, ¯fth order and sixth order.3Srivastava: Ramanujan's Mock Theta FunctionsProduced by The Berkeley Electronic Press, 2005166 B. SRIVASTAVA4. Main Summation FormulaWe give a proof of the following identity(4.1)pX°=0®°¯° = ¯p+1pX°=0®° +pXm=0(¯m ¡ ¯m+1)mX°=0®° :Proof. The proof is a simple rearrangement of seriespXm=0(¯m ¡ ¯m+1)mX°=0®° = (¯0 ¡ ¯1)®0 + (¯1 ¡ ¯2)1X°=0®°+ (¯2 ¡ ¯3)2X°=0®° + ¢ ¢ ¢+ (¯p ¡ ¯p+1)pX°=0®°= ®0¯0 + ®1¯1 + ¢ ¢ ¢+ ®p¯p¡ ¯p+1(®0 + ®1 + ¢ ¢ ¢+ ®p)=pX°=0®°¯° ¡ ¯p+1pX°=0®° ;which proves (4.1). ¤5(a). Relation between Tenth Order Partial Mock ThetaFunctions and Tenth Order Mock Theta Functions(i) Taking ®° = q°(°+1)=2(q;q2)°+1, ¯° = q°+1 in (4.1), we have(5.1) ªRp(q) = qp+2©Rp(q) + (1¡ q)pXm=0qm+1©Rm(q):(ii) Letting p!1 in (i), we have(5.2) ªR(q) = (1¡ q)1Xm=0qm+1©Rm(q):(iii) Taking ®° = q°(°+1)=2(q;q2)°+1, ¯° = (q; q2)°+1 and letting p ! 1 in (4.1), wehave(5.3) ª(q) = (q; q2)1©R(q) +1Xm=0(q; q2)m+1q2m+3©Rm(q);whereª(q) =1Xn=0qn(n+1)=2:4Mathematical Journal of Okayama University, Vol. 47 [2005], Iss. 1, Art. 15http://escholarship.lib.okayama-u.ac.jp/mjou/vol47/iss1/15RAMANUJAN'S MOCK THETA FUNCTIONS 167(iv) Taking ®° = q(°+1)(°+2)=2(q;q2)°+1, ¯° = 1q°+1 in (4.1), we have(5.4) qp+2©Rp(q) = ªRp(q)¡ (1¡ q)pXm=0qp¡mªRm(q);(v) NowXRm(q) =mX°=0(¡1)°q°2(¡q; q)2°=mX°=0(¡1)°q°2(1 + q2°+1)(¡q; q)2°+1=mX°=0(¡1)°q°2(¡q; q)2°+1 + ÂRm(q):(5.5)Taking ®° =(¡1)°q°2(¡q;q)2°+1 , ¯° = q2°+1 in (4.1) and using (5.5), we havepXm=0q2m(XRm(q)¡ ÂRm(q)) = ¡ q2p+21¡ q2 (XRp(q)¡ ÂRp(q))(5.6)+1q(1¡ q2)ÂRp(q):(vi) Making p!1 in (v), we haveÂR(q) = (1¡ q2)1Xm=0q2m+1(XRm(q)¡ ÂRm(q)):(5.7)5(b). Relation between Tenth Order Mock Theta Functionsand Sixth Order Mock Theta FunctionsIn this section we connect all the four tenth order mock theta functionswith sixth order mock theta functions.(i) Taking ®° =(¡1)°q°2 (q;q2)°(¡q;q)2° , ¯° =1(q;q2)°in (4.1), we have(5.8) XRp(q) =©Lp(q)(q; q2)p+1¡pXm=0q2m+1(q; q2)m+1©Lm(q):(ii) Letting p!1 in (i), we have(5.9) XR(q) =©L(q)(q; q2)1¡1Xm=0q2m+1(q; q2)m+1©Lm(q):5Srivastava: Ramanujan's Mock Theta FunctionsProduced by The Berkeley Electronic Press, 2005168 B. SRIVASTAVA(iii) Taking ®° =(¡1)°q(°+1)2 (q;q2)°(¡q;q)2°+1 , ¯° =1(q;q2)°in (4.1), we have(5.10) ÂRp(q) =ªLp(q)(q; q2)p+1¡pXm=0q2m+1(q; q2)m+1ªLm(q):(iv) Letting p!1 in (iii), we have(5.11) ÂR(q) =ªL(q)(q; q2)1¡1Xm=0q2m+1(q; q2)m+1ªLm(q):(v) Taking ®° =q°(°+1)=2(¡q;q)°(q;q2)°+1, ¯° = 1(¡q;q)° in (4.1), we have(5.12) ©Rp(q) =½Lp(q)(¡q; q)p+1 +pXm=0qm+1(¡q; q)m+1 ½Lm(q):(vi) Letting p!1 in (v), we have(5.13) ©R(q) =½L(q)(¡q; q)1 +1Xm=0qm+1(¡q; q)m+1½Lm(q):(vii) Taking ®° =q(°+1)(°+2)=2(¡q;q)°(q;q2)°+1, ¯° = 1(¡q;q)° in (4.1), we have(5.14) ªRp(q) =¾Lp(q)(¡q; q)p+1 +pXm=0©qm+1(¡q; q)m+1ª¾Lm(q):(viii) Letting p!1 in (vii), we have(5.15) ªR(q) =¾L(q)(¡q; q)1 +1Xm=0©qm+1(¡q; q)m+1ª¾Lm(q):5(c). Relation between Tenth Order Mock Theta Functionsand Third Order Mock Theta Functions(i) Taking ®° = q°(°+1)(¡q;q2)°+1 , ¯° =1(q;q2)°+1in (4.1), we have(5.16) ©Rp(q2) =°p(q)(q; q2)p+2¡pXm=0q2m+3(q; q2)m+2°m(q):(ii) Letting p!1 in (i), we have(5.17) ©R(q2) =°(q)(q; q2)1¡1Xm=0q2m+3(q; q2)m+2°m(q):6Mathematical Journal of Okayama University, Vol. 47 [2005], Iss. 1, Art. 15http://escholarship.lib.okayama-u.ac.jp/mjou/vol47/iss1/15RAMANUJAN'S MOCK THETA FUNCTIONS 169(iii) Taking ®° =(¡1)°q°2(¡q2;q2)° , ¯° =1(¡q;q2)° in (4.1), we have(5.18) XRp(q) =©p(¡q)(¡q; q2)p+1 +pXm=0q2m+1(¡q; q2)m+1©m(¡q):(iv) Letting p!1 in (iii), we have(5.19) XR(q) =©(¡q)(¡q; q2)1 +1Xm=0q2m+1(¡q; q2)m+1©m(¡q):(v) Taking ®° =(¡1)°q°2(¡q;q2)° , ¯° =1(¡q2;q2)° in (4.1), we have(5.20) XRp(q) =1(¡q2; q2)p+1ªp(¡q) +pXm=0q2m+2(¡q2; q2)m+1ªm(¡q):(vi) Letting p!1 in (v), we have(5.21) XR(q) =1(¡q2; q2)1ª(¡q) +1Xm=0q2m+2(¡q2; q2)m+1ªm(¡q):6. Relations between Fifth Order Mock Theta Functions andThird Order Mock Theta Functions and Their Partial Sums(i) Taking ®° = q°2(¡q;q)° , ¯° =1(¡q;q)° in (4.1), we have(6.1) fp(q) =1(¡q; q)p+1 f0p(q) +pXm=0qm+1(¡q; q)m+1 f0m(q):(ii) Letting p!1 in (i), we have(6.2) f(q) =1(¡q; q)1 f0(q) +1Xm=0qm+1(¡q; q)m+1 f0m(q):(iii) Taking ®° = q°2(q;q2)°, ¯° = q° in (4.1), we have(6.3) F0p(q) = q(p+1)2ªp(q) +pXm=0nqm2(1¡ q2m+1)oªm(q):(iv) Letting p!1 in (iii), we have(6.4) F0(q) =1Xm=0qm2(1¡ q2m+1)ªm(q):7Srivastava: Ramanujan's Mock Theta FunctionsProduced by The Berkeley Electronic Press, 2005170 B. SRIVASTAVA7. Expansion of Tenth Order Mock Theta Function in Termsof Partial Mock Theta Function of Tenth Order(i) Taking ®° = q°(°+1)=2(q;q2)°+1, ¯° = q°+1 in (4.1), we have(7.1) ªRp(q) = qp+2©Rp(q) + (1¡ q)pXm=0qm+1©Rm(q):(ii) Letting p!1 in (i), we have(7.2) ªR(q) = (1¡ q)1Xm=0qm+1©Rm(q):(iii) Taking ®° =(¡1)°q°2(¡q;q)2° , ¯° =q2°+11+q2°+1in (4.1), we haveÂRp(q) =q2p+31 + q2p+3XRp(q)(7.3)+ (1¡ q2)pXm=0q2m+1(1 + q2m+1)(1 + q2m+3)XRm(q):(iv) Letting p!1 in (iii), we have(7.4) ÂR(q) = (1¡ q2)1Xm=0q2m+1(1 + q2m+1)(1 + q2m+3)XRm(q):8. Continued Fraction RepresentationWe shall prove two lemmas and then with the help of these lemmas expressthese tenth order mock theta functions as continued fractions.We de¯neA(®; q) =1Xn=0(¡1)nqn2®2n(¡®q; q)2n ;(8.1)B(®; q) =1Xn=0qn(n+1)=2®n(q; q2)n+1:(8.2)We observeA(1; q) = XR(q) and qA(q; q) = (1 + q)ÂR(q):B(1; q) = ©R(q) and qB(q; q) = ªR(q):Lemma 1.(8.3) A(®; q) +®2q1 + ®qA(®q; q)¡ ®3q3(1 + ®q)(1 + ®q2)A(®q2; q) = 1:8Mathematical Journal of Okayama University, Vol. 47 [2005], Iss. 1, Art. 15http://escholarship.lib.okayama-u.ac.jp/mjou/vol47/iss1/15RAMANUJAN'S MOCK THETA FUNCTIONS 171Proof.A(®; q) = 1 +1Xn=1(¡1)nqn2®2n(¡®q; q)2n= 1¡1Xn=0(¡1)nq(n+1)2®2n+2(¡®q; q)2n+2= 1¡ q®21 + ®q1Xn=0(¡1)nqn2+2n®2n(¡®q2; q)2n+1So1 + ®qq®2[A(®q; q)¡ 1] +A(®q; q) = q2®1Xn=0(¡1)nqn2+4n®2n(¡®q2; q)2n+1=q2®1 + ®q2A(®q2; q):Simplifying we obtain the lemma. ¤For ® = 1, we have the relation(8.4) A(1; q) +q1 + qA(q; q)¡ q3(1 + q)(1 + q2)A(q2; q) = 1:Lemma 2.(8.5) B(®; q) = ®qB(®q; q) + qB(®q2; q) + 1:Proof.qB(®q; q) =1Xn=0q(n+1)(n+2)=2(q; q2)n+1®n=1®1Xn=1qn(n+1)=2(q; q2)n®n =1®1Xn=1qn(n+1)=2(q; q2)n+1(1¡ q2n+1)®n=1®1Xn=0qn(n+1)=2(q; q2)n+1®n ¡ q®1Xn=0q(n2+5n)=2(q; q2)n+1®n ¡ 1®=1®B(®; q)¡ q®B(®q2; q)¡ 1®;which proves the lemma. ¤For ® = 1, we haveB(1; q) = qB(q; q) + qB(q2; q) + 1:9Srivastava: Ramanujan's Mock Theta FunctionsProduced by The Berkeley Electronic Press, 2005172 B. SRIVASTAVAContinued Fractions for the Mock Theta Functions(a) By Lemma 1A(®; q) +®2q1 + ®qA(®q; q)¡ ®3q3(1 + ®q)(1 + ®q2)A(®q2; q) = 1;and we have the continued fractionA(®; q)A(®q; q)= S(®; q) +T (®; q)A(®q; q)A(®q2; q)= S(®; q) +T (®; q)S(®q; q) +T (®q; q)A(®q2; q)A(®q3; q)= S(®; q) +T (®; q)S(®q; q) +T (®q; q)S(®q2; q) +T (®q2; q)A(®q3; q)A(®q4; q)whereS(®; q) =·1A(®q; q)¡ ®2q(1 + ®q)¸T (®; q) =®3q3(1 + ®q)(1 + ®q2)Putting ® = 1, we haveA(1; q)A(q; q)= S(1; q) +T (1; q)S(q; q) +T (q; q)S(q2; q) +T (q2; q)A(q3; q)A(q4; q)ButA(1; q) = XR(q) and qA(q; q) = (1 + q)ÂR(q);10Mathematical Journal of Okayama University, Vol. 47 [2005], Iss. 1, Art. 15http://escholarship.lib.okayama-u.ac.jp/mjou/vol47/iss1/15RAMANUJAN'S MOCK THETA FUNCTIONS 173and so we have the continued fraction representation for the tenth ordermock theta functions XR(q) and ÂR(q).(8.6)q1 + qÂR(q)XR(q)= S(1; q) +T (1; q)S(q; q) +T (q; q)S(q2; q) +T (q2; q)A(q3; q)A(q4; q)(b) By Lemma 2B(®; q)B(®q; q)=µ®q +1B(®q; q)¶+qB(®q; q)B(®q2; q)= S(®; q) +qB(®q; q)B(®q2; q)= S(®; q) +qS(®q; q) +qB(®q2; q)B(®q3; q)= S(®; q) +qS(®q; q) +qS(®q2; q) +q:(8.7)whereS(®; q) = ®q +1B(®q; q):Putting ® = 1, we haveB(1; q)B(q; q)= S(1; q) +qS(q; q) +qS(q2; q) +q:ButB(1; q) = ©R(q) and qB(q; q) = ªR(q);11Srivastava: Ramanujan's Mock Theta FunctionsProduced by The Berkeley Electronic Press, 2005174 B. SRIVASTAVAand so we have the continued fraction representation for the tenth ordermock theta functions ©R(q) and ªR(q).(8.8)q©R(q)ªR(q)= S(1; q) +qS(q; q) +qS(q2; q) +q:Conclusion. This study of mock theta functions is interesting in the sensethat a mock theta function has been expressed in a series of mock thetafunction of di®erent order. Moreover we feel that a number theoretic inter-pretation can be given of these expansions.References[1] G. E. Andrews, The ¯fth and seventh order mock theta functions, Trans. Amer.Math. Soc. 293(1986), 113{134.[2] G. E. Andrews, Ramanujan's ¯fth order mock theta functions as constant terms,Ramanujan Revisited (G. E. Andrews, R. A. Askey, B. C. Berndt, K. G. Ramanathanand R. A. Rankin) Academic Press, Boston, 1988, 47{56.[3] G. E. Andrews and F. G. Garvan, Ramanujan's \Lost"Notebook VI: The mocktheta Conjectures, Adv. in Math. 73(1989), 242{255.[4] G. E. Andrews and D. Hickerson, Ramanujan's \Lost"Notebook VII: The sixthorder mock theta functions, Adv. in Math. 89(1991), 60{105.[5] Youn-Seo Choi, Tenth order mock theta functions in Ramanujan's \Lost" Notebook,Invent. Math. 136(1999), 497{569.[6] S. Ramanujan, \Collected Papers", Cambridge University Press, London/New York,1927 (reprinted by Chelsea, New York).[7] D. B. Sears, On the transformation theory of basic hypergeometric functions, Proc.London Math. Soc. 53(1951), 158{180.[8] G. N. Watson, The ¯nal problem: An account of the mock theta functions, J. LondonMath. Soc. 11(1936), 55{80.[9] G. N. Watson, The Mock theta functions (2), Proc. London Math. Soc. 42(1937),274{304.Bhaskar SrivastavaDepartment of MathematicsLucknow UniversityLucknow, India(Received February 12, 2004 )12Mathematical Journal of Okayama University, Vol. 47 [2005], Iss. 1, Art. 15http://escholarship.lib.okayama-u.ac.jp/mjou/vol47/iss1/15

Ramanujan's Mock Theta Functions

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