ON A Q-SERIES IDENTITY OF COOGAN - ONO

International Journal of Advanced Trends in Engineering and Technology (IJATET) International Peer Reviewed - Refereed Research Journal, Website: www.dvpublication.com Impact Factor: 5.965, ISSN (Online): 2456 - 4664, Volume 10, Issue 2, July - December, 2025 98 ON A Q-SERIES IDENTITY OF COOGAN - ONO R. Sivaraman*, J. Yaljá Montiel-Pérez** & J. López-Bonilla*** * Department of Mathematics, Dwaraka Doss Goverdhan Doss Vaishnav College, Chennai, Tamil Nadu, India ** Centro de Investigación en Computación, Instituto Politécnico Nacional, Av. Juan de Dios BátizMiguel Othón de Mendizábal S/N, Nueva Industrial Vallejo CP 07738, CDMX, México *** ESIME-Zacatenco, Instituto Politécnico Nacional, Edif. 4, 1er. Piso, Col. Lindavista CP 07738, CDMX, México Cite This Article: R. Sivaraman, J. Yaljá Montiel-Pérez & J. López-Bonilla, “On a Q-Series Identity of Coogan - Ono”, International Journal of Advanced Trends in Engineering and Technology, Volume 10, Issue 2, July - December, Page Number 98-99, 2025. Copy Right: © DV Publication, 2025 (All Rights Reserved). This is an Open Access Article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium provided the original work is properly cited. DOI: Abstract: We show an elementary deduction of certain q-series identity obtained by Coogan and Ono. Key Words: Coogan-Ono‟s q-series identity, Z-transform, q-binomial series. 1. Introduction: Coogan-Ono [1, 2] obtained the q-series identity: (z ; q)n (−z q ; q)n ∞ n=0 zn= 1 + z (−1)n∞ n=0 qn2z2n; (1) Here we exhibit an elementary approach to motivate (1). It is possible to generalize this identity, in fact [1]: (z ; q)n (−zq ; q)n ∞ n=0 tn=1 1 − t (z ; q)n(−t ; q)n (−zq ; q)n (tq ; q)n ∞ n=0 1−z t q2n (−1)n zn tn qn2, (2) Which reproduces (1) if t = z. 2. Z-Transform: We wish to write the left side of (1) in powers of q and z, then: (z ; q)n (−zq ; q)n ∞ n=0 zn= bn ∞ n=0 qh(n)zn, (3) And we must determine bn and h n . From (3) when q→1 : bn ∞ n=0 zn= ( z (1 − z) 1 + z )n∞ n=0 =z + 1 z2 + 1 , (4) Therefore, the Z-transform of the sequence { bn } is given by Z bn =z (z + 1) z2 + 1 , and its inversion is immediate: bn = 1, 1, −1, −1, 1, 1, −1, −1, … , n = 0, 1, 2, … (5) Which generates the expansion: bn ∞ n=0 zn= 1 + z −z2−z3+ z4+ z5−z6−z7+ z8+ z9−⋯ , (6) = 1 + z 1−z2+ z4−z6+ z8−⋯ = 1 + z (−1)n ∞ n=0 z2n. The expression (6) suggests that instead of (3) we consider a relation with the following structure: (z ; q)n (−zq ; q)n ∞ n=0 zn= 1 + z (−1)n∞ n=0 qf(n)z2n. (7) 3. q-Series Identity: We know the q-series [3]: (z ; q)n= a(n, k) ∞ k=0 zk, a n, k = n k q(−1)kqk(k−1)/2 , (8) 1 (−zq ; q)n = b(n, k) ∞ k=0 zk , b n, k = n + k −1 k q(−1)k qk, Hence: (z ; q)n (−zq ; q)n= c(n, r) ∞ r=0 zr , c n, r = a n, j r j=0 b n, r −j , (9) Therefore: c m, 0 = 1, c 0, m = 0, m = 0, …,4,…, c 1,1 =− 1 + q , c 1,2 = q 1 + q , c 1,3 =−q2 1 + q , c 2,1 =− 1 + q 2, c 3,1 =− 1 + q + q2 1 + q , (10) c 2,2 = q2 1 + q + q2 (1 + q), … Besides, from (7) and (9): (z ; q)n (−zq ; q)n ∞ n=0 zn= Qm ∞ m=0 zm, Qm= c j, m −j , m j=0 (11) = (−1)n ∞ n=0 qf n z2n + z2n+1 , That is: Q0= Q1= qf(0) = 1, Q2= Q3=−qf 1 =−q, Q4= Q5= qf 2 = q4, Q6= Q7=−qf 3 =−q9,… Which suggests that f n = n2, then (7) implies the q-series identity (1), q.e.d. International Journal of Advanced Trends in Engineering and Technology (IJATET) International Peer Reviewed - Refereed Research Journal, Website: www.dvpublication.com Impact Factor: 5.965, ISSN (Online): 2456 - 4664, Volume 10, Issue 2, July - December, 2025 99 References: 1. G. H. Coogan, K. Ono, A q-series identity and the arithmetic of Hurwitz zeta functions, Proc. Am. Math. Soc. 131, No. 3 (2003) 719-724. 2. J. Wang, A general q-expansion formula based on matrix inversions and its applications, Ramanujan J. 53, No. 28 (2020) 1-24. 3. Hei-Chi Chan, An invitation to q-series. From Jacobi‟s triple product identity to Ramanujan‟s „most beautiful identity‟, World Scientific, Singapore (2011).