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Complex Method on Octagonal Number S International Journal of Recent Technology and Engineering (IJRTE) ISSN: 2277-3878, Volume-8 Issue-4S5, December 2019 Complex Method on Octagonal Number S. Sudha, A. Gnanam Abstract: In Number theory Study of polygonal numbers is rich in varity. In this paper we establish a Complex Octagonal B. Theorem: Number using Continued Fraction algorithm. For 푛 ≥ 3, 푂푛 푂푛+2 + Keywords: Algorithm, Continuedfraction, complex Number. 푂푛+1 푂푛+3 푛 푛 + 1 Octagonal Number. 0; 1, , 8푛 + 0; 1, , 8(푛 + 2) 푤ℎ푒푛 푛 푠 표푑푑 = 2 2 푛 (푛 + 2) 0; 1, − 1,1,1,2푛 − 1,2 + 0; 1, − 1,1,1,2(푛 + 2) − 1,2 푤ℎ푒푛 푛 푠 푒푣푒푛 I. INTRODUCTION 2 2 Proof: A Simple continued fraction [1] is an expression of the Case (i):- n is odd form Let n = 2k-1, Where 3≤ 푘 ≤ 푛 푏 Then 푎 + 0 0 푏 푂2푘−1 푂2푘+1 2푘 − 1 푎 + 1 + = 0; 1, , 8(2푘 − 1) 1 푏 푂 푂 2 푎 + 2 2푘 2푘+2 2 ⋱ + 0; 1, 푘 , 8(2푘 + 1) Where the 푎 are a possibly infinite sequence of integers Next we have to prove that n = 2k+1 such that 푎1 is non-negative and the rest of the sequence is To find the continued fraction of positive. We write 푎1; 푎2, 푎3 … … . The above fraction 푂2푘+1 푂2푘+3 also calls them Regular continued fractions. + 푂2푘+2 푂2푘+4 II. CONTINUED FRACTION ALGORITHM A. Real Part:-[3] 푂 4 2푘 + 1 2푘 + 1 − 1 + 1 Suppose we wish to find continued fraction expansion 2 2푘+1 = of 푥 ∈ 푅. 푂2푘+2 4 2푘 + 2 2푘 + 2 − 1 + 1 16푘 2+8푘+1 Let 푥0 ∈ 푥 and set 푎0 = 푥0 , = 2 1 1 16푘 +2푘+9 2 Define 푥1 = and set 푎1 = 푥1 and 푥2 = ⇒ 16푘 +8푘+1 푥0− 푥0 푥1− 푥1 푥0 = , 푎0 = 0 1 16푘 2+2푘+9 푎 = 푥 … . 푥 = ⇒ 푎 = 푥 … 16푘+8 2 2 푘 푥 −[푥 ] 푘 푘 푘−1 푘−1 푥1 ⇒ 1 + 2 ⇒ 푎1 = 1 This process is continued infinitely or to some finite stage till 16푘 +8푘+1 1 an 푥 ∈ 푁 exists such that 푎 = 푥 . 푥 ⇒ 푘 + ⇒ 푎 = 푘 2 16푘 + 8 2 III. OCTAGONAL NUMBER 푥3 ⇒ 16푘 + 8 ⇒ 푎3 = 16푘 + 8 =8(2k+1) A. Definition: Centered Octagonal Number 3 푂 The Number 1,9,25,49,81,121,…….. are called centered ∴ 2푘+1 = 0; 1, 푘, 8(2푘 + 1) 푂 octagonal numbers. The number that represents associate 2푘+2 in nursing polygonal shape with a dot within the center B. Imaginary part:- and every one dots different dots encompassing the middle 푂 4 2푘 + 3 2푘 + 3 − 1 + 1 2푘+3 ⇒ dot in associate in nursing polygonal shape lattice . 푂2푘+4 4 2푘 + 4 2푘 + 4 − 1 + 1 The 푛푡ℎ centered octagonal number is given by the 4[4푘2+6푘−2푘+6푘+9−3]+1 = formula 4[4푘2+8푘−2푘+8푘+16−4]+1 2 푂푛 = 4푛 푛 − 1 + 1 4[4푘 +4푘+6푘+6]+1 = 4[4푘2+14푘 +12]+1 4[4푘2+10푘+6]+1 = 4[4푘2+14푘 +12]+1 16푘2+40푘+24+1 = 16푘2+56푘+48+1 2 푂2푘+3 16푘 + 40푘 + 25 ⇒ 2 푂2푘+4 16푘 + 56푘 + 49 2 16푘 +40푘+25 푥0 = ; 푎0 = 0 16푘 2+56푘+49 Then 16푘 2+56푘+49 16푘+24 Revised Manuscript Received on December 5, 2019. 푥 ⇒ = 1 + ⇒ 푎 = 1 1 16푘2+40푘+25 16푘 2+40푘+25 1 S. Sudha, Department of Mathematics, Nazareth College of Arts and Science, Chennai – 62, Research Scholar, Department of Mathematics, Government Arts College, Trichy- 22. A. Gnanam, Department of Mathematics, Government Arts College, Trichy- 22. Retrieval Number: D10051284S519/2019©BEIESP Published By: DOI:10.35940/ijrte.D1005.1284S519 Blue Eyes Intelligence Engineering 11 & Sciences Publication Complex Method on Octagonal Number 16푘2 + 40푘 + 25 1 Since 푥2 ⇒ = 푘 + 1 + ⇒ 푎2 푂 푂 16푘 + 24 16푘 + 24 2푘 + 2푘+2 = 0; 1, 푘 − 1,1,1,4푘 − 1,2 = 푘 + 1 푂2푘+1 푂2푘+3 16푘 + 24 + 0; 1, 푘, 1,1,4푘 + 3,2) 푥 ⇒ ⇒ 16푘 + 24 3 1 Hence by the result is true for all value of n where n is ⇒ 8(2푘 + 3) even. 푂 Since case (i) & case (ii) for each 푛 ≥ 3, the continued ∴ 2푘+3 = 0; 1, 푘 + 1 , 8(2푘 + 3) 푂 fraction expansion of 2푘+4 푂푛 푂푛+2 푂 푂 + 2푘+1 2푘+3 푂푛+1 푂푛+3 ∴ + 푛 푛 + 1 푂2푘+2 푂2푘+4 0; 1, , 8푛 + 0; 1, , 8(푛 + 2) 푤ℎ푒푛 푛 푠 표푑푑 = 2 2 = 0; 1, 푘, 8(2푘 + 1) 푛 (푛 + 2) 0; 1, − 1,1,1,2푛 − 1,2 + 0; 1, − 1,1,1,2(푛 + 2) − 1,2 푤ℎ푒푛 푛 푠 푒푣푒푛 + 0; 1, 푘 + 1 , 8(2푘 2 2 + 3) By the results is true for all values of n when n is odd. Case (ii):- n is even IV. ILLUSTRATION Let n = 2k – 2 Let 푛 = 3, Then 푂3 푂5 25 81 푂2푘−2 푂2푘 2푘 − 2 + = + + = 0; 1, − 1,1,1,2 2푘 − 2 푂4 푂6 49 121 푂2푘−1 푂2푘+2 2 A. Real Part:-[3] − 1,2 푂3 25 = , 푠표 푎0 = 0 2푘 − 2 + 2 푂4 49 + 0; 1, − 1,1,1,2 2푘 − 2 25 2 ∴ 푥 = − 1,2 0 49 = 0; 1, 푘 − 2,1,1,2 2푘 − 2 − 1 49 24 푇ℎ푒푛 푥1 = = = 1 + ⇒ 푎1 = 1 1,2+0;1,푘−1,1,1,22푘−2−1,2 푥0 − 푥0 25 25 1 25 1 푥2 = = = 1 + ⇒ 푎2 = 1 푥1− 푥1 24 24 Next we have to prove that n = 2k 1 24 푥3 = = = 24 ⇒ 푎3 = 24 To find the continued fraction of 푥2− 푥2 1 푂2푘 푂2푘+2 25 + ∴ = 0; 1,1,24 푂2푘+1 푂2푘+3 49 B. Imaginary Part:- C. Real Part:-[3] 푂 81 5 = , 푎 = 0 푂 121 0 푂2푘 6 81 = 0; 1, 푘 − 1,1,1,4푘 − 1,2 푥 = , 푎 = 0 푂2푘+1 0 121 0 D. Imaginary part:- Then 푂2푘+2 4 2푘 + 2 2푘 + 2 − 1 + 1 1 121 40 = 푥1 = = = 1 + ⇒ 푎1 = 1 푂2푘+3 4 2푘 + 3 2푘 + 3 − 1 + 1 푥0 − 푥0 81 81 4[4푘 2+2푘+4푘+2]+1 = 4[4푘 2+4푘+6푘+6]+1 1 81 1 4[4푘 2+6푘+2]+1 푥2 = = = 2 + ⇒ 푎2 = 2 = 푥1 − 푥1 40 40 4[4푘 2+10푘+6]+1 1 40 16푘 2+24푘+8+1 푥 = = = 40 ⇒ 푎 = 40 = 3 3 16푘2+40푘+24+1 푥2 − 푥2 1 16푘2 + 24푘 + 9 81 = = 0; 1,2,40 16푘2 + 40푘 + 25 121 2 푂2푘+2 16푘 + 24푘 + 9 Hence ⇒ 25 81 2 + = 0; 1,1,24 + 0; 1,2,40 푂2푘+3 16푘 + 40푘 + 25 49 121 16푘 2+24푘+9 푥0 = ; 푎0 = 0 16푘2+40푘+25 V. ILLUSTRATION Then 16푘2+40푘+25 16푘+16 푥 ⇒ = 1 + ⇒ 푎 = 1 Put n = 4 1 16푘 2+24푘+9 16푘 2+24푘+9 1 푂 푂 49 121 16푘2 + 24푘 + 9 8푘 + 9 4 + 6 = + 푥2 ⇒ = 푘 + ⇒ 푎2 = 푘 푂5 푂7 81 169 16푘 + 16 16푘 + 16 A. Real Part:-[3] 16푘 + 16 8푘 + 7 푥 ⇒ ⇒ 1 + ⇒ 푎 = 1 푂4 49 3 8푘 + 9 8푘 + 9 3 = ⇒ 0; 1,1,1,1,7,2 8푘+9 2 푂 81 푥 ⇒ ⇒ 1 + ⇒ 푎 = 1 5 4 8푘+7 8푘+7 4 B. Imaginary part:- 8푘+7 1 푥 ⇒ ⇒ (4푘 + 3) + ⇒ 푎 = 4푘 + 3 5 2 2 5 2 푂6 121 푥6 ⇒ ⇒ 푎6 = 2 = 1 푂7 169 푂2푘+2 ∴ = 0; 1, 푘, 1,1,4푘 + 3,2) 121 푂2푘+3 퐿푒푡 푥 = , 푎 = 0 0 169 0 Then Retrieval Number: D10051284S519/2019©BEIESP Published By: DOI:10.35940/ijrte.D1005.1284S519 Blue Eyes Intelligence Engineering 12 & Sciences Publication International Journal of Recent Technology and Engineering (IJRTE) ISSN: 2277-3878, Volume-8 Issue-4S5, December 2019 1 169 48 푥1 = = = 1 + ⇒ 푎1 = 1 푥0− 푥0 121 121 1 121 25 푥2 = = = 2 + ⇒ 푎2 = 2 푥1 − 푥1 48 48 1 48 23 푥3 = = = 1 + ⇒ 푎3 = 1 푥2 − 푥2 25 25 1 25 2 푥4 = = = 1 + ⇒ 푎4 = 1 푥3 − 푥3 23 23 1 23 1 푥5 = = = 11 + ⇒ 푎5 = 11 푥4 − 푥4 2 2 1 푥6 = = 2 ⇒ 푎6 = 2 푥5 − 푥5 121 = 0; 1,2,1,1,11,2 169 Since 49 121 + 81 169 = 0; 1,1,1,1,7,2 + 0; 1,2,1,1,11,2 VI. CONCLUSION In this paper we have identified complex octagonal number using continued fractions. REFERENCES 1. https://crypto.stanford.edu 2. International Journal of Engineering Science, Advanced Computing and Bio –Technology “Ratios of Polygonal Numbers as Continued Fractions” – A. Gnanam and S. Krithika. 3. The International Journal of analytical and experimental modal analysis “ Algebraic Approach on Octagonal Numbers”- A.Gnanam and S. Sudha AUTHORS PROFILE S. Sudha She received M.Sc. (Maths) Degree in Seethalakshmi Ramasamy College, Trichy in 2001, M.Phil (Maths) from ST. Joseph’s College, Trichy in 2007, presently she is pursuing Ph.D. from Bharathidasan University, Trichy and working as Assistant Professor at Nazareth College of Arts & Science, Avadi since December 2007. She started her career as Assistant Professor at Valluvar College of Science & Management, Karur from March 2007 to Oct 2007. She has teaching experience of 13 years. She has handled Allied Mathematics, Algebraic Structures, Numerical Analysis, Operations Research, Linear Algebra, Transform Technique and Analytical Geometry. She has participated in 19 and presented 7 papers in various National and International level Conference and Seminars. She has published 2 books with ISBN and one paper in International journal. A.
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