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Logarithmic Spirals and Right-Angled Triangles

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Logarithmic Spirals and Right-Angled Triangles Logarithmic Spirals and Right-angled Triangles Rahul Gohil Student, B.E(Computer Engineering), Thadomal Shahani Engineering College, Mumbai, India Affiliated to University of Mumbai Abstract This article presents some recurrent functions which we can use to make Right-angled Triangles and through their special arrangement we can obtain logarithmic spirals, and their discrete form. The recurrent functions are obtained through recurrence relations which can be expressed as a linear combination of fibonacci numbers. Keywords: Logarithmic Spirals, Kepler Triangles, Discrete Logarithmic Spirals, Recurrent Functions 1. Introduction G. Anatriello, et al[1] discussed about continue triangles and how they discretize a logarithmic spiral by polygon chains and obtained different discrete spirals known as (r; k)-male spirals which discretize well known logarithmic spirals. Here we give a different approach for discretization of logarithmic spiral using right-angled triangles. 1.1. Recurrence relations which are linear combination of fibonacci numbers Let's take the Fibonacci sequence, f 0(n) = f 0(n − 1) + f 0(n − 2), where f 0(0) = 0 & f 0(1) = 1. " # 1 1 It's easy to see that for A = 1 0 " # f 0(n + 1) f 0(n) An = f 0(n) f 0(n − 1) Email address: [email protected] (Rahul Gohil) Let's take a sequence of recurrence relations fa;b;c such that a a a fa;b;c(n) = fa;b;c(n − b) + fa;b;c(n − c) , where ffa;b;c(0); fa;b;c(1); : : : ; fa;b;c(c − 1)g is given, a > 0, c > b > 0 are integers We can define this sequence in matrix form as " a # " #" a# fa;b;c(n) 1 1 fa;b;c(n − b) a = a fa;b;c(n − b) 1 0 fa;b;c(n − c) Let h be a positive integer such that the following relation holds. h " a # " # " a # fa;b;c(n) 1 1 fa;b;c(n − hb) a = a fa;b;c(n − b) 1 0 fa;b;c(n − (h − 1)b − c) Therefore " a # " 0 0 #" a # fa;b;c(n) f (h + 1) f (h) fa;b;c(n − hb) a = 0 0 a (1) fa;b;c(n − b) f (h) f (h − 1) fa;b;c(n − (h − 1)b − c) a 0 a 0 a We get that fa;b;c(n) = f (h + 1)fa;b;c(n − hb) + f (h)fa;b;c(n − (h − 1)b − c) a a 0 a 0 a fa;b;c(n − b) + fa;b;c(n − c) = f (h + 1)fa;b;c(n − hb) + f (h)fa;b;c(n − (h − 1)b − c) 0 a 0 a Substituting fa;b;c(n − b) = f (h)fa;b;c(n − hb) + f (h − 1)fa;b;c(n − (h − 1)b − c) 0 a 0 a a f (h)fa;b;c(n − hb) + f (h − 1)fa;b;c(n − (h − 1)b − c) + fa;b;c(n − c) = 0 a 0 a f (h + 1)fa;b;c(n − hb) + f (h)fa;b;c(n − (h − 1)b − c) a 0 a 0 a fa;b;c(n − c) = f (h + 1)fa;b;c(n − hb) + f (h)fa;b;c(n − (h − 1)b − c) 0 a 0 a − f (h)fa;b;c(n − hb) − f (h − 1)fa;b;c(n − (h − 1)b − c) a 0 a 0 a fa;b;c(n − c) = f (h − 1)fa;b;c(n − hb) + f (h − 2)fa;b;c(n − (h − 1)b − c) a Let's look at the equation of fa;b;c(n − c) for 2 values of h, k and k + 1 and equate. 0 a 0 a f (k − 1)fa;b;c(n − kb) + f (k − 2)fa;b;c(n − (k − 1)b − c) = 0 a 0 a f (k)fa;b;c(n − (k + 1)b) + f (k − 1)fa;b;c(n − kb − c) 0 a 0 a 0 a f (k −1)fa;b;c(n−kb) +f (k −2)fa;b;c(n−(k −1)b−c) = f (k −1)fa;b;c(n−(k +1)b) 0 a 0 a + f (k − 2)fa;b;c(n − (k + 1)b) + f (k − 1)fa;b;c(n − kb − c) 2 a a a Since fa;b;c(n − kb) = fa;b;c(n − (k + 1)b) + fa;b;c(n − kb − c) 0 a 0 a f (k − 1)fa;b;c(n − kb) + f (k − 2)fa;b;c(n − (k − 1)b − c) = 0 a 0 a f (k − 1)fa;b;c(n − kb) + f (k − 2)fa;b;c(n − (k + 1)b) 0 a 0 a f (k − 2)fa;b;c(n − (k − 1)b − c) = f (k − 2)fa;b;c(n − (k + 1)b) Therefore we can say that n − (k − 1)b − c = n − (k + 1)b n − kb + b − c = n − kb − b c = 2b Therefore a a a a fa;b;c(n) = fa;b(n) = fa;b(n − b) + fa;b(n − 2b) under the relation mentioned at (1). A. Fiorenza et al[2] discussed about the limit of consecutive terms of generalized homogeneous recurrences and proved that the limit exists. Now, if 0 ≤ n−(h+1)b < n b, the integer solution of h is h = − 1. b h " a # " # " a # fa;b(n) 1 1 fa;b(n − hb) a = a fa;b(n − b) 1 0 fa;b(n − (h + 1)b) " a # " 0 0 #" a # fa;b(n) f (h + 1) f (h) fa;b(n − hb) a = 0 0 a fa;b(n − b) f (h) f (h − 1) fa;b(n − (h + 1)b) n Let λ = , h = λ − 1 n b n " a # " 0 0 #" a # fa;b(n) f (λn) f (λn − 1) fa;b(n − hb) a = 0 0 a fa;b(n − b) f (λn − 1) f (λn − 2) fa;b(n − (h + 1)b) Let mn = n mod b = n − λnb " a # " 0 0 #" a# fa;b(n) f (λn) f (λn − 1) fa;b(b + mn) a = 0 0 a fa;b(n − b) f (λn − 1) f (λn − 2) fa;b(mn) Therefore a 0 a 0 a fa;b(n) = f (λn)fa;b(b + mn) + f (λn − 1)fa;b(mn) 3 Theorem 1.1. 1=a fa;b(n + p) d 0 0 lim = cp0;q0 ' p ;q n!1 fa;b(n + q) a a 'fa;b(b + mp0 ) + fa;b(mp0 ) 0 0 such that cp ;q = a a 'fa;b(b + mq0 ) + fa;b(mq0 ) ( h − k h1 = k1 = 0 or fh1; k1g 2 Zb − f0g dp0;q0 = λp0 − λq0 = h − k − 1 h1 = 0; k1 2 Zb − f0g or h1 2 Zb − f0g; k1 = 0 0 0 where p = n + p = hb − h1 j hb ≥ p > (h − 1)b; h1 2 Zb 0 0 and q = n + q = kb − k1 j kb ≥ q > (k − 1)b; k1 2 Zb Proof. 1=a 0 a 0 a ! fa;b(n + p) f (λp0 )fa;b(b + mp0 ) + f (λp0 − 1)fa;b(mp0 ) = 0 a 0 a fa;b(n + q) f (λq0 )fa;b(b + mq0 ) + f (λq0 − 1)fa;b(mq0 ) 0 11=a 0 ! 0 f (λp0 ) a a 0 0 0 Bf (λp − 1) 0 fa;b(b + mp ) + fa;b(mp ) C B f (λp0 − 1) C = B C B 0 ! C B f (λq0 ) C 0 0 0 a 0 a @ f (λq − 1) 0 fa;b(b + mq ) + fa;b(mq ) A f (λq0 − 1) 1=a 0 a a! f (n + p) f (λ 0 − 1) 'f (b + m 0 ) + f (m 0 ) lim a;b = p a;b p a;b p 0 a a n!1 fa;b(n + q) f (λq0 − 1) 'fa;b(b + mq0 ) + fa;b(mq0 ) a a 'fa;b(b + mp0 ) + fa;b(mp0 ) 0 0 Let cp ;q = a a 'fa;b(b +pmq0 ) + fa;b(mq0 ) 0 λ 0 −1 λ 0 f (λ 0 − 1) ' p 5 ' p p = p = 0 λ 0 −1 λ 0 f (λq0 − 1) ' q 5 ' q Let dp0;q0 = λp0 − λq0 n + p n + q λ 0 = , λ 0 = , we can choose such h and k that, p b q b n + p n + q = hb − h , = kb − k , b 1 b 1 0 0 where p = n + p = hb − h1 j hb ≥ p > (h − 1)b; h1 2 Zb 4 0 0 and q = n + q = kb − k1 j kb ≥ q > (k − 1)b; k1 2 Zb ( n + p h h = 0 = 1 b h − 1 h1 2 Zb − f0g ( n + q k k = 0 = 1 b k − 1 k1 2 Zb − f0g Therefore, ( n + p n + q h − k h = 0; k = 0 or fh ; k g 2 Z − f0g − = 1 1 1 1 b b b h − k − 1 h1 2 Zb − f0g; k1 = 0 or h1 = 0; k1 2 Zb − f0g Therefore, 1=a fa;b(n + p) d 0 0 lim = cp0;q0 ' p ;q n!1 fa;b(n + q) 1.2. Geometric Interpretation Let (x1; y1) be any point on the 2D Plane.Let (fa;b(0)+x1; y1) and (fa;b(0)+x1; fa;b(1)+ y1) be points on the plane, by connecting them we get the triangle. Let I0 = (x1; y1) and I1 = (fa;b(0) + x1; y1), we define a function P which is point on the 2D Plane.
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