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 {fa,b,c(0), fa,b,c(1), . . . , fa,b,c(c − 1)} 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 ϕ p ,q n→∞ fa,b(n + q) a a ϕfa,b(b + mp0 ) + fa,b(mp0 ) 0 0 such that cp ,q = a a ϕfa,b(b + mq0 ) + fa,b(mq0 ) ( h − k h1 = k1 = 0 or {h1, k1} ∈ Zb − {0} dp0,q0 = λp0 − λq0 = h − k − 1 h1 = 0, k1 ∈ Zb − {0} or h1 ∈ Zb − {0}, k1 = 0 0 0 where p = n + p = hb − h1 | hb ≥ p > (h − 1)b, h1 ∈ Zb 0 0 and q = n + q = kb − k1 | kb ≥ q > (k − 1)b, k1 ∈ 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 )
1/a 0 ! 0 f (λp0 ) a a 0 0 0 f (λp − 1) 0 fa,b(b + mp ) + fa,b(mp ) f (λp0 − 1) = 0 ! f (λq0 ) 0 0 0 a 0 a f (λq − 1) 0 fa,b(b + mq ) + fa,b(mq ) f (λq0 − 1)
1/a 0 a a! f (n + p) f (λ 0 − 1) ϕf (b + m 0 ) + f (m 0 ) lim a,b = p a,b p a,b p 0 a a n→∞ fa,b(n + q) f (λq0 − 1) ϕfa,b(b + mq0 ) + fa,b(mq0 )
a a ϕfa,b(b + mp0 ) + fa,b(mp0 ) 0 0 Let cp ,q = a a ϕfa,b(b +√mq0 ) + fa,b(mq0 ) 0 λ 0 −1 λ 0 f (λ 0 − 1) ϕ p 5 ϕ p p = √ = 0 λ 0 −1 λ 0 f (λq0 − 1) ϕ q 5 ϕ 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 | hb ≥ p > (h − 1)b, h1 ∈ Zb
4 0 0 and q = n + q = kb − k1 | kb ≥ q > (k − 1)b, k1 ∈ Zb
( n + p h h = 0 = 1 b h − 1 h1 ∈ Zb − {0} ( n + q k k = 0 = 1 b k − 1 k1 ∈ Zb − {0} Therefore, ( n + p n + q h − k h = 0, k = 0 or {h , k } ∈ Z − {0} − = 1 1 1 1 b b b h − k − 1 h1 ∈ Zb − {0}, k1 = 0 or h1 = 0, k1 ∈ Zb − {0}
Therefore,
1/a fa,b(n + p) d 0 0 lim = cp0,q0 ϕ p ,q n→∞ 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. Let P (0) = (fa,b(0) + x1, fa,b(1) + y1), the P function has 2 class variables(a class variable is used to denote a variable associated with the object of that class in programming languages), namely x and y, which denote the x co-ordinate and y co-ordinate of that point respectively. We define P (n) as (P (n − 1).x − fa,b(n + 1),P (n − 1).y) n ≡ 1 mod 4 (P (n − 1).x, P (n − 1).y − f (n + 1)) n ≡ 2 mod 4 P (n) = a,b (P (n − 1).x + f (n + 1),P (n − 1).y) n ≡ 3 mod 4 a,b (P (n − 1).x, P (n − 1).y + fa,b(n + 1)) n ≡ 0 mod 4
We define a function L which is a line on the 2D Plane. Let L(0) = (I0,P (0)), this means that L(0) is a line between points I0 and P (0),
5 L(1) = (I1,P (1)). L function has 5 class variables, namely point1, point2, slope, constant, length denoting the first point in the tuple, the second point in the tuple, slope of the line, the constant associated with it and the length of the line respectively. We define L(n) as L(n) = (L(n − 2).point2,P (n)) Essentially L(n) is the hypotenuse of the nth triangle. We also define a function L0 which describes base and height of a triangle which has same class variables as L. 0 0 Let L (0) = (I0,I1) and L (1) = (I1,P (0)).
L0(n) = (L0(n − 1).point2,P (n − 1))
Let’s also define a function for a triangle T , T (0) = (L0(0),L0(1),L(0)). T function has 3 class variables line1, line2 and line3, denoting base, height and hypotenuse respectively. T (n) = (T (n − 1).line2,L0(n + 1),L(n)) For each n, L0(n) is the base, L0(n + 1) is the height and L(n) is the hypotenuse of triangle T (n). Through out this paper, we are going to assume n being an integer such that n ≡ 0 mod 4 Theorem 1.2. lim P (n).x − x1 = %a,bfa,b(n − 2b) n→∞
2/a b−1 ϕ X i λ 0 −λn+2 1/a where % = (−1) (c 0 0 ϕ 2i ) a,b ϕ2/a − (−1)b 2i ,2b i=0
Proof. P (n).x can alternatively be defined as
n/2 X i P (n).x = x1 + (−1) fa,b(n − 2i) i=0 . n/2 X i P (n).x − x1 = (−1) fa,b(n − 2i) i=0
6 P (n).x − x Let g(n) = 1 fa,b(n − 2b)
n/2 X (−1)if(n − 2i) g(n) = i=0 fa,b(n − 2b) n/2 X (−1)if(n − 2i) = f (n − 2b) i=0 a,b n/2 X f(n − 2i) lim g(n) = lim (−1)i n→∞ n→∞ f (n − 2b) i=0 a,b Using 1.1 n/2 X i d 0 0 1/a lim g(n) = lim (−1) (c2i0,2b0 ϕ 2i ,2b ) n→∞ n→∞ i=0
n/2 X i d 0 0 1/a Let S(n) = (−1) (c2i0,2b0 ϕ 2i ,2b ) i=0 n − 2b n d 0 0 = λ 0 − λ 0 = λ 0 − = λ 0 − + 2 = λ 0 − λ + 2 2i ,2b 2i 2b 2i b 2i b 2i n n/2 X i λ 0 −λn+2 1/a S(n) = (−1) (c2i0,2b0 ϕ 2i ) i=0 S(n) can be visualized as,
0 λ 0 −λn+2 1/a b λ 0 −λn+2 1/a 2b λ2(2b)0 −λn+2 1/a S(n) = (−1) (c00,2b0 ϕ 0 ) + (−1) (c2b0,2b0 ϕ 2b ) + (−1) (c2(2b)0,2b0 ϕ ) + ...
1 λ 0 −λn+2 1/a b+1 λ2(b+1)0 −λn+2 1/a + (−1) (c20,2b0 ϕ 2 ) + (−1) (c2(b+1)0,2b0 ϕ ) + ...
. + .
b−1 λ2(b−1)0 −λn+2 1/a 2b−1 λ2(2b−1)0 −λn+2 1/a + (−1) (c2(b−1)0,2b0 ϕ ) + (−1) (c2(2b−1)0,2b0 ϕ ) + ...
7 So, it is split into b rows and λn/2 + 1 columns, the last column contains mn/2 terms and b−mn/2 zeros. Each row represent a geometric series, let’s see the common ratio for each row. For the first row,
b λ 0 −λn+2 1/a 1/a (−1) (cb0,2b0 ϕ b ) = (−1)b ϕλ2b0 −λn+2−λ00 +λn−2 0 λ 0 −λ +2 1/a (−1) (c00,2b0 ϕ 0 n ) 1/a = (−1)b ϕλ2b0 −λ00 1/a = (−1)b ϕλn−2−λn = (−1)bϕ−2/a
For the second row,
b+1 λ2(b+1)0 −λn+2 1/a (−1) (c2(b+1)0,2b0 ϕ ) 1/a = (−1)b ϕλ2(b+1)0 −λn+2−λ20 +λn−2 1 λ 0 −λ +2 1/a (−1) (c20,2b0 ϕ 2 n ) 1/a = (−1)b ϕλ2(b+1)0 −λ20 1/a = (−1)b ϕλ20 −2−λ20 = (−1)bϕ−2/a
For the bth row,
2b−1 λ2(2b−1)0 −λn+2 1/a (−1) (c2(2b−1)0,2b0 ϕ ) 1/a = (−1)b ϕλ2(2b−1)0 −λn+2−λ2(b−1)0 +λn−2 b−1 λ2(b−1)0 −λn+2 1/a (−1) (c2(b−1)0,2b0 ϕ ) 1/a = (−1)b ϕλ2(2b−1)0 −λ2(b−1)0
n − 4b + 2 n − 2b + 2 λ 0 − λ 0 = − 2(2b−1) 2(b−1) b b n + 2 n + 2 = − 4 − + 2 b b = −2
So, the common ratio for bth row = (−1)bϕ−2/a
8 Therefore S(n) can be reduced to ! b −2/a λn/2+1 0 λ 0 −λn+2 1/a 1 − ((−1) ϕ ) S(n) = (−1) (c 0 0 ϕ 0 ) 0 ,2b 1 − (−1)bϕ−2/a
! b −2/a λn/2+1 1 λ 0 −λn+2 1/a 1 − ((−1) ϕ ) + (−1) (c 0 0 ϕ 2 ) 2 ,2b 1 − (−1)bϕ−2/a
. + .
! b −2/a λn/2+1 b−1 λ 0 −λn+2 1/a 1 − ((−1) ϕ ) + (−1) (c 0 0 ϕ 2(b−1) ) 2(b−1) ,2b 1 − (−1)bϕ−2/a
! b−1 b −2/a λn/2+1 1/a 1 − ((−1) ϕ ) X i λ 0 −λn+2 S(n) = (−1) c 0 0 ϕ 2i 1 − (−1)bϕ−2/a 2i ,2b i=0 b−1 1/a 1 X i λ 0 −λn+2 lim S(n) = (−1) c2i0,2b0 ϕ 2i n→∞ 1 − (−1)bϕ−2/a i=0 ! b−1 2/a 1/a ϕ X i λ 0 −λn+2 lim S(n) = (−1) c2i0,2b0 ϕ 2i n→∞ ϕ2/a − (−1)b i=0
lim S(n) = %a,b n→∞
lim g(n) = %a,b n→∞
P (n).x − x1 lim = %a,b n→∞ fa,b(n − 2b) Therefore, lim P (n).x − x1 = %a,bfa,b(n − 2b) n→∞
Similarly we can prove that,
lim P (n).y − y1 = %a,bfa,b(n − 2b + 1) n→∞
lim P (n + 1).x − x1 = −%a,bfa,b(n − 2b + 2) n→∞
9 2. Some Properties
Here, we will focus on b = 1,
1. As n → ∞, angle between L(n) and L(n + 1) tends → 90◦
Proof. Let A(n) be angle between L(n) and L(n + 1) ! −1 L(n + 1).slope − L(n).slope A(n) = tan 1 + (L(n + 1).slope · L(n).slope)
Let 1 + (L(n + 1).slope · L(n).slope) = s L(n + 1) = (P (n − 1),P (n + 1)) and L(n) = (P (n − 2),P (n)) s − 1 L(n + 1).slope = L(n).slope P (n + 1).y − P (n − 1).y (s − 1)(P (n).x − P (n − 2).x) = P (n + 1).x − P (n − 1).x P (n).y − P (n − 2).y
P (n + 1) = (P (n).x − fa,1(n + 2),P (n).y) P (n) = (P (n − 1).x, P (n − 1).y + fa,1(n + 1)) P (n − 1) = (P (n − 2).x + fa,1(n),P (n − 2).y) P (n).y − P (n − 2).y (s − 1)(P (n).x − P (n − 2).x) = P (n).x − P (n − 2).x − fa,1(n + 2) − fa,1(n) P (n).y − P (n − 2).y
P (n).x = P (n − 1).x = P (n − 2).x + fa,1(n), therefore P (n).x − P (n − 2).x = fa,1(n) P (n).y = P (n − 1).y + fa,1(n + 1) and P (n − 1).y = P (n − 2).y, therefore P (n).y − P (n − 2).y = fa,1(n + 1) f (n + 1) (s − 1)(f (n)) a,1 = a,1 fa,1(n) − fa,1(n + 2) − fa,1(n) fa,1(n + 1) −f (n + 1)2 a,1 = s − 1 fa,1(n + 2)fa,1(n) −f (n + 1)2 Let g0(n) = a,1 , using 1.1, fa,1(n + 2)fa,1(n) ϕ1/a lim g0(n) = − n→∞ ϕ1/a
10 lim g0(n) = −1 n→∞ lim g0(n) = s − 1 n→∞ −1 = s − 1 s = 0 Therefore, lim 1 + (L(n + 1).slope · L(n).slope) = 0 n→∞ 1 Since, tan−1 = 90◦ 0 lim A(n) = 90◦ n→∞ Therefore, as n → ∞, angle between L(n) and L(n + 1) tends → 90◦
2. As n → ∞, P (n + 2) tends to lie on L(n)
Proof. Let’s satisfy P (n + 2) in L(n) equation. S0(n) = P (n + 2).y − L(n).slope · P (n + 2).x − L(n).constant L(n) = (P (n − 2),P (n)) P (n).y − P (n − 2).y f (n + 1) L(n).slope = = a,1 as seen above P (n).x − P (n − 2).x fa,1(n) f (n + 1) L(n).constant = P (n).y − L(n).slope · P (n).x = P (n).y − a,1 P (n).x fa,1(n) P (n + 2) = (P (n + 1).x, P (n + 1).y − fa,1(n + 3)) P (n + 1) = (P (n).x − fa,1(n + 2),P (n).y) ! 0 fa,1(n + 1) fa,1(n + 1) S (n) = P (n).y − fa,1(n + 3) − (P (n).x − fa,1(n + 2)) − P (n).y − P (n).x fa,1(n) fa,1(n)
fa,1(n + 2)fa,1(n + 1) = − fa,1(n + 3) fa,1(n) f (n + 2)f (n + 1) f (n + 3)f (n + 2) = a,1 a,1 − a,1 a,1 fa,1(n) fa,1(n + 2) Using 1.1 0 1/a 1/a lim S (n) = ϕ fa,1(n + 2) − ϕ fa,1(n + 2) n→∞ lim S0(n) = 0 n→∞ Therefore, as n → ∞, P (n + 2) tends to satisfy L(n)
11 3. As n → ∞, all P (n) with even n tend to lie on the same line, same is true for all P (n) with odd n.
Proof. L(n) = (P (n − 2),P (n)), therefore, for some n,(P (n − 2),P (n),P (n + 2)), all lie on the same line. Similarly P (n+4) satisfies L(n+2), where (P (n),P (n+2),P (n+4)) lie on the same line. Since P (n − 2) and P (n) lie on the same line, P (n − 2) and P (n + 4) must lie on the same line. Therefore (...,P (n − 2),P (n),P (n + 2),P (n + 4),...) lie on the same line.
4. As n → ∞, T (n) is similar to T (n + 1)
Proof. T (n) = (T (n − 1).line2,L0(n + 1),L(n)) T (n − 1) = (T (n − 2).line2,L0(n),L(n)) Therefore, T (n) = (L0(n),L0(n + 1),L(n))
L0(n + 1).length p(P (n).x − P (n − 1).x)2 + (P (n).y − P (n − 1).y)2 = L0(n).length p(P (n − 1).x − P (n − 2).x)2 + (P (n − 1).y − P (n − 2).y)2 0 L (n + 1).length fa,1(n + 1) 0 = L (n).length fa,1(n) Using 1.1 L0(n + 1).length lim = ϕ1/a n→∞ L0(n).length
L(n).length p(P (n).x − P (n − 2).x)2 + (P (n).y − P (n − 2).y)2 = L0(n).length p(P (n − 1).x − P (n − 2).x)2 + (P (n − 1).y − P (n − 2).y)2
L(n).length fa,1(n + 2) 0 = L (n).length fa,1(n) Using 1.1 L(n).length lim = ϕ2/a n→∞ L0(n).length Therefore, as n → ∞ L0(n).length L0(n + 1).length L(n).length : : = 1 : ϕ1/a : ϕ2/a L0(n).length L0(n).length L0(n).length
12 So as we scale T (n) by ϕ1/a we get T (n + 1). We can also use that angle between L(n) and L(n + 1) tends → 90◦ and SAS test on the corresponding sides.
A Kepler triangle is a right angled triangle whose edge-lengths are in geometric √ √ progression with ϕ as the common ratio and can be written as 1 : ϕ : ϕ[3]. So when a = 2 we get that T (n) is a kepler triangle.
3. Logarithmic Spiral
Equation of a logarithmic spiral in polar form is,
r = tekθ where t and k are constants[4]. To determine t and k, consider the equations,
kθn rn = te
kθn+1 rn+1 = te
θn can be defined as, b n c2π + tan−1(L(I ,P (n)).slope) n ≡ 0 mod 4 4 0 n −1 θn = b 4 c2π + tan (L(I0,P (n)).slope) + π n ≡ 1 mod 4 or n ≡ 2 mod 4 n −1 b 4 c2π + tan (L(I0,P (n)).slope) + 2π n ≡ 3 mod 4 where L(I0,P (n)) is a line which passes through I0 and P (n). rn can be defined as,
p 2 2 rn = (P (n).x − x1) + (P (n).y − y1)
13 Now, rn+1 = ek(θn+1−θn) rn r ln n+1 r k = n θn+1 − θn ! p(P (n + 1).x − x )2 + (P (n + 1).y − y )2 ln 1 1 p 2 2 (P (n).x − x1) + (P (n).y − y1) = n+1 −1 n −1 b 4 c2π + tan (L(I0,P (n + 1)).slope) + π − b 4 c2π − tan (L(I0,P (n)).slope)
! p(P (n + 1).x − x )2 + (P (n).y − y )2 ln 1 1 p 2 2 (P (n).x − x1) + (P (n).y − y1) = P (n + 1).y − y P (n).y − y tan−1 1 − tan−1 1 + π P (n + 1).x − x1 P (n).x − x1 Using 1.2 ! p(−% f (n − 2 + 2))2 + (% f (n − 2 + 1))2 ln a,1 a,1 a,1 a,1 p 2 2 (%a,1fa,1(n − 2)) + (%a,1fa,1(n − 2 + 1)) lim k = ! ! n→∞ % f (n − 2 + 1) % f (n − 2 + 1) tan−1 a,1 a,1 − tan−1 a,1 a,1 + π −%a,1fa,1(n − 2 + 2) %a,1fa,1(n − 2) ! pf (n)2 + f (n − 1)2 ln a,1 a,1 p 2 2 fa,1(n − 2) + fa,1(n − 1) = ! ! f (n − 1) f (n − 1) tan−1 a,1 − tan−1 a,1 + π −fa,1(n) fa,1(n − 2) v ! u f (n)2 u a,1 u 2 + 1 u fa,1(n − 1) u ln u ! u f (n − 2)2 t a,1 + 1 2 fa,1(n − 1) = ! ! f (n − 1) f (n − 1) tan−1 a,1 − tan−1 a,1 + π −fa,1(n) fa,1(n − 2)
14 Using 1.1 v u ϕ2/a + 1 u ln t ϕ−2/a + 1 = −1 tan−1 − tan−1 ϕ1/a + π ϕ1/a ln ϕ1/a = 1 − tan−1 − tan−1 ϕ1/a + π ϕ1/a ln ϕ1/a = π π − 2 ln ϕ2/a k = π
2/a kθ ln ϕ θ 2θ Now, r = te = te π = tϕ aπ
2θ − t = rϕ aπ
2θn − = rnϕ aπ P (n).y − y1 n −1 2b c2π + 2 tan 4 P (n).x − x1 − aπ = rnϕ
15 P (n).y − y1 −1 2 tan n P (n).x − x1 − + a aπ = rnϕ P (n).y − y1 −1 2 tan P (n).x − x1 − aπ rn = n ϕ ϕ a
P (n).y − y1 −1 2 tan P (n).x − x1 − aπ s (P (n).x − x )2 + (P (n).y − y )2 = 1 1 ϕ ϕ2n/a Using 1.2 ! %a,1fa,1(n − 1) −1 tan % f (n − 2) a,1 a,1 − π s (% f (n − 2))2 + (% f (n − 1))2 = a,1 a,1 a,1 a,1 ϕ ϕ2n/a Using 1.1 −1 1/a 2 tan ϕ − s aπ 2 2 2/a fa,1(n − 2) % (1 + ϕ ) = a,1 ϕ ϕn −1 1/a 2 tan ϕ − aπ fa,1(n − 2) p = % 1 + ϕ2/a ϕ ϕn/a a,1
16 −1 1/a 2 tan ϕ − aπ 0 a 0 a 1/a f (n − 2)fa,1(1) + f (n − 3)fa,1(0) p = % 1 + ϕ2/a ϕ ϕn a,1 −1 1/a 2 tan ϕ − aπ 0 a a 1/a f (n − 3)(ϕfa,1(1) + fa,1(0) ) p = % 1 + ϕ2/a ϕ ϕn a,1 −1 1/a 2 tan ϕ − 1/a aπ a a ! ϕfa,1(1) + fa,1(0) p 2/a t = √ %a,1 1 + ϕ ϕ ϕ3 5
Therefore, we obtained the constants
−1 1/a 2 tan ϕ − 1/a aπ 2/a a a ! ϕ ϕfa,1(1) + fa,1(0) p 2/a k = ln and t = √ %a,1 1 + ϕ ϕ π ϕ3 5
The equation of logarithmic spiral is 2θ r = tϕaπ
Theorem 3.1. As n → ∞ P (n) satisfies the logarithmic spiral
Proof. Consider cartesian equation of logarithmic spiral,
y = r sin θ and x = r cos θ
yn = rn sin θn and xn = rn cos θn p 2 2 rn = (P (n).x − x1) + (P (n).y − y1) , using 1.2 p 2 2 rn = %a,1 fa,1(n − 2) + fa,1(n − 1)
17 p 2/a rn = %a,1fa,1(n − 2) 1 + ϕ 1/a n−3 ! ϕ a a p 2/a rn = √ ϕfa,1(1) + fa,1(0) %a,1 1 + ϕ 5
Let s = L(I0,P (n)).slope n nπ + 2 tan−1(s) θ = b c2π + tan−1(L(I ,P (n)).slope) = n 4 0 2 ! 1 − x2 Using the trigonometric relation 2 tan−1(x) = cos−1 1 + x2
−1 1−s2 nπ + cos 1+s2 θ = n 2
2 −1 1−s2 −1 (−1)(s2−1) −1 (s2−1) Since s > 1, 1−s < 0, therefore cos 1+s2 = cos 1+s2 = π−cos 1+s2
−1 (s2−1) (n + 1)π − cos 1+s2 θ = n 2 −1 s2−1 −1 s2−1 (n + 1)π − cos 1+s2 (n + 1)π − cos 1+s2 sin θ = sin and cos θ = cos n 2 n 2
x r1 − cos(x) x r1 + cos(x) Using the trignometric relations sin = and cos = 2 2 2 2 v v u 2 u 2 u −1 s −1 u −1 s −1 u1 − cos (n + 1)π − cos 1+s2 u1 + cos (n + 1)π − cos 1+s2 sin θ = t and cos θ = t n 2 n 2 Using the trignometric relation cos(A − B) = cos(A) cos(B) + sin(A) sin(B) −1 s2−1 −1 s2−1 cos (n + 1)π − cos 1+s2 = cos((n + 1)π) cos cos 1+s2 + ! s2 − 1 sin((n + 1)π) sin cos−1 1 + s2
18 Using cos(nπ) = (−1)nand sin(nπ) = 0, cos((n + 1)π) = (−1), sin((n + 1)π) = 0
! ! s2 − 1 s2 − 1 cos (n + 1)π − cos−1 = (−1) 1 + s2 1 + s2
Therefore, v v u s2−1 u s2−1 u1 + 1+s2 u1 − 1+s2 sin θ = t and cos θ = t n 2 n 2
P (n).y − y1 1/a s = L(I0,P (n)).slope = , using 1.1 and 1.2, limn→∞ s = ϕ P (n).x − x1 v v u 1/a u 1/a u1 + ϕ −1 u1 − ϕ −1 t 1+ϕ1/a t 1+ϕ1/a lim sin θn = and lim cos θn = n→∞ 2 n→∞ 2
s ϕ2/a r 1 lim sin θn = and lim cos θn = n→∞ ϕ2/a + 1 n→∞ ϕ2/a + 1 Using 1.2,
lim P (n).x − x1 = %a,1fa,1(n − 2) n→∞ 0 a 0 a1/a = %a,1 f (n − 2)fa,1(1) + f (n − 3)fa,1(0) 1/a n−3 ! ϕ a a lim P (n).x − x1 = %a,1 √ ϕfa,1(1) + fa,1(0) n→∞ 5 1/a n−3 ! ϕ a a 1/a lim P (n).y − y1 = %a,1 √ ϕfa,1(1) + fa,1(0) ϕ n→∞ 5
Now, 1/a s n−3 ! 2/a ϕ a a p 2/a ϕ lim rn sin θn = √ ϕfa,1(1) + fa,1(0) %a,1 1 + ϕ n→∞ 5 ϕ2/a + 1 1/a n−3 ! ϕ a a 1/a lim rn sin θn = %a,1 √ ϕfa,1(1) + fa,1(0) ϕ n→∞ 5
19 Figure 1: Plot of P (n) Figure 2: Logarithmic Spiral
lim rn sin θn = P (n).y − y1 n→∞ Similarly, lim rn cos θn = P (n).x − x1 n→∞ For x1 = 0 and y1 = 0
lim yn = P (n).y and lim xn = P (n).x n→∞ n→∞ Therefore, as n → ∞ P (n) satisfies the logarithmic spiral
Figure 4 at 3 describes plot of L0(n) as a discrete form of the logarithmic spiral. The following plots are visualized using GFKTv1.0.4[7], it is made using Python3 and uses manimlib, available at https://github.com/rahul-gohil/GFKT, archived at doi:10.5281/zenodo.3779050
4. Conclusion, Remarks and Extras
We generalized a formula for recurrence relations that can be expressed as a linear combination of fibonacci numbers. Note that their limit can be modified under cer- tain condition, let’s make use of the fact that ϕn = ϕn−1 + ϕn−2 [8].
20 Figure 4: L0(n) as discrete form of logarith- Figure 3: Plot of L0(n) mic spiral
" # " #" # ϕn 1 1 ϕn−1 = ϕn−1 1 0 ϕn−2 Therefore, " # " #" # ϕn f 0(n) f 0(n − 1) ϕ = ϕn−1 f 0(n − 1) f 0(n − 2) 1
a a n 0 0 ϕfa,b(b + mp0 ) + fa,b(mp0 ) 0 0 ϕ = f (n)ϕ+f (n−1), recall from theorem 1.1, cp ,q = a a , ϕfa,b(b + mq0 ) + fa,b(mq0 ) a 0 a if the initial conditions are set such that fa,b(b + mp0 ) = f (b + mp0 ), fa,b(mp0 ) = 1/a 0 m 0 −m 0 fa,b(n + p) d 0 0 +m 0 −m 0 f (b + mp0 − 1) then, cp0,q0 = ϕ p q and lim = ϕ p ,q p q n→∞ fa,b(n + q) To prevent the initial triangles from being skewed, the initial numbers should at least be within the same order of magnitude. Although all conditions still hold, the initial triangles appear a bit skewed if that is not the case. We also convey that the functions fa,1, produce a logarithmic spiral with equation,
r = tϕ2θ/aπ
21 Figure 5: Plot of T (n) Figure 6: Complete Plot
−1 1/a 2 tan ϕ − 1/a aπ a a ! ϕfa,1(1) + fa,1(0) p 2/a where t = √ %a,1 1 + ϕ ϕ ϕ3 5 So, we get logarithmic spirals whose growth factor is similar to ϕ4/a, of course fac- toring in the constant t. The logarithmic spiral is found in nature at numerous places, due to its property of self-similar structure. Such as but not limited to spiral galaxies[5], spiral beaches[6].
References [1] G. Anatriello, G. Vincenzi, Logarithmic spirals and continue triangles, Jour- nal of Computational and Applied Mathematics, Volume 296, 2016, Pages 127-137, ISSN 0377-0427, doi:10.1016/j.cam.2015.09.004 URLhttp://www. sciencedirect.com/science/article/pii/S0377042715004562
[2] A. Fiorenza, G. Vincenzi, (2011), Limit of ratio of consecutive terms for general order- k linear homogeneous recurrences with constant coefficients. Chaos Solitons and Fractals - CHAOS SOLITON FRACTAL. 44. 145-152. doi:10.1016/j.chaos.2011.01.003
[3] L. Mario (2002). The Golden Ratio: The Story of Phi, The World’s Most Aston- ishing Number. New York: Broadway Books. p. 149. ISBN 0-7679-0815-5.
22 [4] P. Hemenway (2005). Divine Proportion: Phi in Art, Nature, and Science. Ster- ling Publishing Co. ISBN 978-1-4027-3522-6.
[5] G. Bertin and C. C. Lin (1996). Spiral structure in galaxies: a density wave theory. MIT Press. p. 78. ISBN 978-0-262-02396-2.
[6] A. T. Williams, A. Micallef (2009). Beach management: principles and practice. Earthscan. p. 14. ISBN 978-1-84407-435-8.
[7] R. Gohil, (2020), rahul-gohil/GFKT: [email protected] doi:10.5281/zenodo.3784306
[8] G. Meisner, Powers of Phi, https://www.goldennumber.net/powers-of-phi/
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