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Chapter 8

# Chapter 8 - CHAPTER 9 Discrete Dynamical Systems 9.1 1...

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CHAPTER 9 Discrete Dynamical Systems 9.1 Iterative Equations ! First-Order Linear Iterative Equations 1. yy nn + =+ 1 1 2 3, y 0 1 = The solution of ya y b + 1 is yb a a n n n F H G I K J 0 1 1 so with a = 1 2 , b = 3, and y 0 1 = , we have (a) y n n n n = F H I K + L N M M O Q P P =− F H I K + 1 2 3 1 5 1 2 6 1 2 1 2 bg (b) 0 0 6 n 10 y n + 1 1 2 y 0 1 = (c) y n n F H I K + 5 1 2 6 As n increases, the orbit approaches 6 from below. 2. + 1 1 2 y 0 1 The solution of y b + 1 is a a n n n F H G I K J 0 1 1 689

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690 CHAPTER 9 Discrete Dynamical Systems so with a = 1 2 , b = 3, and y 0 1 =− , we have (a) y n n n n F H I K + L N M M O Q P P F H I K + 1 2 3 1 7 1 2 6 1 2 1 2 bg (b) n 10 y n –1 7 yy nn + =+ 1 1 2 3, y 0 1 (c) y n n F H I K + 7 1 2 6 As n increases, the orbit approaches 6 from below. 3. + + 1 1 2 3, y 0 1 = The solution of ya y b + 1 is yb a a n n n F H G I K J 0 1 1 so with a 1 2 , b = 3, and y 0 1 = we have (a) y n n n n F H I K + −− L N M M O Q P P =− − F H I K + 1 2 3 1 1 2 2 1 2 3 2 (b) 0 0 3 n 10 y n + + 1 1 2 3, y 0 1 = (c) As n increases, the orbit approaches 2, with successive y n alternately above and below this value. Thus, ne →= 2.
SECTION 9.1 Iterative Equations 691 4. yy nn + =− + 1 1 2 3, y 0 1 The solution of ya y b + =+ 1 is yb a a n n n F H G I K J 0 1 1 so with a 1 2 , b = 3, and y 0 1 we have (a) y n n n n =− − F H I K + −− L N M M M O Q P P P F H I K + 1 2 3 1 2 1 3 2 3 1 2 2 ej (b) –2 4 n 10 y n + + 1 1 2 3, y 0 1 (c) y n n F H I K + 3 1 2 2 As n increases, the orbit approaches 2, with successive y n alternately above and below this value. 5. + 1 23 , y 0 1 = The solution of y b + 1 is a a n n n F H G I K J 0 1 1 so with a = 2, b = 3, and y 0 1 = we have (a) y n n n n F H G I K J =× − 21 1 42 3 (b) 0 0 5000 n 10 y n + 1 , y 0 1 = (c) y n n As n increases, the solution grows with- out bound.

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692 CHAPTER 9 Discrete Dynamical Systems 6. yy nn + =+ 1 23 , y 0 1 =− The solution of ya y b + 1 is yb a a n n n F H G I K J 0 1 1 so with a = 2, b = 3, and y 0 1 we have (a) y n n n n =− + F H G I K J 21 1 (b) 0 0 2500 n 10 y n + 1 , y 0 1 (c) y n n As n increases, the solution grows with- out bound. 7. + + 1 , y 0 1 = The solution of y b + 1 is a a n n n F H G I K J 0 1 1 so with a b = 3, and y 0 1 = we have (a) y n n n () + F H G I K J = 3 1 for all n . (b) 0 0 1.25 n 10 y n + + 1 , y 0 1 = (c) The orbit starts at y 0 1 = and remains there, because this is a fixed point. It is, however, a repelling fixed point, so at any other value for y 0 the solution is unbounded. See Problem 8.
SECTION 9.1 Iterative Equations 693 8. yy nn + =− + 1 23 , y 0 1 The solution of ya y b + =+ 1 is yb a a n n n F H G I K J 0 1 1 so with a 2, b = 3, and y 0 1 we have (a) y n n n n =−− () + F H G I K J =− − + 21 3 22 1 (b) 1500 n 10 y n –2500 + + 1 , y 0 1 (c) As n increases, the solution displays larger and larger oscillations with suc- cessive y n alternately above and below zero. 9. + 1 3, y 0 1 = The solution of b + 1 is n b n 0 so with b = 3, and y 0 1 = , we have (a) y y nb n n =+= + 0 13 (b) 0 0 n 10 y n 35 + 1 y 0 1 = (c) As n increases, the solution grows with- out bound.

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694 CHAPTER 9 Discrete Dynamical Systems 10.
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Chapter 8 - CHAPTER 9 Discrete Dynamical Systems 9.1 1...

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