1. Chapter 11 Section 1 Introduction to Difference Equations

2. Basic Idea Have a list of numbers that follow a certain pattern. The next number depends on the value of the previous number. The Difference Equation is used to calculate the next number in the list based on the previous number in the list.

3. Example List: 4, 7, 16, 43, 124, … Pattern [Next number in the list] = 3 · [Previous number in the list] – 5 Pattern written as a Difference Equation y n = 3 · y n – 1 – 5

4. General Form List:y 0, y 1, y 2, y 3, y 4, … Difference Equation: y n = a · y n – 1 + b Where y n represents the next number in the list y n – 1 represents the previous number in the list y 0 represents the initial value

5. When given a Difference Equation and an Initial Value 1.Generate the first few terms: y 0, y 1, y 2, y 3, y 4 2.Graph the terms: (0, y 0 ), (1, y 1 ), (2, y 2 ), (3, y 3 ), (4, y 4 ) 3.Solve the difference equation (known as the Solution of the Difference Equation ).

6. Solutions of the Difference Equation When a is not equal to 1 (pages 527 and 533): y n = [ b/(1 – a) ] + ( y 0 – [ b/(1 – a) ] ) · a n When a = 1 (page 533): y n = y 0 + b · n

7. Exercise 7 (page 532) For y n = ½ y n – 1 – 1, y 0 = 10 (a) Generate y 0, y 1, y 2, y 3, y 4. (b) Graph these first few terms. (c)Solve the difference equation.

8. Exercise 7 (a) Solution Recall: y n = ½ y n – 1 – 1, y 0 = 10 y 0 = 10 y 1 = ½ (10) – 1 = 4 y 2 = ½ (4) – 1 = 1 y 3 = ½ (1) – 1 = – ½ y 4 = ½ (– ½) – 1 = – 5/4 = – 1.25 Answer:10, 4, 1, – ½, – 5/4, …

9. Exercise 7 (b) Solution Graph (0, 10), (1, 4), (2, 1), (3, -1/2), (4, -5/4) 1 n ynyn 2 34 2 4 6 8 10 – 2

10. Exercise 7 (c) Solution Solve the difference equation (i.e. find the solution of the difference equation). Recall: y n = ½ y n – 1 – 1, y 0 = 10 So : a = ½, b = –1, b/(1 – a ) = – 2 Since a ≠ 1 y n = [ b/(1 – a) ] + ( y 0 – [ b/(1 – a) ] ) · a n = ( – 2 ) + ( (10) – ( – 2 ) ) (1/2) n y n = – 2 + 12 · (1/2) n

11. Exercise 7 Additional Question Use the solution of the difference equation to find y 8. Solution: y n = – 2 + 12 · (1/2) n y 8 = – 2 + 12 · (1/2) 8 y 8 = – 2 + 12 · (1/256) y 8 = – 2 + 3/64 y 8 = – 128/64 + 3/64 y 8 = – 125/64 (which is approx. – 1.953) Answer: – 125/64