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Introduction to Difference Equations: Understanding Patterns and Solutions

Explore how Difference Equations help predict patterns in lists of numbers, and learn to solve equations with initial values. Example problems and solutions provided.

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Introduction to Difference Equations: Understanding Patterns and Solutions

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  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 yn = 3 · yn – 1 – 5

  4. General Form • List: y0 , y1 , y2 , y3 , y4 , … • Difference Equation: yn = a· yn – 1 + b • Where yn represents the next number in the list yn – 1 represents the previous number in the list y0 represents the initial value

  5. When given a Difference Equation and an Initial Value • Generate the first few terms: y0 , y1 , y2 , y3 , y4 2. Graph the terms: (0 , y0) , (1 , y1) , (2 , y2) , (3 , y3) , (4 , y4) 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): yn = [ b/(1 – a) ] + ( y0 – [ b/(1 – a) ] ) ·an • When a = 1 (page 533): yn = y0 + b· n

  7. Exercise 7 (page 532) • For yn = ½ yn – 1 – 1 , y0 = 10 (a) Generate y0 , y1 , y2 , y3 , y4 . (b) Graph these first few terms. (c) Solve the difference equation.

  8. Exercise 7 (a) Solution Recall: yn = ½ yn – 1 – 1 , y0 = 10 y0 = 10 y1 = ½ (10) – 1 = 4 y2 = ½ (4) – 1 = 1 y3 = ½ (1) – 1 = – ½ y4 = 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) yn 10 8 6 4 3 4 2 n 1 2 – 2

  10. Exercise 7 (c) Solution • Solve the difference equation (i.e. find the solution of the difference equation). • Recall: yn = ½ yn – 1 – 1 , y0 = 10 • So : a = ½ , b = –1 , b/(1 – a ) = – 2 • Since a ≠ 1 • yn = [ b/(1 – a) ] + ( y0 – [ b/(1 – a) ] ) ·an = ( – 2 ) + ( (10) – ( – 2 ) ) (1/2)n yn = – 2 + 12 · (1/2)n

  11. Exercise 7 Additional Question • Use the solution of the difference equation to find y8 . • Solution: yn = – 2 + 12 · (1/2)n y8 = – 2 + 12 · (1/2)8 y8 = – 2 + 12 · (1/256) y8 = – 2 + 3/64 y8 = – 128/64 + 3/64 y8 = – 125/64 (which is approx. – 1.953) Answer: – 125/64

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