2. LECTURE 2 OF 4 7.0 NUMERICAL METHODS 7.2 Solutions of Non-Linear Equations

3. At the end of the lesson, students should be able to: a)To discuss the iteration method by writing f(x)=0 in the form of x=g(x). The iteration scheme is n=1,2,3…. b)Identify that the iteration method fails when in the neighborhood of the roots of the equation LEARNING OUTCOMES:

4. Numerical Method ( ITERATION METHOD) can be used to find approximate roots / solutions. (2) Solve x 3 – e x = 0 We cannot find the exact root of the above equation Example: (1) Solve x 2 – 4x – 5 = 0 => ( x + 1 )( x – 5) = 0 x = -1 or x = 5 The roots of the equation

5. 7.2 (1) : ITERATION METHOD 1. Find an initial approximate value x 1. The steps are : 2. Rewrite the equation in the form : x = g(x), g(x) is called the iteration function 4. This method fails if So we are looking for 3. Find g ’ (x)

6. How to find an initial approximate value ? 2 Methods GraphicalAlgebraic

7. Sketch the graph y = f(x).The real root the point where the graph intercept at x-axis OR Rewrite f(x) = 0 to a new form : F(x) = G(x) Sketch the graph y = F(x) and y = G(x). The real root the point of intersection between the two graphs. Graphical Method Algebraic Method Find two values a and b such that f(a) and f(b) have different signs At least one root must lie between a and b if f(x) is continuous.

8. Example Locate the approximate value of the equation by using the graphical method. Solution y x 43 y = - x + 4y = ln x The intersection is approximately at x = 2.9

9. Example 1 Show that the equation x 3 – 6x + 7 = 0 has a root between -3 and -2. Find the root of the equation correct to 3 d.p. using the Iteration Method. Solution Let f(x) = x 3 – 6x + 7 f(-3) = (-3) 3 -6(-3) + 7= -2 < 0 f(-2) = (-2) 3 – 6(-2) + 7= 11 > 0 different sign Since f(-3) < 0 and f(-2) > 0, thus one of the the roots is between -3 and -2

10. Let x 1 = -2.5 the initial approximate value Determine g(x) by writing f(x) = 0 to a new form : x = g(x)

11. CHECK g(x) is not an iteration function Therefore, g(x) is an iteration function The required solution is x= - 2.900

12. USING THE CALCULATOR Key in MODE 1 (COMP) ALPHA Y ALPHA = (6 ALPHA X – 7) CALC (X?) -2.5 = (-2.80204..) CALC Ans = ( -2.87696…) CALC Ans = ( -2.89495…) CONTINUE THE PROCESS UNTILL YOU GET To find the approximate root for x 3 – 6x + 7 = 0 x 1 = -2.5, iteration function : THE SAME ANSWER FOR AT LEAST TWICE

13. Example 2 Show that the equation Has a root between 0.2 and 0.3. Taking 0.2 as the first approximation, find the root of the equation correct to 3d.p. Solution At least one root must lie between 0.2 and 0.3

14. Therefore, g(x) is an iteration function

15. The required solution is 0.246

16. Example 3 Using the iteration method, find the solution of correct to 3 d.p. Solution: x=g(x) Given x 1 = -0.5,

17. The required solution is - 0.567