CHAPTER 5 Solving Linear Equation Math

B3001/UNIT 10/1

SOLVING NON LINEAR EQUATIONS

Prepared by : Siti Sharmila Osmin/Nithya A/P Periasamy Page 1

Unit 10 SOLVING NON LINEAR EQUATION

General Objective : To solve non linear equations.

Specific Objective : Upon completion of this module, you will be able to:

1. Solve non linear equationS using fixed point

iteration.

2. solve non linear equationS using Newton-Raphson

method

Objectives

B3001/UNIT 10/2



10.0 INTRODUCTION

In this module we consider one of the most basic problems of numerical

approximation, the root finding problem, It involves finding the root x of an equation of

the form f(x) = 0, for a given function f. We will solve non linear equations using the

Fixed Point Iteration and Newton-Raphson method. Sometimes the equations given to

us are in a form of polynomials.

10.1 FIXED POINT ITERATION

A fixed point for a given function g is the number p for which g(p) = p. In thus

section we will consider the problem of finding solutions to fixed-point problems and the

connection between these problems and the root-finding problems we wish to solve.

Root finding problems and fixed-point problems are equivalent classes. Given a

root finding problem f(p) = 0 we can define a function g with a fixed point at p in a

number of ways. i.e.. g(x) = x - f(x) or as g(x) = x + 3 f(x).

Our first task is to become comfortable with this new type of problem and to

decide when a function has fixed points and how the fixed points can be determined.

INPUT

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Let have a polynomial f(x). We rewrite in the form of x = p(x) to find

approximate iteration that focus to a root (solution)

Iteration steps:

1. Find an approximate root. Put this value to the right to find new root,

x1 = p(xo).

2. This new approximation solution is substitute to the right to find another

approximate root

x2 = p(x1).

3. This process is repeatedly as to find the most appropriate root

Xn+1 = p(xn)..( 10.1 )

(10.1) is a fixed-point iteration. If xn+1 xn diminished when n increases.

The root will converge to = p (),

Example 10.1:

Find the root for 5xex = 1 . Give the root to the nearest 0.0001, using fixed point

iteration method.

Solution:

Write the equation in the form of

)(5

xpe

xx

Let xo = 1, fixed point iteration will list the results as;

x1 = 0.07357

x2 = 0.18581

x3 = 0.16608

x4 = 0.16939

x5 = 0.16883

x6 = 0.16892

After the sixth iteration we found that .0001.01 nn xx

0.16892 0.16883

= 0.00009 < 0.0001

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The first three iteration shows that the points are approximately nearing the real

root

Fig 10.1

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ACTIVITY 10.1

10.1 Use fixed point iteration, solve for the equation to the nearest 10-3

.

a. x sin x = 0

b. x2 + 5x 3 = 0

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FEEDBACK 10.1

10.1 a. 0.0

b. 0.541

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10.2 NEWTON-RAPHSON METHOD

The Newton-Raphson or simply Newtons method is one of the most powerful

and well known numerical methods for solving a root-finding problem f(x) = 0. There are

many ways of introducing Newtons method, we are going to start with a fixed point

nearest to the root, and we have to draw a tangent from that fixed point to the function.

That tangent cuts the x-axis at a second fixed point called x. We have to repeat the

process until it is nearest to the exact root. (Refer to fig 10.2),

Rajah 10.2

Let the first point (x0,y0) to the function y0 = f(x0). The tangent equation will be

y - f(x0) = f(x0) (x x0 ).

If tangent cuts the x-axis at x1, then

0 - f(x0) = f(x0) (x1 x0 )

INPUT

Refer Fig. 10.2 :

When x = x1 ; y = 0, then

equations can be find as

0 - f(x0) = f(x0) (x x0 )

B3001/UNIT 10/8



010

0

)('

)(xx

xf

xf

Then

)('

)(

0

001

xf

xfxx .(10.2)

From fig.7.2, we find that is the real root, then x1 is between and x0. In this

context x1 is approximately a better root than x0. This process repeats itself when

x1 substitute x0 to give a root x2, this new root is between and x1. These

processes will be repeated until too approximate roots having the same degree of

wanted place value.

Example 10.2:

Solve the following equation

3x3 + 2x 4 = 0

Correct to the third decimal place..

Solution:

Rewrite the equation

y = 3x3 = 4 - 2x

Given the the functions of y = 3x3

and y = 4 2x (Fig. 9.3), we will have to draw

two graphs. The point of intersection.is the root (solution).

Construct a table, for a few values of f(x).

x 0 1

f(x) -4 1

The solution or the root is between x = 0 and x = 1.

Newton-Raphson

Method

For x = 1,

f(x) = 3x3 + 2x 4

f(1) = 3(1) + 2(1) 4 = 1

B3001/UNIT 10/9



Rajah 10.3

We first choose initial approximations x0 through method of False Position as below:

11

40

41

10

x

= 0.8

This value xo is substituted using Newton-Raphson method (10.2)

f(x0) = f(0.8)

= 3(0.8)3 + 2(0.8) 4

= -0.864

f(x) = 9x2 + 2

f(x0) = f(0.8)

= 9(0.8)2 + 2

= 7.76

Using false position

formulae :

22

11

12

0

1

yx

yx

yyx

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Using (10.1) , we will find

76.7

864.08.01

x

= 0.91134

Successive of approximations are

4748.9

09339.091134.02 x

= 0.90148

and

3140.9

000767.090148.03 x

= 0.90139

When x2 and x3 are rounded to the third decimal place both x2 and x3 is the same

number. When this is true thats mean the root of the equation is x = 0.901

(correct to third decimal place). In other words, the iteration stops when

.0001.023 xx

Example 10.3:

Calculate estimation of a positive root correct to the third decimal place for the

following equation.

Kos x x2 = 0

Solution :

f(x) = kos x x2

x 0.0 0.5 1.0

f(x) 1.0 0.6275 -0.4597

3(0.91134)3 + 2(0.91134) 4

= 0.09339

9(0.91134)2 + 2 = 9.4748

B3001/UNIT 10/11



Using False Position method,.

6275.00.1

4597.05.0

4597.0627.0

10

x

0872.1

7735.0

= 0.7114

the root is in between 0.5 dan 1

Using Newton-Raphson iteration,

f(0.7114) = kos (0.7114) 0.71142

= 0.2513

f(x) = -sin x 2x

f(0.7114) = -sin (0.7114) 2(0.7114)

= -2.0757

The next approximation is

0757.2

2513.07114.01

x

= 0.8325

f(0.5903) = -0.0200

f(0.5903) = -2.4046

4046.2

0200.08325.02 x

= 0.8242

f(0.8242) = -0.00016

f(0.8242) = -2.3824

3824.2

00016.08242.03 x

= 0.8241

Therefore, the real root correct to three decimal place is 0.824.

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ACTIVITY 10.2

10.1 Use Newton-Raphson method to find solutions accurate to within 10-3

for

the following problems.

a. 3 kos x 5x + 4 = 0

b. x2 2x 3 + ex = 0

c. sin x x2 + 3 = 0

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FEEDBACK 10.2

1 a. 1.0818

b. 1.3544

c. 1.4183

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SELF ASSESSMENT

10.1 The equation x3 + 2x

2 5x 1 = 0 has a approximation solution x = 1.4. Use

Newton-Raphson method to find solution accurate to three decimal place.

10.2 The equation 2x3 7x2 x + 12 = 0 has a approximation solution x = 1.5. Use

Newton-Raphson method to find solution accurate to three decimal place

10.3 Use Newton-Raphsons method to approximate, within 10-4, the value of x that

produces the point on the graph of ex + x 2 = 0

B3001/UNIT 10/15



FEEDBACK

10.1

x xn f(xn) f(xn) )('

)(1

n

nn

xf

xfx

0 1.4 -1.336 6.48 1.606

1 1.61 0.3075 9.2163 1.5766

2 1.577 0.01075 8.7688 1.5758

3 1.5758 0.00023 8.7526 1.57577

The solution correct to 10-3

is x = 1.576.

10.2


)(1

n

nn

xf

xfx

0 1.5 1.5 -8.5 1.676

1 1.676 0.0769 -7.610 1.6861

2 1.6861 0.000307 -7.5478 1.68614


is x = 1.686.

B3001/UNIT 10/16



10.3


)(1

n

nn

xf

xfx

0 0.4 -0.108 2.49 0.4434

1 0.443 0.000372 2.557 0.44286

2 0.4429 0.000117 2.5572 0.44285


is x = 0.4429.

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CHAPTER 5 Solving Linear Equation Math