MAT495_Chapter 10

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Part 2 APPROXIMATION METHODS MAT 295

135

Chapter 10

Bisection

MethodAt the end of this chapter, students should be able to:

Define the location at which a non-linear function equals zero

Derive and apply Bisection method

10.1 Introduction

Chapter 10 shall discuss one of the most basic problems in numerical

analysis: finding values of a variable x that satisfy the equation 0)( xf . Asolution to this problem is called a zero of )(xf or a root of )(xf . In this

chapter, the focus of our discussion will be on solving non-linear equations.

Before further discussion, lets take a look at some classes of functions:

(a) linear functions:

baxxf )(

(b) polynomials (non-linear):

012

22

21

1 ...)( axaxaxaxaxaxf n

nn

nn

n

, 1n

(c) transcendental functions (non-linear):

xxxf tan)(

xexf x 2ln3)(

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10.2 Root Finding

In any algebra books we learn that the real roots of a quadratic function can

be found either by factorizing or by using standard formula i.e. quadratic

formula. Even if the roots are irrational, they can always be found.The roots of a polynomial of higher degree can also be found by using the

factor theorem (only when the roots are integers or simple rational fractions).

There are many equations whose roots cannot be evaluated exactly by any

methods. The approximate value of the roots of such equations can be found

either by a graphical approach or by one of a number of methods using

successive numerical approximations or by a combination of these processes.

Basically when we are discussing about finding roots, we want to find out

where

(i) the function )(xf crosses thex-axis or

(ii) two equations intersect each other (Figure 10.1).

((aa))ff((xx))ccrroosssseessxx--aaxxiiss ((bb))ttwwooeeqquuaattiioonnssiinntteerrsseecctt

eeaacchhootthheerr

FFiigguurree1100..11

Definition

Given an equation 0)( xf , the function )(xf is non-linear if it is not of the

form bax .

Definition

If a function )(xf is continuous on the interval ),( ba and if )(af and )(bf

have different signs (one positive and one negative) or 0)()( bfaf , then

there exists at least one real root on the interval ),( ba .

x

y

0

root

y=f(x)

r

x

y

0

root

y=f(x)

y=g(x)

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Take a note that the definition demands that the function )(xf be continuous

on the given interval. Figure 10.2 shows two cases that may occur

if 0)()( bfaf .

((aa))ff((xx))iissccoonnttiinnuuoouuss ((bb))ff((xx))iissnnoottccoonnttiinnuuoouussFFiigguurree1100..22

Example 1

Show that 3)( xxf has a root in the interval [-1, 2].

Identify the interval

[-1, 2] a= -1 and b= 2

Identify the function and compute )(af and )(bf

3)( xxf or 03 x

1)1( f 8)2( f

Determine if 0)()( bfaf

08)8)(1()1()2( ff

Since 0)1()2( ff it can be concluded that there exists a root in [-1, 2].

Solut ion

Steps : Root finding

Identify the interval ),( ba

Identify the function )(xf and compute )(af and )(bf

Determine if 0)()( bfaf then ],[ bar

x

y

0

ba

f(b)

f(a)

x

y

0 a b

f(b)

f(a)

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Example 2

Given 013 xx , show that one of the solutions of the equation lies

between -2 and 1.

Identify the interval

[-2, 1] a = -2 and b = 1

Identify the function and compute at aand b

013 xx or 1)( 3 xxxh

5)2(h 1)1(h

Determine if 0)()( bhah

05)1)(5()1()2( hh

Hence, there exists a root in the interval [-2, 1].

Example 3

Let 02xex . Determine the interval of the root of the function of

interest.

Identify the function :

02 xex 2)( xexf x

In order to determine the interval of the root, lets plot the graph. There

are two options to do this.

(i) Plot 2)( xexf

x

or

(ii) Break 2)( xexf x to two simpler functions, i.e.

)()(

2

02

xgxh

xe

xe

x

x

Solut ion

Solut ion

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(a) 2)( xexf x (b) xexh )( and xxg 2)(

Figure 10.3

Based on Figure 10.3 it can be shown that there exists a root in [0, 1].

Example 13 showed the existence of a root between given intervals. The

examples did not provide the procedure to determine the root. There are a

number of procedures that can be used to locate the desired root.

Determine the existence of a root for the following functions in a given

interval.

(i) 166)( 3 xxxf ; [1, 2]

(ii) xxxf 3coscos)( ;

2,

2

(iii) 01

1ln2

x

x

The procedure of solving a non-linear equation numerically is different from

the procedure of solving the equation analytically. Analytical method of root

finding attempted to find exactsolution to the equation. Not all equations can

be solved analytically. In many instances in real applications, it is impossible

or cumbersome to find the root analytically, thus the equation is solved

Warm up exercise

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numerically. Numerical method of solving non-linear equation begins with a

rough estimate of the root. The information is used to produce a better

estimate. The process is repeated until a desired accuracy is achieved. The

repetition process is also known as iterative method of root finding. In this

chapter, we are going to discuss two numerical methods of root finding

namely

(i) Bisection method

(ii) Newtons method

1100..33Bisection Method

The simplest numerical technique that shall first be discussed here is the

Bisection method or sometimes called the method of halving of intervals. The

method is based on the Intermediate Value theorem and attempts to locate a

solution to the equation 0)( xf in a sequence of intervals of decreasing

size.

Figure 10.4

In order to apply the Bisection method, we have to assume the function )(xf

is continuous on the interval ),( ba such that 0)()( bfaf . Hence, there

exists at least one value r in the interval ),( ba in which 0)( rf (ris the root).

The interval is then halved or divided into two subintervals by the midpoint of

aand b. Checking is done to locate in which of the two smaller subintervals

the root lie. The process is repeated until a desired accuracy is achieved

(Figure 10.4).

x

y

0 ba

f(b)

f(a)

r1 r2 r3

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Example 4

The solution of the equation 03 x is known to exist between the values -

1 and 2. Apply the Bisection method with 3 iterations to reduce the interval

of the root.

Identify the equation and determine )(xf

03 x or 3)( xxf

Identify the existence of a root

[-1, 2] a = -1 and b = 2

3)( xxf 1)1( f 8)2( f

08

)8)(1()2()1(

ff

Compute2bac

5.02

21

c

and 125.0)5.0( f

Determine

0)125.0)(1()5.0()1( ff

0)8)(125.0()2()5.0( ff

Since 0)5.0()1( ff then )5.0,1(r

Solut ion

Steps : Bisection Method


Identify the existence of a root : 0)()( bfaf

Compute2

bac

and )(cf

Determine if

0)()( cfaf then ),( car

0)()( cfbf then ),( bcr

Repeat until desired iteration/accuracy

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Repeat until desired iterations/accuracy

i x3)( xxf Root in Bisection

0 -1 -1.000

1 2 8.000 [-1, 2] 0.5

2 0.5 0.125 [-1, 0.5] -0.25

3 -0.25 -0.016 [-0.25, 0.5] 0.125

Hence, after 3 iterations, the estimated root is 0.125.

Example 5

Given 1)( 3 xxxh , justify that one of the roots of this function lies

between -2 and 1. Hence, use the Bisection method to calculate the

approximation to 2 significant digits.

Identify the equation

013 xx or 1)( 3 xxxh


[-2, 1] a = -2 and b = 1

1)( 3 xxxh 5)2( h 1)1( h

05

)1)(5()1()2(

hh

Compute2

bac

5.02

12 c and 375.1)5.0( h

Determine

0)375.1)(5()5.0()2( hh

0)1)(375.1()1()5.0( hh

Then )5.0,2( r .

Solut ion

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Repeat until desired iteration/accuracy

i x 1)( 3 xxxh root in Bisection

0 -2 -5.000

1 1 1.000 [-2, 1] -0.5

2 -0.5 1.375 [-2, -0.5] -1.250

3 -1.250 0.297 [-2, -1.25] -1.625

4 -1.625 -1.666 [-1.625, -1.25] -1.438

5 -1.438 -0.536 [-1.438, -1.25] -1.344

6 -1.344 -0.084 [-1.344, -1.25] -1.297

The root of 1)( 3 xxxh approximated to 2 significant digits is 3.1r .

Example 6

The equation xex 2 has one solution. Estimate the solution to 2

decimal places.


xex 2

02 xex or 2)( xexf x


There exists a root in [0, 1]. (Refer Example 3)

Compute2

bac

and f(c)

5.0210 c and f(0.5) = 0.149

Determine

0)5.0()0( ff

0)1()5.0( ff

Then )5.0,0(r

Solut ion

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3. Determine the first four approximations of the actual root of the equation

02

13 x in the interval [0,1] by the Bisection method.

4. By the method of Bisection determine the first three approximations of the

following equations

(a) 01sin 3 xx ; [-2, 0]

(b) 03ln xx ; [4, 5]

(c) xxxf 3coscos)( ;

2,

2

(d) xx tan2 ; [0, 3]

5. Show that 023

xx has a root between 1 and 2. Use the Bisectionmethod to approximate the root accurate to two decimal places.

6. Using the Bisection method, solve 3ln xx , given that the root is close to

the value 2. Obtain the root correct to two decimal places.

7. Use the Bisection method to approximate the zero of 21)( xxxf in [0,

1]. Give your answers accurate to 2 significant digits.

8. Solve xexcos2 accurate to one decimal place using the Bisection method.

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