bracketing method

8/13/2019 bracketing method

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Lecture 2 Numerical Analysis

2Eng. Malek Abuwarda

Roots of Equations

Bracketing methodsBisection Method

False position method

Open methods

Simple fixed point iteration

Newton Raphson

Secant Method

Modified Newton Raphson

System of nonlinear equations

Roots of polynomials

Using computers

Mullers Method


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Roots of Equations

What is the root of equation?It is the value of the equations variable which makes the equation equal to

zero.

Direct method

2( ) 0f x ax bx c= + + =

What is the root for the equation below?

2 4

2

b b acx

a

=

We can use the quadratic formula

( ) xf x e x=

Another example

( ) tan( )f x x x= or

Can you find the roots??

Exact solution but it is not

always available.


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Roots of Equations

Graphical Solution

2( )f x ax b=

What is the root for the equation below?

f(x)

x

roots

f(x)=0 f(x)=0


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Graphical Solution

ExampleThe parachutist velocity is

What is the drag coefficient c needed to reach

a velocity of 40 m/s if m=68.1 kg, t =10 s, g=9.8 m/s2

)(tm

c

e1c

mgv

=

40)1(38.667)(

)1()(

146843.0 =

=

c

tm

c

ec

cf

vec

mgcf

-8.40120

-2.26916

6.06712

17.6538

34.1154

f(c)c


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Graphical Solution Graphical techniques are of limited

practical value because they are not

precise. However, graphical methods

can be utilized to obtain rough

estimates of roots.

Graphical methods are important

tools for understanding the properties

of the function.


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Graphical Solution

Use of computer graphics tolocate roots

( ) sin 20 cos3x x x= +


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Bracketing Methods

Two initial guesses (xl andxu) are required for the root which

bracketthe root (s).

If one root of a real and continuous function, f(x)=0, is bounded by

values xl, xu then f(xl).f(xu)


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Bracketing Methods

Bisection Method

Generally, iff(x) is real and continuous in the intervalxl toxuand

f(xl).f(xu)


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Bisection Method

Step 1: Choose lowerxland upperxuguesses for the root such that:

f(xl).f(xu)


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Bisection Method

Example

0.146843667.38( ) (1 ) 40cf c ec

=

What is the root of the equation below?

From the graphical solution we can identify that

the root lies in the interval (12 , 16)


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Bisection Method

1612 13 14 15-0.425

1.569

-2.268

6.067

1514.87514.75

-0.425

0.059

-0.062 -0.184

(14+16)/2=15 -0.425 (14, 15)

(12+16)/2=14 1.569 (14, 16)

16 -2.268 (12, 16)

12 6.067 -

(14.75+14.875)/2=14.812 -0.062 (14.75, 14.812)

(14.75+15)/2=14.875 -0.184 (14.75, 14.875)

(14.5+15)/2=14.75 0.059 (14.75, 15)

(14+15)/2=14.5 0.552 (14.5, 15)

c (c) Interval

(14.75+14.812)/2=14.781 -0.001

1514 14.25 14.5 14.75-0.425

1.569

0.059

0.552

bracketing method

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