The Islamic University of Gaza

Faculty of Engineering

Civil Engineering Department

Numerical Analysis

ECIV 3306

Chapter 5

Bracketing Methods

PART II ROOTS OF EQUATIONS

Roots of Equations

Bracketing Methods

Bisection method

False Position Method

Open Methods

Simple fixed point iteration

Newton Raphson

Secant

Modified Newton Raphson

System of Nonlinear

Equations

Roots of polynomials

Muller Method

Study Objectives for Part Two

ROOTS OF EQUATIONS

• Root of an equation: is the value of the equation variable which

make the equations = 0.0

• But

a

acbbxcbxax

2

40

22 −−

=⇒=++m

?0sin

?02345

=⇒=+

=⇒=+++++

xxx

xfexdxcxbxax

ROOTS OF EQUATIONS

• Non-computer methods:

- Closed form solution (not always available)

- Graphical solution (inaccurate)

• Numerical systematic methods suitable for

computers

Graphical Solution

roots

• The roots exist where f(x) crosses the x-axis.

f(x)

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

• Plot the function f(x)

Graphical Solution: Example

• The 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

)(t

m

c

e1c

mgv

−

−=

40)1(38.667

)(

)1()(

146843.0−−=

−−=

−

−

c

tm

c

ec

cf

vec

mgcf

Check: F (14.75) = 0.059 ~ 0.0

v (c=14.75) = 40.06 ~ 40 m/s

Numerical Systematic Methods

I. Bracketing Methods

f(x)

x

roots

f(xl)=+ve

f(xu)=+ve

xl xu

No roots or even

number of roots

f(x)

x

roots

f(xl)=+ve

f(xu)=-ve xl xu

Odd number of roots

Bracketing Methods (cont.)

• Two initial guesses (xl and xu) are required for the

root which bracket the 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) <0.

(The function changes sign on opposite sides of the root)

Special

Cases

Effect of computer scale

resolution

Bracketing Methods

1. Bisection Method

• Generally, if f(x) is real and continuous in the interval xl to xu

and f (xl).f(xu)<0, then there is at least one real root between

xl and xu to this function.

• The interval at which the function changes sign is located.

Then the interval is divided in half with the root lies in the

midpoint of the subinterval. This process is repeated to

obtained refined estimates.

f(x)

x xu xl

f(xu)

f(xu)

xr1

f(x)

x xu xl

f(xu)

xr2

xr = ( xl + xu )/2

f(xu)

f(xr1)

f(xr2)

(f(xl).f(xr)<0): xu = xr xr = ( xl + xu )/2

Step 1: Choose lower xl and upper xu

guesses for the root such that:

f(xl).f(xu)<0

Step 2: The root estimate is:

xr = ( xl + xu )/2

Step 3: Subdivide the interval according to:

– If (f(xl).f(xr)<0) the root lies in the

lower subinterval; xu = xr and go to

step 2.

– If (f(xl).f(xr)>0) the root lies in the

upper subinterval; xl = xr and go to

step 2.

– If (f(xl).f(xr)=0) the root is xr and stop

Bisection Method - Termination Criteria

• For the Bisection Method εa > εt

• The computation is terminated when εa

becomes less than a certain criterion (εa < εs)

%100

:

×−

=

true

eapproximattrue

tX

XX

ErrorrelaiveTrue

ε

1

:

100%

100% (Bisection)

n n

r ra n

r

u la

u l

Approximate relative Error

X X

X

X X

X X

ε

ε

−−

= ×

−= ×

+

Bisection method: Example

• The 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

f(c)

c

)(t

m

c

e1c

mgv

−

−=

40e1c

38667cf

ve1c

mgcf

c1468430

tm

c

−−=

−−=

−

−

)(.

)(

)()(

.

f(x)

x 16 12

-2.269

6.067

14

f(x)

x 14 16 -0.425 -2.269

1.569

1.569

(f(12).f(14)>0): xl = 14

1. Assume xl =12 and xu=16

f(xl)=6.067 and f(xu)=-2.269

2. The root: xr=(xl+xu)/2= 14

3. Check f(12).f(14) = 6.067•1.569=9.517 >0;

the root lies between 14 and 16.

4. Set xl = 14 and xu=16, thus the new root

xr=(14+ 16)/2= 15

5. Check f(14).f(15) = 1.569•-0.425= -0.666 <0;

the root lies bet. 14 and 15.

6. Set xl = 14 and xu=15, thus the new root

xr=(14+ 15)/2= 14.5

and so on…...

15

Iter. Xl Xu Xr εεεεa% εεεεt%

1 12 16 14 5.279 --

2 14 16 15 6.667 1.487

3 14 15 14.5 3.448 1.896

4 14.5 15 14.75 1.695 1.204

5 14.75 15 14.875 0.84 0.641

6 14.74 14.875 14.813 0.422 0.291

• In the previous example, if the stopping criterion is εt =

0.5%; what is the root?

Bisection method: Example

Bisection method

Flow Chart –Bisection Start

Input: xl , xu , εs, maxi

f(xl). f(xu)<0

i=0

εa=1.1εs

False

while

εa> εs &

i <maxi

2

1

u r

r

x xx

i i

+=

= +

False

Stop

Print: xr , f(xr ) ,εa , i

xu+xl =0

100%u l

a

u l

x x

x xε

−=

+

True

Test=f(xl). f(xr)

εa=0.0 Test=0

xu=xr Test<0

xl=xr

True

True

False

Bracketing Methods

2. False-position Method

• The bisection method divides the interval xl to xu in

half not accounting for the magnitudes of f(xl) and

f(xu). For example if f(xl) is closer to zero than f(xu),

then it is more likely that the root will be closer to

f(xl).

• False position method is an alternative approach

where f(xl) and f(xu) are joined by a straight line; the

intersection of which with the x-axis represents and

improved estimate of the root.

2. False-position Method

• False position method is an

alternative approach where

f(xl) and f(xu) are joined by

a straight line; the

intersection of which with

the x-axis represents and

improved estimate of the

root.

f(x)

x xu xl

f(xu)

f(xl)

xr

f(xr)

)()(

))((

)()(

ul

uluur

ur

u

lr

l

xfxf

xxxfxx

xx

xf

xx

xf

−

−−=

−=

−

False-position Method -Procedure

Step 1: Choose lower xl and upper xu guesses for the

root such that: f(xl).f(xu)<0

Step 2: The root estimate is:

Step 3: Subdivide the interval according to:

– If (f(xl).f(xr)<0) the root lies in the lower

subinterval; xu = xr and go to step 2.

– If (f(xl).f(xr)>0) the root lies in the upper

subinterval; xl = xr and go to step 2.

– If (f(xl).f(xr)=0) the root is xr and stop

)()(

))((

ul

uluur

xfxf

xxxfxx

−

−−=

False-position Method -Procedure

False position method: Example

• The 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

f(c)

c

)(t

m

c

e1c

mgv

−

−=

40)1(38.667

)(

)1()(

146843.0−−=

−−=

−

−

c

tm

c

ec

cf

vec

mgcf

f(x)

x 16 12

-2.269

6.067

14.91

False position method: Example

1. Assume xl = 12 and xu=16

f(xl)= 6.067 and f(xu)= -2.269

2. The root: xr=14.9113

f(12) . f(14.9113) = -1.5426 < 0;

3. The root lies bet. 12 and 14.9113.

4. Assume xl = 12 and xu=14.9113, f(xl)=6.067 and

f(xu)=-0.2543

5. The new root xr= 14.7942

6. This has an approximate error of 0.79%

False position method: Example

Flow Chart –False Position Start

Input: xl , x0 , εs, maxi

f(xl). f(xu)<0

i=0

εa=1.1εs

False

while

εa> εs &

i <maxi

( )( )

( ) ( )

1

u l ur u

l u

f x x xx x

f x f x

i i

−= −

−

= +

False

Stop

Print: xr , f(xr ) ,εa , i

i=1

or

xr=0

0 100%r r

a

r

x x

xε

−=

True

Test=f(xl). f(xr)

εa=0.0 Test=0

xu=xr

xr0=xr

Test<0

xl=xr

xr0=xr

True

True

False

False Position Method-Example 2

False Position Method - Example 2

Pitfalls of the False Position Method

• Although a method such as false position is often superior to

bisection, there are some cases (when function has significant

curvature that violate this general conclusion.

• In such cases, the approximate error might be misleading and

the results should always be checked by substituting the root

estimate into the original equation and determining whether

the result is close to zero.

• major weakness of the false-position method: its one

sidedness That is, as iterations are proceeding, one of the

bracketing points will tend stay fixed which lead to poor

convergence.

Modified Fixed Position

• One way to mitigate the "one-sided" nature of

false position is to make the algorithm detect

when one of the bounds is stuck. If this occur,

the function value at the stagnant bound is

divided in half. This is thought to fasten the

convergence.