The Islamic University of GazaFaculty of Engineering

Civil Engineering Department

Numerical Analysis

ECIV 3306

Chapter 5Bracketing Methods

PART II ROOTS OF EQUATIONS

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

aacbbxcbxax

240

22

?0sin?02345

xxxxfexdxcxbxax

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)

xf(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

mc

e1c

mgv

40)1(38.667)(

)1()(

146843.0

c

tmc

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)=-vexl 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)

xxuxl

f(xu)

f(xu)

xr1

f(x)

xxuxl

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)<0Step 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 a > t

• The computation is terminated when a becomes less than a certain criterion (a < s)

%100

:

true

eapproximattruet X

XXErrorrelaiveTrue

1

:

100%

100% (Bisection)

n nr r

a nr

u la

u l

Approximate relative Error

X XX

X XX X

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

mc

e1c

mgv

40e1c

38667cf

ve1c

mgcf

c1468430

tmc

)(.)(

)()(

.

f(x)

x1612 -2.269

6.067

14

f(x)

x14 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 a% t%1 12 16 14 5.279 --2 14 16 15 6.667 1.4873 14 15 14.5 3.448 1.8964 14.5 15 14.75 1.695 1.2045 14.75 15 14.875 0.84 0.6416 14.74 14.875 14.813 0.422 0.291

• In the previous example, if the stopping criterion is t = 0.5%; what is the root?

Bisection method: Example

Bisection method

Flow Chart –BisectionStart

Input: xl , xu , s, maxi

f(xl). f(xu)<0

i=0a=1.1s

False

whilea> s & i <maxi

21

u rr

x xx

i i

False

Stop

Print: xr , f(xr ) ,a , i

xu+xl =0

100%u la

u l

x xx x

True

Test=f(xl). f(xr)

a=0.0Test=0

xu=xrTest<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)

xxuxl

f(xu)

f(xl)

xr

f(xr)

)()())((

)()(

ul

uluur

ur

u

lr

l

xfxfxxxfxx

xxxf

xxxf

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

mc

e1c

mgv

40)1(38.667)(

)1()(

146843.0

c

tmc

ec

cf

vec

mgcf

f(x)

x1612

-2.269

6.067

14.91

False position method: Example1. 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 PositionStart

Input: xl , x0 , s, maxi

f(xl). f(xu)<0

i=0a=1.1s

False

whilea> s & i <maxi

( )( )( ) ( )

1

u l ur u

l u

f x x xx x

f x f xi i

False

Stop

Print: xr , f(xr ) ,a , i

i=1 or

xr=0

0 100%r ra

r

x xx

True

Test=f(xl). f(xr)

a=0.0Test=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.