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The Islamic University of GazaFaculty of Engineering
Civil Engineering Department
Numerical AnalysisECIV 3306
Chapter 5Bracketing Methods
PART II ROOTS OF EQUATIONS
Roots ofEquations
Bracketing MethodsBisection method
False Position Method
Open Methods
Simple fixed point iteration
Newton Raphson
Secant
Modified Newton Raphson
System of NonlinearEquations
Roots of polynomials
Muller Method
STUDY OBJECTIVES FOR PART TWO
• Understand the graphical interpretation of a root• Know the graphical interpretation of the false-position method
and why it is usually superior to the bisection method.• Understand the difference between bracketing and open
methods for root location.• Understand the concepts of convergence and divergence.• Know why bracketing methods always converge, whereas open
methods may sometimes diverge.• Know the fundamental difference between the false-position
and secant methods and how it relates to convergence
ROOTS OF EQUATIONS
• Root of an equation: is the value of the equation variable whichmake 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 forcomputers
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 of40 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
c
f(c)
c=14.75 Check: F (14.75) = 0.059 ~ 0.0
v (c=14.75) = 40.06 ~ 40 m/s
Numerical Systematic MethodsI. Bracketing Methods
f(x)
x
roots
f(xl)=+ve
f(xu)=+ve
xl xu
No roots or evennumber 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)
Bracketing Methods1. 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(xl )
xr1
f(x)
xxuxl
f(xu)
xr2
xr = ( xl + xu )/2
f(xu)
f(xr1)
f(xr2)
(f(xl).f(xr)<0): xu = xrxr = ( 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 tostep 2.
– If (f(xl).f(xr)>0) the root lies in theupper subinterval; xl = xr and go tostep 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 abecomes 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
40)1(38.667)(
)1()(
146843.0 c
tmc
ec
cf
vec
mgcf
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=16f(xl)=6.067 and f(xu)=-2.269
2. The root: xr=(xl+xu)/2= 14f(14)=1.569
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 rootxr=(14+ 16)/2= 15
f(14)=-0.4255. 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 ----- 5.2792 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
Bracketing MethodsBisection
Example;Use bisection method to find the root of
Continue the iterations until the approximate error fallsbelow a stopping criteria ( s ) = 0.5%
104)( 23 xxxf
Flow Chart –BisectionStart
Input: xl , xu , s, maxi
f(xl). f(xu)<0
i=0a=1.1 s
False
whilea> s &i <maxi
False
Stop
Print: xr , f(xr ) , a , i
12
ii
xxx ulr
xu+xl =0
100%u la
u l
x xx x
True
Test=f(xl).f(xr)
a=0.0Test=
0
xu=xr
Test<0
xl=xr
True
True
False
Bracketing Methods2. False-position Method
• The bisection method divides the interval xl to xu inhalf not accounting for the magnitudes of f(xl) andf(xu). For example if f(xl) is closer to zero than f(xu),then it is more likely that the root will be closer tof(xl).
2. False-position Method
• False position method is analternative approach wheref(xl) and f(xu) are joined bya straight line; theintersection of which withthe x-axis represents animproved estimate of theroot.
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 theroot such that: f(xl).f(xu)<0Step 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 andf(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=1.1 s
False
whilea> s &i <maxi
( )( )( ) ( )
1
u l ur u
l u
f x x xx xf x f x
i i
False
Stop
Print: xr , f(xr ) , a , i
i=1or
xr=0
0 100%r ra
r
x xx
True
Test=f(xl). f(xr)
a=0.0Test=0
xu=xrxr0=xr
Test<0
xl=xrxr0=xr
True
True
False
False Position Method-Example 2
False Position Method - Example 2
Roots of Polynomials: Using SoftwarePackages
MS Excel:Goal seekf(x)=x-cos x