numarical metoud lecture 02
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ECEG-2112 : Computational MethodsLecture 02:
Solution of Nonlinear Equations
1
ECEG-2112 : Computational MethodsLecture 02:
Solution of Nonlinear Equations
Read Chapters 5 and 6 of the textbook
Lecture 02Solution of Nonlinear Equations
( Root Finding Problems )
Definitions Classification of Methods
Analytical Solutions Graphical Methods Numerical Methods
Bracketing Methods Open Methods
Convergence Notations
Reading Assignment: Sections 5.1 and 5.2
2
Definitions Classification of Methods
Analytical Solutions Graphical Methods Numerical Methods
Bracketing Methods Open Methods
Convergence Notations
Reading Assignment: Sections 5.1 and 5.2
Root Finding ProblemsMany problems in Science and Engineering areexpressed as:
0)(such that value thefind
,functioncontinuousaGiven
rfr
f(x)
3
0)(such that value thefind
,functioncontinuousaGiven
rfr
f(x)
These problems are called root findingproblems.
Roots of Equations
A number r that satisfies an equation is called aroot of the equation.
)1()3)(2(181573i.e.,
.1,3,3,2:rootsfourhas
181573:equationThe
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xxxxxxx
and
xxxx
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2.tymultipliciwith(3)rootrepeatedaand
2)and1(rootssimple twohasequationThe
)1()3)(2(181573i.e.,
.1,3,3,2:rootsfourhas
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2234
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and
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Zeros of a FunctionLet f(x) be a real-valued function of a realvariable. Any number r for which f(r)=0 iscalled a zero of the function.
Examples:2 and 3 are zeros of the function f(x) = (x-2)(x-3).
5
Let f(x) be a real-valued function of a realvariable. Any number r for which f(r)=0 iscalled a zero of the function.
Examples:2 and 3 are zeros of the function f(x) = (x-2)(x-3).
Graphical Interpretation of Zeros
The real zeros of afunction f(x) are thevalues of x at whichthe graph of thefunction crosses (ortouches) the x-axis.
f(x)
6
The real zeros of afunction f(x) are thevalues of x at whichthe graph of thefunction crosses (ortouches) the x-axis.
Real zeros of f(x)
Simple Zeros
)2(1)( xxxf
7
)1at xoneand2at x(onezerossimple twohas
22)1()( 2
xxxxxf
Multiple Zeros
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121)( 22
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Multiple Zeros3)( xxf
9
0at x3ymuliplicit withzeroahas
)( 3
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Facts Any nth order polynomial has exactly n
zeros (counting real and complex zeroswith their multiplicities).
Any polynomial with an odd order has atleast one real zero.
If a function has a zero at x=r withmultiplicity m then the function and itsfirst (m-1) derivatives are zero at x=rand the mth derivative at r is not zero.
10
Any nth order polynomial has exactly nzeros (counting real and complex zeroswith their multiplicities).
Any polynomial with an odd order has atleast one real zero.
If a function has a zero at x=r withmultiplicity m then the function and itsfirst (m-1) derivatives are zero at x=rand the mth derivative at r is not zero.
Roots of Equations & Zeros of Function
)1and,3,3,2are(Which
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Solution MethodsSeveral ways to solve nonlinear equations arepossible:
Analytical Solutions Possible for special equations only
Graphical Solutions Useful for providing initial guesses for other
methods Numerical Solutions
Open methods Bracketing methods
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Several ways to solve nonlinear equations arepossible:
Analytical Solutions Possible for special equations only
Graphical Solutions Useful for providing initial guesses for other
methods Numerical Solutions
Open methods Bracketing methods
Analytical MethodsAnalytical Solutions are available for specialequations only.
a
acbbroots
cxbxa
2
4
0:ofsolutionAnalytical2
2
13
Analytical Solutions are available for specialequations only.
a
acbbroots
cxbxa
2
4
0:ofsolutionAnalytical2
2
0:foravailableissolutionanalyticalNo xex
Graphical Methods Graphical methods are useful to provide an
initial guess to be used by other methods.
0.6
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2
1
Numerical MethodsMany methods are available to solve nonlinearequations: Bisection Method Newton’s Method Secant Method False position Method Muller’s Method Bairstow’s Method Fixed point iterations ……….
These will be coveredin ECEG-2112
15
Many methods are available to solve nonlinearequations: Bisection Method Newton’s Method Secant Method False position Method Muller’s Method Bairstow’s Method Fixed point iterations ……….
These will be coveredin ECEG-2112
Bracketing Methods In bracketing methods, the method starts
with an interval that contains the root anda procedure is used to obtain a smallerinterval containing the root.
Examples of bracketing methods: Bisection method False position method
16
In bracketing methods, the method startswith an interval that contains the root anda procedure is used to obtain a smallerinterval containing the root.
Examples of bracketing methods: Bisection method False position method
Open Methods In the open methods, the method starts
with one or more initial guess points. Ineach iteration, a new guess of the root isobtained.
Open methods are usually more efficientthan bracketing methods.
They may not converge to a root.
17
In the open methods, the method startswith one or more initial guess points. Ineach iteration, a new guess of the root isobtained.
Open methods are usually more efficientthan bracketing methods.
They may not converge to a root.
Convergence Notation
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if to tosaidis,...,...,,sequenceA 21 converge
18
Nnxx
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if to tosaidis,...,...,,sequenceA 21 converge
Convergence Notation
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19
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Speed of Convergence We can compare different methods in
terms of their convergence rate. Quadratic convergence is faster than
linear convergence. A method with convergence order q
converges faster than a method withconvergence order p if q>p.
Methods of convergence order p>1 aresaid to have super linear convergence.
20
We can compare different methods interms of their convergence rate.
Quadratic convergence is faster thanlinear convergence.
A method with convergence order qconverges faster than a method withconvergence order p if q>p.
Methods of convergence order p>1 aresaid to have super linear convergence.
Bisection Method The Bisection Algorithm Convergence Analysis of Bisection Method Examples
Reading Assignment: Sections 5.1 and 5.2
21
The Bisection Algorithm Convergence Analysis of Bisection Method Examples
Reading Assignment: Sections 5.1 and 5.2
Introduction The Bisection method is one of the simplest
methods to find a zero of a nonlinear function. It is also called interval halving method. To use the Bisection method, one needs an initial
interval that is known to contain a zero of thefunction.
The method systematically reduces the interval.It does this by dividing the interval into two equalparts, performs a simple test and based on theresult of the test, half of the interval is thrownaway.
The procedure is repeated until the desiredinterval size is obtained.
22
The Bisection method is one of the simplestmethods to find a zero of a nonlinear function.
It is also called interval halving method. To use the Bisection method, one needs an initial
interval that is known to contain a zero of thefunction.
The method systematically reduces the interval.It does this by dividing the interval into two equalparts, performs a simple test and based on theresult of the test, half of the interval is thrownaway.
The procedure is repeated until the desiredinterval size is obtained.
Intermediate Value Theorem Let f(x) be defined on the
interval [a,b].
Intermediate value theorem:if a function is continuousand f(a) and f(b) havedifferent signs then thefunction has at least one zeroin the interval [a,b].
f(a)
23
Let f(x) be defined on theinterval [a,b].
Intermediate value theorem:if a function is continuousand f(a) and f(b) havedifferent signs then thefunction has at least one zeroin the interval [a,b].
a b
f(b)
Examples If f(a) and f(b) have the
same sign, the functionmay have an evennumber of real zeros orno real zeros in theinterval [a, b].
Bisection method cannot be used in thesecases.
a b
24
If f(a) and f(b) have thesame sign, the functionmay have an evennumber of real zeros orno real zeros in theinterval [a, b].
Bisection method cannot be used in thesecases.
a b
The function has four real zeros
The function has no real zeros
Two More Examples
a b
If f(a) and f(b) havedifferent signs, thefunction has at leastone real zero.
Bisection methodcan be used to findone of the zeros.
25
a b
If f(a) and f(b) havedifferent signs, thefunction has at leastone real zero.
Bisection methodcan be used to findone of the zeros.
The function has one real zero
The function has three real zeros
Bisection Method If the function is continuous on [a,b] and
f(a) and f(b) have different signs,Bisection method obtains a new intervalthat is half of the current interval and thesign of the function at the end points ofthe interval are different.
This allows us to repeat the Bisectionprocedure to further reduce the size of theinterval.
26
If the function is continuous on [a,b] andf(a) and f(b) have different signs,Bisection method obtains a new intervalthat is half of the current interval and thesign of the function at the end points ofthe interval are different.
This allows us to repeat the Bisectionprocedure to further reduce the size of theinterval.
Bisection Method
Assumptions:Given an interval [a,b]f(x) is continuous on [a,b]f(a) and f(b) have opposite signs.
These assumptions ensure the existence of atleast one zero in the interval [a,b] and thebisection method can be used to obtain a smallerinterval that contains the zero.
27
Assumptions:Given an interval [a,b]f(x) is continuous on [a,b]f(a) and f(b) have opposite signs.
These assumptions ensure the existence of atleast one zero in the interval [a,b] and thebisection method can be used to obtain a smallerinterval that contains the zero.
Bisection AlgorithmAssumptions: f(x) is continuous on [a,b] f(a) f(b) < 0
Algorithm:Loop1. Compute the mid point c=(a+b)/22. Evaluate f(c)3. If f(a) f(c) < 0 then new interval [a, c]
If f(a) f(c) > 0 then new interval [c, b]End loop
b
f(a)
c
28
Assumptions: f(x) is continuous on [a,b] f(a) f(b) < 0
Algorithm:Loop1. Compute the mid point c=(a+b)/22. Evaluate f(c)3. If f(a) f(c) < 0 then new interval [a, c]
If f(a) f(c) > 0 then new interval [c, b]End loop
ab
f(b)
c
Bisection Method
b0
29
a0
b0
a1 a2
Example
+ + -
+ - -
30
+ - -
+ + -
Flow Chart of Bisection MethodStart: Given a,b and ε
u = f(a) ; v = f(b)
c = (a+b) /2 ; w = f(c) no
31
is
u w <0
a=c; u= wb=c; v= w
is(b-a) /2<εyes
yesno Stop
no
Example
Answer:
[0,2]?intervalin the13)(
:ofzeroafindtomethodBisectionuseyouCan3 xxxf
32
Answer:
usedbenotcanmethodBisection
satisfiednotaresAssumption
03(1)(3)f(2)*f(0)and
[0,2]oncontinuousis)(
xf
Example
Answer:
[0,1]?intervalin the13)(
:ofzeroafindtomethodBisectionuseyouCan3 xxxf
33
Answer:
usedbecanmethodBisection
satisfiedaresAssumption
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[0,1]oncontinuousis)(
xf
Best Estimate and Error LevelBisection method obtains an interval thatis guaranteed to contain a zero of thefunction.
Questions: What is the best estimate of the zero of f(x)? What is the error level in the obtained estimate?
34
Bisection method obtains an interval thatis guaranteed to contain a zero of thefunction.
Questions: What is the best estimate of the zero of f(x)? What is the error level in the obtained estimate?
Best Estimate and Error LevelThe best estimate of the zero of thefunction f(x) after the first iteration of theBisection method is the mid point of theinitial interval:
35
2
2:
abError
abrzerotheofEstimate
Stopping CriteriaTwo common stopping criteria
1. Stop after a fixed number of iterations2. Stop when the absolute error is less than
a specified value
How are these criteria related?
36
Two common stopping criteria
1. Stop after a fixed number of iterations2. Stop when the absolute error is less than
a specified value
How are these criteria related?
Stopping Criteria
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38
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)2log(
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39
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Example
11
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Example Use Bisection method to find a root of the
equation x = cos (x) with absolute error <0.02(assume the initial interval [0.5, 0.9])
41
Question 1: What is f (x) ?Question 2: Are the assumptions satisfied ?Question 3: How many iterations are needed ?Question 4: How to compute the new estimate ?
42
Bisection MethodInitial Interval
a =0.5 c= 0.7 b= 0.9
f(a)=-0.3776 f(b) =0.2784Error < 0.2
43
Bisection Method
0.5 0.7 0.9
-0.3776 -0.0648 0.2784Error < 0.1
44
0.7 0.8 0.9
-0.0648 0.1033 0.2784Error < 0.05
Bisection Method
0.7 0.75 0.8
-0.0648 0.0183 0.1033 Error < 0.025
45
0.70 0.725 0.75
-0.0648 -0.0235 0.0183 Error < .0125
Summary Initial interval containing the root:
[0.5,0.9]
After 5 iterations: Interval containing the root: [0.725, 0.75] Best estimate of the root is 0.7375 | Error | < 0.0125
46
Initial interval containing the root:[0.5,0.9]
After 5 iterations: Interval containing the root: [0.725, 0.75] Best estimate of the root is 0.7375 | Error | < 0.0125
A Matlab Program of Bisection Methoda=.5; b=.9;u=a-cos(a);v=b-cos(b);
for i=1:5c=(a+b)/2fc=c-cos(c)if u*fc<0
b=c ; v=fc;else
a=c; u=fc;end
end
c =0.7000
fc =-0.0648
c =0.8000
fc =0.1033
c =0.7500
fc =0.0183
c =0.7250
fc =-0.0235
47
a=.5; b=.9;u=a-cos(a);v=b-cos(b);
for i=1:5c=(a+b)/2fc=c-cos(c)if u*fc<0
b=c ; v=fc;else
a=c; u=fc;end
end
c =0.7000
fc =-0.0648
c =0.8000
fc =0.1033
c =0.7500
fc =0.0183
c =0.7250
fc =-0.0235
ExampleFind the root of:
root thefind tousedbecanmethodBisection
0)()(1)1(,10*
continuousis*
[0,1]:intervalin the13)( 3
bfaff)f(
f(x)
xxxf
48
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Example
Iteration a bc= (a+b)
2f(c)
(b-a)2
1 0 1 0.5 -0.375 0.5
2 0 0.5 0.25 0.266 0.25
49
2 0 0.5 0.25 0.266 0.25
3 0.25 0.5 .375 -7.23E-3 0.125
4 0.25 0.375 0.3125 9.30E-2 0.0625
5 0.3125 0.375 0.34375 9.37E-3 0.03125
Bisection MethodAdvantages Simple and easy to implement One function evaluation per iteration The size of the interval containing the zero is reduced
by 50% after each iteration The number of iterations can be determined a priori No knowledge of the derivative is needed The function does not have to be differentiable
50
Advantages Simple and easy to implement One function evaluation per iteration The size of the interval containing the zero is reduced
by 50% after each iteration The number of iterations can be determined a priori No knowledge of the derivative is needed The function does not have to be differentiable
Disadvantage Slow to converge Good intermediate approximations may be discarded
Newton-Raphson Method Assumptions Interpretation Examples Convergence Analysis
51
Assumptions Interpretation Examples Convergence Analysis
Newton-Raphson Method(Also known as Newton’s Method)
Given an initial guess of the root x0,Newton-Raphson method uses informationabout the function and its derivative atthat point to find a better guess of theroot.
Assumptions: f(x) is continuous and the first derivative is
known An initial guess x0 such that f’(x0)≠0 is given
52
Given an initial guess of the root x0,Newton-Raphson method uses informationabout the function and its derivative atthat point to find a better guess of theroot.
Assumptions: f(x) is continuous and the first derivative is
known An initial guess x0 such that f’(x0)≠0 is given
Newton Raphson Method- Graphical Depiction - If the initial guess at
the root is xi, then atangent to thefunction of xi that isf’(xi) is extrapolateddown to the x-axisto provide anestimate of the rootat xi+1.
53
If the initial guess atthe root is xi, then atangent to thefunction of xi that isf’(xi) is extrapolateddown to the x-axisto provide anestimate of the rootat xi+1.
Derivation of Newton’s Method
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Example
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Examplek (Iteration) xk f(xk) f’(xk) xk+1 |xk+1 –xk|
0 4 33 33 3 1
1 3 9 16 2.4375 0.5625
58
2 2.4375 2.0369 9.0742 2.2130 0.2245
3 2.2130 0.2564 6.8404 2.1756 0.0384
4 2.1756 0.0065 6.4969 2.1746 0.0010
Convergence Analysis
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Convergence AnalysisRemarks
When the guess is close enough to a simpleroot of the function then Newton’s method isguaranteed to converge quadratically.
Quadratic convergence means that the numberof correct digits is nearly doubled at eachiteration.
60
When the guess is close enough to a simpleroot of the function then Newton’s method isguaranteed to converge quadratically.
Quadratic convergence means that the numberof correct digits is nearly doubled at eachiteration.
Problems with Newton’s Method
• If the initial guess of the root is far from
the root the method may not converge.
• Newton’s method converges linearly near
multiple zeros { f(r) = f’(r) =0 }. In such a
case, modified algorithms can be used to
regain the quadratic convergence.
61
• If the initial guess of the root is far from
the root the method may not converge.
• Newton’s method converges linearly near
multiple zeros { f(r) = f’(r) =0 }. In such a
case, modified algorithms can be used to
regain the quadratic convergence.
Multiple Roots
21)( xxf
3)( xxf
62
1atzeros0at xzeros
twohas threehas
-x
f(x)f(x)
Problems with Newton’s Method- Runaway -
63
The estimates of the root is going away from the root.
x0 x1
Problems with Newton’s Method- Flat Spot -
64
The value of f’(x) is zero, the algorithm fails.
If f ’(x) is very small then x1 will be very far from x0.
x0
Problems with Newton’s Method- Cycle -
65
The algorithm cycles between two values x0 and x1
x0=x2=x4
x1=x3=x5
Newton’s Method for Systems ofNon Linear Equations
2
2
1
2
2
1
1
1
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211
11
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IterationsNewton
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Example Solve the following system of equations:
0,1guessInitial
05
0502
2
yx
yxyx
x.xy
67
0,1guessInitial
05
0502
2
yx
yxyx
x.xy
0
1,
1552
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5
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xF
yxyx
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Solution Using Newton’s Method
2126.0
2332.1
0.25-
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25.725.1
11.5
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50
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1
50
5
50
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1
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xyx
xF
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68
2126.0
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1
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1
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ExampleTry this
Solve the following system of equations:
0,0guessInitial
02
0122
2
yx
yyx
xxy
69
0,0guessInitial
02
0122
2
yx
yyx
xxy
0
0,
142
112',
2
1022
2
Xyx
xF
yyx
xxyF
ExampleSolution
0.1980
0.5257
0.1980
0.5257
0.1969
0.5287
2.0
6.0
0
1
0
0
_____________________________________________________________
543210
kX
Iteration
70
0.1980
0.5257
0.1980
0.5257
0.1969
0.5287
2.0
6.0
0
1
0
0
_____________________________________________________________
543210
kX
Iteration
Secant Method Secant Method Examples Convergence Analysis
71
Secant Method Examples Convergence Analysis
Newton’s Method (Review)
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0
10
20
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Example
001.0
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78
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10
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errorwith
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Examplex(i) f(x(i)) x(i+1) |x(i+1)-x(i)|
-1.0000 1.0000 -1.1000 0.1000
-1.1000 0.0585 -1.1062 0. 0062
79
-1.1062 0.0102 -1.1052 0.0009
-1.1052 0.0001 -1.1052 0.0000
Convergence Analysis
The rate of convergence of the Secant methodis super linear:
It is better than Bisection method but not asgood as Newton’s method.
iteration.iat theroot theofestimate::
62.1,
th
1
i
i
i
xrootr
Crx
rx
80
The rate of convergence of the Secant methodis super linear:
It is better than Bisection method but not asgood as Newton’s method.
iteration.iat theroot theofestimate::
62.1,
th
1
i
i
i
xrootr
Crx
rx
Comparison of RootFinding Methods
81
Advantages/disadvantages Examples
SummaryMethod Pros ConsBisection - Easy, Reliable, Convergent
- One function evaluation periteration- No knowledge of derivative isneeded
- Slow- Needs an interval [a,b]containing the root, i.e.,f(a)f(b)<0
82
Newton - Fast (if near the root)- Two function evaluations periteration
- May diverge- Needs derivative and aninitial guess x0 such thatf’(x0) is nonzero
Secant - Fast (slower than Newton)- One function evaluation periteration- No knowledge of derivative isneeded
- May diverge- Needs two initial pointsguess x0, x1 such thatf(x0)- f(x1) is nonzero
Example
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83
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Solution_______________________________
k xk f(xk)_______________________________
0 1.0000 -1.00001 1.5000 8.89062 1.0506 -0.70623 1.0836 -0.46454 1.1472 0.13215 1.1331 -0.01656 1.1347 -0.0005
84
_______________________________k xk f(xk)
_______________________________0 1.0000 -1.00001 1.5000 8.89062 1.0506 -0.70623 1.0836 -0.46454 1.1472 0.13215 1.1331 -0.01656 1.1347 -0.0005
Example
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85
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1
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3
k
kk
xf
xx
x
xxxf
Five Iterations of the Solution k xk f(xk) f’(xk) ERROR ______________________________________ 0 1.0000 -1.0000 2.0000 1 1.5000 0.8750 5.7500 0.1522 2 1.3478 0.1007 4.4499 0.0226 3 1.3252 0.0021 4.2685 0.0005 4 1.3247 0.0000 4.2646 0.0000 5 1.3247 0.0000 4.2646 0.0000
86
k xk f(xk) f’(xk) ERROR ______________________________________ 0 1.0000 -1.0000 2.0000 1 1.5000 0.8750 5.7500 0.1522 2 1.3478 0.1007 4.4499 0.0226 3 1.3252 0.0021 4.2685 0.0005 4 1.3247 0.0000 4.2646 0.0000 5 1.3247 0.0000 4.2646 0.0000
Example
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87
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x
xf
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Example
0.0000-1.5671-0.00000.5671
0.0002-1.5672-0.00020.5670
0.0291-1.5840-0.04610.5379
0.46211.3679-0.6321-1.0000
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1)(',)(
:ofrootafind toMethodsNewton'Use
k
kkkk
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xf xf xf x
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88
0.0000-1.5671-0.00000.5671
0.0002-1.5672-0.00020.5670
0.0291-1.5840-0.04610.5379
0.46211.3679-0.6321-1.0000
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ExampleEstimates of the root of: x-cos(x)=0.
0.60000000000000 Initial guess0.74401731944598 1 correct digit0.73909047688624 4 correct digits0.73908513322147 10 correct digits0.73908513321516 14 correct digits
89
Estimates of the root of: x-cos(x)=0.
0.60000000000000 Initial guess0.74401731944598 1 correct digit0.73909047688624 4 correct digits0.73908513322147 10 correct digits0.73908513321516 14 correct digits
ExampleIn estimating the root of: x-cos(x)=0, toget more than 13 correct digits:
4 iterations of Newton (x0=0.8) 43 iterations of Bisection method (initial
interval [0.6, 0.8]) 5 iterations of Secant method
( x0=0.6, x1=0.8)
90
In estimating the root of: x-cos(x)=0, toget more than 13 correct digits:
4 iterations of Newton (x0=0.8) 43 iterations of Bisection method (initial
interval [0.6, 0.8]) 5 iterations of Secant method
( x0=0.6, x1=0.8)