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Open Methods
Lecture NotesDr. Rakhmad Arief SiregarUniversiti Malaysia Perlis
Applied Numerical Method for Engineers
Chapter 6
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Review of previous method
In the previous chapter, the root is located within an interval
Lower bound Upper bound
Repeated application if these methods always result in closer estimates of true value of the root
These methods are said to be CONVERGENT because they move closer to the truth as computation progresses
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Open Methods
In contrast, the open methods are based on formulas that require only a single or two of x.
Do not bracket the root. It is called as diverge or move
away from the true root as the computation progresses. Fig (b)
However, when it is converge (Fig. c), they do not so much more quickly than the bracketing method
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Simple Fixed-Point Iteration
This method also called as, simple fixed-point iteration, or one point iteration or successive substitution method
Open methods employ formula to predict the root. The formula can be developed by rearranging the function f(x)
= 0 so that x is on the left-hand side of the equation:
Example: can be rearranged to:
Example: can be rearranged to:
)(xgx
0322 xx2
32 x
x
0sin x xxx sin
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Simple Fixed-Point Iteration
The utility of is to provides a formula to predict a new value of x as function of an old value of x
Thus given an initial guess at the root xi, it can be used to compute a new estimate xi+1 as expressed by the iterative formula:
Thus the approximation error (as in the previous chapters) can be calculated as
)(xgx
%1001
1
i
iia x
xx
)(1 ii xgx
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Ex. 6.1 Simple Fixed-Point Iteration
Use simple fixed-point iteration to locate the root of
Solution The function can be separated directly and
expressed in the form of
With an initial guess of x0 =0, then the iterative equation can be calculated as
xexf x )(
ixi ex 1
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Ex. 6.1 Simple Fixed-Point Iteration
an initial guess of x0 =0, true root: 0.56714329
ixi ex 1
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Convergence
Why this result can be called as convergence?
Reading assignment: Read section 6.1.1 pp. 135
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Ex. 6.2 Two-curve graphical method
Separate the equation into two parts and determine its root graphically
Solution Reformulate the equation as: and
xexf x )(
xy 1xey 2
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In graphical methods The intersection of the
two curves indicates a root estimate of approximately x =0.57
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Exercise
Use simple fixed-point iteration to locate the root of
Use an initial guess of x0 = 0.5
xxxf sin2)(
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Solution
The first iteration is
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Solution
The remaining iterations are
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The Newton-Raphson Methods
The most widely used of all root-locating formula is the Newton-Raphson equation.
If the initial guess at the root is xi, a tangent can be extended from the point [xi, f(xi)]
The point where this tangent crosses the x axis usually represents an improved estimate of the root.
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The Newton-Raphson Methods
The Newton-Raphson method can be derived on the basis of this geometrical interpretation
The first derivative at x is equivalent to the slope:
which can be rearranged to yield
1
)()(
ii
ii xx
xfxf
)(
)(1
i
iii xf
xfxx
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Ex. 6.3 Newton-Raphson Method
Use the Newton-Raphson method to estimate the root of
Employing an initial guess of x0 = 0 Solution The first derivative of the function can be evaluated as
Which can be substituted along with the original function
xexf x )(
1)( xexf
)(
)(1
i
iii xf
xfxx
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i
i
xi
x
ii e
xexx
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Ex. 6.3 Newton-Raphson Method
Starting with an initial guess of x0 = 0, this iterative equation can be applied to compute:
The approach rapidly converges on the true root. Faster that by using simple fixed-point iteration
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Termination Criteria and Error Estimates
The approximate percent relative error can also be used as a termination criterion.
Other criteria can also be obtained such as discussed in Example 6.4
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Pitfalls of the Newton-Raphson Method
Although the Newton-Raphson method is often very efficient, there are situations where it performs poorly.
A special case – multiple roots – will be addressed later
However, even when dealing with simple roots, difficulties can also arise, as in the following example
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Ex. 6.5 Slowly Converging Function with Newton-Raphson
Determine the positive root of f(x) = 1010-1 using the Newton-Raphson method and an initial guess of x = 0.5
Solution The Newton-Raphson formula for this case is
Which can be used to compute:
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10
1 10
1
i
iii x
xxx
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Iteration results
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Pitfalls of the Newton-Raphson Method
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Pitfalls of the Newton-Raphson Method
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The Secant Method
A potential problem in implementing the Newton-Raphson method is the evaluation of the derivative.
For polynomials and many other function this is not convenient
There are certain function the derivative may be extremely difficult or inconvenient to evaluate
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The Secant Method
For these cases, the derivative can be approximated by a backward finite divided difference
The secant method can be formulated as:
ii
iii xx
xfxfxf
1
1 )()()(
)(
)(1
i
iii xf
xfxx
)()(
))((
1
11
ii
iiiii xfxf
xxxfxx
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Graphical depiction of the secant method
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Ex. 6.6
Use the secant method to estimate the root of f(x) = e-
x – x. Start with the estimates of xi-1 = 0 and x0 = 1.0. Solution (true root = 0.56714329…) First iteration
Xi-1 = 0 f(xi-1)=1.0 X0 = 1 f(x0) = -0.63212
Calculate t = 8.0%
)()(
))((
1
11
ii
iiiii xfxf
xxxfxx
61270.0
)63212.0(1
)10(63212.011
x
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Ex. 6.6
Solution (true root = 0.56714329…) Second iteration
X0 = 1 f(x0)= - 0.63212 X1 = 0.61270 f(x1) = - 0.07081
)()(
))((
1
11
ii
iiiii xfxf
xxxfxx
56384.0)07081.0(63212.0
)61270.01(07081.061270.02
x
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Ex. 6.6
Solution (true root = 0.56714329…) Third iteration
X0 = 0.61270 f(x1)= - 0.07081 X1 = 0.56384 f(x1) = 0.00518
)()(
))((
1
11
ii
iiiii xfxf
xxxfxx
56717.0)0518.0(07081.0
)56384.061270.0(00518.056384.03
x
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Ex. 6.6x
Use the secant method to estimate the root of f(x) = e-x – 2x. Start with the estimates of xi-1 = 0 and x0 = 1.0.
Solution (true root = xxxx) First iteration
Xi-1 = 0 f(xi-1)=xx X0 = 1 f(x0) = xxxx
Calculate t = 8.0%
)()(
))((
1
11
ii
iiiii xfxf
xxxfxx
61270.0
)63212.0(1
)10(63212.011
x
32
Next week
Reading assignment: Read pp. 146 – 157 Read pp. 160 – 174
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Quiz (60 Minutes)
What do you know about mathematical model in solving engineering problem? (10 marks)
Use zero through third order Taylor series expansions to predict f(2.5) for f(x) = ln x using a base point at x = 1. Compute the true percent relative error for each approximation. (15 marks)
Determine the real root of f(x)= 5x3-5x2+6x-2 using bisection method. Employ initial guesses of xl = 0 and xu = 1. iterate until the estimated error a falls below a level of s = 15% (20 marks)
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The difference between the Secant and False-Position Methods
Similarity of Secant and False-Position Methods The approximate root equation are identical on a term by term
basis
Both of them use two initial estimates to compute an approximation of the slope.
Critical difference: How in both methods the new estimate replaces the initial value
)()(
))((
1
11
ii
iiiii xfxf
xxxfxx
)()(
)(
ul
uluur xfxf
xxxfxx
False Position Method
Secant Method
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36
Ex. 6.7
Use the false-position and secant methods to estimate the root of f(x) = ln x.
Start the computation with value of xl = xi-1 = 0.5 xu = xi = 5.0
Solution
The false-position method
The secant method
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Modified Secant Method
)()(
))((
1
11
ii
iiiii xfxf
xxxfxx
)()(
)(1
iii
iiii xfxxf
xfxxx
ii
iii xx
xfxfxf
1
1 )()()(
i
iiii x
xfxxfxf
)()(
)(
Modified Secant Method
Original Secant Method
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Ex. 6.8
Use the secant method to estimate the root of f(x) = e-x – x. Use a value of 0.01 for and start with x0 = 1.0.
Solution (true root = 0.56714329…) First iteration
x0 = 1 f(x0) = -0.63212 x1+ x0 = 1.01 f(x1+ x0) = -0.64578
Calculate t = 5.3%
)()(
)(1
iii
iiii xfxxf
xfxxx
537263.0
)63212.0(64578.0
)63212.0(01.011
x
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Ex. 6.8
Solution (true root = 0.56714329…) Second iteration
x0 = 0.537263 f(x0) = -0.047083 x1+ x0 = 0.542635 f(x1+ x0) = -0.038579
Calculate t = 0.0236%
)()(
)(1
iii
iiii xfxxf
xfxxx
56701.0
)047083.0(038579.0
)047083.0(005373.0537263.01
x
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Ex. 6.8
Solution (true root = 0.56714329…) Third iteration
x0 = 0.56701 f(x0) = 0.000209 x1+ x0 = 0.572680 f(x1+ x0) = -0.00867
Calculate t = 2.365x10-5 %
)()(
)(1
iii
iiii xfxxf
xfxxx
567143.0
)000209.0(00867.0
)000209.0(00567.056701.01
x
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Multiple Roots
)1)(1)(3()( xxxxf )1)(1)(1)(3()( xxxxxf
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Multiple Roots
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Modified Newton-Raphson
A modifications have been proposed by Ralston and Rabinowitz (1978)
)(
)(1
i
iii xf
xfxx
)()()(
)()(21
iii
iiii
xfxfxf
xfxfxx
Original Newton-Raphson
Modified Newton-Raphson
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Ex. 6.9
Use both the standard ad modified Newton-Raphson methods to evaluate the multiple root of
f(x) = (x-3)(x-1)(x-1)
with initial guess of x0 = 0.
Solution (true value: 1.0) First derivative: f’(x) = 3x2 - 10x + 7 Second derivative f’’(x) = 6x - 10
)(
)(1
i
iii xf
xfxx
)()()(
)()(21
iii
iiii
xfxfxf
xfxfxx
7103
3752
23
1
ii
iiiii xx
xxxxx
)106)(375(7103
)7103)(375(2322
223
1
iiiiii
iiiiiii
xxxxxx
xxxxxxx
Original Newton-Raphson
Modified Newton-Raphson
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Original Newton-Raphson Method
)(
)(1
i
iii xf
xfxx
7103
3752
23
1
ii
iiiii xx
xxxxx
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Modified Newton-Raphson Method
)()()(
)()(21
iii
iiii
xfxfxf
xfxfxx
)106)(375(7103
)7103)(375(2322
223
1
iiiiii
iiiiiii
xxxxxx
xxxxxxx
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Ex. 6.9
For both methods to search for single root at x = 3 using initial guess of x0 = 4
Both methods converge quickly, with the standard method being somewhat more efficient
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Modified secant method
For multiple root secant method can be modified as:
)()(
))((
1
11
ii
iiiii xfxf
xxxfxx
)()(
))((
1
11
ii
iiiii xuxu
xxxuxx
)(
)()(
xf
xfxu i
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System of Nonlinear Equations
We have discussed on the determination of roots of single equation
How about locating roots of a set of simultaneous equations
Nonlinear equations is algebraic and transcendental equations that do not fit following format
0...)( 2211 bxaxaxaxf nn
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By Fixed-Point Iteration
Non linear equations:
Two simultaneous nonlinear equation with two unknown, x and y.
102 xyx
573 2 xyy
010),( 2 xyxyxu
573),( 2 xyyyxv
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6.10
Use fixed-point iteration to determine the root of Eq. (6.16). Note that a correct pair of roots is x = 2 and y = 3. Initiate the computation with guesses of the x = 1.5 and y = 3.5.
Solution
102 xyx
573 2 xyy
i
ii y
xx
2
1
10
21 357 iii yxy
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6.10
Solution On the basis of the initial value x = 1.5 and y = 3.5
21429.25.3
)5.1(10 2
1
ix
37516.24)5.3)(21429.2(357 21 iy
i
ii y
xx
2
1
10
21 357 iii yxy
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6.10
Solution On the basis of the initial value x = 1.5 and y = 3.5
010),( 2 xyxyxu
573),( 2 xyyyxv
xyx 10
x
yy
3
57
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6.10
Solution On the basis of the initial value x = 1.5 and y = 3.5
17945.2)5.3)(5.1(10 x
86051.2)17945.2(3
5.357
y
xyx 10
x
yy
3
57
56
6.10
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By Newton-Raphson
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Assignment
Solve problem 6.2
