The Islamic University of Gaza
Faculty of Engineering
Civil Engineering Department
Numerical Analysis
ECIV 3306
Chapter 6
Open Methods
Open Methods
• Bracketing methods are based on assuming an interval of the function which brackets the root.
• The bracketing methods always converge to the root.
• Open methods are based on formulas that require only a single starting value of x or two starting values that do not necessarily bracket the root.
• These method sometimes diverge from the true root.
Open Methods- Convergence and Divergence Concepts
Converging increments
f(x)
xi xi+1x
f(x)
xi xi+1x
Diverging increments
1. Simple Fixed-Point Iteration
• Rearrange the function so that x is on the left side of the equation:
)(
)(0)(
1 ii xgx
xxgxf
• Bracketing methods are “convergent”.
• Fixed-point methods may sometime “diverge”, depending on the stating point (initial guess) and how the function behaves.
Simple Fixed-Point Iteration
Examples:1.
2. f(x) = x 2-2x+3 x = g(x)=(x2+3)/23. f(x) = sin x x = g(x)= sin x + x3. f(x) = e-x- x x = g(x)= e-x
xxg
or
xxg
or
xxg
xxxxf
21)(
2)(
2)(
02)(2
2
Simple Fixed-Point Iteration Convergence
• x = g(x) can be expressed as a pair of equations:y1= x
y2= g(x)…. (component equations)
• Plot them separately.
Simple Fixed-Point Iteration Convergence
1
1
( )
( ) 1
Suppose that the true root:
Subtracting 1 from 2
( ) ( ) (3)
2r
i i
r i r i
r
x g x
x x x
x
x
g
g
x
g
Simple Fixed-Point Iteration Convergence
Derivative mean value theorem: If g(x) are continuous in [a,b] then there exist at
least one value of x= within the interval such that:
i.e. there exist one point where the slope parallel to the line joining (a & b)
' g b g ag
b a
Simple Fixed-Point Iteration Convergence
1
'
'
'1
', 1 ,
'
'
( ) ( )
Let and
1.0 theerror decreases with each iteration
1.0 theerror increases with each iteration
r i r i
i r
r i
r i
r i r i
r i r i
t i t i
x x g x g x
a x b x
g x g xg
x x
g x g x x x g
then x x x x g
E g E
If g
If g
Simple Fixed-Point Iteration Convergence
• Fixed-point iteration converges if :
( ) 1 (slope of the line ( ) )g x f x x
• When the method converges, the error is roughly proportional to or less than the error of the previous step, therefore it is called “linearly convergent.”
Simple Fixed-Point Iteration-Convergence
Example: Simple Fixed-Point Iteration
1. f(x) is manipulated so that we get x=g(x) g(x) = e-x
2. Thus, the formula predicting the new value of x is: xi+1 = e-xi
3. Guess xo = 0
4. The iterations continues till the approx. error reaches a certain limiting value
f(x)
Root x
f(x)
x
f(x)=e-x - x
g(x) = e-x
f1(x) = x
f(x) = e-x - x
Example: Simple Fixed-Point Iteration
i xi g(xi) % t%
0 0 1.01 1.0 0.367879 100 76.32 0.367879 0.692201 171.8 35.13 0.692201 0.500473 46.9 22.14 0.500473 0.606244 38.3 11.85 0.606244 0.545396 17.4 6.896 0.545396 0.579612 11.2 3.837 0.579612 0.560115 5.90 2.28 0.560115 0.571143 3.48 1.249 0.571143 0.564879 1.93 0.70510 0.564879 1.11 0.399
Example: Simple Fixed-Point Iteration
i xi g(xi) % t%
0 0 1.01 1.0 0.367879 100 76.32 0.367879 0.692201 171.8 35.13 0.692201 0.500473 46.9 22.14 0.500473 0.606244 38.3 11.85 0.606244 0.545396 17.4 6.896 0.545396 0.579612 11.2 3.837 0.579612 0.560115 5.90 2.28 0.560115 0.571143 3.48 1.249 0.571143 0.564879 1.93 0.70510 0.564879 1.11 0.399
Flow Chart – Fixed Point
Start
Input: xo , s, maxi
i=0a=1.1s
1
Stop
1
whilea< s & i >maxi
i=1 or
xn=0
x0=xn
100%n oa
n
x x
x
Print: xo, f(xo) ,a , i 0
1nx g x
i i
False
True
2. The Newton-Raphson Method
• Most widely used method.• Based on Taylor series expansion:
)(
)(
)(0
g,Rearrangin
0)f(x when xof value theisroot The
...!2
)()()()(
1
1
1i1i
2
1
i
iii
iiii
iiii
xf
xfxx
xx)(xf)f(x
xxfxxfxfxf
Solve for
Newton-Raphson formula
The Newton-Raphson Method
• A tangent to f(x) at the initial point xi is extended till it meets the x-axis at the improved estimate of the root xi+1.
• The iterations continues till the approx. error reaches a certain limiting value.
f(x)
Root x
xixi+1
f(x) Slope f /(xi)
f(xi)
)(
)(
)()(
/
/
i
ii1i
1ii
ii
xf
xfxx
xx
0xfxf
Example: The Newton Raphson Method
11)(
)(/1
x
x
ix
x
ii
iii e
xex
e
xex
xf
xfxx
• Use the Newton-Raphson method to find the root of e-x-x= 0 f(x) = e-x-x and f`(x)= -e-x-1; thus
Iter. xi t%0 0 1001 0.5 11.82 0.566311003 0.1473 0.567143165 0.000024 0.567143290 <10-8
Flow Chart – Newton Raphson
Start
Input: xo , s, maxi
i=0a=1.1s
1
Stop
1
whilea >s & i <maxi
i=1 or
xn=0
x0=xn
100%n oa
n
x x
x
Print: xo, f(xo) ,a , i
00 '
0
1
n
f xx x
f x
i i
False
True
Pitfalls of The Newton Raphson Method
3. The Secant Method
The derivative is replaced by a backward finite divided difference
)()(
))((
i1i
i1iii1i xfxf
xxxfxx
Thus, the formula predicting the xi+1 is:
/ 1
1
( ) ( )( ) i i
ii i
f x f xf x
x x
/ ( )if x
The Secant Method
•Requires two initial estimates of x , e.g, xo, x1. However, because f(x) is not required to change signs between estimates, it is not classified as a “bracketing” method.
•The scant method has the same properties as Newton’s method. Convergence is not guaranteed for all xo, x1, f(x).
Secant Method: Example
• Use the Secant method to find the root of e-x-x=0; f(x) = e-x-x and xi-1=0, x0=1 to get x1 of the first iteration using:
Iter xi-1 f(xi-1) xi f(xi) xi+1 t%
1 0 1.0 1.0 -0.632 0.613 8.0
2 1.0 -0.632 0.613 -0.0708 0.5638 0.58
3 0.613 -0.0708 0.5638 0.00518 0.5672 0.0048
)()(
))((
i1i
i1iii1i xfxf
xxxfxx
Comparison of convergence of False Position and Secant Methods
False Position Secant Method
Use two estimate xl and xu Use two estimate xi and xi-1
f(x) must changes signs between xl and xu
f(x) is not required to change signs between xi and xi-1
Xr replaces whichever of the original values yielded a function value with the same sign as f(xr)
Xi+1 replace xi
Xi replace xi-1
Always converge May be diverge
11
1
( )( )
( ) ( )i i i
i ii i
f x x xx x
f x f x
( )( )
( ) ( )u l u
r ul u
f x x xx x
f x f x
Comparison of convergence of False Position and Secant Methods
• Use the false-position and secant method to find the root of f(x)=lnx. Start computation with xl= xi-1=0.5, xu=xi = 5.
1. False position method
2. Secant method Iter xi-1 xi xi+1
1 0.5 5.0 1.8546
2 5 1.8546 -0.10438
Iter xl xu xr
1 0.5 5.0 1.8546
2 0.5 1.8546 1.2163
3 0.5 1.2163 1.0585
False Position and Secant Methods
xi-1
xixu
xl
Although the secant method may be divergent, when it converges it usually does so at a quicker rate than the false position method
See the next figure
• Comparison of the true percent relative Errors Et for the methods to the determine the root of
f(x)=e-x-x
Flow Chart – Secant Method
Start
Input: x-1 , x0,s, maxi
i=0a=1.1s
1
Stop
1
whilea >s & i < maxi
i=1 or
Xi+1=0
Xi-1=xi
Xi=xi+1
1
1
100%i ia
i
x x
x
Print: xi , f(xi) ,a , i11
1
( )( )
( ) ( )
1
i i ii i
i i
f x x xx x
f x f x
i i
False
True
Modified Secant Method
Rather than using two initial values, an alternative approach is using a fractional perturbation of the independent variable to estimate
1
( )
( ) ( )i i
i ii i i
x f xx x
f x x f x
is a small perturbation fraction
/ ( ) ( )( ) i i i
ii
f x x f xf x
x
/ ( )if x
Modified Secant Method: Example
• Use the modified secant method to find the root of f(x) = e-x-x and, x0=1 and =0.01
0 0
0 0 0 0
1 1
1 1
1 1 1 1
2 1
1 0.63212
1.01 0.64578
( )0.537263
First Iteration
Second Iteration
5.3%( ) ( )
0.537263 0.047083
0.542635 0.038579
( )
(
i ii i t
i i i
i ii i
x f x
x x f x x
x f xx x x
f x x f x
x f x
x x f x x
x f xx x x
f x
0.56701 0.0236%
) ( ) ti i ix f x
Multiple Roots
x
f(x)= (x-3)(x-1)(x-1) = x3- 5x2+7x -3
f(x)
1x
3
Double roots
f(x)= (x-3)(x-1)(x-1)(x-1) = x4- 6x3+ 125 x2- 10x+3
f(x)
1 3
triple roots
Multiple Roots
•“Multiple root” corresponds to a point where a function is tangent to the x axis.
•Difficulties- Function does not change sign with double (or even number of multiple root), therefore, cannot use bracketing methods.- Both f(x) and f′(x)=0, division by zero with Newton’s and Secant methods which may diverge around this root.
4. The Modified Newton Raphson Method
• Another u(x) is introduced such that u(x)=f(x)/f /(x); • Getting the roots of u(x) using Newton Raphson
technique:
)()()(
)()(
)]([
)()()()()(
)(
)(
//2/
/
1
2/
/////
/1
iii
iiii
i
iiiii
i
iii
xfxfxf
xfxfxx
xf
xfxfxfxfxu
xu
xuxx
This function has roots at all the same locations as the original function
Modified Newton Raphson Method: Example
Using the Newton Raphson and Modified Newton Raphson evaluate the multiple roots of f(x)= x3-5x2+7x-3 with an initial guess of x0=0
)106)(375()7103(
)7103)(375(
)()()(
)()(
2322
223
//2/
/
1
iiiiii
iiiiii
iii
iiii
xxxxxx
xxxxxx
xfxfxf
xfxfxx
7x10x3
3x7x5xx
xf
xfxx
2i
i2i
3i
ii
ii1i
)(
)(/
•Newton Raphson formula:
•Modified Newton Raphson formula:
Newton Raphson Modified Newton-RaphsonIter xi t% iter xi t%0 0 100 0 0 1001 0.4286 57 1 1.10526 112 0.6857 31 2 1.00308 0.313 0.83286 17 3 1.000002 000244 0.91332 8.75 0.95578 4.46 0.97766 2.2
•Newton Raphson technique is linearly converging towards the true value of 1.0 while the Modified Newton Raphson is quadratically converging.•For simple roots, modified Newton Raphson is less efficient and requires more computational effort than the standard Newton Raphson method
Modified Newton Raphson Method: Example
Systems of Nonlinear Equations
• Roots of a set of simultaneous equations:
f1(x1,x2,…….,xn)=0
f2 (x1,x2,…….,xn)=0
fn (x1,x2,…….,xn)=0
• The solution is a set of x values that simultaneously get the equations to zero.
Systems of Nonlinear Equations
Example: x2 + xy = 10 & y + 3xy2 = 57
u(x,y) = x2+ xy -10 = 0
v(x,y) = y+ 3xy2 -57 = 0
• The solution will be the value of x and y which makes u(x,y)=0 and v(x,y)=0
• These are x=2 and y=3• Numerical methods used are extension of the open
methods for solving single equation; Fixed point iteration and Newton-Raphson.
Systems of Nonlinear Equations: 1.Fixed Point Iteration
1. Use an initial guess x =1.5 and y =3.52. The iteration formulae:
xi+1=(10-xi2)/yi and yi+1=57-3xiyi
2
3. First iteration,
x=(10-(1.5)2)/3.5=2.21429y=(57-3(2.21429)(3.5)2=-24.37516
4. Second iteration:
x=(10-2.214292)/-24.37516=-0.209y=57-3(-0.209)(-24.37516)2=429.709
5. Solution is diverging so try another iteration formula
Systems of Nonlinear Equations: 1.Fixed Point Iteration
1. Using iteration formula:
xi+1=(10-xiyi)1/2 and yi+1=[(57-yi)/3xi]1/2
First guess: x=1.5 and y=3.52. 1st iteration:
x=(10-(1.5)(3.5))1/2=2.17945y=((57-(3.5))/3(2.17945))1/2=2.86051
3. 2nd iteration:
x=(10-(2.17945)(2.86051))1/2=1.94053y=((57-(2.86051))/3(1.94053))1/2=3.04955
4. The approach is converging to the true root, x=2 and y=3
Systems of Nonlinear Equations: 1.Fixed Point Iteration
The sufficient condition for convergence for the two-equation case (u(x,y)=0 and v(x,y)=0) are:
1
1
u v
x x
and
u v
y y
Systems of Nonlinear Equations: 2. Newton Raphson Method
• Recall the standard Newton Raphson formula:
1
( )
'( )i
i ii
f xx x
f x
• which can be written as the following formula
1
( )
'( )
'( ) ( )
i i i
ii
i
i i i
x x x
f xwhere x
f x
f x x f x
• By multi-equation version (in this section we deal only with two equation) the formula can be derived in an identical fashion:
• u(x,y)=0 and v(x,y)=0
1
i i
i i
i ii i
i i
i i
i ii i
u u
x ux y
y vv v
x y
u u
x ux y
y vv v
x y
Systems of Nonlinear Equations: 2. Newton Raphson Method
• And thus
1
1i i i i
i i i ii i i i
u u v ux y y y
u v v uv v v ux y x yx y x x
1
i ii i
i ii i i i
v uu v
y yx x
u v v ux y x y
1
i ii i
i ii i i i
v uu v
x xy yu v v ux y x y
Systems of Nonlinear Equations: 2. Newton Raphson Method
• x 2+ xy =10 and y + 3xy 2 = 57 are two nonlinear simultaneous equations with two unknown x
and y they can be expressed in the form:
2
2 ,
3 , 1 6
u ux y x
x y
v vy xy
x y
i xi yi Ui vi Ui,x ui,y Vi,x Vi,y a,x
a,y
0 1.5 3.5 -2.5 1.625 6.5 1.5 36.75 32.5
1 2.03603 2.84388 -.06435 -4.7560 6.91594 2.03603 24.26296 35.74135 26.3 23.1
2 1.9987 3.00229 1.87 5.27
Systems of Nonlinear Equations: 2. Newton Raphson Method