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8/10/2019 t 5 Open Methods
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Motilal Nehru National Institute of Technology
Civil Engineering Department
Computer Based Numerical Techniques
CE-401
Roots of the equations: Open Methods
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Open Methods
Bracketing methodsare based on assuming aninterval of the function which brackets the root.
The bracketing methodsalways convergeto theroot.
Open methodsare based on formulas thatrequire only a single starting value of x or twostarting values that do not necessarily bracket
the root. These method sometimes divergefrom the true
root.
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1. Simple Fixed-Point Iteration
Rearrange the function so that x is onthe 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) andhow the function behaves.
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Simple Fixed-Point Iteration
Examples:1.
2. f(x) = x 2-2x+3 x = g(x)=(x2+3)/2
3. f(x) = sin x x = g(x)= sin x + x
3. f(x) = e-x- x x = g(x)= e-x
x
xg
or
xxg
or
xxg
xxxxf
21)(
2)(
2)(
02)(
2
2
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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.
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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 erroris roughly proportional to or less than the
error of the previous step, therefore it is
called linearly convergent.
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Simple Fixed-Point Iteration-Convergence
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Steps of Simple Fixed Pint Iteration
1. Rearrange the equation f(x) = 0 so that x ison the left hand side and g(x) is on the righthand side.
e.g f(x) = x2-2x-1 = 0x= (x2-1)/2
g(x) = (x2-1)/2 2. Set xiat an initial guess xo.
3. Evaluate g(xi)
4. Let xi+1= g(xi) 5. Find a=(Xi+1xi)/Xi+1, and set xi at xi+1 6. Repeat steps 3 through 5 until |a|
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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 theapprox. 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
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Example: Simple Fixed-Point Iteration
i xi g(xi) % t%
0 0 1.01 1.0 0.367879 100 76.3
2 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 3.83
7 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
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Example: Simple Fixed-Point Iteration
i xi g(xi) % t%
0 0 1.01 1.0 0.367879 100 76.3
2 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 3.83
7 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
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Example
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Flow ChartFixed Point
Start
Input: xo, s, maxi
i=0
a=1.1s
1
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Stop
1
while
amaxi
xn=0
x0=xn
100%n oan
x xx
Print:xo, f(xo) ,a, i 01
nx g x
i i
False
True
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2. The Newton-Raphson Method
Most widely used method.
Based on Taylor series expansion:
)(
)(
)(0
g,Rearrangin
0)f(xwhenxofvaluetheisrootThe
...!2)()()()(
1
1
1i1i
2
1
i
iii
iiii
iiii
xf
xfxx
xx)(xf)f (x
x
xfxxfxfxf
Solve for
Newton-Raphson formula
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The Newton-Raphson Method
A tangent tof(x)at theinitial pointxiis extended
till it meets the x-axis at
the improved estimate of
the rootxi+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
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Example: The Newton Raphson Method
11)(
)(/1
x
x
ix
x
i
i
iii
e
xex
e
xex
xf
xfxx
Use the Newton-Raphson method to find theroot of e-x-x= 0f(x) = e-x-xand f`(x)= -e-x-1;thus
Iter. xi
t%
0 0 100
1 0.5 11.8
2 0.566311003 0.147
3 0.567143165 0.00002
4 0.567143290
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Flow ChartNewton Raphson
Start
Input: xo, s, maxi
i=0
a=1.1s
1
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Stop
1
while
a>s &
i
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M-file for Newton Raphsons Method
function [root,ea,iter]=newtraph(func, dfunc, xr, es, maxit, varargin)
% newtraph: Newton-Raphson root location zeroes
% [root,ea,iter]=newtraph(func,dfunc,xr,es,maxit,p1,p2,...):% uses Newton-Raphson method to find the root of func
% input:
% func = name of function
% dfunc = name of derivative of function
% xr = initial guess% es = desired relative error (default = 0.0001%)
% maxit = maximum allowable iterations (default = 50)
% p1,p2,... = additional parameters used by function
% output:
% root = real root% ea = approximate relative error (%)% iter = number of iterations
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M-file for Newton Raphsons Method
if nargin
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Pitfalls of The Newton Raphson Method
Cases where Newton Raphson method diverges or exhibit poor
convergence.a) Reflection point b) oscillating around a local optimum
c) Near zero slope , and d) zero slope
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Example Newton Raphsons Method
Evaluate 29 to five decimal places by Newton-Raphsons iterative method
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Solution
Evaluate 29 to five decimal places by Newton-Raphsons iterative method
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Solution
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3. The Secant Method
The derivative is
sometimes difficult to evaluateby the computer program. It
may be replaced by a backward
finite divided difference
)()(
))((
i1i
i1iii1i
xfxf
xxxfxx
Thus, the formula
predicting thexi+1is:
/ 1
1
( ) ( )( ) i ii
i i
f x f xx
x x
/ ( )i
x
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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 asNewtons method. Convergence is not
guaranteed for all xo, x1,f(x).
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Secant Method:Example
Determine a root of the equation sinx + 3 cos x
2 = 0using the secant method. The initial approximations
x0 and x1are 0 and 1.5.
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Comparison of convergence of False Position and
Secant Methods
False Position Secant Method
Use two estimatexland xu Use two estimatexiand xi-1
f(x) must changes signs betweenxl
and xu
f(x) is not required to change signs
betweenxiand xi-1
Xrreplaces whichever of the original
values yielded a function value with
the same sign as f(xr)
Xi+1replace xiXi replace xi-1
Always converge May be diverge
Slower convergence than Secant in
case the secant converges.
If converges, It does faster then False
Position
1
1
1
( )( )
( ) ( )
i i i
i i
i i
f x x xx x
f x f x
( )( )
( ) ( )
u l u
r u
l u
f x x xx x
f x f x
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Comparison of convergence of False Position and
Secant Methods
Use the false-position and secant method to find the root off(x)=logex. Start computation with xl=xi-1=0.5, xu=xi= 5.
1. False position method
2. Secant methodIter 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
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Comparison of
the true percentrelative Errors Et
for the methods
to the determine
the root off(x)=e-x-x
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Flow ChartSecant Method
Start
Input: x-1, x0,s, maxi
i=0
a=1.1s
1
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Stop
1
while
a>s &
i < maxi
Xi+1=0
Xi-1=xiXi=xi+1
1
1100%
i ia
i
x x
x
Print:xi, f(xi) ,a, i1
1
1
( )( )
( ) ( )
1
i i ii i
i i
f x x xx x
f x f x
i i
False
True
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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 i
i i i
x f xx x
f x x f x
is a small perturbation fraction
/ ( ) ( )( ) i i iii
f x x f xxx
/ ( )i
f x
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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%
) ( ) t
i i ix f x
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Multiple Roots
x
f(x)= (x-3)(x-1)(x-1)= x3- 5x2+7x -3
f(x)
1x
3
Doubleroots
f(x)= (x-3)(x-1)(x-1)(x-1)
= x4- 6x3+ 125 x2- 10x+3
f(x)
1 3
tripleroots
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Multiple Roots
Multiple root corresponds to a pointwhere 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 withNewtons and Secant methods which maydiverge around this root.
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4. The Modified Newton Raphson Method by
Ralston and Rabinowitz
Another u(x)is introduced such that u(x)=f(x)/f/(x);
Getting the roots of u(x) using Newton Raphsons
technique:
)()()(
)()(
)]([
)()()()(
)(
)(
)(
//2/
/
1
2/
/////
/1
iii
iiii
i
iiiii
i
iii
xfxfxf
xfxfxx
xf
xfxfxfxf
xu
xu
xuxx
This function has rootsat all the same locations
as the original function
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Modified Newton Raphson Method: Example
Using the Newton Raphson and Modified Newton
Raphson evaluate the multiple roots off(x)= x3-5x2+7x-3with an initial guess ofx0=0
)106)(375()7103(
)7103)(375(
)()()()()(
2322
223
//2/
/
1
iiiiii
iiiiii
iii
iiii
xxxxxx
xxxxxx
xfxfxfxfxfxx
7x10x33x7x5xx
xfxfxx 2
i
i
2
i
3
ii
i
ii1i
)()(/
Newton Raphson formula:
Modified Newton Raphson formula:
M difi d N t R h M th d E l
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Newton Raphson Modified Newton-Raphson
Iter xi t% iter xi t%
0 0 100 0 0 100
1 0.4286 57 1 1.10526 11
2 0.6857 31 2 1.00308 0.31
3 0.83286 17 3 1.000002 00024
4 0.91332 8.7
5 0.95578 4.46 0.97766 2.2
Newton Raphson technique is linearly converging towards thetrue value of 1.0 while the Modified Newton Raphson is
quadratically converging.For simple roots, modified Newton Raphson is less efficientand requires more computational effort than the standard
Newton Raphson method
Modified Newton Raphson Method: Example
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Systems of Nonlinear Equations
Roots of a set of simultaneous equations:
f1(x1,x2,.,xn)=0
f2 (x1,x2,.,x
n)=0
fn (x1,x2,.,xn)=0
The solution is a set of x values that
simultaneously get the equations to zero.
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Systems of Nonlinear Equations
Example:x2
+ xy = 10& y + 3xy2
= 57u(x,y)= x2+ xy -10 = 0
v(x,y)= y+ 3xy2 -57 = 0
The solution will be the value of x and y which makesu(x,y)=0and v(x,y)=0
These arex=2and y=3
Numerical methods used are extension of the openmethods for solving single equation; Fixed point
iteration and Newton-Raphson. (we will only discuss
the Newton Raphson)
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Systems of Nonlinear Equations:
2.Newton Raphson Method
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Systems of Nonlinear Equations:
2.Newton Raphson Method
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x 2+ xy =10 and y + 3xy 2 = 57are two nonlinear simultaneous equations with two unknown x and y they canbe expressed in the form: use the point (1.5,3.5) as initial guess.
2
2 ,
3 , 1 6
u u
x y xx y
v vy xy
x y
i xi yi Ui Vi ui,x ui,y vi,x vi,ya,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
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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
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Problem
The polynomial f (x) = x3 -6x2 +11x- 6.1 has a real root
between 3 and 5. Apply the Newton-Raphsonsmethod to
this function using an initial guess ofx0= 3.5. Explain your
results
Problem
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Problem
The polynomial f (x) = x3 -6x2 +11x- 6.1 has a real root
between 3 and 5. Apply the Newton-Raphsonsmethod to
this function using an initial guess of x0= 3.5. Explain your
results
Problem
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Problem
The polynomial f (x) = x3 -6x2 +11x- 6.1 has a real root
between 3 and 5. Apply the Newton-Raphsonsmethod to
this function using an initial guess of x0= 3.5. Explain your
results
Problem
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Problem
The polynomial f (x) = x3 -6x2 +11x- 6.1 has a real root
between 3 and 5. Apply the Newton-Raphsonsmethod to
this function using an initial guess of x0= 3.5. Explain your
results
Problem
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Problem
Find the root of polynomial f (x) = x3 -6x2+11x- 6.1. Apply
the Secantsmethod to this function using an initial guess
of x-1= 2.5 and x0= 3.5.
Problem
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Problem
Find the root of polynomial f (x) = x3 -6x2+11x- 6.1. Apply
the Secantsmethod to this function using an initial guess
of x-1= 2.5 and x0= 3.5.
Problem
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Problem
Find the root of polynomial f (x) = x3 -6x2+11x- 6.1. Apply
the Secantsmethod to this function using an initial guess
of x-1= 2.5 and x0= 3.5.