Post on 02-Mar-2018
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Chap6/1
CHAPTER 6 : ROOTS OF EQUATIONS:Open Methods
LESSON OUTCOMES
To calculate roots of equation using
Open Methods:1. Newton-Raphson Method
2. Secant Method
3. Fixed-point Iteration
y
x
- when a function is zero
- when a function crossesx-axis
roots
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Chap6/2
OPEN METHODS : INTRODUCTION
Open methods are based on
formulas that require only onesingle or two starting value(s)
of x that do not need to
bracket the root.
Not always work as
sometimes it can diverge,
depending on the initial guess
When it converges, it
reaches solution much more
faster than Bracketing
method.
Bisection method Open method
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Chap6/3
NEWTON-RAPHSON METHOD
Also called as Newtons method, it is a powerful and
widely used method.
Based on Taylor series expansion; truncate the series after
1st order derivative term:
)(
)(
)(0
Re
0)(whenofvaluetheisrootThe
...)()()(
1
1
11
1
i
iii
iiii
ii
niii
xf
xfxx
xx)(xf)f(x
arranging,
xfx
Rxxfxfxf
Newton-Raphson formula
Solve forxi+1
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Chap6/4
NEWTON-RAPHSON METHOD
Convenient method for functions
whose derivatives can be evaluatedanalytically.
Not suitable for functions whose
derivatives cannot be evaluatedanalytically
It converges very fast as it is
quadratically convergent (seebook in Box 6.2, p141)
Pitfalls: cant find multiple roots, may not
converge if guess is near zero slope !
)(
)(1
i
iii
xf
xfxx
Fig. 6.5
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Chap6/5
NEWTON-RAPHSON METHOD
Rootis estimated byextending a tangent atx1
down tox-axis
)(
)(
0)()(
1
1
i
iii
ii
ii
xf
xfxx
xxxfxf
:Tangent
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Chap6/6
NEWTON-RAPHSON METHODALGORITHM
i
iii
xf
xfxx
1
STEP 1: Assume initial value forxi x0
STEP 2: Calculate the value ofxi+1using equation
STEP 3 :Calculate the approx. error |a|
STEP 4: Repeat STEP 2until a< spre-specified value,
then root =xi+1
1001
1
i
iia
x
xx
i= 0,1,2,3,
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Chap6/7
Newton-Raphson method Poor Convergence
Inflection point,f '' = 0 inthe vicinity of a root
Oscillation of slopes
Guess is near zero slope
Zero slope is encountered
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Chap6/8
CLASS ACTIVITY
Determine the highest real root off(x) =x3 6x2 + 11x 6.1
using Newton-Raphson Method. Consider three iterationswith guessxo=3.5. Also calculate the estimated error, aafter
each iteration.
i
iii
xf
xfxx
1
1001
1
i
iia
x
xx
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Chap6/9
ii
ii xf
xfxx
1
i x(i+1) xi f(xi) f'(xi) epsilon(a) %
1 3.1913 3.5000 1.7750 5.7500
2 3.0687 3.1913 0.3994 3.2576 4.00
3 3.0473 3.0687 0.0519 2.4264 0.70
SOLUTION:
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Chap6/10
SECANT METHOD
For functions whose derivatives are difficult to evaluate,
the derivatives f '(x) can be approximated byBackwardFinite-Divided Differencing scheme :
1
1)()(
)(
ii
ii
i xx
xfxf
xf
ii
ii
xf
xfxx
1:RaphsonNewton
)()(
))((:methodSecant
1
1
1
ii
iii
ii
xfxf
xxxfxx
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Chap6/11
SECANT METHOD
Requires 2 initial guesses:xi&
xi-1 that do not necessarilybracket the root. Hence, NOT abracketing method !
Similarity to false-positionmethod:
)()())((
1
11
ii
iiiii
xfxfxxxfxx
Definition: Secantis astraight line that intercepts acurve at 2 or more points
2initial guesses ofxi
andxi-1
uL
uLuur
xfxf
xxxfxx
Fig. 6.7
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Chap6/12
SECANT METHOD
Secant method requires
TWOguesses:x1andx2
Chord / Secant
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STEP 1: Assume 2 initial guesses forxi andxi-1 ati=0
STEP 2: Calculate the value ofxi+1using
STEP 3: Calculate the error |a|
STEP 4: RepeatSTEP 2until a< pre-specified value, then
set root =xi+1
NOTE: Convergenceis not guaranteed for allxoChap6/13
SECANT METHODALGORITHM
)()(
))((
1
11
ii
iiiii
xfxf
xxxfxx
1001
1
i
iia
x
xx
i= 0,1,2,3,
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Chap6/14
CLASS ACTIVITY
Use three iterations to determine the root off(x) =exx by
using Secant Method with initial estimate of x-1=0 andx0=1.0. Also calculate the estimated error, a after eachiteration.
1001
1
i
iia
x
xx
)()(
))((
1
11
ii
iiiii
xfxf
xxxfxx
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Chap6/15
f(x) = exx
x-1=0 andx0=1.0 )()(
))((
1
1
1ii
iii
ii xfxf
xxxf
xx
iter x(i+1) x(i-1) xi x(i-1)-xi f(xi) f(i-1) eps(a) %
1 0.6127 0.0000 1.0000 -1.0000 -0.6321 1.0000
2 0.5638 1.0000 0.6127 0.3873 -0.0708 -0.6321 8.67
3 0.5672 0.6127 0.5638 0.0489 0.0052 -0.0708 0.59
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Chap6/16
BRACKETING VS OPEN METHOD
BRACKETING Method
Bracketing methods
can alwaysconvergeto
a root.
Bracketing method is
slowto converge
OPEN Method
Open method might
divergeor move away
from the true root
depending on the initialguess (see pitfalls)
When it converges, it
reaches results fasterthan the bracketing
methods.
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Rearrange the function so thatxis on the left side of the
equation: x= g(x) For example: x2 x 2 = 0 can be manipulated to yield
x=x2 2
or x= (x+ 2)1/2
or x= (x+ 2)/x
From an initial estimate ofxi, the next estimate,xi+1can be
ITERATIVELYcomputed by the function ofxi+1= g(xi) The approximate percent error:
Chap6/17
FIXED-POINT ITERATION
To obtain x= g(x)
%1001
1
i
iia
x
xx
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Chap6/18
EXAMPLE
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Chap6/19
CLASS ACTIVITY
Use fixed-point iteration to find the root of x2 - 3x+2=0
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Chap6/20
Xi eps(a)
1 0.0000
2 0.6667 100.00
3 0.8148 18.18
4 0.8880 8.24
5 0.9295 4.47
6 0.9547 2.64
7 0.9705 1.63
8 0.9806 1.03
9 0.9872 0.67
SOLUTION
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x=g(x) can be expressed as a
pair of equations:f1(x) =x and
f2(x) = g(x)
Plot these two equationsseparately the intersectionof the two curves will be the
root! The method is slower to
converge because it is
linearly convergent Chap6/21
FIXED-POINT ITERATION CONVERGENCE
Figure 6.2
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Fixed-point methods may sometime diverge, depending
on the starting point (initial guess) and how the functionbehaves (see Sec 6.1.1).
It will converge if the absolute slope
When the method converges, the error is roughly
proportional to the error of the previous step, therefore it
is called linearly convergent.Chap6/22
FIXED-POINT ITERATION CONVERGENCE
1)( xg the absolute slope ofy = g(x) is lessthan the slope ofy=x, see Fig 6.3 (p137)
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Chap6/23
FIXED-POINT ITERATION CONVERGENCE
|g(x)|1
diverge
|g(x)|>1
|g(x)|
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