root finding 2

8/8/2019 Root Finding 2

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1

Root Finding 2


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Open MethodsChapter 6

Open methods

are based on

formulas that

require only asingle starting

value of x or two

starting values

that do notnecessarily

bracket the root.

Figure 6.1


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Simple Fixed-point Iteration

...2,1,i,given)(

)(0)(

1 oii xx

gx

xxgxf

Bracketing methods are convergent.

Fixed-point methods may sometime diverge,depending on the starting point (initial guess) and

how the function behaves.

Rearrange the function so that x is on the left side

of the equation:


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Two curve graphical method

x=g(x) can be expressed

as a pair of equations:

y1=xy2=g(x) (component

equations)

Plot them separately.

Figure 6.2


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Convergence

Fixed-point

iteration linearly

converges if

itit EgE ,1, )(

1)( xg


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6


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7

Newton-Raphson Method

Most widely used method.

Based on Taylor series expansion:

)(

)(

)(0

g,Rearrangin

0)f(xwhenxofvaluetheisrootThe

!2)()()()(

1

1

1i1i

3

2

1

i

iii

iiii

iiii

xf

xfxx

xx)(xf)f(x

xOx

xfxxfxfxf

Newton-Raphson formula

Solve for


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8

A convenient method for

functions whosederivatives can be

evaluated analytically. It

may not be convenient

for functions whosederivatives cannot be

evaluated analytically.

And, error is as quadratic

convergence.

Fig. 6.5

2

,1,)(2

)(it

r

rit E

xf

xfE



9

Example 6.3: Use the Newton-Raphson method to estimate the root of

employing an initial guess ofx0 =0. (True root is 0.56714329...)

Ans:

xexfx)(

I xi t(%)

01234

00.5000000000.5663110030.5671431650.567143290

10011.80.1470.0000220



10

Fig. 6.6

Cases with poor

convergence



11

The Secant Method

A slight variation of Newtons method forfunctions whose derivatives are difficult toevaluate. For these cases the derivative can beapproximated by a backward finite divided

difference.

,3,2,1)()(

)(

)()()(

1

11

1

1

ixfxf

xxxfxx

xfxf

xxxf

ii

iiiii

ii

iii



12

Requires two initial

estimates of x , e.g, xo,

x1. However, becausef(x) is not required to

change signs between

estimates, it is not

classified as a

bracketing method.

The secant method has

the same properties as

Newtons method.

Convergence is not

guaranteed for all xo,

f(x).

Fig. 6.7



13

Fig. 6.8

Difference between the Secant and False-Position Methods



14

Example 6.6: Use the Secant method to estimate the root of

start with initial estimates ofx-1 =0 and x0 =1.0. (True root is 0.56714329...)

Ans: 0.61270 0.56384 0.56717

t 8.0% 0.58% 0.0048%

xexfx)(



Figure 6_09.jpg



16

Multiple Roots

None of the methods deal with multiple roots

efficiently, however, one way to deal with problems

is as follows:

)()()]([

)()()(

)(xfindThen

)(

)()(Set

21

1i

iii

iiii

i

ii

i

ii

xfxfxf

xfxfxx

xu

xux

xfxfxu This function has

roots at all the same

locations as the

original function



17

Multiple root corresponds

to a point where a function istangent to the x axis.

Difficulties

Function does not change

sign at the multiple root,therefore, cannot use

bracketing methods.

Both f(x) and f(x)=0,

division by zero with

Newtons and Secant

methods.

Fig. 6.13



18

Systems of Nonlinear Equations

0),,,,(

0),,,,(0),,,,(

321

3212

3211

nn

n

n

xxxxf

xxxxfxxxxf

Two simultaneous nonlinear equations with two

unknowns, x, and y:

0),(

0),(

yxv

yxu Fix-point iteration

Newton-Raphson



19

Taylor series expansion of a function of more than

one variable

)()(

)()(

111

111

iii

iii

ii

iii

iii

ii

yy

y

vxx

x

vvv

yyy

uxx

x

uuu

The root of the equation occurs at the value of x and y

where ui+1 and vi+1 equal to zero.

y

vy

x

vxvy

y

vx

x

v

y

uyx

uxuyy

uxx

u

ii

iiii

ii

i

ii

iiii

ii

i

11

11



20

xv

yu

yv

xu

x

uv

x

vu

yy

x

v

y

u

y

v

x

u

y

uv

y

vu

xx

iiii

ii

ii

ii

iiii

ii

ii

ii

1

1

Determinant of

theJacobian of

the system.

A set of two simultaneous equations with two

unknowns that can be solved for.


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Example 6.12: Use the multiple-equation Newton-Raphson method to determine

roots of equations

Note that a correct pair of roots isx=2 andy=3. Initiate the computation with

guesses ofx=1.5 andy=3.5.

Ans: 2.03603 2.84388

0573),(

010),(

2

2

xyyyxv

xyxyxu

root finding 2

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