3. Roots of Equations Contd

8/12/2019 3. Roots of Equations Contd

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Roots of Equations: Open

methods Newton-Raphson

Secant Method Muller methods


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Newton-Raphson Method


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Newton-Raphson Method

)(x f ) f(x

-= x xi

iii 1

f(x)

f(x i)

f(x i-1)

xi+2 x i+1 x i X

ii x f x ,

Figure 1 Geometrical illustration of the Newton-Raphson method. http://numericalmethods.eng.usf.edu3


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Derivation

f(x)

f(xi)

xi+1 x i X

B

C A

)()(

1i

iii x f

x f x x

1

)()('

ii

ii

x x x f

x f

AC AB

tan(

Figure 2 Derivation of the Newton-Raphson method. 4http://numericalmethods.eng.usf.edu


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Algorithm for Newton-RaphsonMethod

5http://numericalmethods.eng.usf.edu


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Step 1

)( x f Evaluate symbolically.

http://numericalmethods.eng.usf.edu6


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Step 2

i

iii x f

x f -= x x 1

Use an initial guess of the root, , to estimate the new value of the root, , asi x 1i x



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Step 3

0101

1 x

- x x =

i

iia

Find the absolute relative approximate error as a



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Step 4

Compare the absolute relative approximate error withthe pre-specified relative error tolerance .

Also, check if the number of iterations has exceeded themaximum number of iterations allowed. If so, one needsto terminate the algorithm and notify the user.

s

Is ?

Yes

No

Go to Step 2 using newestimate of the root.

Stop the algorithm

sa



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Example 1

A polynomial is expressed as

423 1099331650 -.+ x.- x x f Use the Newtons method of finding roots of equations to finda) The root x. Conduct three iterations to estimate the root of the above

equation.

b) The absolute relative approximate error at the end of each iteration, andc) The number of significant digits at least correct at the end of each iteration.



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Example 1 Cont.

423 1099331650 -.+ x.- x x f


To aid in the understanding of how thismethod works to find the root of an

equation, the graph of f(x) is shown to theright, where

Solution

Figure 4 Graph of the function f(x)


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Example 1 Cont.


x- x x f

.+ x.- x x f -

33.03'

10993316502

423

Let us assume the initial guess of the root of is . 0 x f m05.00 x

Solve for x f '


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Example 1 Cont.

06242.0

01242.00.05

109

10118.1

0.05

05.033.005.03

10.993305.0165.005.005.0

'

3

4

2

423

0

001

x f x f

x x


Iteration 1The estimate of the root is


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Example 1 Cont.

14http://numericalmethods.eng.usf.eduFigure 5 Estimate of the root for the first iteration.


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Example 1 Cont.

%90.19

10006242.0

05.006242.0

1001

01

x x x

a


The absolute relative approximate error at the end of Iteration 1 isa

The number of significant digits at least correct is 0, as you need an absoluterelative approximate error of 5% or less for at least one significant digits to becorrect in your result.


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Example 1 Cont.

06238.0

104646.406242.0

1090973.8

1097781.306242.0

06242.033.006242.03

10.993306242.0165.006242.006242.0

'

5

3

7

2

423

1

1

12 x f

x f x x




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Example 1 Cont.

17http://numericalmethods.eng.usf.eduFigure 6 Estimate of the root for the Iteration 2.


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Example 1 Cont.

%0716.0

10006238.0

06242.006238.0

100

2

12

x

x xa



The maximum value of m for which is 2.844. Hence,the number of significant digits at least correct in the answer is 2.

m

a

2105.0


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Example 1 Cont.

06238.0

109822.406238.0

1091171.81044.406238.0

06238.033.006238.03

10.993306238.0165.006238.006238.0

'

9

3

11

2

423

2

223

x f x f

x x




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Example 1 Cont.

20http://numericalmethods.eng.usf.eduFigure 7 Estimate of the root for the Iteration 3.


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Example 1 Cont.

%0

10006238.0

06238.006238.0

100

2

12

x

x xa



The number of significant digits at least correct is 4, as only 4 significantdigits are carried through all the calculations.


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Advantages and Drawbacks ofNewton Raphson Method



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Advantages

Converges fast (quadratic convergence), if itconverges.

Requires only one guess



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Drawbacks


1. Divergence at inflection pointsSelection of the initial guess or an iteration value of the root that is closeto the inflection point of the function may start diverging away fromthe root in the Newton-Raphson method.

For example, to find the root of the equation .

The Newton-Raphson method reduces to .

Table 1 shows the iterated values of the root of the equation.

The root starts to diverge at Iteration 6 because the previous estimate of0.92589 is close to the inflection point of .

Eventually after 12 more iterations the root converges to the exact value of

x f

0512.01 3 x x f

2

33

113

512.01

i

iii

x x

x x

1 x

.2.0 x


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Drawbacks Inflection Points

IterationNumber

x i

0 5.0000

1 3.6560

2 2.7465

3 2.1084

4 1.60005 0.92589

6 30.119

7 19.746

18 0.2000 0512.01 3 x x f 25http://numericalmethods.eng.usf.edu

Figure 8 Divergence at inflection point for

Table 1 Divergence near inflection point.


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2. Division by zeroFor the equation

the Newton-Raphson methodreduces to

For , thedenominator will equal zero.

Drawbacks Division by Zero

0104.203.0 623 x x x f


ii

iiii

x x

x x x x

06.03

104.203.02

623

1

02.0or0 00 x x Figure 9 Pitfall of division by zeroor near a zero number


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Results obtained from the Newton-Raphson method may oscillateabout the local maximum or minimum without converging on a rootbut converging on the local maximum or minimum.

Eventually, it may lead to division by a number close to zero and maydiverge.

For example for the equation has no real roots.

Drawbacks Oscillations near localmaximum and minimum

022 x x f


3. Oscillations near local maximum and minimum


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Drawbacks Oscillations near localmaximum and minimum


-1

0

1

2

3

4

5

6

-2 -1 0 1 2 3

x

f(x)

3

4

2

1

- - 3.142

Figure 10 Oscillations around localminima for . 22 x x f

Iteration Number

0123

456789

1.00000.5

1.75 0.30357

3.14231.2529

0.171665.73952.69550.97678

3.002.255.0632.092

11.8743.5702.02934.9429.2662.954

300.00128.571476.47

109.66150.80829.88102.99112.93175.96

Table 3 Oscillations near local maxima andmimima in Newton-Raphson method.

i x i x f %a


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4. Root JumpingIn some cases where the function is oscillating and has a number of roots,one may choose an initial guess close to a root. However, the guesses may jumpand converge to some other root.

For example

Choose

It will converge to

instead of-1.5

-1

-0.5

0

0.5

1

1.5

-2 0 2 4 6 8 10

x

f(x)

-0.06307 0.5499 4.461 7.539822

Drawbacks Root Jumping

0sin x x f


x f

539822.74.20 x0 x

2831853.62 x Figure 11 Root jumping from intendedlocation of root for

. 0sin x x f


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Secant Method


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31

Secant Method Derivation

)(x f ) f(x

-= x xi

iii 1

f(x)

f(x i)

f(x i-1)

x i+2 x i+1 x i X

ii x f x ,

1

1 )()()(ii

iii x x

x f x f x f

)()())((

1

11

ii

iiiii x f x f

x x x f x x

Newtons Method

Approximate the derivative

Substituting Equation (2) intoEquation (1) gives the Secantmethod

(1)

(2)

Figure 1 Geometrical illustration of theNewton-Raphson method.


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32

Secant Method Derivation

)()())((

1

11

ii

iiiii x f x f

x x x f x x

The Geometric Similar Trianglesf(x)

f(x i)

f(x i-1)

xi+1 xi-1 xi X

B

C

E D A

11

1

1

)()( ii

i

ii

i

x x x f

x x x f

DE

DC

AE

AB

Figure 2 Geometrical representation ofthe Secant method.

The secant method can also be derived from geometry:

can be written as

On rearranging, the secant methodis given as


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http://numericalmethods.eng.usf.edu

33

Algorithm for Secant Method


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34

Step 1

0101

1 x - x x = iiia

Calculate the next estimate of the root from two initial guesses

Find the absolute relative approximate error

)()())((

1

1

1ii

iii

ii x f x f x x x f

x x


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35

Step 2

Find if the absolute relative approximate error is greater thanthe prespecified relative error tolerance.

If so, go back to step 1, else stop the algorithm.

Also check if the number of iterations has exceeded themaximum number of iterations.


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36

Example 1 Cont.

Use the Secant method of finding roots of equations to find the x Conduct three iterations to estimate the root of the above equation. Find the absolute relative approximate error and the number of significantdigits at least correct at the end of each iteration.

A polynomial is given as

423

1099331650-

.+ x.- x x f


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http://numericalmethods.eng.usf.edu 37

Example 1 Cont.

423 1099331650 -.+ x.- x x f

To aid in the understanding of how thismethod works to find the root of an

equation, the graph of f(x) is shown to theright,

where

Solution

Figure 4 Graph of the function f(x).


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Example 1 Cont.

Let us assume the initial guesses of the root ofas and

0 x f 02.01 x


06461.0

10993.302.0165.002.010993.305.0165.005.0

02.005.010993.305.0165.005.005.0

423423

423

10

10001 x f x f

x x x f x x

.05.00 x


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Example 1 Cont.

The absolute relative approximate error at the end of Iteration1 is

%62.22

10006461.0

05.006461.0

1001

01

x x x

a

a

The number of significant digits at least correct is 0, as you needan absolute relative approximate error of 5% or less for onesignificant digits to be correct in your result.


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Example 1 Cont.

Figure 5 Graph of results of Iteration 1.


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Example 1 Cont.


06241.0

10993.305.0165.005.010993.306461.0165.006461.0

05.006461.010993.306461.0165.006461.006461.0

423423

423

01

01112 x f x f

x x x f x x


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Example 1 Cont.

The absolute relative approximate error at the end ofIteration 2 is

%525.3

10006241.0

06461.006241.0

1002

12

x x x

a

a

The number of significant digits at least correct is 1, as you needan absolute relative approximate error of 5% or less.


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Example 1 Cont.



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Example 1 Cont.


06238.0

10993.306461.0165.005.010993.306241.0165.006241.0

06461.006241.010993.306241.0165.006241.006241.0

423423

423

12

12223 x f x f

x x x f x x


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Example 1 Cont.

The absolute relative approximate error at the end ofIteration 3 is

%0595.0

10006238.0

06241.006238.0

1003

23

x x x

a

a

The number of significant digits at least correct is 5, as you needan absolute relative approximate error of 0.5% or less.


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Iteration #3



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Advantages

Converges fast, if it converges Requires two guesses that do not need to bracket

the root


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Drawbacks

Division by zero

10 5 0 5 102

1

0

1

2

f(x) prev . guessnew guess

2

2

0

f x( )

f x( )

f x( )

1010 x x guess1 x guess2

0 xSin x f


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Drawbacks (continued)

Root Jumping

10 5 0 5 102

1

0

1

2

f(x)x'1, (first guess)x0, (previous guess)Secant linex1, (new guess)

2

2

0

f x( )

f x( )

f x( )

secant x( )

f x( )

1010 x x 0 x 1' x x 1

0Sinx x f


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50

Mller Method

M llers method obtains a root estimate by projecting a parabola to the x axis through threefunction values.


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51

Mller Method

c x xb x xa x f )()()( 22

22

The method consists of deriving thecoefficients of parabola that goes through thethree points:

1. Write the equation in a convenient form:


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52

2. The parabola should intersect the three points [x o , f(x o )],[x 1 , f(x 1 )], [x 2 , f(x 2 )]. The coefficients of the polynomial

can be estimated by substituting three points to give

3. Three equations can be solved for three unknowns, a, b,c. Since two of the terms in the 3 rd equation are zero, itcan be immediately solved for c=f(x 2 ).

c x xb x xa x f

c x xb x xa x f

c x xb x xa x f ooo

)()()(

)()()(

)()()(

222222

212

211

22

2

)()()()(

)()()()(

212

2121

22

22

x xb x xa x f x f

x xb x xa x f x f ooo


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53

)(

)()(

)()()()(

x-xhx-xh

If

2111

1

112

11

11

2

11

12

121

1

1

121o1o

x f cahbhh

a

hahbhhhahhbhh

x x x f x f

x x x f x f

o

o

oooo

o

oo

Solved for aand b


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54

Roots can be found by applying an alternative form ofquadratic formula:

The error can be calculated as

term yields two roots, the sign is chosen to agree with b . Thiswill result in a largest denominator, and will give root estimatethat is closest to x 2.

acbbc x x

42223

%1003

23

x x x

a


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55

Once x3 is determined, the process is

repeated using the following guidelines:1. If only real roots are being located, choose the

two original points that are nearest the new rootestimate, x3.

2. If both real and complex roots are estimated,employ a sequential approach just like in secantmethod, x1 , x 2, and x3 to replace xo , x 1, and x2.


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Example


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