Lectures 3 4 5

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users.isc.tuc.gr/~adelis/ENG407.html

Lectures 3-4

Dr A.I. Delis TUC

2012Part 2

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Roots of Equations

Motivationart

02

=++= cbxaxxf )( a

acbb

x 2

42 =

3sin log 4 0 ?xBut how about e x x x + + =

pprox ma on so u on pre-compu er :

f(x)

Plot method

Trial and error

x

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An Example of Roots of Equation in Engineering

Parachutists velocity: )()/( tmce

c

gmv = 1

Problem: determine the drag coefficient cfor a parachutist of a given

mass m to attain a prescribed velocity in a set time period, that is,

Given v, m,g, and t, find c.

Approach 1: try to represent c as an explicit function c =f(v, m, g, t)(fails most of time)

Approach 2: express the formula in an explicit form and solve for the

zero root

01 == vec

cf )()(

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Nonlinear Equation

Solvers

Bracketing Graphical Open Methods

Bisection Newton Raphson

a se os on

(Regula-Falsi) Secant

All Iterative

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Bracketing Methods

, Chapter 3

wo n a guesses or e rooare required. These guesses

must bracket or be on eithers e o t e root.

==

If one root of a real andcontinuous function, f(x)=0, is

bounded by values

x=x , x =x thenf(xl).f(xu)

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No answer No root

Nice case (one root)

ops wo roo s

work for a while!!)

Dr A.I. Delis TUC

2012Part 2

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Two roots( Might

work for a while!!)

function. Need

special method

Dr A.I. Delis TUC

2012Part 2

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MANY-MANY roots. What do we do?

f(x)=sin 10x+cos 3x

Dr A.I. Delis TUC

2012Part 2

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= ,

1. Pick xl and xu such that they bound the root of

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If f(xl). f[(xl+xu)/2]>0, root lies

in the upper interval, then xl=(xl+xu)/2, go to step 2.

If f(xl). f[(xl+xu)/2]=0, then + ul xxx l u .%100

2

+ ul xx

.s

a

(s tolerance)+

ul xx

or

. a s, s op. erw se repea

the process. %1002

+ ul

u

xx

Dr A.I. Delis TUC

2012Part 2

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Dr A.I. Delis TUC

201Part 2

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Step 1

Choose x and xu as two guesses for the root such that

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Step 2

Estimate the root, xm of the equation f (x) = 0 as the mid

f(x)

u

x

xm =2

xux

m

13Figure Estimate of x

m

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Step 3

Now check the following

a) If , then the root lies between x

and xm;

= =

( ) ( ) 0ml xfxfthen x

= xm; xu = xu.

0=xx m.algorithm if this is true.

14

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Find the new estimate of the root

xx

m =xu +

2

oldnew xx

Find the absolute relative approximate error

=new

m

ax

m

where

new

rootofestimateprevious=oldmx

15

m

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Compare the absolute relative approximate error with the pre-specified errora

Go to Step 2 using new

.s

Is ?

.

sa >

No Stop the algorithm

Note one should also check whether the number of iterations is more than themaximum number of iterations allowed. If so, one needs to terminate the algorithm

and notify the user about it.

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Three ways in which the interval may bracket the root:

(a) True value lies at center

(b) and (c) True value lies near the extreme

Discrepancy between true value and solution never

Dr A.I. Delis TUC

2012Part 2

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exceeds the interval length

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Error estimate equivalence

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2012Part 2

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Pros

Easy

Cons

Slow Always find root Know a and b that

required to attain an Multiple roots

computed a priori. o accoun s a en o

f(xl

) and f(xu

), if f(xl

) is

c oser o zero, s e y

that root is closer to xl .

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2012Part 2

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Length of the first Interval Lo=b-a

= o

After 2 iterations L2=Lo/4

= k k o

saa

Dr A.I. Delis TUC

2012Part 2

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o ve pro em . or a proo

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EXAMPLE

If the absolute magnitude of the error is

4

10

=s

and Lo=2, how many iterations will you

have to do to get the required accuracyin the solution?

2 44 = k .2k

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2012Part 2

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Dr A.I. Delis TUC

2012Part 2

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Algorithm

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2012Part 2

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Dr A.I. Delis TUC

2010Part 2 (Chapters 5-6-7)

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Exam le 1You are working for DOWN THE TOILET COMPANY thatmakes floats for ABC commodes. The floating ball has aspec c grav y o . an as a ra us o . cm. ou areasked to find the depth to which the ball is submerged when

floatin in water.

Figure Diagram of the floating ball

25

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Exam le 1 Cont.

The equation that gives the depth x to which the ball is

010993.3165.0423

=+

xxa) Use the bisection method of finding roots of equations

to find the depth x to which the ball is submerged under

water. Conduct three iterations to estimate the root ofthe above equation.

end of each iteration, and the number of significant

digits at least correct at the end of each iteration.

26

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Example 1 Cont.From the physics of the problem, the ball would besubmerged between x= 0 and x= 2R,

where R= radius of the ball,

that is

20 x

( )055.020 x.

Diagram of the floating ball

27

E l 1 C t

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Example 1 Cont.

Solution

To aid in the understanding of how this

method works to find the root of an

e uation the ra h of f x is shown to the

right,

where

( ) 423 1099331650 -.x.xxf +=

28Graph of the function f(x)

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Example 1 Cont.

Let us assume 00.0

=

=x

.u

Check if the function changes sign between x

and xu .

( ) ( ) ( ) ( )4423

4423

10662.210993.311.0165.011.011.0

10993.310993.30165.000

=+==

=+==

x

fxf l

Hence

( ) ( ) ( ) ( ) ( ( 010662.210993.311.00 44

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Example 1 Cont.

Graph demonstrating sign change between initial limits

30

C

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Example 1 Cont.

055.011.00

=+

=+

=uxxx

Iteration 1

The estimate of the root is22

23

( ) ( ) ( ) ( ) ( )( ) 010655.610993.3055.00......

54>==

=+==

ffxfxf

x

ml

m

Hence the root is bracketed between xm and xu, that is, between 0.055 and 0.11.

,

11.0,055.0 == ul xx,

as we do not have a previous approximation.a

31

E l 1 C t

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Example 1 Cont.

Estimate of the root for Iteration 1

32

E l 1 C t

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Example 1 Cont.

0825.011.0055.0

=+

=+

=uxxx

Iteration 2

22

( ) ( ) ( ) ( )

010655.610622.10825.0055.0

10622.110993.30825.0165.00825.00825.0

54

4423

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Example 1 Cont.

34

Estimate of the root for Iteration 2

E l 1 C t

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Example 1 Cont.

The absolute relative approximate error at the end of Iteration 2a

is

oldnew

xx =new

m

ax

1000825.0

..

=

%333.33=

one o e s gn can g s are a eas correc n e es ma e roo

of xm = 0.0825 because the absolute relative approximate error is

35

.

E l 1 C

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Example 1 Cont.

0825.0055.0=

+=

+=

uxxIteration 3

The estimate of the root is .22

m

5423

( ) ( ) ( ) ( ) ( )( ) 010563.510655.606875.0055.0......

55

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Example 1 Cont.

37Estimate of the root for Iteration 3

Example 1 Cont

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Example 1 Cont.

The absolute relative approximate error at the end of Iteration 3

isa

old

m

new

m xxnew

m

ax

10006875.0

..

=

%20=

Still none of the significant digits are at least correct in theestimated root of the equation as the absolute relative

approximate error is greater than 5%.

38

even more era ons were con uc e an ese era ons are

shown in Table 1.

Table 1 for Example 1

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Table 1 Root of f(x)=0 as function of number of iterations for bisection method.

Iteration x

xu xm a % f(xm)

1 0.00000 0.11 0.055 ---------- 6.65510

5

2

3

0.055

0.055

0.11

0.0825

0.0825

0.06875

33.33

20.00

1.622104

5.563105

4

5

0.055

0.06188

0.06875

0.06875

0.06188

0.06531

11.11

5.263

4.48410

2.593105

5

7

8

.

0.06188

0.06188

.

0.06359

0.06273

.

0.06273

0.0623

.

1.370

0.6897

.

3.176106

6.49710

7

9

10

0.0623

0.0623

0.06273

0.06252

0.06252

0.06241

0.3436

0.1721

1.265106

3.0768107

39

Table 1 Cont

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Table 1 Cont.

Hence the number of significant digits at least correct is given

b the lar est value orm for which

105.02

m

a

105.01721.0

2

2

m

m

( ) 23442.0log.

m

..og =m

2=m

The number of si nificant di its at least correct in the estimated

40root of 0.06241 at the end of the 10

th

iteration is 2.

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See also example 3-1: for a MATLAB

realization!!

Dr A.I. Delis TUC

2012Part 2 41

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The False-Position Method

(Regula-Falsi) If a real root isl u

f(x)=0, then we canapproximate the

linear interpolationbetween the points [xl,f x and x f x tofind the x

r

value suchthat l(xr)=0, l(x) is thelinear approximationof f(x).

==

Dr A.I. Delis TUC

2012Part 2 42

False Position Method: derivation

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False-Position Method: derivation

Based on two similar triangles, shown in Figure,one gets:

x x

r l r ux x x x=

The signs for both sides of Eq. (1) is consistent, since:

( ) 0; 0l r lf x x x< >

( ) 0; 0u r uf x x x> 0 then xu=xr == > fu=f(xr)

r need go no further!

Dr A.I. Delis TUC

2012Part 2 46

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. w xl xu u

convergence to be declared. If they are not go back

.

Why this method?

Faster but not alwa s

Always converges for a single root.

Note: Always check by substituting estimated root in the

r .

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Part 2 (Chapters 5-6-7) 48

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An example of slow convergence

10==


Part 2 49


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010993.3165.0423

=+=

xxxfTable 2: Root offor False-Position Method.

Iteration x x mx % x1 0.0000 0.1100 0.0660 N/A -3.1944x10-5

2 0.0000 0.0660 0.0611 8.00 1.1320x10-5

3 0.0611 0.0660 0.0624 2.05 -1.1313x10-7

4 0.0611 0.0624 0.0632377619 0.02 -3.3471x10-10

50

2

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m2

m

a

2105.002.0

.

m

21004.0

2)04.0log(

m

)3979.1(2

.

m

3

3979.3

=

mSo

m

The number of significant digits at least correct in the

51

.

is 3.

Open Methods

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p

based on formulas

that re uire onl asingle starting valueof x or two startingvalues that do notnecessarily bracket

e roo

Open methods can

e er verge orconverge (rapidly)


Part 2 52

Sim le Fixed oint Iteration

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Sim le Fixed- oint IterationRearrange the function so that x is on the

0 == xx

e s e o e equa on:

...2,1,k,given)( 1 == okk xxgx

Bracketing methods are convergent.

Fixed-point methods may sometimes

diver e de endin a on the statin oint(initial guess) and (b) how the function


Part 2 53

.

Exam le:

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Exam le:

xxxxf 02)( 2 =

xxg 2)( 2 =

or

or

xg

2

1)(+=

x


Part 2 54

Conver ence

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Conver ence

x=g x can e expresse

as a pair of equations:

y1=x

= x com onent

equations) .


Part 2 55

Conclusion

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Fixed-point iteration converges if

x)f(x)linetheofslopethethanless(i.e.1)( = xg

When the method converges, the error is

ee a so sec on . pages -

roughly proportional to or less than the error ofthe revious ste therefore it is called linearl

convergent.

100%=approx.current

approx.previous-approx.currenta sa

quadratic convergence (if

n a guess su c en y

close to root)

for functions whose

derivatives cannot beevaluated analytically.



Cases where the NR exhibits poor

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p

convergence

(a) f(x) has an inflection point near root

(b) NR oscillates around local min

(c) Jump from away from root

"ero s ope =



Example 1:NR Method

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The floating ball has a specific gravity of 0.6 and has aradius of 5.5 cm. You are asked to find the depth to which

e a s su merge w en oa ng n wa er.

The equation that gives the depth x to which the ball is submerged under water is given by

423

a) Use the NR method of finding roots of equations to find the depth x to which the ball issubmerged under water. Conduct 3iterations to estimate the root of the above equation.

.. =

62

,significant digits at least correct at the end of each iteration.

Example 1 Cont.

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Solve for ( )xf'

( ) .+x.-xxf -

'

1099331650

2

423=

x-xx .=

Let us assume the initial guess of the root of is

. . This is a reasonable guess (think why

0=xf

0 0.05mx = 0=x

and are not good choices) as the extreme values

of the depth x would be 0 and the diameter (0.11 m) of the

m11.0=x

ball.

63

Example 1 Cont.

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The estimate of the root is

( )' 00

01 =xf

xxx

( ) ( ) 10.993305.0165.005.005.0 2423

+=

10118.10.05

...

3

4

=

( )01242.00.05 =

.=

64

Example 1 Cont.

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65

Estimate of the root for the first iteration for the NR method.

Example 1 Cont.

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

xx1

=x

a

10006242.0

..

=

%90.19=

e num er o s gn cant g ts at east correct s , as you neean absolute relative approximate error of 5% or less for at least one

66

.

Example 1 Cont.

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Iteration 2The estimate of the root is

( )

'

1

12=

xfxx

( ) ( ) 10.993306242.0165.006242.0

423

1

+=

( ) ( )

1097781.3

06242.033.006242.03.

7

2

1090973.8.

5

3

=

06238.0..

=

67

Example 1 Cont.

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68

.

Example 1 Cont.

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The absolute relative approximate error at the end of Iteration 2 is

a

12

=

xx

2

xa

100

06238.0

..=

.=

The maximum value ofm for which is 2.844.

m

a

2105.0

Hence, the number of significant digits at least correct in the

answer is 2.

69

Example 1 Cont.

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Iteration 3The estimate of the root is

( )

'

223 =

xfxx

( ) ( ) 10.993306238.0165.006238.0

423

2

+=

( ) ( )06238.033.006238.03

.

11

2

1091171.8

.06238.0

3

=

06238.0

..

=

=

70

Example 1 Cont.

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Estimate of the root for the Iteration 3 for the NR method.

71

Example 1 Cont.e a so ute re at ve approx mate error at t e en o terat on

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e a so ute re at ve approx mate error at t e en o terat on

isa

10012

=xx

a

06238.006238.0

2

06238.0

=

=

The number of significant digits at least correct is 4, as only4 significant digits are carried through all the calculations.

72

On the Drawbacks of NR1 Divergence at inflection points

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1. Divergence at inflection points

Selection of the initial guess or an iteration value of the root that is close to

root in the Newton-Raphson method.

3

==, .

The Newton-Raphson method reduces to .

.

( )2

33

1

512.01 +=

+

iii

xxx

Next table shows the iterated values of the root of the equation.

ix

.

close to the inflection point of .

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

1=x

.2.0=x

73

NR Drawbacks Inflection Points

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Table : Divergence near inflection point.

Iteration

Number

xi

0 5.0000

1 3.6560

.

3 2.1084

.

5 0.92589

6 30.119

7 19.746

18 0.20003

Divergence at inflection point for

. =+= xx

74

NR Drawbacks Division by Zero

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2. Division by zero

For the equation

( )0104.203.0 623 =+= xxxf

the Newton-Raphson method

reduces to

ii xxxx104.203.0 623 +

=

ii xx .

For , the

denominator will e ual zero.

02.0or0 00 == xx

75

NR Drawbacks Oscillations near

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Results obtained from the Newton-Ra hson method ma

.

oscillate about the local maximum or minimum without

converging on a root but converging on the local maximum or

minimum.

,

and may diverge.

For example for the equation has no real

roots.

022 =+= xxf

76

NR Drawbacks Oscillations near

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5

6f(x)

Iteration

Newton-Raphson method.

4

Number

0 1.0000 3.00

ix i a

2

3

2

1

1

23

0.5

1.750.30357

2.25

5.0632.092

300.00

128.571476.47

0

-2 -1 0 1 2 3

x4

-1.75 -0.3040 0.5 3.142

4

5

6

3.1423

1.2529

0.17166

11.874

3.570

2.029

109.66

150.80

829.88-

Oscillations around local minima for ( ) 22 += xxf78

9

5.73952.6955

0.97678

34.9429.266

2.954

102.99112.93

175.96

77

NR Drawbacks Root Jumping4 Root Jumping

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4. Root Jumping

In some cases where the function is oscillating and has a( )xfnum er o roo s, one may c oose an n a guess c ose o a

root. However, the guesses may jump and converge to some1.5

.1

For example

0

0.5

-2 0 2 4 6 8 10

x( ) 0sin == xxf

Choose

-1

-0.5

- . . . .

539822.74.20 == x

=

instead of-1.5

2831853.62 == x

78

oo ump ng rom n en e oca on o roo or

.( ) 0sin == xxf

The Secant Method

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A slight variation of Newtons method for

evaluate. For these cases the derivative can be

difference.

1 1 xx ii

1

xx iii

,3,2,1)()(

)(1

11 =

=

+ i

xfxfxfx

ii

iiiii


Part 2

79

Re uires two initial

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estimates of x , e.g, xo,x1. However, because

f(x) is not required to

change signs between

,classified as a

bracketin method.

The secant method has

Newtons method.

Convergence is not

guaranteed for all xo,

f(x).


Part 2

80

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Part 2

81

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Part 2

82

Multiple Roots

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Example: f(x) =(x-3)(x-1)(x-1)(x-1)

Pose some difficulties:

.even roots (can use only open

2. In NR and secant f(x) goes to

3. NR and secant are now linearly


Part 2

83

Multi le Roots

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None of the methods deal with multiple roots, ,

problems is as follows:

)()(Set

i

i

ixf

xu

=roots at all the same

locations as the

)(

-

ixuxx =

original function

)('')()(')('

'

i

xfxfxfxf

xu

=

)(')(

)]('[ 2

iixfxf

xf


Part 2

84)('')()]('[ 2

1

iii

iixfxfxf

+

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original equation to determine whether f(xr) 0.

ASSIGMENT 1 PROBLEMS:

Dr A.I. Delis TUC2012 Part 2

85

m f n-Lin r E i n

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0),,,,( 3211 =nxxxxf

0),,,,( 3212 =nxxxxf

0),,,,( =xxxxf


86

ay or ser es expans on o a unc on o more an

i bl

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one variable

)()( ii yyu

xxu

uu

+

+=

ii vv

yx

11111 iiiiii

yx

=+++

The root of the equation occurs at the value of x

i+1 i+1 .


87

uuuu ii

iiii

ii

i11

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yyxxuyyxxiiiii

++=+++ 11

vy

vxvy

vx

v ii

i

iii

i

i

i +

+=

+

++

11 yxyx

-

unknowns that can be solved for.


88

i

i

i

i

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yvyu ii

x

vuv

x

u iiiiii

+1

uv

vu ii

vuvu

xxyyiiiiii

=+

1

xyyx Determinant of

e aco an othe system.


89

Roots of Pol nomials

Ch t 7

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Chapter 7

nxaxaxaax ++++= 2

Follow these rules:

1. For an nth order equation, there are n real or

com lex roots.

2. Ifn is odd, there is at least one real root.

3. If complex root exist in conjugate pairs (that is,

+ i and - i , where i=s rt -1 .


90

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The efficac of bracketin and o en methodsdepends on whether the problem being solved

involves com lex roots. If onl real rootsexist, these methods could be used. However,

open and bracketing methods, also the openmethods could e susceptible to divergence.

Special methods have been developed to find

Mller and Bairstow methods.


91

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Mllers method obtains a root estimate

projecting a parabola to the x axis through three

function values.


92

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The method consists of deriving the

coefficients of arabola that oes throu h thethree points:

1. Write the equation in a convenient form:

cxxbxxaxf ++= )()()( 22

22


93

2. The parabola should intersect the three points

[xo, f(x

o)], [x

1, f(x

1)], [x

2, f(x

2)]. The coefficients of the

l i l b ti t d b b tit ti th

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o o 1 1 2 2polynomial can be estimated by substituting three

points to give

cxxbxxaxf ooo ++= )()()( 22

2

cxxbxxaxf ++= )()()(

2

21

2

211

2. Three equations can be solved for three unknowns,

22222

a, b, c. Since two of the terms in the 3r equation

are zero, it can be immediately solved forc=f(x2).

)()()()( 212

2121

222

xxbxxaxfxf

xxxxaxx ooo

+=

+=


94

xxhxxh ==

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x-xhx-xh 121o1o ==

)()()()( 121

1 xfxfxfxf oo

=

=

)()( 112

11

121

hhahhbhh oooo

o

+=++

11

2

11 hahbh=

and b

211

1xcahha o

==+=


95

quadratic formula:

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quadratic formula:

xx2

23

+=

The error can be calculated as

%10023 xx

a

=3

, .This will result in a largest denominator, and will give root

estimate that is closest to x .

Dr A.I. Delis TUC

2012 Part 2

96

Once x3 is determined the process isi h f ll i i li

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Once x3 is determined, the process isr in h f ll in i lin

1. If only real roots are being located, choose the

estimate, x3.

. o rea an comp ex roo s are es ma e ,

employ a sequential approach just like in secantme o , x1, x2, an x3 o rep ace xo, x1, an x2.

See

Dr A.I. Delis TUC

2012 Part 2

97

7/28/2019 Lectures 3 4 5

98/98

Dr A.I. Delis TUC

2012 Part 2

98

Lectures 3 4 5

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