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Ramin Shamshiri EGM6341, HW #3 Due 05/2/09

Homework #3

Due 05/2/09

2- Write a program implementing the algorithm Bisect given in section 2.1. Use the program to calculate

the real roots of the following equations. Use an error tolerance of =10-5

a. 32 = 0

b. 3

= 2

+ + 1c. =

1

0.1+2

d. = 1 + 0.3cos ()

Solution (a)

Root1= -0.45897

Root2= 3.733080864

Initial Guess

a=4, b=-4 a=4,b=-10 1.5

-2 2.75-1 3.375-0.5 3.6875-0.25 3.84375-0.375 3.765625-0.4375 3.7265625-0.46875 3.74609375-0.453125 3.736328125-0.4609375 3.7314453125-0.45703125 3.73388671875-0.458984375 3.732666015625-0.4580078125 3.7332763671875-0.45849609375 3.73297119140625-0.458740234375 3.73312377929688-0.4588623046875 3.73304748535156-0.45892333984375

3.73308563232422-0.458953857421875 3.73306655883789-0.458969116210938 3.73307609558105

3.73308086395264

Solution (b)

Root= 1.83927917480469

Initial Guess

a=2, b=-2

0

11.5

1.75

1.875

1.8125

1.84375

1.828125

1.8359375

1.83984375

1.837890625

1.8388671875

1.83935546875

1.839111328125

1.8392333984375

1.839294433593751.83926391601563

1.83927917480469

Solution (c)

Root=

0.64971923828125

Initial Guess

a=1, b=-1

00.5

0.75

0.625

0.6875

0.65625

0.640625

0.6484375

0.65234375

0.650390625

0.6494140625

0.64990234375

0.649658203125

0.6497802734375

0.64971923828125

Solution (d)

Root= 1.1298828125

Initial Guess

a=2, b=1

0.5

1.25

0.875

1.0625

1.15625

1.109375

1.1328125

1.12109375

1.126953125

1.1298828125

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4- Implement the algorithm Newton given in section 2.2. Use it to solve the equations in problem 2.

Solution (a)

Root1= -0.458962274194841

Root2= 0.910007572548888

Root3=3.7330790654949

=

Initial Guess X0=0 Initial Guess X0=1

-1 0.914155281832543

-0.586656659702033 0.910017665783406

-0.469801907724523 0.910007572548888

-0.459053916955023

-0.458962274194841

Initial Guess X0=3

6.31543546409326

5.47414948273021

4.751646063932224.20113467567426

3.8687230259069

3.74791688762659

3.73327895349399

3.7330790654949

Solution (b)

Root=

1.83928676250499

= + +

Initial Guess

X0=0

-1

-0.5

0.666666666666667

-1.14814814814815

-0.637079608343976

0.0516071519600516

-0.910874140183681

-0.405089904362404

2.3240180836761.96143817945071

1.85002218085057

1.83938071768498

1.83928676250499

Solution (c)

Root=

0.64975084244802

=

. +

Initial Guess

X0=00.763832077797239

0.665436995401239

0.650088831737976

0.64975084244802

Solution (d)

Root= 1.1284251543001

= + . ()

Initial Guess

X0=0

1.3

1.12952765547397

1.1284251543001

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9- Use the secant method to solve the equations given in problem 2.

Solution (a)

Root1= 0.910007571538623

Root2= -0.458962267538054

Root2=3.73307902863279

= Initial Guess

X0=1, X1=1 X0=4, X1=0

0.780202717105698 -0.714521757423027

0.902866735744908 -0.349878802434762

0.910623538896086 -0.439082896657181

0.910004960093375 -0.460866176241738

0.910007571538623 -0.458931851237079

0.780202717105698 -0.458962221693513

0.902866735744908 -0.458962267538054

0.910623538896086

0.910004960093375

0.910007571538623

X0=3, X1=4

3.51170436247579

3.68065825616918

3.74559850251939

3.73246065052258

3.73307193240732

3.73307903268198

3.73307902863279

Solution (b)

Root=

1.83928675488903

=

+ +

Initial Guess

X0=2, X1=11.66666666666667

2.125

1.8014938236139

1.83156305015595

1.83953463989082

1.83928516681885

1.83928675488903

Solution (c)

Root=

0.649750681680885

=

. +

Initial Guess

X0=0, X1=10.311406831047801

0.496402289269508

0.761987170302371

0.625110291746252

0.646106965788709

0.649879522701252

0.64975002022824

Solution (d)

Root=

= + . ()

Initial Guess

X0=0, x1=1

1.14244605487164

1.128326304321

1.12842502375615

1.12842509299257

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Figure 1: Computer Application for Problem 2, 4 and 9. Author Ramin Shamshiri, EGM6341, Homework #3, 5/Feb/2009

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X0=0.25, c=0.2

1.703125000

0.944278717

1.035261085

0.979598124

1.012489169

0.992600476

1.004472485

0.997328529

1.0016071610.999037254

1.000578204

0.999653278

1.000208105

0.999875163

1.000074912

0.999955056

1.000026967

0.999983820

X0=0.5,c=0.5

1.312500000

1.161743164

1.041356946

1.002600964

1.000010156

X0=0.8, c=0.9

0.940800000

0.961915801

0.973399024

0.980612830

0.985498534

0.988963873

0.991498738

0.993393570

0.9948324380.995937926

0.996794832

0.997463573

0.997988214

0.998401492

0.998728089

0.998986837

0.999192240

0.999355553

0.999485564

0.999589166

0.999671788

0.999737721

0.999790363

0.9998324090.999866003

0.999892851

0.999914312

0.999931469

0.999945188

0.999956159

0.999964932

0.999971949

0.999977561

0.999982050

0.999985641

0.999988513

0.999990811

0.999992649

0.9999941190.999995296

0.999996237

0.999996989

0.999997591

0.999998073

0.999998459

0.999998767

0.999999013

0.999999211

0.999999369

0.999999495

0.999999596

0.999999677

0.999999741

0.999999793

0.999999834

0.999999868

0.999999894

0.5,0.4

1.350000000

1.094150000

0.992140894

1.001645746

0.999674103

1.000065307

0.999986944

1.000002611

0.999999478

1.000000104

X0=0.5,c=0.5

1.3125

1.161743

1.0413571.002601

1.00001

X0=0.8,c=0.5

1.056

1.004792

1.000034

X0=0.7,c=0.5

1.121500000

1.023040182

1.000802390

1.000000966

X0=0.9,c=0.5

1.0145000001.000316899

1.000000151

X0=0.6,c=0.5

1.208

1.069395

1.007391

1.000082

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Prob. 24.a

=2, P=1, c=2

x error

1.5 0.5

1 1

2 0

2 0

Prob. 24.b

=1.44225, P=2, c=1

x error

3 1.55775043

2.111111 0.66886154

1.631784 0.18953457

1.463412 0.02116242

1.442554 0.00030455

1.44225 6.4294E-08

1.44225 2.8866E-15

1.44225 0

1.44225 0

1.44225 0

1.44225 0

1.44225 0

Prob. 24.c

=3, P=1, c=0.75x error

2.5 0.5

3.428571 0.42857143

2.709677 0.290322583.234783 0.234782612.833676 0.16632444

3.130155 0.13015533

2.90546 0.094540273.072622 0.07262162

2.946505 0.053494993.040665 0.04066509

2.969808 0.030191883.022816 0.02281612

2.982985 0.01701504

3.012816 0.01281579

2.990419 0.00958115

3.007203 0.00720311

2.994607 0.005392623.00405 0.00404993

2.996966 0.00303437

3.002278 0.002277512.998293 0.00170716

3.001281 0.00128092

2.99904 0.00096038

3.00072 0.00072046

2.99946 0.00054025

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pp.117-127 of Atkinsons textbook

#2 Bisection method

Solutions:

a) f(x) = exp(x)-3x2 = 0

Finding r1

n x0 x1 f0 f1 x2 f2 |xn-xn-1|

1 0 -1 1 -2.6321206 -0.5 -0.14347

2 0 -0.5 1 -0.1434693 -0.25 0.591301 2.500E-01

3 -0.5 -0.25 -0.1434693 0.59130078 -0.375 0.265414 1.250E-01

4 -0.5 -0.375 -0.1434693 0.26541428 -0.4375 0.07143 6.250E-02

5 -0.5 -0.4375 -0.1434693 0.07142978 -0.46875 -0.0334 3.125E-02

6 -0.4375 -0.46875 0.0714298 -0.0333957 -0.453125 0.019672 1.563E-02

7 -0.46875 -0.453125 -0.0333957 0.01967188 -0.4609375 -0.0067 7.813E-03

8 -0.453125 -0.4609375 0.0196719 -0.006698 -0.45703125 0.006528 3.906E-03

9 -0.4609375 -0.4570313 -0.006698 0.00652786 -0.45898438 -7.5E-05 1.953E-03

-8

-6

-4

-2

0

2

4

-1 0 1 2 3 4

f

x

-10

0

10

20

30

40

50

-1 0 1 2 3

exp(x)

3*x^2

x

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10 -0.4570313 -0.4589844 0.0065279 -7.485E-05 -0.45800781 0.003229 9.766E-04

11 -0.4589844 -0.4580078 -7.485E-05 0.00322907 -0.45849609 0.001578 4.883E-04

12 -0.4589844 -0.4584961 -7.485E-05 0.00157775 -0.45874023 0.000752 2.441E-04

13 -0.4589844 -0.4587402 -7.485E-05 0.00075161 -0.45886230 0.000338 1.221E-04

14 -0.4589844 -0.4588623 -7.485E-05 0.00033842 -0.45892334 0.000132 6.104E-05

15 -0.4589844 -0.4589233 -7.485E-05 0.00013179 -0.45895386 2.85E-05 3.052E-05

16 -0.4589844 -0.4589539 -7.485E-05 2.8474E-05 -0.45896912 -2.3E-05 1.526E-05

17 -0.4589539 -0.4589691 2.847E-05 -2.319E-05 -0.45896149 2.64E-06 7.629E-06

Finding r2

n x0 x1 f0 f1 x2 f2 |xn-xn-1|

1 0 1 1 -0.2817182 0.5 0.898721

2 1 0.5 -0.2817182 0.89872127 0.75 0.4295 2.500E-01

3 1 0.75 -0.2817182 0.42950002 0.875 0.102 1.250E-01

4 1 0.875 -0.2817182 0.10200029 0.9375 -0.08313 6.250E-02

5 0.875 0.9375 0.1020003 -0.0831293 0.90625 0.011157 3.125E-02

6 0.9375 0.90625 -0.0831293 0.01115658 0.921875 -0.03556 1.563E-02

7 0.90625 0.921875 0.0111566 -0.0355608 0.9140625 -0.0121 7.813E-03

8 0.90625 0.9140625 0.0111566 -0.0120951 0.91015625 -0.00044 3.906E-03

9 0.90625 0.91015625 0.0111566 -0.0004425 0.908203125 0.005364 1.953E-03

10 0.9101563 0.90820313 -0.0004425 0.00536378 0.909179688 0.002462 9.766E-04

11 0.9101563 0.90917969 -0.0004425 0.00246234 0.909667969 0.00101 4.883E-04

12 0.9101563 0.90966797 -0.0004425 0.00101036 0.909912109 0.000284 2.441E-04

13 0.9101563 0.90991211 -0.0004425 0.00028405 0.91003418 -7.9E-05 1.221E-04

14 0.9099121 0.91003418 0.0002841 -7.918E-05 0.909973145 0.000102 6.104E-05

15 0.9100342 0.90997314 -7.918E-05 0.00010245 0.910003662 1.16E-05 3.052E-05

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16 0.9100342 0.91000366 -7.918E-05 1.1636E-05 0.910018921 -3.4E-05 1.526E-05

17 0.9100037 0.91001892 1.164E-05 -3.377E-05 0.910011292 -1.1E-05 7.629E-06

Finding r3

n x0 x1 f0 f1 x2 f2 |xn-xn-1|

1 1 4 -0.2817182 6.59815003 2.5 -6.56751

2 4 2.5 6.59815 -6.567506 3.25 -5.89716 7.500E-01

3 4 3.25 6.59815 -5.8971601 3.625 -1.89715 3.750E-01

4 4 3.625 6.59815 -1.8971518 3.8125 1.657987 1.875E-01

5 3.625 3.8125 -1.8971518 1.65798743 3.71875 -0.27446 9.375E-02

6 3.8125 3.71875 1.6579874 -0.2744588 3.765625 0.650897 4.688E-02

7 3.71875 3.765625 -0.2744588 0.65089669 3.7421875 0.178278 2.344E-02

8 3.71875 3.7421875 -0.2744588 0.17827846 3.73046875 -0.05054 1.172E-02

9 3.7421875 3.73046875 0.1782785 -0.0505415 3.736328125 0.063251 5.859E-03

10 3.7304688 3.73632813 -0.0505415 0.06325148 3.733398438 0.006201 2.930E-03

11 3.7304688 3.73339844 -0.0505415 0.00620129 3.731933594 -0.02221 1.465E-03

12 3.7333984 3.73193359 0.0062013 -0.0222085 3.732666016 -0.00801 7.324E-04

13 3.7333984 3.73266602 0.0062013 -0.0080132 3.733032227 -0.00091 3.662E-04

14 3.7333984 3.73303223 0.0062013 -0.0009083 3.733215332 0.002646 1.831E-04

15 3.7330322 3.73321533 -0.0009083 0.00264587 3.733123779 0.000869 9.155E-05

16 3.7330322 3.73312378 -0.0009083 0.00086861 3.733078003 -2E-05 4.578E-05

17 3.7331238 3.733078 0.0008686 -1.991E-05 3.733100891 0.000424 2.289E-05

18 3.733078 3.73310089 -1.991E-05 0.00042434 3.733089447 0.000202 1.144E-05

19 3.733078 3.73308945 -1.991E-05 0.00020221 3.733083725 9.12E-05 5.722E-06

Need 16, 16 & 19 iterations for the error |xn-xn-1| to reach below tolerance 0.00001 for the three roots.

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b) f(x) = x3 - x2 x -1 = 0

n x0 x1 f0 f1 x2 f2 |xn-xn-1|

1 2 -1 1 -2 0.5 -1.625

2 2 0.5 1 -1.625 1.25 -1.85938 7.500E-01

3 2 1.25 1 -1.85938 1.625 -0.97461 3.750E-01

4 2 1.625 1 -0.97461 1.8125 -0.14331 1.875E-01

5 2 1.8125 1 -0.14331 1.90625 0.386871 9.375E-02

6 1.8125 1.90625 -0.14331 0.386871 1.859375 0.111721 4.688E-02

7 1.8125 1.859375 -0.14331 0.111721 1.8359375 -0.01827 2.344E-02

8 1.859375 1.835938 0.111721 -0.01827 1.8476563 0.046101 1.172E-02

9 1.835938 1.847656 -0.01827 0.046101 1.8417969 0.01376 5.859E-03

10 1.835938 1.841797 -0.01827 0.01376 1.8388672 -0.00229 2.930E-03

11 1.841797 1.838867 0.01376 -0.00229 1.840332 0.005723 1.465E-03

12 1.838867 1.840332 -0.00229 0.005723 1.8395996 0.001712 7.324E-04

13 1.838867 1.8396 -0.00229 0.001712 1.8392334 -0.00029 3.662E-04

14 1.8396 1.839233 0.001712 -0.00029 1.8394165 0.00071 1.831E-04

y

-10

-5

0

5

10

-2 -1 0 1 2 3

y

x

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15 1.839233 1.839417 -0.00029 0.00071 1.839325 0.000209 9.155E-05

16 1.839233 1.839325 -0.00029 0.000209 1.8392792 -4.1E-05 4.578E-05

17 1.839325 1.839279 0.000209 -4.1E-05 1.8393021 8.37E-05 2.289E-05

18 1.839279 1.839302 -4.1E-05 8.37E-05 1.8392906 2.11E-05 1.144E-05

19 1.839279 1.839291 -4.1E-05 2.11E-05 1.8392849 -1E-05 5.722E-06

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c) f(x)=exp(x)-1/(0.1+x2)

n x0 x1 f0 f1 x2 f2 |xn-xn-1|

1 0 1 -9 1.809191 0.5 -1.20842

2 1 0.5 1.809191 -1.20842 0.75 0.607566 2.500E-01

3 0.5 0.75 -1.20842 0.607566 0.625 -0.16997 1.250E-01

4 0.75 0.625 0.607566 -0.16997 0.6875 0.242489 6.250E-02

5 0.625 0.6875 -0.16997 0.242489 0.65625 0.043119 3.125E-02

6 0.625 0.65625 -0.16997 0.043119 0.640625 -0.06158 1.563E-02

7 0.65625 0.640625 0.043119 -0.06158 0.648438 -0.00879 7.813E-03

8 0.65625 0.648438 0.043119 -0.00879 0.652344 0.017276 3.906E-03

9 0.648438 0.652344 -0.00879 0.017276 0.650391 0.004272 1.953E-03

10 0.648438 0.650391 -0.00879 0.004272 0.649414 -0.00225 9.766E-04

11 0.650391 0.649414 0.004272 -0.00225 0.649902 0.001013 4.883E-04

12 0.649414 0.649902 -0.00225 0.001013 0.649658 -0.00062 2.441E-04

13 0.649902 0.649658 0.001013 -0.00062 0.64978 0.000198 1.221E-04

14 0.649658 0.64978 -0.00062 0.000198 0.649719 -0.00021 6.104E-05

15 0.64978 0.649719 0.000198 -0.00021 0.64975 -6.2E-06 3.052E-05

16 0.64978 0.64975 0.000198 -6.2E-06 0.649765 9.58E-05 1.526E-05

17 0.64975 0.649765 -6.2E-06 9.58E-05 0.649757 4.48E-05 7.629E-06

y

-10

-6

-2

2

6

-3 -2 -1 0 1 2

y

x

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d) f(x)=x-1-0.3cos(x)

n x0 x1 f0 f1 x2 f2 |xn-xn-1|

1 0 2 -1.3 1.124844 1 -0.16209

2 2 1 1.124844 -0.16209 1.5 0.478779 5.000E-01

3 1 1.5 -0.16209 0.478779 1.25 0.155403 2.500E-01

4 1 1.25 -0.16209 0.155403 1.125 -0.00435 1.250E-01

5 1.25 1.125 0.155403 -0.00435 1.1875 0.075306 6.250E-02

6 1.125 1.1875 -0.00435 0.075306 1.15625 0.035418 3.125E-02

7 1.125 1.15625 -0.00435 0.035418 1.140625 0.015517 1.563E-02

8 1.125 1.140625 -0.00435 0.015517 1.1328125 0.005578 7.813E-03

9 1.125 1.132813 -0.00435 0.005578 1.12890625 0.000612 3.906E-03

10 1.125 1.128906 -0.00435 0.000612 1.12695313 -0.00187 1.953E-03

11 1.128906 1.126953 0.000612 -0.00187 1.12792969 -0.00063 9.766E-04

12 1.128906 1.12793 0.000612 -0.00063 1.12841797 -9.1E-06 4.883E-04

13 1.128906 1.128418 0.000612 -9.1E-06 1.12866211 0.000301 2.441E-04

14 1.128418 1.128662 -9.1E-06 0.000301 1.12854004 0.000146 1.221E-04

15 1.128418 1.12854 -9.1E-06 0.000146 1.12847900 6.85E-05 6.104E-05

16 1.128418 1.128479 -9.1E-06 6.85E-05 1.12844849 2.97E-05 3.052E-05

17 1.128418 1.128448 -9.1E-06 2.97E-05 1.12843323 1.03E-05 1.526E-05

18 1.128418 1.128433 -9.1E-06 1.03E-05 1.12842560 6.42E-07 7.629E-06

y

-3

-2

-1

0

1

2

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

y

x

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#4 Newtons method

Solutions:

a) f(x) = exp(x)-3x2

= 0

Finding r1

n r1 |xn-xn-1|

0 -1

1 -0.58665666 0.4133433

2 -0.469801908 0.1168548

3 -0.459053917 0.010748

4 -0.458962274 9.164E-05

5 -0.458962268 6.658E-09

Finding r2

n xn |xn-xn-1|

0 2

1 1 1

2 0.91415528 0.085844718

3 0.91001767 0.004137616

4 0.91000757 1.00932E-05

5 0.91000757 6.0179E-11

Finding r3

n xn |xn-xn-1|

0 3

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1 6.315435464 3.315435

2 5.474149483 8.413E-01

3 4.751646064 7.225E-01

4 4.201134676 5.505E-01

5 3.868723026 3.324E-01

6 3.747916888 1.208E-01

7 3.733278953 1.464E-02

8 3.733079065 1.999E-04

9 3.733079029 3.686E-08

b) f(x) = x3

- x2 x -1 = 0

n xn |xn-xn-1|

0 3

1 2.3 0.7

2 1.95170399 3.483E-01

3 1.84847377 1.032E-01

4 1.83935569 9.118E-03

5 1.83928676 6.894E-05

6 1.83928676 3.925E-09

c) f(x)=exp(x)-1/(0.1+x2)

n xn |xn-xn-1|

0 1

1 0.58610873 0.41389127

2 0.645119782 0.059011052

3 0.649727346 0.004607564

4 0.649750681 2.33349E-05

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5 0.649750682 5.89759E-10

d) f(x)=x-1-0.3cos(x)

n xn |xn-xn-1|

0 -2

1 1.953676238 3.953676

2 1.119933931 0.833742

3 1.128428782 0.008495

4 1.128425093 3.69E-06

#9 Secant method

Solutions:

a) f(x) = exp(x)-3x2

= 0

Finding r1

n x0 x1 x2 f0 f1 f2 error

1 0 -1 -0.2753213 1 -2.63212 0.5319228

2 -1 -0.27532126 -0.3971505 -2.63212 0.531923 0.1990472 1.22E-01

3 -0.2753213 -0.39715054 -0.4699999 0.53192 0.199047 -0.037697 7.28E-02

4 -0.3971505 -0.46999990 -0.4583999 0.19905 -0.0377 0.0019031 1.16E-02

5 -0.4699999 -0.45839993 -0.4589574 -0.0377 0.001903 1.653E-05 5.57E-04

6 -0.4583999 -0.45895739 -0.4589623 0.0019 1.65E-05 -7.37E-09 4.88E-06

Finding r2


1 0 1 0.78020272 1 -0.28172 0.3557657

2 1 0.78020272 0.90286674 -0.28172 0.355766 0.0211592 1.23E-01

3 0.78020272 0.90286674 0.91062354 0.35577 0.021159 -0.001834 7.76E-03

4 0.90286674 0.91062354 0.91000496 0.02116 -0.00183 7.774E-06 6.19E-04

5 0.91062354 0.91000496 0.91000757 -0.00183 7.77E-06 2.827E-09 2.61E-06

Finding r3

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1 2 4 2.82271483 -4.61094 6.59815 -7.080698

2 4 2.82271483 3.43212293 6.59815 -7.0807 -4.396142 6.09E-01

3 2.82271483 3.43212293 4.43006988 -7.0807 -4.39614 25.060725 9.98E-01

4 3.43212293 4.43006988 3.58105651 -4.39614 25.06072 -2.560436 8.49E-01

5 4.43006988 3.58105651 3.65975864 25.0607 -2.56044 -1.329536 7.87E-02

6 3.58105651 3.65975864 3.74476739 -2.56044 -1.32954 0.2293184 8.50E-02

7 3.65975864 3.74476739 3.73226200 -1.32954 0.229318 -0.015846 1.25E-02

8 3.74476739 3.73226200 3.73307027 0.22932 -0.01585 -0.00017 8.08E-04

9 3.73226200 3.73307027 3.73307904 -0.01585 -0.00017 1.282E-07 8.76E-06

b) f(x) = x3 - x2 x -1 = 0


1 3 4 2.517241379 14 43 6.096765

2 4 2.517241 2.272275572 43 6.096765 3.296784 2.45E-01

3 2.517241 2.272276 1.983845303 6.096765 3.296784 0.888218 2.88E-01

4 2.272276 1.983845 1.87747957 3.296784 0.888218 0.215574 1.06E-01

5 1.983845 1.87748 1.843390652 0.888218 0.215574 0.022526 3.41E-02

6 1.87748 1.843391 1.839412969 0.215574 0.022526 0.000691 3.98E-03

7 1.843391 1.839413 1.839287182 0.022526 0.000691 2.33E-06 1.26E-04

8 1.839413 1.839287 1.839286755 0.000691 2.33E-06 2.43E-10 4.27E-07

c) f(x)=exp(x)-1/(0.1+x2)


1 1 3 0.80082071 1.809191 19.97565 0.8784118

2 3 0.800821 0.69966549 19.97565 0.878412 0.3168178 1.01E-01

3 0.800821 0.699665 0.64259974 0.878412 0.316818 -4.815E-02 5.71E-02

4 0.699665 0.6426 0.6501283 0.316818 -0.04815 2.522E-03 7.53E-03

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5 0.6426 0.650128 0.64975362 -0.04815 0.002522 1.961E-05 3.75E-04

6 0.650128 0.649754 0.64975068 0.002522 1.96E-05 -8.016E-09 2.94E-06

d) f(x)= x-1-0.3cos(x)


1 1 3 1.1318299 -0.16209 2.296998 0.00432867

2 3 1.13183 1.12830271 2.296998 0.004329 -0.0001556 3.53E-03

3 1.13183 1.128303 1.12842507 0.004329 -0.00016 -2.669E-08 1.22E-04

4 1.128303 1.128425 1.12842509 -0.00016 -2.7E-08 1.6509E-13 2.10E-08

#13 Newtons method is the commonly used method for calculating square roots on a computer. To use

Newtons method to calculate square root of a, an initial guess x0 must be chosen, and it would be most

convenient to use a fixed number of iterates rather than having to test for a convergence. For definiteness,

suppose that the computer arithmetic is binary and that the mantissa contains 48 binary bits. Write = 2 ,

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40

212

80

72

22

32

1

2

1||

xxx

e

x

ee ,

8041223

160

153

23

4

2

1

2

1||

xxxx

e

x

ee

Since4

1

0 x ,2

1 bxn for n>0, and

24

1|| 0 e , it follows that

20

8715164 1012.2

)4

1()

2

1(

1

2

1

24

1||

e < 482

=15

106.3

Thus 4 iterations are definitely enough to reduce the error on the 4th

iteration to less than machine

epsilon.

On the other hand, since 3x will have only half of the significant digits comparing with 4x , it is likely

that || 3e will be on the order of )10(10

O or a little smaller. (In reality, through trial-and-error, we

can find that the maximum || 3e occurs at b=0.5075 with || 3e =13

100.8

which is higher than

the specified tolerance). Thus 3 iterations are NOT enough to bring the error down to48

2 level.

Hence 4 iterations are required number of Newtons iteration.

Improvement of x0:

Consider 00 xbe . In order to reduce the initial error 0e , 0x should be chosen to follow

b . The guess given by )12(3

10 bx gives correct values at b=1/4 and 1 since )12(

3

10 bx is a

linear interpolation between (b, x0) = (1/4, ) and (1, 1). To improve the initial guess, we can use a quadratic

fit to match three points:

(b, x0) = (1/4, ), ( , 0.70710678), and (1, 1).

The following fit )5.0)(25.0(0.32352092)25.0(50.828427125.00 bbbx

match those three points exactly.

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Of course, you can use different points such as (b, x0) = (1/4, ), (5/8, 0.790569415), and (1, 1) to improve the

initial guess.

#21 Solution:

a) For x = x + c f(x) = g(x) to converge

we need |g(x)| =|1 + c f(x)|

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Solution:

First we observe from the given iteration that

3( ) 2 (1 )g x c x cx

Thus the first order derivative of g(x) is:

2'( ) (1 ) 3g x c cx

Since the iteration converges to =1, the derivative at the root is

'( ) (1 ) 3 1 2g c c c

Convergence requires | '( ) | 1g so that

1 1 2 1c

Thus 0 1c

In order to obtain quadratic convergence, we need '( ) 0g which requires

1 2 0c

so that c= .

#24 Solution:

a) g(x) = -16 + 6x + 12/x, r=2

g(x) = 6-12/x2

=> g(2) = 6-12/4=3>1 => will NOT converge.

b)2

1

3

2)(

xxxg , r =31/3

g(x) = 2/3 2/x3

=> g(31/3

) = 2/3- 2/3 = 0 will CONVERGE.

Furthermore, because g(r)=0, it will converge quadratically (p=2).

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c) g(x) = 12/(1+x) r=3

g(x) = -12/(1+x)2 => g(3) = -12/16=-0.75. => will converge.

Rate of convergence near root is=| g(3)| = 0.75.

Since g(r)0 (which is a reasonable thing to assume for finding square root), we can see that

from (1),

1x > 0.

Assume it is true for kx (i.e. kx > 0), then 1kx > 0 follows from (1).

Second, since

ax

axa

ax

axxax

n

n

n

nnn

2

3

2

2

13

)(

3

)3((2)

if ax 0 , then ax 1 based on (2). Hence axn using induction.

If ax 0 then ax 1 based on (2). Hence axn using induction.

Third, since

ax

xaxxx

n

nnnn

2

2

13

)(2we get

nn xxa 1 if ax 0 ,

axx nn 1 if ax 0 .

Thus the limit of nx exists.

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Forth, to find the limit (assumed to be L) we get

aL

aLLL

2

2

3

)3(

It can be easily found that L= a .

b. (2) =>

ax

ee

n

nn

2

3

13

=>

axe

e

nn

n

23

1

3

1

In the limit n becomes large, axn so thatae

e

n

n

4

1

3

1

. Thus the order of convergence is 3.

#39 Solution:

Results of the Newtons method using single precision:

n zn f(zn)

1 (2.462611, 4.645821) (2.810165, -0.9090271)

2 (2.470614, 4.640501) (4.7531128E-03, 1.1505127E-02)

3 (2.470639, 4.640533) (-7.6293945E-05, 4.5776367E-05)

4 (2.470639, 4.640533) (1.2969971E-04, 4.5776367E-05)

egm6341_sol_hw_03

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