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Homework4.docx

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Homework4.docx

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Homework4.docx

Homework4

Acanintheshapeofarightcircularcylinderistobeconstructedtocontain1000cm3.Thecirculartopandbottomofthecanmusthavearadiusof0.25cmmorethantheradiusofthecansothattheexcesscanbeusedtoformasealwiththeside.Thesheetofmaterialbeingformedintothesideofcanmustalsobe0.25cmlongerthanthecircumferenceofthecansothatasealcanbeformed.Find,towithin10-4,theminimalamountofmaterialneededtoconstructthecan.

Solution:

Theareaofthiscanisaboutthebelowformula

S=

Andthevolumecanbewriteas

V=

h

AndV=1000,wecancombinethistwoformulas

S（r）=

TofindtheminimumofS,wemustfindthezeropointofderivative.

First,weneedtogetknowaboutthisfunction,sowecanuseMATLABtolookforitsgraph.

Wecanseefromthisgraphthatthisfunctiondecreaseatfirstbutincreasesmonotonicallyfrom0to100.Sowhenthisderivativeequals0,thevalueoffunctionisthesmallest.Andforseekingmoreaccurately,wechangethefootstep.

Wecanseethatthesolutionisbetween5and6.

Thus,wecanstartourprogrambyusingmethodsasfollows:

Beforeallprograms,wefirstdefinethisfunctionanditsderivate.

functiony=f（x）

y=4*pi*（x+0.25）-2000/x^2-500/（pi*x^3）;

end

functions=df（x）

s=4*pi+6000/x^3+2000/（pi*x^4）;

end

Andtheareaofmaterial

functiony=s（x）

y=2*pi*（x+0.25）^2+2000/x+250/（pi*x^2）;

end

1BisectionMethod

functionp=zqh2f（a,b,TOL,N）

i=1;

FA=f（a）;

whilei<=N

p=a+（b-a）/2;

FP=f（p）;

ifFP==0||（b-a）/2

fprintf（'Theprocedurewassuccessfulafter%diterations',i）

return

end

i=i+1;

ifsign（FA）*sign（FP）>0

a=p;

FA=FP;

else

b=p;

end

fprintf（'Methodfailedafter%diterations.',N）

end

Sowecangettheansweronthecommandwindow

>>x=zqh2f（5,6,10^-4,50）

Theprocedurewassuccessfulafter14iterations

x=

5.3638

>>s（x）

ans=

573.6490

2Newton’sMethod

functionp=new（p0,TOL,N）

i=1;

whilei<=N

p=p0-f（p0）/df（p0）;

ifabs（p-p0）

fprintf（'Theprocedurewassuccessfulafter%diterations',i）

return

end

i=i+1;

p0=p;

end

fprintf（'Methodsucceededafter%diterations.',i）

end

functiony=f（x）

y=4*pi*（x+0.25）-2000/x^2-500/（pi*x^3）;

end

Sowecangettheansweronthecommandwindow

>>x=new（5.5,10^-4,50）

Theprocedurewassuccessfulafter6iterations

x=

5.3639

>>s（x）

ans=

573.6490

3SecantMethod

functionp=zqhsec（p0,p1,TOL,N）

i=2;

q0=f（p0）;

q1=f（p1）;

whilei<=N

p=p1-q1*（p1-p0）/（q1-q0）;

ifabs（p-p1）

fprintf（'Theprocedurewassuccessfulafter%diterations',i-1）

return

end

i=i+1;

p0=p1;

q0=q1;

p1=p;

q1=f（p）;

end

fprintf（'Methodfailedafter%diterations.',N）

end

>>x=zqhsec（5,6,10^-4,50）

Theprocedurewassuccessfulafter4iterations

x=

5.3639

>>s（x）

ans=

573.6490

4MethodofFalsePosition

functionp=zqhFP（p0,p1,TOL,N）

i=2;

q0=f（p0）;

q1=f（p1）;

whilei<=N

p=p1-q1*（p1-p0）/（q1-q0）;

ifabs（p-p1）

fprintf（'Themethodsucceededafter%diterations',i-1）

return

end

i=i+1;

q=f（p）;

ifq*q1<0

p0=p1;

q0=q1;

end

p1=p;

q1=q;

end

fprintf（'Methodfailedafter%diterations',N）

end

>>x=zqhFP（5,6,10^-4,50）

Themethodsucceededafter5iterations

x=

5.3639

>>s（x）

ans=

573.6490

5Muller’sMethod

functionp=muller（p0,p1,p2,TOL,N）

h1=p1-p0;

h2=p2-p1;

o1=（f（p1）-f（p0））/h1;

o2=（f（p2）-f（p1））/h2;

d=（o2-o1）/（h2+h1）;

i=3;

whilei<=N

b=o2+h2*d;

D=（b^2-4*f（p2）*d）^.5;

ifabs（b-D）

E=b+D;

else

E=b-D;

end

h=-2*f（p2）/E;

p=p2+h;

ifabs（h）

fprintf（'Themethodsucceededafter%diterations',i-2）

return

end

p0=p1;

p1=p2;

p2=p;

h1=p1-p0;

h2=p2-p1;

o1=（f（p1）-f（p0））/h1;

o2=（f（p2）-f（p1））/h2;

d=（o2-o1）/（h2+h1）;

i=i+1;

end

fprintf（'Methodfailedafter%diterations',N）

end

>>x=muller（5,5.5,6,10^-4,50）

Themethodsucceededafter3iterations

x=

5.3639

>>s（x）

ans=

573.6490

Theresultcanbesummarizedasfollows:

Method

Bisection

Newton’s

Secant

MethodofFalsePosition

Muller

Initialinput

56

5.5

56

55.56

Numberofiterations

14

6

4

5

3

r

5.3638

5.3639

S

573.6490

Wecanseethatallmethodsgettherightanswer,however,it’sbecause

Ihavegavesomeaccurateinitialinput,butinfact,itseemsimpossibletoseekfortheaccurateinitialguessabouttherealproblems,sowecanchangetheinitialguess.

Method

Bisection

Newton’s

Secant

MethodofFalsePosition

Muller

Initialinput

1100

1

1100

150100

Numberofiterations

20

14

failed

44

r

5.3639

failed

5.3639

S

573.6490

failed

573.6490

Method

Bisection

Newton’s

Secant

MethodofFalsePosition

Muller

Initialinput

10100

10

10100

1050100

Numberofiterations

failed

20

37

58

11

r

failed

5.5638

5.5639

5.3643

5.3639

S

failed

573.6490

Method

Bisection

Newton’s

Secant

MethodofFalsePosition

Muller

Initialinput

50

50100

175100

Numberofiterations

failed

23

r

failed

5.3639

S

failed

573.6490

SowecanconcludethatMuller’smethodperformedwellcomparingwithothermethods.

Asequenceconvergeswithrateif

=C,forsomefinitenonzeroconstantC.

Andwecantakethelogarithmofthis,

ln

-r

=lnC

ln

-r

=lnC

Combinethistwoformulas,wecangettheC

C=

/ln

.

Thus,nowwecancalculatetherateofconvergenceofthesemethods.

Anexample:

[m,n]=size（t）;

fori=1:

n-1

E（i）=norm（t（i）-t（n）,inf）;

end

forj=1:

n-3

ratew（j）=log（E（j+2）/E（j+1））/log（E（j+1）/E（j））;

end

[r,t]=new（5,TOL,ITER_max）;

Method

Bisection

Newton’s

Secant

MethodofFalsePosition

Muller

1

12.9557727

0.50290755

Inf

2

-0.6569398

1.37796699

-6.902630032

-6.904234927

-3.749273

3

-0.6153895

1.19964438

-0.233047526

-0.233139472

-0.354789

4

1.60420236

1.07575008

2.735474494

0.737597777

1.1663

5

0.188049

1.02830821

-0.996504932

0.820872915

1.3336679

6

6.39329123

1.03270313

-0.003496286

0.871196322

0.5711671

7

-0.5183499

1.12430688

293.0388991

0.904640953

2.6134839

8

-1.2678488

-0.982197711

0.928136772

1.4523304

9

1.63579036

-0.017833374

0.945223662

2.210862

10

-0.564575

55.41324306

0.957928826

1.7028477

11

-1

-0.977759576

0.967514815

12

-0.022314426

0.974819451

13

44.25184599

0.980425851

14

-0.969958453

0.984754232

15

-0.030232323

0.988114905

16

32.55685727

0.990740836

17

-0.957104149

0.992808853

18

-0.043432001

0.994454219

19

22.5702687

0.995781083

20

-0.933540696

0.996870232

21

-0.068227894

0.997784967

22

14.30701432

0.998575634

23

-0.883677468

0.999283147

24

-0.123955278

0.999941779

25

7.904257453

1.000581397

26

-0.753554402

1.001229315

27

-0.29973831

1.00191189

28

3.564601712

1.002655975

29

-0.307098009

1.003490354

30

-2.018977475

1.004447245

31

1.622817733

1.005564026

32

1.065575756

1.006885325

33

1.833419663

1.00846569

34

1.593124794

1.010373148

35

1.616710475

1.012694116

36

1.015540397

37

1.01905945

38

1.023449961

39

1.028986286

40

1.036058386

41

1.045240146

42

1.057412959

43

1.074005077

44

1.097497392

45

1.132622697

46

1.189708459

47

1.29662631

48

1.566476855

Thus,wecanapproximatethatBisectionmethodislinearlyconvergent,

Secant,FalsePositionmethodandtheMuller’smethodaresuperlinearconvergent.However,Newton’smethodshouldquadraticallycoverage,butit’snotforthistime,maybesomefaulthasexistedinmycodes.