Solutions manual for an introduction to numerical analysis 1st edition by mayers and endre pages deleted

Thenumberof correct guresin theapproximationxnjg: From1.24forNewton'smethod wehave j x n+1j j x nj2 !jf00 j 2jf0 j: where B=log 10jf00 j 2jf0 j: IfB issmall thenD n+1 iscloseto2D n,but i

Thexed pointsarethesolutionsof

2(x2

n+c)

 =1

2(

2

1+c);

sob subtraction

x

n+1

 =1

2(x2

n

 )=1

2(x

n+ )(x

n

 ):

It follows that if jx

n+ j < 2, then jx

0+ >0

Hencex

1

iscloserto thenwas x

0,soalso0x

1

< Aninduction

argumentthenshowsthateach x

nsatises0x

dependentofthesignofx

0,anditfollowsthatx

n

! forallx

0such

(x)<0andf (x)>0in( 1;0)andthesameargument

showsthatthemethodconvergestothenegativerootifx

For thisfunctionf,Newton'smethod gives

x

n+1

=x

nexp(x

n) x

n2

exp(x

n) 1

=x

n

1 (x

n+2)exp( x

n)

1 exp(x

n)

isverysmall indeed

Inthesamewa ,whenx

n2

andconvergencebecomesquadraticandrapid About100iterationsare

requiredto giveanaccuratevaluefortheroot

However,when x

0

= 100, x

1

is very close to 2, and is therefore

veryclosetothenegativeroot Three,orpossiblyfour,iterationsshould

givethevalueoftheroot tosixdecimalplaces

(x

n):

Toa oidcalculatingthederivativewemightconsiderapproximatingthe

derivativeb

f0

(x

n)

f(x

n+Æ) f(x

n)

()+1

2(x

n

2

f ()+O(x

n

3

=

0

+1

2

2

0

+1

2

2

f ) f(+)

=(f0

) +1

2

2

[3f0

f +(f0

2

f(x+n+f(x

n)) f(x

n)

=

2

[(f0

) +0

f ]

[(f0

) +1

2

0

(3+f0

)f00

]+O()

=2

f (1+f

0

)

f0+O() :

Thisshowsthatifx

0

 issuÆcientlysmall theiterationconverges

quadratically The analysis here requires that f

000

is continuous in a

neighbourhoo of,tojustifythetermsO(

3

) Amorecarefulanalysis

mightrelax thistorequireonlythecontinuityoff

The leading term in x

n+1

 is very similar to that in Newton's

method,but withanadditionalterm

Theconvergenceofthismethod,startingfromapointclosetoaroot,

isverysimilartoNewton'smethod Butifx

veryp or,andthebehaviourmaybeverydierent Fortheexample

quitecloseto 2,andthenrapidconvergencetothenegativeroot.

However, starting from x

0))

is about 10

9500

; thedierence betweenx

0and x

1

is excessively small

Althoughtheiterationconverges,therateofconvergenceissoslowthat

foranypracticalpurpose itis virtually stationary Evenstarting from

x

0

=3manythousandsofiterationsarerequiredforconvergence

Thenumberof correctguresin theapproximationx

njg:

From(1.24)forNewton'smethod wehave

j x

n+1j

j x

nj2

!jf00

()j

2jf0

()j:

where

B=log

10jf00

()j

2jf0

()j:

IfB issmall thenD

n+1

iscloseto2D

n,but ifB issignicantlylarger

2(e0:0001

wherethelastcolumnshowsthenumberofcorrectdecimalplaces

Therootis =0:000099998333361to 15decimalplaces

n) f (

n)

2f0

(x

n):

Nowf

0

()=0,so b theMeanValueTheorem

f0

(x

n) f0

()=(x

n

)f00

(

n)

forsomevalueof

nbetween andx

n Hence

 x

n+1

=( x

n)f00

(

n)

2f00

(

n):

j x

n+1

j<Kj x

nj;

where

K=M

n

!

Then !,f (

n)!f ()andf (

n)!f () This showsthat

1

2

ateachstep

0

()=f ()=0wegetfromthedenitionofNewton's

method,providedthatf

f(x

n)

f0

(x

n)

= x

n+1

6(x

n

3

f000

(

n)

1

2(x

n

2

f000

(

n)

=( x

n)



1f000

(

n)

3f000

(

n)

hood,andNewton's methodconvergesto Also,

j x

n+1j

j x

nj

!2

3

;

sothatconvergenceislinear,withasymptoticrateofconvergenceln(3=2)

TheprooffollowscloselytheproofofTheorem1.9

From (1.23) it follows that x

n+1

< , provided that x

nlies in the

interval I = [X;] Since f is monotonic increasing and f() = 0,

f(x) < 0 in I Hence if x

0

2 I the sequence (x

n) lies in I, and is

monotonicincreasing Asit isbounded abo eb , it converges;since

 is theonly rootof f(x)=0in I, the sequenceconvergesto  Since

f iscontinuousitfollowsthat

 x

n+1

( x

n)

=

f (

n)

2f0

(x

n)

!

f ()

2f0

()

;

sothatconvergenceisquadratic

Neglectingtermsofsecondorderin "weget

is not exact, it is clear that for suÆciently

small"thesequenceconvergesto 1

With x

0

and x

1interchanged, thevalueof x

n 1x

n 1

f(x

n) (f(x

n 1

f(x

n))

Inthelimit asx

n

! bothnumeratorand denominator tendto zero,

soweapplyl'Hopital'sruletogive

n 1f0

(x

n)+f0

(x

n))

f0

(x

n)(x

n 1f0

n 1f0

(x

n 1f0

()

f0

(x

n 1)(x

n 1

 +(f(x

n 1

f()):

Wemustnowusel'Hopital'sruleagain,togivenally

n 1

 +f0

(x

n 1+f0

():

Nowthelimitof'doesnotdependonthewa inwhichx

nandx

():

Nowassumethat

! A1=q

;

Thisgivesaquadraticequationforq, andsinceweclearlyrequirethat

ispositiveweobtainq=

1

2(1+p

5),givingtherequiredresult

Fig 1.6 showsatypicalsituation withf (x)>0, sothe graphoff

liesbelowtheline PQ Here P and Qare thepointscorrespondingto

u

n

and v AlsoR is thepointcorrespondingto , so that f() <0

Hencein thenextiterationu

and soon Thus if f > 0in [u

N

;v

N], and f(u

N) <0< f(v

N) then

N)< 0wesee in thesame

wa thatu

n

=u

NforallnN

Similar results are easily deduced if f < 0 in [u

N

;v

N]; it is only

necessarytoreplacef b thefunction f

NowreturningtothesituationinFig 1.6,thepointv remainsxed,

and the points u

nare monotonically increasing Hence the sequence

N

f(+Æ) (f(v

N) f(+Æ))

Æ(f(v

N) f(+Æ))

:

InthelimitasÆ!0thenumeratoranddenominatorbothtendtozero,

soweapplyl'Hopital'sruletogive

Nf0

Nf0

Hencethesequence(u

n)convergeslinearlyto,andtheasymptoticrate

ofconvergenceis

ln



1(v

N

)f0

()

f(v

N)

()

f0

(

N)

0

(),andthemorerapidlytheiterationconverges

Asymptoticallythismethodconvergesmoreslowlythanthestandard

secantmethod Its advantageis that iff(u

0andf(v

0haveopposite

signstheiterationis guaranteedto convergeto aroot lying in [u

0

;v

0];

themethodis thereforerobust However,it iseasyto drawasituation

where v

0

isfar from , andwhere thebisection method is likelyto be

moreeÆcient

The sequence (x

n) converges to the two-cycle a;b if x

2n

! a and

x

2n+1

!b, orequivalently witha and b interchanged Soa and b are

xed points of the composite iteration x

n+1

= h(x

n), where h(x) =

g(g(x)), and we dene astable two-cycleto beonewhich corresponds

toastable xedpointofh Now

h0

(x)=g0

(g(x))g0

(x

n);

andthecorrespondingfunctiong isdenedb

(x)=

f(x)f00

(x)

[f0

(x)]

2:

Hence,if

f(a)f00

(a)

[f0

(a)]

2

f(b)f00

(b)

[f0

(b)]

2

<1

thetwo-cycleisstable

Newton's method for the solution of x

a

3a2

1

a= a

a3

+a

3a2

1:

Theseequationshavethesolution

a=1

p

5:

(a)

[f0

(a)]

2

f(b)f00

(b)

[f0

(b)]

2

=36;

Multiplicationb Qontherightreversestheorder ofthecolumnsof

A Hence,writingB=QAQ,

Nowsuppose that Ais ageneralnn matrix,that B =QAQ, and

that B can be written as B =L

1U

1, where L is unit lowertriangular

andU isuppertriangular Then

A=QBQ

=QL

1U

1Q

=(QL

1Q)(QU

1Q)

,andthiscan bedoneifalltheleadingprincipal submatrices

of B are nonsingular The requiredcondition is therefore that all the

\trailing"principalsubmatricesofAarenonsingular

Thefactorisationdoesnotexist,forexample,if

is uppertriangular Nowlet

D be the diagonal matrix whose diagonal elements are the diagonal

1

U



;thenA=LDU

asrequired,sinceU isunit uppertriangular

Thegiven conditiononA ensuresthat u



ii6=0fori =1;: : n The

procedureneeds tobemodied slightlywhen u

Ifthe factorisation A =LU is known, wecan usethis procedure to

ndD andU suchthat A=LDU ThenA

T

=UT

DLT

,whichcanbe

written

AT

=(UT

)(DLT

isuppertriangular Thisis

thereforetherequiredfactorisationofA

T

Suppose that therequired resultis truefor n=k, sothat any

non-singularkkmatrixAcanbewrittenasPA=LU Thisisobviously

truefork=1

Nowconsideranynonsingular(n+1)(n+1)matrixApartitioned

accordingto the rst row and column We locate the element in the

rstcolumnwithlargestmagnitude,oranyoneofthemifthereismore

thanone,and interchange rowsifrequired Ifthe largestelementis in

rowr weinterchangerows1andr Wethenwrite

P(1r)

A=



wT

p C



1mT

=wT

pmT

+C=B:

Thisgives

m=1

w;

and

C=B

1

pmT

:

Notethatif=0thisimpliesthatalltheelementsoftherstcolumn

ofAwere zero,contradictingourassumptionthatAisnonsingular

Now det (A)=det(C), andso thematrix C is also nonsingular,

andasitisannnmatrixwecanusetheinductivehypothesistowrite

A=



10T

0P





0T

P





P(1r)

PA=



0T

0 U





;

whichis therequiredfactorisationofA

Thetheorem thereforeholdsforeverymatrixofordern+1,andthe

wherethersttwoequationsareclearlyincompatible Theonlypossible

permutation matrix P interchanges the rows, and the factorisation is

obviouslystillimpossible

Partitioningthematricesb therstkrowsandcolumns,theequation

Ly=bbecomes



L

1O

wherewehaveusedthefact that therst krowsof barezero

Multi-plyingoutthis equationgives

L

1y

1

=0

Cy

1+L

2y

2are nonsingular

Hencetherstkrowsofyarezero

Columnj oftheinverseofL isy

(j)

,thesolutionof

Ly(j)

=e(j)

;

where e

(j)

is column j of the unit matrix and has its only nonzero

element in row j Hence the rst j 1 elements of e

(j)

are zero, and

b what we have just proved the rst j 1 elements of y

(j)

are zero

Thusineachcolumn oftheinverse allthe elementsabo ethediagonal

elementarezero,andtheinverseislowertriangular

TheoperationsonthematrixB are:

(i) Multiplyingalltheelementsin arowb ascalar;

(ii) Addingamultipleofonerowto another

Eachoftheseisequivalenttomultiplyingontheleftb anonsingular

matrix Evidentlytheeectofsuccessiveoperations(i)isthatthe

diag-onalelementsin therstncolumns areeach equalto1,and theeect

ofsuccessiveoperations(ii)isthatalltheodiagonalelementsareequal

tozero Hence thenal resultintherstncolumnsistheunitmatrix

Hencethecombinedeectofalltheseoperationsisequivalentto

mul-tiplyingBontheleftb A

atanystage isthe

determi-nantof theleadingprincipal rrsubmatrix of A, multiplied b each

oftheprecedingscalingfactors1=b

j Hence ifalltheleadingprincipal

submatricesofAarenonsingular,noneofthediagonalelementsb

jused

atanystagearezero

Ateachstage,thescalingoftheelementsinrowjrequires2n

multipli-cations Thecalculationofeachtermb

ikb

ijb

jkinvolvesone multipli-

cation,andthereare2n(n 1) suchelements,asrowj isnotinvolved

Thus each stage requires 2n

However, in stage j the rst j 1 columns and the last n j +1

columnsarecolumnsoftheunitmatrix Hencethescalingofrowjonly

involvesnnonzeroelements,andinthecalculatingofthenewelements

b

ik

, half of the factorsb

jkare zero This reduces the total number of

multiplicationsfrom2n

3

ton3

InthelimitasỈ!0thenumeratoranddenominatorbothtendtozero,

soweapplyl''Hopital''sruletogive

Nf0

Nf0

Hencethesequence(u

n)convergeslinearlyto,andtheasymptoticrate

ofconvergenceis...

Asymptoticallythismethodconvergesmoreslowlythanthestandard

secantmethod Its advantageis that iff(u

0andf(v

0haveopposite

signstheiterationis guaranteedto convergeto aroot lying in [u

0...

Thetheorem thereforeholdsforeverymatrixofordern+1,andthe

wherethersttwoequationsareclearlyincompatible Theonlypossible

permutation matrix P interchanges the rows, and the factorisation