44 C H A P T E R 2 Solutions ofEquations ofOneVariable
17. The particle inthefigure starts atrestonasmooth inclined plane whose angle 6ischanging ata
constant rate
Attheendoftseconds ,theposition oftheobject isgiven by
-sincur
^.
Suppose theparticle hasmoved 1.7ftin1s.Find,towithin 10-5,therate IDatwhich 0changes .
Assume thatg=-32.17 ft/s2.*(0=8
2w2\(e°*-e**
0(t)
2.4Newton 'sMethod
Isaac Newton (1641-1727 )was
oneofthemost brilliant scientists
ofalltime.Thelate17thcentury
wasavibrant period forscience
andmathematics andNewton ’s
work touches nearly every aspect
ofmathematics .Hismethod
forsolving wasintroduced to
findarootoftheequation
y3—2y—5=0,Although he
demonstrated themethod only for
polynomials ,itisclear thathe
realized itsbroader applications .TheBisection andSecant methods both have geometric representations thatusethezero of
anapproximating linetothegraph ofafunction /toapproximate thesolution tof(x)=0.
Theincrease inaccuracy oftheSecant method over theBisection method isaconsequence
ofthefactthatthesecant linetothecurve better approximates thegraph of/thandoes the
lineused togenerate theapproximations intheBisection method .
Thelinethatbest approximates thegraph ofthefunction atapoint onitsgraph isthe
tangent linetothegraph atthatpoint.Using thislineinstead ofthesecant lineproduces
Newton ’smethod (also called theNewton -Raphson method ),thetechnique weconsider
inthissection .
Newton 'sMethod
Suppose that poisaninitial approximation totheroot poftheequation f(x)=0andthat
/'exists inaninterval containing alltheapproximations top.Theslope ofthetangent line
tothegraph of/atthepoint (po,/(po))is/'(po),sotheequation ofthistangent lineis
y-/(Po)=f\po)(x-Po).
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Figure 2.6
Newton 'sMethod
Example 1
Program NEWTON 24
implements Newton ’s
method .This tangent linecrosses thex-axiswhen the^-coordinate ofthepoint onthelineis0,
sothenext approximation ,putopsatisfies
0-f(po)=f'(po)(Pi-Po),
which implies that
Pi Po-f(Po)
f'(Po)'
provided that f'(po)^0.Subsequent approximations arefound forpinasimilar manner ,
asshown inFigure 2.6.
y
-
Slope/'(p,)/^y=/(x)
/iPiJ (Pi))
po p/fSlope/'(po)
X
'^^KPuJiPo ))
Theapproximation p„+\toarootoff(x)=0iscomputed from theapproximation p n
using theequation
Pn+1f(Pn)
f'(Pn)'
provided that f'(pn)i10.
UseNewton ’smethod withp0=1toapproximate therootoftheequation x3-f4x2-10=0.
Solution WewilluseMATLAB tofindthefirsttwoiterations ofNewton ’smethod with
Po=1.Wefirstdefine/(*),/'(*),and pOwith
f-inline (»x-3+4*x~2-10»,»x1)
fp=inlineC,3*x"2+8*x>,’x*)
p0=l
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Thefirstiteration ofNewton ’smethod gives p1=1.454545454545455 using thecommand
pl=pO-f(pO)/fp(pO)
Thesecond iteration isp2=1.368900401069519 using
p2-pl-f(pl)/fp(pl)
andsoon.Theprocess canbecontinued togenerate theentries inTable 2.4.This table was
generated using p0=1,TOL=0.0005 ,andNo=20intheprogram NEWTON 24.Note
thatwehave
i p-P*\**io-10.
Ifwecompare theconvergence ofthismethod with those applied tothisproblem
previously ,wecanseethatthenumber ofiterations needed tosolve theproblem byNewton ’s
method islessthan thenumber needed fortheSecant method .Recall thattheSecant method
required lessthan halftheiterations needed fortheBisection method .
Table 2.4n Pn f i P n )
1 1.4545454545 1.5401953418
2 1.3689004011 0.0607196886
3 1.3652366002 0.0001087706
4 1.3652300134 0.0000000004
Convergence Using Newton 'sMethod
Newton ’smethod generally produces accurate results injustafewiterations .With theaid
ofTaylor polynomials wecanseewhy thisistrue.Suppose pisthesolution tof(x)=0
andthat f"exists onaninterval containing both pandtheapproximation pn.Expanding
/inthefirstTaylor polynomial atpngives
f” (t\
fM=fipn )+f'(pn)(x-P n)+ -pn)2,
andevaluating atJC=pgives
f"(£\
0=f i p)=f i P n )+f i P n )(p-P n)+ ~P^>
where fliesbetween pnand p.Consequently ,iff'(pn)^0,wehave
Since
thisimplies that,f(P n)p-Pn+-r,/'(Pn) 2/'(P)'^(p-Pn)2-
P"+'~P" f’iPnY
P~Pn+l=P-Pn+/(P»)
/'(Pn)/"($)
2/'(Pn)(P“ Pn)2-
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Ifapositive constant Mexists with|/"(JC)|<Monaninterval about p,andifpnis
within thisinterval ,then
M
The important feature ofthisinequality isthatif/'isnonzero ontheinterval ,then the
error|p-pn+1|ofthe(n+l)stapproximation isbounded byapproximately thesquare
oftheerror ofthenthapproximation ,|p— pn|.This implies thatNewton ’smethod has
thetendency toapproximately double thenumber ofdigits ofaccuracy with each successive
approximation .Newton ’smethod isnot,however ,infallible ,aswewillseelater inthis
section .
Example 2Find anapproximation tothesolution oftheequation x=3xthatisaccurate towithin
io-8.
Solution Asolution tox=3xcorresponds toasolution of
0=f(x)=x-3“ \
Since/iscontinuous with/(0)=— 1and/(1)= asolution oftheequation liesinthe
interval (0,1).Wehave chosen theinitial approximation tobethemidpoint ofthisinterval ,
P o=0.5.Succeeding approximations aregenerated byapplying theformula
__„ /0»«)
Pn+l~P"f{Pn)Pn~3-*
1+3-?nIn3’
These approximations arelisted inTable 2.5,together with differences between successive
approximations .Since Newton ’smethod tends todouble thenumber ofdecimal places of
accuracy with each iteration ,itisreasonable tosuspect that p$iscorrect atleast tothe
places listed.
Table 2.5n Pn1i
0 0.500000000
1 0.547329757 0.047329757
2 0.547808574 0.000478817
3 0.547808622 0.000000048
Thesuccess ofNewton ’smethod ispredicated ontheassumption thatthederivative of
/isnonzero attheapproximations tothezero p.Iffiscontinuous ,thismeans thatthe
technique willbesatisfactory provided that f\p)^0andthatasufficiently accurate initial
approximation isused.Thecondition f\p)^0isnottrivial ;itistrueprecisely when p
isasimple zero .Asimple zero ofafunction /occurs atpifafunction qexists with the
property that,forx^p,
f(x)=(x— p)q(x)% where limq{x)± 0.
x->p
Ingeneral ,azero ofmultiplicity mofafunction /occurs atpifafunction qexists with
theproperty that,forxp,
/(*)=(*-P)mqix ), where Urnq{x)#0.x-*P
Soasimple zero isonethathasmultiplicity 1.
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Bytaking consecutive derivatives andevaluating atpitcanbeshown that
 Afunction /with mderivatives atphasazero ofmultiplicity matpifandonly if
o=/(p)=/'(/>)=   =butf(p)#0.
When thezero isnotsimple ,Newton ’smethod might converge ,butnotwith thespeed
wehave seen inourprevious examples .
Example 3Let f(x)=e*— x— 1.(a)Show that/hasazero ofmultiplicity 2atx=0.
(b)Show thatNewton ’smethod with p o=\converges tothiszero butnotasrapidly
astheconvergence inExamples 1and2.
Table 2.6Solution (a)Wehave
n Pn
0 1.0
1 0.58198
2 0.31906
3 0.16800
4 0.08635
5 0.04380
6 0.02206
7 0.01107
8 0.005545
9 2.7750 x10-’10 1.3881 x10~3
11 6.9411 x10~4
12 3.4703 x104
13 1.7416 x104
14 8.8041 x10“ 5
15 4.2610 x105
16 1.9142 xlO6f(x)=e x-x-1,
so
/(0)=e°-0-1=0./'(*)=e x-\
/'(0)=e°-1=0and f'(x)=e x,
and/"(0)=e°=1.
This implies that/hasazero ofmultiplicity 2atx=0.
(b)Thefirsttwoterms generated byNewton ’smethod applied to/with p0=1are
andPi P o/(Po)
/'(po)1--— ?«0.58198 ,e— 1
P2=P\~/(Pi)
/'(Pi)%0.58198 -0.20760
0.78957%0.31906 .
Thefirsteight terms ofthesequence generated byNewton ’smethod areshown inTable 2.6.
Thesequence isclearly converging to0,butnotasrapidly astheconvergence inExamples 1
and2.Thegraph of/isshown inFigure 2.7.
Figure 2.7
/«,
1-
(L e-2)e— 2
(-1.*’1)
V.-
i1H1 II
’s\
1— -1“ 1 ^
1X
One method forimproving theconvergence toamultiple root isconsidered in
Exercise 8.
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E X E R C I S E S E T 2 . 4
l.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.Letf(x)=x2-6andp0=1.UseNewton ’smethod tofindp2.
Let/(x)=-x3-cos*andp0=-1.UseNewton ’smethod tofind p2.Could p0=0beused for
thisproblem ?
UseNewton ’smethod tofindsolutions accurate towithin 104forthefollowing problems .
a.x3— 2x2— 5=0,on [1,4]
b.x3+3x2-1=0,on [-3,-2]
c.x— cos*=0,on [0,7i/2)
d.x-0.8-0.2sinx=0,on [0,TT/2]
UseNewton ’smethod tofindsolutions accurate towithin 105forthefollowing problems .
a.2xcos2x-(x-2)2=0,on[2,3]and[3,4]
b.(x— 2)2— lnx=0,on[1,2]and [e,4]
c.ex— 3x2=0,on[0,1]and[3,5]
d.sinx— e~x=0,on[0,1],[3,4],and[6,7]
Use Newton ’smethod tofindallfour solutions of4xcos(2x)— (x— 2)2=0in[0,8]accurate to
within 10"5.
UseNewton ’smethod tofind allsolutions ofx24-10cosx=0accurate towithin 10-5.
UseNewton ’smethod toapproximate thesolutions ofthefollowing equations towithin 10~5inthe
given intervals .Inthese problems ,theconvergence willbeslower than normal because thezeros are
notsimple .
a.x2— 2xe~*+e-2*=0,on [0,1]
b. COS(X+V^)+J:U/2+V2)=0,on [-2,-1]
c.x}-3*2(2-')+3x(4-*)+8'=0,on [0,1]
d.e<“ +3(ln2)V'-(ln8)e4'-(ln2)\on[-1,0]
Thenumerical method defined by
f(Pn-l)f\Pn-1)
Pn^ [f'(Pn-l))2~/(p*-,)/"(p*-,)’
forn=1,2,...,canbeused instead ofNewton ’smethod forequations having multiple zeros.
Repeat Exercise 7using thismethod .
UseNewton ’smethod tofindanapproximation toy/3correct towithin 104,andcompare theresults
tothose obtained inExercise 9ofSections 2.2and2.3.
UseNewton ’smethod tofindanapproximation toV^25correct towithin 10~6,andcompare theresults
tothose obtained inExercise 10ofSections 2.2and2.3.
Newton ’smethod applied tothefunction /(x)=x2— 2with apositive initial approximation p0
converges totheonly positive solution ,V2.
a.Show thatNewton ’smethod inthissituation assumes theform that theBabylonians used to
approximate V2:
b.Usethesequence in(a)with p0=1todetermine anapproximation thatisaccurate towithin
io-5.
InExercise 14ofSection 2.3,wefound thatfor/(x)=tannx-6,theBisection method on[0,0.48 J
converges more quickly than themethod ofFalse Position with p0=0andp,=0.48.Also ,the
Secant method with these values ofpoandp\does notgiveconvergence .Apply Newton ’smethod to
thisproblem with (a)po=0and(b)po=0.48.(c)Explain thereason foranydiscrepancies .
UseNewton ’smethod todetermine thefirstpositive solution totheequation tanx=x,andexplain
why thisproblem cangive difficulties .
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14.
15.
16.
17.
18.
19.
20.UseNewton ’smethod tosolve theequation
* 1 1 7 , 1 ., 710=-+-x— xsinx— -cos2x,with p{)=— .2 4 2 2
Iterate using Newton ’smethod until anaccuracy of105isobtained .Explain why theresult seems
unusual forNewton ’smethod .Also ,solve theequation with p0=5TTand p0=10TT.
Player Awillshut out(winbyascore of21-0)player Binagame ofracquetball with probability
P=l± £( PV1
2U-P+P V'
where pdenotes theprobability that Awillwinanyspecific rally (independent oftheserver ).(See
(K,JJ,p.267.)Determine ,towithin 10\theminimal value ofpthatwillensure thatAwillshut out
Binatleast halfthematches they play.
Thefunction described by/(x)=ln(x:+1)— e0AxCOSTTX hasaninfinite number ofzeros.
a.Determine ,within 10'6,theonly negative zero.
b.Determine ,within 106,thefour smallest positive zeros.
c.Determine areasonable initial approximation tofindthenthsmallest positive zero of/.[Hint:
Sketch anapproximate graph of/.]
d.Use(c)todetermine ,within 10"6,the25thsmallest positive zero of/.
Theaccumulated value ofasavings account based onregular periodic payments canbedetermined
from theannuity  dueequation ,
A-^ia+iy-i].
I
Inthisequation ,Aistheamount intheaccount ,Pistheamount regularly deposited ,andiisthe
rateofinterest perperiod forthendeposit periods .Anengineer would liketohave asavings account
valued at$750,000upon retirement in20years andcanafford toput$1500 permonth toward this
goal.What istheminimum interest rateatwhich thisamount canbeinvested ,assuming that the
interest iscompounded monthly ?
Problems involving theamount ofmoney required topayoffamortgage over afixed period oftime
involve theformula
A=^[l-(l
I
known asanordinary annuity equation .Inthisequation ,Aistheamount ofthemortgage ,Pisthe
amount ofeach payment ,andiistheinterest rateperperiod forthenpayment periods .Suppose that
a30-yearhome mortgage intheamount of$135,000isneeded andthattheborrower canafford house
payments ofatmost $1000 permonth .What isthemaximum interest ratetheborrower canafford to
pay?
Adrug administered toapatient produces aconcentration intheblood stream given byc(/)=Ate~,f3
milligrams permilliliter thours after Aunits have been injected .Themaximum safe concentration
is1mg/ml.
a.What amount should beinjected toreach thismaximum safeconcentration andwhen does this
maximum occur ?
b.Anadditional amount ofthisdrug istobeadministered tothepatient after theconcentration falls
to0.25 mg/ml.Determine ,tothenearest minute ,when thissecond injection should begiven .
c.Assuming that theconcentration from consecutive injections isadditive andthat75%ofthe
amount originally injected isadministered inthesecond injection ,when isittime forthethird
injection ?
Let/00=33i+l-7 S2*.
a.UsetheMATLAB function fzero totrytofindallzeros of/.
b.Plot/(x)tofindinitial approximations toroots of/.
c.UseNewton ’smethod tofind roots of/towithin 10-16.
d.Find theexact solutions of/(x)=0algebraically .
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2.5 Error Analysis andAccelerating Convergence
Intheprevious section wefound thatNewton ’smethod generally converges very rapidly ifa
sufficiently accurate initial approximation hasbeen found .This rapid speed ofconvergence
isduetothefactthatNewton ’smethod produces quadratically convergent approximations .
Order ofConvergence
Suppose thatamethod produces asequence {p„ }ofapproximations thatconverge toa
number p.
 Thesequence converges linearly if,forlaige values ofn,aconstant 0<Mexists with
\P~Pn+\\<M\p-pn\.
 Thesequence converges quadratically if,forlarge values ofn,aconstant 0<Mexists
with
\P~Pn+\\<M\p-pn|2.
Theconstant Miscalled anasymptotic error constant .
Thefollowing illustrates theadvantage ofquadratic over linear convergence .
Illustration Suppose that {pn}%Loislinearly convergent to0with
\Pn+\\lim '=0.5\p„\
andthat {p*};Ji0isquadratically convergent to0with thesame asymptotic error constant ,
M— 0.5,so
1I 1IC\Clim =0.5.
n~¥0°\Pn\2
Forsimplicity weassume thatforeach nwehave
*0.5 and^r*0.5.
\Pn\ IPnV
Forthelinearly convergent scheme ,thismeans that
\Pn-0|=\p„\*0.5|p„ _,|«(0.5)2|p„ .2|*   «(0.5)” |/Jol.
whereas thequadratically convergent procedure has
\Pn-0|=\pn\«0.5|p„ _,|2*(0.5)[0.5|/>„ _
2|2]2=(0.5)3|p„ .2|4
*(0.5)3[(0.5)|p„ _312]4=(0.5)7|p„ _318
«-.-«(0.5)2'’-1|A,|2".
Table 2.7illustrates therelative speed ofconvergence ofthesequences to0if|/?oI=Ipol=1-
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