082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017
220 Chapter 6 Sequences and serieswhenthesevaluesaresubstitutedintotheequationbothsidesareequal.Equationswherethe unknown quantity, x, occurs only to the first power are called linear equations.Otherwise an equation is non-linear. A simple way of finding the roots of an equationf (x) = 0 istosketch a graph ofy = f (x)as shown inFigure 6.8.The roots are those values of x where the graph cuts or touches the x axis.Generally, thereisnoanalytical wayofsolvingtheequation f (x) = 0andsoitisoftennecessary to resort to approximate or numerical techniques of solution. An iterativetechnique is one which produces a sequence of approximate solutions which may convergeto a root. Iterative techniques can fail in that the sequence produced can diverge.Whetherornotthishappensdependsupontheequationtobesolvedandtheavailabilityofagoodestimateoftheroot.Suchanestimatecouldbeobtainedbysketchingagraph.The technique we shall describe here is known as simple iteration. It requires that theequation berewrittenintheformx =g(x). Anestimateoftherootismade,sayx 0,andthis value is substituted into the r.h.s. ofx =g(x). This yields another estimate,x 1. Theprocess isthen repeated. Formally weexpress this asx n+1=g(x n)This is a recurrence relation which produces a sequence of estimates x 0,x 1,x 2,....Under certain circumstances the sequence will converge to a root of the equation. Itis particularly simple to program this technique on a computer. A check would be builtinto the program totestwhether ornot successive estimates areconverging.Example6.20 Solve the equation f (x) = e −x −x = 0 by simple iteration.Solution Theequationmustfirstbearrangedintotheformx =g(x),andsowewritee −x −x = 0asx=e −xIn this example we see that g(x) = e −x . The recurrence relation which will produceestimates of the rootisx n+1= e −x nTable6.1Iterativesolutionofe −x −x=0.nx nf(x)Figure6.8A rootof f (x) = 0 occurswhere thegraph touches orcrossesthexaxis.x0 01 12 0.3683 0.6924 0.5015 0.6066 0.546... 0.567
6.6 Sequences arising from the iterative solution of non-linear equations 221Suppose weestimatex 0= 0.ThenThenx 1=e −x 0 =e −0 =1x 2= e −x 1 = e −1 = 0.368Theprocessiscontinued.ThecalculationisshowninTable6.1fromwhichweseethatthe sequence eventually converges to 0.567 (3 d.p.). We conclude thatx = 0.567 is arootof e −x −x = 0.Notethatiftheequationtobesolvedinvolvestrigonometricfunctions,angleswillusuallybemeasured inradians and not degrees.Itispossibletodeviseatesttocheckwhetheranygivenrearrangementwillconverge.Fordetailsofthisyoushouldrefertoatextbookonnumericalanalysis.Thereareothermore sophisticated iterative methods for the solution of non-linear equations. One suchmethod, the Newton--Raphson method, isdiscussedinChapter 12.EXERCISES6.61 Showthat the quadratic equationx 2 −5x −7 = 0 canbe written in the formx = √ 7+5x.Withx 0 =6locate arootofthisequation.2 Forthe quadraticequationofQuestion1showthat analternative rearrangement isx = x2 −7.Withx5 0 = 0.6 findthe secondsolution ofthisequation.3 Forthe quadraticequationofQuestion1showthatanotherrearrangement isx = 7 +5. Trytoxsolvethe equation usingvarious initialestimates.Investigatefurtheralternative arrangements oftheoriginal equation.4 Showthat onerecurrence relationforthe solution ofthe equatione x +10x−3=0isx n+1 = 3−ex n10Withx 0 = 0 locate aroot ofthe given equation.5 (a) Show thatthe cubicequationx 3 +3x −5 = 0can be writtenas(i)x= 5−x3 ,3(ii) x = 5x 2 +3 .(b) By sketching agraph forvalues ofxbetween 0and 3 obtain aroughestimate ofthe rootoftheequation given in part (a).(c) Determine which, ifeither, ofthe arrangementsin part (a)converges more rapidly to the root.Solutions1 6.142 −1.145 (b) 1.15(c) Arrangement (ii)converges more rapidly4 0.18
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6.6 Sequences arising from the iterative solution of non-linear equations 221
Suppose weestimatex 0
= 0.Then
Then
x 1
=e −x 0 =e −0 =1
x 2
= e −x 1 = e −1 = 0.368
Theprocessiscontinued.ThecalculationisshowninTable6.1fromwhichweseethat
the sequence eventually converges to 0.567 (3 d.p.). We conclude thatx = 0.567 is a
rootof e −x −x = 0.
Notethatiftheequationtobesolvedinvolvestrigonometricfunctions,angleswillusuallybe
measured inradians and not degrees.
Itispossibletodeviseatesttocheckwhetheranygivenrearrangementwillconverge.
Fordetailsofthisyoushouldrefertoatextbookonnumericalanalysis.Thereareother
more sophisticated iterative methods for the solution of non-linear equations. One such
method, the Newton--Raphson method, isdiscussedinChapter 12.
EXERCISES6.6
1 Showthat the quadratic equationx 2 −5x −7 = 0 can
be written in the formx = √ 7+5x.Withx 0 =6
locate arootofthisequation.
2 Forthe quadraticequationofQuestion1showthat an
alternative rearrangement is
x = x2 −7
.Withx
5 0 = 0.6 findthe second
solution ofthisequation.
3 Forthe quadraticequationofQuestion1showthat
anotherrearrangement isx = 7 +5. Tryto
x
solvethe equation usingvarious initialestimates.
Investigatefurtheralternative arrangements ofthe
original equation.
4 Showthat onerecurrence relationforthe solution of
the equation
e x +10x−3=0
is
x n+1 = 3−ex n
10
Withx 0 = 0 locate aroot ofthe given equation.
5 (a) Show thatthe cubicequationx 3 +3x −5 = 0
can be writtenas
(i)x= 5−x3 ,
3
(ii) x = 5
x 2 +3 .
(b) By sketching agraph forvalues ofxbetween 0
and 3 obtain aroughestimate ofthe rootofthe
equation given in part (a).
(c) Determine which, ifeither, ofthe arrangements
in part (a)converges more rapidly to the root.
Solutions
1 6.14
2 −1.14
5 (b) 1.15
(c) Arrangement (ii)converges more rapidly
4 0.18
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