Advanced Calculus 3rd Edition - Taylor Angus & Wiley.Fayez

Reclaus€TaylorWeRebelMann 5=ADVANCED=Q,CALCULUS aol Wiley \—$_—____— hid Edition —___ Copyright ©1955, 1972, 1963, byJohn Wiley &Sons All rights reserved. Published simultaneously inCanada Reproductionortranslationofanypartof this work beyond that permitted by Sections 107and 108 ofthe 1976 United States Copyright ‘Act without the permission ofthe copyrieht ‘owner isunlawful. Requests forpermission ‘orfarther information should be addressed to the Permissions Department, John Wiley &Sons Library ofCongress Cataloging inPublication Data: Taylor, Angus Fis, 1911- ‘Advanced calulus Includes index. Calculus. 1.Mann,W.Robert(Wiliam Robert, 120. HLTitle. Qaws72 198 SIS siete ISBN: ouTIERS 666AACR? Printed inthe United States ofAmerica 0 a Jev[as Jaco canned on Nov/as /2oo xiv CONTENTS 17/FUNDAMENTAL THEOREMS ON CONTINUOUS FUNCTIONS 17, Purpose oftheChapter 527 17.1 Continuity and Sequential Limits 527 17.2 The Boundedness Theorem 529 17.3. The Extreme-Value Theorem 529 174 Uniform Continuity 529 17.5 Continuity ofSums, Products, and Quotients 532 17.6 Persistence ofSign 532 17.7 The Intermediate-Value Theorem 533 18 /THE THEORY OF INTEGRATION 18. The Nature oftheChapter 535 18.1 The Definition ofIntegrability 535 18.11 The Integrability ofContinuous Functions 539 18.12 Integrable Functions with Discontinuities $40 18.2 The Integral asaLimit ofSums $42 18.21 Duhamel’s Principle $45 18.3. Further Discussion ofIntegrals 548 18.4 The Integral asaFunction ofthe Upper Limit S48 18.41 The Integral ofaDerivative 550 18.5 Integrals Depending onaParameter 551 18.6 Riemann Double Integrals 554 18.61 Double Integrals and Iterated Integrals $57 18.7 Triple Integrals 559 18.8 Improper Integrals $59 18.9 Stieltjes Integrals 560 19/INFINITE SERIES 19. Definitions and Notation 566 19.1 Taylor's Series 569 19.11 ASeries forthe Inverse Tangent 572 19.2 Series ofNonnegative Terms 573 19.21 The Integral Test 577 19.22 Ratio Tests 579 19.3. Absolute and Conditional Convergence $81 19.31 Rearrangement ofTerms 85 19.32 Alternating Series 587 19.4 Tests forAbsolute Convergence 590 195 The Binomial Series 597 19.6 Multiplication ofSeries 600 19.7 Dirichlet’s Test 604 1/INTRODUCTION Acourse inadvanced calculus must build upon the presumption that students studying thesubject have already gained some knowledge ofelementary cal- culus. We shall therefore begin bytaking abackward look over those parts of calculus with which the reader ofthis book should have facility and ameasure ofunderstanding. Our object insuch aretrospect isnot toconduct asystematic review. The purpose is,rather, toestablish acommon point ofview forstudents whose training incalculus, uptothis point, must inevitably reflect awide variety ofpractices inteaching, choice ofsubject matter, and distribution ofemphasis between the acquisitions ofproblem-solving skills and mastery offundamental theory. As we survey the field ofelementary calculus we shall stress the conceptual aspect ofthe subject: fundamental definitions and processes which underlie alltheapplications. Inafirst course incalculus itisoften thecase that the fundamental notions are introduced through the medium ofparticular geometrical orphysical applications. Thus, tothe beginner, the derivative may betypified by,oreven identified with, the speed ofamoving object, while the integral isthought ofasthe area under acurve. We now seek totake amore general, orabstract, view. Differentiation and integration are processes which are carried out upon functions. We need tohave aclear understanding ofthe definitions ofthese processes, quite apart form their applications. Another aspect ofour survey will beour concern with thelogical unfolding ofthe fundamental principles ofcalculus. Here again westrive totake amore mature point ofview. We wish toindicate inwhat respects itisdesirable and necessary tolook more deeply into the derivations ofrules and proofs of theorems. There are places inelementary calculus, asusually taught tobegin- ners, where the development isnecessarily inadequate from the standpoint oflogic. Inmany places thereasoning leans heavily onintuition oronone sort or another ofplausibility argument. That this state ofaffairs persists ispartly dueto adeliberate placing ofemphasis: wemake our primary goal the attainment of skill inthemanipulative techniques ofcalculus which lend themselves readily to applications atanelementary level inphysics, engineering, and thelike, This kind ofskill (up toacertain point) can beimparted without paying much attention toquestions oflogical rigor. Butitisalso true that there arelogical inadequacies inafirst course incalculus which cannot bemade good entirely within thecustomary time limits ofsuch acourse (two orthree semesters), even 1 4 FUNDAMENTALS OFELEMENTARY CALCULUS: cht dependent variable y.Here, however, welook toward understanding theprin- ciples ofcalculus asthey apply tofunctions which arearbitrary except insofar asthey arerestricted byspecified hypotheses. We shall indue course have todeal with functions of more than one variable. The general notion ofafunction isstill that ofacorrespondence. Areal function Foftwo real variables x,yisacorrespondence which assigns a number F(x, y)asthevalue ofthefunction corresponding tothepair ofvalues x,yofthe two independent variables. The use offunctional notation and the designation ofthefunction bythesingle letter Frequire nodetailed comment, since thebasic ideas arenodifferent from those already explained. ‘The characteristic feature ofcalculus isitsuse oflimiting processes. Differentiation and integration involve certain notions ofpassage toalimit. A fuller discussion ofideas about limits ispresented later oninthis chapter (§§1.6-1.64). Here wewish totouch ononly one limit notion, that ofthelimit of areal function ofone real variable. This notion isfundamental inthe definition of aderivative. Suppose fisafunction which isdefined forallvalues ofxnear thefixed value Xo,and possibly, though notnecessarily, atx»aswell. Wewish toattach a clear meaning tothestatement: f(x) approaches A(ortends tothelimit A)asx approaches xo.The symbolic form ofthestatement is limf(x)=A. (i) The symbol Aisunderstood tostand forsome particular real number. The arrow isused asasymbol fortheword “approaches.” Sometimes (1.1~1) isexpressed intheform f(x)»Aasx—>Xo.Herearethreetypicalexamplesofstatements of this kind: (a)x°>8 asx2, (b)(x=1)'7>3 asx10, (c)logox—+2 as x 100, Definition. Theassertion (1.1-1) means that wecaninsure that theabsolute value (f(x) —Alisassmall asweplease merely byrequiring thattheabsolute value |x~xo) besufficiently small, and different from zero. This verbal statement isexpressible interms ofinequalities asfollows: Suppose €isanypositive number. Then there issome positive number 6such that fx) Al<e if,O<|x-x <8. (1-2) Note that0<[x~x9]isthesame asx#x9.Note also that[f(x)~ Al<eisthe same asA~€ <f(x)<A+e,and|x~xo]<6thesameas9-8<x<x9+5. Wecangive ageometrical portrayal oftheinequalities (1-1-2). Letthepoints (x,y)with y=f(x) belocated onarectangular co-ordinate system; also locate the point (xp,A).Forany€>0drawthetwohorizontal linesy=A*€.Now (11-1) means that, bychoosing 5small enough, those points ofthegraph of y=f(x) which liebetween thetwo vertical lines x=x9*6andnotonthelineX=4X9will also liebetween the horizontal lines y=A+e. Fig. |shows a 14 FUNCTIONS 5 specimen ofthis situation. The diagram also shows y how 5may have tobemade smaller as€becomes smaller. A+9Itistobeemphasizedthat(1-1-1)placesno|by| restrictions whatever onthevalue offatxo,incaseATIN itisdefined atthat point. al iadAppreciation oftheformaldefinition ofthe pi ® meaning of(1.1-1) takes time andexperience. The m5 tot8 formal definition isthe basis forexact reasoning on Fig. 1. matters involving the limit concept. But itisalsoQuiteimportant todevelopanintuitiveunderstanding ofthenotionofalimit.Thismay bedone byconsidering alarge number ofillustrative examples and by observing theway inwhich thelimit concept isused inthedevelopment ofcalculus. ‘One needs tolearn byexample how afunction f(x) may failtoapproachalimitasx approaches Xo. ‘The variable xmay approach Xofrom either oftwo sides. Letususex>xot toindicate that xapproaches x»from theright, and x+x»~ toindicate approach from the left. The conditions forlim,.,,f(x)= Aare then that f(x)+A as X-+Xo+ and also f(x)+A asx—+>x9~. Interms ofinequalities themeaning of f(A asx—>x9+ isthis: toany €>0 corresponds some &>0 such that f(x)=Al<eifx9<x<xo+.Themeaning off(x)+A asx+x»— maybe expressed inasimilar way. Example 5.The limit off(x) asx+x»may failtoexist because: (a)The limits from right andleftexist butarenotequal. This isthecase with f(xy=14 i wheref(x)>2asx+0+andf(x)+0asx40.(b) The values off(x) may get larger and larger (tend toinfinity) asx+x9 from one side ortheother, orfrom both sides. This isthecase with f(x) =I/xas x0. (c)The values off(x) may oscillate infinitely often, approaching nolimit, This isthecase with f(x) =sin(1/x) which oscillates infinitely often between ~1 and +1asx0 from either side. Thegraphs ofthethree foregoing functions areshown inFigs. 2a,2b,and2c, respectively. Example 6.Iff(x)=e~"!",thenlim,.of(x)=0.To“see”thecorrectness of thisresult, onemust have clearly inmind thenature oftheexponential function. When xisnear zero, —1/x? islarge andnegative; now ¢raised toalarge negative power isasmall positive number. Hence e~' isnearly 0when xisnearly 0,and f(x)+0asx+0. This isanexample ofarough intuitive argument leading toa conclusion about acertain limit. Itisinstructive tosee how the intuitive argument ismade precise by , uw FUNCTIONS 1 Letussupposethat€<1,sothatlog(1/e)>0.Thenfurtherequivalent formsare ret, [1I” o<x'<ipgrey’ °SI1<ljogares) * Itnow appears that, if0<e <1, wecan choose 1 78[paar] andthen (1.1-3) will hold, asrequired, Even invery obvious situations itisworth while topractice finding a8 corresponding toagiven €,just todrive home anappreciation ofthemeaning of the definition ofalimit, Example 7.Given ¢>0, find 8sothat [f(x)—4| <eif0<|x~2|<6, where f(x) =x°=x2,This willshow that lim,..(x°~x°)= 4.Wehave Wa4(x=D++2), Tobegin with, letusconsider only values ofxsuchthat|x~2|<1,or1<x<3. Forsuch xwecertainly have 4<x?+x+2<14, and hence jox34]514fx-2), Now weseethat |x’~x?~4]<eprovided14|x-2|<e, or|x—2|<e/14. Hence wechoose for6any positive number such that both 61 and 6=«/14. This choice meets therequirements. Reasoning with limits isfacilitated byvarious simple theorems. Among the most important such theorems are the following rules, which we state here informally: Suppose that limf(x) =Aand limg(x)=B;then fim[f(x) +g(x) =A+B, (1) lim[f(x)g(x)] =AB, (1-3) im£4 prolimf(a)BrProvidedBx0. (1-6) Formal proofs ofthevalidity ofthese three rules aremade in§1.64. Meanwhile ‘weaccept them and use them. Closely related tothelimit concept istheconcept ofcontinuity. Definition. Suppose thefunction fisdefined atxeandforallvalues ofxnear Xo, Then thefunction issaid tobecontinuous atx»provided that limf(x) =f(a). (1-7) 8 FUNDAMENTALS OFELEMENTARY CALCULUS cht Most ofthefunctions which wedeal with incalculus arecontinuous; points ofdiscontinuity are exceptional, but may occur. Afunction may fail tobecontinuous atx9eitherbecausef(x)doesnotapproach anylimitatallasx—>xe,orbecause itapproaches alimit which isdifferent from f(x). Example 8The function f(x)=[x](seeExample2)isdiscontinuous atxif Xpisaninteger, butiscontinuous atxifXoisnotaninteger. Weobserve inthiscase that lim,.. f(x)does notexist, forwhen xisnear 2, f(x)=Lifx<2andf(x)=2ifx>2.Thesituation issimilaratotherintegers. x “2-1 TY . Oh i4 fy=t Fig. 3. ‘The graph ofy=f(x) isshown inFig. 3.Atthebreaks inthegraph when xisan integer n,thevalue off(n) isindicated byaheavy dot. Example 9.Suppose wedefine afunction by f(x) =[x]+2x x)=1 Direct inspection shows thefollowing: s)=1, f)=0 ifO<x<1, f)=1, fixy=0 if 1<x<2, fQ)=1. x rrtyrrt 3-2-1 for 2 3* fisimie)#1211 Fig. 4. a FUNCTIONS: 9 Consequently, lim,.. f(x)=0;butf(1)=1,andsofisnotcontinuous atx=1. ‘Thegraphofy=f(x)isindicated inFig.4.Fromthedefinition itmaybeseen that f(n)=1ifnisaninteger andf(x)=0ifn<x<n+1. Example 10.Letusdefine f(x)=(sinx)/x ifx#0.This definition off(x)has nomeaning ifx=0,since division by0isundefined. However, letusmake the additional definition /(0)=1.With thisdefinition,fiscontinuous atx=0.For,as welearn inelementary calculus, Jim2%=1, (1-8) ox Since wehave defined f(0) =1,(1.1-8) shows that lim,of(x)=f(0);therefore f iscontinuous atx=0,bythedefinition. We have based theconcept ofcontinuity directly upon theconcept ofa limit. Acondition forcontinuity ofafunction may begiven directly interms of inequalities, just aswedefined alimit interms ofinequalities. Thus, iffis defined throughout some interval containing x»andallpoints near x»,fiscontinuous at2iftoeachpositive€corresponds somepositive8suchthat [fC)- f(x0)|<€ whenever |x—x<6. (11-9) This form ofthecondition forcontinuity isequivalent totheoriginal definition. Many common words areused inmathematics inaspecialized way. Usually the mathematical meaning ofaword has some relation tothecommon meaning oftheword; butmathematical meanings areprecise, whereas common meanings arebroad orvariable. The adjective “continuous” isaword ofthis kind, with a restrictive and precise mathematical meaning. Experience shows that students tend toread more, intheway ofpreconceived notions about themeaning ofthe term, into theword “continuous” than isimplied bythedefinition. Inanalytic geometry and calculus webecome familiar with thegraphs ofmany functions, and there isatendency toassociate the term “continuous function” with the picture ofasmooth, unbroken curve. Now itistrue that iffiscontinuous at each point ofaninterval, thecorresponding part ofthegraph ofy=f(x) will be an unbroken curve. But itneed not be smooth. Smoothness isrelated to differentiability; themore derivatives fhas, thesmoother isitsgraph. Afunction may becontinuous without having aderivative. Inthat case thegraph ofy=f(x) might besocrinkly, s0devoid ofsmoothness, astomake correct visualization of itquite impossible. EXERCISES Where the square-bracket notation occurs inthese exercises, [f(x)] denotes the algebraically largest integer which is=f(x) (see Example 2). 44 FUNCTIONS u where thecoefficients ds,dy,....ds areconstants, and nis aninteger 20. Ifn=0, P(x) is constant invalue, and thedegree ofP(x) issaidtobezero.Ifn=1andao#0,wesaythatthe degree ofpolynomial isn,Prove that P(x) iscontinuous atevery point Xo,Usetheresult of Exercise 6,What other result about limits doyou use? 8.Byarational function ofxwemean afunction defined byanexpression =P)Ri) PGy where p(x) andP(x) arepolynomials (see Exercise 7).The function isdefined except when P(x) =0.Show that itiscontinuous atxoifitisdefined there. Use definition (1.1-7) and state exactly what appeal you make tofacts about limits stated inthetext orestablished in Previous exercises. int L)pena: 2) ye : 9Ifflxy=sin}, (a)findf(E), m=1.2.2.3 (0)find{(,2.), m=1,5,%. 005 (ord (2), n=3.7.tases G)Mowdoesthederivative /(2)behaveas+0? 10.(a)How does 2" behave asx-+0+? (b)asx-+0-? (c)What can you say nl» aboutigTea™ 11,Graph each of the following functions: (a)(xl, ()eI}, (©) e+2) (@)|x"|, (e)[1~x*|. Doanyofthese functions have anypoints ofdiscontinuity? 12.Graph thefunction x~[x]and discuss itsdiscontinuities. 13.Whichofthefollowingfunctionsiscontinuous atx=0?(a)[x?+2},(b)[4~x"}, (©[2=1}, Graph each function when 15x 1. 1416a)=BIB) Ga)nd(0,-0.1, D.LDICD.. (0)Withoutusing thesquare brackets, write expressions forf(x)if0<x <}andif-1<x <0. (¢)What is Kim, 0f(x)? 1s.ffx)=H (ayfindFED,$0,10.$F, FG.(O)Express f(x)without absolute values ifx>1; if0<x<1. (¢)What canyou sayabout lim... f(x)? 16,1rg(ay=2XAEBI=2, a)find(1),(-D.1-2.10). (0)Wrteanexpression forf(x) without absolute values if0<x; if-2<x<0. (c)What canyousayabout Hime of)? 17.Iff(x)=(72°14, (a)findFO),F(1),$2),£6,$0),JG. ()Isfcontinuous at x=0? ()Isitcontinuous atx=1?(d)atx=V2? 18.Iff(x)=[sinx},(a)find(0),f(-w/2),f(—12), f(HI4),f(—214).(0)Doeslimf(x) exist? (c)What does f(x)approach asx-+(m/2)-? (4)asx-+0-? 19.Prove thatlim,..(x? +2x)=3byfinding 6interms ofagiven positive €sothat |x?+2x-3)<eifx=1]<5. 20,showthat|r!45—de|=dale3H2<x<4.Hence,forany>0,find sothat , 1 lthe-Al<eitfx-31<2,thusprovingdirectlythatkim=p7g™4 21.Showthat|(1+x°)—|S7|x|if-1<x<1,Thenprovebythedefinition(1.1-2)that lim,ofl+x)’=1(i.e,findasuitable 6foranygivenpositive ¢). a at DERIVATIVES B Inaddition tothenotation f'(x) forthederivative wefrequently usethe notation &f(x), Example 1.Using form (111-2), calculate f'(x9) iff(x)=x7.Here $(8)~fee)=P=x5=(xHox+x); $40) =lim(x+9) =2 = Example 2.Using form (1.11-4), calculate f'(x) iff(x)=Ux.Here a MOM$0)" eth tg)«time _-_4PC)=Nimhye Definition. Afunction which has aderivative atacertain point issaid tobe differentiable atthat point. Inthedefiniton (1.11-2) wewere assuming that fwas defined inaninterval extending some distance oneach side ofthepoint x.Itisunderstood that xmay approach xyfrom either side, and that thelimit ofthedifference quotient isthe same when xapproaches x»from theleft aswhen theapproach isfrom theright. Itis useful todefine one-sided derivatives. Using thenotation forlimits from theright and left, respectively, asexplained just prior toExample $in$1.1, we define the right-hand derivative f:(14) and the left-hand derivative f*(x0) as follows: £549) =timLQ=LE0), (ies) bin ae £540)=timED=L0), (11-6) in x0 provided thelimits exist. Ifindiscussing afunction weconfine our attention wholly toaninterval ax, then weshall understand that (a) means f(a), and that f'(b) means J¥(b). Ifa<x9<b, however, and ifthefunction isdifferentiable atxy,then we must have f(x.)=f(x).Thederivativef'(x)isthenthecommonvalueofthe two one-sided derivatives. For anexample ofacase inwhich thetwo one-sided derivatives exist but areunequal, see Exercise 12. Example 3,Let (x)= V3i—cos 2x). Show that this function isnot differentiable atthepoints x=0, 7,2m... andfindtheone-sided derivatives atthese points. Werecall thetrigonometric identity L-cos 2x=2sin* x. 16 FUNDAMENTALS OFELEMENTARY CALCULUS cnt Tofixtheideas precisely, suppose thatgisdefined when a<<, andthat the values ofthefunction satisfy theinequality a<g(t)<b. Suppose that fisdefinedwhena<x<b.Then,replacing xbyg(t),weobtainthecompositefunction Fdefined by(1.11-11). THEOREM IL.Suppose gisdifferentiable atapoint tooftheinterval a<1<. Letx= g(ts), and suppose that fisdifferentiable atxo.Then thecomposite function Fisdifferentiable atto,and F'(to)=f'Cxa)g'(t0). (11-12) Inelementary calculus this theorem isoften expressed symbolically ina different way, bywriting ¥=f), x=g(0. ‘Then dy_dy_ dx dt dx" dt" Example4.Suppose f(x)=x",g(t)=t—C.Then FQ)=(- 0)" and F(t) =170~PY~28). We accept Theorem IIasknown from elementary calculus. The proof isa somewhat delicate matter, however, and the student who wishes tostudy the proof will find adiscussion ofitinExercise 26attheend ofthis section. Throughout calculus there are two aspects ofthe development ofthe subject. Ontheone hand weformulate concepts and rules applicable toarbitrary functions having certain properties. Theorems Iand Ilareofthis type. Onthe other hand there are the particular functions which wedeal with asillustrations and inallpractical applications, e.g., (1—x°)"%, sin2x,tan“ x,logx,€"°°, and many others. We assume that thestudent knows theformulas fordifferen- tiation ofthestandard elementary functions, and ingeneral weshall regard all such functions asbeing available forillustrative purposes. Inorder toillustrate thepossibility ofvarious kinds ofsituations which do notordinarily arise with thestandard elementary functions, wesometimes resort tothe contrivance offunctions specifically defined soastoexhibit some peculiarity. Such specially contrived functions serve tohelp thestudent ap- preciate thegenerality oftheconcept ofafunction. They also teach him tobe wary oftacitly assuming more than isimplied inagiven definition orhypothesis. Example $.Let afunction bedefined asfollows: f(x)=xsingifx#0, £0) =0. 12 MAXIMA ANDMINIMA 23 limitations f(x) might notattain any absolute extreme values, asweseeby Examples 1-3. Inpractice thefunctions weareinterested inareusually differentiable atall points oftheinterval (there may sometimes beisolated exceptional points). Ifan absolute extreme Value occurs ataninterior point oftheinterval, itisalso a relative extreme value inthesense ofTheorem III,and therefore wemust have J()=0 atthe point, provided fisdifferentiable. We therefore have the following guiding principle insearching forpoints atwhich f(x) can attain an absolute maximum orminimum value: Suppose fisdifferentiable onthegiven interval, except perhaps atafinite number ofpoints, and suppose itisknown that anabsolute maximum (orminimum) value isactually attained. Then thepoint of attainment iseither (a) apoint where f'(x) =0, (b)@point atoneendoftheinterval, or (c)apoint where fisnotdifferentiable. Inthecommon type ofproblem studied inelementary calculus, thesolution is usually found under (a). Infact, itusually happens with physical orgeometrical problems that there isonly one interior point ofthe given interval where J'(x)=0.Solutions under(b)dooccursometimes, eveninphysicalproblems, and acarefully reasoned solution should always take account ofthe situation at theends ofthe interval, perhaps even before computing f'(x). The situation (c) may occur also, but this will bemore rare incommon practice. Example 4,Findanumber xbetween 0and1suchthatf(x)=2+ Sisas small aspossible. We observe that f(x)>0 when 0<x <1;fiscontinuous inthe specified open interval. Also, f(x) becomes very large (infact f(x)>+2)asxapproaches, cither endoftheinterval. Weconclude that thegraph ofy=f(x) near theends of theinterval has anappearance somewhat asshown inFig. 10.Itfollows from this reasoning that ifwechoose aclosed interval a=x5b,with a>0 and very near 0,and b<1and very near 1,thefunction fwill have smaller values inthe interior oftheinterval [a,b]than ithasintherest oftheinterval (0,1).Since fis, continuous onthe finite closed interval (a,b],itmust attain avalue atsome point of(a,b)whichisanabsoluteminimumamongallthevaluesoccurringon theinterval. Thisabsolute minimum willalsobeanyabsolute minimum among allthevalues offoccurring on theopen interval (0,1).Now fisdifferentiable in(0,1); hence therequired point ofabsolute minimum must bea point atwhich f'(x) =0,We therefore proceed tocom- pute thederivative and solve theequation f'(x) =0: gyandy 8, 6x24x=2PO= Sq = rane 3x°42x-180,x=) orx=-1. Fig.10. 26 FUNDAMENTALS OFELEMENTARY CALCULUS cht (a)Ifthere isaboat available atA,what combination ofrowing and walking will take him toBintheleast possible time? (©)Discuss theproblem incase therowing andwalking speeds are, respectively, uand © miles perhour. Check your results carefully inthespecial case u=2, v=4, 17. Consider aand basfixed, with b<a. Let ¢beavariable such that b<¢ <a. Let¢betheacuteanglebetweenthetangentstothecirclex"+y?=c? andtheellipseb*x?+a*y*=ab"atapointofintersection. Findtanwhencischosensothatisgreatest. 18,Write out theproof ofTheorem IIIforthecase ofarelativeminimum. Show that, if{has arelative minimum atx.,and g(x) =~f(x), then ghasarelative maximum atxe.Hence deduce theproof forthecase ofaminimum from thefacts already established for amaximum 1.2 /THE LAW OFTHE MEAN (THE MEAN-VALUE THEOREM FOR DERIVATIVES) ‘The theorem which goes bythename ofthelaw ofthemean isone ofthemost important theoretical results inthesubject ofdifferential calculus. Itisused asa tool inmany places inthelater developments ofcalculus, both differential and integral, particularly inconnection with proofs. We wish toemphasize very strongly that thestudent ofadvanced calculus needs togain anappreciation of thepower ofthelawofthemean asaninstrument ofsystematic reasoning. The first step should betobecome thoroughly familiar with thecontent ofthe law itselt. \THEORENEINS(Thelawofthemean.)Letfbeafunctionwhichiscontinuous at ‘each point ofthe closed interval asx =b, and letithave aderivative at each point oftheopen interval a<x <b. Then there isapoint x=Xinthe ‘open interval (a<X <b) such that f(b) ~fla) =(b~a)f(X). (2-1) The theorem has ageometrical interpretation. Represent the function graphically bythe curve y=f(x), and letA,Bbethe points onthe curve corresponding tox=a,x~b,respectively. The formula (1.2-1) states thatthere issome point onthecurve, with abscissa x=X,atwhich thetangent isparallel totheline AB. There may bemore than one suitable value ofX;theessential thing isthat there isalways atleast one (see Fig. 11). Itisworth noting that (1.2-1) remains true ifwe exchange aand b,forboth sides merely change sign B when thisisdone. Thus, suppose x),x2aretheend-points 7 ofaninterval onwhich theconditions ofthelaw ofthe AR“ mean are satisfied. Then we can write f02)~ f00) =Ga xf), (1.22) —¥ + where x= issome point between x,and x). In Fig. I. 30 FUNDAMENTALS OFELEMENTARY CALCULUS cht Example 7.Suppose that fsatisfies theconditions ofthelawofthemean on the interval aSxb, and that f'(x)>0 when a<x<b. Show that f(x) in- creases as xincreases. We are toshow that x)<xz implies f(x)<f(x)whenevera5x,<xy5b. The law ofthe mean tells usthat there issome £such that x,<£<x; and $2) ~f(x) =Ga~xf"). Since (x2— x1) >0and f'(€)> 0,weinfer that f(x:)— f(x)>0;thisisequivalent to(x1)<f(x). Example 8.Suppose that fisdefined and differentiable onanopen interval containing thepoint xo.Suppose that {"(xs) =0and that forallxsufficiently near Xo,{"(X) >0when x<xp and f"(x) <0 when x>xo.Show that these conditions aresufficient toguarantee that fhas arelative maximum atXo. The argument isbased onExample 7.Asxincreases, f(x) increases when $x) >0. Bysimilar reasoning f(x) decreases when xincreases iff’(x) <0. Inthe present case weseethat thegiven conditions imply that forsome small number hhf(x) isincreasing asxgoes from x9—h toXo,and decreasing asxgoes from Xo toXo+h. Hence f(x) must attain arelative maximum atx. From this argument itwill beapparent tothe student how one may formulate sufficient conditions forarelative minimum atx. EXERCISES 1.Use thelawofthemean toshow that }<V66~8 <i 2.Prove that there isnovalue ofmsuch that x’~3x-+m =0 has two distinct roots intheinterval0x=1.UseRolle’sthoerem.3.Iff(x)=x"~3x+24, a=0, h=4, find asuitable value of@inthe formula 2. 4.For what values ofCisCx ~sinxanincreasingfunctionofx(forallx)? 5.Show that 2/r <(sin0)/@<1if0<0</2. Hnwt:Examinethesignofthe derivative of(sin 6)/8. 6.Prove thefollowing inequalities, using thelaw ofthemean. (a)VIE <4+(x-15)/8 ifx>15 ()tan"!x<(w/4)+(0-1)if<x. (oFt <tan'r<Z-154 it0<x<1 (@) hil +h?)<tanh<h ifO<h 7.Prove that theinequality (1.2-S) also holds if~1<h <0, Explain thereasoning about inequalities with care, noting that if0<A<1 and B<0, then AB>B. 8.Prove thefollowing inequalities: (a)(Ey<lor+4)<x,if1<x<00r0<x. (b)Veg VIFRCI, if-1<x<00r0<x, tSegglym<t-gqigmif- 1<x<0or0<x 32 FUNDAMENTALS OFELEMENTARY CALCULUS cht (©)Show, conversely, that if(x) increases asxincreases, thecondition in(a)issatisfied ‘whenever x1#x2,(Use thelaw ofthemean.) (@)Under theconditions in(¢)show that thecurve y=f(x) between anytwoofitspoints liesentirely below thechord joining those points. Begin byshowing that ananalytic expression ofthis state ofaffairs is {)~ fos)_fs)-fx) =m oe whenever 11<x <1. 1.3 /DIFFERENTIALS The notion ofadifferential isclosely related tothat ofaderivative. For functions ofasingle independent variable this relationship isvery close indeed, and very simple. Forfunctions ofseveral independent variables therelationship isless simple. Atthis point weareconcerned only with functions ofasingle variable. Wepresume that thestudent isacquainted with differentials and their uses inthe formal procedures ofelementary calculus. Our purpose here is mainly todefine differentials carefully and demonstrate the fundamental pro- perty upon which much oftheusefulness ofdifferentials depends. Suppose that fisafunction oftheindependent variable x,and letusassume that fisdifferentiable for certain values ofx(i.e., that "(x) exists for these values). Definition. Let dxdenote anindependent variable which may take onany value whatsoever. Then thefunction ofxand dxwhose value isf'(x)dx iscalled the differential off.Observe that thedifferential isahomogeneous linear function of dx; that is,forafixed value ofx,thedifferential has asitsvalue afixed multiple ofdx. Ifwewrite y=f(x), and iffisdifferentiable foraparticular value ofx,itis customary towrite dy=f'(x)dx, 31 sothat dyisthe value ofthe differential offfor assigned values ofxand dx. Ifweregard xasfixed, dyisadependent vari- yy dy able whose value depends ontheindependent H variable dx.Thevariables dxanddyareoften refer- {yey redtoasthedifferentials ofxandy,respectively. H s 4TheadjacentFig.12illustratesgeometrically 5Citnayde thefunctional dependence ofdyondx,aswellas onde the relation tothe function fitself, The xy- a~* co-ordinate axes and the graph ofthe function 2Gora) tt y=f(x) are shown inunbroken fines. Asecond veh co-ordinate system isshowing with itsorigin ataFig.12 13 DIFFERENTIALS 33 typical point (x,y)ofthecurve y=f(x). Theaxes inthissystem arescales forthe measurement ofthevariables dx,dy.Theequation (1.3-1)hasasitsgraphastraight lineofslope f'(x). This lineis,ofcourse, thetangent tothecurve y=f(x) atthe origin ofthedx-dy co-ordinate system. From (1.3-1) wehave thequotient relation a=f”) M1352), whenever dx# 0.The d-notations dxand dygoback toLeibniz’s work inthe seventeenth century, but Leibniz did not define thederivative bythelimit ofa quotient aswedidin(1.11-4). Itistobeemphasized that there isnoneed fordx and dytobesmall in(1.3-2). Example 1.Ify=f(x)=sin.x, calculate the value ofdyfor x=13, dx= 16. Here f'(x) =cosx,sody=cosxdx.Evaluating, weobtain z)2_7 dy=(cos) =F. Probably themost important feature oftheformula (1.3-1) isthat itstruth is unaffected bytheintroduction ofanew independent variable. Example2.Suppose y=x?andx=1+,sothaty=(+1)=184200. If weregardxasanindependent variable,thendy=2xdx,by(1.3-1).Heredxis anindependent variable. Butifweregard tasanindependent variable, then both xand yaredependent ont,and thenotations dy,dxacquire new meanings: xaP+tde=Gr+ Idt, yattt2+, dy=(61°+81)+21)dt. But even with these new meanings, itisstill true that dy=2xdx,We verify this bywriting 2xdx =20)+HBE+1)dt=(61+81+21)dt=dy. What we have verified here inaparticular case may bedemonstrated in general by appealing tothe rule for differentiating acomposite function (Theorem I,$1.11). Suppose y=f(x) and x=g(t), sothat y=F(t), whereF(1)=f(g(t)).Then,withtasindependent variable, dy=F(t)dt,dx=g(t)dt. 33) ByTheorem ITwehave FO=f@x. ECHR: Hence combining (1.3-3) and (1.34), dy=f'(x)g'() dt=f'(x) dx, sothat(1.3+1) holds, even though xanddxarenolonger independent variables. ‘The useofdifferentials isagreat convenience inalgebraic manipulations 34 FUNDAMENTALS OFELEMENTARY CALCULUS cn. which are incidental tomuch work incalculus. Using differentials rather than derivatives, one isoften enabled toretain adesirable symmetry bynotforcing a decision astowhich variable isindependent. The differential formula forarc length ofaplane curve illustrates this point. The formula is ds?=dx?+dy’; MLS theco-ordinates (x,y)ofapoint onthecurve arefunctions ofsome parameter, and the arc length s,measured from some chosen initial point onthe curve, likewise depends onthe parameter. But the formula (1.3-5) holds (granted suitable conditions onthecurve) nomatter what theparameter may be. ‘The fact that #"(x) isthe ratio ofdytodxmomatter what variable is independent isofgreat usefulness when wewish tocompute theslope ofacurve defined paramettically. EXERCISES 1.@)Ify=(1-x)/01+ 2°),computedywhenx=1,dx=2. (b)If x=tan(t!2), compute dxwhen t=x/2, dt=2. (©)Ifxin(a)isreplaced byitsvalue interms oftfrom (b), show that, onsimplification, y=cos. From this formula compute dywhen t=n/2, dt=2.Compare theanswers to (©)and (c)with your result inpart (a) 2.From x=reos®, y=rsin®, and ds*=dx*+dy? derive the formula ds*= ars de. 3.(@)Ify=f(a),y=(4),and80on,whyisdy"=y"dx? (b)What aredy”andd(y’? when expressed with dxasafactor? (©Suppose that x=f(t), y=g(t), and write x’=f'(), "= #'(t), andsoon.Show that 4(ay) xy-yeal) 4.For aplane curve C,construct the tangent ata typical point P(x, y,andletangles ¢,o@beasindicated onFig.13, 0that forthegeneral case =0+y+nmwherenisanin- teger (n= 0inFig. 13), From this equation and therelations Satand.2= tant 3 show that gy tany=24¥=yde _do OI =xdxtydy~ dr a Fig.13. 8.From ciny=25(sceExercise4)finddyintermsoe 1aandat ofrr. and d0, wherer=#6,and=4, 6.The curvature ofaplane curve y=f(x) isdefined asK=déds, where dydx = tan6andds?=dx?+dy?,Derivetheformula Kay 14 ‘THE INVERSE OFDIFFERENTIATION 37 learns todifferentiate inelementary calculus. Thus, forexample, theequations yee Yi Vigeen aeVFX have nosolution within this class. Itiswell topause atthispoint andreflect upon themeaning oftheword “function.” Inelementary differential calculus practically allour experience is with functions ofafew basic types: algebraic, trigonometric and inverse trigonometric, exponential and logarithmic, and rather simple compounding of these types. Itturns out that within this class of“elementary” functions, differentiation always leads tofunctions which areagain intheclass. Such isnot the case with the inverse ofdifferentiation, however; there are elementary functions which arenotderivatives ofelementary functions, e.g., e~* and V1+x'. Now thegeneral theorems ofcalculus deal with functions which are arbitrary except for requirements ofdifferentiability orcontinuity, and which certainly need not beelementary inthe sense ofthefirst part ofthis paragraph. ‘Once we have rejected the limitation ofour considerations to“elementary” functions, wemay well ask: What nonelementary functions doweknow? Ife isnotthederivative ofanyelementary function, howarewetofindsolutions of theequation dy/dx = e-"?Evidently itisnecessary insome fashion toacquire a supply ofnonelementary functions ofwhich weknow thederivatives. There are several very important methods forbuilding such asupply. One method isthat ofintegration ofknown functions. Starting with agiven con- tinuous function f(x) defined onsome interval, weform Fan= fsa, (rs) where aand xbelong totheinterval, and aiskept fixed. Another method isthat offorming infinite series whose terms aregiven functions ofx: F(x) =w(x) +260) +WO) We shall later beable toshow that fVietdr isafunction F(x) such that F'(x)= V14 x,and that exe oY xo Sa 7 isafunction F(x) such that F'(x)=e"". . The study offunctions defined byinfinite series willconcern usinalater chapter ofthisbook. Our immediate interest willbeconfined tofunctions ofthe type (1.4-7) defined byintegration. Weshall presently learn (see theend of $1.52) that iff(x) iscontinuous onagiven interval a<x=b, the general solution oftheequation dy/dx =f(x) onthatinterval isy=F(x) +C, where Cis 15 DEFINITE INTEGRALS 39 Step3.Find thevalue ofthefunction ateach ofthepoints chosen inStep 2, and form the sum FOR)Ax+ACE)Baat+FOG)Bye (5-2) ‘This iscalled anapproximating sum. ‘Step 4.Find thelimit ofthesums (1.5-2) asnisincreased and themaximum ofthenumbers Ax;,...,Ax, ismade toapproach zero. This limit is,by definition, thedefinite integral (1.5-1), sothat ff(a)dx=lim3fox)Ax, (5-3) InChapter 18,weshall study thetheory ofintegration systematically. At thattime weshall prove that thesums (I.5-2) doactually approach alimit inthe case ofany continuous function. For the present wetake for granted the existence ofthis limit, and itsuniqueness. Afurther discussion ofthe limit concept associated with theintegral will befound in§1.63. Adefinite integral isthus defined asthe limit ofacertain kind ofsum associated with thefunction. Ageometrical interpretation oftheintegral can be made interms ofthearea under thecurve y=f(x) from x=a tox=b.Each term intheapproximating sum (1.5-2) isthearea ofoneoftheshaded rectangles inFig. 14.The area under thecurve isthelimit ofthesum oftheareas ofthese ¥ ALIN |i7, Of ans, fav nad alk Fig. 14. rectangles. We assume that the student isalready familiar with this geometrical interpretation oftheintegral, and with theextension oftheinterpretation (bythe concept ofnegative area) tothose situations where the curve y=f(x) goes below the x-axis. We emphasize, however, that wedonot define the definite integral asthe area under the curve. The area interpretation ismerely a convenient method ofbringing our intuition into play toaidusingrasping the nature ofthedefinition (1.5~3). We “feel” that the area exists, and that agood approximation toitcanbeobtained bythesums (1.5-2), provided wetake allthe subintervals short enough. Actually, thearea isdefined asbeing equal tothe limit in(1.5-3). 1s DEFINITE INTEGRALS 43 There isaconvenient formula forthesumofthesquares oftheintegers from Iton (see Exercise 6): Paes gtMOEMOND, ss) Combining theforegoing observations, weseethat * 3 [far fim+DORDP as9 5= 6n Now nin+D2n+1)_ 4,3,1,eer ee aeaad 5-10) sothat, asm>, theexpression ontheright in(1.5-10) approaches 2asalimit. Hence, from (1.5-9), * gra2BLbfea Since bwas arbitrary, itfollows that fvdee> ‘Therefore fdeaPoe, EXERCISES 1.(a)Usingfourequalsubintervals, calculateupperandlowersumsfortheintegralPCP30+)ds, (b) Repeat (a) using eight equal subintervals. (©Calculate thevalue oftheapproximating sum (1.5-2), using four equal subintervals, nd taking xitobethemidpoint ofthekth subinterval. 2.(a)Calculate thevalueoftheapproximating sum(1.5-2)fortheintegral f'4, using sixequal subintervals and taking xitobethemidpoint ofthekth subinterval. ()Caleulate upper and lower sums fortheintegral in(a),using sixequal subintervals. A table ofreciprocals will befound convenient forthisexercise, 3.Follow theinstructions ofExercise 1asapplied totheintegral fe(4x?~ 12x+ 10)dx. 4.Apply thedefinition (1.5-3) tofind thevalue oftheintegral ff(x) dxiff(x) =6, where cisaconstant. 5.(a) Let An)=1+24>0-+m, NotingthatAdn)=n+(n~1)+---+4, show that 24m)=nin+De (©)Using theformula forA,(n) found in(a),calculate {2xdxbyamethod likethatused inExample 2 6,LetAsn)=P+2+-+--+n®, ObtainaformulaforAin)asfollows:Startwith (+ Wp’ 3p?+3p +1. 46 FUNDAMENTALS OFELEMENTARY CALCULUS Chet continuous. Such existence ismade plausible byintuitive consideration ofthe variation invalue ofacontinuous function, An indubitable proof ofthe exis- tence ofXmust await our systematic consideration ofthe properties of continuous functions (inChapter 3).The remarks which wemade about the existence ofmand Minconnection with theproof ofRolle's theorem (51.2) apply equally here totheexistence ofX. ‘The number j.defined by(1.51-3) iscalled theaverage value ofthefunction (x) ontheinterval [a,b].The sense inwhich this concept ofaverage value isan extension ofthe simple notion ofthe arithmetic mean asanaverage value is indicated inExercise 1 EXERCISES 1.Let [a,b] bedivided into nequal parts, and lety;bethe value off(x) atthe midpoint oftheithsubinterval. The arithmetic mean ofys... Yeis Agatebitye, Show that j= limy- An _2. Byinterpreting the integral asanarca, calculate theaverage value off(x)= Va" =x"ontheinterval aSxSa. 3.Aright circular cone ofaltitude Hand radius ofbase Rhas itsaxis along the x-axis, For agiven value ofxletA(x) denote thearea ofcross section ofthecone bya plane perpendicular tothe x-axis atthat point, What isthe average value ofA(x), x ranging over allvalues for which the plane cuts the cone? 4.InTheorem VIitwas asserted that anXcan befound onthe closed interval (a,b] such that (151-1) holds. Iff(x) isconstant on{a,b}, sayf(x) C,then Xmay betaken asany point oftheclosed interval, forinthat case ff(x)dx=C(b~a),andC=f(X), nomatter how wechoose X.Hence certainly wecan choose Xsothat a-<X <b. Show that this canalso bedone iff(x) isnotconstant on[a,b]. State precisely what you are taking forgranted about continuous functions. 1.52 /VARIABLE LIMITS OF INTEGRATION Before coming tothe main subject ofthis section itwill bewell toconsider a matter ofnotation. Inthesymbolic expression [toa werefer toxasthe variable ofintegration. The value ofthe integral does not depend upon the letter which isused for the variable ofintegration. For example, [iva-fvar=ffwdu. Incases where thelimits ofintegration areliteral symbols itisimportant to avoid using thesame letter foralimit ofintegration andalso forthevariable of 153 THEINTEGRAL OFADERIVATIVE 49 EXERCISES 1.The functions f,g,hare assumed tobecontinuous forallvalues oftheir independent variables. Complete each ofthefollowing equations: @£froa- wfaa- ©G1y)=fmdt,10)= 2A) =O" ds find (a)$0), (b)SUD. (©6%). ={sn @)F(Z “(® (a aiFey= [2%dy,ndaw)F(Z), F(Z), PO. 4.IfGx)= fs] ds,find @GID, ©)GO, ©GR, @Ga. 5.(a)IfF(x)= {6t(t— De"dt, find thepoints ofrelative maxima and minima of F(x). (b)What isthe value ofFO)? (¢)For what values ofxisF'(x)>0, and for what values ofxisFG) <0? 6.IfF(x) =6Pe” dt,find theabsolute minimum value ofF(x). 1.53 /THE INTEGRAL OF ADERIVATIVE ‘The theorem which we shall prove inthis section isfundamental, for it establishes the standard technique whereby definite integrals are calculated in practice. The four-step defining process ofarriving atadefinite integral, asset forth in§1.5, isdifficult toapply. For alarge and important class ofintegrands thefollowing theorem provides aconvenient method offinding thevalue ofthe integral ‘THEOREM VIII. Let fbeagiven function continuous ontheclosed interval a,b). Suppose that Fisany differentiable function such that F'(x)=f(x) when a=x 5. Then *fferax=FO)- Feo. assay Proof. We arebyhypothesis given afunction F(x) whose derivative isf(x). ByTheorem VII (81.52) weknow another function with this same derivative, namely f“feadt. ‘Thus the function Ge=Fo)- [foat isconstant, byTheorem V(§1.2), since itsderivative iszero. Now Gla)=Fla)~0, by(1,52-2), Also, G)=Fo)fdt. 50 FUNDAMENTALS OFELEMENTARY CALCULUS cnt Since G(x) isconstant wehave G(b)=Ga), * F(b)~[#0dt=F(a) This result isequivalent to(1.53-1), sotheproof iscomplete. Example 1.Find thevalue oftheintegral fysinxdx. Applying Theorem VIII, weseekafunctionofxwhosederivativeissinx.Such afunction is~cos x.Therefore as : Jsinxdx =~cos$+c0s0= 1, Wearenow inaposition toseeclearly theconnection between differentiation and integration. Asconcepts, bytheir definition, these processes are quite independent ofeach other. Itturns out, however, that each process isina certain sense inverse tothe other, The two aspects ofthis mutual inverseness aredisplayed byTheorems VII and VIIL. Ifwewant afunction defined when 5x 5b and having asitsderivative acertain given continuous function f(x), theclass ofallfunctions satisfying ourwant isthefamily J:f(t)dt+C.If,on the other hand, wewish tointegrate agiven continuous function f(x), wecan doso bytheformula fftepae= Fe)-Fe@ provided wecan find afunction F(x) having f(x) asitsderivative atallpoints of theinterval fa,b] Example2.Evaluatetheintegral{° We seek afunction whose derivative is1/x when —105 x=~2. The familiar formula Atogx=4 (153-2)felon =2 . will notquitedo,forlogxisnotdefinedifx<0.But,ifx<0,log(-x)isdefined, and d ebLep-tHog -+c-t (53-3) Hence, byTheorem VIII, with f(x) =I/x, F(x) =log(-x), wehave [iS-tx-0|7 =log2—log10=log|. ‘The formulas (1.53-2) and (1.53-3) can becombined inthesingle formula a 1,4ggxe . 534) Aroixl=+itx¥0. (15344) 2a ‘THEFIELD OFREAL NUMBERS B ‘Throughout this section the word “number” will beunderstood tomean “real number,” and symbols a,b,c,x,... willstand forreal numbers. There aretwonumbers with special properties, namely 0and 1.The special Properties areexpressed bythelaws a+0=a and a-l=a 1-1) forevery number a. ‘The number 0isspecial foraddition, while 1isspecial formultiplication. ‘The operations ofsubtraction and division may bedefined with theaidofthese special numbers inthefollowing way: Toevery number acorresponds its negative, —a, which isthe “additive inverse” ofa.Bythis wemean that x=—a satisfies theequation atx=0. (21-2) Likewise every number aexcept 0has amultiplicative inverse, denoted bya”'. Thatis,ifa#0,x=a'satisfiestheequation ax=1. 21-3) We then define thesubtraction ofbfrom abytheequation a-b=a+(-b). 14) Similarly, wedefine thedivision ofabybas a ,: pra).(1-8) The properties ofthereal numbers which wehave just been discussing are summed upbriefly inthe language ofmodern algebra bysaying that the real numbers form afield. The word “field” here has aspecial technical meaning. When wesay that asystem ofnumbers Fconstitutes afield wemean the following: 1.Ifa.andbareinF,thena+bandabareinF. 2.The commutative, associative, and distributive laws hold. 3.Fcontains distinct special numbers 0and 1with theproperties (2.1-1). 4.Equation (2.1-2) hasasolution inFforeach a,and (2.1-3) has asolution inFfor each a4 0. Inabstract algebra itisshown how the other familar laws ofelementary algebraarededucible fromthelawsgoverning afield.Amongtheimportant rulesthatcanbeprovedare:a-0=0and(~a)(—b) =ab.Weshallnotundertake anysystematic deductions ofthis kind. We mention, however, therule: Ifab=Oandb¥0,thena=0. Q.1-6) This isproved asfollows: Since b#0,there isanumber b~!such that bb~' =1.Fromab=0weconclude thata(bb“')=0-b-',ora-1=0, ora=0.Note that thesystem ofintegers (positive, negative, and zero) isnotafield, although itfails tobeoneonly through thefactthata~'need notbeaninteger 24 ‘THEAXIOMOFCONTINUITY Ta such that pSpo+1.Weassert that noisthesmallest integer inS.For, ifnis any member ofS,wehave pp<n. Let m=n— po,orpo+m=n. Here misa positive integer. Therefore, mopo+1Spo+m =n, ornoSn. This completes theargument. EXERCISES 1.Prove byinduction that, forevery natural number n,either 1=nor 1<n. 2.Prove thevalidity ofthefollowing form oftheprinciple ofmathematical in- duction, resting your argument ontheform enunciated inthetext. Let B(n) denote a Proposition associated with theinteger n.Suppose B(n) isknown (orcan beshown) tobe true when n=no,and suppose thetruth ofB(n +1)canbededuced ifthetruth ofB(n) is assumed. Then B(n) istrue forevery integer nsuch that non SUGGESTION: Let A(n) betheproposition B(ne+ n~1) 2.4 /THE AXIOM OF CONTINUITY ‘The facts expressed inthestatement that thereal numbers form anordered field arequite familiar. Wearenow going todiscuss amuch less familiar property of thereal number system. Most students beginning acourse inadvanced calculus will have had noexperience inmaking useofthis property, and quite possibly may never have heard ofit.We call itthe axiom ofcontinuity. The Axiom ofContinuity. Suppose that allreal numbers are separated into two collections, which wedenote byLand R,insuch away that 1.every number iseither inLorinR. 2.each collection contains atleast one number. 3.ifaisinLandbisinR,thena<b, Then there isanumber csuch that all numbers less than ¢are inLand all numbers greaterthanareinR.(Thenumber¢itselfmaybelongeithertoLortoR,depending ontheparticular way inwhich Land Rareformed.) Itisconvenient tohave aname foraseparation ofallreal numbers into collections Land Rmeeting thespecifications (1)-(3). Wecallsuch aseparation ‘acut; thenumber cisthen called the cutnumber. The cut number correspond- ingtoaparticular cutisunique. For suppose agiven cuthasthedistinct cut numbers ¢;and c:,One ofthem isthegreater, say c)<¢:. Consider thenumber =ate,bat, which lieshalfway between c,and ¢2:¢; <b<¢3. Now ¢,<b implies that bisin R,byone oftheproperties ofthecut number cy.Likewise b<¢: implies that b isinL.Hence bisinboth Land R.This isimpossible, however, forbythe specification (3)Land Rcannot have any members incommon. The assumption ofdistinct cut numbers has led toacontradiction. Therefore, we conclude that ‘any cuthas but one cutnumber. 80 THEREALNUMBERSYSTEM ch.2 point asanorigin, anarbitrary direction aspositive, and —+—+—+—+-+—- anarbitrary unitoflength. Wethen mark offsegments ~2~1 °128 ofunit length oneither side ofthe origin, thus obtaining Fig. 19. the points which welabel asshown inFig. 19.There isaone-to-one correspondence between the real numbers and thepoints ontheline. This entitles us,forbrevity, tospeak of“the point a” insteadof“thepointcorresponding tothenumber a.”Theinequality a<bhasthegeometrical interpretation thatbliesinthepositive direction fromaalongtheline. Wecall theline, thus regarded asageometrical representation ofthereal number system, theaxis ofreals, orthereal number scale. 2.7 /LEAST UPPER BOUNDS Byasetofrealnumbers wemeananaggregate orclassofnumbers. Itmaybeformed according toany rule, and the number ofitsmembers may befinite or infinite. Ifthe conditions laid down for determining the setare such that no number satisfies them, thesetissaid tobeempty. Itisvery convenient tohave a brief symbolism toindicate that anumber belongs toagiven set. The statement that the number sbelongs tothe setSisexpressed symbolically inthe form 5€S (read sisamember ofS).The symbolic form ofthestatement that sdoes not belong toSiss€S.Thus, ifSisthe setofprime positive integers, 3€S and 82S. IfSisaset ofnumbers, and ifMisanumber such that s=Mfor each 8€S, wesay that Misanupper bound ofS,Evidently any number larger than Misalsoanupperbound ofS.IfAisanupperboundofSandifthereisnonumber smaller than Awhich isalso anupper bound forS,wecall Atheleast upper bound ofS.Obviously asetcannot have more than one least upper bound. Example. The setSofnumbers oftheform n/(n +1),forallpositive integers n,consists of1/2, 2/3, 3/4, 4/5... etc. Evidently |isanupper bound ofS.But more istrue; 1isthe(unique) least upper bound ofS.Toverify this wemust show that ifc<1,¢cannot beanupper bound ofS,i.e., that there issome n such that ¢<n/(n +1).Togetsuch annweappeal toTheorem Iin§2.4, which tells us that there exists an nsuch that 1<n(1—c). But ¢<1, and soc< n(1—¢)=n~ne, ornc+c<n.Butthen(n+Ile<n,andso¢<nj/(n +1).This completes theargument. . The following theorem isoffundamental importance: THEOREM II.IfSisasetofreal numbers which isnotempty and which hasan upper bound, then ithas aleast upper bound. Proof. Weappeal totheaxiom ofcontinuity. LetLbethesetofallnumbers xsuchthatx<sforsome sinS,andletRbethesetofallnumbers ysuchthat 27 LEAST UPPER BOUNDS 81 8Syforevery sinS.Clearly LandRtogether comprise allreal numbers. If sES, then s—1EL; AER ifAisanupper bound ofS.Thus neither LnorR isempty. IfxEL and yER, wehave x<s forsome sinS.But sSy, and therefore x<y.We have therefore defined acut. Let ¢bethe cut number. We shall prove that cistheleast upper bound ofS.Itcertainly isanupper bound. For ifwesuppose c<s forsome sinS,wecan choose anumber zbetween c and s.Then zER since ¢<z, and z€Lsince z<s; thus we have acontradic- tion, foranumber cannot belong toboth Land R.Ifbisany number smaller than c,theproperties ofthecut number insure that b©L,and hence b<s for some sinS.Thus bcannot beanupper bound ofS.The proof that cistheleast upper bound ofSisnow complete. ‘Theorem IIexpresses aproperty ofthereal number system which isadirect consequence ofthe axiom ofcontinuity. Itiseasily demonstrated that ifthe statement ofTheorem IIistaken asanaxiom concerning thereal numbers, the truth oftheaxiom ofcontinuity may bededuced (making itatheorem instead of anaxiom). For akey tothis demonstration see Exercise 1.Thus theaxiom of continuity and the existence ofleast upper bounds asstated inTheorem ITare equivalent propositions. Hereafter, inarguments where wecould tean equally well either ontheaxiom ofcontinuity oronTheorem II,weshall usually appeal tothe latter. ‘As animmediate application we shall prove Theorem XIII ofChapter 1($1.62). We reword itslightly. THEOREM Ill. Let {x,} be@sequence such that x;5X25 °+SXy SXer Seo and suppose that thesetofnumbers x.has anupper bound: x,=Mforevery n.Then thesequence isconvergent, itslimit being theleast upper bound of the numbers Xp. Proof. Let Abethe least upper bound ofthenumbers x,.Then if¢>0 we have A~€ <x, forsome n,say n= N,and x,5Aforevery n,Since xw%%,for every n=N (byvirtue oftheassumption that x,=x,,.), weseethat A~€< X_SA ifNSn. Thus bydefinition lim... x,=A.This proves thetheorem. The notion oflower bound ofaset, and ofthe greatest lower bound, are defined inexactly thesame way asupper bound and least upper bound, except that thenotions of“less than” and “least” arereplaced throughout by“greater than” and “greatest.” We may summarize thedefining properties oftheleast upper bound and greatesi lower bound asfollows: The setShastheleast upper bound Aif55A forevery sinSandif,€ being any positive number, A~e <5foratleast one sinS. The setShasthegreatest lower bound BifBs forevery sinSand, € being anypositive number, s<B+.foratleast onesinS. ‘THEOREM IV.IfSisasetofrealnumbers which isnotempty andwhich hasa lower bound, then ithasagreatest lower bound. “N 3/CONTINUITY In$1.12 and §1.2 wepointed out the need toknow that ifafunction is continuous atallpoints ofafinite closed interval itactually attains anabsolute maximum and anabsolute minimum atpoints ofthe interval. Again, in$1.51, we saw that another property ofcontinuous functions occupies akey position inthe proof ofthe mean-value theorem for integrals. After our study ofthe real number system inChapter 2weareprepared toprove that continuous functions doinfact possess theproperties referred toabove. The definition ofacontinuous function was given in§1.1. We repeat the definition. Definition. Letfbeafunction which isdefined insome interval containing thepoint Xoeither inside oratone end. We say that fiscontinuous atx»provided that Tim sn,f(x) =f(x). IfXoisatoneendoftheinterval, xmust approach xfrom one side only. Wesaythat fiscontinuous onaninterval ifitiscontinuous ateach point ofthe interval. Often itisconvenient toexpress thedefinition ofcontinuity inanalternative butequivalent way, using inequalities: fiscontinuous atXoprovided thattoeach positive number €corresponds some positive number 6such that |f(x)— f(x0)| whenever |x—xo]<6andxisintheintervalonwhichfisdefined.Observe that |f(x)~—f(xo)|<e€ isequivalent tothedouble inequality f(x) ~«<f(x)< f(x) +€.Thechoice ofthenumber 6will asarule depend both on€and onx» (and ofcourse ontheparticular function f), Among the important theorems about continuity isthefollowing assertion about the continuity ofsums, products, and quotients: THEOREM L.Let fandgbefunctions definedonthesameinterval. Iff(x)andg(x) arecontinuous atapoint x=xp,soaref(x)+ g(x) and f(x)- g(x). If a(x)#0,thequotient£22jsalso satx= g(Xq)#0,thequotient 2G) isocontinuow: X=Xo. Theproof stems directly from thefundamental limit theorem (Theorem XIV, $1.64). We have, forexample, jim120.FO_forogx)~tim(ny800" 85 88 CONTINUOUS FUNCTIONS: ch3 bisecting 1,.Onatleast one ofthese closed subintervals (denote such aone by 1)fmust fail tobebounded. We proceed tobisect I;,obtaining anew subinterval I;onwhich ffails tobebounded. Byrepetition ofthis process we generate asequence I,ofclosed intervals oneach ofwhich fisnotbounded. ‘The length of1,is(b~ a)/2*"'. Hence itisclear that I,isanest, asdefined in $2.8. ByTheorem VIof§2.8 there isasingle point, sayx=c,which isineach of theintervals 1,,and hence intheinterval (a,b].Now, asshown atthebeginning ofourfirst proof, fisbounded onsome interval containing thepoint c.Denote such aninterval byJ.Since thelength of1,tends tozero asnincreases, andsince¢isinI,itisclearthatJmustcontainJ,whennissufficiently large.Butthis involves acontradiction, forfisnotboundedonI,anditisboundedonJ. Because ofthis contradiction, our initial assumption that thetheorem isfalse must berejected. We have thus completed theproof. EXERCISES L.Let f(x)=2xsin(1/x)—cos(1/x). Isthisfunctionboundedontheinterval0<x= re 2.Consider thefunction tan 'x,defined forallvalues ofx.Isitbounded? 3.Which ofthefollowing functions arebounded ontheindicated intervals? ex i @ * -t<x<t; @Zt, a@) risxsh ©ayqay StS? wmHhocrct @0<xs% ©tsinZocest. 4.Without attempting tofind exact absolute maxima, find numbers M_such that [f()| 5Montheintervals indicated ineach ofthefollowing cases: (a)f(x)=x"6x"45x7=2, “1xhs =3sin?x-2cosx sin}cos*,05x 52m; (b)f(xy=3 2cos sin5cosposS2n; ©fey tsx52. 3.2 /THE ATTAINMENT OF EXTREME VALUES. Suppose wearegiven afunction f,and suppose weknow that thefunction is bounded onacertain given interval. Let mand Mbethegreatest lower bound and least upper bound, respectively, ofthevalues off(x) ‘onthegiven interval. Isitnecessarily thecasethatf(x) y actually takes onthevalues mandMontheinterval? A Examples show that theanswer tothis question isnegative. H Example 1.Suppose wedefine f(x) =x?if0Sx <1, H f(x)=0ifx=1(seeFig.22).ThisfunctionhasM=1for H theinterval0=x=1,butthereisnoxontheintervalsuch i-s that f(x)=1. Note, however, that thefunction isnot Y continuous atx=1. Fig. 22, 22. Letfbeafunction which isdefined forallx,continuous atx=0,and such that f(x+y)=f(x)+f(y)forallvaluesofxandy.Showthatf(x)=Cx,whereC=f(1).Begin byproving (a)thatf(m/n) =(m/n)f(1) ifmandnarepositive integers, (b)thatf(—x)= f(x), and (c)that f(0)=0. Then note Exercise 3and apply Exercise 4tothefunction f(x)- xf0). y , 4a TAYLOR'SFORMULAWITHINTEGRAL REMAINDER ” 4.2 /TAYLOR’S FORMULA WITH INTEGRAL REMAINDER Consider apolynomial P(x) ofdegree n: POR)=box"+DAT!byBO, G20 whereby#0.Ifwechooseanyparticular valueofx,sayx=a,itispossibletoexpress P(x) asasum ofpowers of(xa), thehighest power being n: P(x) =ex a)"+ex a) FH ew (42-2) That this isthecase may beseen asfollows: Itis clearly true ifn=0,forinthat case thetwo expressions forP(x) areidentical inform. For n= 1wehave @ linear function P(x) =box+by,and wewish toexpress itintheform ca(x ~a)+ 1.Choosing co by,wehave P(x) bx —a)=by+abo, sothat P(x)=bolx—a)+¢) with¢,=b,+aby.Ingeneral,weproceedbyin- duction, assuming that the desired type ofrepresentation ispossible with polynomials ofdegree <n~1,where n&I.Then forthepolynomial (4.2-1) we choose cy=bo,sothatP(x)~bu(x~a)"isapolynomialofdegreeatmostn—1. Hence wecan express this polynomial asasum ofpowers of(x~a): P(x) bx=a)=ex=a4oe This isequivalent to(4.2-2), and completes the induction proof. ‘Once weknow that therepresentation (4.2-2) ispossible, itisvery easy to find convenient formulas for coy...,¢e Let usdifferentiate (4.2-2) ktimes, where 05k Sn [ifk=0, P(x) means P(x)]. After doing this wesetx=a.In this process the only term ofP(x) which leads toanon-zero result is Gu-a(x a), Thereforemq)={4 Pmay= [Heleva aI} =kena where, according totheusual convention, 0!=1 Consequently 6 Ny NO oonGe=Pa), sothat (4.2-2) may bewritten intheform . pea P(x) =P(a)+ Prayer—a)+PL(x—a)?+++PMxay, where we have reversed the order ofthe terms tosuit our convenience Now letusask whether there isany counterpart ofthis formula when the polynomial P(x) isreplaced byafunction f(x) which isnotapolynomial. That is,letusask what relation theexpression Hla)+FCay(x—a)+2D(a4 LO(x—ayn as OTHER FORMS OFTHE REMAINDER 99 THEOREM II.Letf(x) and itsfirst n+|derivatives (n=0)becontinuous ina closed interval containing x=a(either inside oratone end), Letxbeany point ofthis interval. Then lx) =fa)+f(ay(x=a)++COx—ay"+Raviy 42-7) theremainder being given by RaatfC=Nery de. (4.2-8) We have indicated the procedure forproving the theorem bysuccessive integration byparts, starting from (4.2-4). Ifone wishes, hemay give the proof more formally bymathematical induction. Formula (4.27) iscalled Taylor's formula with remainder. Various formulas fortheremainder may begiven, asweshall see in$4.3. The size oftheremainder may sometimes beestimated from (4.2-8). Thus, forexample, if|f"'"(t)| SMwhena=tSx,wecanseethat M[yg -Ma ayt!Revd [)ce—pra=Meer 4.3 |OTHER FORMS OF THE REMAINDER Itispossible toobtain (4.2-7) with adifferent formula forR,,,, under slightly lessstringent assumptions. Itwillsuffice toassume merely thatf*""(x) exists on aninterval, without necessarily being continuous. THEOREM III. Let fand itsfirst nderivatives becontinuous when aSx Sb (where a<b), and letthe(n+ 1)stderivative f*(x) exist when a<x<b. Then there isavalue x=X, a<X<b, such that f(b)=fla+flaytb~a)++++D(H—aye+O (Haye,nt (n+Dt (43-1) The same formula holds incase b<a, alltheinequalities then being reversed. This isageneralization ofthelawofthemean, andactually coincides with thelawofthemeaninthespecial casen=0. Itisdifficult togive aproof of(4.3-1) which willseem well motivated and freefrom artifice. Thefollowing proof hasbeen discovered asaresult ofcareful study and acertain amount oftrial and error. We define two functions F(x)=f(b)-fe)feb -EO b—m, 43-2) and =»oO. (4.3-3) GoGD 43-3) 100 EXTENSIONS OFTHELAWOFTHEMEAN cna Observe that F(b) =G(b) =0.Incalculating thederivative ofF(x) wefind that agreat deal ofcancellation occurs between terms arising from thedifferentiation oftheright member of(4.3-2). Thus Aifenb-91=-fonb= 4/00. ‘The f(x) here cancels thederivative ofthe previous term, ~f(x); the term =f"Gx)(b —x) iscanceled byone oftheterms coming from thedifferentiation of -ce(b—x).Thefinalresult,whichthestudentshouldverify,is Pay= Op —ay, (434) We also have Gy=-O= 43-5) Let usnow apply Cauchy's form ofthelaw ofthemean (4.1-1). Since F(b) and G(b) are zero, itreads F(a) _F(X) _FX)Gla” Gixy %PO= Gx) G- Taking account of(4.3-3), (4.3-4), and (4.3-5), this may bewritten apor =ay"F=f Toye (43-6) Ifwenow put x= ain(4.3-2) and use (4.3-6), weobtain the desired formula (43-1). This proof isvalid whether a<b orb<a, because Cauchy's formula (4-1-1) isunaffected byaninterchange ofaand b. ‘The formula inTheorem IIIiswritten inavariety ofdifferent ways by changes innotation. One important form commonly occurring intheliterature is obtained byputting b=a+h, where hmay beeither positive ornegative. The number Xbetween aand a+h may then bewritten inthe form X=at Oh, where @issome number such that 0<0 <1. Thus we have fla+h)= f(a)+flayh+>EO Oe pe43-72) Another form results byputting b=xin(4.3-1). Inthis form wemay write Fla)=flay+Fay a)+--+Mex ay+Res, (43-8) with £2) geet - Rav(n+1!(x=ay", (43-9) where Xlies between xand a. Formula (4.3-9) iscalled Lagrange’s form oftheremainder. 114 EXTENSIONS OFTHE LAW OFTHE MEAN cna 5.Suppose that fandghave continuous derivatives ofthefirst norders inaclosed interval aSx5b.Furthermore, assumethatf(a)=f(a)= +++=f*"(a)=0,g(a)= a(a)=---=g""%(a)=0, andthat g(a) #0,Without using "Hospital's rule, show that im£2)_fa),afasFa) Use Taylor's formula with remainder. 6Whatcanyouconclude inTheorem V1,iftim£12}failstoexisttherasadefinite ‘numerical limit oras+=or—%? Give your answer after acareful examination ofthe limits sin! sin! -mae tmtinSE. 7.Evaluate each ofthefollowing limits: (a)timsint- (tim sfsin?re, ptiLPPEEdt.a)timSOE, 8.Give aproof ofCase |ofTheorem VI,assuming that ¢isareal number, and utilizing thefollowing suggestions: Letbbeapoint oftheinterval, I,and consider the functions f,gontheclosed interval with end-points band c,after defining {(c)=g(c)= 0.Now apply Theorem I,$4.1, and make thededuction of(4.5-2) from (4.5~1). Explain theargument carefully. Why dowedefine f(c) =g(c) =0? MISCELLANEOUS EXERCISES: 1.Suppose that fisdefined and differentiable inaninterval containing x=a(the point x=amay beatone end oftheinterval, inwhich case derivatives atx=aaretobe considered asone-sided limits). Suppose also that {*(a) exists, but assume nothing else about second derivatives. Show that Pe)=igD=L0)==af). aa 2.Suppose that fsatisfies theconditions ofExercise 1,and that x= aisan interior point ofthe interval inquestion. Show that, iffhasarelativeminimumatx=a,then {'(a)z0, while f"(a)0 iffhas arelative maximum atx=a.These arenecessary conditions forarelative extreme atx=a.Now assume that f'(a) =0and f"(a)>0, and prove that fmust have arelative minimum atx=a.These aresufficient conditions fora felative minimum. State aset ofsufficient conditions fora relative maximum atx=a,and prove thesuficiency oftheconditions 3.Generalize theresult ofExercise 1,obtaining on$(2)~fla)(x=ayf(a)~+FFfray $%a)=tim5pa 5/FUNCTIONS AND THEIR REGIONS OF DEFINITION Thus farinthis book we have dealt with functions ofasingle independent variable. But wedonot gofarineither pure orapplied mathematics until we have occasion toconsider functions oftwo or more variables. We assume that the student has some familiarity with the concept ofafunction ofseveral independent variables ‘One ofthe first things that claims our attention when we begin tostudy functions ofseveral variables isthe nature ofthe region ofdefinition ofsuch a function. The functions ofone variable which westudy incalculus are usually defined onintervals ofthe real axis. There are only afew different types of intervals. Ifthe interval isfinite, itmay contain both itsend-points, orjust one, orneither. Ifthe interval isinfinite, but isnot the entire axis, ithas just one end-point, and this may ormay not becounted asbelonging tothe interval There ismuch more variety inthe case offunctions ofseveral variables. We shall give some illustrative examples, taking thenumber ofindependent vari- ables tobe two. Example1.f(x,y)=log(1~x?y°).Thefunction isdefined only when x7+ y?<1, since otherwise thelogarithm isundefined. The region ofdefinition istheinterior oftheunit circle with center attheorigin. InFig. 28thecircle isdashed toindicate that theboundary ofthe circular area does not belong tothe region ofdefinition ¥ Y y Z D>, Fig. 28. Fig. 29. Example 2.F(x, y)= Vx"+"1 +log(4—x7-y*), Here wemust have x7+y2 1inorder forthesquare root tobereal, while wemust have x°+ y?<4 forthelogarithm tobedefined. The region ofdefinition 116 sa POINT SETS 117 ofF(x,y)istheannularregionbetweenthecirclesx*+y?=1andx°+y?=4, The inner circumference ispart ofthe region ofdefinition, while the outer circumference isnot(see Fig. 29). Example 3.g(x,9)=Sez. The function isdefined except when the denominator iszero, that is, ‘everywhere except atthepoints oftheparabola y=4x (see Fig. 30). ¥ y } yoo j A Fig. 30. Fig. 31. Example 4.G(x, y)=Ve y+Very 1. The region ofdefinition here isdefined bytheinequalities x?= y?,x?+ y= 1. ‘The lines x~y=0,x+y=0divide theplane into four quadrants. The inequality x? y?states that thepoint (x,y)liesin(orontheedge of)oneofthose twoof the four quadrants which contain the x-axis. The other inequality states that (x,y)liesoutside oronthecircle x’+y?= 1.Hence theregion ofdefinition of G(x, y)isthat part ofthexy-plane which isshaded inFig. 31. Similar examples might begiven for functions ofthree independent vari- ables. The region ofdefinition might bethe interior ofacube, theinterior and boundary ofanellipsoid, the space between two concentric spheres, orthe interior ofasurface formed like theinner tube ofabicycle tire. Because ofthegreat variety ofpossible regions ofdefinition ofafunction of two ormore variables, itisdesirable todevote some attention tomatters of terminology about configurations ofpoints intheplane. Not only will this make iteasier forustostate things clearly, butitwill eventually become absolutely indispensable indeveloping parts ofour subject. We shall sometimes use the word “domain” for “region ofdefinition,” and bythe range ofafunction we shall mean the set ofvalues which the function takes on. 5.1 /POINT SETS In§2.7 weexplained themeaning ofthephrase “asetofreal numbers.” Since weidentify real numbers with points ontheaxis ofreals, wemay equally well 84 POINT SETS. 119 thepointsontheparabola y?=4x.InExample 3,C(S)istheexteriorofthecircle, ie.,allpoints such that x?+y?> 1, Definition. AsetSiscalled closed ifitscomplement isopen. The setofExample 3isclosed (the student should verify this tohisown satisfaction); thesets ofExamples 1and 2arenotclosed. Asetconsisting ofany finite number ofpoints isclosed. Asetmay beneither open nor closed, aswe seeinthenext example, Example 4,LetSbethesetofallpoints forwhich 1$.x?+ y?<4. This setis theregion ofdefinition ofthefunction F(x, y)ofExample 2,§5.Itisnotopen, because apoint ofthecircle x’+y*=1 has nocircular neighborhood which belongs entirely toS(see Fig. 29). The v complement ofShastwoparts:thesetofallpointsforwhichyyyedhecewyy x?y?<1,andthesetofallpointsforwhichx?+y?Z4.Itis j easily seen that C(S) isnot open, forapoint onthecircle a x'+y?=4 hasnocircular neighborhood which belongs ~| * entirely toC(S). Therefore Sisnotclosed. The setC(S) is Ya shown inFig. 34. 5 Z IfSisaset,thecomplement ofC(S)isSitself.Hence, bydefinition, C(S) isclosed if$isopen. Thus, ifoneofthe FisM4. two sets S,C(S) isopen, theother isclosed. Definition. IfSisapoint set, apoint Piscalled aboundary point ofSifevery neighborhood ofPcontains atleast one point ofSand one point ofthe complement C(S). The collection ofallboundary points ofSiscalled the boundary ofS.Wedenote itbyB(S). ‘The sets introduced inExamples 1-4have thefollowing boundaries: Example 1,B(S) isthecircle x?+y*=1. Example 2.B(S) istheparabola y?=4x. Example 3,B(S) isthecircle x’+y?=1. Example 4.B(S) consists ofthetwocircles x*+y*=1andx*+y*=4, Itisclear from thedefinition ofboundary that asetSand itscomplement C(S) have thesame boundary. IfasetSisopen, noboundary point ofSis actually inS.IfSisclosed, B(S) ispart ofS.These statements may beverified byreferring tothedefinitions. InExample 4,B(S) ispartly inSand partly in cS). Definition. Apoint PofasetSiscalled aninterior point ofSifthere issome circular neighborhood ofPwhich belongs entirely toS.Theinterior ofasetSis thesetconsisting ofallinterior points ofS. ‘ 54 POINT SETS 121 entirely ofinterior points. Forarectangle with each side parallel toacoordinate axis, tosaythat Risopen means that there arenumber pairs a,bwith a<band c,d with c<d such that Risthesetofallpoints (x,y)forwhich a<x <b and ¢<y <d. The same thing isfrequently said more briefly asfollows: {R=(x,y):a<x<b and c<y<d}. Aproperty ofopen rectangles which weshall find useful later inproving the inverse function theorem (inChapter 12)isbrought out inthe following: Assertion: Noopen rectangle Ristheunion oftwo nonempty disjoint open subsets. When weundertake tojustify this very plausible assertion, wesee that the key toaproof isagood understanding oftheproperty ofopenness. Let usreason bycontradiction and begin bysupposing that Ristheunion oftwo nonempty disjoint open subsets, Aand B.Now consider theline segment [PQ] connecting apoint PinAtoa point QinB.Since Risanopen rectangle, theline segment [PQ] obviously lies entirely inR.Since the subset Atowhich Pbelongs is open, there issome positive number rsuch that thedisc ofradius rcentered atPiscontained inA.Therefore thereissomeintervalextending along(PQ],fromPtoward Q,which lies inA.Since Balso isopen, the same argument shows that allpoints oftheline segment [PQ] which aresufficiently close toQmust, like Qitself, lieinB. Now letDbethesetofallnumbers which aredistances from Pofpoints on [PQ] which belong toA.Then Disasetofnonnegative numbers which is bounded above, since thedistance from PtoQisobviously one upper bound. Bythe least upper bound property ofthe real numbers (§2.7), Dmust have a least upper bound; call itd.Bythepreceding paragraph, weknow that disa positive number less than thedistance from PtoQ.From now on,weshall concentrate our attention onthat point Cof[PQ] which isatthe distance d from P.Obviously Cmust belong toR,yetweshall soon see that itcannot belong toeither AorB.This contradiction will prove the Assertion. Suppose first that Cbelongs toA.Then Cmust befarther than any other point ofA,that isalso in[PQ], from thepoint P.Since Aisopen there must be some disc centered atCand contained inA.But this implies that there are points ofAonthesegment (PA] which arefarther than Cisfrom P,which isa contradiction. Now tryassuming that Cbelongs toB.Clearly nopoint between Cand Qcan belong toA.Since Bisopen, there issome disc ofpositive radius xcentered atCandcontained inB.Butthiswould imply that d—1isanupper bound for D,contradicting the fact that disthe least upper bound. This completes theproof. EXERCISES 1.The setSconsists ofallpoints (x,y)such that x°+y°<1 and x<0ify=0. Describe Singeometrical language, with theaidofafigure. IsSopen, closed, orneither? What istheboundary ofS? / 122 FUNCTIONS OFSEVERAL VARIABLES chs 2.The setSconsists ofallpoints (x,y)such thateither x°+y?=1 ory=0 and 05x 51.Does this sethave any interior points? Isitclosed? 3.ThesetSconsistsofallpoints(x,y)suchthaty=x?andy31.Drawafigure,and‘describe $ingeometrical language. Is$open, closed, orneither? What istheboundary ofS? 4.Theset$consists ofallpoints (x,y)suchthat0<xy1andx>0.Isthisset ‘open, closed, orneither? What isB(S)? 5.The setSconsists ofallpoints (x,y)such that y=sin(I/x) and x>0. Does this set have any interior points? Itisclosed? What isB(S)? 6.ThesetSconsists ofallpoints (x,y)forwhich x+y?<4 andy>0except forthe points with 0<yS1 and x= In, n=1,2,...5 i€., except for the points ofacertain infinite sequence ofline segments each one unit long. Isthis setSopen? What isthe boundary of$?Are there any points ofB(S) which arealso inS?IsSa region? Isits complementary setC(S) aregion? 7.Letf(x,9)= (y—sin+) ",thefunction beingdefined whenever thisexpression hhas ameaning, but nototherwise. Isthesetofpoints where fisdefined aregion? What istheboundary oftheset? 8,Letf(x,y)=logsinx+y"",thefunctionbeingdefinedwheneverthisexpression has ameaning (real numbers only aretobeconsidered). Describe, with theaidofa diagram, thesetofpoints (x,y)where fisdefined. Isthesetopen, closed, orneither? Of what does itsboundary consist? 5.2 /LIMITS We wish todefine what ismeant bythe statement “f(x, y)approaches Aasa limit when thepoint (x,y)approaches (xo,¥o).” The statement iswritten inthe form aol JG)=A. (2-1) Ingivingthedefinition weshallassumethatthefunctionfisdefinedinaregionRandthat(xo,ys)iseitheraninteriorpointofRorontheboundary ofR.Thepoint (x, ye)may, but need not, belong toR.IfitisinR,themeaning of(5.2-1) has nothing whatever todowith thevalue f(xo, yo)atthe point (Xo,ys).The statement (5.2-1) isnow defined tomean that if¢isanypositive number, there is some neighborhood of(xo,ya)such that if(x,y)isintheneighborhood, inR,and different from (x,y«),then [f(x, y)—A]<e.Thisdefinitionmaybecompared with that forfunctions ofone variable in§§1.1, 1.61. The limit notion can be expressed verbally asfollows: The meaning of(5.2-1) isthat f(x,y)isina prescribed neighborhood ofAontherealaxis provided (x,y)isanypoint other than (xo,yo)inasuitably chosen (sufficiently small) neighborhood of(x,ys)in theplane. With theadoption oftheterm neighborhood weobtain aunification ofthe limit concept forfunctions ofone, two, orthree independent variables. The extension tomore than three variables causes notrouble and involves nonew a 52 LimiTs 123 principle. Wecontinue tousegeometric language; themeaning ofa“spherical neighborhood” inaspace offour variables ismade clear bytheinequality (=xa)+(y=yo)+(2=20)?+Ow—Wo)?<8% The fundamental theorems about limits carry over tofunctions ofseveral variables. Wecite particularly Theorems X($1.61) and XIV ($1.64). When wesay that f(x, y)>Aas(x,y)>(to,yo),itmustbestressedthatthe limit must exist and bethesame, nomatter how (x,y)approaches (xo,yg).The student will recall that, forafunction ofone variable, f(x)Aasxx)means that f(x)>Aasxxy+andalsoasx>x)~.Butinthecaseoftwovariables, (x,y)can approach (xo,yo)inainfinite number ofways. Ifitispossible tofind two different modes ofapproach to(xo,yo)such that f(x, y)approaches different limits inthe two cases, ornolimit atallinatleast one ofthe cases, then Tim, stm f(s¥)does notexist. Example1.Letf(x,y)=reactThisfunctionisdefinedexceptattheorigin. Let usshow that the limit off(x, y)as(x,y)>(0, 0)does not exist. If(x,y)> (0,0) along the x-axis, wehave f(x,0) =I(x 0).If(x,y)>(0, 0)along the y-axis, wehave f(0, y)=—1(y# 0).Thus the limits for the two modes ofapproach are 1 and 1respectively. This shows that f(x, y)has nolimit as(x,y)(0, 0). To prove directly that acertain function approaches acertain limit as(x,y+Go,yo),Wehavetoworkwithinequalities. Thefollowing example willillustrate thetechnique. Itisnotourintent,atthisstageofastudent’s training,tohave him cultivate extensively the technique ofworking exercises ofthe type represented bytheexample. The purpose ismerely tomake clearer theessentialcontentofthedefinition ofalimit Example 2.Show that im 282 Yedithseey? =o: 62-2 Interms ofinequalities, this means that if¢isany positive number, wehave to show that another positive number 6(depending one)can befound, such that 2x=y'| og 2495sP|Fe|<<ito<ee +y)<8; (5.2.3) inotherwords,denoting thefunction underconsideration byf(x,y),wehavetoshow that, if€>0, there issome circular neighborhood ofthe origin (whose radius wedenote by)such that |f(x, y)—0|<e if(x,y) isinthespecified neighborhood oftheorigin butnotactually attheorigin. Weproceed tofindsuch anumber 8,considering ¢asgiven. A “ Now 2x?ys2xLy?=2axlx?+Lyly?, one Also, [x]S074 y)"? and [ylSO?+yy". 124 FUNCTIONS OFSEVERAL VARIABLES chs Therefore (2x?—ys(x?+yx?+y4)5207+2), and PSP]s20e+ytiro<etey Itisnow clear that (5.2-3) willhold if8ischosen inany manner such that 0<5 5el2.Thus (5.2-2) isproved. Inwork ofthis kind thestudent willfind thesimple inequalities Jatbl <lobfe ab]Sa?+b?, (5.2-4) tea-b}<10)+40 la|+|b]sV3(a"+by? (52-5) quite useful. See Exercise 6forremarks about these inequalities. EXERCISES 1,Findthelimitof7%;as(x,»)approaches (0,0)alongtheliney=x;alongthe line y=mx. 2.Does,in,4zA2s exist?Givereasons 3Examine thebehavior ofey as(x,y)approaches (0,0)alongvarious straight lines. Then consider what happens forapproach tothe origin along the curve y=xIsthere alimit as(x,y)—(0,0) without restriction? 4.Show ineach case that the given function does not approach alimit as (x,y)+(0,0), byexamining thebehavior ofthefunction foratleast two modes of approach x= e+os ote xy? xt+3x7y?+2xy?OF OS 5.Define afunction bysetting f(x,y)=0 ifyO orify=x’, andf(x,y)=1 if 0<y <x", Show thatf(x,y)=0 as(x,(0,0) along anystraight linethrough theorigin, Find acurve through theorigin along which f(x, y)=1(except attheorigin). 6.IfAand Bare nonnegative numbers, theinequality A=B isequivalent to A’=B’.Usethisfacttoprovethecorrectness of(5.2-4); thenshowthat(5.2-5) is correct. 7.Letfessy)=ay(%F2). Showthatf(x,y}54a"+7),andhenceprovethat f(x,¥)approaches alimitas(x,y)>(0,0). &Letfo=325If€>0,find8sothat0<(x?+y?)"<8impliesIf(x,y)|<«. 9.If€>0,show that[2x?~6xy +5y"|<e when (x?+9)!<(@/13)"", 10,Showthat35+)<eit0<x"+y*<6%forasuitablychosen8dependingon« COMP Cao4 showsob? 53 ‘CONTINUITY 125 1,Showthat|x’y"](x?+y")"". 12,Does2#272 approach alimitax,3)0,0)? 5.3 /CONTINUITY The notion ofcontinuity depends onthenotion oflimit, aswas pointed outat thebeginning ofChapter 3. Definition. Letf(x, y)bedefined inaregion R,and let(xo,ys)beapoint ofR.We saythat fiscontinuous atthispoint if celim,fla9)=fx0. If(%o,90)isaninterior point ofR,themode ofapproach of(x,y)to(x.yo)is unrestricted inthis definition. But, if(x,x)isaboundary point ofR,there isthe restriction that (x,y)must remain inR.We say that fiscontinuous inRifitis, continuous ateach point ofR. Iffand garedefined inthesame region R,and each iscontinuous atapoint (%,yp)ofR,then thesum and product functions fOsy)+R0% ys $05 yBO ¥) arealso continuous at(xo,yo). The quotient function fe.» ayy) iscontinuous at(x,yo)provided g(xo, 3)#0. These assertions are direct generalizations ofTheorem I,§3.They may beextended tofunctions ofmore than two variables. The theorem ofChapter 3allhave important analogues for functions of several independent variables. We donot wish atthis point toprove allthese analogous theorems, but we shall discuss the statements ofcertain theorems which will beused inthechapters immediately following. Indealing with theanalogues ofTheorems II($3.1) and III($3.2) ofChapter 3itisnecessary tointroduce theconcept ofabounded point set. Definition. Apoint set$intheplane iscalled bounded ifallitspoints areinside some suficiently large circle. Forapoint setinspace thedefinition issimilar; we write “sphere” instead of“circle.” Examples. Theinterior andboundary ofatriangle form abounded point set. ‘The setofallpoints between thelines y=0, y=1isnotabounded point set. Wenow state two important theorems. ‘THEOREM I.Ifafunction iscontinuous ateach point ofaclosed andbounded 7 126 FUNCTIONS OFSEVERAL VARIABLES chs region R,the function isbounded onthe region (ie., the values ofthe function form abounded setofreal numbers). THEOREM Il.Let fbecontinuous onaclosed and bounded region R.Let m andMbethegreatest lower bound andleast upper bound ofthevalues of onR.Then ftakes oneach ofthevalues m,Matleast once inR. Proofs ofthese two theorems will beconsidered later, in$§17.2, 17.3. Theorem VofChapter 3(§3.3) hasthefollowing analogue, Westate itfor thecase oftwo independent variables. THEOREM Ill. Let fbedefined inanopen setcontaining the point (xs, yo) Suppose that fiscontinuous atthepoint and that f(xo, ys)#0.Then there is aneighborhood of(Xo,ys)throughout which f(x, y)has thesame sign asat (Xo, Yo). The proof islefttothestudent. There isafeature ofthecontinuity ofafunction f(x, y)which deserves notice. Ifwefixy,say y= yo,f(&, ya)isafunction ofxalone. Likewise f(xo, y)is afunction ofyalone. Itcan happen that each ofthese functions ofasinglevariable iscontinuous, andyetthatf(x,y)isnotcontinuous. Anillustration ofthis possibility isgiven inExercise 3. There isanother theorem which will beneeded later. Itdeals with composite functions, and may beroughly stated inthe form: Acontinuous function of continuous functions iscontinuous. The number ofvariables isimmaterial. Examples. F(z) =sinzisacontinuous function ofz,andf(x,y)=(1+xy)is continuous function ofx,y.Therefore F(f(x, y))=sin(1 +xy? isacontinuous function ofx,y.Or, again, FQyy,2=x?+y? +2?and f(xy)=x( tx+yy? arecontinuous functions ofx,y,zand x,y,respectively. Therefore 2y4y? x FO,yf = ++TH isacontinuous function ofx,y. Weformalize one such theorem about composite functions. THEOREM IV.LetF(x, y,2)becontinuous inanopen setBofspace. Let (x,y)becontinuous inanopen setRofthexy-plane. Writing z=f(x,y). ‘suppose thatthepoint (x,y,z)isinBwhen (x,y)isinR.Then thecomposite function F(x, y,f(x,y))iscontinuous inR. Fltay)) hyolswhine FOI etsonDadHere? Pvek sa MODES OFREPRESENTING AFUNCTION 127 We shall notgive aproof here. This theorem isaspecial case ofthe Theorem IIwhich isproved in$11.7, EXERCISES Lff(x,y)=8G*P?whenx'+¥*40,howmust/(0,0)bedefinedsoas10 make fcontinuous at(0,0)? 2.Letusdefinef(x,y)=92)ifx40,andf(x,y)=yifx=0.Doesfhaveany points ofdiscontinuity? 3.Ifwedefine f(x,»)= xy/(x"+ y°)when x°+ y#0,andf(0,0)= 0,show that fis discontinuous at(0,0), butalso that f(x,0)andf(0, y)arecontinuous functions ofxand y, respectively, with noexceptions. 4.Letf(x, »)=(x?+y’)tan“"(y/x) ifx40,anddefinef(0,0)=0,butdonotconsider Jfdefined ifx=@and y#0. (a)Isfcontinuous at(0,0) according tothedefinition inthetext?(b)sitpossibletodefinefattheoneadditionalpoint(0,1)sastomakeitcontinuousthere? 5.Letf(x, y)=xylog(xy) ifxy>0, and define f(x, y)=0ifxy=0,Where, ifatal, isfdiscontinuous? M6, Letf(x,y)= (Sx+y)Mx— y).Show directly bythedefinition that fiscontinuous at(4,1) byproving that, if«>0, f(x,9)4,1}<eprovided(x,y)isinasufficiently small neighborhood of(4,1). Start byshowing that Wes,9)=104,9]52]—4]+Bly—1) atthepointsofthesquare3<x<5,0<y<2. 7.Iff(x,y) =e" when x#y,how must fbedefined when x=ysoastomake itcontinuous atallpoints oftheplane? 8Let usdefine f(x,y) =0 ifyO orifx°Sy, and f(x,y)=4ylx?~ypla* if o<y<x’ (a)Isfcontinuous at(0,0)? (b)Discuss possible discontinuity atother points ontheline y=Oor thecurve y=x",9.Letfix,y)=x(1—x"~ 9",theregionRofdefinitionbeingdefinedbyx°+y°<1. Ibis possible toaddthesingle point (0,1) toRanddefinefs0astomakeit continuous atthat point? Consider values offonthecitcle x*+y"—y =0,and also at other points inRnear (0,1.10,Iff(x,y,2)=xy210e"+92422)when49°42" 40,isitpossibletodefine{10,0,0) soastomake fcontinuous attheorigin? 5.4 /MODES OFREPRESENTING AFUNCTION Thestandard method ofrepresenting afunction ofonevariable isbygraphing in rectangular co-ordinates. We write y=f(x) and plot thepoints (x,y).Iffis continuous onaninterval, thegraph will beacurve intheplane. - The corresponding procedure forthecase oftwo independent variables is familiar. Wewrite z=f(x,y),andplotthepoints (x,¥,z). Iffiscontinuousina region Rofthexy-plane weobtain asurface inspace (see Fig.36). ni che ie 128 FUNCTIONS OFSEVERAL VARIABLES chs y 2 1 Co ' a}ose + { 2} 65 8 oT oe + y tfweodod506 f z ca ores 46 Fig.36. M7. When wegotothree independent variables there isnosatisfactory analogue ofthe foregoing methods ofgraphical representation, forwecannot draw upon any familiar geometric intuition tovisualize w=f(x, y,z) asdefining a configuration inspace offour dimensions. There is,however, another mode of representation which ishelpful. Itisavailable aswell inthe case oftwo independent variables, andsince thefigures areeasier todraw, webegin with that case. WhenfisdefinedinaregionR,wecanthinkofeachpointofRasbeinggiven alabel, namely, thevalue f(x, y)atthat point. Agood example isobtained bythinking ofthe xy-plane asamap onwhich elevations above sea level are marked atvarious locations, f(x,y)beingtheelevationinfeetat(x,y)(seeFig. 37). Tocarry this example further, imagine that the map isatopographic map with contour lines drawn in,showing lines ofequal elevation. Each line is labeled; there isaline for 500 feet above sea level, others for 400, 600, and so ‘on. Inthe aggregate, the configuration ofthese lines, together with their numbering, gives usagood visual representation oftheelevation asafunction ofxand y. This “topographic map” idea can becarried over toany function f(x, y). Instead ofcontour lines weconsider curves along which f(x, y)isconstant in value. Such acurve iscalled alevel curve ofthe function. Ifthe constant value is C,theequation ofthelevel curve isf(x, y)=C.SeeFig38andFig.39.Isobars ¥ 2 Fig.38.Level curves off(x,y)=x+y". 54 MODES OFREPRESENTING AFUNCTION 129 y Fig.39.Level ofcurves off(x,y) =x°+y°-2. (curves ofequal atmospheric pressure) onameteorological chart furnish another good example oflevel curves ofafunction. The three-dimensional analogue ofthis mode ofrepresentation isnow easily grasped. Instead oflevel curves we shall have level surfaces f(x, y,2)=C. Common physical examples offunctions ofthree variables which are con- veniently visualized inthis way are density and temperature inagas orother medium. 6 PARTIAL DERIVATIVES Bt these derivatives, ifwewrite u=f(x,y),arethefollowing: a(*)-fu (24)=au ax\ax)~Gay \ax)~Byax’ (2)ee ax(ay) “axay’ ay(ay)=35 Example 2.For thefunction ofExample 1wehave Fu teFHmays yte™, Fu se? Ze”ayax=2x+3xy’e Bye”, eu Se)—Byte, aeayTTB Bye, eu 2yte-) 3Sore oxy? —6xye™. We observe that eu eu dyax axay en inthis example. We shall ordinarily find that therelation (6-1) holds true forthe functions wemeet inpractice, forthe relation isvalid atapoint provided both thesecond derivatives aredefined inaneighborhood ofthepoint and continuous atthepoint. This will beproved in§7.2 (Theorem III). Apartial derivative off(x,y)isagainafunction ofx,y.Todenote thevalue ofZatthepoint(xo,yo)wemayuseoneoftheexpressions (2) afAx}09)"—OXLoer0" These notations are rather awkward, however; itisdesirable tohave astandard functional notation for partial derivatives. For afunction f(x,y) oftwo in- dependent variables weshall write =4, =H.fay=t fay=7 Forthevalue ofapartial derivative atapoint wethen have expressions such as =at) =(2 ficuyo= (2). ftad)= (2), 132 ‘THE ELEMENTS OFPARTIAL DIFFERENTIATION cn For second derivatives we use the notations a1 (L =2(f =2(a at(£ feed=2(FO), potnr= 5(2). Observe theordering ofthenumerical subscripts inrelation totheorder of carrying outthedifferentiations; Irefers toxand 2refers toy. ‘The notation isextended inanobvious way toderivatives oforder higher than thesecond, and also tofunctions ofmore than two independent variables. ‘Thus forexample, 2a fates=25(2), and ai, 212) w= gatened [2(%8)]. 6.1 /IMPLICIT FUNCTIONS We often deal with functions which are defined implicitly asthe solution of certain equations. Inordinary practice wecan find thepartial derivatives ofsuch afunction bythesame procedures which welearn inelementary calculus. Example 1.Find2fromtheequation yt eye ye -retreat, 6.1-) contheunderstanding that zisdependent and x,yare independent. Wehave dx228216* 9ax and so az 9x #9 (6.1-2) The equation (6.11) actually defines two functions of(x,y),corresponding to thetwo choices ofsign in 2 ye =23(1-2-¥)”. :z=23(1-5-2) (61-3) Bysubstituting (6.1-3) in(6.1-2) weobtain thepartial derivative foreach of these two functions: a_53x(,_2_y)™eoFte(-t6- 5) GIy 61 IMPLICIT FUNCTIONS 133 The result (6.1-4) could also have been obtained bydifferentiating (6.1-3) directly. The procedure can also beapplied inthecase offunctions defined by simultaneous equations. Example 2.Ifuand varedefined asfunctions ofx,ybytheequations ucosn-x=0 1-5) usinv~y=0, findthe partialderivatives Mt,%.Los Wesax"ax Method I.One method ofprocedure istoattempt tosolve foru,vinterms ofx,y.Ifthis can beaccomplished, wecan then calculate the required partial derivatives directly. From (6.1-5) wehave uw?cos?p=x?,usin?v=y?, Now add these equations and use afamiliar trigonometric identity. The result is Wextty) oru=tVxF (6.1-6) Next, going back to(6.1-5), wesubstitute thevalue just found foru.Wefind x ,cosv=ay sinsWS 1-7) thesame sign being taken before theradical inboth cases. We might also write =2 7 tan v= (6.1-8) incases x#0.We seethat there areingeneral two possible determinations ofu from (6.1-6); forvthere areaninfinite number ofpossible determinations from (6.1-7), differing bymultiples of2m.The derivatives ofumay befound from (6.1-6): i ox Vee Yr Infinding thederivatives ofvitiseasier towork from (6.1-8). Wehave 2922=F,secty= 1+tanto=1+2y sec?»SP=Srsec?v=1+tan?o1+8y x,-a- =a. ax” xTsec"y x+y) This result could have been obtained byexpressing vasaninverse tangent and then differentiating. Itshould, however, benoted that visnotnecessarily the principal value oftheinverse tangent ofy/x. / 62 GEOMETRICAL SIGNIFICANCE OFPARTIALDERIVATIVES 135 6.2 /GEOMETRICAL SIGNIFICANCE OF PARTIAL DERIVATIVES Just asthe ordinary derivative ofafunction ofone variable has itsgeometric realization inthe slope ofaline which istangent toacurve, sothe partial derivatives ofafunction oftwo variables have ageometrical significance in connection with aplane which istangent toasurface. Our purpose inthis sectionistoshowhowthepartialderivatives 2and2f,whenx=aandy=b, are related tothe plane which istangent tothe surface z=f(x, y)atthepoint (a,b,c). Inthissection weshall notgive aformal definition ofthetangent plane. We reserve full discussion ofthis matter to$6.4, because the concept ofthe tangent plane isthegeometrical counterpart oftheconcept ofthedifferential of afunction oftwo variables. Let $bethe surface z=f(x, y),and let(a,b,c) beapoint onS.Then c=f(a, b).Consider theline through thepoint (a,b,c)parallel tothez-axis. Let usvisualize various planes containing this line, each such plane cutting the surface Sinacurve. One such plane, cutting the y-axis perpendicularly aty=, isshown inFig. 40. Inthediagram, thecurveoftheintersection ofSandthis j//| plane, y=b,isrepresented ashavingatangentlineLat J the point (a,b,c). One can also imagine aplane x=a, passing through (a,b,c) and intersecting the x-axis \ perpendicularly at(a,0,0). More generally, one can imagine aplane different from either ofthese two,but im soplaced thatitpasses through (a,b,c) andisparallel rey tothez-axis,Thestudent shouldconstruct forhimself a”diagram similar toFig.40,with aplane through (a,b,c) ,“— Fab) parallel tothez-axis butcutting neither thex-axis nor fig,gp, the y-axis atright angles. This exercise ingeometrical visualization will behelpful forthefollowing discussion. ‘Suppose that each plane through (a,b,c)and parallel tothez-axis cuts the surface S$inacurve which has atthe point (a,b,c)atangent line which isnot parallel tothez-axis. Suppose further that allofthese tangent lines, correspond- ingtothedifferent planes ofthetype described, lieinasingle plane. Then this single plane must surely bethetangent plane tothesurface Sat(a,b,c),ifindeed there issuch atangent plane. Fig. 40shows four different curves onS,together with their tangent lines, allintersecting at(a,b,c). Assuming now thatthere isatangent plane to$at(a,b,c) notparallel tothe z-axis, letusseehow tofinditsequation. Ifcosa,cosB,cosyarethedirection cosines ofalinewhich isnormal (perpendicular) tothisplane, theequation of theplane can bewritten (cosa)(x~a)+(cosB)(y~b)+(cosyz~¢)=0. Since theplane isnotparallel tothez-axis, weknow that cosy#0; wecan therefore solve forz~¢bydividing bycosy,thus obtaining anequation ofthe 4 «2 GEOMETRICAL SIGNIFICANCE OFPARTIAL DERIVATIVES 137 for the direction cosines ofthe normal toaplane are proportional tothe coefficients ofx,y,and 2respectively, intheequation oftheplane. Here isa result toberemembered The line normal tothe surface z=f(x,y) atagiven point has direction ratios a%.,@.-1, thepartialderivatives beingevaluated atthepointinquestion. Example 1.(a) Find the equation ofthe plane tangent tothe paraboloid 482=2x?+3y? atthepoint (3,2,)). (b)Find thedirection cosines ofthenormal tothesurface atthepoint. (a)Wehave a ze.aE ads, 48 =6y; atthepointinquestion,therefore, 22=1,2=1,andtheequationofthetangent wepointinquestion, therefore,$==jp$==a e plane is ofMa-3)+Ky-2),oF2x+2y-82=5. (b)Toobtain thedirection cosines from theratios 4:1:~1, wefirstcompute 2=3V2 (ey?e+yy?=V2. s a 3v2 ‘Thedirection cosines arefound bydividing 4,1,-1by—Z=. They are, new, 1. 1,4 accordingly, 1. 1, 4. inelys53 IVE Wi Example 2.Show that atevery point ofintersection ofthe two surfaces 2=2x'+y)), 82=17-(x'+y"), the normals tothetwo surfaces are per- pendicular. (Because ofthis wesay that thesurfaces intersect orthogonally.) ‘The student will readily find that thesurfaces are paraboloids ofrevolution intersecting each other allalong thecircle x°+y=1intheplane z=2.There arenoother intersections. Now, atapoint ofthefirst paraboloid thedirection ratios ofthe normal tothe surface are axidy:-1y forthesecond paraboloid thedirection ratios ofthenormal arefound tobe -X.2.-Rede, ‘The condition forperpendicularity ofthese two normals atapoint which is common tothe two surfaces istherefore =)5ay(2)a1= ax() +4y(G)+ 1-0, fengend ae! Yin t ‘ y‘A K 138 ‘THE ELEMENTS OFPARTIAL DIFFERENTIATION cn.6 or~(x"+ y°)+1=0. Since thisequation issatisfied along theintersection ofthe surfaces, thedemonstration ofperpendicularity iscomplete. EXERCISES 1,Find theequation ofthetangent plane tothesurface z=e™*siny (a)atx=0, y=7/2, (b)atx=0, y=, (€)atx=0, ¥=0. (€)Make asgood a diagram asyou canofthesurface for 05y$x, x>0. 2.Find theequation oftheplane tangent tothesurface x?+ 2xy?—72?+3y +1=0at G1. 3.Prove that theplane tangent tothesurface 2=x?—y? atthepoint (a,b,c) is pierced bythez-axis atthepoint forwhich z=~c. 4.Find thepoints ofthepraraboloid z=.x"+y?—1 atwhich thenormal tothe surface coincides with the line joining theorigin tothe point. What isthe acute angle between thenormal and the2-axis atthese points? 5.Ifa’#b®,provethatnonormaltothesurfacez=(x"/a")+(y"/b")~¢, atapoint forwhich x#0 and y#0, can pass through theorigin, 6.Prove that thespheres x74y?+z*=16, x?+(y—5)'+27=9 intersect orthogon- ally, using themethod ofExample 2. 6.3 /MAXIMA AND MINIMA We sometimes have occasion toinquire about the largest orsmallest value attained byafunction under specified circumstances. Inspeaking about maxi- ‘mum (orminimum) values itisvery important todistinguish between arelative maximum and anabsolute maximum. Suppose weare dealing with afunction f(x,y)definedinaregionRofthexy-plane. Definition. Wesay that thefunction fhas arelative maximum atthepoint (a,b) ifthere issome neighborhood of(a,b) such that f(x, y)= f(a,b)forallpoints (xy) ofRwhich are inthis neighborhood. We may express the definition otherwise bysaying that thevalue offat(a,b)isatleastasbigasatanyofthe points (x,y)around (a,b)and not toofaraway. ‘Thus forinstance, within agiven range ofmountains, theelevation oftheland surface above sealevel attains arelative maximum atthesummit ofany particular peak intherange. Definition. Letfbedefined inaregion R,andletSbeanypart ofR(i.e., any point-set inR).Inparticular, Smight beallofR.Suppose there isinSapoint (a,b) such that f(x,y)3f(a,b)forallpoints(x,y)inS,Wethensaythatonthe setSthefunction fhasanabsolute maximum at(a,b). Observe that, onagiven setS,fcanhave arelative maximum which isnot anabsolute maximum, Observe also that afunction may failtohave anabsolute ‘maximum onagiven set(think ofthefunction 1/(xy) inthefirst quadrant), Similar definitions are made for relative and absolute minima ofafunction. to) @ 63 MAXIMA AND MINIMA 139 Inproblems where wehave tofind theabsolute maximum ofafunction ona given setweusually find that itisconvenient tobegin bylooking forrelative maxima. Ifthere areonly afew ofthelatter wemay beable easily toselect one which furnishes anabsolute maximum. Hence itisuseful tohave criteria for locating relative extrema. THEOREM I.Letfbedefined onaregion R,andletthefunction have arelative extreme (maximum orminimum) atthepoint (a,b) ofR.Suppose further that (a,b)isaninterior point ofR(not ontheboundary), and that fhasfirst partial derivatives at(a,b). Then these derivatives arezero atthat point: f(a,b)=0, —fxla,b)=0. 63-1) Proof. This theorem should becompared with Theorem IIIof§1.12. The proof isbased onthis earlier theorem. Consider f(x, b);this isafunction ofthe single variable x,itsvalues being those ofthe function f(x, y)along the line y=b.Asafunction ofx,f(x,b)hasarelativeextremeatx=a.Moreover, the derivative off(x,b) atx=a isf,(a,b). Therefore, byTheorem III, §1.12, we conclude that f,(a,b) =0.Inthe same way, applying this earlier theorem tothe function f(a, y)ofthesingle variable y,weconclude that f,(a, b)=0. The hypothesis that (a,b)isaninterior point ofRisessential. Arelative extreme can occur ataboundary point ofR,and inthat case equations (6.31) may not hold. Itisimportant torealize that, under theconditions stated inTheorem I,the vanishing ofthe first partial derivatives isanecessary, but not sufficient, condition forarelative extreme. Ifthesurface z=f(x, y)hasatthepoint x=a, y=b atangent plane which isparallel tothe xy-plane, then equations (6.3-1) hold; but zneed notbearelative extreme atsuch apoint. A“saddle-point” ofa surface isanillustration ofsuchasituation. The foregoing definitions and Theorem Iextend tofunctions ofthree or more variables inanobvious manner. Asinelementary calculus, sufficient conditions forarelative maximum or minimum can beformulated byadding to(63-1) certain conditions onthe second derivatives offatthepoint (a,b).Wediscuss such conditions in$7.6. Forthepresent, however, weproceed toillustrate some uses ofTheorem I. Example 1.Find thepoint oftheplane 2x—3y—4z=25which isnearest to the point (3,2,1). ‘ IfDisthedistance from thepoint (x,y,z)oftheplane to(3,2, 1),wehave D?=(x—3) +(y—2)°+ (z= I)andz=(2x—3y—25).Hence, eliminating z, DP=(x—3) +(y~27+Gxly 3, Weseek theminimum value ofD?asx,yrange through allpossible values. In thiscase allpoints areinterior points oftheregion (namely thewhole xy-plane), andD*haspartial derivatives atallpoints, Wetherefore look forpoints atwhich 4 2 140 ‘THEELEMENTS OFPARTIALDIFFERENTIATION on 2 ) 2D)_AP0.Theequations tobeconsidered are x” ay Ax-3)+2xly—B)=0, Ay=2)+2x—ly—8)@)=0. Onsimplifying, weobtain 10x—3y =53, 6x +25y =-55. The solution isfound tobex=5, y=~. Substituting intheequation ofthe plane, wefind z=~3. Wenow argue asfollows: The function D?certainly has anabsolute minimum (from the geometrical nature ofthe problem). This absolute minimum isalso arelative minimum, and the conditions ofTheorem I apply. But weobtain aunique point atwhich the two first partial derivatives vanish. Hence, this point must furnish thedesired absolute minimum. Example 2.Locate the points which might furnish relative maxima and minima ofthe function S(x,y) =2xy<a?=yy? intheclosed region x?+y? 1(which istheregion ofdefinition ofthefunction). Hence, find the absolute maximum and minimum values ofthe function. We first apply thecriterion ofTheorem I.We have te -y-yynBarys3x(1—x=)", af. _g-yeBeare early The interior points oftheregion arethose forwhich x°+y?<1. The interior points which might furnish arelative maximum orminimum are among those which wefind bysolving theequations 3a(1—x7— y= —2y, . . » 32) By(I =x?=y?)!"=2x, Anobvious solution ofthese equations isx=0, y=0.Ifneither xnor yiszero wemay divide one equation bytheother and obtain theresult xy, 2a y?,pay ortays Hence, substituting back inthefirst equation of(6.3-2) after squaring both sides, we obtain 9x71 =2x7)=4x7,oF918x?=4. Thuswefindx?=y?=jj.Goingbackagainto(6.3-2)wehave(1—x?- y?)"?=} @ 63 MAXIMA AND MINIMA 141 and hence 3x()=—2y, orx=—y. Note that x=y isruled out. There are therefore three points inallwhich satisfy (6.3-2). They are Po=(0,0), Pi=(VP -IV3), Ps=CAVE IV. The accompanying table ofvalues may now beconstructed: Point Value off Py =i PyandP:—~23/27 We emphasize that Theorem Idoes notassert that thefunction has relative extrema atallthree ofthese points; itonly states that ifany relative extrema occur atinterior points, such extrema are found among these three points Before drawing any conclusions about absolute extrema wemust investigate the behavior ofthefunction ontheboundary oftheregion, Ontheboundary we have x?+y?= Iandtherefore f(x,y)=2xy. Tolook forextreme values offon theboundary wemight solve fory:y=+V1— x",andlook fortheextremes of +2xV1— x’bythemethods ofelementary calculus. Wefindsuch extremes when x?=|.Amore elegant procedure istointroduce theparametric equationsx=cos6, y=sin@fortheboundary circle (here @istheusual angle ofpolar co-ordinates). Then 2xy =2cos 6sin@=sin26,and wesee that thevalues range between 3xTm), 7St ~1(at6=32or7Z)and+1(at¢=orF). Wehave now found four more points which must beconsidered along with the original three when Welook forthe absolute minimum and maximum values of the function inthe closed region. When we compare the values +1with the values atthepoints listed inthetable, wesee that thefunction has theabsolute maximum value +1,andtheabsolute minimum value —1.Themaximum occurs atthetwoboundary points (V2/2, V2/2), (~V2/2, —V2/2).Theminimum occurs atthe interior point Pyand atthe two boundary points (V2/2, ~V2/2, (-V2/2, V2/2). Ourwork hasnotsettled thequestions astowhether theinterior points P,,P,arepoints ofrelative extremaorsaddlepoints.Theyareinfactsaddle points, asmay beshown byanexamination ofthefunction inpolar co-ordinates. Example 3,Ashelter foruseatthebeach istobebuilt intheform ofa box-like space with canvas covering onthetop, back, andends. If96square feet ofcanvas are available, what should bethe dimensions ofthe shelter togive it maximum cubic content? Lettheshelter beyfeetbetween ends, xfeetfrom front toback, andzfeet high. Itsvolume isV=xyz. The area tobecovered bycanvas is A= 2dxztxyt yz Since A= 96,wecan usethis lastequation toeliminate one variable, sayy: 96-2x2 yy_4,B=xzyore VOR YG: © 142 ‘THE ELEMENTS OFPARTIAL DIFFERENTIATION ch.6 Here Visexpressed interms oftheindependent variables x,z;wecan set av _al . nisi2¥-2¥<9andsolveforxand2.Analternative procedure whichisinsome ways preferable isthefollowing: Differentiate both oftheequations V=xyz, A=2xz tay tyz with respect toxand z,regarding yasafunction ofxandz.Differentiation with respect toxgives = ®o=ay x. OnMayet ar2onrtx ttyt2e Wenoweliminate 5%between these twoequations YoY oe = =» de-yty-B=0,2x=y. Since xand zenter symmetrically, weinfer that 2z=yalso,andhencethat x=z, Togetthevalues ofx,y,2wereturn totheformula forA.We now have 96=2x(x)+x(2x)+Qx)x=6x", Hence x*= 16,x=z=4, y=8. The volume oftheshelter ofthese dimensions is 128 cubic feet. ‘One logical issue still remains tobesettled intheforegoing “solution” ofthe problem posed inExample 3.How doweknow that wereally found the dimensions which yield maximum volume? Our method was based ontwo assumptions: First, that there isashelter ofmaximum volume under thegiven conditions, and second, that when Visexpressed asafunction ofthe in- dependent variables x,z,theequations SY=$Y~0aresatisfied whenVattains itsmaximum. Ifwecanjustify these two assumptions, oursolution will befully established. Let usthen consider Vasafunction ofxand z.The formula is 48—xz. V=2nSe, (63-3) we donot consider allvaluesofxandz,however,butonly, those which have ameaning fortheproblem under consid- 4 eration. Thuswemusthavex20,2=0.We musthavexzs J48also,sinceanegative volumewouldhavenomeaning. LATheregion Rinwhich weconsider thevalues ofVisthus |,composed ofallpointsofthexz-planeforwhichx20,2=|>>0,xz548,excepttheonepointx=0,z=0.Thispointis—}7m,ruled out,since Visnotdefined there. Theregion Ris©! shown inFig. 41. Fig. 41. \ 63 MAXIMA AND MINIMA 143, Let usnow establish thefact that among allpossible values ofVinthe region R,there isanabsolute maximum value, and that this maximum occurs at aninterior point ofR.Observe that Vispositive inthe interior ofRand that V=0 atallpoints oftheboundary ofRexcept theorigin (where Visnot defined). Now, from thefact that xz548 inRweseethat 48 1vaso lag and hence V-+0 asavariable point (x,2)ofRmoves insuch away that either x2 or2%, Finally, V-+0 as(x,2)-+(0,0). For, 2xzSx?+2? and x+2@ Vix" +2"aretrue inequalities (the latter when xand zarenon-negative), and therefore 96(x2+2°) seo 0s Vs)-ovrFZ, Vitter sothat V-+0 as(x,z)-»(0, 0).From theforegoing arguments itisnow clear that ifweform anew region Rybyexcluding from Rthepoints forwhich x?+z7<3? andx?+2*> 1/8?,where4issufficiently small,thevaluesofVattheseexcluded points will allbesmaller than some ofthevalues ofVinthe remaining region Ro.Since Roisabounded and closed region inwhich Viscontinuous, Vmust have amaximum value inRo(Theorem II,§5.3). The maximum value ofVinRy will also beamaximum value ofVinrelation toallpoints ofthelarger region R. Since this maximum ispositive, itmust occur ataninterior point ofR. WenowapplyTheorem Itodrawtheconclusion thatSY=2¥=9atthe point where Visamaximum. Since there turned out tobeonly one point inR, namely x=z=4, atwhich these conditions are satisfied, this point must bethe point where Visamaximum, EXERCISES 1,Find thepoint oftheplane x+4y+4z=39nearest thepoint (2,0, 1. 2.Find the greatest value ofthe function xy(e—x—y) inthe closed triangular region with vertices (0,0), (c,0), and(0,¢).Assume ¢>0. 3.Find theabsolute maximum of144x?y*(I—x-y) inthefirst quadrant ofthe xy-plane. 4.Does f(x,y)=x74Ixy+y2+(S761x)+(S76ly)haveanyabsoluteextremainthe region x>0, y>0? Ifso,find where such extrema occur, and thetype (maximum or ‘minimum). Give allthesupporting details ofyour argument. '.Find the absolute minimum value of 2gyay(2ATa=by)? SexyaareyesPASO b2) where A,a,b,andcarepositive constants. Allvalues ofxandyareadmitted. How do ‘you know that aminimum exists? 6, Find the absolute extreme values ofthe function f(x, y)=2xy+(I—x?—y°)"*in theregion x+y" 1. Vs TEELEMENTOFPARTIALDiFERENTATION one 7.Find the absolute extreme values ofthe function f(x,y)=xy~(1—x7-y')"”in theregion x°+y'S 1 8.(a)Introducing polarco-ordinates, showthatthefunctionofExample2becomes f(a,y)= F(r,0)=Fsin20~(1°),(b)FindtheextremevaluesofFfor-15751, @unrestricted, considering Fasdefined inaregion ofther@plane. (c)Atthepoints of theré-plane whichcorrespond tothepointsP,,P:ofExample 2,show2E=o,2F=o, £F-<0,28>0,Fromthesefactsexplain whythesepointsaresaddlepointsandnotpoints ofrelative extrema. 9.Solve Exercises 6and 7bytheintroduction ofpolar co-ordinates asindependent variables. 10.Find thegreatest value ofthefunction sinxsinysin(x+y)intheclosedtrian- ‘gular region with vertices (0,0), (7,0), 0,7). 11.Find theabsolute maximum value ofthefunction (x*+2y")e~"*””, considering allpossible values ofxand y. 12,Findtheabsoluteextremaofthefunction3x°~Sxy—4y?+2x+I6yinthesquare0sx52,05ys2 13.Find theabsolute extrema ofthefunction x"+y?+3xy?~ 15x~I5yinthesquare 05x53,05y53 ~14,Find theminimum value ofthefunction (12/x) +(18/y)-+.xy inthefirst quadrant. How doyou know there isaminimum? 15.Find themaximum value ofthefunction (xy~4y ~8x)/x*y? inthefirst quadrant. How doyou know there isamaximum? 16,Arectangular box without atophaslength x,width y,anddepth z.The combined area ofthesides and bottom isfixed asSsquare feet. ExpressthevolumeVoftheboxasa function ofx,y,and show that Visgreatest when x=y=(S)3)", z=x/2.Justify your sotution completely. 17,Consider thefunction Viv+y"+V(x=1)'+y". (a)Explaincarefullywhythis function must have anabsolute minimum atsome point ofthe plane. (b)What isthe minimum and where does itoccur? (¢)Are alltheminimum points found bysetting the partial derivatives equal tozero? 18,Consider thefunction f(x,y)=[y|+Vx7+ 0—D*-(a)Atwhatpointsdooneorbothofthe first partial derivativesofffailtoexist? (b)Findall points where they both exist and areequal tozero. (c)What istheabsolute minimum value off,and atwhat points does itoccur? 19.Forwhat position ofthepoint (x,y)isthesum ofthedistance from (x,y)tothe xcaxis andtwice thedistance from (x,y)tothepoint (0,1) aminimum? 20.Let f(x,y, 2)bethesum ofthethree distances: from (x,y,2) tothe y-axis, from(x,y,2)tothez-axis,andfrom(x,y,2)tothepoint(1,0,0).Findtheabsoluteminimumvalueoff,and where itoccurs. 6.4 /DIFFERENTIALS Our purpose inthis section istodefine what ismeant bysaying that a real-valued function ofseveral real variables isdifferentiable and todefine the ® a 64 DIFFERENTIALS 145 differential ofsuch afunction. These definitions areneeded inorder that wemay derive thechain rules fordifferentiating composite functions (Theorem III,§6.5 and Theorem V,§7.3). Westress thecase oftwo variables, buttheideas apply to functions ofthree ormore variables, asweshall sce. in§1.3wedefined thedifferential ofafunction foftheindependent variable xasthefunctionofxandanindependent variabledxwhosevalueisf’(x)dx.Thedifferential offisdefined ateach point xwhere fhasaderivative.Theindependent variabledxcanhaveanyvalue.Forafixedvalueofxthevalueofthedifferential is ‘amultiple ofdx,themultiplier being thevalue /"(x) ofthederivative offatx.Itis Useful torestate thedefinition ofdifferentiability asfollows: thefunction fiscalled differentiable atxifitisdefined atxand allpoints near xand ifthere exists a number Csuch that jimU(x+Ax)=fx)=CAx|_ 7tim, Taal 0. MOAT) Wecan rewrite (6.4-1) intheequivalent form imEADJO)_|. tin c=% ‘and this new form isinturn equivalent totheassertion limL*+4)-10) . 642)= ax But (6.4-2) isthe same astheassertion that fhas aderivative atx,with value f@)=C. Letthefunctional symbol forthedifferential bedf,anddenote bydf(x; dx)the value ofdfasafunction ofxand dx.We useasemicolon rather than acomma to separate xand dxinorder toemphasize thefact that dependence ofdfondxisin general different from itsdependence onx;dfisalinear function ofdx.Wecan replace dxbyother symbols, such asAxorh.We can write (64-1) inthe form lime+W)—fla)=dsMN9, 6a) This isbecause CAx=f'(x)Ax=df(x;x);wesimplyreplaceAxbyh.The characteristic features ofthedifferential are: (1)that df(x; h)is(for fixed x)a multiple ofhand (2)that thelimit relation (6.-3) isvalid. This limit relation can be described bysayingthatdf(x;h)isagoodapproximation tof(x+h)—f(x)inthesense that the difference Sx +h)~ f(x) ~dfx; h) issmall incomparison with |h|as|h|+0. The foregoing discussion ofthe one-variable case provides uswith a motivation for the definitions ofdifferentiability and the differential for the case ofafunction oftwo variables. For afunction fofxand ywewant the a , 146 ‘THEELEMENTS OFPARTIALDIFFERENTIATION ch. differential df(x, y;dx,dy)tobealinear combination Adx+Bdy,where Aand Bare numbers determined byfand (x,y), and wewant alimit relation analogous to(6.4-3). Inspeaking ofpoints near (x,y)werepresent them by (x+h, y+k), where thedistance from (x+h,y+k)to(x,y)isV7+R. Definition. The function fof(x,y) iscalled differentiable at(x,y) ifitis defined forallpoints near (x,y){that is,inaneighborhood of(x,y)]andifthere exists numbers A,B{depending onfand (x,y)]such that (x+hy +k)~ f(xy)(Ah+Bho) 3 limUethy+b=f(y)(Ah+BR)_9, wey erst Vit+k The differential offat(x,y) isthen defined tobethefunction dfof(x,y) and (dx, dy) with thevalue f(x,y: dx,dy)=Adx+Bdy. F6HES) Observe that thedifferential isalinear function ofdxand dy,that is,alinear combination ofthem. The variables dxand dycan beassigned any values whatsoever. Ifwesetz=f(x,y),thevalue ofthedifferential isoften denoted by dzasgiven bytheformula =La st si de= dx+ dy. 6.46) Asanimmediate consequence ofthe definitions wecan see that when fis differentiable at(x,y)thepartialderivatives ©and2existat(x,y)andare siven bytheformulas Lajuny=a Leposy=B. eazy iy Take k=0, hx0in(6.4-4) and leth+0. The result is im|e+hyf») Ab|9, oy h which isequivalent to im{+hy)~fsy) is h “a Therefore, bydefinition, Aisthepartial derivative offat(x,y).The result for B x,isobtained inthesame way. Itfollows thatwhen fisdifferentiable at(x,y)the pairofnumbers A,Bsatisfying (6.4-4) isunique. Itis important toobserve that therequirement onfofbeing differentiable at point isstronger than the requirement that fhave partial derivatives with respect toxand yatthe point. This fact isillustrated bythe functions in «Exercises 2and 7.Ineach case there thefunction isnot differentiable at(0,0) and yethas first partial derivatives with respect toxand y,respectively, there. \ 4 DIFFERENTIALS 147 Itwill beproved later that ifthefirst partial derivatives offexist throughout aneighborhood ofapoint and arecontinuous atthat point, then fisdifferenti- able atthepoint. (See Theorem II,§7.1.) This simple criterion assures usthat most ofthe functions weordinarily encounter aredifferentiable atmost points. The following theorem isuseful (inthe proof ofTheorem V,§7.3, for example). THEOREM Il.Ifafunction isdifferentiable atapoint, itiscontinuous there. Proof. Letusdefine afunction uof(h,k) bytheformula =fethey+b)f(x,y)~(Ah+ Bk) (hk)VTE oy when (h,k)#(0,0). Then we see from (64-4) that the limit ofu(h,k) as (h,k)+(0,0)is0.Fromthedefinition ofuitfollowsthat Sct hyy +k) fx y)=Ah+Bk+u(h, IVETE and from this we see atonce that lim|[fc+hy+k)SOsy=0. oy But by§5.3 this means that fiscontinuous at(x,y). Itfollows from thetheorem that iffisnot continuous at(x,y)itcannot be differentiable there. ‘Afunction can becontinuous without being differentiable, asthefollowing ‘example shows. Example 1.Letf(x,y) =Vx"+y". This function iscontinuous atallpoints,(0,0)included. Butitisnotdifferentiable at(0,0).Infact,itdoesnotevenhavefirst partial derivatives at(0,0). Toverify this consider theratio (for h#0) £(h.0)— f0.0) _Vie=0 _|hj, h hh ‘This ratiois1ifh>Oand—1ifh<0.Henceithasnolimitash-+0,andthepartial derivative 2doesnotexistat(0,0).Theargument isthesameasregards$E,by symmetry. ‘Weshall presently give anexample toillustrate theproperty ofthedifferential expressed in(6.44). Forthedetails oftheexample andother problems itisuseful to know thefollowing inequalities: 2ab|=a?+b?; OREDS VatFB"sal+|b]3V2Va" +B (6:4210)* 4 148, ‘THEELEMENTS OFPARTIALDIFFERENTIATION cn The proof of(6.4-9) issimple: 0({a|~|b))? =Jaf’~2ja]|b| +|b=a*—2Jab| +b*. Ontransposing 2jab| weobtain (6.4-9). The first part of(6.4-10) follows from the ‘obvious fact that a?+b?s a+2)al|b|+b?=(Ja\+|b)). The second part of(6.4-10) isobtained with theaidof(6.4-9): (a|+|b))?=a?+2jab|+b?sa?+(a?+b*)+b?=Aa"+b’). Now extract square roots ateach oftheends oftheforegoing inequality. Example 2.Let f(x,y)=3x"y+2xy"+1.Findthedifferential atx=1,y=2 after verifying that thelimit relation (6.4-4) holds forthis case. We use (6.4-7) tofind Aand B.We have He 2 Foggaoa t2y, Pearttaey, Evaluating at(1,2), wefind f(1, 2)= 15,fi(1,2)=20,f.(1,2)=11,soweformthe expression {(1+h,2+k)~ (1,2)~Qh+11k) =3(1+WYQ+ k)+21+HY2+ky+115~Oh+11k). Upon expansion and simplification, wefind that theexpression reduces to 2Gh?+Thk+k?)+hk@h+2k). Thus, forthis case, theexpression ontheleft side of(6.4-4) becomes 24+Thk+ tim2GHE+Thk+k2)+hkGh+2h))60-00 Vit +k Now, clearly, 3h?+k? 3(h?+k’), Also, by(6.4~9), 2|hk| sh? +k.Therefore, the fraction whose limit weareconsidering isnot larger than (2+(13+Yh+2k)Vivek which equals VIFF RE(13+3h+2k). This clearly approaches 0as(h,k)->(0,0), sowe are through with the verification. The differential at(1,2) is df(1,25 h,k)=20h +11k. We saw inFig. 12of§1.3 that forthefunction f,when weconsider thedifferential offat(x,y),therelationship betweendxanddyisthis:asdxvaries, ° 64 DIFFERENTIALS 149 thepoint(x+dx,y+dy)movesalongthelinetangenttothegraphat(x,y).There isasimilar relationship inthecase ofafunction oftwo variables. Letus assume that fiscontinuous, and consider the surface Sdefined byz=f(x, y). Tosay that fisdifferentiable ataparticular point (xo,ys)can beinterpreted geometrically. The implication ofdifferentiability isthat atthepoint (Xo,Yo.24), where 29=f(x»,Yo),thesurface$hasatangentplanenotparalleltothe2-axis. The equation ofthis plane is 2—29=A(x~Xo)+BCy~Yo) M641 whereAandBarethevaluesof2£and2.respectively, at(Xo,ya).Toshow that this plane isindeed thetangent plane, weneed adefinition ofwhat ismeant bysaying that aplane istangent toasurface atacertain point. Let Pybethe point (xo,yo,20)and letMbeaplane containing Ps.We define Mtobethe tangent plane toSatPpif,when Pisapoint ofSdifferent from Pp,theangle between the line PoP and the plane Mapproaches 0asPapproaches Py.We shall show that this condition isfulfilled bythe plane (6.4-11) asaconsequence ofthedifferentiability offat(xo, ya). For this purpose weanticipate aresult from Chapter 10that may already be familiar tothe reader—that the cosine ofthe angle @between two vectors is equal tothe dot product ofthe two vectors divided bythe product oftheirlengths.See(10.2-3).Wetake6tobethenonobtuse anglebetweenthelinePsPand the normal tothe plane (6.4-I1) atPy.Because @isthecomplement ofthe anglebetween PoPandtheplane,wehavetoshowthat@>5+ orequivalently, that cos @+0. Now, letPbethepoint onSdetermined byx=xo+h, y= yo ky where hand karesmall and notboth zero. The vector PyP hascomponents h,k, Az, where Az=flay+h,yoK)—SUX,Yo. 64-12) Aunit vector normal totheplane (6.4-11) hascomponents A/d, B/d, ~1/d, where d=2(A'+B +1)", and the sign ofdischosen sothat the angle between PyP and the normal is nonobtuse. Then (using formula (1021-3) forthedot product), Ah +Bk~A2| . £0808EBS DSR TT” ual Because Aand Bareconstants weseefrom (6.4-13) that wehave toprove that an [A+ Bk-A2| ditohEFREzy= seine Now, considering (6.4-12), weseethat thefraction in(6.4-14) isnotlarger than (9+hyyo+k)~(Xo.yo)=(Ah+BR), , +ky (64-15) oa DIFFERENTIALS 151 Just asinthe two-variable case, we see that, when fisdifferentiable at (x)..-.0%)sf has first partial derivatives given by Baa teA2ecem 46418) Ifweuseadependent variable wtodenote thevalue off,then duisdefined asafunction ofx),...,X, and dx),...,dx, bytheformula = etLh tu=Fass 4ZLee sGebe9) ‘Asamatter oftechnique inusing differentials, itis important toknow the standard formulas d(u +v)= du+dv, d(uv) =udv +vdu, u)_vdu-udo_ap)=" Here uand vmay represent differentiable functions ofseveral independent variables. Suppose, forexample, that w= flx,y),0= 84,9). Then, bydefinition, aMay4 du=oxdx+aydy, age4 do=Fede+5dy, =Hue)gy5Hue) d(uv)=axdx+‘aydy. But (un) _vy, ,ax“ ax+eax withasimilarformula for=..Withtheserelations beforehimthestudentcan readily write outthework necessary toverify therelation d(uv)=udv+vdu. We likewise have such formulas as de* =e*du, dsinu=cosudu, ty =fdtan'u lew whereuisanydifferentiable function ofseveral variables. Sincetheseareall 6s COMPOSITE FUNCTIONS AND THE CHAIN RULE 155 We shall presently prove this formula under suitable hypotheses. Itisvery important forthestudent tolearn thestructure ofthis formula; asapart ofsuch learning, hemust grasp clearly therole ofthe variables inthe notation ofeach termin{6.5-4). Intheterm34,wandxarerelatedbySED}; wisdependent, andxisoneofthethreeindependent variables x,y,z.Intheterm4¥,xand¢ arerelated bythefirstofequations46.S42F; xisdependent andtisindependent. Intheterm4",uandtarerelated byW853); wisdependent andtis independent. Analternative notation for(6.Si8)'is 4G_oFdf,aFdg,aFdh, dt~axdtaydtazdt ests This notation isclearer and less subject tomisunderstanding. However, both methods ofwriting theformula arewidely used, and thestudent should become familiar with both ofthem. There are other varieties ofcomposite functions inaddition tothe type presented in(6.53), The function Fmay depend onadifferent number of variables (e.g., 2,or4,5,...). Also, the variables x,y,zmay depend upon more than one variable. Suppose, forexample, that wehave u=F(x,y), (65-6) X= f65,0.y =865.0. Then, under suitable differentiability assumptions, ifwe write G(s,1)= F(f(s, 1),g(s, 1), we have the differentiation formulas 2G_aF af,aFa8, asaxas”ayas 65-5 8G_aFaf,oFag, at” ax att ay at ‘Thegeneral rulecovering allformulas such as(6.5=5) and(6577 isoften called thechain rule, orthecomposite-function rule. Todescribe the situation generally, letususe the term first-class variables fortheindependent variables onwhich Fdepends, and theterm second-class variables forthevariables onwhich Gdepends. Observe that Gisformed by replacing each first-class variable inFbyafunction ofthe second-class variables. Inthe differentiation formulas such as(6.5-7) we have asmany different formulas asthere arevariables ofthesecond class; each formula has as many terms asthere arevariables ofthefirst class. Weshall now formulate and prove atheorem about formulas such as(6.5-5) or(6.5-7). Forsimplicity weassume thesituation isthat represented in(6.5-6). THEOREM Ill. Let F(x,y)bedefined insomeregionRofthexy-plane having 166 ‘THEELEMENTS OFPARTIALDIFFERENTIATION chs Likewise, du Laud au,ay"2as 2a 652-7 Differentiating again, ude (#)+32(%)- ox?” 2ax Vas)” ax Vat an2 Now, by(6.52-6), with inplace ofu, (2H) 4.2(4)+42(2): ae(Sr)25($e)+34Gs a(au), Wefind2(24)inthesameway.Thus ®u laut au 1aw tatu wedas*4oras Aasataae 652-8)au_au Weshallassumeenoughaboututoinsurethat24=24.Thestudentshould show for himself that @u_lau 1au tat ay4as"DanasS48a cay Subtraction of(6.52-9) from (6.52-8) now gives theresult uu auSet aa (652-10) The basic point tobenoted inusing thechain rule inconnection with second derivatives isthis: Suppose uisafunction offirst-class variables x,y,...and of second-class variables s,1,.... Then, ifwe have written down achain-rule formula foroneofthederivatives 34.%,....thissameformula isvalidifwe replacewthroughout byanyoneofthederivatives $4.#....Wearethusable toexpress symbols tike2(34)entirely intermsofsecondderivatives ofwwith respect tothe first-class variables x,yy... EXERCISES2648G, 1.Find formulas comparable to(6.52-4) for“$and“Gintheproblem of Example 1.Hence show that PO106,126 aFeR oF ar Pa” a ay 12 ‘THE ELEMENTS OFPARTIAL DIFFERENTIATION cn. ,work outthe result analogous tothat ofExercise 3.(b)What isthe result forthree dimensions comparable tothat ofExercise 4? (€)Develop generalizations oftheresults of(a)and(b)forthecase ofndimensions, taking r=x}-+++++ x2. 6.Suppose that F(x, y)isdefined anddifferentiable inanopen region R,andsuppose that xaF/@x +ydFJay =nF(x,y)ateachpointoftheregion.Then,if(x,y)isinR,the relationF(tx,ty)="F(x,y)holdsinanyintervalfo<t<1,(Wherefe=0)providedthat =1isinthisintervalandprovidedthat,forallsucht,thepoints(tx,ty)areinR.Toprove thisconverse ofEuler's theorem, letf(t)=F(tx, ty),where (x,y) isafixed point of R.Use the hypothesis onFtoprove that if'(t)=nf(1).Fromthis,inferthatf(0)t-*isa constant (depending onx,y).Then complete theproof. 6.6 /DERIVATIVES OF IMPLICIT FUNCTIONS In§6.1 wedealt with the differentiation ofimplicit functions inavariety of particular situations. Wedidnotattempt todeal with general cases inwhich the functions were merely indicated bysome functional symbol. Inpractice itis necessary tohave formulas todeal with implicit functions interms ofgeneral notation. Asimple buttypical case isthat arising when zisdefined asafunction ofx, ybyanequation ofthe form FQ, y,2)=0. (66-1) Suppose, forinstance, that theequation is x+2xz+2? yz-1=0, sothat inthis case F(x,y,2)=x?+2xz +2?yz-1 (66-2) Proceeding asin§6.1, wehave a az_ az 2x42x$2422+222-yZao, a,5,8_yas weHtrFoy Gzno. az 2e+2z a Ix+2e-y (66-3) [on aay” 2xF22—y Now letusobserve that ifweregard x,y,zasindependent in(6.6-2), then oF. oF, FL -SetaktAGenaGrmWety. 67 EXTREMAL PROBLEMS WITH CONSTRAINTS 177 7.Let G(x, ¥,2,0) =0have solutions x=f(y, 2,0), y=g(%, 2,0), 2=h(x, y,v), and bymeans ofone ofthese equations atatime letu=F(x, y,2) become afunction of (9,2), (2,0), and (x,y, 0)respectively. Show that, subject tocertain conditions, uy (au) (au) (a2) _(au) (ax(3). G)..°G2)., &)..- Ge),G@). ‘Anexample isfurnished byxyp—z=0,w=x7+y*+2", andthestudent should check the meaning oftheproblem interms ofthis special case ifhefeels anillustration tobe necessary. 8Suppose w=f(x,y) isasolution ofF(x,y,u)=0, and that y=g(x,2) isa solutionofG(x,y,2)=0.LetH(x,2)=f(x.8(%,2)). ShowthatF\G:H:= F:G,andFG:H, =F:G,~ F\G: are identities inxand = 9.(a)Starting from (6.6-4), show that 2: _IF.~2F\Fot FiFo, a Fi ‘ #2 git (b)Derive analogous formulas for32-5and55. 10.Ifz=f(x, y)satisfies anequation ofthe form z=F(ax +by+cz), where a,b, and¢areconstants, show thatb2=a$2. a ay 11.Ifz=f(x,y) satisfies anequation oftheform F(x+y+2,x°+y"+2*)= 0,show a az that9-2) + DE+G-0)E =O, 12.Suppose that the function z=f(x,y) satisfies anequation ofthe form F(ax+by+62,Ax?+By'+ C2’)=0,wherea,b,¢andA,B,Careconstants, Showthat2=OFi+ 2AxF;, ax eR+2C2F: 13,If =(4 y)satisfies theequation F(f(x, y,2), (8, ys2))=0, show that a__Fifs+ Fags ayFifstFigs 14MEGilssx29), Gelenas.y) and fO,22) are given, and ifgi(xi.x2)= Gils,Xs,fs.) (E=1,2),showthat Agung)_AGG),afGs.G:),afG..G:) 204,42)8G4,%2) ”axr4G.) 2AC,y)” with yreplaced byf(x,,x:) after thedifferentiations. This formula isused inthetheory offirst-order partial differential equations. 6.7 /EXTREMAL PROBLEMS WITH CONSTRAINTS Many interesting maximum orminimum problems arise insuch aform that we arerequired tofind anextremal value ofafunction, say F(x, y,z), where the variables x,y,2arenotindependent ofeach other, butarerestricted bysome relation existing between them, this relation being expressed byanequation G(x y,2) =0. ‘7 EXTREMAL PROBLEMS WITH CONSTRAINTS 179 Wethen seek tomake thequantity w=Foxy. f(x,y) amaximum orminimum. Accordingly wewant tosolve theequations au au0,50, (61-3) Now au_aF ,aF au_aF,aFafaxax}a2Oxayay*azay” (67-4) where zisreplaced byf(x, y)after thedifferentiations areperformed. Wealso have theidentity Gx, yf)-k=0, from which itfollows bydifferentiation that IG,0Gaf93G,0Gaf_BG to 26449 (67-5) Ifwesolvetheseequations forzand2andsubstitute in(6.7~4),weobtainthe equation aFaG_ ak3G ou_oxoz azx, ax 36 a andasimilarequation for2Equations (6.7-3)nowtaketheform aF4G_aFaG aPAG_AFaG_oeoe9OeayOFOEBy (6.7-6) inwhich zisreplaced byf(x, y)after thedifferentiations areperformed, Now themethod ofimplicit functions forthis extremal problem with constraint may bedescribed asfollows: We donot actually solve for zatthe outset. Instead, wecarry the work along and arrive atequations (6.7-6) asequations inallthree variables. These two equations, together with theconstraint (6.71), give usthree equations which wesolve assimultaneous equations inx,y,z.The required points ofextreme value will beamong thepoints found inthis way. This general assertion issubject tosome qualifications torule out exceptional cases. We could, ofcourse, think ofyasafunction ofxand z,orofxasafunction ofy and z,the functional relation ineach case being determined bythe constraint. ‘These alternatives would give uspairs ofequations different from (6.7-6), but equivalent tothem. The implicit-function method was used inthesolution oftheproblem of 6a LAGRANGE’S METHOD 183 Solve these three equations along with theequation ofconstraint Gx y,z)=k (683) tofind thevalues ofthefour quantities x,y,z,A.More than onepoint (x,y,2) ‘may befound inthis way, butamong thepoints sofound willbethepoints of extremal values ofF. Tounderstand thereason forthevalidity ofLagrange’s method, observe that theequations (6.8-2) areprecisely F,+AG,=0,FitAG:=0,Fy+AG)= 0. 68-4) Here Aisacertain constant. These equations state, therefore, that atapoint where they aresatisfied, F\,F:,and F;are proportional toG;, G:,and G:. But weknow from Theorem Vthat such proportionality occurs atthepoints ofthe surface G(x, y,z)=kwhereFhasanextremevalue.Thusthepointsofextreme value will beamong those found bysolving the four simultaneous equations (68-3) and (6.8-4). Thus Lagrange’s method isjustified inthis type ofproblem. The parameter Aoccurring inLagrange’s method iscalled Lagrange’s multiplier, One ofthegreat advantages ofLagrange’s method over the method of implicit functions orthe method ofdirect elimination isthat itenables usto avoid making achoice ofindependent variables. This issometimes very im- portant; itpermits theretention ofsymmetry inaproblem where thevariables enter symmetrically atthe outset. Example 1.Find thedimensions ofthebox oflargest volume which can be fitted inside theellipsoid Byy eoathed=h (68-5) assuming that each edge ofthebox isparallel toaco-ordinate axis. Each oftheeight corners ofthebox will lieontheellipsoid. Let thecorner inthefirst octant have co-ordinates (x,y,2); then thedimensions ofthebox are 2x,2y,2z,and itsvolume isV=8xyz. Wewish tofind theabsolute maximum of Vsubject totheconstraint (6.8-5). Bythe remarks attheend of§6.7 weknow that anabsolute maximum exists. Following Lagrange’s method, weset _ ae ytw=8xy2+a(5+5+5). ‘The equations (6.8-2) inthis case are x 8yz+2a 5=0, 82x+2ajs=0, (68-6) Bxy+2Ara=0. 186 ‘THEELEMENTS OFPARTIALDIFFERENTIATION ch.6 EXERCISES 1.Arectangular box lies inthefirst octant, with one corner attheorigin and the diagonally opposite corner onthe plane (x/a)+ (s/b)+(zic)= I(a, b,c >0). Find the maximum possible volume ofthebox. 2.Apply Lagrange’s method infinding theextreme values ofx°+y*+2°subject to theconstraint (x'/a")+(y"/b?)+(2"/c*)=1,wherea>b>c>0. 3.Atriangleissuchthattheproductofthesinesofitsanglesisamaximum.Show that thetriangle isequilateral 4.Find themaximum value ofxyzi(a’x +by+2) subject totheconditions ayz=A’,x,y,2>0(a, b,€,Aall>0). S.Findtheminimumofx+y+zsubjecttotheconditions(alx)+(bly)+(clz)=1, x,y, 7>0(a, b,€and x,y,zall>0), ‘6.Theperimeter ofatrianglehasaprescribed value2s.Determine thesidesofthetriangle soastomaximize thearea. 2.LetD=|X ¥].Findthemaximum valueofD*subjecttotheconditions x'ty?= a’,w+ 0"=b?, where a>0, b>0. Solve theproblem intwo ways: (1)by Lagrange’s method, and (2)bysetting x=acos, y=a.sin @,w= bcos, v=bsindandusing@,¢asindependent variables.8.Find theminimum value ofx°+y"+2° forpositive x,y,and 2,ifitisrequired that ax-+ by+cz =1,where a,b,¢arepositive constants 9.Suppose a,b,¢arepositive constants. Ifx,y,zarepositive and ayz+bzx+ exy =Babe, show that xyz 5abe. 16,Solve the following problems by Lagrange’s method: (a)Example 1, $6.3; (b)Example 3,$6.3; (c)Example 2,$6.7; (d)Example 3,$6.7. 11,Let the lengths ofthe sides ofafixed triangle ofarea Abea,b,c.From an interior point Odraw theperpendiculars tothesides ofthetriangle, and lettheir lengths bex,y,zcorresponding toa,b,¢.Ifnow aparallelepiped isconstructed with edges x,y, and volume V,show that Vis'a maximum when the lines from Otothe vertices ofthe triangle divide itinto three equal areas. What isthemaximum value ofV? 12. For the situation described inExercise 11, show that (2+ yee Malt b+ ez 4a® 13.Find the minimum distance from (0,0,c) tothe cone z*=(x"/a*)+(y"/b*). Assume ¢>0 and 0<b <a. 14.Given theellipse b°x +ay? =a°b findthepoints (x,y)ontheellipse sothat theline normal totheellipse at(x,y)passes asfaraspossiblefrom theorigin.Whatisthisgreatestdistancefromtheorigintoanormal? 4 18.Find, byLagrange’s method, anextreme value ofxyz subject totheconditions(I/x)+(Hy)+(2)=6.x.yzall>0(ca positive constant). Isthis value anabsolute maximum,anabsolute __|iP ‘minimum, orneither? H 16.Aparticle istotravelfromAtoPandthence toB,bya Nig broken lineasindicated inFig. 46.Velocity from AtoPisvy andfrom PtoBisv3.Show byLagrange’s method thatwhen the B time oftravel isleast, (sin @,)/(sin 63)=v/v Fig. 46. ss ‘QUADRATIC FORMS 191 must besatisfied byx’= 1,y’=0.Substituting thesetwovaluesintothesecond equation reveals that B’=0and therefore that wehave hitupon arotation which reduces Q(x, y)tothecanonical form QW y= ACTF COT, and thetransformed equation oftheconic isthecanonical form (6.9-4). Itis the invariance ofthe discriminant which now allows ustofind A’and C’ easily. Notice that the Lagrange function, L(x, y)=Ax?+2Bxy + Cy? A(x? +y?)associated with Q(x, y)isitself aquadratic form having as itsdiscriminant (A~A)(C ~A)~ B®.This can bewritten asthedeterminant 4=AB B c-al ‘Wehave just seen that therotation which transforms theco-ordinates (x,,y,)into (1,0) transforms L(x, y)into LIQ, y)=AP+CUP=MAP +O which isaquadratic form with discriminant (A’— A)(C’~A). By(6.9-5) wehave that A-k Bi ven[Aa coals anc». ‘The discriminant ofL(x,y)isclearlyasecond-degree polynomial in4whose leading coefficient is1.Each such polynomial has the unique factorization (=n)Q=n) where 7;and r:are the zeros ofthe polynomial. From this it follows that the coefficients A’and C’inthe canonical form ofthe quadratic form Q(x, ¥)aresimply thetwo roots ofthequadratic equation A-A B| |B. cual7® (69-7) These two roots can ofcourse befound assoon asequation (6.9-1) isgiven, and theequation oftheconic reduced immediately tocanonical form, without going through again each time ourfairly lengthy argument which justifies theprocess. Since there are two roots, the reader might wonder which tocall A’and which C’. The answer isthat itdoesn’t matter—there are two equally good canonical forms towhich every quadratic form intwo variables (and every central conic) can bereduced. This ambiguity was presaged when webegan by picking apoint (x,y:)where oneoftheextreme values ofQ(x, y)ontheunit circle was taken on. For our purposes atthat time—the elimination ofthe xy-term—it made nodifference whether this point gave amaximum ora minimum value toQ.Theonly difference itmakes nowisthatifthispoint, whose co-ordinates intheprimed system arex’=1,y'=0, isapoint ofmaxi- mum, then ATisthelarger ofthetwo roots. Ifthispoint (x;,y,)towhich werotate thepositive axisofabscissas gives aminimum, then A’isthesmaller ofthetwo roots. Weseethisfrom theobvious fact that ifa>, then themaximum value 192 ‘THEELEMENTS OFPARTIALDIFFERENTIATION cn ofax’+By?ontheunitcircleisa,andthisvalueistakenonat(1,0)—aswellas at(=1,0)ofcourse.If@<Bthenaistheminimum value,anditistakenonat(1,0) and (1,0). The following example shows how easy itistofind the canonical form oftheequation ofacentral conic bythis method. Example 1.Find thedimensions oftheellipse 73x? +72xy +S2y?= 100, Here A= 73, B=36, C=52. ‘The equation (6.9-7) becomes N= 125d +2500=0, with roots A= 25, A=100, This means, therefore, that theequation oftheellipse can beputintheform 25x" 100y""= 100,or4+y=1, byarotation ofaxes. The principal semiaxes oftheellipse aretherefore 2and 1. The generalization tothecase ofthree variables isnow arather easy matter. ‘The essential statement ofthegeneralization isthis: Given thequadratic form Q(x, y,2)asin(69-2), itis possible tomake arotation oftheco-ordinate axes so that Q(x, y,2)becomes G(x’,y'2!)=Ax?+Ary?+Az, (6.9-8) and the A’saretheroots oftheequation A-k DE*D*peaF|e eon (EB F CHA Aproof ofthis statement isindicated inExercise 14.Ifone wishes, matters may bearranged sothat Ay=A:2 As. Example 2.Reduce thequadratic form xy+.xz~ yztothe form (6.9-8) and soidentify thesurface xy+xz~yz=~. The equation (6.9-9) becomes “A 44 =04}o-a-t|4 o-f al inthiscase. Onexpansion and simplification this becomes “14-2 =0, theroots ofwhich arefound tobe4,|,—1.Thus, after asuitable rotation, our equation becomes fx? ly?— 21, This defines ahyperboloid oftwo sheets with circular cross sections per- pendicular tothe2’-axis (for |z'|> 1). 68 ‘QUADRATIC FORMS 193 Amore symmetrical notation forquadratic forms isinsome ways extremely desirable. Ifwewrite x),x2,X)instead ofx,y,2,aquadratic form inx),x3,x5will have terms ofallpossible types xix},with iandjassuming thevalues 1,2,3.If wewrite ajforthecoefficient ofxj,thequadratic form willhave theap- pearance F(tyX20)=duxt+ankitaaxxs +@y%X) +anx} +ax; (6.9-10) +GyXX) +One +anxt. Since xix2= x:x;, weagree tomake ay: az=half thetotal coefficient ofxxx; similarly foraysand as. The discriminant oftheform is,bydefinition, the determinant ay ay ayJesaaf (19 ay a ay The determinant appearing in(6.9-9) isseen tobethediscriminant oftheform QUx y2)=AGP+y+24), EXERCISES 1.Findthedimensions oftheellipse41x*~24xy+34y?=25. 2.Show that F(x, y)=x7—4xy—2y*=1istheequationofahyperbola.Finda point onthe unit circle atwhich F(x, y)isamaximum. Then draw the xy-axes, the axes ‘ofsymmetry ofthehyperbola, and thehyperbola itself, 3.Find themaximum and minimum values ofF(x, y)=9x 6xy+y? onthecircle 22+ y?= 1.IAs isthemaximum value, show that F(x, y)=Avistheequation oftwo lines, each tangent tothecircle atapoint where themaximum occurs. 4.Let F(x, y,2)=y?+2?—V3xy +Vixz+2yz,Findthemaximumandminimum values ofthis function onthesurface oftheunit sphere. What does (6.9-8) become inthis §.Reduce each ofthefollowing quadratic forms tothestandard form (6.9-8); (a)y242?Vay—V3xz+2y2. (®)13x+13y24102?+Buy4x2—4yz 6.(a)Put F(x, y,2)=xy+yz+2xintheform(6.9-8). (b) What isthe maximum value ofFonthe unit sphere? (©)Atwhat points does itoccur? 7.Determine thesigns ofA,,Aa,Asforeach ofthefollowing quadratic forms, and ‘soidentify thetype ofeach quadric surface. You need notfindtheactual value oftheA's. (@)tay tyz=l (b) yztxz—ay=1. (@)x+2y?+32?—Day=yz=2.8Find maximum and minimum values of17x?~30xy +17y" when Sx?~6xy + sy'=4, 194 ‘THEELEMENTS OFPARTIALDIFFERENTIATION ch 9.Find theminizium value ofx’+y?+xysubject tothecondition 2x?+6xy+2y?= 9°. 10.Suppose that the locus ofAx’+2Bxy+Cy"=1 isanellipse. Consider the problem oflocating themaximum andminimum values ofx’+y"ontheellipse. ApplyLagrange’s method,startingwiththeexpression x°+y?—(Ax?+2Bxy+Cy"),Showthat,intheresulting equations forlocating theextreme values, Amust bearoot ofthe equation 1-,A Ba| [cat als and that theroots ofthisequation aretheextreme values ofx*+ y?,What istherelation between these roots and the semiaxes ofthe ellipse? 11.Ifx,y,4aresolutions ofthesimultaneous equations A-A)x+ By =0,(Aree By xtyd, Bx+(C-A)y=0, show that Ax?+2Bxy +Cy?= A.Hence show that, inthenotation used inthetext, Ayand Aare respectively themaximum and minimum values ofF(x, y)when x°+y?= 1. 12,(a)Assume that F(x, y)=k(k>0) defines afamily ofellipses. Let Avand Azbe theroots of(6.9-7), with Ay=As.Show that theellipse F(x, y)=Asisexternally tangent to thecircle x°+y?= Iattheends oftheminor axis oftheellipse, andthat theellipse F(x, y)=Asisinternally tangent tothecircle attheends ofthemajor axis oftheellipse. Draw afigure showing these twoellipses, thecircle, andthex’y-axes. What canyouinfer from thefigure about maximum and minimum values ofF(x, y)onthecircle? (b)Assume that F(x, y)= kdefines afamily ofhyperbolas, IfAyand A3aretheroots of (6.9-7) with A:= As,explain why A:>0and Az<0. Draw afigure showing thex’y'-axes, theunit circle, thehyperbolas F(x, y)= Ai,F(x, y)™ Aa,and other members ofthefamily, 13. Prove (6.9-5)byactuallysubstituting (6.9-3)in(6.9-1)andcomputingA’,B’,C’. Also prove that A’+C'= A+C. 14,(a)SupposethatQ(xy,2)in(69-2)becomesanewformG(x’,y'.2’)withcoefficients A’,B’,...,F’ when weshift tonew axes x’y’2" obtained byarotation from the xyz-system, Write the equations which correspond to(6.9-6) for the problem of making G(x’, y’,2’)amaximum ontheunit sphere. Ifthenew axes arechosen sothat this maximum occurs atx'=1, y'=2'=0, show that D'= E'=0, and that G(x’,y',2))= Aux? Bly?+C'z"+2F'y'z", whereAyisthemaximumvalueofGontheunitsphere. Now expiain how itispossible tochoose anew setofaxes x",y",2",byarotation from the x'y'z'-system, rotating about the x’-axis, insuch away that the form becomes Aux"+Any"+As2™,whereAandAsarerespectively themaximumandminimumvalues ‘ofGsubjecttothetwoconstraintsx’=0,y"+27=1. (®)Ttmay beproved algebraically that thediscriminant ofaquadratic form Q(x, y,2) is ‘equal tothediscriminant ofthenew form G(x’. y’,2)which isobtained from Q(x, y,2)by arotation ofaxes. Use this fact toprove that thenumbers Ai,Az,Axdescribed inpart (a) Ofthis problem are the roots ofthe cubic equation (6.9-9). MISCELLANEOUS EXERCISES, 1.Choose aandbsothatfi(Vx~a~ bx)"deisassmallaspossible;a+bxis then called a“least-square” approximation toVxintheinterval (0,1). 7/PRELIMINARY REMARKS This chapter isnot primarily concerned with the technique ofpartial differen- tiation, but with statements and proofs ofsome ofthe important theorems about differentials and partial differentiation. We have separated the material ofthe chapter from that ofChapter 6foranumber ofreasons. Instudying thesubject ofpartial differentiation thestudent needs first ofalltogetacquainted with the new ideas which thesubject presents tohim. Heneeds toassimilate these ideas through the medium ofexamples and problems. He will want a‘reasoned development ofthe subject, but he will be more interested inmastery of technique and appreciation ofsome applications than inthedetails ofthelonger proofs, particularly asregards the proofs oftheorems which heisquite willing to take forgranted intheearly stages. The theoremsof$87.1,7.2and7.3havebeen referred toinChapter 6.The student needs toknow these theorems, but hecan very well gothrough Chapter 6without studying their proofs. Sections 7.4and 7.5areonasomewhat different footing. Every student who uses advanced calculus islikely tohave need ofTaylor's formula for afunction ofseveral variables. One meets references totheformula frequently intheliter- ature ofapplied mathematics and invarious branches ofhigher analysis. The law ofthemean ismerely aspecial case ofTaylor's formula. We have putthis mater- ialhere rather than inChapter 6because itsapplications are not soimmediate. The final section ofthechapter deals with tests formaximum orminimum values ofafunction interms ofthesecond partial derivatives. These tests areof great importance—both practical and theoretical. Within the limits oftime ofanordinary year course inadvanced calculus the instructor may wish tomake only aselection from Chapter 7with such adegree ofemphasis on the proofs asheorshe sees fit. The various sections are practically independent ofeach other, except that §§7.4 and7.5gotogether. The definition ofdifferentiability given in§6.4 can bereformulated slightly in equivalent ways that weshall find useful. We shall give thereformulations fora function ofnvariables. We use the notation Wh=he +na? ly introduced inconnection with 4-17). 196 1” ‘SUFFICIENT CONDITIONS FOR DIFFERENTIABILITY 197 Afunction fofnvariables that isdefined inaneighborhood ofthepoint (1.05%) isdifferentiable at(x;,...,.%) ifand only if(1)allofthefirst partial derivatives offexist at(xi...) and (2)wecan write theformula HO Magee XnFB) FRI6«eySa)=Dfine oySa) e+elle, WD where €isavariable quantity depending onhy,...,hy with value 0when hy= +++=hy=0, and such that €-+0 when |h|+0. This formulation isobviously equivalent tothat stated inconnection with G11), because theA,'s inGIT must ofnecessity begiven byA;= FilXIy «9Xe) [Se€ (6.4-18)]. Another equivalent formulation isobtained ifwereplace ihin(7-2) byhl, where Walla=[hes]+==+[al a3) The reason fortheequivalence isthat Uhl=lhe=Vnlihh, ue) or,inmore explicit form. (p+ +D'S[hy]++[tg]SsVinhd+=+"? es) The first of these inequalities iseasily proved, for ifwe calculate ({hy++++ [hgl)?, Weobtain alloftheterms hi,...,hiplusadditional terms,none ofthem negative. The second inequality in(7-5) isaconsequence ofCauchy's in- equality, given inExercise 29,§6.8. (Inthat inequality put a;=|h,| and b,=1.) Because oftheforegoing wecan seethat itdoes notmatter whether wewrite <¢lhi oreile in(7.2) (with €«+0as[he0). Forweseeby(7.4) thatfhl|>0 is equivalent to[hhl4>0, andifellh||= eyiihlje, weseethat€->0 isequivalent to €4+0 because €=€4Vn and €4=¢ (with €=€4=0 ifallthehjsare0). 7.4 /SUFFICIENT CONDITIONS FOR DIFFERENTIABILITY Theconcept ofdifferentiability forafunction ofseveral variables wasdefined in §6.4. Tobedifferentiable atagiven point afunction must have first partial derivatives atthatpoint. Butthisalone isnotenough. Wemay have afunction f(x, y)such that f(0,0) andf,(0,0)exist,andyetsuchthatfisnotdifferentiable at(0,0); foranillustration seeExercise 7,§6.4. The following theorem deals with sufficient conditions fordifferentiability: THEOREM I.Suppose thefunction f(x,y)isdefined insome neighborhood of thepoint(a,b).Suppose oneofthepartialderivatives, say2E,existsateach point oftheneighborhood andiscontinuous at(a,b),while theother partial derivative isdefined atleast atthepoint (a,b). Then fisdifferentiable at (a,b). 198 GENERAL THEOREMS OFPARTIAL DIFFERENTIATION ch.7 Proof. We shall use one ofthe formulations ofdifferentiability from the preceding section. We shall show that wecan write fla+h, b+k)~f(a, b)=fi(a, b)h+fla, b)k+e(\h| +|k), KIB where €-+0 as(h,k)+(0,0). We work withsmall values ofhand kyand weexpresstheleftsideof(te)asthesumoftwoUerences?—eshny f(a +h, b+k) f(a, b)= fla +hb +k)—f(a,b+k)+fla,b+k)fla,b).7.12) Next weapply the law ofthe mean (Theorem IV, §1.2) tof(x,b+k) asa function ofxalone. The result isthat there isapoint between x=a and x=a+h, which wemay denote byx=a+6h,where 0<6<1,suchthat flat+h,b+k)— fla, b+k) =fila +Oh,b+kh. Because f,isassumed tobecontinuous at(a,b),wecan write f(a +0h, b+k) =fila,b)+e, where €-+0 ash+0 and k+0. Thus wehave flat h,b+k)—fla,b +k) =f(a, bh +eh, and wecan puttheexpression ontheright oftheequality sign here inplace of thefirst difference ontheright in(7.1-2). Fortheother difference ontheright in (7.1-2) weuse thedefinition off(a,b)asthelimitofaquotienttowrite(when k40) fabsbifid)thle 2D).£4,b)+e where €;-+0 ask+0. Wedefine €;tobe0ifk=0.This permits ustoreplace the second difference ontheright in(7.1-2) byf(a, b)k +ek. The result isthat we have f(a+h,b+k)~fla,b)=f(a,byh+fla,b)k+eh+ek. Tobring this equation into theform of(7.1-1) wedefine ohtek ,TET ifInitikizo and =0ifjh]+|k|=0.Then,if|h|+[ki#0, or lelSleatt lela=e+le because, clearly, |h|~[h|+[k|and|k|<[h|+kl.Butnowitisevidentthat€+0 when |h|+|k|+0, because €,-+0 and€:-+0. The proof isnow complete. Itwill beobserved that theconditions ofthetheorem arenotsymmetrical as regards xandy.Wemight equally well have assumed themere existence of f(a, b),while assuming thecontinuity offx(x,y)at(a,b). Ingeneral, fora 12 CHANGING THEORDER OFDIFFERENTIATION 199 function ofmore than two variables, weassume themere existence ofone ofthe first partial derivatives, andthecontinuity oftheother first partial derivatives. Wemay then conclude thatthefunction isdifferentiable. Theproof issimilar to that ofTheorem I.For most practical purposes the following statement is sufficient: ‘THEOREM Il.Afunction ofseveral variables isdifferentiable atapoint ifthe function and allitsfirst partial derivatives aredefined in’some neighborhood ofthepoint and ifthese derivatives arecontinuous atthepoint. EXERCISES 1.Letf(x,y)=(x*+ yi(x’+y’)ifx?+y?#0,anddefinef(0,0)=0.Showthatfhas first partial derivatives atallpoints, satisfying theinequalities [f(x,y]5Ox.fx,YS 6ly|. Isfdifferentiable at(0,0)? 2.Letf(x,y)=(x—y"Wx"+y’)ifx°+y?#0,anddefinef(0,0)=0.Showthatfhas first partial derivatives atallpoints, but that these derivatives arediscontinuous at(0,0) ‘The function isnot differentiable at(0,0). To prove this, show first that ifitwere differentiable, one could write aprecytedalt hyde where €-+0 as(x,y)-+(0,0). Then show that this isimpossible. SUGGESTION: Consider thesituation when y= —x. 3.Letfxy)=(x7+y")sinweifx’+y°#0,anddefineg(0,0)=0. Showthatf hasfirstpartial derivatives atallpoints, butthat these derivatives arediscontinuous at(0,0). ‘Show that [f,(x, y)|=2Ix| +1. Prove that fisdifferentiable at(0,0). This example shows that thehypotheses inTheorems Iand I]aresufficient, butnot necessary, conditions for differentiability. 4.Prove that the function foon=Geo if40.0 10,0) =0 satisfiesLaplace's equation, 3h+£40,everywhere, butthatfisnotevencontinuous (letalone differentiable) attheorigin. 7.2 /CHANGING THE ORDER OF DIFFERENTIATION We mentioned attheoutset ofChapter 6that weordinarily find therelation ay ay =ayaxdxay aay tobevalid forthefunctions f(x, y)which wemeet ineveryday useofcalculus. ‘Therelation (7.2-1) may befalse inparticular cases, however, andsoitiswell to know something oftheconditions sufficient toguarantee itsvalidity. 14 ‘THE LAW OFTHE MEAN 205 Observe that the point (a+0h,b+k)isapointof (a+hb+k) thesegment Lsomewhere between itsends (see Fig. 48). When asethastheproperty that foreach pair of points belonging toitthestraight line segment connec- (a+0h,b+0k) tingthem consists entirely ofpoints which alsobelong to %a,) theset,thenthesetissaidtobeconvex. Fig.48,Ifthedomain Rofthe function FinTheorem VIis both open and convex, then clearly (7.4-2) holds for every pair ofpoints (a,b) and (a+h,b+k) inR.It follows immediately that ifboth first partial derivatives ofFvanish throughout R,thenFhasthesamevalueat(a+h,b+k)asat(a,b).Byholdingaandbfixedwhile letting hand kvary, (a+ h,b+k) can bemade torepresent each point inR. Therefore weseethat Fmust take onthesame value ateach point ofitsdomain— inother words, Fmust beconstant, This result isananalogue, forfunctions oftwo variables, ofTheorem V,$1.2. Actually, wecan getamore general analogue ifwereplace thecondition of convexity byaweaker condition called connectedness. For our purpose here it will suffice todefine connectedness forsets that areopen. §Definition. Anopen setSintheplane (orinspace ofthree dimensions) iscalled connected incase each pair ofpoints inScan bejoined byapath consisting ofa finite number ofstraight line segments joined end toend consecutively, thewhole path lying entirely inSand not crossing itself anywhere. Such apath may be called apolygonal arc. Later, in§17.7, weshall give ageneral definition ofconnectedness, applic- able toany set, open ornot. When that definition isapplied toopen sets, itturns out tobeequivalent tothedefinition which weare using here. Asanexample ofanonconnected set, letSconsist ofall points oftheplane forwhich x?>1,thatis,allpoints except ye those forwhich ~1 x$1.Plainly Sconsistsoftwoseparated parts (seeFig.49).Two points, oneineach part, cannot be & 8 joined byabroken-line path lying entirely inS.This particu- —y 7 larsetScomes naturally toour attention ifwestudy the 7 function Z fqy)=y+VET Fig.49. Now we come tothe theorem. "THEOREM VII. Let F(x, y)beafunction which isdefined and differentiable throughout aconnected open set S,and suppose that the first partial derivatives ofFvanish ateach point ofS.Then F(x, y)isconstant in s. 206 GENERAL THEOREMS OFPARTIALDIFFERENTIATION ch. ‘ProofiSuppose AandBareanytwopoints ofS.Let them bejoined byapath consisting ofsegments AP, PiP2,.. -,Pa-1Pay PaB, alllying inS(see Fig. $0).By P thecomment just after theproof ofTheorem VIwesee that Fhas the same value atAasatP;,the same value at P P,asatP;,and soon,sothat Fhasthesame value atBas atA.This means that Fisconstant inS. Fig. 50. Theorems VIand VII admit ofimmediate extension tofunctions ofmore than two independent variables. The extension ofthelaw ofthemean forthree independent variables is F(a+h,b+k,+1)F(a,byc)=hE(8,5,2)+KFAR,9,2)+UPAR,9,2s 4-6) where $=ath,F=b+0k,2=c+Ol Formula (7.4-2) can also bewritten inthe form P(x, y)=Fla,b)=F(X,Y)(xa)+FAX,¥y~b), Oa) where (X,Y)isacertain point ontheline joining (a,b)and(x,y). In§2.7 wehad occasion torefer tothe fact that asetmay beempty. For logical reasons weneed tobeaware that, even though asetisempty, itmay have certain properties “by default.” For example, wecite the fact that the empty setisopen. The logic ofthesituation isthat asetSiscalled open iffor every point PinSthere isaneighborhood ofPcontained inS.If isempty there isnopoint PinSand hence the requirement about aneighborhood ofP has noforce asarestriction onS.We say that the requirement “issatisfied vacuously,” which means that itissatisfied bydefault. Hence Sisopen. Similarly, anempty setisconnected. EXERCISES 1.Let F(x, y)=xy*~x°y. Find the appropriate value of@in(7.4-2) if(a)a=b= O.h=1,k=2;(b)a=b=0,h=3, k=2:()a=b=1,h=3, k=? 2.Let F(x,)bethequadraticfunctionAx?+2Bxy+Cy*.Showthat(7.4-7)holds with X={(x+ a),¥=4(y +6). What does thismean about thevalue of@in(7.4-2)? 3.TakingF(x,y)=sinxcosy,provethatforsome@between0andIitistruethat325 05%cos20—*sin™sin7% 3+EcoscosPFsinsin 4.(a)Write outformula (7.4-2) forF(x, y)=log(xe”), with a=1,b=0,h=e~1, k=1. ()Write out (7.4-7) forthis same function, with a,b,x,yarbitrary, except that a>0,x>0. 5.Ifx#ain(7.47), showthatY=b+2=5(X—a). Henceshowthat,under suitable conditions, F(1,0)— FQ, 1)=FAX, I~X)~ FAX, 1-X)- 6.Let F(x, y)=(1—2xy +x)", Asaresult ofconsidering F(1,0)~ F(O, 1),show 18 ‘TAYLOR'S FORMULA AND SERIES 207 that there isanumber @such that 0<@ <1 and 1-V2= 21-301 -26 +30". 7.Let F(x, y,2)= xyz. Find theappropriate value of@in(7.4-6) if(a) a=b=c= Oke kta 1;(b)a=b=0,c=1,h=k=1l=-1; @)a=c=0,b=1halal, k=0. 8.Show thattheopen disc {(x,y):x*+y*<I]isconvex. HINT: Begin byshowing that theline segment determined byany two points (a,b)and(a+h,b+k)inthediscisthesetofall pointsoftheform{a+th,b+tk}where0=1. (Actually, westill have @convex setifwejoin tothis open disc some orallofits boundary.) 9.Explain why theempty setisconnected, and why every setconsisting ofjust one point isconnected 7.5 /TAYLOR'S FORMULA AND SERIES Just aswe extended the ordinary law ofthe mean tofunctions ofseveral variables, sowemay extend theversion ofTaylor's formula given in$4.3. The method isthesame asthat employed intheproof ofTheorem VI,$7.4. Wewrite f(t)=F(a+th,b+tk) 5-1) andapply Taylor’s formula tof(1), using thetwo values ¢=0,1=1.From (4,3-7) with a=0,h =1we have J)=f)+O)+--+LOL, gcgci, (75-2) at (sD! ‘The assumptions arethat Fand itspartial derivatives oforders 1toninclusive aredifferentiable atallpoints along theline joining (a,b)and (a+h,b+k). The main problem now isthat ofcalculating thehigher derivatives offfrom (7.5-1). ‘The first derivative isgiven by(7.4-4) intheprevious section. Working from that formula, we see that $1) =hUAFy, +kFux)+kIhF2,+KF), where allthepartial derivatives onthe right areevaluated at(a+ th,b+tk). Since Fy:=F:,(Theorem IV,$7.2),wehave(0)=WFy+DhkF+PF. This issometimes written inthe form maya (hee 2yir=[(m2red) FOr] ae itbeing understood that a,,ay, _p@e OF pF,(netkay)F(x,y)=WE+hk5+ ‘Theanalogy with thepattern ofthebinomial expansion isnow evident, Wehave 75 ‘TAYLOR'S FORMULA AND SERIES 209 From F(x, y)=x"!y"! wereadily find BFL _yty OFLay ax Yay PFayy OFLay OFayot yaxay Yoaye OE Itisnow easily seen that (7.54) becomes a eeeTECTED IthOktR =ie hk K Ro"TFOmCT+OR)”>OATOY*TFOHTA The series (7.5-6) begins 1nH 14(h-4(CWItheeTERETE 7+ P+hk B+. Detailed verification should besupplied bythestudent ashereads this example. Ifwewrite x=1+h,y=—I+k, thelast formula becomes 1 iste -y-gr Ttle-)-OFD) +1 =P+=DOFDF DF Weshall notinvestigate theprecise limitations on(x~1)and (y+1) which are necessary inthis expansion. There are situations inwhich weneed theTaylor series with remainder for functions ofmore than two variables. Suppose forexample that wewish toexpandafunction F(x),X2,-..5%4) aboutthepointa),d3,....da.Onewayofwriting theexpansion toterms ofdegree k,followed byaremainder, is F(a,+By3Bay y+By)=FCM3,yg)+ ae 3 beth Yi+Daltagthagt than)P+ 1 oop 2 aytap(nathngetothe) B where allthederivatives upthrough those oforder kareevaluated atthepoint (ai,42...54q), and those of order k+1 are evaluated at a)+0h,a:+ Ohs,...., +h, where 0issome number properly between 0and 1.The line through (4, ds,.... dq)and (a+hy,d)+ hy,» dy+hy)has parametric equations xy=ai+thy, X3= a+ thy... X=dy+thy,Thereforethepointatwhichthe (k+I)storder partial derivatives intheremainder term areevaluated isbetweenthe two points where ¢=0 and t=1. Another way ofsaying the same thing results from letting x;=a,+hy,X2=43+yy..5%e=dy+hy,Replacing theh’sbytheirequivalents intermsof 76 SUFFICIENT CONOITIONS FOR ARELATIVE EXTREME 21 10,(a) Find alinear function ofxand ywhich isagood approximation for Fox,y)=tan-'(7=%) whenxandyaresmall (©)Write theconstant andlinear terms inTaylor's series ofF(x) inpowers ofx—3and yok 11,Write outinfulltheexpression 1 (yop 2)(taytky)Peay What does theexpression become (a)ifF(x, y)=x*=x°y"+ y's (b)ifF(x,y)=sinxy andifonesetsx=y=(72)"” after doing thedifferentiation? 12.(a)Carry onthework oftheillustrative example inthetext, showing that oF I"(n— p)!p!a-ptrceBeareCDPeleste and that thepolynomial ofdegree minh andkintheTaylor's series is CU ek he HDL (©) Assuming that [h| <1and [k|<1, write at enya kyTiyeieHO-w) AURORE gLEKER ‘Then multiply thetwo series together term byterm, andcollect together theterms oflike degree. Compare with theresult found in(a). 7.6 /SUFFICIENT CONDITIONS FOR ARELATIVE EXTREME In$6.3 we discussed relative maxima and minima for afunction f(x,y).In Theorem Iofthat section wereached theimportant conclusion that iffattains a relative extreme value ataninterior point ofitsregion ofdefinition, then necessarily thepartialderivatives 2andSLvanishatthepoint(provided these derivatives exist, ofcourse). The conditions Hig A.too 76-1) atapoint donotinthemselves guarantee arelative extreme, however. Inthis section wewish todevelop criteria which, taken together with conditions (7.6-1), will guarantee arelative extreme, and enable ustodistinguish arelative maximum from arelative minimum. Itwill behelpful ifwebegin byreviewing briefly the analogous con- siderations forafunction ofone variable. Suppose wehave afunction y=f(x) defined onsome interval having x=aasaninterior point. Wesuppose ftobe differentiable ontheinterval, and weassume that thesecond derivative exists at x=a 214 GENERAL THEOREMS OFPARTIAL DIFFERENTIATION cn Observe that week Ah?+2Bhk +Ck?=°'G(¢). , : . , Ath)This shows that thesign ofG(#)isthesameasthesign yan ¥ ofthequadratic function Y iy: : Bee f(h,k)=Ah?+2Bhk+Ck. h Now letusregard h,kasrectangular co-ordinates in mayb asystem with origin atthepoint x=a, y=. Let h', k’denote rectangular co-ordinates inasystem Mis:St. obtained from thehk-system byarotation about the origin ofthesystem (see Fig. $1). Aswesaw in$6.9, itispossible tochoose therotation insuch away thatf(h,k)becomes Ain? +ok =PSU), 76-6) where A;and A;areroots oftheequation A-v’ OB A+ B= |Bcoal-M AFOn+Ac-BP=0. 6-7) Here h?+k?=h" +k?=1°.Observe that theproduct oftheroots of(7.6-7) is Ar= ACB, (76-8) and that the sum is Ate AFC, (76-9) Everything now depends onthesign oftheexpression (7.6-6). Itisclear that if Ayand A;areboth positive, G(¢) ispositive, and wehave thecase ofaminimum at(a,b),whereas ifAyand Azareboth negative, wehave thecase ofamaximum at(a,b).Let usnow consider cases (i)and (ii)ofthetheorem. The hypothesis B*~ AC<0implies thatAandCareofthesame sign, andalso, by(7.6-8), that Arand Azare ofthe same sign. Consequently, from (7.6-9) we see that B?~ AC<0andA>0 imply that A;and Azarepositive, whereas B?— AC<0 and A<0imply that A;and A;arenegative. The conclusions incases (i)and (ii) aretherefore established bytheforegoing arguments. Incase (iii), B?—AC>0impliesthatA;andA;areofoppositesigns.Now wecanchoose ¢soastomake h’=0, k’#0, r'G()=Ask”,andwecanalso choose @soastomake h’#0, k'=0, PG($)= Ash”. Thus G() can change sign, and sowehave neither amaximum noraminimum at(a,b). Finally, ifB*— AC=0,atleast oneoftheroots A,andAziszero, by(7.6-8). No conclusion about amaximum orminimum can bedrawn inthis case. The reasons forthis areclear from (7.6-6) and (7.6-5). Asexamples wecite thethree functions woyyxteyh 220 GENERAL THEOREMS OFPARTIALDIFFERENTIATION ch. (ii)Anondegenerate criticalpointwhichisneitherastrictlocalminimumnorastrictlocal maximum isasaddle point, where thefunction has astrict local maximum from some directions and astrict local minimum from others. There are extensions ofGundelfinger’s rule which enable ustoclassify degenerate critical points, where themaxima and minima arenonstrict, butwe shall not include these details. EXERCISES 1.Find allthecritical points ofeach ofthefollowing functions. Test each critical point byTheorem IX,and state your conclusion, (a)y?43x44 1207424, (b)x2—12y?+ 4y"+ By (©)x¢4y*=2x?+xy29°,(@) Py?=5x°=Bry=5y*, (©)xy(12-3x-4y). ®Py(a~x-y),a>0. (®)(1-21 y—D. (hyPyQ4-x-yP 14 8 0S-yw ()Wo+y?)~18x—Day+SVT"+250. (h)Sex?+y?)24x~32y—GOVE+1000. (12x siny~2x?siny+x?siny0y. 2.If@and bare positive, show that (alx)+(bly)+xyhasaminimumatitsonly critical point. What isthesituation ifaand bareboth negative? ifthey have opposite signs? 3.How many critical points hasthefunction (ax?+by?)e™**”ifb>a>0?Discuss the nature ofeach ofthem. 4.Find theshortest distance from thepoint (1,~1, 1)tothesurface z=xy. Setup thesquared distance asafunction ofx,y,find thecritical points ofthefunction, and test them byTheorem IX. 8.Discuss theproblem offinding theshortest distance from thepoint (0,0, a)tothe surface z=xy,where a>0. Proceed asdirected inExercise 4,Separate the case 0<asland1<a 6.Ifzisdefinedasafunctionofx,ybytheequation2x°+2y"+2"+8xz—2+8= 0,find thepoints (x,y,2) atwhich zhas arelative extreme, and test bysecond derivatives for amaximum orminimum ineach case. 7.Proceed asdirected inExercise 6,starting from theequation x°+2y?+32*—2ay -2yz =2. 8.Suppose that Fisdefined intheneighborhood of(a,b),and that Fi(a, b)=0, F(a, b)<0. Why isitimpossible forFtohave arelative minimum at(a,b)? 9.Locate thecritical points ofthefunction xyz(x +y+2~1). Show that there are sixlines allofwhose points aredegenerate critical points, andonenondegenerate critical 18 SUFFICIENT CONDITIONS FOR ARELATIVE EXTREME 21 point, Isthisamaximum oraminimum point? Can youanswer thislastquestion without second derivative tests? 40.Showthateverycriticalpointofthefunction “= isdegenerate, 11.Locate thecritical points ofthefunction F(x,y,2)=(ax?+by?+ exe", where a>b>c>0. Show that there aretwo points ofmaximum value ofF,onepoint of ‘minimum value, and four critical points atwhich there isneither amaximum nor a ‘minimum. 12.Study thefunction F(x, y)=(y?-x?Xy?= 2x"). Show that there arefour lines which divide teplane into eight regions, ineach ofwhich Fhasaconstant sign. Discuss thecritical point ofthefunction. Are there any relative extrema? 13,Discuss thesignofthefunction F(x, y)=(2x?~y)(x°~ y)atvarious points ofthe plane, byappropriate consideration oftheregions into which theplane isdivided bythe two parabolas y=x", y=2x7. Discuss thecritical points ofthefunction. Show that, along every straight line through theorigin, thevalues ofFreach aminimum at(0,0) butthat F has neither amaximum nor aminimum at(0,0). MISCELLANEOUS EXERCISES 1.Generalize TheoremIof$6.3tofunctionsofnvariables.2.Suppose f(x, ¥)isdifferentiable at(a,b), with A=fi(a,b), B=f:(a,b). Let Fr, 6)=f(a+rcos6,b+rsin 8).Then F\(0, #)exists and isequal toAcos 6+Bsin6,forevery8.Provethisdirectlyfromthedefinition ofdifferentiability offandthefactthat FiO, #)=lim, .0(1/r) {F(r, #)~ FO, ®)). 3. F(xy)=(1-0~-yety—D, a=b=k write Taylor's series for F(a +h, b+k). What doyou conclude about thesign ofthedifference F(a+ h,b+k)— F(a,b) when hand kare small? 4,Define f(x,y)= (x= y"VO2+ y?)ifx+y? #0, and f(0,0)=0. Ifweintroduce cylindrical co-ordinates (r,@, 2)intheusual way, thesurface z=f(x, y)isrepresented by z=r(cos’ @~sin’), Observe that thesurface consists ofabundle ofhalf-lines; the hhalf-line corresponding toafixed value of@starts attheorigin andpasses through the cylinder x?+y?=1atapointforwhichz=cos?@~sin’@. Byplotting thecurve z=cos’ @—sin’@with @andztreated asplane rectangular co- ‘ordinates, andthen rolling uptheplane toform acylinder, onecanvisualize thesurface. Dothis. Does thesurface have atangent plane attheorigin? '.Suppose thatfand¢arefunctionsofasinglevariable,andthateachfunctionhas continuous first and second derivatives. We shall suppose that (a) =c#0 and that $a) 40. Let Fx, y,2)= f(x) +{09 +2), GOR, y,2) =&G)4()4(2). Consider the extremal problem forF(x, y,2) subject totheconstraint G(x, y,2)=¢?. Show that a possible solution oftheproblem occurs when x= y=z=a,and that theextreme will bea relative minimum if gyPC) BAN &puronSa) Gian<P ‘andarelative maximum iftheinequality isreversed. Asinstances consider: first, fx) =x7,6(4) =x,a>0; andsecond, f(x) =e", 6(4) =X. oe] ‘THEFUNDAMENTAL THEOREM ns THEOREM I.LetF(x, y,z)beafunction defined inanopen setScontaining the point (Xo,Yo,Za)Suppose that Fhascontinuous first partial derivatives inS. Furthermore assume that F(X, Yo,Zo)=0,Fi(Xo, YouZo)#0. Under these conditions there exists abox-like region defined bycertain inequalities |x~xd<a,ly~yol<b,|z-2<e, lying intheregion Sand such that thefollowing assertions aretrue: LetRbetherectangular region |x—xo]<a,|y—yol<binthexy-plane. Then 1.For any (x,y)inRthere isaunique zsuch that |z=zl<e andF(x,y,z)=0. Let usexpress this dependence ofzon(x,y)bywriting z=flxy) pany 2.The function fiscontinuous inR. 3.The function fhas continuous first partial derivatives given by _Finy.2), _Fit y.2), $09)=BGayHE)=Egy wherezisgivenby(8.11). Proof. The first part ofthe proof isconcerned with determining suitable values forthepositive constants a,b,c which arementioned inthetheorem. Let Abearectangular parallelpiped (box)withcenter at(xo,Yo,Z)suchthatthe whole ofAisentirely inthe region S,and such that, moreover, Fy(x, y,z)has everywhere inAthe same sign which ithas at(xo, yo,20). This choice ofAis possible since Sisopen and F;iscontinuous (see Theorem III,§5.3). For definiteness letusassume F,>0 inA.Consider the top and bottom faces ofthebox A.Ifwedenote theheight ofthebox by2c,these faces willlie intheplanes z=z»+c. Since F,>0, thevalue ofFincreases aswegoupward along anylineparallel tothez-axis. Since F(X, yo,Zo)=0,itfollows that F(%, Yo,2+¢)>0 and F(X,Yo,Ze—€)<0. Because ofthecontinuity ofFweseethat Fwill bepositive inasmall rectangle with center at(Xo, yo,20+¢) inthe plane z=zo+c, and negative inasmall rectangle withcenter at(xo,yo,2»—¢) intheplanez=z~c. Letuschoosepositive numbers a,b,sothatthese rectangles aredetermined bytheinequalities [x—xo]<a,ly—yol <b. We also take care tochoose aand bsothat the box Bdefined bythe Ds IMPLICIT FUNCTION THEOREMS che uswhat wecan becertain of,under appropriate conditions, inspeaking ofa function 2=F Be) defined implicitly byanequation oftheform PO,X354%2)=0, ‘We use geometrical language inspeaking ofthe regions ofdefinition ofthe above functions. ‘THEOREM 12Let F(x,,...,X,,2) bedefined inan(n+1)-dimensional neigh- borhood ofthepoint (a;,...,4dy,¢). Suppose that Fhas continuous partial derivatives inthis neighborhood, and furthermore, assume that FCG), «4 GqyC)=0,Fyity «voyAny€)#0. Under these conditions there exists abox-like region defined bycertain inequalities fxrail<Ag,«op[a—iy]<Aw[2—|<C, lying intheab.eneighborhood, and such that thefollowing assertions are true: Let Rbethe n-dimensional region [xia] <Ai,sosbXe—a]<An inthespace ofthevariables x),...,X. Then 1.For any (xy,..05%,) inRthere isaunique 2such that Je e|<C and F(X... %m2)=0. Let usexpress this dependence ofzon(x1,....X,) bywriting aCe 2.The function fiscontinuous inR. 3.The function fhascontinuous first partial derivatives given by a 2 FG te2)GaP ont)=re f=Ay where 2=f(x... 5%): EXERCISES: 1.Itistrue that thepart ofthelocus defined byx+y +z~sin xyz=0near the point (0,0,0) can berepresented intheform z=f(x, y)? 2.How canyou besure that theequation e*(x*+y*+z*)~V1F2+y=0 hasa solution 2=f(x,»)which iscontinuous atx=1,y=0, with f(1,0) =0?Using thetangent plane asanapproximation tothesurface, calculate f(1+h, k)approximately when hand kare small 232 IMPLICIT FUNCTION THEOREMS che Jacobian determinant jar oF HFG) _|auovau.0)~ |aG_aG| (83-4) au oe Itwill beseen byreferring back to$6.6 that this same Jacobian arises inthe denominators ofthe expressions for the partial derivatives ofuand vas functions ofx,y,2,assuming that such functions are defined byequations (831). The foregoing considerations indicate that ifweexpect tosolve equations (83-1) for u,v,weshould make the assumption that the Jacobian (83-4) is different from zero. Itmust bekept inmind that inthegeneral (nonlinear) case weareconcerned notwith theactual solution foru,vintheelementary sense of expressing u,vinterms ofx,y,zbymore orless simple formulas, but with the solution inthe theoretical sense ofknowing certainly that there exist functions (8.3-2) satisfying equations (8.3-1). Aqualified guarantee ofthe existence of solutions inthis theoretical sense isfurnished bythefollowing theorem: \THEOREM DMELetSbeaneighborhood ofthepointPo:(Xo,Yo,Z0,Uo,vo)inthe S-dimensional space oftheco-ordinates x,y,z,u,v.Suppose that the functions F,Goccurring inthesystem (8.3-1) are continuous and have continuous first partial derivatives inS,Also assume that both functions vanish atthepoint Pobut that theJacobian (8.3~4) does not vanish atthe point, Under these conditions there exists abox-like region lying inS,defined bycertain inequalities [x=xo]<a,ly~yol<b,|2=zo]<e, (83-5)|u=wl<ex,|v~vol<By (83-6) such that thefollowing assertions aretrue: LetRbetheregion defined, inthe3-dimensional space oftheco-ordinates x,¥,2bytheinequalities (8.3-5). Then 1.Toany (x,y,z) inRthere corresponds aunique pair ofvalues u,v such that theinequalities (8.3-6) aresatisfied and thefunctions, F,Gvanish (ie., equations (8.3-1) aresatisfied). This correspondence defines uand vas functions ofx,¥,2,say ={0%¥2).0=BOG2). 2.The functions f,garecontinuous inR. 3.Thefunctions f,ghave continuous partial derivatives given by Hf_1AFG) ag, _1(FG) ax FAG,v)axTFA(x) oo 234 IMPLICIT FUNCTION THEOREMS che then apply Theorem IItothe equation H(x, y,z,u)=0 toobtain asolution u=f(x, y,2). Finally, substituting in weobtain vasafunction ofx,y,2: v=B(x, ¥,2)=H(%,¥,2,FX,Ys2) Weshall omit theexact details ofthelimitation ofthemagnitudes ofthe differences x~x9,y~yo,... inorder tovalidate alltheforegoing arguments. In applying Theorem IIweareassured that thefunctions ¢,fhave continuous first partial derivatives. The function g,asacomposite function, will then have continuous first partial derivatives also. The formulas (8.3-7) for the partial derivatives offandghave already been obtained (see (6.6-11)). Wecanappeal to this earlier derivation now that wehave proved theexistence offandgandthefact that they dopossess continuous partial derivatives. We shall not take space tostate formally the analogue ofTheorem IIIfor systems ofmore than two equations. The nonvanishing ofthe appropriate Jacobian isthekey condition. The proof ofthegeneral theorem forrequations may bemade bymathematical induction onr.The proof for r=1 isthat of ‘Theorem II;hence allthat isnecessary istomake thestep from rtor+1.This isnotdifficult, andmay bepatterned after theproof ofTheorem III,which isthe step from r=1 tor=2. Suggestions forthis work are contained inExercises 10, 11. EXERCISES 1,Dothere exist functions f(x,y), 8(% y),continuous inaneighborhood of(0,1), such that f(0, 1)= 1,(0, 1)==1, and such that Ux,y)P+xg(x,y)—y=0,Lets, y)P+ yfley y)—x =0? Explain your answer. 2.Suppose that thethree equations wt oht wat =O, +ot yF= Out Wea Z=O aresatisfied byaparticular setofvalues (xo, yo,Zo,to,to,wo)ofthevariables. (a)What condition onthis setofvalues issufficient toinsure that all“nearby” sets (x,ysZs ©,W) satisfying the three equations are given byequations u=f(x,y,z), v=g(% ¥.2). w= h(x,y,2), where f,gh are single-valued and continuous, with values wo,to.Wo respectively at(xo,Yo20)? _(b)Solve thegiven equations explicitly foru,v", w’,andfrom thesolutions sofound explain what happens ifthesufficient condition inpart (a)isnot satisfied. 3.Suppose that(xo,Yo,2s,MaBo,We)satisfy theequations wrettwe LieSeSerreyta What aresufficient conditions which guarantee that all“nearby” sets satisfying these ‘equations can berepresented intheform w=$53,201, 0=BOK9.20)? 9/INTRODUCTION InChapter 8wediscussed implicit function theory, which deals with problems inwhich we have asystem of m simultaneous (not necessarily linear) equations innunknowns, where n>m.The question is,“When can thesystem besolvedtoexpresssomemofthevariables asfunctions oftheothern—mIn§8.1 we gave detailed consideration tothe case inwhich m=1 and n=3. In$8.2 weextend this tothecase ofone equation inn+1variables, and inTheorem Illof§8.3 wewent asfarasshowing how the same methods could provide ananswer for asystem oftwo equations, each having five variables From this point, one could infer what thegeneral theorem must beforsystems ofmequations innvariables. Inthis chapter weswitch our attention slightly towhat iscalled inverse function theory. The problem here arises when we have asystem ofnsimul: taneous equations which transform one ordered n-tuple ofnumbers into another, that is,asystem ofthe following kind FOCKayoeSe)= FOC Kaye) =Yo ony FON Xap Ka) =Yn When wesubstitute anordered n-tuple ofnumbers into the functions onthe left, the nvalues which weget form another ordered n-tuple ofnumbers onthe right. ‘The system (9-1) transforms the n-tuple (x), .X2,..+.%_) into (V1, Yas. Yn)s and weshall find itconvenient tospeak ofsystem (9-1) asatransformation. The problem which we wish toconsider is, “Given transformation (9-1), when itis possible, intheory atleast, tosolve forthe x'sinterms ofthe y's?” Notice that wealready have theanswer inthespecial case where thegiven functions on the left are linear. The transformation then reduces to FOC eeKa) =GuXtFOrto+ake=Ye $CyoXa)=aX+Ankat+Maas=2 49.2) FMCaoa)=Wyk+aaa°FMaka=Ya 237 238 ‘THE INVERSE FUNCTION THEOREM WITH APPLICATIONS cn Cramer's rule tells usthefollowing: Ifthecoefficient determinant ay ay s+ dy oe a isdifferent from zero, then for allvalues ofthe given y's, the x-values are uniquely determined. Furthermore, thex'sturn out tobelinear combinations of the y's, just asin(9-2) the y'sare linear combinations ofthe x's. This means that the solution has the form buystbiayot>>+DinYa=Xbuysbrave++Bae=X ey DaYitDaaYat +-*+Dana =Xe ‘The transformation represented by(9.4) iscalled theinverse ofthat represented by(9-2). Ifone isinterested inknowing just what values the b's must have in (9-4), they can befound bytaking the solution of(9-2) given byCramer's rule, that is, eM ges ade,wet naB, as, where Aq isthe determinant obtained from Abyreplacing the mth column of&bythecolumn ofy'sontheright in(9-2). Expanding each ofthe4,byminors ofthe y'sinthemth column gives us(9-4). SoCramer’s rule isthebest known example ofaninverse function theorem, and itis themost satisfactory, inthesense that itgives acomplete answer tothe inverse function question, inthose cases towhich itapplies. Butunfortunately it applies only tothose cases of(9-1) inwhich allthefunctions arelinear, that is, inthose cases where (9-1) reduces to(9-2). Before leaving thisvery special case, itisimportant tonotice acertain important feature which itexhibits. In(9-2) ‘ifp=x.4,Therefore, thecoefficient determinant, 4,canalsobewrittenin thefollowing form, PQ geWP yP Fe We pm ee ph whichisseentobe2Z7L7s-=2L°), thatis,theJacobian ofthef’swithrespect tothex's. This same Jacobian determinant isgoing toturn upagain inthe nonlinear case. ° INTRODUCTION 239 Mathematicians frequently trytosolve problems invery general language so that their solutions will apply toaslarge aclass ofproblems aspossible. We have been discussing the linear version ofthe inverse function theorem inlanguage sufficiently general toinclude anyfinite number ofvariables. Weshall now illustrate what wehave been saying inanextremely simple case, with n=2. Example 1.Does thetransformation 2Qu+3v=x @-5) ut2oey have aninverse? Ifso, find it. The coefficient determinant, A,is 23. [I[a4-se1e0. ‘The fact that aninverse transformation does exist follows immediately from the fact that A#0. Weeasily find theinverse tobe 2x-3y=u 06) -xt2y=0. Inthis example, wetried todistinguish between two problems—showing that aninverse transformation exists and then actually finding it.The reason isthat when wecame tothe nonlinear case, weshall frequently find ourselves inthe position ofbeing able toprove that aninverse transformation exists, but not having any idea how tofind it. Before leaving Example 1,itisworthwhile toremind ourselves what it ‘means tosay that inequations (9-6) wehave solved thegiven equations (9-5) for wand vinterms ofxand y.Itmeans that ifwe substitute inequations (9-5) for wand vtheir values asgiven by(9-6) theresult isanidentity. Ifwemake these substitutions weget 22x ~3y)+3(-x +2y)= x, and on (2x~3y) 42x +29) =y. ‘These equations immediately reduce to xex and 1-8) yay which istheidentity transformation inthexy-plane. In(9-7) wehave performed thecomposition ofthetransformations (9-5) and (9-6) byfirst performing (9-6) ‘onapoint (x,y)andthen performing (9-5) ontheresult. Equations (9-8) show that weended upright where westarted—with thepoint (x,y),andtherefore (9-6) followed by(9-5) isreally justtheidentity transformation. Thestudent will now finditeasy toshow thatifheorshestarts with apoint (u,v)andtransforms 92 MAPPINGS 247 with 6<x <14, intersects Ainasegment thatiscarried (byT;') intoanarcofa circle inthefirst quadrant ofthewv-plane. The arcisthat part ofthecircle inthe interior oftherectangle B.When 6<x<10, thecircle cuts thetwo adjacent sides v=0 andw=VI2 ofB;when 10<x<14, thecircle cuts thesides u=V20 andv=V8.SeeFigures 57aand57b. Similarly, asyvaries from 2to 10,each ofthelines onwhich yisconstant, with 2<y<10, intersects Aina segment thatistransformed byT;'intoanarcofhyperbola cutting through the interiorofB.When2<y<6,thehyperbolic arcintersects thesidesu=V2andv=V8 ofB,andwhen 6<y<10, thearcintersects thesides v=0 and u=V20. Now, tofindthepart ofBthatcorresponds toR(wedenote itbyE), wemust locate inBjust those portions ofthecircular arcs that correspond tothesegments ofthelines x=constant that cutthrough theinterior ofR.Itis clear that these portions ofthe circular arcs will becut offbetween the two hyperbolic arcs that correspond tothelines y=constant which form thetopand bottom ofR.Inthis way wesee that the set Eisthe interior ofacurvilinear quadrilateral lying inBand bounded bytwo circular arcs and two hyperbolic arcs. Figure 57b shows the case ofthe Ecorresponding toRwhen Risa square. Incidentally, one sees that the part ofBnot inEoronitsboundary consists offour separate pieces, each ofwhich istheimage ofatriangular piece ofAexterior toR, EXERCISE Intheequations x=f(u, v),y=g(u, 6)regard u,vasfirst-class variables and x,yas second-class variables, the relation between the classes being that expressed by(9.i-1). From this point ofview the chain rule gives aFfou,af0, \iuax*a6ax" aw aFynga aG 4 js where 3means 2Fand3%means 5,Carry onwiththechain ruleinthismanner to obtain three other equations, and then solve them, thus obtaining equations (9.1-5). 9.2 /MAPPINGS ‘Suppose that x=Slur), y= (uv) 92-1) issome transformation and u=Fxy), v=G(xy) (92-2) isitsinverse. Intheprevious section wehave interpreted (x,y)asrectangular co-ordinates inoneplane, and(u,»)asrectangular co-ordinates inanother plane. We sometimes find itconvenient tosay that the point (u,v) ismapped into the point (x,y).Equations (9.2-1) aresaid todefine amapping, orapoint trans- formation. Ifthefunctions f,garedefined inacertain region oftheuv-plane, we saythatthisregion ismapped intothexy-plane. Theconfiguration inthesecond 22 MapPiNGs 249 ifyo= 7/3, thepoint (u,v)corresponding to(x,ys)isgiven by use, 0=Met sothatuandvarealwayspositive. Themapping oftheliney=7/3ontotheray isindicated inFig. 58,the part ofthe raycorresponding tox<0isshown bya dotted line. ’ los) yi en/a®) f } / O] = O] “ Fig. 58 Ifwenow consider thestrip inthexy-plane between thelines y=0,y=2m, weseethat theimage ofthestrip istheentire uv-plane with theexception ofthe origin. Line segments x=xcrossing thestrip map into circles centered atthe origin, ofradius less than one ifx»<0, and ofradius greater than one ifx»>0. Lines y=yomap into rays, theangle between therayand thepositive u-axis being yo.The origin inthe uv-plane isnot obtained astheimage ofany point in thexy-plane, but(u,v)->(0,0) asx->— =.The nature ofthemapping issug- gested byFig. 59a and Fig. 59b inwhich certain corresponding areas are indicated bysimilar shading. ’ —Yy(a> a {SKT/ =bil ‘(iie = aZo i=o l Fig. $90 Fig. 590, 92 MAPPINGS 251 Py » wal FHvery v=] a 3 vee ra) = oj1 4 Fig. 610. Fig. 616. Wenow seethat certain cells with curved sides inthexy-plane aremapped into rectangular cells inthe uv-plane (see Fig. 61a, 61b). The Jacobian ofthemapping (9.2-5) is BCU, 2)_ geGuay) +y). Hence themapping islocally one-to-one except intheneighborhood oftheorigin inthexy-plane. The notion ofapoint transformation, ormapping, isnotlimited totwo- dimensional problems. We may, for example, speak ofmappings from the xyz-space into uww-space. Observe that inthecase ofmappings from (x,y)to(u,v)which arelocally one-to-one, wemay interpret uand vascurvilinear co-ordinates inthexy-plane, orwemay interpret xand yascurvilinear co-ordinates intheuv-plane. Similar remarks apply when there are more variables ineach set. The sign oftheJacobian has asignificance which deserves mention. Con- sider amapping with Jacobian notzero at(xo,Yo),Sothat themapping is,atleast locally, one-to-one. IfCisasmall closed curve enclosing thepoint (Xo,yo)inthe xy-plane, the image ofCinthe uv-plane will beasmall closed curve C’ enclosing thepoint (u9, vs)which corresponds to(xo,ys). Ifthepoint (x,y)goes. around Cinthecounterclockwise sense,theimagepoint(u,v)willgoaround C'. Butwill(u,»)gocounterclockwise also? The answer depends onthesign ofthe Jacobian ofthemapping. IfJ>0,(u,v)willgoaround C’inthesame sense that (x,y)goes around C;butifJ<0, (u,v)will gointhesense opposite tothat of (x,y).Weshallnotprovethisfactjustnow.SeeExercise 7and$15.32. EXERCISES 1.Consider themapping x=au,y=be,wherea>0,b>0.Findoutwhatregionin theuv-plane corresponds totheregion inthexy-plane bounded bytheellipse (x"/a*)+ Orb)=1. 2.Let Rbethe region inthe xy-plane bounded bythe lines xy =0, x+y =0, x~2y =2, Find theregion R’inthewo-plane onto which Rismapped bytheequations u=2x-y,v=x-2y, 93 SUCCESSIVE MAPPINGS 253 Then fer-y, nets The single mapping which isproduced bycarrying out two successive transformations iscalled theresultant, orproduct, ofthetwo transformations. Intheabove example, wehave auc), Em,FT) aCr En) aay) asthestudent should verify forhimself. Observe that vou=-2y. Hence HE.)ACs2)(y—yy(—2)=ay=20)‘au,6)a¢x,y)~~I=4Y=yy This illustrates ageneral truth which wenow state formally. THEOREM I.Let T,denote atransformation from the xy-plane into the uv-plane, and letT;denote atransformation from theuv-plane into the &x-plane. Let the resultant transformation from thexy-plane into theEn- plane bedenoted byTs.Then theJacobian ofT;istheproduct ofthe Jacobians ofT;and Ts,that is, Gm) _al.) (u,v) aGx,y)~(u,v)3YY : Itis assumed that thetransformations T;and T;arecontinuously differenti- able. Itisfurther assumed that the transformation T;isdefined forpoints (u,v) ‘obtained byapplication ofthetransformation T;topoints (x,y)insome region R ofthexy-plane. Proof oftheTheorem. Weregard &,7asfunctions ofthefirst-class variables u,v,which areinturn functions ofthesecond-class variables x,y.Thus a_afau,aEa0axauax*avax 3€,9ou,OEavay*auayavay" withsimilarequations for2,3Itisthenamatterofstraightforward oa TRANSFORMATIONS OFCO-ORDINATES 255 6.Suppose Fi,...,Fq are continuously differentiable functions ofxy,...,X5 and thatSF:F)40,wherepisafixedinteger,1p<n.SupposetheequationsWy=FURokedooeoslp©Figo Xs)areSolved£0FX15...5XpimtermsOfty.tpandpots.+5%qandthatthesevaluesofx,,...,%arethensubstituted intoF,.s,---»Fosgiving risetofunctions (tis. -.UppXprteo+ey%a)s imP+Hy. .4M.Show that (Fs... Fa)_AUF. Fp)Opes) Occ e¥e) BRangHp)BCkpeteeeeXe)” Where itis assumed that u),..-,p arereplaced by FAG.tdaesFolinsoe) after calculating the derivatives inthe last Jacobian. SUGGESTIONS OF METHOD: Note first ofallthat 4(Fi.-+-5Fos Xystese+e%a)"Fxi,..2a)=p+ly..-snLetTyandTsbedefinedasfollows:Tr T ee) mae f= Folie) ue fonsByes pot=PorilbieeB ban te=Wales ba Now apply Theorem II(for thecase ofnvariables) and Exercise 5 9.4 /TRANSFORMATIONS OF CO-ORDINATES ‘The formulas connecting the rectangular co-ordinates (x,y)ofapoint and the rectangular co-ordinates (x,y')ofthesame point inarotated system (see Fig. 62) are x=x'cos —y'sing, vf P04-1) ceedy=x’sinb+y'cos6. ee ‘Thetransformation (9.4-1) hastheinverse ei, a x=xcos6+y sind, 042 me yl==xssin d+ycos4 asthestudent should verify forhimself. The student should also verify that atey)ayy 043)ey) OGY) ‘The familiar formulas connecting rectangular and polar co-ordinates inthe plane are x=rcos, y=rsing. (44) 9s CURVILINEAR CO-ORDINATES 259 or xaeoauto,wee . provided uand varenotboth zero. Ifthis result issubstituted in(9.5-2), weget yews? ory=tur. There arethus two possible differentiable transformations, xewo=, y=uv, (95-3) and xew-0, y=—uv, 05-4) which may beused todetermine apoint (x,y)bygiving values ofu,v.Either transformation allows ustousewand vascurvilinear co-ordinates. Suppose, for example, that weuse (9.5-3). There issome arbitrariness inthechoice ofsigns foruand v,but theproduct uvmust have thesame sign asy.We may, for instance, use u20 allthetime; then wemust use »>0 when y>0 and v<0 when y<0. Figure 65has been labeled inaccord with this choice. Other choices arepossible, however. From (9.5-3) wefind BOY) 90424 plStacey 2+ 8. This iszero only when u=v=0,which by(9.5-3) isequivalent tox=y=0. The origin iscalled asingular point ofthewo-curvilinear co-ordinate system. Itisnot possible touse uand »asco-ordinates, throughout aregion having theorigin as aninterior point, insuch away astohave aone-to-one correspondence between (x,y) and (u,v) with the transformation from (u,v) to(x,y)and the inverse transformation from (x,y)to(u,v)both continuous throughout the region, Now letusconsider the general theory ofcurvilinear co-ordinates inthe plane. Suppose wehave given acontinuously differentiable transformation x=flv), y=gu,v). (9.5-5) such that the Jacobian y=ihe) Au, v) isnotequal tozero foracertain pair ofvalues uo,ve.Denote thecorresponding values ofxand ybyxo,Yo.The following discussion will relate topairs (x,y) sufficiently near (x,yo)and pairs (u,v) sufficiently near (Ww,v9)Sothat wecan use the conclusions described inTheorem 1,§9.1. Inparticular, there isa continuously differentiable inverse transformation u=Foyy), v=G(xy). (9.5-6) Let usdenote the Jacobian ofthe inverse transformation byj.By(9.1-6) of ‘Theorem Iweknowthatj=f. 9s CURVILINEAR CO-ORDINATES 261 three-dimensional systems can begenerated bystarting with atwo-dimensional system and rotating the plane about aline inthe plane. The two families of curves inthe plane generate surfaces inspace. Half-planes through theaxis of rotation form athird setofsurfaces. Spherical co-ordinates arederived from plane polar co-ordinates inthis way. Example 2.Consider thetransformation X=uvcos#, y=uvsind, z=u—v, (95-10) with Jacobian 2) roy?+ _Saaray72ue(u?+0°. (95-11) Letusassume u=0, v20, and write r=(x?+y?)'"= uv,Then weseefrom (9.5-10) that x=rcos0, y=rsind. The equations rau, 2=W—0? (9.5-12) can beregarded asdefining asetofplane curvilinear co-ordinates (u,v)inthe rz-plane. With rotation about the z-axis, using r,@aspolar co-ordinates inthe xy-plane, weobtain equations (9.5-10), which wecanusetoestablish u,v,@as curvilinear co-ordinates inspace. The u-curves and v-curves intherz-plane are parabolas (compare with equations (9.5-3) and (9.5-1), (9.5-2)), afew ofwhich are shown inFig. 66. Inthe three-dimensional system, the u-surfaces and v-surfaces areparaboloids ofrevolution about thez-axis, andthe@-surfaces are half-planes with the z-axis asedge (see Fig. 67). The transformation has a continuously differentiable inverse intheneighborhood ofany point notonthe z-axis, forsuch points areobtained only when neither «nor viszero, and the Jacobian (9.5-11) isthen not equal tozero. Allpoints ofthe z-axis are singular points ofthewv8-coordinate system. z fen v-surface ene weaurface 7 v whe wel # Fig. 66. Fig. 67. 9.6 IDENTICAL VANISHING OFTHE JACOBIAN. FUNCTIONAL DEPENDENCE 263 (©)Solve for r,d,0interms ofx,y,2,assuming r>0, 0<6< 2/2, 0<0<z/2 for convenience. Compute 51%2:0} andveritythatitisthereciprocal oftheJacobian found in 9.Discuss thethree-dimensional system ofcurvilinear co-ordinates (u,v, ),where x=rcos0, y=rsin8, p-—sine p= sinhcosh u+cose Soak w+ eos and P=.2+y*, Begin bydiscussing and sketching theu-curves and v-curves inan re-plane. The relevant equations tobeobtained are P42? 2retahus1=0, P+2+2zctno-1=0. Then rotate around the z-axis, Find thesingular points oftheco-ordinate system. The co-ordinates (u,¢,8)are known astoroidal, oFring, co-ordinates. Describe the u-surfaces and v-surfaces byname. 10.What istheanalogue of (9.1-6) foratransformation x=f(u) where there isjust ‘one variable ineach set? 9.6 /IDENTICAL VANISHING OF THE JACOBIAN. FUNCTIONAL DEPENDENCE Inthis section weinquire into the state ofaffairs when the Jacobian determinant ofatransformation isequal tozero throughout aregion, First we consider briefly the significance ofthis phenomenon inthe linear case. After that we move tononlinear mappings from theplane totheplane, with reasoning which could beused toextend thetheorems tohigher dimensions. The two linear functions ax+byand cx+dyaresaid tobelinearly dependent incase there are two numbers k;and k;,not both ofwhich are zero, such that k\(ax +by)+ky(ex +dy)=0. Those who have had some linear algebra will recall that anecessary and sufficient condition for this isthat jablleal“° ‘This determinant isthe Jacobian ofthe linear transformation u=ax+by, v=ext dy. Sothe vanishing ofthe Jacobian ofthe transformation isequivalent tothe existence oftwo numbers kyand k:such that ku+kyw =0 forall (x,y). 6-1) 9.6IDENTICAL VANISHING OFTHEJACOBIAN. FUNCTIONAL DEPENDENCE 265 ProofofTheorem If.Consider theequation u=F(x,y).Suppose 240at (x,yo); then theimplicit-function theorem guarantees theexistence ofasolution x=f(u,y) giving alltriples (x,yu) near (xo,yo,uo) for which u=F(x, y). Moreover, ar a)on 06-6) ax Allthis applies when thedifferences x—x», ¥—Yo,U~up are sufficiently small. Consider the function G(x,y)asafunctionofwandy,withx=f(u,y).The partial derivative with respect toyis aF.G) 2Gf9G. AY) 9 axay ay aF ax because of(9.6-6) and (9.6-3). Thus Gif(u, y),y)isactually independent ofy. Let uswrite 4(u)= Gf(u,y),y).Sincex=f(u,y)isequivalent tou=F(x,y) forthevalues ofthevariables here inquestion, #(u) =G(f(u, y),y)isequivalent to(9.6-4).IfweassumeSF10insteadofFF#0,similarreasoning againleads to(9.6-4). Example 2,The argument isillustrated byF(x,y)=x°y°, G(x, y)=—2xy, xo=Yo=I,flu,y)=Vuly,o(u)=—Vu.Observe that,withthesesamefunc-tions F,G,butwith x»=—1, yo= 1,weobtain f(u,y)=— Vuly, (u)=Vin. ‘Theorem IIIcan begeneralized tonfunctions ofnvariables. For n=3the hypotheses that AFG, H)_‘aGyz)~° oon inaneighborhood of(xo,yor20)and that atleast one oftheJacobians AFG) (FG) a(F,G) ay)" 8G,2)" 9G) isnot zero atthis point, leads toaconclusion oftheform H(x, y,2)=(FG ¥,2),GU ¥s2)), (06-8) where (u,v) isdefined near uy=F(xo, Yo.20),Yo=G(Xor You20): Theorem IIIsays that under certain conditions theidentical vanishing ofthe Jacobian implies that neighborhoods (two-dimensional subsets) inthexy-plane faremapped into curves (one-dimensional subsets) intheuv-plane. Our next theorem will besomewhat like aconverse ofthis. We shall assume that forall 9.8 IDENTICAL VANISHING OFTHE JACOBIAN. FUNCTIONAL DEPENDENCE 267 xand yinterms ofu,v, 2,and substitute thesolutions forxandyinH(x,y,2).Observe that theresult isindependent ofz,From this work find thefunction (u,v) such that (96-8) hoids. (b)Show that thislineofreasoning isapplicable inthegeneral case, provided HFG)2.9),aay) *° 4.Show ineach case that thefunctions arefunctionally dependent, and find theway inwhich thethird function depends onthefirst two. Use themethod ofExercise 3. (@)waxtytzomrytyetowesttytt? ()w=xi(y~2),0 =y=2),w=ZIG). 5.Without using theinverse function theorem, prove thevariation onTheorem 1V obtained byreplacing thethird sentence ofthat theorem bythefollowing: Let®(u, e)be adifferentiable function inR’forwhich {+3>0 ateach point ofR’.HINT: Allyou need isthe chain rule and the fact that ahomogeneous linear system has anontrivial solution ifand only ifthe coefficient determinant iszero. MISCELLANEOUS EXERCISES 1.(a) Find the inverse ofthe transformation waSiende5,wherePaateytert, (®) What are the u-surfaces? Iculate 2¢4% (©)Calculate S81) 2It =rcos4,x=rsin$cosy,¥=rsin6sin¥cos8,z=rsin6sin#sin6,showthatP4224yh2PadfindBEEI, This indicates how spherical co-ordinates may beintroduced infour-dimensional space ‘3.The notations inthis exercise arethose ofthediscussion preceding Example 2in $9.1. Show that thesetE,which istheimage oftheopen rectangle Runder theinverse transformation, isconnected. HINT: Let Adenote that subset ofEconsisting of(Ue, t) andthose points ofEwhich canbereached from (ue,ts)byapolygonal path made upof finitely many linesegments andlying altogether inE.Aisnotempty. LetBbethepart of Ewhich isnotinA.We want toshow that Bisempty. Show that Aisopen and Bmust beopen. Then T(A) and T(B) aretwo disjoint open sets whose union isR.By§5.1, Example 5,they cannot both benonempty. Therefore T(B) isempty, soBisempty andEFisconnected. 104 VECTORS INEUCLIDEAN SPACE 269 There is,ofcourse, apoint ofview about Euclidean geometry (ofthree dimensions, letussay, tobespecific), inwhich the geometric theory iscon- structed without anorigin and aco-ordinate system, the fundamental notions andtheorems being developed from assumptions about points, lines, planes, and theuse ofdistance and theconcepts ofparallelism and perpendicularity. Inthis aspect ofEuclidean geometry there isnoparticular point tobegiven special recognition as“the origin” and there arenoparticular lines tobegiven special status asco-ordinate axes. We shall use this “‘co-ordinate-free” point ofview from time totime and weshall take advantage ofourcommon familiarity with what wemay call the“physical reality” ofthree-dimensional space, thespace in which welive and about which wehave some useful intuitive perceptions based ‘onexperience. But we shall build our systematic treatment ofvectors in Euclidean space onthefoundation provided byR’asasetofthings called either points orvectors. The same thing canbedone forR*asthebasic model for Euclidean plane geometry (when treated analytically). And then itisreadily seen how tomake thegeneralization toR",where ncan beany positive integer. Letusdenote elements ofR’byA=(Ay, Az,A:), B=(Bs, Bs,By), and soon. Weshall call them vectors, using boldface type forthesymbols. Wecall Aj,A, Asthecomponents ofA.They aresimply theco-ordinates ofAifwethink ofit asapoint (which weare free todo,ofcourse). The reason forusing the word vector isthat wearegoing tointroduce definitions ofalgebraic operations onthe elements ofR?which make itinto what iscalled avector space inthetechnical terminology oflinear algebra. Itisacommon usage torepresent Avisually asan arrow from (0,0, 0)to(Ai, Az,As). IfPisthetipofthearrow (see Fig. 68), then PisAwhen wethink ofAasapoint. There isanalgebra ofvectors inR’which rests onaddition andsubtraction ofvectors and multiplication ofvectors byreal numbers (which areoften called scalars). The vector (0,0, 0)iscalled thezero vector, and denoted by0. The vector sum ofAand Bisdefined by A+B=(Ay,Az,Ay)+(By,Bs,By)=(Ay+By,A+Bs,Ay+By).(10.1-1) ’ D 1 las © i, id im a Fig. 68. 10.4 VECTORS INEUCLIDEAN SPACE 271 We see from (10.1-12) that (A+AJ? =(Ap+A}+AR)? isthelength ofthe vector A,that is,thedistance from theorigin tothetipofA (ortoA,thought ofasapoint). Weshall denote thelength ofAbyAl]andcallit the norm ofA. Thus JA}=(A- Ay”? (10.1-16) The only vector with zero length is0.We say that each nonzero vector determines (orhas) adirection inR’,IfA#0andB#0, wesaythat AandB have the same direction ifone isapositive multiple oftheother B= cA, where c>0. When Bisanegative multiple ofAwesaythat Aand Bhave opposite directions. Ifwesubject every vector inR?toatransformation byadding toeach vector thesame nonzero vector C,sothat every Aistransformed into A+C, wecall this atranslation ofthespace. Every point iscarried into anew point which isa distance [CIinthedirection of€fromtheoriginalpoint. There isaconvenient way tovisualize addition and subtraction ofvectors. Let Aand Bbetwo nonzero vectors, and picture them asarrows emanating from theorigin. IfBisnot amultiple ofAthetwo vectors determine aunique plane. ‘The sum C=A+B isthen thevector emanating from theorigin which forms thediagonal oftheparallelogram ofwhichAandBareadjacent sides.Another wayofdescribing thesituation intuitively isasfollows: Displace Bbyatranslation which brings itsinitial point (the origin) into coincidence with theterminal point (tip) ofA.The sum A+B isthen the vector from the initial point ofAtothe terminal point ofthedisplaced vector B(see Fig. 69). This mode ofrepresenting D . le ° A Fig. 09. vector addition gives rise tothename “the parallelogram law ofaddition.” IfBhas thesame oropposite direction aA,theaddition ofBtoAcan bevisualized bythesame 8 process ofdisplacement ofB:A+B extends from the initial pointofAtothetipofthedisplaced vector B. a| Vector subtraction canalsobedisplayed visually by 4 ' useoftheparallelogram law. SeeFig.70andrecall that i A=B isthatvector which, when added toB,gives A. apa! Multiplication byscalars canlikewise bedisplayed visu- ally. See Fig. 71. Fig. 70. 10.11 ORTHOGONAL UNIT VECTORS INR? 273 ‘ordered pairofpoints suchthatPQandRShavethesamelength anddirection,PGandRSaresaidtobeequal(or,sometimes, equivalent) vectors.Thismeans,thatRScanbebrought intocoincidence withPQbyatranslation ofthewhole spacewhereby Rismoved alongastraight linetoP,andSislikewise moved, in thesame direction, toQ.Inphysics avector isoften called afree vector ifitis considered tobethesame after any translation ofthespace. Inour presentation ofvectors inR’,however, wethink ofallvectors ashaving their initial point at theorigin. With this mode ofthinking about vectors wecall R’aEuclidean vector space. Asmentioned previously, wefind itconvenient torefer toavector ‘Aas either avector orapoint, sothat itisneedless tohave aseparate notation forthepoint that isthetipofA. 10.11 /ORTHOGONAL UNIT VECTORS INR* Let afixed rectangular co-ordinate system bechosen with origin O.Itis conventional, particularly indealing with physical applications, towork with right-handed co-ordinate systems, and weshall ordinarily adhere tothis con- vention. Now leti,j,kbevectors, each ofunit length, inthe directions ofthe positive x,y,and zaxes, respectively (see Fig. 73). Itisclear that ifAisany vector, wecan express itinthe form A=AjitArj+Ask, (10.11-1) where Aj, A;,Ayarethecomponents ofA(see Fig. 74). We call i,j,kthe 2 k 2 9 y i v ws AG Fig. 73. Fig. 74 fundamental orthonormal triad associated with thisparticular co-ordinate sys- tem. The word “orthonormal” isacombination of“orthogonal” and “normal.” The vectors i,j,kform anorthogonal set; that is,they are mutually per- pendicular. Avector issaidtobenormalized, ornormal, ifitisofunitlength. The orthonormal character ofthetriad i,j,kisexpressed bytherelations ibej-jek-k=1, ijajk=k-i=0. (10.11-2) 40.2 ‘THE VECTOR SPACE R” 275 Thelengthofx,calleditsnorm,isdenotedby[ji]anddefinedby I=G++ +x (10.12-2) From (10.12-1) wesee that BP=x-x. (10.12-3) With (0,0,...,0) asthe vector 0,the same algebraic laws hold forR*as those given forR?in(10.1-4) to(10.1-11) inclusive and (10.1-13) to(10.1-15) inclusive. The following inequality, known asCauchy’s inequality, isvery important: ‘ ra ye [Sam] (S0)" (Si). (10.12-4) One way ofobtaining Cauchy's inequality was indicated inExercise 29,§6.8. For another way see Exercise 4attheend ofthis section. We can rewrite (10.12-4) as fe+ylSibeli (10.12-5) From this itiseasy toshow that e+ylSl +Il. (10.12-6) We leave the derivation of(10.12-6) tothe student; see Exercise 5.This inequality iscalled the triangle inequality. Itcorresponds tothe geometric assertion that inatriangle thelength ofone side isnever larger than thesum of thelengths oftheother two sides. Wedefine distance inR*bythenatural generalization oftheformula inR?. The distance d(x,y)between xandy(thought ofaspoints)isdefined as dexy)= [3os-wF] =bev. 0.12-7) Two vectors x,yinR*aresaid tobeorthogonal ifx-y=0.The naturalness ofthis definition isseen from thediscussion accompanying (10.1-19). There isasetofnmutually orthogonal vectors ofunit length inR*entirely analogous tothevectors i,j,kinR?that were introduced in§10.11. Wedefine e,=(1,0,0,...,0) = 0,1,0,....0) (10.12-8) =0.0,...0,D. Itisclear that je,l|=1ande,-e;=0ifi¥j,sowecalle,...,€, anorthonormal set. We see atonce that wecan write Kemet mee (10.12-9) sothat each vector xisalinear combination ofthee,'s, thecoefficients being the components ofx.This representation ofxistheanalogue oftherepresentation ofAin(10.11-1). v0.12 ‘THEVECTORSPACER” 277 VisYrs+05¥4s Weobtain thesystem ofequations civ temy vt tay m=0 cave vit cna vat eaves ve=0 (10.12-13) civ bee vate tev sv 20. Ontheother hand, ifequations (10.1213) aresatisfied byasetofc;'s, then soare theequations that result when wemultiply both sides ofthefirst equation bycy, thesecond by¢:,and soon.But, byuseofthealgebraic rules governing dot products, theequations thus obtained can bewritten ew (ei toe Hey) =O ext (emi tetea) =0 eave (eit +evi)=0. Onadding these equations, weconclude that (east eav) (erst +n) =0, and hence that equation (10.12-12) isvalid, foravector iszero ifand only itsdot product with itself iszero (see (10.12-2) and (10.12-3)). We see, therefore, that anequation oftheform (10.12-12) holds true ifand only ifthe system ofhomogeneous linear equations (10.12-13) (with the c's regarded asunknowns) issatisfied. But, bythealgebraic theory ofsimultaneous linear equations, thehomogeneous system (10.1213) issatisfied byasetofc's that are not allzero ifand only ifthedeterminant ofthesystem (which isthe determinant Gin(10.12-11) isequal tozero. This proves thetheorem. The determinant Giscalled the Gram determinant, orthe Gramian, inhonor ofJ.P. Gram, amathematician ofthenineteenth century. We shall presently see hisname again. The members ofalinearly independent set ofvectors need not beunit vectors, ofcourse, and they need notbemutually orthogonal, butifwearegiven alinearly independent setofkvectors Ai,.... Ax itispossible byasystematic process toconstruct anorthonormal setofkvectors ¥,¥3...-.¥i, each ofwhich isalinear combination oftheA,’s. (Here weareassuming k=2.)This process is called theGram-Schmidt process, inrecognition ofthe work ofJ.P.Gram and E.Schmidt, a f°-€ je mathematician oftheearlytwentieth century. | The idea oftheprocess isquite simple; the i basic method can beinterpreted geometrically by ' dealingwithtwolinearly independent vectors that a2 arenotorthogonal. Wedenote them byCand D © ce and picture them inaplane. See Fig. 75. Fig.75. 10.12 ‘THE VECTOR SPACE 2° 279 weseethat vj,v2,vyform anorthonormal set. Moreover, v;isamultiple ofAj,v2 isalinear combination ofA;and Az,and vsisalinear combination ofAy,As,and Ay.Itisnow clear how tocontinue theprocess ofobtaining more v,"saslong as there arestill A,'s from which tosubtract their projections on¥j,...,¥i-t- Example. Show that the following four vectors inR‘are linearly in- dependent. Then apply theGram-Schmidt process toconstruct anorthonormal setfrom thegiven vectors. The vectors are: AR(LLID A= (,-1,0,-0,As=(0.0,1,D, c=(1.—2,-2,0). (20.12-15) We illustrate theapplication ofTheorem I.First wecalculate Ar Ard, Ay Ap=—2, Aye Ay= 2, Are A= 3, AsAr=2, 0AyeAy=—1 AnsAg=2, Ay+As=2, Ayt Age —2 Aer A= 9 The Gramian is |4202-3G-|2 2-12 . |2-1 2-2 |-3 2-29 Bystandard methods for calculating the value ofadeterminant wefind that G=25.Because G#0weconclude that thefour vectors arelinear independent. Therefore wecan proceed with the Gram-Schmidt process. We listtheresults bystages, leaving thedetailed calculations tobeverified bythestudent. waGhhd, B= Ar+vi=(,-4,4,-0, v=Br, BreAyws Chv=By, By=Activi~$24vs,Be=(-4 49, IBY vem Ghd. EXERCISES: 1.Show directly bythedefinition that Aj,...,Ax isalinearly dependent setof vectors ifatleast one ofthe vectors iszero. 2.Show directly bythe definition that asetofnonzero and mutually orthogonal vectors islinearly independent. 3.Show that the vectors Ay=2i, Az=3+4j, As=i+2j+3k are linearly in- dependent, andapply theGram-Schmidt process tothem, 4.(a)Deduce theinequality (10.12-4) intheform (10.12-S) with theaidofthe 102 CROSS PRODUCTS INR? 281 ofmagnitude AXB JA Bl|= [Al[BIsin@ B whoselineisperpendicular totheplaneofAandB,and se whose direction issuch that A,B,and AxB forma Kk} ssright-handed system(seeFig.76).IfthevectorsA,Bliealong athesame line,theydonotdetermine aplane. Inthiscase aM sin@=0,however andsoAxB=0,Notethatthemagnitude pig76ofAXBisequaltotheareaoftheparallelogram ofwhichA, aBareadjacent sides. The motivation for this definition, and itsusefulness, will bebetter under- stood bythe student after hehas seen the occurrence ofthe cross product in physical applications and inlater mathematical developments. Ithas only very limitedanalogies withtheordinary product oftwonumbers. Moreover, thecross product issomething peculiar tovectors inthree dimensions, having noanalogue forvector spaces ofdimension other than three. ‘The principal algebraic rules governing thecross product are AXB= -(BXA), (102-1) (cA)xB=c(AXB), (10.2-2) AX(B+C)= (AxB)+(AxC). (10.2-3) Multiplication isnotcommutative, butanticommutative, asweseeby(10.2-1). ‘This lawisapparent {rom thedefinition ofthecross product, since B,Aand —(AxB)formaright-handed system. Thelaw(10.2-2) isalso apparent fromthe definition ofthecross product. The rule AX (cB) =c(A B) (10.2-4) can bededuced from (10.2-1) and (10.2-2). Now consider the proof ofthe distributive law(10.2-3). The law obviously holds ifA=0,so weconsider theproof ontheassumption that B A0.IfBisanyvector, letB’denote thevector adprojection ofBontheplaneperpendicular toA. (2 - through theorigin (see Fig. 77). Clearly [B'|= A {BIsin6,and therefore AxB=AxB’.Now, pro- jecting inthis manner, weseethat B’+C’ isthe projection ofB+C.Therefore, instead ofproving Fig.77. (10.2-3), itissufficient toprove AX(B'+C)=AXB+AXC. (10.2-5) The advantage here isthat thevectors B’,C’and B'+C’ either are0orare perpendicular toA. 103 RIGID MOTIONS OFTHE AXES 283 rows, respectively. Accordingly, asamemory device, wesometimes write |ij k| AxB=|A, AyAy). (10.2-7) |B) By By} EXERCISES 1.Find theindicated cross products. (a)G+) +W)x0+ 25+ 3h); (b)(k=34k)x2h5}+310;(©)@-2)-W x3)+48), 2.Find thearea oftheparallelogram ofwhich thevectors A=i—J+2k and B= 21+ 4j—kareadjacent sides. 3.(a)Let A,B,Cbenoncoplanar vectors from ©with terminal points P,Q.R respectively. Explain why \B—A)x(C A) isavector perpendicular totheplane of PQR, and oflength equal tothearea ofthetriangle PQR. (b)Find thearea ofthe triangle formed bythepoints (1,1,—2), (2,-1, 1),(1,3, =D. 4.Find A«(BxC)andB-(AxC)if (a) A=U-3)+ 5k, B=-i+ 4j+2k,C= 4+3); ()A=2143)+k,Bat+2}+Sk,C=-21+4) +38 5S.(a) IfA=(Aj,Aa,Ay),ete.,showthat iaA:i A-(@BxC)=|B, Bs Bs}. CG (b) How does itfollow from (a)that A+(BX C)=(Ax B)-C? (©)What isthevalue ofA(Bx C)ifany two ofthevectors areequal? (4)Explain why thenumerical value ofthedeterminant in(a)isequal tothevolume of theparallelepiped having thevectors A,B,Casconcurrent edges. (@)I£A,B,CarepermutedinallpossiblewaysintheproductA-(BxC),howmanydifferentvalues ¢an beobtained? (O Find the values ofD-(B~ A)and D-(C~ A), where D=AXB+BXCHEXA, 6.Let thepairs, A,Band C,Deach determine aplane. Write anequation involving dot and cross products expressing thecondition and these two planes beperpendicular 10.3 /RIGID MOTIONS OF THE AXES Byarigidmotion oftheaxeswemeanashiftfromarectangular co-ordinatesystem xyzwith origin Otoanother rectangular co-ordinate system x'y'z’ with origin O',both systems having thesame unit ofdistance, and both systems having thesame orientation (i.e., both being right-handed orboth left-handed). ‘Such ashift canbeaccomplished intwo stages: byatranslation toanew system with origin O’andaxes parallel totheoriginal axes, followed byarotation about 0.The equations foratranslation oftheco-ordinate system arevery simple, and need notconcern usright now, since weregard thevector space with origin 103 RIGIDMOTIONS OFTHEAXES 285 These aretheequations oftransformation fortherotation oftheco-ordinate system. The inverse transformation can befound inexactly the same way, starting from (10.3-3). The equations are xs hx’+hy'+h2" y= myx!+my!+mz" (10.3-5) z=mx'+ my’ +ns2', Consider now avector Aofthevector space with origin O.This vector will have components Aj, Az, A;inthe xyz-system, and Aj, AS, Ajinthe x'y'2'- system. Since thecomponents ofAare merely theco-ordinates oftheterminal point ofA,wesee that the two sets ofcomponents are related inexactly the same way that xyz and x’y'z' are related, that is, Aj=Art mArtmAs AS= LA,+mAr+mAs (10.3-6) AS=bAi+ mAr+ mAs, with aninverse setofrelations corresponding to(10.3-5). The two sets of relations areeasily kept inmind byatable similar to(10.3-1): Ar A: As Ab] bom om AS|bom: otsA} boom om Itfollows from these laws oftransformation ofcomponents that ifweknow the components ofavector inone co-ordinate system, wecan find itscomponents in any system obtained byarotation oftheaxes. EXERCISES 1.Show that F=Litmj+mk,1=Liby+bk, and obtain four other allied relations. Start from the fact that, ifAisany vector, AS (ADEADIHAWK. 2.What isthenumerical value ofi+ (xk)?Express thisproduct interms ofi, K’ bytheresults ofExercise 1,anddeduce that hob b ‘my mz ms|=1. moms omy} See Exercise Sa, $10.2 3.Observe that ifonesolves (10.3-5) forx’byCramer's rule, anduses theresult of Exercise 2,one finds x!=(many —msn) +(nabs~malay+(lamy~bms)z. 104 INVARIANTS 287 between thecomponents ofvectors inthetwo systems isexplained in§10.3. The fact that thecomponents ofAalong the axes oftherotated system are Aj,Ai, Ajenables ustowrite A= Ali’+Aij’+ Aik’, There isasimilar representation ofB.Therefore A+ B=(Aji'+Aij’+Aik)(Bil+BS’+Buk). We can calculate this dot product bythealgebraic rules governing dot products. Because ofthefact that thevectors i,j, k’form anorthonormal system wehave relationships such asi'+i/=1, i'-j'=0, and soon, When we complete the calculations we find that A+B= AjBi +AiBS+ ASB. (0.4.1) From the definition ofA- Bin(10.1-12) we now see that AiBy+AzB2+AyBy=AjBi+ASBi+ASBS. (10.4-2) ‘Thus we see that the formula for A-B isthe same interms ofthe primed components asitisinterms oftheunprimed components. That iswhat wemean when we say that the expression onthe left in(10.4-2) isaninvariant, oris invariant with respect toarotation oftheco-ordinate axes. We could anticipate this result, ofcourse, because offormula (10.1-19), which expresses A-B in geometric terms, using the length ofthe vectors and the angle between them, which wearealready accustomed tothink ofasbeing independent ofthechoice ofaparticular co-ordinate system. Another example ofinvariance isprovided bythecross product oftwo vectors. Bythis wemean that the cross product AxBcan beexpressed inthe form AXB =(ASB~ASB9W +(ASBi—AjBj’+(AjBS— ASBik’, (10.4-3) which hasthesame form as(10.26) except that everything (components and the set oforthonormal vectors) isreferred tothe x'y'z'-co-ordinate axes. The procedure used toderive (10.2-6) started from ageometrical definition ofAxB, without theexplicit involvement ofaco-ordinate system. The derivation of (10.2-6) depended onexpressing AandBaslinear combinations ofi,jandkand then making useofthealgebraic rules (10.2-1), (10.2-2), (10.2-3) andthenine special formulas forthevarious cross products such as1X1, 1Xj,1X...as exhibited in§10.2. The same method will lead ustothe formula (10.4-3) ifwe express AandBaslinear combinations ofi,j,andk’,andifwemake useofthe special formulas forthevarious cross products such as ¥x¥=0, Fxy=k, ixk =f and soon.The correctness ofthese formulas isapparent from thegeometric definition ofcross products. 104 INVARIANTS, 289 aslinear combinations ofi,j,kasfollows: =hitmy+nik j= hit my+mk (104-9) K’= bit mj+mk. Also, because i’,j',andk’arelinearly independent andR”isthree dimensional, any vector AinR?isalinear combination ofi’,’andk’,thecoefficient of being A-i’, and soon. Inthis way we can see with the aid of(104-9) that islet bythe j=mi+maf!+mk! (10.4-10) k= ni’+ng’+mk’. For example, thecoefficient ofjintheexpression forkisk-j’ =m:,aswecan see from the second equation in(10.49). There are various relationships between the elements inthe determinant \hmomD=/|b mz m (10.4-11) [my om Forexample, I}+13+1=1 istheexpression ofthefact that i= 1,and Ils+ ‘mim;+nn=0istheexpression ofthefactthati«j=0. Now letusfind the values ofx1’, XJ’, Xk’, and soon. Itisclear from thevery definition (10.4-4) that ix1’=j'xj'=k’xk’=0, sowecanconcentrate onVx’, {XK ki" and then obtain j'X1', etc., byusing the first result in (10.4-5). Weobserve that thex’and y’components ofixj’are(x) and (i<j) +f,both ofwhich arezero because, aswesaw from (10.4-7), (AX B)-C =0 ifany two ofthe three vectors are the same. The z'component of1X,’ is (x) +k,and this isD,aswesee from (10.47), (10.4-9), and (10.411). Thus ixj= Dk’. We can find j’xk’and k’xi’bythe same method. We display the results: ixj'=Dk yxk’= DY. (10.4-12) Kxi'= Dy’ Wedon't yetknow thevalue ofD,but wewill find it,Let uscalculate jk, using thesecond and third formulas in(10.4-10): DK=(mi+maf+mk!)x(nal+nai+mk) When this isworked out using therules (10.4-5) and (10.4-6) and theresults (10.4-12), wefind BK=(mgm,myn) +(mgm,—myn)DY’+(mynz—mam)Dk. 105 SCALAR POINT FUNCTIONS 291 ‘scheme (10.3-7) becomes Av A: Ay Alo 1 0 As -1 0 0 AlO oo 1 ‘Then, show byanexample that, ifAand Barevectors, A,B) isnotascalar invariant. Likewise show that thetriple (A:B1, A2B:, AyB3) does notdefine avector invariant, thatis,thatingeneralthevectorhavingcomponents (A,B),A2B2,AsBs)inthexyz-systemdoes nothave components (AiBi, AiB3, AsB\) inthex’y’z'-system. Consider, e8., A=1, Beity. 2.IsA+ Ast Aya scalar invariant if A=Ad+Asj-+ Ask? Justify your answer. 3.Let thenine direction cosines inthetable (10.3-1) bespecified asfollows: aj v2|v2 afar] a v3|v3|v3 aft |-2Vé|V6|Vo LetAwi+j,Bai-j+k. (a)Calculate AxBdirectly interms off,j,k. (b)Calculate AandBandthen AB interms offy,” (€)Reconcile the results in(a)and (b)byconverting AB from (b)tothe answer in(a) byexpressing i,J’,kin terms ofi,jk. 10.5 /SCALAR POINT FUNCTIONS ‘The word “scalar” isused tocontrast with the word “vector.” Itiscustomary, in any context where vectors and real numbers areboth being discussed, torefer to real numbers asscalars. Thus, inthis book, ascalar isareal number. The word may also beused asanadjective. Let usrecall thegeneral meaning oftheword “function.” Afunction isa correspondence between two classes ofobjects; these two classes arecalled respectively thedomain ofdefinition ofthefunction and therange ofvalues of the function, or,more briefly, the domain and the range ofthe function. The function itself isthe correspondence whereby toeach object inthedomain is assigned acorresponding object intherange. Let usnow consider acase in which thedomain isaclass ofpoints and therange isaclass ofreal numbers. In such acase weshall call thefunction ascalar point function. Iffdenotes the 1051 VECTOR POINT FUNCTIONS. 293 tiability for scalar point functions, without reference toco-ordinate systems. Alternatively, however, the definitions may bemade with reference tosome arbitrarily chosen rectangular co-ordinate system. Thus, iff(P)= F(x, y,z)in that system, wecan saythat fiscontinuous atPpifFiscontinuous at(x0, Yor 2),with asimilar definition fordifferentiability. Although these definitions are made with reference toaparticular rectangular co-ordinate system, they are actually independent ofthechoice ofthat system. For example, ifF(x, y,z)= G(x’, y',2'), where the two systems are related byarotation, and ifFis differentiable, then Gisdifferentiable also (byTheorem V,$7.3). 10.51 /VECTOR POINT FUNCTIONS ‘The concept ofavector point function issimilar totheconcept ofascalar point function inthe matter ofbeing independent ofparticular choices ofco-ordinate systems. The difference isthat thefunction values arevectors instead ofscalars.‘Thedomainofdefinition ofavectorpointfunctionissomesetofpointsP.Therange ofvalues ofthefunction issome setofvectors. Letfdenote thefunction, and letFdenote thevector corresponding toP.Then wewrite F=f(P) where J(P) depends just onPitself and notontheco-ordinates wehappen tobeusing. Example 1.LetR=OP,andletAbeafixed vector. Then each ofthe expressions R, (A-RR, AXR, (RIOR defines avector point function. Itisoften convenient tointroduce aco-ordinate system inorder todeal with ‘avector point function. We shall beconcerned mostly with rectangular co- ordinate systems. Ifwehave anxyz-system with origin O,leti,j,kbethe fundamental orthonormal triad associated with thexyz-system. Suppose wehave afunction F=f(P); letthecomponents ofFbedenoted byFi(x, ¥,2),Fu(X, y,2), FAQ y,2).Then F=F(x, y,201+ Puls, y,2))+ Gs y,2k. (10.51-1) Inanother rectangular system x’y'z', obtained from thexyz-system byrotation, Fwillhaveadifferent setofcomponents, andwiththetriad1’,j',k’forthenewsystem, Fwill beexpressed intheform F=Fix’, y',200+ Fa’, y's207+ FAG, y's29K. (10.51-2) ‘The components F;,F3,Fjwill berelated toF),F:,F;inthesame way that Aj, Af,Ajarerelated toAy,Az,Asinequations (10.3-6). Also, thetwo orthonormal triads arerelated inthemanner indicated bytable (10.3-1).‘ThevalueFof thevector point function isavector invariant. Note, however, that anindividual component ofF,such asF(x, y,2),isnotascalar invariant, foringeneral F(x, y,2)4Fix’, "52. 108 ‘THE GRADIENT OFASCALAR FIELD 295 ‘Thus thegravitational force Fontheunit mass atPis r=-k Mr (10.51-4) Aportrayal ofthis vector field issuggested byFig. 81. EXERCISES 1.The equations oer)x=vit yd , 1 ste'tyythy,y=ty)Va ze-le'tyye teev2 define arotation ofaxes. (a)IfF(x, y,z)=2x?—y?~2? istherepresentation ofascalar point function inthe xyz-aystem, find therepresentation G(x’, y'.2°ofthefunction inthex’y'z"-system, (b)What isthetable (10.3-7) forthis rotation ofaxes? Find thexyz-representation F(x, y,2)forthescalar point function forwhich G(x’, y’,2!)=x’+y'+ V2z", (©)Findthecomponents ofthevectorfieldF=Vizi+(y +2))+Vixk inthex'y’2'- system, (@)Express thevector fieldF=V2y't' +z'j'+x'k’ inthexyz-system. 2.The equations x= |Qx+3y +62), y'= 4x6) +22), 2=K6x+2y—32) define arotation ofaxes (a)Express x,y,zinterms ofx’,y’,and 2’. (b)Find theexpression ofx+y?— 2”inthexyz-system. (6)Express the vector field ~y-+x} inthe x'y'z-system. (d)Doxi+2yj+32kandx‘+2y'j' +32'k’represent thesamevectorfield?Justifyyour answer. 10.6 /THE GRADIENT OF ASCALAR FIELD Ascalar point function fmay bethought ofbyimagining each point Patwhich Jisdefined ascarrying alabel with thevalue f(P) ofthefunction atthat point (see thediscussion ofthesecond mode ofrepresenting functions in§5.4). When ascalar function isrepresented inthis way, itisoften called ascalar field. Letfbeascalar field defined throughout some region R,and suppose that f isdifferentiable inR.Wearegoing todefine theconcept ofthegradient ofthe field. The gradient ofascalar point function isavector point function. Itis 296 VECTORSANDVECTORFIELDS cn10 convenient touse arectangular co-ordinate system intheprocess ofdefining the gradient. However, wemust take care tobecertain that thedefinition gives usa result which isindependent ofthe choice ofthe particular rectangular co- ordinate system. Let asystem ofrectangular co-ordinates xyz, with origin O,beselected arbitrarily. Inthis co-ordinate system letPhave co-ordinates (x,y,2), and letf(P) = F(x, y,z) bethe representation ofour point function. Form the vector field whose representation inthexyz-system is aF,,oF,aFrie ee (10.6-1) This vector field, orvector point function, iswhat weshall call thegradient of thegiven scalar function f. Presently weshall give aninterpretation ofthegradient which enables usto think ofitapart from the co-ordinate system. But first letusshow that the definition ofthegradient yields thesame vector field nomatter what rectangular co-ordinate system ischosen. Ifasecond co-ordinate system isselected, thetwo systems are related insuch away that one can beobtained from theother by either atranslation orarotation, orboth. Let usconsider the case where both systems have thesame origin, and arerelated byarotation ofaxes. This isthe case ofprincipal importance forourdiscussion ofthegradient. The case where a translation may beinvolved isconsidered inExercise 13. Let the co-ordinates ofthe two systems bexyz and x'y'z’, related asin (10.3-4) and (10.3-5), and letthetwo representations ofthescalar field be f(P)=F(x9,2)=GO,4.2 We wish toshow that aF,OF),FyWGy,AG»,aGy, 3SebGeDtGekmSatdtGok (10.6-2) Once this isdone, itwill beclear that thedefinition ofthegradient offby expression (10.6-1) isinvariant under arotation oftheaxes. Toprove(10.6-2) wemustshowthatthetriple(9E.2F,2)jgrelatedto Laba ax ay az) ° (2G9G|G) K i thetriple(36,26.2G) justasthetriple(Ay,AnAs)isrelatedtothetriple (Aj, Ai,Ad) in(10.3-6) or(10.3-7). Now, bythechain rule, 3G_OFax,OFay,aFaz,ax’ xax’ ayax’ azOx withsimilarequationsforgsand36.From(10.3-5)weseethat a 3 =,=m3 =m. 107 THEDIVERGENCE OFAVECTORFIELD 301 Also, bythechain rule, AF,_Fax|OF,ayAF,a2 ax” axax’ ayax’? a2Ox! 22Fig99 =Em En SE withsimilar formulas for2?ana2°,ax’adi Inthesame way, OFS1OFym,2Fy9,2%, Sei SetmsGE+mS 2F4,2Figy,Fg93,Fea mE mE andsoon,Thestudent shouldhimself writeoutalltheformulas. From2£!we obtain nine terms: BFFigjp,Fogyn,HE ant WagFhmgythm Ge mds +mjE+man aFy OF), 29F,nuhGetmamGetmhSe OFgFi ; , , ‘Theformulas forSF}and$F)areobtained byadvancing thesubscripts to2and 3respectively on|,m,and n.Now Henen=t, ym, +lam:+hms=0, and soon, Hence itmay beseen that aF\,FS,aF}_aP, ,aP:,ay Fi,aF},OF}_aFi,AF,Fi, 7 ax’oy’tae" ax*ay*az (107-4) This proves theinvariance of(10.7-2). Definition. The scalar function (10.7-2) iscalled thedivergence ofthevector field (10.7-1). Itisdenoted bydivF: ep=OF.OF:,aF, div8=SE4 (107-5) Observe that divFisascalarfieldassociatedwiththevectorfieldF.We shall nottryatthispoint todisplay themethematical orphysical importance of 10.7 ‘THEDIVERGENCE OFAVECTOR FIELD 303 issometimes expressed as wiby pte ed.vids pind Inthisform ¥(which isread as“del") iscalled avector differential operator. Wesaythatthecomponents oftheoperator¥inthisparticular co-ordinate system are co a ay Recalling theformula A+B=A\By+A.Bs+ A\By for the dot product oftwo vectors, we see that appearances would lead usto write pet r+trrip,Veen Sr+2r+or ‘The expression ontheright here isinfact divF,ifthe“product” ofthesymbols LFisunderstood tomeanthederivative 1andsoon.Wethushave@ justification ofthe notation (10.7-6). Particular interest attaches tothedivergence ofthegradient ofascalar field. Ifthescalar field isu=f(P) weseethat . ou, au, audivigeaduw)=T54544TH. (10.7-8) Inthe ¥notation divigrad u)=9>Vu. Itiscustomary towrite V-¥= Vso that oyuu, Puvu4THoH (10.7-9) ‘The left member of(10.7-9) isread as“del-squared ofu,"or“del-squared u.” ‘The equation au, ew, ou ‘ax*ay!at? isoffundamental importance inmany branches ofapplied mathematics. Itis known asLaplace's equation, inhonor oftheresearches ofthefamous French mathematician Pierre Simon deLaplace (1749-1827). Accordingly theexpression ¥'u isoften called theLaplacian ofu. Since thegradient and thedivergence are both invariants with respect to rigid motions oftheaxes, itfollows from (10.7-8) thatV'uisascalar invariant 11/INTRODUCTION The development ofvector mathematics presented inthepreceding chapter took place mostly during thesecond half ofthenineteenth century. One ofthe leading contributors tothedevelopment was the American genius Josiah Willard Gibbs ofYale University. The first part ofChapter 10,which deals with the ways vectors combine with other vectors and with scalars (real numbers), iscalled vector algebra. During the twentieth century, vector algebra expanded vastly into asubject called linear algebra, which has important applications inmuch of mathematics, especially advanced calculus. This chapter will present enough linear algebra toenable ustouse the subject tounify and extend the results obtained inthe last several chapters. Prerequisite tofullunderstanding ofthis chapter and thenext issome knowledge ofsimultaneous linear systems involving nequations innunknowns. Inparti- cular, weshall use the following fact: Anecessary and sufficient condition that such asystem have aunique solution isthat the determinant ofthe coefficient matrix bedifferent from zero. InChapter 12weoccasionally use some ofthe most elementary rules forcomputing determinants. Abig step inthe transition from vector algebra tolinear algebra isthe realization that functions from avector space toavector space are inthem- selves examples ofvectors. The first part ofthis chapter will bedevoted to explaining indetail what this means. We can begin with the once popular question, “What isavector?” The common reply used tobethat avector isa quantity having both magnitude and direction. We shall soon see that neither magnitude nor direction isessential tovectors, and that most things having both (trains, forexample) arenotvectors. Nosatisfactory answer was arrived atuntil itwas realized that one should nottrytodefine avector asanobject having certain qualities, but rather asamember ofafamily ofobjects governed by certain rules. Inparticular, the important considerations are how tocombine a vector with other vectors and with scalars through certain operations. The problem ismuch likethat ofdefining achecker. Wemight naturally begin bysaying that itisawooden orplastic disk, butsuch abeginning does notlead to anything satisfactory. Ifone were asked whether thetopofasoft drink bottle is achecker, one’s first inclination would probably betosay no. Yet many games ofcheckers have been played with these objects. The important fact which emerges from allthisisthatthere isnoproperty intrinsic toathing considered in isolation which can settle thequestion ofwhether itisorisnot achecker. The 309 " INTRODUCTION 311 We pause here torecall the use ofthe setmembership symbol €,already introduced in§2.7. If#isany set, thenotation a©Fmeans “aisamember (or element) of9." Itcan bevariously read as“abelongs to" or“aisin” Other slight variations ofthe verbal rendering ofthe notation are useful. For instance, we may write one ofthe foregoing rules about vectors asfollows: (ab)x =a(bx) foreach a,b ER and each x€Y.Inthis context €may beread as“in” or“belonging to.” ‘The entire mathematical structure made upofV,R,thetwo operations of vector addition and multiplication ofavector byascalar, and the rules governing them isproperly called avector space. Toavoid prolixity, however, this term isfrequently applied just totheVwhen therest ofthestructure is understood. The expressions linear space and linear vector space arefrequently used assynonyms forvector space. Example 1,Let Vdenote thesetR"ofallordered n-tuples ofreal numbers, with addition and scalar multiplication defined asfollows: Ifx=(x1, 3, 5%) and y=(¥i5 You ++5Yq)then XFS FHF Yap et Ie andif¢isanyscalar,x=(cx,€X3,...,€%,).ItcaneasilybeverifiedthatVtogether with these operations constitutes avector space. We discussed "in $10.12. Remember that thezero vector 0is(0,0,...,0). InChapter 12weshall study functions with domain inR"and range inR", where nand mmay ormay not be the same. Example 2.Here, Visthesetofallreal-valued continuous functions defined on(0,1].The reader already knows how toadd two functions and how to multiply one byascalar. Healso knows that these operations, performed on continuous functions, always give continuous functions. We merely wish to point out that this long-familiar structure isavector space, even though wehave said nothing about “magnitude and direction” ofthe vectors. Example 3.Let Vstand for the collection ofall2x2 matrices. We add matrices simply byadding theelements incorresponding positions, and inorder tomultiply amatrix byascalar, wemultiply each element ofthematrix bythat scalar. Even though wehave not associated any magnitude ordirection with these matrices, they are nonetheless vectors. ‘Anexplanation ofwhy R®iscalled n-dimensional was given in§10.12. For thegeneral (abstract) case wecall avector space finite-dimensional ifthere exists afinite set of vectors uj,...,u, such that every vector xcan be represented inexactlyonewayasalinearcombination oftheu's: X= C(tootCathe The uniqueness ofrepresentation means that the coefficients ¢j,...¢, are uniquely determined byx.This implies that thec'sareall0ifx=0. Italso implies that nou,is0,forifsome u,=0, thechoice of¢;cannot beuniquely 13 MATRICES AND LINEAR TRANSFORMATIONS 315 which means that y=(¥1, Yo..++5 Ym) Where eX Fa t+ aeke =Yr aki tanks t+++aea%y =Y2 eek +ae ++ +ane =Yow This canbewritten asAx=y,where theelements ayofthematrix Aaredefinedbyay=ay.Thiscanbeexpressed bysayingthatthemxnmatrix,A,isthetranspose ofthe nxmcoefficient matrix ofthe system (11.3-1). The transpose ofamatrix Misamatrix obtained byinterchanging therows and columns ofM. Wedenote thetranspose ofMbyMT. In§12.8 weshall usethese facts: the transpose oftheproduct oftwo matrices istheproduct ofthetransposes in reverse order; the transpose ofthe sum isthe sum ofthe transposes; and the transpose ofthe transpose ofamatrix isjust theoriginal matrix. The student is asked toverify these properties inExercise 4. Observe that the zero element of£(R",R") isrepresented bythe mxn matrix inwhich all the entries are zero. ‘The matrix Awhich represents Twas constructed with the help ofwhat were called thestandard bases inthedomain and range spaces, R"and R".There are other bases and ifwe had used them we would have obtained other matrix representations ofT.This suggests that Amight more properly bereferred toas thestandard representation ofT.But since these are the only bases which we shall use, weshall sometimes speak only of“the” matrix representation ofa linear transformation. Itispossible tosummarize theresults ofthis section uptothis point bythe following theorem, which isofbasic importance, especially inthe next few sections. THEOREM I.Each mxn matrix isthe standard representation ofaunique linear transformation from R"toR", and conversely, every element of G(R", R") has aunique standard representation asanmXnmatrix. IfT€4(R",R") and L©£(R",R”), then wecan define afunction L°Tfrom R™ toR” asfollows: (LeT)x)= LITO] for allKER". This isacomposite function—the composition ofLand T—and iseasily proved tobelinear (Exercise 2).Therefore, LeT€£(R", R°). Itisimportant tobeable toexpress thematrix which represents LT interms ofthematrices represent- ingLand T.This can bedone inthefollowing straightforward way. Let Abe themn matrix representing Tand letBbethepXm representation ofL. Then, (LeT)(x)=L{Tx]=B(Ax)=By, 16 METRICS 319 whence bx}-by]Sx+yl. (15-1) Ifweexchange xandyhere and bear inmind that y+x=x+y, weobtain Wy xl=x+yl (Ls-2 Since |lx\)~lIyll iseither [x|—lyl| orfly|—lx], depending onthesignofthe difference, we conclude that [ht—hydSx+yl. (ILS-3) Because |-yl= ly,wecan change x+y tox—y ineach ofthelast three inequalities. Actually, the triangle inequality (4)isderivable from (11.5-1). See Exercise 8.On this account we shall also refer to(11,S-1) asatriangle inequality. Likewise for(11.5-2) and (11.5-3). If|lx|=1 wecall xaunit vector. Ifxisany nonzero vector, asuitable multiple ofxwill beaunit vector. The right multiplier isI/|xJ, asweseeby property (3)ofthe norm. Inany vector space with anorm, the unit sphere is defined tobethesetofallunit vectors; therefore theequation oftheunit sphere isIx=1. The norm of§10.12: x-(3x) iscalled the Euclidean norm inR*. We could also define other norms asfollows: bel= Ix, or [hi]=maximum of|x),[xs--+a Showing that these last two definitions satisfy theconditions (1)to(4)isleft for the exercises. 11.6 /METRICS Inpreparation for$11.7, where weshall discuss point sets invector spaces, and continuity offunctions with domains ofdefinition andranges ofvalues invector spaces, weshall now discuss measurements ofdistance inavector space provided with anorm. Indealing with any vector space, itiscommon practice tousethewords point and vector interchangeably. Thus wemay speak either ofthevector xor thepoint x,Inanyvector space provided with anorm wecallJx—yi|thedistance between xandy.Forvectors intheplane theappropriateness ofthisdefinition is shown inFig.83,which shows diagrammatically how theequal lengths ofthe 320 LINEAR TRANSFORMATIONS cht x y a \ “ \ a \ va \ my ° vx Fig. 83. vectors x~y and y~x isthe same asthedistance between theends ofthe vectors xand y. Just asweabstracted the notion ofthe length ofavector bycalling the length ofxitsnorm lj]and listing thefour conditions that thenorm must satisfy, sowemay abstract the idea ofdistance between pairs ofpoints and listfour conditions that we want tobesatisfied byadistance function, Adistance function isafunction that gives usavalue d(x,y)forthedistancefromxtoy. We require thefollowing four axioms tobesatisfied: Dy:d(x,y)Z0forallxandy. Dz:d(x,y)=0ifandonlyifx=y. Dy: d(x,y) =d(y,x) forallxand y. Da: d(x,y)+d(y,2)d(x,2). Axiom Dyiscalled thetriangle inequality because itexpresses thefact that ina configuration ofthreepoints, whichwemaythinkofasforming atriangle, the distance along one leg ofthe triangle isnever greater than the sum ofthe distances along theother two legs. Our definition d(x,y)=xyl]ofdistanceinavectorspacewithanorm satisfies the four conditions D,~ Dy; Dyissatisfied asaconsequence ofthe triangle inequality (4)in$11.5. Adistance function satisfying conditions D,—D,iscalledametric. Wehave seen how touse anorm todefine ametric. The idea ofametric need not be confined tovector spaces, however, forthere are noreferences toaddition of vectors orscalar multiples ofvectors inthe axioms D,~ D,. Infact, there are metrics which do not come from norms, See Exercises 11and 12. 11.7 /OPEN SETS AND CONTINUITY IfEisasetandGisaset,thenthesetconsisting ofallobjects whichbelong toEortoG,ortoboth,iscalledtheunionofEandG,whichwedenote byEUG.The setconsisting ofallmembers ofboth EandGisdenoted byEG andis called theintersection ofEand G.Ifp(x) denotes some proposition involving x, then{x:p(x)}denotes thatsetconsisting ofallxforwhichtheproposition p(x)istrue. Forexample, {x:x-a=3}isthesetofallvectors whose inner product with thevector ais3.Similarly, {x©E:p(x)}isthatsetofvectorsbelongingtotheset Eforwhich the proposition p(x) istrue, 7 OPEN SETS AND CONTINUITY 321 For any vector space having anorm, wedefine the sphere ofradius r centered atthevector a,denoted byS(a, r),asfollows: Sta, r)={x:x—all =r}. ‘The open ball, B(a, r),ofradius r,centered ata,isgiven by Bla, r)={x:|x—al]<r}. The corresponding closed ball, B(a, r),istheunion ofthese two sets; that is, Bla,r)=Bla,1)USa,7)={x:|x—all Sr). Notice that inR’,S(a, r)isaspherical surface andB(a, r)consists ofthesetof points lying inside this surface. B(a,r) iswhat might becalled the solid sphere made upofthesurface, together with thesetofpoints lying inside. How would you describe S(a, r),B(a, r),and B(a, r)inR°?inR?Notice that S(a, r)0Bla, r) istheempty set. IfEisany subset ofanormed vector space, wedefine aninterior point ofE tobeapoint which belongs toEand which isthe center ofsome open ball contained inE.Notice that this isjust amore general statement ofour earlier definition in$5.1. Anopen setcan still bedefined asone consisting entirely of interior points, and aclosed setisstill thecomplement ofanopen set. Byaneighborhood ofapoint, wesimply mean anopen setcontaining the point. This isageneralization ofourearlier usage where thesets which wecalled neighborhoods were open intervals, open disks, oropen rectangles. The really essential property ofwhat wecall aneighborhood ofapoint isthat itisasetfor which thepoint isaninterior point. The definition which weuse isthe simplest way togetthis property. Asubset Eofanormed vector space issaid tobebounded incase there is some number Msuch that |x] <Mforallx€ E. Let {x*}i-: denote asequence inavector space ¥having anorm |[|The ‘superscript isnotanexponent but issimply anindex giving theorder oftheterm inthe sequence. Asone would expect from the theory ofsequences ofreal numbers (see $1.62), has themeaning that if€isany positive number, then there exists apositive integer Nsuch that x—all<e whenever N-Sk. This isalso expressed by saying that thesequence {x*}7., converges toa.Interms ofour recently introduced notation, itcan beexpressed bysaying that if€isany positive number, thenallbutatmostafinitenumber ofthetermsofthesequence liein the e-ball, B(a, ¢),centered ata. The definition ofcontinuity still makes sense inallvector spaces having norms, Suppose that#andYarenormed vector spaces andfisafunction from some subset Dof#toY,Tosaythatfiscontinuous atthepoint aofitsdomain Dmeans that if€isany positive number, then there exists some positive "7 OPEN SETS AND CONTINUITY 323 iLYy A . Fig. 88. radius €>0, centered at,there exists some 6>0such that g(y)€B(c,€)forall yinB(b,5)atwhichgisdefined.SeeFig.85. Since fiscontinuous ata,there issome p>0such that f(x)€B(b,5)forallx€B(a,p)atwhichfisdefined. Therefore, forallthosexinB(a,p)atwhich isdefined, (x)€Bie,€). Observe that thepoints ofB(a, p)atwhich isdefined arethose points xat which fisdefined forwhich f(x) belongs tothedomain ofg. ‘THEOREM III. IfTisalinear transformation from R"toR™, then there exists some number Msuch that [Tx]=Mjx} forall xR". Proof. Letthestandard matrix representation ofTbethemxnmatrix Aas in$11.3. Then for each xER", ay ays am) i) [yn adn1dan\L2 TafAl-{"|. (amyar +++dan!\a)\ Ym ¥=(Ju Yors++s Ym)isthat vector inR™whose ithcomponent y;isgiven by BytheCauchy inequality, (see (10.12~4)) Bay! yy ay inl=(3a3)“(Sa3) =(Sai) he. Now, letQdenote themaximum ofthemnumbers a \"(Sai) =1,2,...,m). Then |y;|=Qlx] foralliand [Tx] =kyl= VytFyeFESsVinQiat =VnQh 326 LINEAR TRANSFORMATIONS chit prove isthat, forallT;and T;in£(R",R"), maxiT,+Tx]STH+TA (118-2) fortheleftsidehereis]T,+TJ.Now,bydefinition, (T,+T.)x=T;x+Tx,andso, forany unit vector x, (T+ Txt =|Tx +Tals [Teal +Tx] SIT+ITS.forcertainly Tx]=|[T\]bythedefinition of|T;),andlikewise forT:.Butthen weseeatonce from theforegoing that (11.8-2) istrue. Wehave now proved that thenorm defined by(11.8-1) does indeed satisfy thetriangle inequality. Animportant property ofthis norm isthat, forallx, (7x1 SITs. (18-3) The proof ofthis fact (Exercise 13)isshort. The reader should pause long enough now todistinguish clearly among the three meanings which the norm ‘symbol has in(11.8-3). The norm which we have defined for linear transformations isuseful even thoughwehavenowayofevaluating itexcept incertain special cases. Itwill be sufficient forourpurposes tohave apractical bound forthenorm, and wehave already obtained one intheprocess ofproving Theorem IIIof$11.7. Itwas found in(11.7-1) that ifT©2(R",R") and Aisanmxnmatrix representing T, then ITs Van, where K=maxJay). Wegetanimportant lower bound forT||byconsidering what Tdoes tothe unit vectors making upthe standard basis inR". Recall that for each r& {1,2,...,m},¢, isthat ordered n-tuple consisting of1inthe rthplace and 0'sin theother n—1places. IfC,isthenorm ofTe,, ITel =[Ald =Kain don... dao=(3a2)=6 C,can bethought ofasthelength ofthevector represented bytherthcolumn in A.Let C=max C,for r€{1,2,...,m}. Inother words, Cisthe length ofthe longest ofthencolumn vectors inA.Since |Tul| =Cforatleast one unit vector, IT&C. From thedefinition ofC,weseethat C,isgreater than orequal to max,|a,|. Hence, C=max, C,isgreater than orequal tomax,(max, |a,|) =K,and so|T|=C=K.Combining thiswithourpreviously obtained upperbound,we have KS|T|s Vmnk. (18-4) Thespace £(R", R)isaninteresting onefrom thepoint ofview ofcomputing norms. Recall that theelements ofthis space arecalled linear functionals and 330 LINEARTRANSFORMATIONS chat invertible. Itiseasily verified that the displayed matrices are indeed mutually inverse. 11.10 /THE SET OFINVERTIBLE OPERATORS Letusdenote by@thesetofinvertible operators belonging tothevector space -£(R"). InChapter 12weshall use theimportant fact that ©isanopen setin £(R"). This means that ifTEM and Lisanoperator such that |T~LIis sufficiently small, then LE. Toprove this we first prove some preliminary results. LEMMA, IfT€£(8") and |||<1, then I~Tisinvertible, ie.(I~ T)EQ, and yet Md-Ty's ay Proof. Consider any x40. We shall show that (I~ T)x4 0,which implies thatI~Tisinvertible. Now (I~ T)x| =x~Tx}=|x—|Tx}], by(11.5-3). But Tx} |T|x|).Therefore Ma T)xh= xia -Tp>0 (1110-1) because ||]<1and{x>0.Therefore(I~T)'exists. ‘Toestimate the norm of(I~ T)"' wecan substitute forxin(11.10-1) the vector (I~ T)'y, where yisanarbitrary vector inR". Onthe left weget N= T)~ TY"yl]= ty]=by Therefore, lyl=|—T)'y\(1—|IT), or \d-Ty'yhs ah forallyER". This gives theinequality for(I~ T)"'] stated inthelemma. COROLLARY. IflI~T\) <1,then Tisinvertible. Wededuce thecorollary asanapplication ofthelemma byputting I—T in place ofTinthelemma, observing thatI~(I~T)=T.Note thatthecorollary states that theentire open ball centered atI,with radius 1,liesin2. ‘THEOREM IV.Thesetofinvertible operators isanopen setin#R"). Infact, if TEM and |T-LI <I/|T 'I,then LEQ. Moreover, \ ry 110-1 Isrrp (11.10-1) Proof. We use the fact that Tisinvertible towrite L=T—(T—L)= T[I~T-'(T~L}}.By(11.9-1)andourhypothesis, (7-7 ~Lys tTWIT-Ly<1. 332 LINEAR TRANSFORMATIONS cna 4.(a)Show that thetranspose ofthesum oftwo matrices isequal tothesum of their transposes (b)Ifthevector xinR™isthought ofasan(mx1)matrix, then itstranspose x”isa (1m) matrix. Show that forany (nxm) matrix Aand any xER", (ay =xtA% Then generalize this toshow that (AB)" =BTAT inallcases where Aand Barematrices such that their product, AB, isdefined.(©)IfAisan(xXm)matrix,xER™andyER,then ("A= y"(Ax), proving that matrix multiplication isassociative inthis very special case (@ Show that for every matrix A, (AT A §.Showthat(11.5-1)andproperty(3)ofanormin$11.5implyproperty(4)ofthat norm bymaking suitable replacements forxand yin(I1-S-D) 6.For each xER", let bb-(3ix") Show that foreach p21, |bisanorm onR*. HINT: Use Minkowski’s inequality (Exercise 32,$6.8). 7.Let[|i and Ja,with thesubscript mstanding for“max,” betwo functions from R10 Rdefined asfollows: bh=5het Fae=maxI. ‘Show that |and faarenorms onR*.For thespecial case where n=2,draw theunit sphere (circle) inthese two norms onthesame co-ordinate axes with theunit sphere inthe Euclidean norm. 8.Prove that iffand garecontinuous, real-valued functions on(a,b), then 1° “if mp yi{ffreoeenar}=(fronax)"(f"ereoas)” ‘What condition isboth necessary and suficient forequality? HINT: Since fDAG)+gGoFdee0forallA, thequadratic equation inA, wfPerdes2a f"peneenars feaac=o cannot have two distinct real roots, and therefore the discriminant cannot bepositive. 1130 ‘THE SET OFINVERTIBLE OPERATORS 333 This inequality, which isreminiscent ofthe Cauchy inequality, isknown asthe ‘Schwarz inequality. Notice that theCauchy inequality can beproved bythemethod used here, starting fromthefactthat55",(Aa,+b,)”=0canbewrittenasaquadratic equation inAwhich obviously cannot have distinct real roots. The similarity between these two. inequalities has ledmany authors tolump them together under thesame name—the ‘Cauchy-Schwarz inequality. ‘Wehave assumed thatf.g,f°,8°,fg,and(Af+g)* areallintegrable, These things are proved inChapter 18. 9.Let©denote thevector space ofcontinuous functions on(0,1](see Example 2, $11) andlet|f,and |;befunctions from €toRdefined asfollows: UhmaxYoo), tb=[foo ax] Prove that||andJ|parenormson€.HINT:For||fpuseExercise8. 10.Prove that if|||isany norm onavector space, Y,then thefunction ddefined by d(x,y)=|x- ylforallxandyin¥ isa metric 1.Prove that ifVisany vector space and disthefunction defined by dix,y)=0 ifx=y, and d(x,y)=1 otherwise, then disametric. Prove thatthere cannot exist anynorm, f,on¥such thatthisparticular metric isgiven byxy forallxand yinY. 12.Letpdenote theusual distance function intheplane, R’,anddefine ete) 1)TEpts. Show that disalso ametric ontheplane. Show that itis notpossible todefine anorm on theplane such that disexpressible interms ofthis norm asinExercise 10. 13.Prove (118-3). 14.Let S(O, 1)denote theEuclidean unit sphere inR",that is $0.1)=(xen:3x2=1) Prove that S(0, 1)isclosed. 1S,Let{x"}Z-1 denote asequence inR*,that is,x*=(xt,x,...,2%). Prove that a necessary and sufficient condition that fimx*=y=(VtYap Fad isthat lima} =y,for=1,2,.-..m. 16.Ifxand yare any two points inR°, then the equation ofthe straight line 334 LINEARTRANSFORMATIONS. cht determined bythem can bewritten rextty-»). For n=2 you may have thought ofthis astheparametric representation oftheline, t being theparameter. Ingeneral, theright-hand side isafunction from RtoR”.Notice that thefunction maps theinterval 011 into that linear segment between xand yinclusive.Thismeansthatwecandefinethelinesegmentdetermined byxandytobethesetofpointsoftheform(I~t)x+tyfort€[0,1].Equivalently, wecandefinethissegmenttobe thesetofpoints Amt Aay, where A120, A220 and Ay+A2= 1.Tosay that asubset Eofavector space isconvex ‘means that ifxand ybelong toE,then theline segment determined byxand yalso lies in E,Prove that every ball—open orclosed—is convex. 17.Suppose that {: A+B; g:B>C; and h:C-+D. Since composition offunctions isabinary operation, weget afunction from AtoDbyforming either h=(g«f) or (hg) +f.Show that these two functions arethesame. Inother words show that although composition offunctions isnotnecessarily commutative, itis always associative. 18,Using Exercise 17show that ifLand Tareinvertible linear operators, then L»T isinvertible, and(L*T)'= T~'s L~',That is,theinverse ofthecomposition oftwo invertible operators isthecomposition oftheir inverses intheopposite order. 19. Show that ifTand Lare two members of/(R") such that TeL=LeT =I, then they must beinvertible and each istheinverse oftheother. 20, Let Aand Bdenote linear operators onR", Prove that ifBisinvertible and A ‘commutes with B,then Acommutes with B~' 21,Show that ifT€4(R") and |T]<1, then A Tyte TAT THE TMT forevery positive integer k. 22,Show that thesetofinvertible operators isnotabounded subset ofZ(R"). 23,Show that thefunction f(T) =T~', defined forT€0(asdefined in$11.10), is ‘notuniformly continuous, That is,show that inchoosing 6sothat [T'~Tol|<6implies |T-'=To'| <¢, where Toand ¢arepreassigned, 6cannot bechosen independently ofTo, 336 DIFFERENTIAL CALCULUS OFFUNCTIONS FROM &*TO cn.t2 where Aisthe mXn coefficient matrix. Itisimportant tothink of(12-2) asa special case of(12-1), and toseethat this special case consists ofallthelinear transformations from R"toR™. This isthevector space which wehave already studied under thename (R", R™). The relevance oflinear algebra todifferential calculus consists inthefact that itispossible toobtain very good local approximations toquite general functions, such as(12-1), byusing linear functions, such as(12-2). We shall see that this enables ustodeduce important information about the behavior ofa nonlinear function near apoint bystudying thelinear (oraffine) functions ofbest approximation atthat point. We begin byextending theidea ofadifferential to our more general setting. 12.1 /THE DIFFERENTIAL AND THE DERIVATIVE Our first objective here istoextend inasuitable way forfunctions from R*to R™ the definition given in$6.4 ofdifferentiability and the differential for functions from R?orR"toR[see (6.4-4) and(6.4-17)]. Our second objective is toextend inasuitable way forfunctions from R”toR™therelationship between differentials and derivatives that exists inthe case ofafunction from RtoR,as setforth in§1.3. Then weshall show that adifferentiable function from RtoR™ iscontinuous, and weshall state and prove thegeneral chain rule fordifferenti- able functions. Inelementary calculus ithas long been customary tointroduce the deriva- tive first, and then the differential. For functions from R"toR", where n>1,it isnatural tobegin with thedifferential and come tothederivative afterward. In fact, the concept ofthe derivative when n> 1ismore sophisticated than the concept ofthe derivative in§1.3; itrequires ustothink ofthederivative asa function from &*toF(R", R™). But when n= m= Ithemore sophisticated point ofview isinfullharmony with theelementary point ofview in$1.3. In§6.4, and later in§7, wediscussed the notion ofthe differential ofareal function ofseveral real variables. The differential ofafunction ffrom R®toRis afunction of(xi,...,%.) and (dx1,..., d%4) whose value is, Beant +Zhdew ay where thepartial derivatives areevaluated at(x\,... .X.): Thus thedifferential is alinear function of(dxi,....dx,) when we Keep (xi...) fixed. But the definition ofthedifferential requires more offthan merely that ithave first partial derivatives with respect toeach ofthevariables x1,...%» Ifweusethe vector notation X=Oke B= (ieee tas thefunction ffrom R*toR,defined inaneighborhood ofx,issaid tobe differentiable atxifthere exist numbers Ay,..., As, depending onfand x,such 342 DIFFERENTIAL CALCULUS OFFUNCTIONS FROM R*TOR™ cn.12 that PoAy=x. (12.2-5) thederivatives being evaluated ata, Conversely, letusstart with theassumption thateach component function is differentiable. This implies that the first partial derivatives allexist and that (12.2-4) issatisfied foreach i,where thenumbers Ayaregiven by(122-5). It follows from this that (12.2-3) issatisfied, because the Euclidean norm ofa vector isnolarger than the sum oftheabsolute values ofitscomponents. (See thefirst inequality in(7-5).] This completes theproof. Aswe know from examples inChapter 7,the mere existence ofthe first partial derivatives ofacomponent function f®atthepoint x=aisnotsufficient tomake {"differentiable ata.However, aswas stated inTheorem IIin§7.1, ifa function from R" toR,defined inaneighborhood ofa,has first partial derivatives, not just ata,but ateach point inthe neighborhood, and ifthese derivatives arecontinuous ata,then thefunction isdifferentiable ata.(Aproof ofthis proposition, forthe special case n=3,isasked forinExercise 3.)ASa consequence, wecan beassured that thefunction finTheorem IIIisdifferenti- able ataifallthe partial derivatives inthe Jacobian matrix (12.2-2) exist throughout aneighborhood ofaand arecontinuous ata.This isauseful way of testing fordifferentiability inpractice. ‘There aretwo particular cases that deserve special mention. One isthecase inwhich n=1;the other isthe case inwhich m=1. When n=1, the vector xbecomes areal variable xand thecomponent functions are real functions ofareal variable. Inthis case the Jacobian matrix (122-2) has just one column. Itselements aretheordinary (not partial) deriva- tives with respect toxofthecomponent functions. Ifthederivatives atx=aof the component functions are Aj,..., Amthederivative f(a) maps the scalar h into the vector (Ayh,...,Ayh) and we can regard f(a) as the vector (An Ande When m= 1wehave ascalar function fofthevector x.Inthis case the Jacobian matrix representing f"a) has one row and ncolumns, and can be regarded asavector (f,(a),.... fa(a)), Where fa)=2evaluated atx=a.ax The differential is df(a,b)=f(ayh=¥fi@dhy (12.2-6) In§10.6 wedefined thegradient ofascalar function defined onanopen set inR?.Itisnatural toextend that definition bydefining thegradient ofascalar function ffrom R"toRasthe vector function grad ffrom R"toR"with 1221 DIRECTIONAL DERIVATIVES AND THEMETHOD OFSTEEPEST DESCENT 343 components given bythepartial derivatives: =(4... aadfox)=(%...,i) 22-7) when fisdifferentiable atx;in(12.2~7) thepartial derivatives areevaluated atx.If weview thegrad f(x) asaone-rowed matrix, weseethat itrepresents f’(x). When £'(a) isapplied toavector xtheresulting scalar canbeviewed asadotproduct: F'G)h=(gradf(x)“hb, (12.2-8) Asamatter ofnotational convenience weshall denote thevalue ofgrad f(x)whenx=aasgradf(a).Inthe next section weshall discuss anapplication ofthe gradient tothe problem offinding apoint where ascalar function attains amaximum or minimum value. 12.21 /DIRECTIONAL DERIVATIVES AND THE METHOD OF STEEPEST DESCENT Just aswesay that anonzero vector determines adirection (the direction ofthe arrow that represents thevector) (inR’), soweshall saythat anonzero vector in R®determines adirection inR",Now consider ascalar function f,defined and continuous onsome open setinR".Because fiscontinuous, thechange inthe value off(x) aswemove inany given direction from aparticular point will be gradual, and will besmall ifthechange indistance issmall. Ifwisaunit vector and aisapoint inthe domain off,wedefine Dafa)=im(2+=f), (12.21-1) provided thelimit exists, asthedirectional derivative offatainthedirection of u,Intaking thelimit in(12.211) itisassumed that 1ispositive and small enough toassure that a+tw isalways inthedomain off.The directional derivative isto beinterpreted astherate ofchange perunit ofdistance, ata,ofthevalue off,in the direction ofu.This isinaccord with the standard interpretation ofthe derivative inelementary calculus, because thedistance between a+tuand ais Ma+tu)—al =fra]=thu}=e Inthecase inwhich fhas afirst partial derivative with respect tox;ata,itis readily seen that this partial derivative isequal tothedirectional derivative ata inthedirection ofthestandard basis vector e,forinthat case theonly difference between a+te;andaisthejthco-ordinates, thejthco-ordinateofa+te, being a)+and that ofabeing a;(where a=(as,...» d)} ‘There isanimportant relationship between directional derivatives andthe gradient. ForR?thiswas mentioned in§10.6. The situation inR”isstated inthe next theorem. 346 DIFFERENTIAL CALCULUS OFFUNCTIONS FROM &”TOR™ ch.12 isaminimum. This obviously occurs when ure gy «sey (1221-5) Toseethatthismethod actually canbecarried out,wegoback toExercise 13of§6.3. Using aprogrammable pocket calculator, aprogram offewer than 200 stéps canbewritten togenerate thesequence (12.214) using thea,'sdetermined by(12.21-5). The program can also have the machine print outf(x», y.)and ligradf(x,ya)|ateachstage.Ifwestartwiththeinitialguess(xe,ye)=(1,1),the following results are obtained. " % yWeradfxsya)flaYn) 0 1 1 10.81665, -3 11.6724 1.4483 5.8338 =28.571221,6899 1.2582 2.64037 -29.557431978 1.10243 1.74862 29.916841.95826 1.01917 0.39027 =29.9912Ss1.99880 1.00946 0.16869 -29.999361.99673 1.00153 0.03104 -29.999971.99992 1.00069 9.01233 —29,9999960281.99977 1.00011 0.00223 -29.9999997291.99999 1.00005 0.00088 -29.99999998101.99998 1.00001 0.00016 -%011 1.999999597 1.000003484 0,0000624022 -30 12 1,999998817 1.000000555 0.0000112465 —30 13 1.999999971 1.000000248 0,000004434 —30. 14 1.999999916 —1.000000039 0.000007991 —30. 15 1.999999998 1.000000018 0.000000315 —30 16 —1.999999994 1,0000000030,000000567 —30 17 2.000000 1.000000001 0.000023 —30 The minimum value of~30 isfound to10digits in10steps, but togetthe critical point with equal accuracy takes 18steps. The convergence here istediously slow. Afaster method will begiven in $12.3. However, with the method ofsteepest descent one can getconvergence even ifthe initial guess isnot very good, whereas the faster method may not converge atallifone starts from abad initial guess. This possibility will be illustrated in§12.3. Inconclusion, wemake theobvious comment that thesame ideas lead easily toamethod ofsteepest ascent which isuseful incase one istrying tofind a maximum rather than aminimum (Exercise 25). Finally, weshould make itclear that formula (12.21-5) isjust one ofvarious methods forchoosing thea's. There 123 NEWTON'S METHOD 349 vector _(2,aF)_ sradFxy)=(TF) =0.0). (12.3.2) Bydirect calculation wefind grad F(x, y)=37+ y?—5, y?+xy 5), and sotheequation (12.3-2) isequivalent tothetwo simultaneous equations xty-S=0, dxyty?-S=0. (12.3.3) These equations happen tobeeasy tosolve byalgebra. The only critical point in theinterior ofthe first quadrant is(2,1).Later, inExercise 27,thestudent can tryoutthemethod ofNewton onthis problem, and observe how itcan lead from afirst guess atthe solution toahighly accurate approximation tothe exact solution. For the general case offfromR"toR",Newton's method proceeds ina manner that isentirely analogous tothespecial case n=1.Astart ismade with a guessed approximation xotothe solution off(x)=0.Then weuse f'(xe(x —xo)as anapproximation tof(x)~ f(x), asiswarranted bythedefinition ofthederiva- tive. Then wesetf(x) =0intheapproximate formula (x) ~fox) =FOu)(~ xd) and solve forx,denoting thesolution byx,.Toachieve thesolution, weassume that thelinear operator f'(xq) hasaninverse, [f(x9)]"', sothat theequation 0~ fx)=£(x0NX1~0) (123-4) leads tothe formula 1=Xo~[F(X] "fC. (12.3-5) ‘Weemphasize that in(12.3-4) theoperator f'(x9) isacting onthevector x,~x9 and that in(12.3-S) theoperator [f'(x9)J' isacting onthevector f(x). Wethen proceed byarepetition ofthe process, obtaining asequence ofvectors x), Xa). 2.)Xm ++» Where Xavi =Xe [F%)T HO). (123-6) Itisassumed that f'(x,) has aninverse foreach x,.Sufficient conditions forthe convergence ofthis vector form ofNewton's method are given inadvanced texts onnumerical analysis. ‘We conclude this section with anapplication ofthis powerful method toa two-dimensional problem, carried out onaprogrammable pocket calculator. Example. Wewish tosolve thesystem —134+x=2y +Sy?y'=0, -M+x-Myty+y=0. 350 DIFFERENTIAL CALCULUS OFFUNCTIONS FROM 2"TO2" cn.12 Notice thatbytheelementary method ofeliminating xandfactoring the resulting cubiciny,wefindimmediately thesolution (5,4),andwealsoseethatthisistheonlysolution which involves onlyrealnumbers. Thinking ofthe system asf(x,y)=0, weget _(1-24 Wy-3y?tean=() Tiaay aay?) The determinant ofthismatrix is6y?~8y~ 12,sofisanonsingular linear ‘operator except onthelines y=(2+ V2), thatis,approximately, y=2.23and y=~0.897, The inverse isgiven by(see $11.9) =1442y43y? 2-10+3y? 1_[6y=8y-12 6y?—8y 12 (tes,»)T(iaeaie3 6y7=By— 126y7—By— 12 soNewton's method, expressed by(12.3-6), gives the sequence ofvectors generated bythefollowing formula Ket) (Xm) _ayy1(~13+x_—2yn+Syk—ye (a)=()teewor(a54eaeeggs) Aprogram togenerate this sequence with apocket calculator can bewritten with fewer than 200 steps. Starting with the initial guess (xo,yo)=(10,8)the following results aregenerated. " Xe Ye o 8 1 ~21.80 3.701 2~22100 4.6503339071 4.1187 4498746 4.00507 5 4.999918991 4.000087 6 5.000000000 4.000000, This shows that convergence from thestarting point (10,8)isquite rapid, giving accuracy to10significant digits injust sixsteps. Such success cannot beassured for allstarting points however. Notice the very different behavior ofthe ‘sequence generated when westart with (xe,ye)=(15, —2). 12.4 /AFORM OF THE LAW OF THE MEAN FOR VECTOR FUNCTIONS Inthis section weobtain (Theorem V)ageneralization ofthelawofthemean (as presented in§1.2 and §7.4), and anapplication ofitintheform ofaninequality (Theorem V1); both areapplicable todifferentiable functions from R"toR™.We 352 DIFFERENTIAL CALCULUS OFFUNCTIONS FROM R*TOR™ on.12 Since¢isjustadifferentiable function fromRtoR,theordinary lawofthe‘mean tells usthat there issome number @that 0<@ <Iand (1)~60)=oO)1-0)=66). Thus wesee that [8(9)—f(u)]w={fu+6—w))(v—w)} -w. Letting £=w+0(v~u) weobtain (12.4-1), thus concluding theproof. THEOREM VI. Under theassumptions ofTHEOREM V,there issome point § ontheline segment connecting wand vsuch that Itc)~tap]sYew—wy), (12.42) where ||||ofcourse denotes theusual Euclidean norm. Proof. If{(u) =f(¥), (12.4-2) isobviously true, because 0=0and thenorm ofeveryvectorisnonnegative. Iff(u)#f(v),westartbytakingtheabsolutevalue ofeach side in(12.4-1) and applying Cauchy's inequality (see (10.12-5)) to theright-hand side. {E4¢v) ~(a)] -w)=[LPCEDLy~w)]-w]SJOEY~wlfw] (12.4-3) ‘Then we choose wasfollows: =fo)=f)©it)=tw) Aswecan see, wi]=1and ~t¢u}|=MODHOOP=egy— {Lt(y)~uy)-we}it)tfiv)~f(u)). Putting these results in(12.4-3) weobtain (12.42). 12.41 /THE HESSIAN AND EXTREME VALUES In(6.9-10) wehadoccasion toconsider aquadratic form inthree variables. Inn variables, stich afunction could beexpressed as Qhy,- 5Bq)=auih]+aiahyhy+++yhyhy Fanhshy+ aah} +--+ daeltaha i (241-1) $gig +=25=gah We can just aswell assume that ay=as,andthisisordinarilydone.Taking advantage ofmatrix notation and therules ofmatrix multiplication, wecan more conveniently represent thequadratic form by QM, yha)=BAB (12.41-2) 126 ‘THEFUNDAMENTAL INVERSION THEOREM 359 So,suppose [x1~all <r.The contradiction towhich weshall come isthat d<d, and we shall arrive atthis byanadroit use ofthe defining property ofthe derivative offatxs,Because x1€ U,T= (x) isinvertible; there is,therefore, a unique vector hsuch that Tih =b~f(x). Evidently h0,because b#fix). We know nothing else about thesize ofji,butwecan make Jthi= [Jk small by‘making#small.Wewishtomakesurethat[xs+th—al<r. Wecandothisasfollows: Choose tsothat 0<t and tf|<r~[x1~al. (Here wemake use ofthe assumption that x,~al <r.)Then xa+thalls[es~all+ehh<r. Because fisdifferentiable atx,itfollows (see (12.1-4)) that jimi th)—feu)—Thy_9 =ehh) 7 Letustherefore impose on1theadditional requirement that itbesmall enough to make (ou+th)=fu)—TAth}tii ni ‘Then teas+th)x)=Tueny)<4 (26-7) ‘The reason forthechoice ofthemagnitude d/2fh intheinequality before (12.6-7) will appear presently. We place one more restriction ont,namely, that <1. ‘These three restrictions arecompatible and achievable. From the definition ofdwe know that <M +th) bh But b= f(x)+Tih,andsoweseethat 45Mas +th)~ta)~Tinh. From thetriangle inequality wesecthat 4.5 Mla +th) ~H0xs)~Tyfas+#h)~Hoan)THEM+|THCEN)~TH}.02.6-8) But{T.th) ~Ty) =[(t— TIN] =(1~ Hi~fwd =(=Nd. Therefore, from (12.6-7) and (12.6-8) weseethat td walt!a<fsa—nd=d(t-3)<4. Thisisthecontradiction wehavebeenseeking,sowemustconcludethatxs~all=r- Wenow return totheconsideration ofll~(x). Wearetrying toshow that itis bounded away from 0.Now tb~f¢x0)= Bo.)+ex)exfa)~fo=d~HD.Butfbf(x)=d=~fla},andMls)—H6xa)=m~4(rI)—al, asweseebyapplying (12.6-5) with x»andxyinplace ofwandv,respectively 27 ‘THEIMPLICIT FUNCTION THEOREM 363 Wehave toremember ofcourse thatxR?andy€R°, andthatftakes itsvalues inR? IfAand Baresets, then thesetofallordered pairs oftheform (a,b), where aA andb€B iscalled theCartesian product ofAand Bandis denoted byAxB.Orderedpairs(x,y)introduced intheprecedingparagraph belong toR?xR’;thus wesaythatfisafunction from some subset ofR?XR?to R?.Since thepoints ofR?R?areordered pairs oftheform (x,y), where xisanorderedpairofrealnumbers andyisanorderedtripleofrealnumbers, (x,y)canbeidentified with anordered quintuple ofreal numbers. Hence, itiscommon to identify R?xR?with R°. Inorder togetclearly inmind what ismeant bysolving (12.7-4) foryasa function ofx,consider thefollowing very simple special cases. f(x,y) =3x+2y-S=0 This can besolved ataglance toget ¥=6(x) =M5 —3x). The essential feature ofthis function 4(x) isthat ifwesubstitute itforyinthe equation f(x, y)=0,wegetanidentity, namely L(G G(x) =3x+2h(X)~S=3x+(S—3x)-S =O. Inthis special instance, theidentity holds forallx.Inmore complicated cases, ‘wemay have tosettle foridentities which arevalid only over some subset ofthe vector space towhich xbelongs. And inthemore general case oftheequation (x,y)=0,tosolveforyintermsofxmeanstofindafunction(x)suchthat Ix,6(0)] =0,atleast forallxbelonging tosome setinR?. Consider thesetofallpairs (x,y)such that f(x,y)=0. We call this the solution setforthegiven equation. Toavoid dealing with asituation which isof nointerest, wemust assume that theequation does have solutions, that is,that thesolution setisnot empty. We are interested inknowing whether this sethas theproperty thatwhen(x;,y:)and(x:,y:)bothbelongandx)~x:,thenneces-sarily y;= yz.Ifitdoes have this property, then yisdetermined asafunction of x. ‘There areinstances inwhich thesolution setdoes notdetermine yuniquely asafunction ofx.Forexample, totake acase inwhich xand yareboth inR, suppose f(x,y) =x?+y*-1.Then(0,1)and(0,-1)arebothinthesolutionset, sothat there are two values ofy(instead ofonly one) corresponding tox=0. But ifwestart with thepair (0,1) and confine attention topairs (x,y)ofthe solution set for which xisclose to0and yisclose to1,wefind that this restricted portion ofthesolution setdoes define yuniquely asafunction ofx, theformula being y=V1— x"(positive square root). This restriction ofattention toallpoints (x,y) ofthesolution setclose toaparticular point (a,b) ofthe solution setisastandard feature ofimplicit function theorems. ‘Now fortheimplicit function theorem. Weshall state and prove itforthe case wehave been discussing ofthree equations with five variables, but only 127 ‘THEIMPLICIT FUNCTION THEOREM 365 where allthepartial derivatives areevaluated at(a,b). Wewish toshow thatthe linear transformation represented bythismatrix isnonsingular. Notice thatthe3x3submatrix inthelower right-hand corner isnonsingular, because itrepresents thederivative off(a,y) asafunction ofyaty=b;inthe upper left-hand corner wehave the2x2 identity matrix having determinant 1. Sothedeterminant ofthe 5x5matrix has thesame value asthat ofthe determinant ofthe3X3 submatrix inthelower right-hand corner. Inother words, thederivative ofFat(a,b) isnonsingular. Weobserve that Fmaps (a,b)into thepoint (a,0)ofR?xR°.Byapplyingthe inverse function theorem (Theorem VIII) toF,wesee that there must exist a neighborhood %of(a,b)inR°such that %iscontained inWand such that F defines aone-to-one mapping ofWonto aneighborhood Vof(a,0) inR’. Moreover, theinverse mapping F"'isofclass Con¥.Itiseasy toprove (see Exercise 21)that every neighborhood of(a,b)contains aneighborhood ofthis point which isaCartesian product DxE,where Disaneighborhood ofainR? and Eisaneighborhood ofbinR’;wecan assume that %itself issuch a Cartesian product, and shall dothis forconvenience. Letuswritepointsof¥intheform(x,z),wherexisinR?andzbelongsto R®.Toeach (x,2) in¥,there corresponds aunique point F(x,2)~(x,y) in% such that f(x, y)=zand F(x,y)=(x,2).Thus,yisdetermined uniquelybyxand 2,andthisdefinesyasafunction ofxand2,sayy=g(x,2).Inparticular,g(a,0)=b.ThenF-'(x,2)=(x,a(x,2)).SinceFisofclassC",itisreadilyseen that gisalso ofclass C”, Let$bethesetofx'ssuchthat(x,0)isinV.Clearly,$iscontained inD,since (x,0)comes from some point inU=DxEbythemapping F.Observe that {a,0) comes from (a,b). Itiseasy toprove (see Exercise 20)that Sisanopen subset ofR*.Letusdefine afunction @onSbytheformula (x)=g(x,0) Observe that (x) isinR?andthat (x,(x)) isin%when xisinS.Inparticular, (a)=band(x)isinE.ObservealsothatF(x,0)=(x,(x).Thismeansthat (x,0)=F(x, (3) =(x,ffx,6(0), and hence, that f(x,(x)) =0.The function 6is ofclass C"”onS,because $(x) =g(x,0)andweknowthatgisofclassC', Finally, toprove thelastassertion inTheorem IX,assume that yisapoint of Esuch that f(x,y)=0forsomexinS.ThenF(x,y)=(x,0) isinV,andhence, since themapping of%onto ¥isone-to-one, weareassured thaty=g(x,0)= 4(x). This completes theproof. Inconclusion, we shall now state amore general form ofthe implicit functiontheorem. Exceptforminorchangesinnotation, theproofisthesameasthat forthespecial case just treated. THEOREM X(THEIMPLICIT FUNCTION THEOREM). Suppose thatWisanopen subset ofR°** (which weshall identify with R?xR*)andletfbea continuously differentiable function from W’toR*.Assume further that there isapoint (a,b) inWsuch that f(a,b)= 0,and such that thederivative of f(a,y)asafunction ofyisnonsingular aty=b.Then theequation f(x,y)=0 12.8 DIFFERENTIATION OFSCALAR PRODUCTS OFVECTOR VALUED FUNCTIONS 367 properties ofthescalar product inExercise 8of$10.12 onecanexpress thefirst dot product ontheright in(12.8-4) asasum ofnine terms, one ofwhich is (x)-g(x).Onlookingcarefully attheothereightterms,wecanseethat (x4 b)— B(x) =£00)-gGDH-+Fh-BO) +[hl{asumoffiveterms} +£GOh =gh. ‘The sum ofthefirst two terms here ontheright oftheequality sign isevidently linear inh,and therefore ofthe type desired for thedifferential of.What is needed, then, istoshow that theremaining expressions ontheright, taken together, have anabsolute value less than orequal tojfiltimes some function of hhthat approaches zero ashdoes. This isnotvery difficult, andweleave ittothe student tocarry out thesteps. (Exercise 28). For purposes ofapplications, itisadvantageous totranslate (12.8-1) from the language ofdifferentials tothe language ofderivatives. This reformulation can bebroken down into several small steps. Since dot multiplication ofvectors iscommutative, wecan write (x) “b=f(x) g'G)h +BOX) “00h, and expressing dotproducts interms ofmatrix multiplication (§11.4) leads to (x) b= [£60)]" [eh] +[0x] [Gh]. Using thefact that matrix multiplication isassociative, (x) ={{160)" 00} +(Ia) F))h =[£60" g(x)+a0)"F@)Ih. InExercise 4,Chapter 11,itisindicated that thetranspose ofthe sum oftwo matrices isthesum oftheir transposes, andthat thetranspose ofaproduct isthe product ofthetransposes inthereverse order. Using these two facts weget 1)b=[e')"£00+£6)"g00)"by andfrom (11.4-2) thiscanbewritten asthedotproduct ofthevector inbrackets with h,that is, deb(x,b)=6'(x)-h=[p'(3)"0x)+£60)"g63)]-h. ‘Comparing this with (12.2-8) weseethat theonly way this can hold forallhisforthevectorinbrackets tobethegradientof¢atx.Therefore 6°09) =grad (x)= B60)" tO)+£60" 80, and wehave arrived atthefollowing reformulation ofTHEOREM XI. ‘THEOREM XI’. Under thehypotheses ofTHEOREM XIwhere (x) =£00) -(3).¢isdifferentiable atxand813)=g()1) +£00)"80). (12.8-5) 132 DEFINITION OFADOUBLE INTEGRAL 379 Ifwecompare thelimits in(13.1-3) and(13.1-7), weseethat they have the same form. Infact, thelimit in(13.1-3) isthespecial case ofthat in(13.1-7) for which f(x,y)=xo(which happens tobeindependent ofy).Limits ofsums having thegeneral form (13.1-6) occur inavariety ofcontexts, with widely different interpretations. The mathematical properties common toallsuch limits furnish uswith astarting point forthegeneral theory ofdouble integrals. 13.2 /DEFINITION OF ADOUBLE INTEGRAL Let Rbeaclosed, bounded region inthexy-plane, and letf(x, y)beafunction defined inR.Inavery general theory ofintegration, wemight seek toplace no more restrictions onthe function fand the region Rthan are absolutely necessary for the development ofthe theory. Inthe interests ofsimplicity, however, weshall make rather severe limitations onR,and we shall assume at theoutset that the function fiscontinuous inR.Later itwill bepossible (and desirable) tobroaden the treatment sothat certain kinds ofdiscontinuities off are permitted. The term “region” was defined in§5.1. We are now concerned with closed, bounded regions. IfRissuch aregion, ithasaninterior and aboundary. Since R isclosed, theboundary ispart oftheregion. The limitations weplace onRarein the nature ofassumptions about the character ofthe boundary. We have in mind, roughly speaking, that theboundary ofRshall consist ofafinite number orarcs ofsmooth curves joined together toform aclosed curve, orpossibly several (but afinite number of)such curves. Asmooth curve isdefined tobea curve with acontinuously turning tangent. Circles, parabolas, and straight lines are among the simplest kinds ofsmooth curves. Itismore difficult than one might suppose tobeprecise indescribing theboundary ofaregion; weshall not, attempt toexpress our assumptions more exactly than inthe above statement. , Hereafter inthis chapter, inspeaking ofaregion Rinconnection with adouble integral, theforegoing assumptions will betaken forgranted without explicit mention. Indefining adouble integral, westart from approximating sums having the appearance of (13.1-6), butthesubregions AA, arechosen in EH eee aprescribed manner, andarenotarbitrary in maya anu shape. Lettwosetsoflinesbedrawn, oneset ay PE)!parallel tothex-axis, theother setparallel to VEREthey-axis(seeFig.92),Thespacingofthelines K¢: eeanneednotberegular, butthespacing should be SEER close enough sothattherectangles formed bythe 5] * intersections ofthe two sets oflines are small in comparison with R.The network thus formed in Fig.92. thexy-plane iscalled arectangular partition; one oftherectangles ofthenetwork iscalled acell. Some ofthecells will belong entirely toR;others will contain points which do 382 DOUBLE AND TRIPLE INTEGRALS ch.13 The subregions Rj,Riareofcourse subject tothesame assumptions asRasfar astheir boundaries are concerned. 13.22 /INEQUALITIES. THE MEAN-VALUE THEOREM Its atonce apparent from thedefinition ofthedouble integral that [ftanarzo itfny20ink (13.2-1) © Hence, iff(x, y)®a(x,9)inR,wehave Jftenaaz ffecnaa. « « Now —LFG, ISFO, 9)Sf WI, and therefore ~ffiscn sans ffsears[fies aa. * = " This result can bewritten |[frenaa]s fffla,99)dA. (13,22-2) . fi Let Abethe area ofR.Then, taking f(x, y)= 1,weseethat JJte.da~timS,DASA Hence, forany constant c, ffcdA=cA. (13.22-3) Suppose now (returning tothecase ofanarbitrary continuous f)that m,M are numbers such that, inRy ms f(x,y) SM. Then mA~ffmarsffso.dasffmaa~Ma,* c = 384 DOUBLEANDTRIPLEINTEGRALS cn.13 This theorem appears tobeintuitively evident from thegeometrical inter- pretation ofthedouble integral asavolume, asexplained in$13.1, Apurely analytical proof may begiven. Inthis proof the property (13.21-3) plays an important role. We forego thedetails. ‘Among other things, this theorem has theconsequence that weare able to define thedouble integral ofacontinuous scalar point function over aregion R; theintegral isindependent ofco-ordinate systems, and istherefore ascalar invariant. Before reading thefollowing brief remarks onthis subject, thestudent will dowell toread thefirst part of$10.5. Let Rbeaplane region ofthetype assumed in$13.2, and letf(P) bea continuous scalar point function defined inR.With anarbitrary choice of rectangular co-ordinates intheplane, lettherepresentation off(P) be f(P) =Foxy), Phaving co-ordinates (x,y).Consider theintegral ffronaa. (03.23-1) asdefined earlier inthis chapter. Ifsome other rectangular co-ordinate system is setupinthe plane, denote the new co-ordinates ofPby(x’,y’), and the new representation off(P) by(x’, y).Then theintegral Jfoc. yaa (1323-2) has thesame value as(13.23-1); for, the approximating sums converging tothe integral (13.23-2), formed for arectangular partition ofthe x'y'-coordinate system, will also converge tothe integral (13.23-1), byvirtue ofTheorem II, since F(x, y)=@(x', y’)when (x,y)and (x’,y’)refer tothesame point. Itfollows that ifwedefine thedouble integral off(P) over Rby Jfrmaa- ffreayaa, (13.23-3) * x then theintegral isascalar invariant. 13.3 /ITERATED INTEGRALS. CENTROIDS We shall now learn how tocalculate thevalue ofadouble integral byperforming. twosuccessive single integrations. Ourinitial explanation ofthismethod rests on thegeometric interpretation ofthedouble integral asavolume, asinthe discussion which culminates informula (13.1-8). 134 USE OFPOLAR CO-ORDINATES 393 Inthese iterated integrals a,Baretheextreme values of@,and a,barethe extreme values ofr,inthe region R.The inner limits Ry,Rz,@1,2areread off from theappropriate one ofthetwo figures, asshown (Fig. 105a orFig. 105b). The use ofpolar co-ordinates may prove advantageous either bysim- plification oftheintegrand, orbysimplification ofthelimits ofintegration in dealing with theiterated integrals. Experience anddiscernment arerequired to beable tojudge whether ornottousepolar co-ordinates. The student’s first task istopractice theuse ofpolar co-ordinates. Example 1.Locate thecentroid oftheplane y region Rshown inFig. 106(above thex-axis and between thecircles ofradii a,b). ‘The centroid isobviously onthe y-axis, so £=0. The area AofRis(m/2\(b*—a*). Hence 2Zim?ady= o«6F(b?—ayiJyda. Nig.106. The boundaries ofRhave very simple equations inpolar co-ordinates. Therefore, weevaluate thedouble integral byaniterated integral inpolar co-ordinates. Here f(x, y= =rsin 0=F(r,8). Also, «=0,B==,R= a,R:= b.Therefore, supplying theextra factor rinthe integrand, wehave oe ffyaa=[dofrsinoar =PSH[sinode=100°, ‘Then _4bina’ 4b+bata’J"3nbi@ 3m b+a ; 4 Forasemicircular region weputa=0.Inthiscase §=30b. Example 2.Findthevolume inside thecylinderx?+(y~a)'= a?andbe- tween theplane z=0andtheparaboloid 4az=x*+y*. The volume inquestion isgiven by -ffLary)v-[fae +y)dA, where Ristheregion inthexy-plane bounded bythecircle x°+(y~a)*= a". Half ofthevolume isshown inFig. 107.Polar co-ordinates areconvenient for 135 APPLICATIONS OFDOUBLE INTEGRALS 395 2.Locate thecentroids ofthe plane regions described asfollows, using double integrals and polar co-ordinates: (a)Inthefirstquadrant, between x’+y?=2axandy=0. (b)Between r=2acos@and rcos @=a,and ontheside ofthelatter curve away from theorigin. (©) Inside the cardioid r=a(1 +sin6). (@)Inside thefirst quadrant loop ofr=asin20. (©)Inthefirst quadrant, inside r=2acos6andoutside r=a, (1)Inside theloop ofr?=2acos2which isbisected bytheray@=0. 3.Find each ofthetwo volumes into which thevolume inExercise 1(e)isdivided by thecylinder x7+y?=a7. 4.Findthevolume insidethesphere x’+(y~a)’+2?=a’andbetween theplanes xeOyeax 5.Find thevolume between theparaboloid z=x?+y?andtheplane z=x. 13.5 /APPLICATIONS OF DOUBLE INTEGRALS In$13.1 weintroduced theconcept ofathin sheet ofmaterial substance, The concept ofadistribution ofmatter without thickness isavery useful one. A plane region which carries such amass distribution iscalled alamina. Alamina isamathematical idealization ofathin sheet, just asaparticle isamathematical idealization ofasmall, concentrated bitofmatter. One may also speak of laminas which are curved surfaces, but here we shall deal only with plane laminas. Wewish tointroduce theconcept ofalamina ofvariable density. Inthecase ofconstant density, thedensity oflamina istheratio ofmass toarea: M o=™, 5A (13.5-1) But wemay imagine alamina inwhich the mass issodistributed that various pieces ofthelamina, although ofequal area, will have different masses. For the general case, the density ofalamina is,bydefinition, anintegrable function a(x, y)such that when itisintegrated over any subregion AR ofthe lamina, it gives themass ofthat portion: am=ffoaa (135-2) Inparticular, thetotal mass is M=ffoaa. (135-3) Weshall consider only thecase ofcontinuous densities. Ifthearea ofARis 135 APPLICATIONS OFDOUBLEINTEGRALS 399 important instatistics andelsewhere. The integrals, JiixdA,f{yd® ® occurring intheformulas (13.3-8) forthecentroid are, bycontrast, called first ‘moments (about they-axis and x-axis, respectively). EXERCISES 1.Ineach oftheparts ofthis exercise alamina ofacertain shape isdescribed, and themanner inwhich itsdensity varies isdefined. Find themass and locate thecenter of mass ofeach lamina, Wherever itoccurs inthis exercise, kdenotes aconstant of proportionality. (a)Triangular lamina with vertices at(0,0),(4,0), (a,b);0=kx. (b) The same lamina asin(a), but with =ky. (©)The lamina occupying theregion defined byx?+y?sia2,x20,y20,with=kx. (@)The lamina of(@),but with o=kxy. (e)Thelamina of(¢),butwith o=k(x? +y?)", (0The lamina inthefirst quadrant, bounded bybx*= a*y, x=0,y=b,with =kx. (g)The lamina of(0),butwith o=k(b—y). (h)The triangular lamina cutfrom thefirst quadrant bytheline x+y=a,with odirectly proportional totheproduct ofthedistances from (x,y)tothesides ofthetriangle. (@The lamina inthefirst quadrant, bounded byr=2acos @and @=0,with «=kr. (DThe lamina of(0),but with o=krsin28. 2.For any distribution ofmass, letI,and I,denote the moments ofinertia ofthe distribution about thex-axis and they-axis, respectively, and letJodenote themoment of inertia about theaxis perpendicular tothexy-plane attheorigin. Show that J>=1, +I, 3.Ineach part ofthis exercise, ahomogeneous lamina isdescribed. Find I,I,,and Joineach case (see Exercise 2). (a)Thecircular lamina bounded byx°+y=a* (b)The annulus bounded bythetwo circles x°+ y?=F? (=1,2,r1<1d. (©)The rectangular lamina bounded byx==a, y=+b. (@)The triangular lamina bounded byy=0,x=a,ay=bx. (©)Theelliptical lamina bounded byb?x*+ ay? =a°b?, (The lamina bounded byy?=2axand x=2a, (2)Thelamina occupying thecircular segment x7+yb?, x=bcosa where 0<a< a2. (8)The lamina occupying thecircular sector 0:5rb, ~P=6:5p,where 0<p =x/2. 4.Foralamina occupying aregion Rinthexy-plane, thedouble integral vs[fonan iscalled theproduct ofinertia ofthelamina with respect totheco-ordinate axes. 4381 POTENTIALS ANDFORCEFIELDS 401 4.Findtheprincipal axes ofinertia forthefollowing laminas: (a)The laminaofExercise4(a),ifa=2,6=1. (b) The lamina ofExercise 4(b). (©)The lamina ofExercise 4d) (@) The lamina ofExercise 4). (©) The lamina ofExercise 4(D. 9.Alamina intheshape ofthecircle x°+y'5a?hasdensity =(x+y). Find its principal axes ofinertia relative toitscenter, andthemoments ofinertia about these axes. 13.51 /POTENTIALS AND FORCE FIELDS Inthetheory ofelectrostatics, theconcepts ofcharge andcharge density are entirely analogous totheconcepts ofmass andmass density, with thisexcep- tion: Charges may beeither positive ornegative, while wehabitually think of masses aspositive. Aparticle ofelectric charge eexerts anelectrostatic force on another particle ofcharge e’according totheinverse-square lawofCoulomb: The magnitude oftheforce isinversely proportional tothesquare ofthe distance between thecharges, anddirectly proportional totheproduct ofthe charges. The force isdirected along thelinejoining thecharges, andlikecharges repel each other, while unlike charges attract. With proper choice ofunits (electrostatic units) theconstant ofproportionality may betaken asunity. ‘The vector form ofCoulomb's law isasfollows: Let ebeatP,e’atP’, and r bethedistance PP’. Then theforce exerted byeoneis, F=S PP’. (13.51-1) This should becompared with theanalogous formula forgravitational attraction between two particles (see (10.51-4)). Next we consider how to deal with the notion of electrostatic force produced byacontinuous distribution ofcharge onaplane lamina, Consider a particle ofunit positive charge atafixed point Q,anywhere inspace, butnoton the lamina. Let @bethe charge density onthe lamina, which we assume occupies aregion Rinthexy-plane. Intheusual manner, wesubdivide Rand consider the force exerted onQbythe system ofpoint charges which is obtained when weconcentrate thecharge Aeofeach part AR ofthelamina ata point Pwithin thepart. The contribution ofthis part tothetotal isaforce ar=4570, where risthedistance PQ (see Fig. 111). Allsuch vectors must beadded, and then wemust carry outthelimiting process. Since Aeisapproximately oAA, the total force exerted bythelamina is F~{[SPQua. (13.51-2) 13.51 POTENTIALS ANDFORCEFIELDS 403 Asystematic study ofthetheory ofelectrostatic fields isgreatly simplified byintroducing theconcept ofthepotential ofthefield. Thepotential atapoint Q,produced byacharge¢atthepointP,isdefined tobe £,wherer=PQ. Forthepotential ofseveral particles, theprinciple ofsuperposition isused, and forcontinuous distributions ofcharge, thestandard integral calculus procedure isemployed. For alamina onthe region R,with charge density oatP,the potential atQisdefined tobe =||po" (13.51-4) 4(Q)iI% The potential isascalar point function. The electrostatic field isavector point function. The relation between the two functions isshown inthe fact that the gradient ofthepotential gives thenegative ofthefield vector: Vou(Q) =F. (13.51-5) The Qonthegradient symbol istoremind one that wemust differentiate with respect totheco-ordinates ofQ.IfPis(x,y,2),and Qis(En, £),wehave P=(PQ)=(=x+(9y+C=2¥, (13.51-6) and war[f2axay, (3.51-7)* F=[f2S2E-oie(—vi+G-Dkldedy. (1351-8) Formula (13.51-5) isthen equivalent to. Fein$=ff2MEnacdy, (13.51-9) and two similar formulas fortheother components ofF. Itisonly invery special instances that thepotential can becomputed in elementary form byintegration. Usually the work leads toelliptic orother nonelementary integrals. Nevertheless, thestudy ofthepotential isvery fruitful. Extensive consideration ofthetheory ofpotential functions isoutside thescope ofthepresent book. Example 2.The lamina bounded bythelines x=0,x= a,y=0, y=binthe xy-plane carries acharge ofdensity o=xy.Find thepotential atthe point Q(0,0, £)onthez-axis. 404 DOUBLEANDTRIPLEINTEGRALS cha The potential is _ x) _f “xd "“at piney =[yy[aeae The firstintegration gives (a+ y+ YP (y?+LVM, so u=fLyla?+y?+2!=y(y?+279")dy, waKarsb+CP?a?+CPPHb+OPPHNC EXERCISES 1.Findthepotential atQinExample 1,andverifythatFy=~24(assuming ¢>0). 2.Find Fy=+k directly inExample 2,and then verify that Fy=ttfromthe answer found inExample 2. 3.FindwandF,atthepointQinExample |if,instead ofconstant density, wehave 4.Find thepotential atacorner ofauniformly charged square lamina ofside b. 5.Find thepotential atapoint ontheedge ofauniformly charged circular lamina of radius b,Itismost convenient totake thepoint inquestion attheorigin. 6.Find thepotential atQ(0,0,b), where b>0, due toauniformly charged square lamina withcorners at(0,0,0),(a,0,0), (a,a,0),(0,4,0).Setuptheintegral inpolarco-ordinates, using thefact that thesquare can bedivided byadiagonal sothat each half contributes thesame amount tothepotential. The integral formula VaTTB™COSG yy tay-1(__b Sin@ er 44jogVEEPTEORTO+ asin8628 Vas bcos" asin willbeuseful. The z-component ofthefield atQmay becomputed from F;=~du/ab but itisperhaps easier tocompute F;directly byintegration, 7,Itcan beshown that du/aé can becomputed from (13.S1-7) bydoing the differentiation under theintegral sign, provided Qisapoint notintheregion Roronits boundary. Proceed from this toverify (13.51-9), using (13.51-6). Thus (13.51-5) isproved. 13.6 /TRIPLE INTEGRALS We shall deal with thedefinition ofatriple integral somewhat more briefly than wedidwith thedefinition ofadouble integral. Webegin with aclosed bounded region Rinthree dimensions, and letf(x, y,z) beafunction defined and continuous inR.As in§13.2 we must make some assumptions about the 138 TRIPLEINTEGRALS 405 character oftheboundary ofR.The precise nature ofthese assumptions need notbemade explicit aslong aswedonotgocarefully into questions of integrability. Weshall forsimplicity think oftheboundary Rasconsisting ofa finite number ofsurfaces, each ofwhich issmooth except possibly atcertain isolated points (e.g., thevertex ofacone) oralong certain curves (e.g, theedgesofacubeortherimsofasolidrightcircularcylinder).Wetake three sets ofplanes, parallel respectively tothex-,y-,and z-axes. ‘The mesh ofrectangular blocks which these planes form inspace iscalled a rectangular partition. Those blocks, orcells, which belong entirely toRare numbered consecutively inany order. Let AV, bethevolume ofthekthcell, and letitsx-,y-,and z-dimensions beAx, Ay, Az. respectively, sothat AV, = AxAy, 42. Finally, let(x,ys,2)beanarbitrarily selected point inthekth cell, Then wedefine thetriple integral ofthefunction fover Rbyfollowing limit, as themaximum dimensions ofallthecells approach zero: JJftex20a=tim&fer.0)Vi (13.6-1) or,inanother notation, Sff$04,y,2)dedydz=tim&f(a,os24)ANAyA(13.6-2) We take forgranted that this limit exists and isindependent oftheparticular method offorming thepartitions and choosing thepoints (x,ys,Z)- The analogue ofTheorem If,§13.23, istrue fortriple integrals; that is,the integral isgiven by(13.6-1) when thesubregions, instead ofbeing rectangular blocks, areformed inany manner (aslong asthey aresufficiently regular in shape). They need notcompletely fillouttheregion R,provided that theamount ofvolume omitted approaches zero inthe limit. These remarks are ofim- portance for the understanding ofwhat happens when weuse cylindrical or spherical co-ordinates. The properties ofdouble integrals explained in$13.21 extend atonce to triple integrals. The same istrue oftheinequalities of$13.22, and themean-value theorem. ‘When itcomes todevising anexplanation oftheevaluation oftriple integrals byiterated integrals, wemust proceed differently than inthecase ofdouble integrals, for nointuitive geometric procedure analogous tothat of§13.3 is available tous(afour-dimensional space would berequired). There isadirect analytical method, however. This method could have been used fordouble integrals aswell. We shall give aheuristic account ofthe method, thus making itsplausibility clear. Afully rigorous account israther long, and itseems advisable toleave thedetails forlater study. Let usfirst state the result. The letters x,y,zcan bewritten insixpossible orders. Corresponding toeach such order there isaniterated integral evaluation 406 DOUBLEANDTRIPLEINTEGRALS cn.13 ofthetriple integral, calling forthree successive single integrations. The main problem oftechnique isthatoflearning how towrite thelimits ofintegration for theiterated integrals. The notation foraniterated integral isillustrated by pee at fayfasf(2+y)dz. (13.63) The integrations in(13.6-3) aretobeperformed intheorder z,x,y. Itwill beenough toexplain thetransition from thetriple integral toan iterated integral forone particular order ofintegration. Suppose this order isfirst with respect toz,then with respect tox,and finally with respect toy.Choosing a typical value ofy,consider thecross section ofRbyaplane y=constant, parallel tothe xz-plane, We assume that Risofsuch ashape that allthe foregoing cross sections are plane regions ofthe type dealt with inour dis- cussion ofiterated integrals intwo dimensions. Asshown inFig. 113, letthe ‘ _ i TO! L_y=b ° H v RRO! /vo> z=Xily) Fig. 113, largest and smallest values ofxinthecross section berespectively Xi(y) and XAy), and letZ,(x, y),Z:(x, y)bethevalues ofzforwhich atypical line parallel tothe z-axis inthecross section cuts theboundary ofR.Finally, lety=aand y=b betheextreme values ofyintheregion R.Then some pe Jfftexnav= Pay[Pax[peay.2ae (136-4) The formula (13.6-4) isthefundamental theorem about evaluating triple integrals byiterated integrals inrectangular co-ordinates. Before giving aheuristic justification ofthe formula wegive anillustrative example. Example. Find thecentroid ofanoctant ofasolid sphere. Letx+y? +z? =a’betheequation ofthesurface ofthesphere. Weconsider 138 ‘TRIPLEINTEGRALS 407 thefirst octant. Evidently £=j=Z,sowefind &only. z Analogous to(13.3-8) wehave vefffoe ma whereVisthevolumeofR.Inthepresentcase/YY¥ V=(x/6)a°. Inthenotation of(13.6-4) weseefrom Fig. 114 that Z)=0,2:= VaEyX=0,K=Vay ™ Fig.114, Hence © ER EITEax-[ af af xd WO), 5 -faf aVOI HPde. The x-integration yields =Ha? y=7P2]VF =Yay. Hence Feat [“(@—yyP ay2: Zab=\[a=ydy=F; weomit thedetails ofthelastintegration. Finally, then X=ja. Now toexplain (13.64). Wegoback tothedefinition (13.6-2). Letussingle ‘out allthecells ofthepartition which belong toRand lieinaparticular column parallel tothez-axis (see Fig. 115). We may choose thepoints (x,ys,2)Sothat the co-ordinates x,y,are the same forallthe points belonging tocells inthe z > rad!ad} fiNLS H { 1ol any H Fe H ( Bau uy> T Fig. 115. 408 DOUBLEANDTRIPLEINTEGRALS ch.13 same vertical column. The values Ax, and Ay, will also bethesame forallthecells inone column, and thearea ofthebase ofthecolumn will beAx,ys. Let us number thecolumns, sayfrom |toN.Suppose AA, isthearea ofthebase ofthe ithcolumn, and suppose thenumber ofcells intheithcolumn ism,Let the points associated with these cells be(xi,¥i,2%) i=ly.+.,ma and lettheir z-dimensions beAzj. Then thesum in(13.6-2) can bewritten intheform xm (3forivis)a2)44. (136-5) The inside sum here isofthe type occurring inthe definition ofadefinite integral with respect toz,The interval ofz-values that isbeing subdivided is approximately from thelower totheupper bounding surface ofR,that is,from 2i(x}, yi)toZ(x;, yi).Hence theinner sum isanapproximation to 248.99f40%Yu2)dz. (13.6-6) For convenience let us write finnBOXy)=f40x,y,2)dz. (13.6-7)een ‘Then theexpression (13.6-6) isg(xi, y/),and(13.6-5) isseen tobeapproximately equal to . Dslr yA (13.6-8) ifthe cell dimensions inthe z-direction are allsufficiently small, This sum, in turn, isofthetype occurring inthedefinition ofadouble integral. IfTisthe plane region obtained byprojecting the points ofRperpendicularly onthe xy-plane, the bases ofthecolumns form arectangular partition ofT.When the dimensions ofthecells ofthispartition aresmall enough, thesum (13.6-8) isvery nearly equal tothedouble integral ffecnaa, * which inturn isequal totheiterated integral fayf*g(xy)dx, (13.6-9) asweseefrom Fig. 113. Wesee, therefore, oncombining (13.6-7) and (13.6-9), that thesum (13.6-5) isanapproximation totheiterated integral fayfaefPsesy.20ae Itmay beshown inmore detail thattheapproximation becomes better andbetter aswetake thelimit defining thetriple integral, sothat (13.6-4) isexactly true. 137 APPLICATIONS OFTRIPLEINTEGRALS 409 13.7 /APPLICATIONS OF TRIPLE INTEGRALS Triple integrals may beused tocalculate thelocations ofcenters ofgravity, the masses ofsolids ofvariable density, moments ofinertia, and other quantities of physical orgeometrical significance. The fundamental principles ofsuch ap- plications arethesame asthose setforth inconnection with double integrals 13.3).WeshallusetheGreekletterxforvolumedensity.Themassofasolidofvariable density 42(x, y,2) occupying aregion Risthen M-[ffuav, The center ofgravity (f,§,2)isfound from theformula x=[ffwav Hf and two other similar formulas. The moment ofinertia about the z-axis is I=fffocs yyw. ‘The product ofinertia relative totheplanes x=0and y=0is Uy=fffomav. Other moments ofinertia 1,I,,and other products ofinertia Uys, Usaredefined byanalogous formulas. Problems ingravitational attraction aremathematically almost identical with problems ofelectrostatic forces, since Newton's law and Coulomb's law are both inverse-square laws. There isadifference insign, since two masses attract each other, whereas two positive charges repel. Newton's lawformass particles mand m’atPand P’,adistance rapart, states that mexerts onm’aforce Fak PP, where kisauniversal constant depending only ontheunits ofmass, distance, and force. Intheoretical work itiscustomary tochoose units such that k=1. Weshall dothis. The force ofattraction onaunit mass atQ,produced byasolid ofdensity 4.occupying aregion R,is F=JfJEOPav, 410 DOUBLE AND TRIPLE INTEGRALS: Ch.13 where r=QP,1isevaluated atP,andintegration iscarried outwith respect to the co-ordinates ofP. The concept ofpotential isuseful inthetheory ofgravitational attraction, The potential atQisdefined tobe a W(Q=JfJbav. The relation between the potential uand the gravitational field force Fis expressed bytheequation F=Vou; ive., thefield isthegradient ofthepotential. The situation iscomparable tothat inelectrostatics (see $13.51); there, however, the field isthe negative ofthe gradient ofthepotential. The difference insign arises from thedifference insign between Newton’s and Coulomb's laws. Example1.Thefirstoctantportionofthesolidinside A thecylinder x?+ y?=a?andbetween theplanes z=0, : 2=hhasdensityo=x.Finditsmass.Wehave LATpeaE Z|hM=[ffxav=['ae [ay[”xax;zy * oSv M-fafKay) dy=[Sdz=ta°h.Jo 3 ’ Thefinding ofthelimits ofintegration isillustrated inpag,44,Fig. 116. Example 2.Find themoment ofinertia about thez-axis ofthehomogeneous tetrahedron bounded bythe planes z=x+y, x=0, y=0, z=1.The integral in this case is tef [faces * L=[face+y)dv oan) <uf'[afoye Ly ‘Thelimits ofintegration arefound byanexamina- + Zo O10)tionofFig.117.Completion ofthe integrationisleft|Z--# asanexercise forthestudent. Theresult is Koo =H.4-% Fig.117. ‘Since thevolume ofthetetrahedron is{,themass isM=4/6,whence 4=6Mand 412 DOUBLE AND TRIPLE INTEGRALS ons The equation 1x?+Ly?+Lz? ~2Unyz— WUas2x —2Uyxy =1 defines what iscalled theellipsoid ofinertia forthebody relative totheorigin O.Asetof axes such that theproducts ofinertia allvanish iscalled asetofprincipal axes ofinertia forthebody. 13.8 /CYLINDRICAL CO-ORDINATES Ifweuse polar co-ordinates inaplane, and arectangular co-ordinate along an axis perpendicular totheplane attheorigin ofthepolar system, the combination iscalled acylindrical co- z ordinate system, Most commonly thepolar co-ordinates aretaken inthexy-plane (seeFig.118), butthere isno P(r02) logical necessity forthis choice. Itisoften convenient to evaluate atriple integral byaniterated integral in olcylindrical co-ordinates. Aswesaw in$13.6, v a 5 2 [ffroneav= [faa [r0.s.20¢H : q Fig. 118.aan 7 Ifweexpress theintegrand incylindrical co-ordinates, sayf(x, y,z)= F(r, 0,z), the double integral in(13.8-1) may beevaluated asaniterated integral inpolar co-ordinates, This leads tothe result orm fffFo.0,2av~ faofrarfF(r,6,2)dz.(13.8-2) Donotfail toobserve the factor rwhich isintroduced into theintegrand ofthe iterated integral. The limits Z,,Z,must beexpressed interms ofrand @;ther and @limits arefound byinspection oftheplane region T,the“shadow” ofRon the xy-plane (see Fig. 115). The result (13.8-2) and others like itmay also be obtained byanargument similar tothat beginning after the Example in§13.6. There are five other possible orders ofintegration. Asystematic method for determining the limits ofintegration for any given order isillustrated inthe following example: Example. Find themoment ofinertia ofahomogeneous right circular cone about itsaxis. Let theradius ofthebase beb,thealtitude beh.The density 4isconstant, soM=y(7/3)b*h. Weplace thecone asshown inFig. 119(we draw only ‘one-fourth thecone). Lettheintegration order ber,z,6.Wemust first setupthe triple integral: 1=[ffucreyrav=n fffrav. 139 ‘SPHERICAL CO-ORDINATES 413 NowpictureasectionoftheregionRmadebyholdingthe zlast integration variable (here @)constant. Inthepresent case thisisthetriangle OAB. Next assign thesecond y,integrationvariablezatypicalvalue,anddeterminetheJ'srange offreedom lefttothefirstintegration variable r. th This process isindicated inFig. 119bythelineCD. Since y OC=z,thevalueofratDisgivenby {£28. Zi ¥ zh o™ Es Therlimits ofintegration aretherefore 0andzb/h. pig.119, Now lettheline CD range inthez-direction asmuch asitmay (from 0toh); these are the z-limits of integration. Finally, let@vary through allvalues necessary tohave the @- sections sweep outtheentire region R.We seethat the0-limits ofintegration are 0and 2x. Therefore (remembering theadditional factor r), ae th phintenfaoa:{’Pdr=Hybth. This may bewritten I=i)Mb*. EXERCISES 1.Forthesolid cone oftheillustrative example find (a)I,; (b)thelocation ofthe center ofgravity ofthefirst octant portion; (¢)theattraction exerted onaunit mass at theorigin; (d)thepotential atapoint (0,0, £),where ¢<0. 2.Find the moment ofinertia ofahomogeneous solid sphere ofradius a,about a diameter. 3.For ahomogeneous solid right circular cylinder ofheight hand radius ofbase a, find themoments ofinertia (a)about theaxis ofthecylinder; (b)about alinethrough thecenterofgravity ofthecylinder, perpendicular totheaxisofthecylinder; (¢)the attraction exerted bythe cylinder onaunit mass atthe center ofone end; (d) the potential atapoint (0,0, ),assuming thecylinder defined byx"+y? a’, 052=h, and assuming {=h. 13.9 /SPHERICAL CO-ORDINATES Toform aspherical co-ordinate system westart from anorigin Oand afixed ray issuing from O.Weshall take therayasthepositive z-axis; there is,however, nonecessity for any one special relation between spherical and rectangular co-ordinates. The spherical co-ordinates are the distance p= OP and the two angles 6,@(see Fig. 120). The angle 8,sometimes called theazimuth ofP,isthe ‘same asthat used inplane polar co-ordinates. The angle isthecolatitude ofP. We always choose @inthe range 0S. For most work pistaken nonnegative. The student should beaware that insome books the roles of@and @are 14/ CURVES AND 14/INTRODUCTION Curves and surfaces are geometric entities with which the student istosome extent familiar. The simplest examples ofthese entities, such astheconic curves intheplane, and spheres, cylinders, cones, and other quadric surfaces inspace, have been encountered repeatedly from analytic geometry through calculus. Geometrical interpretations offunctions ofone ortwo independent variables have led the student tothink ofcurves and surfaces inquite general terms. In this chapter wepropose tomake acareful study ofthe means bywhich we render our intuitive notions about curves and surfaces amenable toprecise mathematical treatment. This isdone partly asanintroduction toabranch of geometry—what isknown asdifferential geometry—and partly aspreparation for thefollowing chapter online and surface integrals. Apoint tobeemphasized isthis: Our intuitive notions about curves and surfaces are allderived from relatively simple examples ofthese things. The general concepts ofcurves and surfaces are very inclusive, however, and inour studies we must remember that when we wish toprove something, we must appeal tothe definitions and previously established theorems, not solely toour intuitions, which may present uswith anoversimplified picture. Direct geometric visualization ofthesubjects ofour discussion is,however, ofgreat value, both for the suggestions wecan derive and for the better understanding and retention ofwhat we learn. 14.1 /REPRESENTATIONS OF CURVES Intuitively wethink ofacurve asaone-dimensional configuration, like thepath ofamoving particle, orassomething wemight obtain bybending and twisting a straight line. Weshall define acurve bysaying that itisanordered configuration ofpoints (x,y,z)given bythree continuous functions ofaparameter: x= fy =e),2=h(t); (141-1) therange oftheparameter istobesome interval (finite orinfinite) ofthereal axis. Wespeak of(14.1-1) asaparametric representation ofthecurve. Acurve may have more than one parametric representation. Ifweinterpret tastime, (14.1-1) may beregarded asdefining thepath ofamoving point. The point may pass through thesame position inspace several times; inthis case thecurve intersects itself. Evidently acurve inthe above sense ofthe word isvery general, andmay notbevery smooth. Imagine, forinstance, thetrack ofatiny particle inBrownian movement over along period oftime. 417 420 CURVES AND SURFACES ch.14 with plane curves. Sometimes thearclength isexpressible interms oftabulated standard integrals, such aselliptic integrals. Example. Consider thefirst octant portion ofthecurve ofintersection ofthe sphere andcylinder vtytada, xt(y-al=a’, (14.2-7) asshown inFig, 124 Itisconvenient tousezasaparameter forthis i curve. Ifwe eliminate xbysubtracting the two equations in(14.27), wefind ( =4a=2,oe DS Substitutingthisresultinthesecondoftheequations am (14.2-7), wefind a ~via?x=pada. # Fig, 124, Asparametric equations ofthecurve, wehave xeZvieae, y-4E pee2a ° 2a Adirectcalculation shows that >_8a?=2? ,5 ds?=ad, The range ofzisfrom 0to2a, sothe length ofthe first-octant portion ofthe curve is pt pates\ 5 =f" §SaS) az (042-8) This integral isimproper atthelimit z=2a,butitisconvergent. Itcan beputin theform ofastandard elliptic integral ofthesecond kind. Forfurther discussion ofthis problem see Exercise 8. EXERCISES: ‘The standard elliptic integral ofthesecond kind isdefined as tk.)f°VI=Siaat, where 0<k <1. If =m/2, theintegral iscalled complete, Values ofthis integral for various values oftheparameters k,dare given inmany books oftables. Insome ofthe exercises itis required that thearclength beexpressed intheform ofsuch astandard integral 14st PRINCIPAL NORMAL. CURVATURE 423 Thequotient AR/As isavector along thelineofthechord asp PP’. SeeFig.126.Since thelength ofARisthelength ofthe a aechordPP’,weseethatwhenP’approaches Pthelimitof|--“7 thelength ofAR/As isunity. Furthermore, thelimitingdirection ofPP’isthatofthetangentatP.There |R|/R+ARfore dR AR oO AR «jim AR =7, dsasAs Fig.126,Differentiation ofRwithrespectto#gives dR_dRds_dsdds dtdtT. (14.3-6) Thisisequivalent toformula (14.3-3). Ifisthetimevariable, 4Fisthevector velocity ofthepoint Pmoving onC. EXERCISES 1.Let Fand Gdenote vector functions ofascalar variable f.Assuming that Fand G atedifferentiable, prove theformulas 4op.gyap 4G,4dF Lec =F 464M G, ©)LaxG= FS ag, using the same method bywhich the rule for differentiating products isderived in elementary calculus 2.IfFinExercise 1isavectorofconstant length,provethatF-4E=0, andthusar +uniess4 conclude that4isperpendicular toFuntess 4F«0, 14.31 /PRINCIPAL NORMAL. CURVATURE Inthis section and the next we continue with the notations used in§14.3, We shall define two more unit vectors, theprincipal normal Nand thebinormal B. which, along with the tangent vector T,form anorthonormal setofvectors associated with thepoint (x,y,z) onthecurve C.These vectors, especially T and N,areimportant inthestudy ofthemotion ofthepoint (x,y,z) along thecurve. The acceleration vector lies inthe plane ofTand N.The component oftheacceleration along thelineofNwillbeshowntodependonthecurvature ofC. Weshallassume thatx,y,zhavesecond derivatives withrespect tos.Prom(14.3-1) we have at_ dx, dy de pt rf +S 1431-1)dsdst! ds) a* ‘ » 432 ‘CURVES AND SURFACES cht and se1(seandean),ay)\~awav*aeau)~~jy Itfollows that aax” dy bob Thus (14.4-4) and (14.4-6) define the same direction; this iswhat wesetout to prove. Once thedirection ofthe normal isknown, itisofcourse aneasy matter to write outtheequation ofthetangent plane. EXERCISES 1.Show that the parametric surface defined byx= asin6cos 0,y= bsin6sin6,2=6086,050527,0565x,isanellipsoid,andthatitisasphereifa=b~c.Thesurface isnot asimple surface clement, however, Which part ofthe definition ofasimple surface element isnot satisfied inthis case? 2.Explain how todivide the surface ofasphere into simple surface elements in several ways. Inparticular, ifthe sphere isx°+y"+z*~ 1,describe amode ofdivision such thatthe points (0,0, +1)aeinterior points ofelements onwhich they lie.Describe a mode ofdivision such that the points (=1,0,0) and (0,*1,0) are interior points ofthe elements onwhich they i. 4.For thecase ofeach ofthefollowing parametric surfaces, obtain anequation of thesurface inrectangular co-ordinates. (a)x=aucosv,y=businv, 2»u(elliptic cone) (b)x=wos 2,y=wsinv,2=ku”(paraboloid ofrevolution). (©)x=asin cosh »,y=bcos ucosh v, z=csinh v(hyperboloid ofone sheet). (8)x=rcos 0,y=rsin@,2=(F'2) sin29(hyperbolic paraboloid). (©)x=aucosv,y=businv,z=wos 20(hyperbolic paraboloid). (x=acoshv,y=coshv608u,2=€cosh8inw 4.Show that the tangent plane totheellipsoid (x/a*)+(y1b?)+(2%lc’)=1 at (a, 29)isCola?)+(yoylb?)+(zazle?)=Ne S.Show that the direction ofthe normal tothe surface inExercise 3e) is~2bucosv:2awsinvab, 6.Describe the parametric surface x=acosu, y~asinu, z=, and find its equation inrectangular co-ordinates. 7.Describe theparametric surface x=2u+v,y= ~t,2=3u,and find itsequation inrectangular co-ordinates. 8.Show that theparametric surface x=u+e, y=u—, z=40"istheparabolic cylinder z=(xy). Show that thetangent plane atthepoint corresponding to(u,») is ox—doy—2=40"9.Ifthe curve y=f(x) inthexy-plane isrevolved around thex-axis, show that the resulting surface can berepresented parametrically inthe form x=u, 9=f(u)cos v, 2=f(u)sine. Assuming that fu) iscontinuous andf(u)>0, show that thedirection of thenormal isfu): ~cos v:~sin v. 440 CURVESANDSURFACES cha calculation showing that EG-F*=jitih+ i (146-7) Inworking problems itwillsometimes befound tobeeasier tocomputeji+j3+jithanEG-F?andviceversa. Example 1.Compute thetotalareaofthetorusx=(a+bcos$)cos6,¥=(a+b cosd)sin8,z=bsind,0<b<a.‘This torus was discussed in$14.5 (see Fig. 129). The part inthefirst octant is asimple surface element corresponding to0505m/2, 0S Sm, and thearea ofthis part isone eighth ofthe total. From (14.54) wesee that, if w=0and v=o, E=(at+bcosdy, F=0, G=b’. Thus VEG ~F*=b(a+bcos ),and thetotal area is 8=8 [do[™o(a+bcos6)d0=An'ab, This isinaccord with thetheorem ofPappus. Ifweuse (14.6-5) instead of(14.6-6) tocalculate the first octant portion of the torus, we find that i= bla +bcos 4)cos6cos4, j= b(a +bcos d)sin8cos&, js= b(a +bcos d)sind, from which i++ i=ba +bcosby. Thus, theidentity (14.6-7) isverified inthis particular case, and the integral for the area iscalculated asbefore. Intheargument leading upto(14.6-5), itwillbeseen that thefact that the region Rwas arectangle intheuv-plane was notessential. IfRisany bounded closed region ofthe uv-plane ofthetype described inthediscussion ofdouble integrals in§13.2, thediscussion leading upto(14.64) applies toanycellina rectangular partition ofthetype shown inFig. 131a and(14.6-4) gives thearea oftheparallelogram which istheimage ofthis cellunder theaffine mapping (14.6-2). Hence, theformulas (14.6-5) and (14.6-6) can beused tofind thearea ‘ofany portion ofasmooth surface which isobtained byaone-to-one and continuously differentiable mapping from theregion Rinthewv-plane. We now consider the special case ofasurface defined byanequation z=f(x,y)forall(x,y)belonging tosomeregionRinthexy-plane. Itwillbeass- umed that fhascontinuous first partial derivatives inR.Wecanthink ofxandy 148 ‘SURFACEAREA 441 asbeing the parameters u,v.This leads ustothefollowing very special case of (144-3). xex y=y z=fuy) (146-8) Bysimple calculations wesee from (14.4-5) that je-L, jet, -henge henge hak Consequently, thearea oftheportion ofthesurface corresponding totheplane region Ris . ty, (ayy? s=ff[r+(Z)+(Z) YPacay. (146-9) Alternatively, ifwewish touse (14.6-6), wecan calculate asfollows: =ae4 de=Fdx+Edy, 2dxtady?(ZaxaFay) ds?=dx?+dy+(Zax+ Fay) ofa(22)]gx?428222 az)ay? =[1+Gz)Jac+23&aay+[1+(2)Jar Interpreting xaswand yasv,wehave azpaaz =14(2). enw(S). Pog O-1+(5) Hence the area ofthe surface is az), (az}” s=ff[t+(S)+(Z) Poaca. (146-10) Example 2.Find thearea oftheupper half ofthesphere x°+y?+2*= a?by using formula (14.6-10). Here =Va-x-y, #2-—, FEN OEE ox Vay withasimilarformula for=‘Thustheintegrand in(14.6-10) becomes 2 2oe [i++ a] =e: ax -y @xy) Vay There isonedifficulty. The hemisphere liesabove theregion Rbounded bythe circle x?+y?=a?inthexy-plane, andweseethatthepartial derivatives ofz 48 ‘SURFACE AREA 443 being obtained asthelimit ofthesum ofareas AAsecy.Itisthis derivation which isusually found inelementary calculus textbooks. Formulas (14.6-6) and (14.6-12) are the standard formulas ofcalculus for dealing with surface area, Where, however, isthedefinition ofsurface area? Are there surfaces which have area, and yetwhich aresuch that thearea cannot be found bytheintegrals mentioned above, perhaps because oflack ofsufficient smoothness? Itislogically and aesthetically desirable tohave adefinition of surface area which isdirectly geometric, and which does not put too many restrictions onthe surface. Agood definition ought not todepend upon the method ofrepresenting thesurface analytically, and should notbelimited to smooth surfaces. The demand for such adefinition poses avery difficult problem, however. Itmay surprise the student toknow that the problem has ‘occupied theattention ofmany able mathematicians over thelast fifty years, and that theend ofresearch onthequestion isnotyetinsight. To present the concept ofsurface area tothe student atthe advanced calculus level, themost satisfactory logical approach seems tobethefollowing: for asmooth simple surface element, with appropriate conditions on its parametric representation, formula (14.6-6) istobetaken asadefinition; the discussion leading uptothe formula isbyway ofmotivation. Itcan beshown that thearea sodefined isindependent oftheparticular parametrization, and is therefore anintrinsic characteristic ofthe surface. This demonstration requires thetheory oftransformation ofdouble integrals, and isdiscussed inChapter 15. Intheparticular case ofsurfaces z=f(x, y),theformula (14,6-12) makes itclear that thearea does notdepend ontheparametrization ofthesurface, butitmust still beshown that the orientation ofthe z-axis isinessential, since the direction ofthis axis plays arole intheformula. EXERCISES 1.Find the area ofasphere,usingtheparametric representation x=asingcos®, y=asindsind, z=acosd. 2.Find thearea ofthepart ofthecylinder x°+z"= a”inside thecylinder y?= a(x +a), 3,Find theareaofthepartofthecone x°+y*= 2"inside thecylinder x°+y?=2ax. 4.Find thearea ofthepartofthesurface z~xyinside thecylinder x°+y*=a*, 5.Find the area of the surface element x=aucosv, y=busine, 2= {wa cos? v+bsin’»),055u1,05v=2n,Identify thesurface andtheportion ofit whose area isfound. _6.Apart ofthesurface 2*=2xy canbeparametrized byx=u", y=0?2=Viwe.(a)Findtheareaofthepartofthesurfaceabovetherectangle 05x=a,0=y Sb. (b)Find thearea ofthepartofthesurface above theregion inthexy-plane between thexy-axes andthecurve x+y"=1.Comparethesolutionsby(14.6-12)and (146-5). 7.Find thearea defined byx=rcos@, y=rsin8, z=8,OrS1, 0S052n. Describe the surface. 15.12 LINEINTEGRALS 447 BuPq Pant or Qt Py Ps Q, Anke Fig. 135, and call ittheline integral ofFwith respect toxalong C.IfPisthepoint (x,y,2),and ifF(P) isdenoted byf(x, y,z),analternative notation fortheline integral is fs0yz)de, Line integrals with respect toyorzaredefined inthesame way, with Sy,orAx replacing Ax in(15.121). Tocompute the value ofafine integral, weuse some parametric represen- tation ofthecurve C.Suppose theparametric equations ofCare x=A(t) y=n(t), z=), astsb, and suppose that x,y,and zhave continuous derivatives with respect tot.We further suppose that the points Aand Bcorrespond tot=a and t=b, respectively, and that (x,y,z)traces out Cfrom AtoBastgoes from atob. Letthepoints PyonCcorrespond topoints f,such that @=fo<ty<++-<t,= (bsletAt=tk—t-1, and letQcorrespond toti,where \.,StiSt. The sum (15.12-1) now takes the form 2LOD, MD,HEDIIAG) ~ADI (15.12-2) Bythelaw ofthemean, M(t)~M(t)=ANC)At where 1issome number between t.-, and 4.Therefore J.f(x,y.2)dx-fJOA),WC),W(EYACD) dt. Indrawing thisconclusion weuseastandard theorem about definite integrals; this theorem appears as(18.21-4), §18.21. Itisaspecial case ofDuhamel’s 18.12 LINE INTEGRALS: 449 representation ofthecurve which isused tocalculate thevalue oftheintegral. Asum ofline integrals with respect tox,y,and 2isoften written with just ‘one integral sign. Thus, fsesyz)dx+g(x,y,2)dy+h(x,y,2)dz means Jf(xy,2)de+fatyz)dy+fh,yz)dz, Example 3.Compute thevalue of [xcdexdy~yede (15.12-4) along theoriented curve shown inFig. 138, consisting : ofaquarter circle inthexz-plane, andlinesegments in 0.1 1,1)thexy-plane andyz-plane, respectively. Denote the (0,1) three parts ofCbyCj,Cs,Cs,respectively. OnC;we choose xasparameter. Then z=VI—x, y=0, 80 Gy dy=0, and y jaed+xdy~yedz =fxzdx+x-0-0-dz (01.0) cses @ 1 £00)=[[xvieP a=) aon ForC;weuseyasparameter; theequations areM83% x=1-y, z=0;sodz=0, and \ [ixdetxdy-yede= 0-de-+xdy-0-0 , eses od .Cyqns !ghd. =[/a-yay=h oy: loneerarin gsHany Finally, using zasparameteronC;,wehavex=0,y=1,dx=0,dy=0,andtheyyc, integral over Cyisjust 9 7 J2-26J.-vae=[[-2ae=-4 ° ‘Thus theline integral (15.12-4) has thevalue f+d-deh EXERCISES: 1.Find thevalues ofthefollowing line integrals. Allthecurves areinthexy-plane (a)foy?dx—xdy, along y=4xfrom (0,0)to(1,2). (b)Je—y dx+xdy, along y*=4xfrom (4,4)to(0,0). 1533 VECTOR FUNCTIONS AND LINE INTEGRALS. WORK 451 11, Prove that the line integral inExercise 10has the same value for allcurves C with initial point at(0,0,0) and terminal point at(1,1,1).HINT: Ifthecurve isexpressed interms ofaparameter t,consider F(t), where F(t) =xy+yz+2x when x,y,z are expressed interms oft 12,Let Cbethe clockwise closed curve bounded bythe lines x= a,x=, the x-axis, and acurve y=f(x), a'S.x5}, assuming that a<b and that f(x) iscontinuous and never negative. Using results from elementary calculus, show (a)that foydx isthe area enclosed byC; (b)that Sey’dxisthevolumegenerated whenthisareaisrevolved a, around thex-axis; (€)that Jcxy dxisthefirst moment ofthisareawithrespecttothey-axis;(4)thatfoly’dxisthefirst..\‘moment ofthearea with respect tothex-axis; (e)thatfox’ydx |__| isthe second moment ofthearea with respect tothey-axis. Itcan beshown later, after wehave learned more about line integrals,thatthesesameinterpretations maybemadefortheforegoingline—f-~--" GintegralsifCisanysectionally smooth,simpleclosedcurveinthe 2 xy-plane (except that in(b)wemust require that thecurve lie ~O} entirely ononesideofthex-axis) ,13,UsingFig.139explainwhyitappearscorrecttosaythat,18:13% ifCis.asimpleclosedcurveorientedcounterclockwise, fcxdyisequaltotheareaenclosedbyC. 14,Using Fig. 139 asaguide, setupaline integral with respect toy,giving the volume ofthesolid generated when thearea enclosed byCisrevolved around thex-axis. ‘Assume, asinthefigure, that thecurve lies entirely above thex-axis. 15.13 /VECTOR FUNCTIONS AND LINE INTEGRALS. WORK Consider aline integral oftheform J.Pax+Qdy+Ras, as.) where P,Q,Rare continuous functions defined along acertain oriented curve C. Such integrals often occur inconnection with vector point functions, and we shall now indicate how theintegral (15.13-1) can beexpressed inadifferent notation bytheuse ofvectors. Let FQ,y,2)=Pi+Q)+Rk bethevector function defined ateach point ofCinsuch away that P,Q,Rare itscomponents inthexyz-co-ordinate system. Let sdenote arclength along C, with s=0attheinitial point ofCand s=Iattheterminal point. We assume that Cissmooth. Then theunit vector tangent toCinthepositive direction ata kiven point is dx, dy, dzTait D+ack 153 GREEN'S THEOREM INTHEPLANE 459 (Bear inmind theco-ordinates oftheinitial and terminal points indetermining thelimits ofintegration.) Thus ‘ [pa--f{P(x,¥)-Px,¥)}de. 5.34) Next consider the double integral, and use the iterated integral formula (13.3-6): aP. ““oP[Faw-f axPEay. e15.3-5) The yintegration may now beperformed with xheld constant. The result, by Theorem VIII, $1.53, is oP.Jpay=Pew,v9~POYo. (15.346) yay ‘On combining (15.36) with (15.3-5) and comparing with (15.34), we see the truth of(15.3-2) foranx-simple region R. Anentirely similar proof may begiven for formula (15.3-3) ifweassume that Risy-simple. The figure forthis case would resemble Fig. 96($13.3). Finally, ifRisboth x-simple and y-simple, wecombine (15.3-2) and (15.3-3) to give (15.3-1). Green's theorem isthus easily proved forregions which areboth x-simple and y-simple. Inparticular, abounded region Risboth x-simple and y-simple ifitsboundary consists ofasingle sectionally smooth convex curve. A rectangle issuch aregion. ‘There are x-simple regions which are not y-simple, and regions which are neither x-simple nor y-simple. Ontheother hand, many regions may bedivided into afinite number ofsubregions, each ofwhich isboth x-simple and y-simple. Forsucharegion itiseasytoproveGreen's theorem. Forinstance, suppose Ris the region bounded between the circle and the large triangle inFig. 145, with axes asshown. This region isneither x-simple nor y-simple, but wecan divide it into four subregions, each ofwhich isboth x-simple and y-simple. The formula ofGreen's theorem therefore holds for each ofthe subregions. Ifwe add corresponding parts ofthefour formulas, thedouble integrals combine togive thecorrect double integral over thewhole ofR.Now consider theline integrals. Individing Rinto parts, ¥ weintroduced four interior connecting lines, Each of these lines occurs twice, but with opposite orienta- tions inthetwo occurrences, since each line belongs tothe boundary oftwo neighbouring subregions.Hence,whenallthelineintegralsareadded,theLN\7\contributions from these interior lines cancel out in pairs, leaving only theline integral around the ~Z = total oriented boundary of R,that is,counter- clockwise around the triangle and ‘clockwise Fig. 145. 462 LINE AND SURFACE INTEGRALS chs therefore <4 ing=A cosa=Gesina=—7 (S.3-11) The correctness of(153-9) isnow apparent, and theproof of(15.3-8) is complete. EXERCISES 1.Use Green's theorem toevaluate thefollowing line integrals: (a)Se2xydx—3xydy,clockwisearoundthesquareboundedbyx=3,x=5,y=1,y=3.(b)fcxy*dx+2x*y dy,counterclockwise around theellipse 4x7+9y*=36.(©)fe?+29)dy,counterclockwise aroundthecircle(x2/'+y=1 (@)fce*sinydx+e*cosydy,around theboundary ofanyregular region. (©)Jex°ydx~y*xdy,counterclockwise aroundtheregionboundedbyy=Va?=¥and y=0. Use polar co-ordinates toevaluate thedouble integral. ©[325%, aroundtheboundary ofanyregularregionnotcontaining theorigin. 2.Calculate the line integrals ofExercise 2,$15.12, parts (a), (d), (D,(g)and (h), using Green's theorem. 3.Let Cbeany sectionally smooth simple closed curve inthexy-plane, oriented counterclockwise. Let Rbetheregion bounded byC,and letRhave area Aand centroid 5). Show that S[eaean [vay=ay, and, ifRisalamina ofconstant unit density, interpret f-xvae and[-yarry*ay asmoments ofinertia, specifying theaxis ofrotation ineach case. 4.LetRbetheregionboundedbytherays0=a,@=andthecurver=(0),as‘shown inFig. 147. Use thethird formula in(15.3~7) toshow that thearea ofRis A=i["Yeonae, ¥ r=fe)Li: z 2 3 Fig. 197 466 LINEANDSURFACEINTEGRALS. ch.15 direction, LetC’denote theboundary ofR’,Weorient C’bytaking thepositivesensealongC’tobethatwhichcorresponds, underthemapping, tothepositivesense along C;thus, as(x,y)moves along Cinthepositive sense, itsimage point (u,b)moves along C’inthepositive sense. With this agreement wehave = 88ay+28] ; [xai.J(u,v)(%du+55dv], (15.32-6) since = =8 cg x=flue) and dy=28du+28do hold forcorresponding points ofCand C’ Next we apply Green's theorem inthe un-plane tothe line integral in (15.32-6). Instead ofPdx +Qdy wehave 28ay+f58du+738de. Therefore, corresponding to 8Q_aP Op8)@(5oe Oncarrying out theindicated differentiations, this latter expression isfound to beprecisely J(u, v).Consequently, oR BayosJ 7 Js3%au+438ao+f)J(u,v)dude. (15.327) ‘The choice ofsign onthe right isdetermined bythe orientation ofC’. Ifthe orientation which wehave given toC’coincides with theusual positive orien- tation oftheboundary ofR’,theplus sign iscorrect; inthecontrary case we must choose the minus sign. Combining (15.32-5), (15.32-6), and (15.327), we see that A=[f+scu. 0)dude. Since Aispositive and Jisalways ofthe same sign, itfollows that the sign chosen in(15.32~7) must bethesame asthesign ofJ.Whichever thesign, formula (15,32-4) iscorrect. ‘The lastremarks enable ustojustify theanswer given tothequestion posedattheendof§9.2.SupposeRisacircularregion.Thepositiveorientation ofitscircumference Ciscounterclockwise. The image ofCwill beasimple closed curve C’,and R’will consist oftheinterior ofC’and C"itself. Hence, theusual positive orientation ofC’will also becounterclockwise. But the mapping ofR onto R’induces acertain orientation ofC’. From the discussion intheforegoing paragraph weseethat theinduced orientation ofC'iscounterclockwise ifand only iftheJacobian ofthemapping ispositive. 478 LINE AND SURFACE INTEGRALS ch.ts (6)(x+y~7)dx+(Sx—By+3)dr.(@)Qay+39)dx+27(@)xe”sinydx+(e"cosy+y)dy. ((xycos.xy +sinay)dx+2°c08xydy. (a)(4x"+10xy*~3y4)dx+(1Sx?y?~ 12xy"+Sydy. (h)(e*sinyy)dx+(€*cosy—x~2)dy. 2.(a) Find afunction wsuch that du=EYae22F2dy, and describe theregion orregions inwhich wisdifferentiable. (b)Find thevalue ofthe line integral from (1,0) to(5,2): from (-3,0) to(~1,4). Ineach case specity any essential limitations ‘onthepath C. 3.(a) Find afunction wsuch that du~—4_- __ede Vyae yWypeteypoa™ and describe theregion orregions inwhich wisdifferentiable. (©)Find thefine integral ofthedifferential form in(a)from (3,5) to(5,13),and specify any necessary limitations onthe path 4.Find afunction of xalone, w= (x), which makes the differential form v(x siny+ycos y)dx+w(x cos y~ysiny)dyexact; then find thefunction ofwhich it isthedifferential, ifthisfunction isequal to0at(0,0) §.LetPybe(1,0),Psbe(~1,0),andPbe(x,9).Let04and0betheanglesbetween thepositivex-axisandP,PandPsPrespectively. Letu~@;+sShowthat du=~(44%)de+(St1h) ay, where ris PsP, r2= PsP. Tomake wasingle-valued function itis necessary tomake some definite agreement about thevalues of@,and 6atallpoints except Piand Ps, (a)Ifitisagreed that ~7<0, 57and 06s<2n, show that uisdiscontinuous ify=0 andx°>1. Bymaking cuts along thelines ofdiscontinuity, wegetasimply connected region inwhich wisdifferentiable (b)Ifitisagreed that 0.5@<2 and05¢;<2x, where arethediscontinuities ofu? (©)Mv =6,~0, and theangles arechosen asin(b), where is discontinuous? 15.5 /FURTHER DISCUSSION OF SURFACE AREA In$14.6 we arrived atthe formula A=[[VEGFdudv ass-1)x 186 ‘THE DIVERGENCE THEOREM 487 ar ; . insteadof2E,andcosainsteadofcosy;acorresponding resultalsoholdsfor zx-simple regions. Noadditional proofs areneeded, since theresults differ from (15.6-2) innotation only, and thelabeling oftheaxes ispurely amatter of notation. Now suppose that Tisaregion which isatonce xy-simple, yz-simple, and zx-simple, andlet$beitssurface. Suppose that P,Q,Rarefunctions which are continuous and have continuous first partial derivatives inT.Atapoint where S issmooth letmbeaunit vector normal toSand extend outward from T,and let nmake angles a,p,yrespectively with thepositive x-,y-,and z-axis. Then by the lemma ev ff J[f%av-ffrcosada, + 5 [JJBav-ffecospaa,ay J[fBav=ffreosyaa. ‘Adding, we obtain the divergence theorem (15.6-1) for aregion Tofthis restricted type. Next weproceed toremove some oftherestrictions onT:Let uscall aregion xyz-simple ifitisatonce xy-simple, yz-simple, and 2x-simple. The region between two concentric spheres (say with centers atO)isnot xyz-simple. But thethree co-ordinate planes divide this region into eight parts, each ofwhich is xyz-simple. Inthis subdivision process, certain additional surfaces are intro- duced as“interior partitions” inT.Each surface element ofsuch aninterior partition isonthe boundary oftwo xyz-simple subregions. Let ussay that a region Tisxyz-standard if,bytheintroduction ofafinite number ofsimple surface efements asinterior partitions, wecan divide Tinto afinite number of xyzsimple subregions. ‘THEOREM VII. Under the stated assumptions onP,Q,R,formula (156-1) holds when Tisanxyz-standard region. Proof, Let Ti,...,T» bethexyz-simple regions composing T,and letS,be theentire surface ofT,‘The surface S,may consist partly ofpieces ofSand partly ofinterior partitions. Bywhat wehave already proved, aP,2Q,AR)gy—JJ + dA. Sfp(E+ee8)av=I(Pcos.a+QcosB+Rcosy)dA. fA‘ 502 LINEANDSURFACEINTEGRALS cnt Here wehave used (15.51-4) and (14.6-7). Next weshow that aP,aP _aPax_aPax az ay8aua0avow (57-4) Infact, aP_aPax,aPay,aPaz, au” axau’ ayou’ a2du withasimilarformula for2P.Therefore, aPa_aPaeaP(aySe99At)AP(2832222), auav”avauayauavavou)*az(awav”avdw and this isequivalent to(15.74). The surface integral in(15.7-2) has now been reduced totheform aPax_aPax JJGraeae5u)dae G ‘Aneasy calculation shows that this isthesame as a (pa)_ a (paxSJ[R(@S)-Z (eB) auae. (057-5) Tothis integral wenow apply Green's theorem inthe uv-plane. Asaresult, (15.7-5) isequal totheline integral ax ax fp dusPEav But this isjust [pax and sowehave completed theproof of(15.72) under theassumptions onSas stated earlier. We have assumed that P,Q,and Rhave continuous partial derivatives insome region containing S. Stoke's theorem may beextended tomore general surfaces byaprocess entirely similar tothat employed intheproof ofGreen's theorem intheplane. ‘The process issuggested byFig. 161. Itconsists individing Sinto afinite number ofsimple surface elements bytheconstruction ofone ormore “cuts,” orinterior dividing lines. We -assumethateachelementanditsboundary takesits aM a>orientation fromtheoverallorientation ofSandC, aSandthateach“‘cut”occursaspartoftheboundary \ff 1aofjust two surface elements, with opposite orienta- tions inthe two cases. Ifnow weadd the formulas Fig. 161. 510 LINE AND SURFACE INTEGRALS. cn.15 (4)Show that, ingeneral, = pared Very-a 4"Wii ata waaVERVE tyaoe) Wey alee ()Atwhat points is¢discontinuous? (©)Atwhat points is4continuous butnotdifferentiable? (@)IfDis theregion consisting ofall ofspace except thez-axis and thecircle C,show thatthe differential form in(x)satisfies conditions (15.8-5) inD. (@)Describe aclosed curve inDwhich isnottheboundary ofanysurface lying inDand towhich Stokes's theorem may beapplied. MISCELLANEOUS EXERCISES 1.IFSis defined by2=f(x,y) with (x,y) ranging over R,and ifthepositive side of Sis chosen sothat cos y>0, show that 4 4Jfeccosa~y.cospyaam ff(yfx2)axay 2.Find afunction whose differential is *cosy-35)dx—(e*siny~7sec* (econ Za) delesinyTae 3.Suppose a>0, b>0.LetPbeapoint ofintersection ofy= —4a’(x ~a°)and y=4b%(x +>), andletRbetheregion bounded bythefirstparabola, thex-axis, andthe lineOP.Showthat{{va=2abbyusingthetransformation x=u?~v%,y=2uv. 4.Use (15.446) with a=b=0tofind afunction usuch that ateyde+41y"—2)dy ey ay The function sofound hascertain discontinuities. Where arethey? 5.Find aregion inthe r-plane which ismapped, byx=rcos0,y=rsin0,into the region Rbetween x'+y?=1, x*+y"=4, and inside x'+y?= 2xHence calculate are 6,Find afirst-quadrant region iotheuo-plane which maps into theregion Rdefined by15x74y°S4,y20,ifthemappingisx=u?~o*,y=2ue.Calculate Se» transforming totheuo-plane, and check your result bycalculating thegiven integral in terms ofpolar co-ordinates. ulate tOX—Y)| axdy,wi isthetriangle boun xe 7.Caleulat[fool5212)| axdy,whereRisthetriangleboundedbyx=,x+y0,andx~2y=2,Usethetransformation u=2x~y,v=x~2y. 14 UNIFORM CONTINUITY 531 definedby0<x51,butitisuniformly continuous onthesetSdefinedbyx=1.The first assertion follows from the fact that ifxo>0 and 4ischosen sothat |t-3]<« ifkx)<6, x” x0 thevalue of§must approach 0asx»->0. Ontheother hand, forthesetS defined byx21 we can take 5=€, because ifxand Xobelong toSand |xxe <ewehave k-2|=atlsixal<e x xl me ‘The essential theorem about uniform continuity will now begiven. THEOREM Y.Suppose Sisaclosed and bounded point set, and suppose the function fisdefined and continuous ateach point ofS.Then fisuniformly continuous onS. Proof. We make use ofthe Heine-Borel theorem. Suppose €>0. Ifx'is any point ofS,thedefinition ofcontinuity assures usthat there issome positive number hsuch that |f(x) ~f(x’)| <e/2 ifxandx’belong toSand |x~x'| <h. The size ofhwill usually vary asx’isvaried. Now consider the open interval x'~(h/2) <x <x'+(h/2). When x’varies over S,thecollection ofallthese open intervals covers thesetS.Bythe Heine-Borel theorem ($16.6) afinite number of these intervals suffice tocover S.Let the centers ofthese intervals bedenoted byXi.-..5» and letthecorresponding values ofhbehi,..., teChoose 8as the smallest ofthe numbers hy/2,..., hy/2. We shall show that this 5will serve asrequired inthedefinition ofuniform continuity. Suppose xand xobelong toS and |x~xo<8. Then xobelongs toone ofthefinite setofopen intervals, say the fone with end points x,+(n/2), $0that \to~x|<h/2. Now [x=]Sxaolfron<8+ But654sandsolx|<hsTheinequalities satisfied byJto—|and|x~xi guarantee that Wx fON<§ and[f(x)-fO0]<5- Therefore [f(x) ~f(a] 5f(x)~f(x)|+[fx—f(40|<e. This completes theproof. The definition ofuniform continuity can beworded soastoapply to functions ofmore than one variable. Itismerely necessary towrite thecondition involving ¢and6intheform“If(P)~f(P,)|<€wheneverPandPoarepointsof Ssuch that d(P, Po)<8." Here d(P, Po)isthedistance between Pand Po. 18a ‘THEINTEGRABILITY OFCONTINUOUS FUNCTIONS 539 ‘The following theorem istheconverse ofTheorem I. THEOREM I.Iffisintegrable on(a,b],and if€>0, there isapartition with upper and lower sums such that S~s <e. ‘The proof isleft asanexercise. EXERCISES 1.Prove (18.1-4) inExample 2bythefollowing steps: First, 1=3; next, J3; and finally, I=J=3.Explain each step fully. 2.Suppose fisdefined asfollows: f(x)=2 if0Sx<1, f(I)=0, f(X)=—1 if 1<x <2, fQ)=3, f(x)=0if2<x<3, {G)=1. (a)Provethatfisintegrable, usingan argument something like that inExample 2,butwith sixsubintervals. (b) Find thevalue ofJ2f(2)dx,using anargument likethatofExercise 1. 3.Suppose fisdefined bytherequirement that f(x) =2ifxisarational number of theform p/28, where pcan take onallthevalues 0,+1,+2,... and qcantake onallthe values 1,2,..., and f(x)= |forallother values ofx.CalculateIandJforthisfunction ‘ontheinterval (0,2},and thus prove that fis notintegrable. 4.Prove Theorem I 5.Iffisintegrable, soistheabsolute-value function [f(x)|. Prove this byshowing that ifs,$refer tof,ands',$"refer toif],then S’—s"= S—s.Then useTheorem I 6.Iffisintegrable over the inverval [a,b], itisalso integrable over any closed interval of(a,6}.Prove this, using Lemma Iand Theorem I. 7.Suppose a<b <c and that fisintegrable over [a,b]andalsoover{b,c}.Prove byTheorems Iand IIthat fisintegrable over (a,€} 18.11 /THE INTEGRABILITY OF CONTINUOUS FUNCTIONS Every continuous function isintegrable. We state this inaformal theorem. THEOREM Ill. Ifafunction fiscontinuous ateach point of[a,b], itis integrable onthat interval. Proof. The argument hinges onTheorem Iandontheuniform continuity of thefunction. Suppose €>0. Choose 6sothat = fea" « 1YO) 10)<555 (a8.11-1) ifx’andx"arepoints of(a,b]suchthat|x’~x"|<8. Thismaybedone, since f isuniformly continuous (Theorem V,$17.4). Now consider any partition (80,Xiso: +Xe)Such that allthesubintervals have length lessthan 8,andlets,Sbethecorresponding loweranduppersums.Theboundedness offisguaranteedbyTheorem II,§3.1. Now, intheinterval [x,-1,xi]wecanchooseapointx’so thatf(x’) isasclose asweliketoM,,andapoint x”sothatf(x") isasclose aswe 182 THEINTEGRAL ASALIMIT OFSUMS, 543 chosen sothatx).Sx} x, i=1,...,m. The following theorem isfundamental inthetheory ofintegration: THEOREM VII. Suppose fisbounded ontheinterval (a,b].Then itisintegrable ifand only ifthesums (18.2-1) approach alimit asthemesh fineness |P| approaches 0.This limit isthen thesame astheintegral defined in$18.1 Inorder toprove Theorem VIIitisbest tobegin byproving thefollowing theorem, usually named after the French mathematician J.G.Darboux (1842— 1917), THEOREM VIII. Suppose fisbounded on(a,b],and lets,Sbethelower and upper sums corresponding toapartition P.Then sapproaches Iand S approaches Jas|P|-0. This means that forany €>0there issome 6>0 such that |s-I|<e and |S-J\<e if|P|<s. Ifwegrant thetruth ofDarboux’s theorem, itisrather easy toprove Theorem VIL. Letussuppose that fisintegrable. Now, ifx,-15 x}5.x, wecertainly have m,5f(x}) 5M,and therefore ss3sada-x 08s. (18.2-2) As|P|+0, Darboux’s theorem asserts that sand Sapproach IandJrespec-tively.ButI=J=f2f(x)dx,andsoweseeby(18.2-2)thatthesums(18.2-1)must approach f?f(x) dxas|P|+0. Ontheother hand, ifweassume that thesums (48.21) approach some limit A,this means that allsuch sums liebetween A~€ and A+ if|P|issufficiently small. But, ifwechoose such apartition and keep itfixed, then byvarying thechoice ofxj,...,x wecanbring thesum (18.2-1) as close asweplease toeither sorS.Consequently wemust have A-eSs and SSAt+e But then S—s 52¢. Since €can bechosen assmall asweplease, weknow by ‘Theorem Ithat fisintegrable. This concludes the proof ofTheorem VII. Westillhave toprove Theorem VIII. This proof isabitintricate indetail. Letusfirst establish thefollowing fact: IfSistheupper sum corresponding toa partition P,and ifS’isthenew upper sum corresponding toapartition P’ obtained from Pbyinserting asingle additional point, then S~S'S2C\P|, (18.23) where Cistheleast upper bound of|f(x)] on[a,b].Toseethis letussuppose fordefiniteness that thenew point €isbetween x9andx,,andusethenotation as intheproof ofLemma I,$18.1. Then $= S'= Mi(xs— x0)~M(E—x0 —Mil8. 574 INFINITE SERIES cn.19 Example 1.The series Tebeleet bees (19.2-1) isdivergent. Itiscalled theharmonic series. We prove thedivergence byshowing that thepartial sums arenot bounded. Let 1 seattlegh Then =s¢— 4 sb 1 TTcreeee since 4,1 tton-delnei*n+2* *3n 7"In 2 With sax>5+4forevery n,itisplainly impossible for{s,}tobebounded. We have SL sahse sth= 2se>sethok and ingeneral n+2 sett? For aseries with negative aswell aspositive terms, convergence ofthe series isnotguaranteed byboundedness ofthepartial sums. For instance, the series I-1+t-1+t-- isnotconvergent, yetforitspartial sums wehave 5,=1ors,=0,depending on theoddness orevenness ofn.Thus these partial sums arebounded. THEOREM I.Let 3a,and 2b, betwo series ofnonnegative terms, and suppose that, forallvalues ofnafter some fixed index N,itistrue that dy=by.Then iftheseries Ebyisconvergent, soisSag, and iftheseries Ea, isdivergent, so isDb. Proof. Indiscussing convergence ordivergence wemay drop theterms with index less than N.Then, for any n>N, Gy+ayy++++OgSby+buytoo+De The proof ofthetheorem isanimmediate consequence ofthisinequality and Theorem I. 584 INFINITESERIES ch.19 (19.3-2), which isthedifference ofthetwo convergent series Jody )o(etade...(tegtate)- (legge): Inthecase oftheseries (19.3-1) each oftheconstituent series lehelees, deledee- isdivergent. Proof ofthe theorem. Suppose that Zu. isabsolutely convergent, with M=ual. Then ua)+lua]+++itSM, nomatter how large nis.Now consider apartial sum oftheseries Za,, say a,+++++a,. Since each a,isapositive term somewhere intheseries Euj,the terms dy,..-, dmalloccur inthesum |u|+++++|u|ifnissufficiently large.But then we see that y+ ayt +--+ an SM. Itfollows byTheorem I(§19.2) that theseries 5a,isconvergent. Inthe same way wesee that b,+-+++ba=M, since each b,isaterm somewhere inthe series E|us|. Thus theseries Eb, isconvergent. Nowsupposethat,inthesum+++++us,thenumberofpositivetermsis Paand the number ofnegative terms isqy.Then Uytootig=(ayt=+ay.)(byt +bg) (19.3-6) Inthecase ofabsolute convergence weletn> and obtain theresult Dun zo-Z,bm Ttmay happen that there areonly afinite number ofa’sorafinite number of b's, ofpossibly none ofone kind orthe other. Inthese cases the series isof course absolutely convergent ifitisconvergent atall,since itsterms from some point onward areallofone sign. Let usthen consider thecase inwhich there are infinitely many terms ofeach sign, sothat p,and q,>% asn+. Let SEUHEyApaH dyBy,=Byte+ay sothat (19.3-6) becomes 5»=Ay,~By. Wealso have ust ltgl=(ayo ap.)+(BietBQ) =Ay+Bu 09.3-7) Now suppose that theseries ¥u,isconvergent, and consider theseries ¥dy, bp. Ifeither ofthese latter series isconvergent, soistheother, byvirtue ofthe relation s,=Ay,~Ba.Forinstance, ifEa, isconvergent, Bg,approachesalimit, since s,and A,,each approach limits, and By,= A,,~%» But tosay that 198 MULTIPLICATION OFSERIES 601 The next question is:How dowemultiply thetwo series together togetanew series? Proceeding just asthough theseries were finite sums, wemight write down the following scheme, which arises bymultiplying the second series successively byeach term ofthefirst series: es ee oe fe bP <b? thet dete i es There will beaninfinite number ofrows, each row being aninfinite series. But ‘weobservethatthereareonlyafinitenumberoftermsofeachdegree,sothatifwecollect together terms oflike degree, weobtain forthefirst few terms Ithe-bP- eet. (19.6-4) tis clear that there ishere asystematic process, butitremains toprove that the process gives aseries which has asitssum theproduct ofthesums ofthetwo original series. There isageneral theorem which justifies theprocess. THEOREM XVII. Suppose that each ofthe series Euy Sv, isabsolutely convergent, with sums Uand Vrespectively: U=wtutuste, (19.6-5) Vetototortes-. (19.6-6) Let Wo= Woto, Wi=Hots +ito, and ingeneral Wa=Mota +ideay Hes #Mabe (19.6-7) Then theseries ¥wyisabsolutely convergent, and itssum isUV: UV =wot wit wrter (19.6-8) Moreover, any infinite series which has asitsterms theproducts ww, (iand j20)arranged inany order, each product occurring once and only once, is absolutely convergent, with sum UV. Proof. Let usconsider thearray Mybe Mobi °° Mate °° Mydo Wyby = Made wee oe w Uybo Wad) °° Wade 608 INFINITESERIES cons 3.Show that 5(1/n)log{1 +(1/n)] isconvergent, 4.Show byTheorem II,orotherwise, that3-2 I/(log n)*isdivergent forallvalues ofc, 5.Expresslogn)"**asapowerofn,andusetheresulttoshowthatGees is convergent. 6.Examine each ofthefollowing series forconvergence ordivergence. @Tore ©Dew(4). Tes @Seeger 7.Show that 24d 1ee en 35---Qnt+) Vael () Classify the values ofxaccording towhether the series lina" isconvergent ordivergent. 8.Findallvaluesofxforwhichtheseries51-3-7-CR—D1 isconvergent. 9.(a)IsEsinx{n-+(1/n)] absolutely convergent? Isitconvergent? (©)Show that 3sin’x{n+(I/n)] isconvergent. (6)For what values of@is£(—1)*(1/n) cost@/n) convergent? (a)Is3{1~cos(zin)} convergent ordivergent? 10,Discuss theconvergence ofeach series, classifying thevalues ofxinto those for which the series converges and those for which itdiverges S (ype bedeQn=1)_2 (yt @Sew ami) Syi2((0= DY?ae ©Gay 5(n+ on ©2a 11.Showthat8210) convergent ifa>eanddivergent if0<a5. 12,(a)Ifx= 1-$45----—112n), show that x++log 2asn+, byusing the definition of Euler's constant (sce Exercise 6,$19.21). HINT: Show that x= Cun ~Cx+log2inthenotationoftheexercisejustmentioned. (©)Prove that thepartial sums oftheseries Le}-beed-tes—- approach +2asn+», Make useofEuler's constant. 13.Prove that, ifue>0and3usisconvergent, soisEh 14,Show that theseries ¥log(n sin(1/n)) isconvergent 204 INTEGRATION OFSEQUENCES ANDSERIES 623 (20.4-2), integrating from —rto1,The result is [Ben [oat[xderit foracre. (204-6) Now dx ' l+r[Ben-twaoftog! + nett[ive ST The latter expression isequal to0ifn+1iseven, and equal to2r""/(n +1) if n+1is odd. Therefore (20.4-6) becomes log{tt=2(r+5+5+--). ‘This result was obtained byadifferent method inExample 1,$19.6. ‘Aswas suggested bythisexample, Theorem IVhasimportant applications in deriving certain series expansions from other series expansions byintegration. The conclusions (20.4-1) and (20.4-2) ofTheorem IVmay befalse ifthe convergence isnotuniform. This isillustrated byExample 3,§20, inwhich the convergence isnot uniform. Tohave (20.4~1) and (20.4-2) itissufficient tohave uniform convergence; but uniform convergence isnot anecessary condition in allcases.Suppose, forexample, thatwemodify Example 3of§20bytakingtheheight ofthetriangle inthegraph ofy=f,(x) tobe2Vn instead of2n.The convergence isstill nonuniform, and limy-. fa(x) =0when 0=x=1.But now ' 1f(x)dx= Jful)dx= and so(20.4-1) istrue inthis case. EXERCISES 1Iff=5S27showthat[sexdx=BP susttyyourreasoning sin3x|sinSx,sin7x ~ 2itfay=AEMBSEOTE...tndaseriesfor[™f(x)dx. 3.Iffalx)=nxe™ andf(x)=lim, f(x), show that thesequence converges nonuniformly ontheinterval 0x51,and that [00a4tim[foeae 4.IE(2)=limfax),wherefala)=72find fsaeandtimffoea. m2 DIFFERENTIATION OFPOWER SERIES 635 THEOREM VIII. Ifafunction f(x) isdefined byapower series (21.2-1) with positive orinfinite radius ofconvergence, thecoefficients arerelated tothe function bytheformulas a,=co (21.2-6) This means that thepower series istheTaylor's series ofthefunction. ‘The formulas (21.2-6) are established bysetting x=0 inthe successive series forf(x), f(x), f"(x), ete. (see (21.2-1)(21.2-4)). Itisclear byinduction that theleading term inthe series forf(x) isnla, and (21.2-6) isadirect consequence. ‘THEOREM IX. Suppose that two power series areconvergent and have thesame sum forallvalues ofxinsome interval |x|<r: >," =Db" -r<x<r Then a,=b,foralln. This theorem iscalled the uniqueness theorem for power series. Itisa corollary ofTheorem VIII. For, letf(x) bethecommon sum ofthe two series. Thenby(21.2-6) weseethata,andbyarebothequalto£0),andtherefore equal toeach other. Example 1.Consider thefunctions Jo(x), J\(x) defined asfollows: xo xt ye Jd)=~Gitgia HEDGhat QL x xt x"Wont tCMagpat e128 Show that they aredefined forallvalues ofx,and that Jdx)=-Jx). (21.2-9) The function Jo(x) iscalled the Bessel function oforder zero offirst kind; Ju(x) iscalled theBessel function oforder one offirst kind. These and other varieties ofBessel functions areofgreat importance because oftheway they arise inmany kinds ofphysical problems. Both series areconvergent forallvalues ofx,asisreadily verified bythe ordinary ratio test(Theorem XIII, $19.4). Toverify (21.2-9) wewrite theseries for Jox) and J\(x) inthe forms - en SeWad=DCWpO) =BODieaRT 212 DIFFERENTIATION OFPOWER SERIES 637 Fromthisrelation wemaydetermine a»,as,a,...successively intermsofanarbitrary ao;likewise as,as,@;,... aredetermined interms ofanarbitrary a).We have 133 3-3 yaTasman a=FFs=0, whence a; ay=ay=+++=0. Fortheeven subscripts, a2= ay a= —Jay ag= +hay anan2@—3q22 FS dae Clearly none ofthese coefficients iszero ifap#0. Now Qn=-39Qn=5)- 1CY-CD Aang duty=OO AE ay+aa when like factors are cancelled from either side ofthis relation we find _ 3 O02 GFHGR (21.2-11) What wehave done thus farshows that ifthere isasolition of(21.2-10) in theform ofaseries ofpowers ofx,thesolution can bewritten 3243pay emg yoate—tafeige type tai: Moreover, the work shows that this really isasolution within the interval of convergence oftheseries, provided there isaninterval ofconvergence. Now the infinite series « 5 .Gn Dean=9* @12-19 isconvergent when |x|<1, asmay readily beverified. Thus wehave found two linearly independent solutions ofthe differential equation (21.2-10): the poly- nomial x~x° and theinfinite series (21.2-12). The coefficients agand a,are arbitrary. EXERCISES 1.The Bessel function oforder moffirst kind (manonnegative integer) isdefined Pa ce tem Jo00)= 30" mr2) Showthat(a)Jax LHEY, O)May, ©HA)= a4 ABEL'S THEOREM 645 series isconvergent when x=R,wecanchoose Nsothat 6<agR™ +OggR™+--++dgaR™™* <€ ifN=m and 0Sk (this isjust theCauchy condition forconvergence). This means, inour present notation, that (21.4-2) issatisfied with m=~«, M=e. Applying thelemma, and noting that —¢S~evo, evo e,weseethat thecon- clusion (21.4-4) ofthe lemma yields (214-6). This completes the proof as regards 0=x3R.The case ofconvergence atx=—R isreduced tothefirst case byconsidering g(x) =f(-x) atx= R. Example 1.Ifthebinomial series (19.5-S) converges atx=1,itssum is2". This assertion may bejustified asfollows: We proved the validity ofthe binomial series expansion for(1+ x)" when |x|<1. Therefore, byTheorem XI, iftheseries converges atx=1,itssum there is fim(1+x)" =2". ‘This always happens ifm>0, bywhat was established inExample 4,§19.4, It may beshown that the series converges atx= 1if-1<m, but diverges if mS ~I(see Exercise 4,$19.5). Next weshow how Theorem XIpermits ustoextend theresult ofTheorem V,$21.1, with respect totheintegration ofapower series. Ifthepower series $02)=5aax® (214-7) isconvergent when |x|<R, then * =F fe peer[[tooae-> 454 (21.4-8) provided theseries ontheright in(21.4-8) isconvergent, irrespective ofwhether ‘ornot theseries in(21.4-7) isconvergent atx=R.Ofcourse, if(21,4-7) isnot convergent atx=R,theintegral in(21.4-8) may beimproper attheupper limit. ‘The proof oftheforegoing assertion issimple. If 0<b <R, wehave . = [soa Seo by(21.1-4). Then, provided theseries in(21.4-8) isconvergent, wehave "*=FGepert tim[foxdx=S585 RO, byTheorem XI. Since itisalso true that imf°sox)ax=["fo)ds timf°400dx=f°fox)a, (21.4-8) isproved. 2 PRELIMINARY REMARKS 655 with finite limits, inwhich thefailure tobeanordinary “proper” integral arises from thebehavior off(x) either asx>aorasx+b,butnotboth. Thus, iff(x) is integrable on(a,c) foreach csuch that a<c<b, butisnotintegrable on[a,b), wesaythat theintegral (22-2) isimproper atx=b.Sometimes wesaythat f(x) hasasingularity atx=b.Wethen define theintegral (22-2) asthelimit tim[sexax ifthelimit exists. The terms convergent and divergent areapplied totheintegral according asthelimit does ordoes notexist. Similar definitions aremade for integrals ofthesecond kind which areimproper atthelower limit ofintegration. “dxf __do__ f'_togx ; Examples,[8 [~=H dx tthe lenos Ieee [ata aeimeroper upper limits, and “(oe!)iax ['han [BE areimproper atthelower limits. Aswith infinite series, itisimportant tobeable totest animproper integral forconvergence ordivergence. There arecertain analogies between thetests for series and tests for integrals, which we shall point out aswe proceed. In practice, however, wedonot need asgreat avariety oftests forintegrals aswe dofor series. Just ascertain functions may berepresented byinfinite series whose terms depend onavariable, socertain functions may berepresented byimproper integrals whose integrands depend on aparameter. As examples, we cite particularly thegamma function I(x), defined by T=flee ‘dt,O<x, integrals oftheform fy=[e*Feat, which areknown asLaplace transforms, andfunctions defined byintegrals of cither ofthe forms EO54)sis 2f v2f(0)sinxtdt,yzJ,£1)cosxtdt, which areFourier transforms. Laplace and Fourier transforms areofgreat importance, both theoretically andpractically. 4 POSITIVE INTEGRANDS. INTEGRALS OFTHEFIRST KIND 657 For theproof wenote that theconvergence ordivergence oftheintegrals is notaffected ifwereplace both lower limits byx=c.Wethen have, ifx>c, Jf(ydt={g(t)dt; the conclusions ofthe theorem now follow atonce from Theorem I. Example1.‘Theintegralftsisconvergent, bycomparison withthe|, View integral [~4»whichisconvergent, asmaybeshowndirectly fromthe definition; for(1+x’)? <x? when x>0. “dx tedi Example 2.Theintegral{”—-“Esm isdivergent, bycomparison withthe integral[”542,whichisdivergent Toseethedivergence ofthesecond integral, note that *dt c >, fpeelostnseasx Now 1< 1Tex “Fx” when x>0, for this inequality isequivalent to 1+x?<(1+x)= 1+3x +3x7+x°, which isobviously correct ifx>0. Thus thefirst integral must diverge, byTheorem II. Toavoid troublesome details ofworking with inequalities inpractice, itis often convenient touse thefollowing theorem rather than tousethecomparison test directly. THEOREM IIL. Suppose Jzf(x) dxand feg(x) dxareintegrals ofthefirst kind with positive integrands, and suppose that thelimit tim£22. p (22.1-3) soemBCX) exists (finite) and isnotzero. Then either both integrals areconvergent, or both are divergent. This isproved inexactly thesame manner asweproved itscounterpart for series, Theorem III, $19.2. Example3.Theintegralf“tS isconvergent. Toprovethis,observe thatforlarge values ofxtheintegrand isabout thesizeof1/(2x"). More exactly, 24 POSITIVEINTEGRANDS. INTEGRALS OFTHEFIRSTKIND 659 . . dx Example 6,Consider theintegral|7psp"wherep>0.Inthiscase,we know that logxincreases moreslowlythananypositivepowerofx.Weapply Theorem IVwith f(x) =(logx)’, g(x) =x?.Then, using l'Hospital’s rule, £0)ogg= fi 1 . HeeG)”2ogxy~HEpowsyCI) This isthe same as x limpion x After acertain number ofapplications ofI'Hospital’s rule wefind that tim£2) 4.00, tea) (One must consider separately thecases inwhich pisorisnot aninteger.) Theorem IVthen assures usthat thegiven integral isdivergent. The student will note that wehave notdeveloped any analogues oftheratio tests of$19.22, Inthe analogy between series and integrals there isnosimple way offormulating acounterpart ofaratio test, because atypical value f(x) of theintegrand has noimmediate successor, inthesense that dys, isthesuccessor ofay. One other difference between infinite series and improper integrals ofthe first kind isworth noting. Ifaseries isconvergent, itstypical term a,approaches zero asn>, But ifanintegral fzf(x) dxisconvergent, itdoes notnecessarily follow that f(x) +0as x2, See Exercise 8,and Example 2,§22.3. EXERCISES 1.Test thefollowing integrals forconvergence ordivergence, using Theorem IIand theknown facts about fz x"dxfora>0 ©foie ©|ee ~_ xdx res » [28 @ [Yhlames OILS *x42 ear ta ©[Ata ny)[tet 4ae, ©seep ©anon 2.Establish thefacts about thevalues oftheexponent pforwhich theintegral fSaiSea¥ isconvergent (a>1.ThenuseeitherTheorem IfoFTheorem IIItotestthe following integrals, using theforegoing integral asastandard, with anappropriate value ofpineach case°de ata of —=*— wo . nn INTEGRALS OFMIXED TYPE 665 tothose of§22.1, ormay bereduced tointegrals with +2asalimit of integration, bythesubstitution x=—u. Example2.Consider fEAS:Weseparatethisinto *xx *_xdx (a)[FE and(oy[SH ‘The integral (b)isconvergent, since Kea wees ifx>0, and fexe™* dxisconvergent (Example 4,§22.1). The integral (a)isalso convergent; for, as x», e*0, and theintegrand behaves like x”. ‘One may set x=~u, and thus get fafo- [tee-[a etx Jeeteu loeu Inthe transformed integral the integrand behaves like uw? as w—>+=, The original integral hasthus been shown tobethesum oftwo convergent integrals offirst kind. EXERCISES: 1.Examine each ofthe following integrals astoconvergence ordivergence, giving a complete analysis oftheconvergence ordivergence ofeach oftheconstituent pure types. se “dx o[Wee ®fPe-y -dx >dxolan ef& ©ferae o»fCBSpen _— ode @[verde ©fe ~ dx "dx ©[aaa 9Ucar 2.Proceed asdirected inExercise 1with each ofthefollowing integrals: @[OEP a)[ae ee mp ©fPee ©[ioe ‘sinh etn= sans ©[se 0fae 4.Ineach ofthefollowing integrals theintegrand contains aparameter. Foreach 22 THEGAMMA FUNCTION 667 or ray=i. (2.2.3) There isavery simple relation between thevalues ofthegamma function at xandx+1.This relation isfound bycarrying outanintegration byparts. We start with ratp=[ret Setting u=t",dv=e"'dt,wehavedu=xt*"'dt,v=—e™', 7 rope fjveta=[-vet] +[artears letting T>, wesee that [feta=-timPe"04%[teat (22.2-4) But limT*e"™ =0, im ‘asweseebyapplying 'Hospital's rulentimestoa.wherenisthefirstinteger greater than orequal tox.Therefore, by(22.2-4), Pe+1)= xP). (22.2-5) From this formula and (222-3) wehave successively, r@=1-rM=1 1Q)=2-P@=2-1 T@)=3-TG)=3-2-1 T(S)= 4-14) =4-3:2-1 Ingeneral wecan write: Pint t=n! 22-6 or Tin) =(n- Dt (22.2-71) Intheordinary elementary sense n!isdefined only ifnisapositive integer. But since T(x) has been defined for every positive x,wesee by(22.2-7) and (22.2-3) that itisnatural tomake the agreement that 0!=1.This iscustomarily done. ‘The gamma function gives usaconvenient method ofinterpolating between the values ofthe factorials n!,and this isone ofthe primary reasons forthe importance ofthegamma function. Just now weshall take forgranted that I(x) isacontinuous function, though wecan prove this later on($22.5, following 668 IMPROPER INTEGRALS cn.22 Theorem VID. Infact, M(x) has continuous derivatives ofallorders, and is analytic. The derivatives are found bydifferentiating with respect tothe parameter xunder theintegral sign in(22.2-1), Recall that 4 sn greet46)Jogtvetlelatt r=ete,2)=logtet"=1"logt ‘Thus Peps[dog neat (22.2-8) This isthe same procedure asthat given in(18.5-2) for proper integrals dependent on aparameter. For improper integrals further justification is required; the problem ismuch the same asthe problem ofjustifying the differentiation ofaseries term byterm. Wereturn tothis problem systematically inTheorem X,$22.5; forthepresent letusproceed with our study ofthegamma function. We can differentiate asecond time, obtaining r=i1"(logte"dt. (22.2-9) ‘The integrals forI"(x) and I"(x) areconvergent integrals ofmixed type, with singularities oftheintegrand att=0,and thesame istrue fortheintegrals giving allthehigher derivatives (see Exercise 1). Itisclear from (22.2-9) that I"(x)>0, and therefore the curve y= T(x) is concave upward forallx>0. We also see that I(x) >0, P(0)= PQ)=1.From these facts weseethat I(x) has just one minimum value, and that this occurs for avalue ofxbetween 1and 2.Tosee how I(x) behaves asx-+0, weobserve that ifwe integrate only from 0to1in(22.2-1), the result isless than T(x). Furthermore, e“isadecreasing function, sothat e~*>ef 0 <1. Therefore etetaset [reared Tay>[tera>et[eta=ak Itfollows from this that P(x)-»-+e asx-+0°. From the information which we have now collected itispossible toshow thegeneral character ofT(x) ona graph. Weleave itforthestudent toprepare such agraph forhimself. ‘The formula (22.2-1) does not define afunction ifx30. Nevertheless we can define T(x) forcertain negative values ofxbyusing formula (22.2-5). If =1<x <0, then 0<x+ I,sothat F(x-+1) hasameaning already defined. We then define I(x) byrequiring rey=FAt0. 222-10) ‘Thus, for instance rey =-21d. Now suppose that -2<x<=1; then -1<x+1<0, sothat P(x +1)isalready 22 ‘THE GAMMA FUNCTION 669 defined. We then define F(x) by(22.2-10), e.g, T-)=--). This process can evidently becontinued, sothat weobtain adefinition ofF(x) forallvaluesofxexcept0,~1,~2,—3,..., andtheequation (22.2-10) holdsforallother values ofx. Itiseasy tosee that P(x) <0when ~1<x <0, and that P(x) asx+0 or x-+~ 1°,We leave itforthestudent tostudy thesituation when ~2<x <~1, -3<x <~2, andsoon.Arough graph should beconstructed. Itwillbeshown later thatPl)=Vz(see(22.41-6)); from thiswemaycalculate P(-}, I(-), ete. EXERCISES 1.Show that Jo"t*"(log 1)"e"" dt isconvergent for n=1,2,...if 0<x. 2.Prepare agraph ofy=I(x), showing thegeneral behavior ofthegamma function for x>0 and inthe intervals -1<x <0, -2<x<~1, ete. 3.Show thatP(x)=26°ute" du ifx>0. 4.Calculate thevalue interms ofVr of (a)exe" dx,(b)fexte”de 5.Ifa>0, show that fox" Ye“dx =a""P(n). ‘What istheimplied restriction onn? 6.Calculate interms ofVz thevalues of (a)Jee dx, (b)fox dx. 7.Show by(22.2-5) that, ifm= 1,2... Pny=PPSOndys, ‘Asaconsequence show that Vat@n +1)=21m +Mn+ D, and Vat Qn)=2'Teyrin +d. These formulas suggest theconjecture that perhaps ValQx)=PTE +) notmerely forx=.n+4 and x=n,where nisapositive integer, butforallx>0. The conjecture iscorrect, ascanbeproved bylaterdevelopments (seeExercise 8,$22.7). owthat23222R=D_ Pint) &ShowMat get Valent) 9.Derivetheformula 1(x)= ['(log+)” "dubyputingw=e*in20.2-1). Then setu=0%,where a>0, and sofind thevalue of ff(od) ovtv, where x>0. 610 IMPROPER INTEGRALS ch.22 10. Usilize the results ofExercise 9toshow that (a)[(ete) a=vie, arent) iz ©(oat) =VF 22.3 /ABSOLUTE CONVERGENCE Animproper integraloffirstkind,J¢f(x)dx,iscalledabsolutely convergent iftheintegral f|f(x)| dxisconvergent. Exactly thesame definition isapplied to integrals ofsecond kind, and tointegrals ofmixed type. The switch from f(x) to f(2)) corresponds exactly totheswitch from ¥a,to¥|a,|indefining absolute convergence oftheinfinite series. Ifanintegral isconvergent, but notabsolute- lyconvergent, itiscalled conditionally convergent. The following theorem corresponds toTheorem IX,$19.3: THEOREM V.Iftheintegral f=|f(x)| dxisconvergent, sois[5f(x) dx. Inother words, ifanintegral isabsolutely convergent, itisconvergent. Proofofthetheorem. Firstofallweobservethat 05fGx)|- foo $f). 23-1) Both parts ofthis double inequality may bechecked byconsidering separately the cases when f(x)=0 and f(x)<0. Now letg(x)=|f(x)|~f(x). Since S2|f(2o|dxisassumedtobeconvergent, theintegralwith2{f(x)|asintegrand is also convergent. Then, by(22.3-1) and Theorem II,$22.1, weseethat Jg(x) dx isconvergent. Butf(x) =|f(x)|~ g(x), and therefore f¢f(x) dxisconvergent, for sums and differences ofconvergent integrals are convergent, asmay beseen directly from thedefinition ofconvergence. (What theorem about limits isused atthis last step intheargument?) The theorem and itsproof apply tointegrals ofthesecond kind; only the limits ofintegration have tobechanged. Totest whether anintegral isabsolutely convergent, wecan apply the methods of§§22.1, 22.11, since the integrand |f(x)| isnever negative. Ifan integral isconditionally convergent, thedemonstration ofitsconvergence is usually amore delicate matter. Many oftheinstances ofpractical importance canbehandled bythefollowing theorem, which isanalogous toDirichlet’s test for series ($19.7). ‘THEOREM VI. Consider animproper integral offirst kind oftheform [leorma, (22.3-2) 23 ABSOLUTE CONVERGENCE on where thefunctions and fsatisfy theconditions: (@) 40 iscontinuous, (0) 50, and lim(1)=0, (b){0 iscontinuous, and theintegral Fay=[pode 23-3) isbounded forallx=a.Then theintegral (22.3-2) isconvergent. Proof. We note that F'(x)=f(x).Therefore, integrating bypartsandnoting that F(a) =0,wehave ffeorma=[ sor@a=se@Fe-f' eorma e234 Let ussuppose that Misabound for |F(x)|, that is,|F(x)|SM. Then |O(2)F(x)| S|(x)|M, andsoo(x)F(x) +0asx,since$(x)-+0 byhypo- thesis. Itthen follows from (22.3-4) that (22.3-2) isconvergent, provided wecan show that theintegral fieorma 235) isconvergent. This integral isinfact absolutely convergent. For, since #(t) 0, |OFO|=- HOIFO|S ~Mew. Itisthen enough toshow that [-Mé"(t)dt (23-6) isconvergent; then (22.3-5) will beabsolutely convergent, byTheorem II,§22.1. Now J-Mor@at=~Macy+Moca)»Moca) asx2, bycondition (a). Thus (22.3-6) isconvergent, and the proof is complete. Example 1.The integral fe(1/t)sintdt isconvergent. (There isnosin- gularity atf=0;seetheremark inExample 2,$22.11.) Here we take o=f f=sine ‘Then Fw sintdt=1~cosx. 5 224 IMPROPER MULTIPLE INTEGRALS. FINITE REGIONS. 673 2Showthat[S22deisconvergentif0<p<2,andabsolutelyconvergentif 1<p<2 3.Showthat["1=£982axisconvergentif1<p<3.Isitabsolutelyconvergent forany ofthese values ofp? 4.Showthat[828050829 gejsconvergent if0<p<4.Isitabsolutelycon- Vergent forany ofthese values ofp? 5.Show that ficos(x*)dxandfexcos(x‘)dxareconvergent. Notethatthein- tegrand inthesecond integral isunbounded. 6.Show that 0 sinx 2ts re ee Usethisresulttoprovethat[”"2dxisnotabsolutely convergent. 22.4 /IMPROPER MULTIPLE INTEGRALS. FINITE REGIONS Consider first thecase ofanintegral Jfsesaa, 224-1) where Risaclosed bounded region, fiscontinuous inRexcept atone point (x,yo),and the behavior offatthat point issuch that thefunction isnot integrable over Rinthe sense of§18.6. The cases ofgreatest practical im- portance are those inwhich f(x, y)either becomes infinite orhas afactor which becomes infiniteas(x,y)>(Xa,Yo,€-8-, fox=4 orfonyy=2B where re(ena +(y~ yo (224-2) Todefine what wemean bythe convergence ordivergence ofthe integral (22.4-1) weproceed asfollows: LetR'bearegionderived fromRbydiscarding asmallregionARhavingthepoint TP (Xe,Yo)initsinterior(R’istheshadedportionofRinFig. I)OY 181).Norestriction isplacedontheshapeofARexcept 7 (YthatitbeaRiemann region inthesensedefined in$18.6. (7 Ofcourse, wealsoassume thatRisaRiemann region, 7 Let dbethemaximum diameterofAR,thatis,thedistance (z0;m0) between twopoints ofARwhich areasfarapart asitis Fig.181. 24 IMPROPER MULTIPLE INTEGRALS. FINITE REGIONS 675 affected bythelocation oftheaxes. Then, with theuseofpolar co-ordinates, 1™Oo(“rtrdp=2mchmarm JfFaaq[a0[reed=2%fe3°") s Since m<2 we see that tim[fean=2%co", rr) r 2-m Thustheintegral(22.44)existsifRisacirclewithcenterat(xo,yo)andm<2,For regions ofother shape, and for(Xo,yo)located ontheboundary ofR,the difficulty can easily beresolved interms ofthecase wehave treated. Example 2.The integral Jfxa (24-5) isabsolutely convergent; for|x~xo=r,by(22.4-2), and so Es]<i. Pir whence, bythe comparison-test principle and the result ofExample 1,the asserted result follows. Similar considerations apply toimproper triple integrals. Improper multiple integrals inwhich the integrand has just one singular point inthe region of integration occur typically inthetheory offorce fields governed bytheinverse- square law, e.g., gravitational orelectrostatic fields. From apurely mathematical point ofview the study ofsuch fields belongs towhat iscalled potential theory. Let Rbeabounded closed region in3-space, and letitbefilled with mass of density u(x, y,z).IfQisthepoint (xo, Yo,Zo),and =[xx0)+(y~yo?+(=20), theNewtonian potential atQproduced bythetotal mass is -{ff# $620.9020=fJfeav. 246) and the gravitational field atQisavector Fwhose first component (inthe x-direction) is Fixf[fu*seav, (24-71 Sav, with similar formulas forF,and F).IfQisapoint ofRthese areimproper na IMPROPER MULTIPLE INTEGRALS. FINITE REGIONS 67 6(a)Istheintegral{{—<—4Aomeconvergent ordi hereRJ)ta=iptesymcomersentordivertent,whereRisthe region 27+? 1? (b)What iftheexponent $isreplaced bym? For whatvi is sot certainly convergent? (©) ForwhatvaluesofpVier ertainlyconvergent’ 7.Let R,f,AR, dhave themeanings used inthediscussion of(22.4-3). (a)Let Wbeasubregion ofRwhich contains (xo,ya)and allthepoints ofRinsome neighborhood of(4ayo.Show thatfff(xy)dA isconvergent ifandonly if {f10.9)4A isconvergent, andthen Jfrenaa= ffranaasffseman, where R~ Wistheregion which results byremoving Wfrom R. ‘SUGGESTION: Consider JJtenaae fftoayaa, where AR issosmall that itis contained inW. (©)Deduce fromtheresult in(a)thatlimJff(x.9)dA=0if{Jf(x,y) dAisconvergent. 8.Suppose g(x,y) isnonnegative and continuous inRexcept at(x¥).Let{W,} bea sequence ofsubregions ofthetype ofWinExercise 7a). Suppose W;contains Ws, WscontainsWs,etc.,and thatdj-»0.asn-»,wheredaisthemaximumdiameterofWx.Finally,itR= R— Wa, assume that te[fetenan ean= Prove that{f(x )dAisconvergent, withvalue ISUGGESTION: Foragiven m,choose AR sosmal that itiscontained inWa. Then choose msothat Ws iscontained inAR. Now show that Jfwasda-1= ffeaamdasff etsyaa-1 and that Jfemndasffetmda~ffacayaa wan m a From here itieasy tocomplete theproof. Write outthewhole argument carefully 2s FUNCTIONS DEFINED BYIMPROPER INTEGRALS 687 We regard xasfixed. This last integral isconvergent, byTheorem VI,$22.3. Now itcanbeshown that theintegral defining G(a) isuniformly convergent when @=0, sothat Giscontinuous for such values ofa,byTheorem VII. Then G(a)> G00) asa-+0°. Now =liAz simG(a)=fimtan*=5 ifx>0. Thus (225-12) isestablished. The assertion about uniform convergence isdiscussed inExercise 20. Itiseasily shown that theintegral in(22.5-12) hasthevalue —m/2if x<0, for theintegral defines anodd function ofx.Thus Fifx>0, [St ae}oitx=o, (22.5-13) mi Fit x<0. From this itis clear that thefunction defined bythe integral isdiscontinuous at x=0. Itmust therefore fail tobeuniformly convergent inany closed interval which contains x=0. EXERCISES 1.1.(a)Let F(x)=Joe"cosxtdt.Assumetheapplicability ofTheoremX,and show that F'(x) =—$2F(). Then find F(x), (b) Bychange ofvariable inthe result of(a)show that *gt =lJFerm, a>[lemeosstat=SyZer™, a>o (©)Bysuitable use ofTheorem VIII, show that theintegral fpte?sinxtat isconvergent uniformly with respect toxforall values ofx,thus justifying theprocedure used in(a). 2.(a) Show that, forall x, [eromrae em bydenoting theintegral byF(x) and showing that F'(x)=-2F(x)whenx>0.The substitution w=x/ isuseful atacertain stage inthework. Explain how you justify the answer when x<0. (b) Deduce from (a)the result *oma?gywh6-2VeRps fle deeVFe™* p>0,an0. 706 IMPROPER INTEGRALS ch.22 2.Examine thefollowing integrals astoconvergence ordivergence: (a)[PLO a,[tog+e)dx A.Suppose f(x) 20, aa=f2"f(4)dx,wherec=xy<xy<x<0++, andx48, Prove thatJZf(x)dxisconvergent ifandonly if©a,isconvergent. 4.LetF(x)=fo"log(t—<°cos?6)do,055x51. (2)Find F’(x) when 05x<1,bydifferentiation under theintegral sign(this isjustified by Theorem XIV, $18.5); evaluate theresulting integral byuse ofstandard tables. Then integrate andfindF(x)=log!*E=*, ateastif052°<1. ()Prove that Fiscontinuous atx=1,byuse ofTheorem VII, and hence deduce from (a)that J@”?logsin0d@=(7/2) log3 5,LetI=fologsin@d0. Show that 12"togsinode=2[~~togcosoe. ‘Then use theformula sin0=2sin(@/2)cos(0/2) todeduce that I=~=log2.6Showthat[”xe"cosbrdx=72Pra(a>0). 7Intheintegral F(x) =fze°*-"dt,assumethatx>0,andmakethechangeof variable u=t—(xit). The resulting integral will beofthe form J(x,u)du.Express f(x) du intheform J6(x, -u)duandinthis way deduce thevalue ofF(x). 8Suppose fef(x)dxisconvergent. Supposealsothat(x)>0,that'(x)is continuous, and 6'(x) 0. Show that {2o(x)f(x) dxisconvergent. 9.UsetheresultofExercise 8toshowthat “£284arisconvergent, to.LetF(x)=2[7ittFindthevalueofFG)foreachx.IsFcontinuousat x=0 I,LetF(y)= fay’e"” dx.Show that F(y)= yforallvalues ofy.Verify that Fo)={,[Zo’e)] arity0,butthatthsifalseity~0.Whatdoyouconcludefrom thisabout theuniform convergence oftheintegral fe"y(3—2y*x)e "dx? 12.Istheequation [fas [[ey-2yre a=[axf°ey-200ay true offalse when a 0? 13. Let f(x,y)=sin(x?+y°). LetRbethesquareregion0x$a,0ySa,andlet Tbetheportion ofthecircular region x°+y?Sa?which liesinthefirstquadrant. Showthat,ifT={ffa,y)dAandJ={Jf(x,y)4A,thenIx/4asaz,butJdoesnot approach any limit 14,(a)LetFex)= [78885dt.ShowthatF°()~F(a)=~2ifx>0,Takefor ranted that F'Gx) canbecalculated bydifferentiating under theintegral. Solve the 72 ANSWERS TOSELECTED EXERCISES CHAPTER 4 $4.3 Pages 103-105 1.x= 814108(x ~3)+54(x-3)°+ 1264-3)"+(x-3), 3.(a)sin?x=x°—}x‘cos2X. Jog(l—x) __,.2f2log(1=X)~3 Pee Pa} cosX(,_ =) =4X (r-¥). xa $45 Pages 112-114 La@t+=; i@es OI 2(0; (©)2;(00; (+e; (Hy. 2)1;Me?Ot40; ©+=; i 7.0)0;(0;@S. Miscellaneous Exercises Pages114-115 6.(a)0;(be.7.Thelimitisa,inthegeneralcase.&1.1.@-+1/(n+2). CHAPTER 5 $5.1 Pages 121-122 1.Sis open. B(S) isthesetdescribed inExercise 2 2.Sisclosed. Ithas nointerior points. 3.Sis closed. B(S) iscomposed oftheline segment y=1,-1=x 51,and theparabolic acy =x7, “15x51, 4.Sisneither open norclosed. B(S) iscomposed ofthepart ofthecurve xy=1inthe first quadrant, and thenonnegative portion ofeach co-ordinate axis. '.$has nointerior points. Itisnot closed. B(S) consists ofSand the line segment x=0,-Isysl. 6.$isnotopen. B(S) consists ofthesemicircular arcy=V4=27, thesegment y=0, ~25xS2, thesegment x=1/n,O0<y =,and thesegment x=0,0<y 51. The points ofthis last segment arein’S. Sisaregion, but C(S) isnot. 7.This setisopen, and therefore aregion. Itsboundary consists ofthey-axis and the curve y=sin(1/x). 8.This set isopen. Itsboundary consists ofthe half-lines x=nz, y=0, and the segments y=0, nz SxS(2n+ I)z,n=0, 21, #2,.... $5.2. Pages 124-125 2No. 88=2Ve. 10,6=Ve2willdo. 12.Yes. $5.3. Page 127 2.No. 4.(a)Yes;(b)no._§.Nodiscontinuities.7.Define f(x,x)=0. 8.(a)No; (b)continuous elsewhere. 9..No. 10.Yes. Define {(0,0,0) =0. INDEX 729 elementary, 16,37 Implicit function theorems, 228,228, 292,364 even 638 365xevaluesofBAR,126,158,177 Improperintegral,889,$77,64,67-674,678Tomogencous, 18-160 9 impicy defined, 13, 222-23, 3 Inequais, 74integrable,$3,385 Infinitedimensionality, 312rltiple-vaued, 2 Infinite serie oa, 638 absolutely convergent, SA2 ofclass C, 384 alternating, 587-588 ofseveral varlables, 16 ondtonally convergent, 3 real analytic, 680 convergent, 867 single-valued, 2 definition, 567 wale of3,283 sivernent 867 Furctioal dependence, 46 comet, 566 Functional natation, 3 harmonic, $74, 78 multiplication of,600-602 Gamma fonction 68,6. Int5noGate,KalFriedrich, 593 Inner prodect270Gauss teat,93,598 dengan 72.78,uaa’nore45 ‘ofabolitevaleofafunction,99Generalizedsoution,48 tscoaienonfmcton,59General solution, 36.37 cofamonotonicfunction,$40 Geometricmean,17-188 ime . Seamusois356 finiofsums,6,$44 Gibbs Joa improper, 59,57,64, 67-674, 68-679%9 rd,386,406,557 Gradient296,297,42-343 iertad,0606557 Gradientfed,506 Imegraltet578 Gram.1-7.27 Intestintoryof388,SSH Gram:Shproces,277 Ineiorpoint,19,321 GeomStet Intermedat-value theorem, 90,544SraviatonalRls,29449-410,678 Intersectionofset12,39 Greatest lower bound8 Imervaofconverges, Greensidentities, 92 tnthelarge,250, Green’theorem, 457,46-468 ate ma,298 Gandeoger's ae,219-230 Invariant 47, ofnearoperator, 327 Harmonic series, 574, 74 Oa transformation, 240, 242-28 Heine-Boret theorem, 523,525, $3 otaiferentasons 33a Heine, Edward, $23 Inverse function theorem, 242, 356-357 Hall, «8 Inverse fonction theory, 337 Hermite polynomial, 639 Inversion theorem, 356-357 Hese, Oto, 354 Inveribie operator, 327-328, 330 Hessian, 353 Trottonal fel, 306 ‘Hlder's inequality, 188 Isomorphic vector spaces, 312 Homogeneous function, 169 erated itera, 6,406,857 omits, 107 Jacobi Catt 178 {dent transformation, 39-241 estan Image, 268 cterminan, 178 Imp fonction concept, 132,22-228 identical vanishing of,264,286