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MATHEMATICAL METHODS for Scientists and Engineers
Donald A. McQuarrie UNIVERSITY OF CALIFORNIA, DAVIS
University Science Books Sausalito, California
University Science Books www.uscibook.s.com
Orde.r Informal.ion: Phone(703)661-1572 Fa.x (703) 661- 150 I Fin:t Printing l1lu.,trator: Mt,rvin Hanson
De,igner: Rohen lshi fl.fanuscript Editor: Jo/111 Murrh.ek Compositor: Windfall Software. 11.1i11g Z:JtX Printer & Binder: £d1rards BrothaJ, Inc. Product.ion Mnnager. Christine• Taylor This book is printed on acid-free paper. Cop) right fl 2003 by University Science Book..; Reproduction or lmnsl::u..ion or any part of this work beyond that pcrmincd by Sect.ion I 07 or 108 of 1hc 1976 United Stutes Copyright Acl without 1.he penni"i(ln of 1he copyright owner is unlawful. Requests for permission or funher infonna1ion ,hould be add.re~scd to Lhe Pennisi.ions Department. Universi1y Science Books. ISBN I -891 389-24-6 ( c lolh cover J. Nonh A mL'Tican Edi I ion ISBN 1-R9 I )89-29- 7 (soft CO\lcr). avai lnble oulsidc of North America only
Library of Congress Ca1aloging-in-Publication Data McQuarric. Donald A. (Donnld All~n) Mmhematical methods for scientists and engineers/ by Donald A. McQuarric. p. cm. Includes bibliographical references and index. ISBN 1-891389-24-6 (cloth). ISBN 1-891389-29-7 (pbk .).
I. Ma1hema1ks. I. TILie. QAJ7.3 .M36 2003 510---dc21
2003043448
Printed in the United S1n1es
or America
10 9 8 7 6 5
3 2
4
I
Contents
Preface
xi
CH PTER
1
Functions of a Single Variable I l E1mctiaos J ') I irniis 9
1.3
(HAPTFR
Continuity 15
1.4
Differentiation 19
1.5 16
Differenti als 2 7
Meao Yahre Ibi--wems 31
1.7
lnlegration 3 7
1.8
Improper Integrals 47
1.9
niform Converg n Re erences 6 I
2 2.1
of Integrals 55
Infinite Series 63 lnfinile Sequence
64
Co nvergen e and Di verg nee of Infinite Series 66
Tests f
, c 70 6 83
101
2.9
Reieceoces
CHAPTER
3
Functions Defined As Integrals 1J 5
3 1 Ibe Gamma Eirocriao 11 5 32
The Bera
Euoc1io □
I 22 al I 3
.7
144 rnoulli Polynomial
149 V
Cuntenl'i
VI
9
4.1
4.2 4.3 4.4
Eul
ompl x Number. 16
4 .6
Powers of Complex Numbers 184
Tri 4.5 The logarithm
s 176 81
Refeceoces I AA fHAPJFK
5 Vectors 5 J 5
J
J 91
Yecracs io Iwo Diroeosioos 19 1 Yecroc Ftioctioos io Iwo Oiroeosioos J92
Vecrocs io Ibcee Diroeosioos 205
·
io Three Diroeosiaos 2J 2 in Space 2.. 1
CHAPTER
6 Functions of Several Variables 6I
fuoc1iaos 232
6.2
Limits and Continuity 239
23 ,
63
Partial Derivatives 244
6.4
Chain Rul e for Partial Differentjation 253 Diffcccorials ,wd 1hr Ialal Diffcrcotial ' 6D
6 5
6 6 The Diceciiaoa l DeciYalive and tbe Gradient 267 6.7
Taylor's Formula in Several Variables 273
68
Maxima aod Minima 229
6.9 The Method oi Lagrange Multipliers 285 6.1 0 Multiple Integrals 291 References 299
Cl:li'e:IEK
z Zl 7.2
7.3
~ctor Calculus 30 1 Vec1or Eield ]OJ Line lnt grals 11 Surface Integrals 322 33 1
7.4
Z5 Referenc s 347
rni'mH
8 Curvilinear Coordinates 8J
349
Plaoe Palac Caacdioates 349 355
8 2
8.3 8.4 85 8.6
(t,cvilioeac Coocdloaies 328 Som r Coordinale Syst
87
References 395
C p r 1ht
Contents
CHAPTER
VII
9
Linear Algebra and Vector Spaces 397
9 I
Derecroioaors 392
9.2
Ga u ia n Elimination 409
9-3
Matrices 418
9.4
Rank of a Malrix 430
9.5 9.6
Vector Spaces 436
9.7
Complex Inner Prucllict Spa(,es 449
Inner Product Spaces 444 References 453
CHAPTER
10
Matrices and Eigenvalue Problems 455
10.1 Orthogonal and Unitary Transformations 455 10.2 Eigenvalues and Eigenvectors 462
10.3 Some Applied Eigenvalue Problems 47 I 10.4 Change of Ba is 481 10.5 DiagonalizaIion of Matrices 491 10.6 Quadratic Forms 499 References 51 J
CHAPTUl
11
Ordinary Differential Equations 515
11.1 Differential Equations of First Order and First Degree 516 11.2 linear first-Order Differential Equations 524 11.3 Homogeneous Linear Differential Equations with Constant Coefficicnls 530 11.4 Nonhomogen ou Lin e•. r Differential EquaIions wi1h Const.·rnt Coefficienls .541 11 ..5 Some Other Types of Higher-Order Differential Equations 556 11.6 Systems o( Linear Differential Equations 563 11.7 Two Jnvaluable Resources for Solutions 10 Differential Equations '170
Ref renc.es 5 7 2
CHAPTER
12
Series Solutions of Differential Equations 575
J 2 J Ibe Power Series Metbad 526 12.2 Ordinary Point nd Singular Points of Differential Equation 58 I 12.3 Series Solutions ear an Ordinary Point: Legend re's Equation 588 I 2.4 Solution N ar R gu lar Singular Point 593 12.5
Referen
CHAPTER
1J I3 J 13 2 13.3 13 4
623
ualitative Methods for Nonlinear Differential E
Ibe Phase Plane 626 Critical Paints io tbe Phase Plane 634 Stability of Critical Points 642 Naolioeac Qscillarars 65 I
13.5 Population Dynamic
Reference._ 665
657
viii
Conr nts
CHAPTER
14 Orthogonal Polynomials and Sturm-Liouville Problems
667
14. 1 Leg ndr Polynomials 668
t 4.2 Orthogona l Polynomial
678 t 4 .3 Sturm-Liouville Theory 687 14.4 Eigen(unction Expansions 696 14 .5 Green's Functions 702 Re(erenc s 710
CHAPTER
15 15 . I
Fourier Series 713 ouri r S ri
as Eigenfu nction E pan ions 714
15.2 Sine and Cosine Series 724 15. I Equation
71 9
References 744
CHAPTER
16
Partial Differential Equations 747
16.1 Some Examples of Partial D ifferenliil I Equations 7 48 16.2 Laplace's EquaIion 755
16.3 The One--Dirnensional Wave Equation 768
16.4 The Two-Dimension I Wave Equation 779 16.5 Th H eat Equation 786 16. 6 The SchrBdinger Equation 796 A. A Particle in a Box 796
B. A Rigid Rotator 798 C. The Electron in a Hydrogen Atom 803 16.7 Th Cla
ifiration of Parti. I Diff r ntial Equation
807
References 8 J 2 CHAPTER
17
Integral Transforms 81 5
17 .1 The Lap lac Transform 816 17 .2 The Inversion o( Laplace Transforms 824 ·17.3 Laplac Tr, n forms and Ordinary Diff r nti al Equatio n 83 2 Lapla e Tra nsforms and Partial Diif ren ti al Equation 839 Fouri r Tran form 84 5 17.6 Fourier Transforms and Partial D ifferential Equations 856 17.7 Th Inv r ion Fom1ula for Lapl e Tran f rm 865 References 867
CHAPTER
18
Functions of a Com lex Variable: Theor
18.1 18.2
: The Cauchy-Ri emann Equat ion ration : Cauchy's Theorem 862 18.4 Cauchy's Integral Formula 894 18. 90 1 ern 911
869 875
18.3
References 9 J 9
Id
Cont ~nts
CHAPTER
IX
19
Functions of a Complex Variable: Applications
92 '1
922 Summat ion of S ric 938 L ation of Zero · 94 5 oniormal Mapping 954 onformal Mapping and Boundary Value Problems 970 Conformal Mapping and Flu id Flow 977
References 903 CHAPJEB
20 Calculus of Variations
9A'l
~0. 1 20.2
995 20 3 Yaciatiaoal Pcoblems witb Caostcainrs JQQJ 20.4 Variational Formulation of Eigenvalue Problems 1006 20.5 Multidimen ional Variational Problems 1015 Referenc CHAPTER
1021
21
Probability Theory and Stochastic Processes I 1 Discrete Random Variables J024
2 1.2 Continuous Random Vari able
1023
1037
71 3 Cbacac1ecistic f11nctians 1044 ?J 4 105 2 21.5 S10 ha tic Pro
e
· 'amples I 063
A. Poi sson Process I 063 B. Th Shor Effect 1067
Refcceoces CHt\PIER
JQZ3
22 Mathematical Statistics
I QZS
22 J Estirnatiao of PiJciJooelecs JQ7') 2.
Thre K y DL!ribution. U
A Ibe Nocmal Dislcibutioo
d in tatisri _al li I
I 085
mas
B. Chi-Squar Oislribution 1087
C. Student I-Distribution I 089 22.3 Confidence Interval!- 1091 A. Co11fiden ,e Intervals for the Mean o( a Nom1JI Dis1rrbution Whose Variance is Known 1093 B. C nfid •n t ~ I ntervil Is for lhe M an of a Normal Di stribution wilh U nkno wn V.:1ri,1nce l 095
Yaciaoce of a No
Answers to Selected Problems 11 3 llluslralion Credits 1154 lodex 1155
· ihu1 ioo JQ9Z
Biographies and Historical Notes
Isaac Newton aod GauCcied I ieboitz xiv Brook Taylor and Colin Maclaurin 62 I eaobacd E11lec I J 4
William Hamilton J 58 T
George Stoke and George
re n 300
Carl Jacobi 348 Ca rl Friedrich Gauss 3%
Arthur Cayley and James Joseph Sylvester 454 Jacob Bernoulli S14 The Weierstrass Function 573
Adrien-Marie Legendre and Wilhelm Bessel 574 Henri Poincare 624
Charles-Franc;ois Sturm and Joseph Liouville 666 Transfin il Num
711
Jo
Why
o Women Mathematicians? 8 13
Pi rr - imon lapla
814
Augustin Louis Ca u hy and Bernhard Riem;inn 868 Harry Nyqui st and Nikolai Joukowski 920 lohann Bernoulli 984
Simeon
De□ i"'
Pnj 0.
= /_ = f(a-)
(6)
A good example of a function tha1 has different right-hand and lcfl-hand limits is the Heavi.side step function (Figure 1.22). defined by
H(x)= {
In !his case, Ii m H ( -E) t!:-,0
~
X
< 0
X>O
(7)
= 0 and t-•O Ii m H (E) = l. As you might have guessed
already. a function will have different righl-hand and left-hand limits at a point where it is discontinuous. The ncxl Example illustrates different right-hand anJ left-hand limit~ and infinite limits. If lf(x)I > N. where N is a positive number. however large, as € ~ 0. 1hen we say I.hat 1/(x)I --+ oo as€--+ 0.
1.2
13
Li111i1~
Example 5: Investigate the behavior of j(x)
= (x + 2)/(x -
'I -r I) near I.he point x
= I.
It is clear that lim ((x) does nor exist. But let ·s look at the .r-1· two one-sided limi1:;.
SOLUTION:
. x +2 I1m - ,-r-x I
= 1.1m
I-
E
+ 2 = -oo
----
,--0l-t:-1
Figure 1.23
. x +2 \' I +( + 2 I1m - - = 1m - - - - = oo I i~(\ 1 + E - I
Figure 1.23 :-hows j(.r).
You should be aware thal a number or the commercially available computer programs. such as Mathematica. Maple. Mat.lab. and MathCad. can evaluate limit'-. For example. the one-line command in Mathematica
r2 -9 ) / x,
X➔ 0
]
gives 6. These programs not only carry out numerical calculations bur can per-
fonn algebraic manipulation~ as well. Consequently. they are often referred to as Computer Algebr.i Syiaems (CAS). We shall refer to them collectively as CAS. To encourage you 10 learn how to use one of Lhese CAS. Problems 15 through 20 ask you 10 use any one of them to evaluate some limits.
1.2
Problems
t. Show 1ha1 a -
,5 < x < a
+ 8 can be wriucn as Ix - o I < 8.
. sin x 2 . Use FiIgure I. 24 to prove mat 11m - .i..
.r--0
:=:
I.
X
C
T
sin x
Figure 1.24 Geometry :is$ociated with lhc proof thul
lim (sin x)/x
,-o
= I.
=
The function f(x) (x + 2)/(x - I) plotted agai11s1 x near che point x = I. The a~~ rnptotc of /(x) i~ indicated by the -.,crtical da._,hcd Ii nc.
.r- I+ X -
Limit [ ( ( 3+x
X
unit circle
14
.h a pll!r I / func tioM~ or a ingle V,ai"iabl •
1- c.o ( ' - 0 us x ...... 0. Notice that this ratio is of the limiting form
3. Use the result of Problem 2 10 show 1ha1
X
0/0.
4. Find 1he following limits: (a)
. sin 3x 11 m - .r-•O
(c)
. sm· x lim ,
. lan lr 11 m - -
(d)
x-
.,
.t-0 X 1/-
. .,
.r-•fl
sin x
lim
(b)
X
Jx
x-•O
S. Find the following limits: (a)
( c)
sin 2x - - as x - O sin x x2 - 9 - - as x X
-3
+J
I + cos ;rx - - - - as x -,.
(b)
,r.r
tan 2 I - cos x
( d)
6. Find the limit of 1J (x + h) - f(x) 1/ II ash (a)
J (x)
(c)
J (x)
as x -
..
0
X"'
= sin x
➔
(b)
J(..--,
= 1/x
(d)
f (x)
= cos x
0 for I.he following tu net.ions:
7. Find the limits of 1he following functions a ' x-,. (a)
Jx+a - Jx + b
. of 8 . F .md lh e 1·mllt
(b) ( Jx+I
-/x+2 - .fi. a.-. x
---+
- Jx)j x + ~
0•
X
9. Dctem1ine the one- ided limits of j(x) .
I
10 . Evaluate lim
.r-o+ I
=
11. Given 1hat 11 11
= x/ lxl .
+ e12
1/ .
+ u,,. 1 has a limi1. find its value.
12. Suppose 1ha1 j(x ) _:;; g(x) ;S li(x) for all values of x in some dektl·d neighbohoud of a. If lim /(xJ r~u lim h(x) = L, prove Lha1 lim g(x) = L. Some author,; call 1his resull lhe sq1u-1'::_1· law. x--u
=
a
13. Prove Lhar lim j(x)g(x) = [ lim f(x>] [ lim g(x>] . .r---a
14. The function y
.r-•a
x-•LJ
= 1/( I -
x)2 is unbounded as x
-> I. Calculnre the value of t5 such 1/1a1 _r > I0 6 if Ix - 11 < 8,
The nO.
discontinuous al x
(x
+ l)(x -
.
= (x 2 -
1)/(.,· - I) is
= I because J (I) is not deli ncd. but using the fac1 thal x 2 - I =
I). we sec !hat lim f(x) x-. J .
=
= x-• lim (x + l) = 2. so tJ1at condition 2 holds. I
In this case we say that x I is a removable discontinuity. A more fom1al definition of con1inui1y is that J (x) is continuous al x = o if ii is possible to find a 8. which may depend upon both E and a, such that lf(x) - f(o)] < E. however small. if Ix - al < 8. Using this 8-€ definition. ii is easy 10 .,;how using Equal.ion 2.2 Lha1 if f(x) and g(.r) are conlinuous au= a. so are f(x) + g(x). f (.x)g(x). and f(x)/g(x) pro\'idcd g(a) t= 0.
f
ln1ui1ively, a discontinui1y is a jump in the graph of the function. For example. the Heaviside step func1ion (Figure 1.22) has a jump di.';con1inuiry al .r 0. and 1he func1ion defined by
=
-I
- I ::: x ::: I
bul
x f= 0
figure 1.25
x=O is discontinuous al x
X
=
The di1>C.ontlnuou~ f"unc1ion f (x ) x 2 + I for - I ::. ., ::.. I hut .r -#; 0 ::ind f U:) 0 for ,1 .-:- 0 plotted ag.ain.,t .r.
= 0 (Figure 1.25).
Another type of discontinuity is displayed by the runction 1/(1 - x) 2 al x (Figure 1.26). We say that 1/( I - .r) 2 has an infinite discontinuity at .x I.
=
=
=I •J
Just as we have right-hand and lef1-hand limi1s, we have continuity from Lhe right and conunui1y from the lefL For example. we say that J(x) is continuous from 1he right at a if lim, J(x) = f(a), or more simply. if /(a+)= j(a). A ,r-,.a, function is continuous at x = a if it is continuous both from the righl and from 1he
left at x
= a. X
We say 1ha1 a function is continuous in an interval if i1 is continuous at all poin1s in the imerval. If the interval is 0 be as smaJI as you wish. TI1en. we need lo find a 8 = ~ (€. r ) such that
Solving rnis inequaJity for,5 gives 8 0. then J. ir f (c) = 0. 1hen 11
/(.t) has a local minimum at 110
x
=c =c
condusion 0 andi one at x = -2 with j"'(-2) = -18 < 0. Thus. there is a local mirumum at .r = J [wi,Lh f( \)1= - I 2J and a local maximum al x -2 [with /(-2) 151. Although we find a local maximum at x = -2, it is not an ohsolllfe max.imum because j(:r:) = 40 al its endpoint x == 3 (see Figure 1.39). The message here is Lhat if /(x) i.s defined over
=
figure 1.36 The funclfon /(x) 1 (x - 1} + 2(.r - I)
=
=
I plo11cd
ngui1r.,1.r.
X
I
=
(a)
=
y
=
a closed interval. then you mus1 cxc1mi11c the bchavior of /(.r;) no1 only at i1s cri1ica\
poin1s. but al its end points as well. X
(b) y
J
f
X
-I
(c)
Figure 1.37
The functions (al, j(.r) = .r..1. (b) g(.rl = .r1. and (c:) /i(x) -x 4 plotted ogninst x.
=
I
X X
Figure 1.38 The funcrion / •.x) - • 21 defined on lhc cl~d inten·al 1-1. I I plotted ::ig.;iins-t x.
Figure 1 .39 TI1c func1ion / (x) 2 • ' - 12.r - 5 defined on the clo d interval [-3. ) I
=
ploue.d ag::iin
1 .1· .
IA
25
Difil'rl'nli.ilion
Example 6:
Find the local exrrema and 1he inflection poin1s of /(x) = x.!(I -
x).!
over
the entire .r axi.s. SOLUTION:
The cquatiOt\
/'(x) = 2x( I - x) 2 - 2l" 2 (1 -
shows that 1here arc critical poinr.s at .r dcriva.1.ivc is J"(x)
.T)
= 2r(l -
x)( I - 2.r)
=0
= 0. x = 1/2. and x = I. The second
= l2x 2 -
12.x
+2
f
The fact that f''(O) > 0 and j"'(I) > 0 and 1ha1 J"(l/2) < U tells us that the critical points x = 0 1md x = I are local minima and that x = 1/2 is a local maximum. The inflection points are given by /" (x) = 0. or at x = (3 ± ./3)/6 (f,igurc 1.40).
-0.5 Note from Figure 1.40 Lhat j(x) = x 2( I - x) 2 is symmelric about the vertical line at .x = 1/2. To see that this i~ so analytically, lei~ x - 1/2, ~o 1ha1 the function 2 2 now reads j(t.) ( , which is an even func1ion in(.
=
= (f ~) ! - ~)
1.4
0.5
Figure 1.40 TI1e function f(:q "" a~;iin,;1
x" t I
x.
Problems
l. Differentiate (b)
sin x
(c)
r
x 2 lan 2x
2. Differentiate
(b)
Jx'- - 3x +
I;
(c)
rr'
3. Differentiate (b)
ln(scc x
+ tan xj;
(cl
x"in .r
4. The t.angen1 line to a curve ul some point (a. b) has the slope 111 = (dy/d.r>x=-u of the 1inc perpendicular lo rhc curve at (fl. b) i.s equal IO - I/ m.
5. Docs f (x)
= J'(aJ. Show that rhe slope
= lxl have a derivative at x = O'?
= x 3 is difforcn1iable in the dosed interval ( 0, I J. The grJph corresponding to 2.r 2 - 2ry + .\ 2 = 4 is an ellipse whmc major axis makes an angle wirh respect
6. Prove that J(x)
7.
to the x arxJs. Plot the func:tir1n. Show thai thi: slopes oft.he Lang,ml lines 10 this curve at 1..he two points where it cros,.cs 1hc x axis arc I.he -.amc. In other wordllfabo/(J and is the basis for a par.1bolic lc11-.; incoming light is focused .it the point F, rhe focu~ of the par:1hola . Thl' equation of 1hc parabola i, \-~ =-l p.r. where p i, the disrancc OF in Figure 1.41.
=
Figure 1.41 lllu$\rntion nf 1hc re Acct ion property of ;.i p;ir;1hol.1. (St'l' Pr.ihll'm 1-1.)
15. No1ice 1.hat J'(a) is defined by Equation I. Normally we evalua1e j'(a) by linding j'(x) ancl then le11ing x =a.bur 1his procedure is 1101 qui1e rhe same as u.sin£ Equaiion I. Consider 1he funclion J(x}
={x
2 sin( 1/x)
0
.\" ¥- () X
=0
The dcrviarive of J(x) al x == 0 is. by defi11itio11.
j'(O>
=
lim f(O + ~x > - f(O)
r- o
6.r
Show lhat /'tO) = 0 in thi ~ ca:,,.e. Now detcmli1ne J'(.~·) and th,m let .r -,. 0 and ~how lhat rhe limit of J' 0. How about Jim x In x? This limit oci:urs fairly often in the l
>0 I
physical chemistry of electrolyte solutions. :-.-uch as aqueous sotu1ions of sodium chloride. In Lhis case. we have the indelemtinale form -0 - oo, so let's look at . I,m , - .o
In x 1/x
-
. = ,· -11m -0
1/.r - 1/x ..,
--
. O = - ., 1,11111 x = o-, 1
Figure 1.46 shows the behavior of x In x as x -,. 0. X
Example 3: Detennine lim xe- · Figure 1.46 T11c beh:iviorof the fun 1ion .f ( r) !JS.I'_..
0.
=x In x
so Lu TI o N: 11,i cxpres:;ion is of the indc1e.m1ina1 · frwm oo • 0. ll becomes an / ,. ._ form by writim! it a~ r l,im .:... e·t
I = ., lim - =0 -'"".\.. e-r
The result of Example 3 i!i- a special ea~ of the limit of x"e-x as x - oo. where 11 is any integer. We can ea.-:ily find this generc1I limit using mathematical induction. We know from Example 3 thatx" e-· Oa-.x -> whenn = I. When using ml.lthematical induction, we assume thal if a slatement is true for some value of 11 L then it must be I rue for n + \ also. So . 11111
.r--:x,
r"
! I
--
e·\
=
• (11 + 1).x" lim - - - -
,r-.'X'
e-r
yll
= (11 + 1) ,-x lim :_ = 0 e·'
(7)
1. 6
35
,\\,',Ill V,l lu, · I heor ffi$
=
=
=
But we know that lim xne-.t 0 as x---.. oo for 11 I. so it must he 1rue for 11 2, 11 = 3. and ~u on. This result. which ts wor1h remembering. says 1ha1 e- .f __,, 0 fa.,ler than any power of x as x - oo. Ano1her limi1 wonh remembering is lnx . I 1m - = 0 I
(8)
,rU
•"-
for any et > 0. This limit says 1ha1 In-~ - oo more slowly than :my positive power of x. no ma11er how small (Problem 7): or cquivalcnt.ly. that xc, In x - 0 a."- .x - 0 for any a > 0 (Problem 8).
Other indetenninatc forms such as o0 • cx:/1• and 1-x can often be handled by tak.ing the logarithm and manipulating the result into the standard indeterminate fonns 0/0 or 00/ 00. For exumple. consider ,!!_~~ x,(. Let y = .\..i·. and then look al lirn In y t'
0
=
lim x In x O
X
Example 4: Dc1cnnine lim
,1 - .....
= 0: and so
- •0
.I
1J •
= I.
:Jr. where p > 0.
=
We'll let y
SOLUTION:
lim y =Lor final l y lim x 1· .I
11
2
p . lake logarithms. and I real
II
as a continuous
variable. . hm ln y I/
'\..
=
. I hm - In p I,
• '\.., JI
=U
So lim r = lim ::fp = 1. Figure 1.47 shows ::/2 plouc-d against n. ,,i •:\.. ,, x,, Problem 12 has you show th.:11 ;j,i - I as 11 - oc.
n figure 1.47 The fun tinn /(x) The line.
11.
3!-)°m,ptote
= ,f j_ plo11ed agains1
i~ shown ~
Whal if you apply I' H6pital's rule and you still get an indctcnninate fonn? Simply apply ii succc--.sivcly Ulillil you no lomgcr obtnin an indetenninatc form. for example.
+ 12x \8x-- + 30x -
2 \'.' - 9x~
.
lim .r-1
2x 3 -
,
5 14
.
= tm,
.i:-1
6x 2 - I 8.x + 12 --,- -- -6x - - 36x + 30
12x - 18
=lim - - - - = .1 I 12.r - 36 4 (See Problem 11. however.)
1.6 Problems I. Use Equaiion 4 to calculme sin{;r /4) to four~place aC(;uracy. Hi111: Realize thal I sin x I _ I and that I cos x I ~ I. 2. Use Equation
4
to calculate the value of e to five-place accuracy. Hi,,,: Use (he fact that e ::c 3.
a
da~hcd
36
Chapter 1 / Fune-lion of a ingl£> \ l;iri .:ibl ·
= x 3 + px + q has one real roo1 if p > 0.
3. Argue that j(x)
4. Use l"Hopital"s rule (a)
lim
ex
I
-
·
lim
x-1
I - cos x X
.r
10
I
lim
(d)
.l
+ cos 2x
rr ~ I - sin 2r
dctem,inc 1hc following limits:
. e' - I 1lffi - .r-•0
6.
-
.
.1-,0
hm
x
S. Use l'Hupital's rule
(c)
(b)
X
.r-,,0
{a)
dctennine 1hc following limits:
l.im - .c-0
(c)
10
sin 2 x
. I - cos X ., I1m
(b)
x-•0
X
~
(d)
I- x
x
.r-
lim ( J :r: 2 + 2..r - .r)
Use l'H6piial's rule 10 de1em1inc lhc following limits: (8)
lim (c.sc x - col x)
(b)
.1~0
(c)
.
l1m
ln sin x ---
(d)
.r-.o+ In tan .t
·1rn I - c~s2 .r 1 .
Jim ln ( I + .r) .r-0
.
7. Show tha1 for every a> 0. I1m
lnx
-
x-oo x«
x-
x-•O
X
= 0.
8. Show that for every a > 0, xa In x -. 0 a.-. x
➔
0.
1
9. Determine lim x /.r_ .i:-oo
. . J i + x:? lo . Dctcnnine 11m - - - . ,t - I. 1hcn we have . I - 1 ) = - -I ( 0 - 1 1 = Illm - I- ( - I - p hi' 1 1- p , l- p
b-")()
lf p < I, then
Um -
1
-(1, 1-p
b-x I - p
-
I) = oo
1.8
lmprop r lnt
If p
49
r.i l
= I. then
J f 1
So we see 1ha1
00
l
f
oo dx . - = lim xf'
,,_
dx
h
-
I
. = lim In J, = h-oo
X
dx convell:!e if p > I and diverge if p :£ I. This result xP
~
is worth rcmcmhering.
Example 2:
1
00
Examine 1he convergence of
c-udx as a func1ion of s.
SOLUTION:
.
= l hm
e- .ra - e-sb
(s > 0) .1·
but equals oo if s= (I or ifs < 0. Thus the integral converges ifs> 0 and ~
diverge~ ifs
0.
Nole that Equal.ion 3 does not say that
!
oc,
'- /(x)dx
For example. consider / = consider lhe two integrals
= c~~ oo
f'
udu
J
-~--
. ·~ I+ u 2 ·
We choose : = 0 in Equation 3. and
. I1m
and Let
11 ;::::
-x in r.hc fln;t integral
10
. 1'
f
so 11c I .mtcgra 1
'
ltm
'X)
-:x.
b-•!lO.
£" n
-11du -I + ,,2
obtain
1
-
(not 1rue)
-c j(.r)dx
xdx " o I + x2
. ., = hrn -I In(!+ a-) 2
=-
11- • -
I ' I h . . -udu -=,, d.1verges. N 1 ow et s ook at t e integral IIom Oto oo. l + u-
l
lim b .
b ud11
--,
,1 I + u·
.
I
= llm 1,
::,;:; 2
ln(I
,
+ h-) = oo
U,.,p!t•r 1 / Fu11ctions of a Single V,1ri.ihl._.
50 Thus, h01h con1ribulions
10 /
diverge. (f we had u.sed
)~":-x.
i
ll
lldll
-a I
Ihen we would nave obt;:iincd a value of zero
+ 11 2 I~
In( I
+ t/) -1 Jn( I + a 2 ) I for the
in1cgral. ll is useful to have snme 1ools 10 detem1ine e,L N(e. x)
(4)
where N is a number tha1 depends upon f and x with x 1 :::: x.::: x~. Equ:nion 4 is the fom1aJ way ot' expressing that F (x) converges for each x in I x 1• x'.! 1- Let ·s suppose. now. 1hat Equation 4 is sillisflcd for a number N(€.) that depends only upon f and not upon x. In other words. suppose that
where N ( ~) is i11depe11de111 of x. In lh is case. we ~ay that F (x) converge~ uni form ly
in
I x 1• x:d- For example. lhe integral
c.:onvcrgcs uniformly to 1/x for x ~ I because
-1 -
X 1
l" O
-n
e · dr
I
,,-.d, ::S"' -h = -t=
X
and e-" will be 0. (See !he previous Example .)
ft~
Before we leave this section (and this chaplcr) we should mcmion some corresponding results for integrals with fini1c limits . I.
If f
J,;'
(x . 1) is ronrin11011s i11 the rc·cta11:~lc a :::: r :::: Ii,
f (x,
i., ('{))/lill/10/IS )or x , .'.:: X
I )di
lim x-•.10
l
b
=
/(x. l)d1
"
!J,
Xz. This thenrem o/1,rn·s
_:'::
F(x)dx
=
JI.\"'" \\"ril
ltm f (x. l)dr
1·r. [J" .Hr. ] r)dt
_rl
-~I
=
o .1-•.ro
2. Under tht· co11di1io11.1· of the previou.l· 1l1t·orem.
1-"
x 1 _:s .t _:s x 1. rhe11 F(x)
dx
11·e
=
U
/rave
fb [ frr· J ~IJ
3. If f (:r.,) and of (.x. 1 )/8:r are co11ti1111011s
i11
(x.
t)dx
]
dr
•XJ
the rl'ctanglt' a
~ t ~
b . .r 1 :5 x
~
x 2 , then
F
, (X)=
[1, Af(:x. l)d t -
• r.
You can easily
1cs1
--
i)x
Lhcse three theorems with F(x)
= .f01 t'-.x,dt
(Problem 10).
1. 9 Problems I. Show 1hat
10...• t·-x dt is unifom,ly l'Onvergeni for .r ~ u
2. Sho"' that
fooc ,,,,. r1dt is unifom1ly cOn\'ergenr for x ~ tJ. > 0 where II a positi\ C integer.
1
>
0.
1
3. We showed 1hat
ix,
ti.-.ridr
= 1/x
is unifonnly convergent for t ;~et> 0 in Problem I. Show 1ha1
Cl
1 0
11
I t'
.!." Id
4. Prove 1ha1
t = -11!- . xn I
fox
,,-.ri cost dt converges unifom1ly (and absolutely) for" :::
x _ b. where O < a < b.
60
C'h,1p1i·r I /
5. Example 2 ~hows
cos O'.\"
1
7. There is
standard Irick 10 evaluate
,X,
.i
- ,-
., .. -1- I
00
9. Does
•
•
fooc c-
1
cos etx d:r. Differentiate with respect 10 et. integrate by parts.
•
1 O
la''° e-a., cos x dx is a continuous dx
---
1 + x:?
of,, Si ng ll' \i,'lri,1hh·
1s uniformly convergem for all real values of Ci..
and notice Lhe result. The answer is (rr 1 -J2) e
8. Show 1hu1
io ns
is uniformly convcrgenl for x ~ a > 0 . Show that
6. Show 1ha1
o
un
=
. hm
1
00
o
u ·O
2 4
1
funclion of a for a > 0.
co-; ax dx ., ----,-. I + .x-
lO. Verif)• the last 1hree theorems ut the end of the section wi1h
ll. Show lhal Jt(sin x)/x dx
F(:r)
=[
. ()
1
e- r,d, and O::: x:::: I.
= rr /2 by wri1ing
10
x , -·i,jn-dx = 1"'.3 sin x (ioc
O
X
t!-.i, dr
) dx
0
and then inten.:hanging orders of inIcgraIion. 12. Show Lhur for:.:, c - ,u (sin x )/x d:r
= co1- 1o by writing
,:.; . .· 1 ,,
-a· r sin .\' d - - ,\'.=
0
1oc
e - n.r
0
X
and inierchanging orders of integr'Jlion .
. . ,, , , 1
13. Eva\ua1e the integral in the previous problem by differentiating with respect to a. 14. Show that /(a. I,)=-=
l.'
-,r. ·- t,· / r· ,1x
7r 1/2
= -' -e- 2" 1' by differemiaring with
2a o :r = b/oJ.. and lh~n integrating with respect to b.
15. Given 1ha1
f..._,,, "cos xu du
Jn
= --1- .,, show 1ha1 I+x-
1= o
ue-u sin x,, du
2
= (I
rcspccl to b. then k:Hing
\'
., .
+ x 2)-
61 References CALCULUS TEXTS: Frank Ayre~. Jr.. and Etlioll Mcndelwn. 1999. Calrnl11s. 4th ed .. Schaum ·s Outline Series. McGr.1w-Hill R. Couranl. 1970. Dif{ercminl and ln1t·Rral Calculus. 2nd ed .. Inicrs.cience, John Wiley C. H. Edwards, Jr., and David E. Pc-nncy. 1998. Calcul11.t and A11aly1ir Gt·nemtry. 5th ed .. Prcn1 ice-Ha.I I Witold Kosmala. l 999. Advanced Calculu.~: A Friemlly Approach. Prentice-Hall Jerrold Marsden and Alan Weinstein. 1985. Calc11/JL1· I, II. and Ill. Springer.Ycrlag Richard Si Ivcnnan. I 989, /:".1.11•111iu/ Colrnlux. Dover Public::uions Murray Spiegel. 1963. Adwm1·1•d Calm/us. Schaum ·s Outline Serie.~. McGraw-Hill Dlivid Widder, 1989. Ad1·a11ced Calc-11/11,1·. 2nd ed .. Dover Publica1ions
MATHEMATICAL TABLES: CRC Standard Matht·nwtical Tal>lcs and Fon1111lai..', JOlh ed., edited by Daniel Zwillingcr. CRC Press (19%) Tables af lnte~mls, Series. n11d Producis. 41.h ed .. I.S. Gradsh1cyn nnd I.M. Ryi.hik, Academic Prcs!i ( 1980)
COMPUTER MATHEMATICAL PACKAGES: !'vlATHEMATICA. Wolfram Re.search. Inc .. Champaign IL. www.wolfram.com MATLAB. The Ma1h\Vorks. Inc., Nnlick MA. www.malhworh.com MAPLE. Wa1erloo Maple. Inc. Waierloo ON. www.\\>alerloomaple.cum MATHCAD. MathSoft. Inc .. Cambridge MA. www.nw1hcad.com
SOME GENERAt MATHEMATICAL WEBSITES: www-history.rncs.sl-and.ac.uk (The www- is correcl.) www. mat h. un,cdu/rnath/math-wch www. mat h. u J) s,, for all 11, 11011im:reasi11g if Sn-I~ s,, for all,, and 111mwro11ically deuensillg ir s11 + 1 < s,, for all 11. Clearly. every bounded monotonic sequence is convcrgcnl. Alrhough every convergent sequence is bounded. 1he converse is not true: there are bounded sequences [for example.
I (-1)" \ J 1hat arc not
convergent. You should realize that the convergence or divergence of a sequence is not affec1ed by adding or deleting a finile number of terms of the sequence. Convergence depends upon the large II behavior of the sequence. or on the far "tail" of the sequence. The criterion given in Equation I explicitly shows that convergence depends only upon the behavior of {s11 I for n > 11 0 •
Example 2: Dc1em1inc whether the sequence large"· De1cm1ine i1s limit as 11
{ '",,"
-
lnx . Let f(x) =-.Now f (x)
SOL Ul ION:
X 11
• 11 the sequence { -ln,-, } decreases tor its limit is zero as 11 - oc·,
2.1
] is incrc:1sing or dccrc:i-,ing for
~
1-lnx = --,x-
< 0 for x?: e. so
3. Usini; . l'H.opI1a . 1· s ru Ie. we see I.hat
Problems 2
J. Show 1ha1 lim 3',
611
11 2 +1
11- 00
r-
2. Show 1..hat lim ~ 11
- ·~
3. Shmv th:it lim
n-•oc
/I + I l/ n
4. Show rh a.1 lim cos ,r-- ......
+ 2 = 3.
= 0.
= I. 11
11
= 0.
5. Show tJiat lim ~ ·= 1) for " > 0. ,,
•: ,C
6. Show that ,, lim ....
~
0, = I.
7. Show rhal 11 !/5" is increasing for n
2"
8. Show that ofx?
11!
-
0 as
11 _,, ;X:> .
5.
r" Can you show that this result gcncrjlii.e1- to :...__ _. 0 as 11 11!
--.
for any value
9. A simple method for showing whether a sequence is increasing or decreasing is 10 u~e a continuou~ function /(x} such that _((11)-.,,. ~,, for 11 = I, 2, . . . and show that either J'(x) > 0 for x ~ I (an increa.-;ing sequence)
66
Chapter 2 / ln1i ni:-, Serie.
or that J'(x) < 0 for x decreasing: (a)
~
I (a decreasing sequence). Determine if the following sequences arc increasing or
4n - I } { 611 + 2
(d)
(c)
r
+ -A) . Show 1ha11f.
. defined by the recur-ion . tormula . l 0. Suppo~e that s11 1s x,,+ 1 = -I ( -~n 2 I=
11+1} ,,
In - -
~
.
Jun s,. " ~
=I e
i~t·. then.
Al/2_
+ F,,- 1defines the Fibonacci sequence. where each tern, is 1he sum of the two preceding 1cm,_,;_ The sequence is I. I. 2, 3. 5, 8. 13. 21. .... The lim F, 1 docs nor cxis1 in 1his case. bur 11-c:,.:, F nssumc that lim ~ does exist. Show thal this limil is equal to ~(I + J5°). " o::: F,. ~
11. The n:..--cursion formula Fn+ 1 == F11
12. Which of the following sequences converge? 11 co. h 11 (..,) " (a) . ,, = -..(b) s,, = -"s1nh 11 11 nl
13. Determine lim __:__ 11-00 11"
14. Use the €-6 nol.ilion 10 prove lhal { ~} convcr_\.!~S to 0. 15. If lim s,. n •
I,
= a and lim n
111
. = b, 1hcn prove th..11 Jim s,,r,, n ·•".
. s,, lim -
= ob and 11
•".
1,,
= -b (I
.
prov11Jcd thol
111
,=-
0
and
,=- 0.
2.2
Convergence and Divergence of Infinite Series
An infinite series is an exprt's~ion of the form oc
L "" =
"1
+ u~ + "J + ...
11,ec]
The pnnial sums of this series are
S3
= " 1 + II 2 + 11 J
and the 111h parrial sum is
If 1hc sequence of partial sums converges. then the series is said to converge. or 10 be convergent. Otherwise. 1he series diverges, or is divcrgenr. If Jim S,, 1/~'X:,
=S
then S is called the .,um of the infinite series.
67 The smndard example of an infinite series is 1he geomelfic series. whose 111h panial sum is
,,
Sn
= L rJ- l = I + r + r 2 + ... + r'i - I j= I
Note that 11 11 + 1 = 11,1 r. h turns out that ii is possible (and easy) to obtain a closed form expression for S,1 • Muhiply S,, by rand subLract the result from S11 10 get
S11
r S11
-
=I-
rn
or I - r" S,,=-1-r
(I)
2
It's ca~y to see here that
lim S,,
It•'\...
= 11 -
lrl
r
=-
IX.-
Jl ~
.t
0
Nore that we start the summation with a.n "
lxl
Cfics. The ndvnmage of t.hc intt'grnl tes1 is that it is usually easier to evalluare-an integral Hurn it is to sum .i. se-rie-s into a closed form.
Integral l'esi·: Let L ''n be n pm:i1in:' .\eries 1,111d /1'1 j(x) he LJ omrimw11,\, positi1·c. decreasing fimctirm :mch that J (11) = 11, 1 for 11 = L 2, .... Thr11.
L,, li 11 t'tJlll 'er,1u·.~ if and rml_1· if {.-x, f Cr )dx cmw1'1g1·s. 2
3 4 (b)
5
6
• I
II
Figure 2.4 A geometric aid 110 the proof of the in 1eg_r.il
ll1e proof of 1hc integral lest is illustrated in Figures 2.4a and 2.4b. \\l'c .:;ec in Figure 2.4a 1h:11 the area under the curve from I 10 N i.'- less than the total- ar~a of 1he rccianglcs. which is 11 1 + 11 2
IC',I.
Ii If
L
11 11
+ · · · + 11 N· Thus. we have
N
J(x)dx
dx
M
We'll usually t::.ikc M ;;::
I. however. ~
Let· s consider 1he series
L ~. The function f
(x)
11-
= I/x'!. equals I/ ~ when 11
11=.I
x is an integer. Punhcnnon:, j(.r) is c:onlinuous and mono1onically decreasing.
Now
f
I p ~ l
(5)
X
I/
converges diverges
L ,,., a.11d L P,, be Jwo positive saie.-. with L 11 11 com·,~r:i,:1 s {f L Un conve~e.~. anJ L u,. rn
73 (Sec Problem 14 for an outline of 1hc proof.) 00
Docs 1he series C
L~ n=I
II"
L
I
I
I
converge? Well. - 1 - - < and we know that 3 ,r ;;;; I 11· + 3 11· + J n 1 - converges (p series wi1h p - 3). and so converges. What if 1 -:i--
~
L - +3
n=I
II"
the series started with then = 0 1crm? We wouldn·t be able 10 apply 1hc above inequaltty for 1he " = 0 term. Docs this make any difference? All we have to do ·,s to wn1e · then= 0. tem1 cxplic11\y - . and then consider . . I ~ I the scncs - ..1.. L ---. 3 11= I l!J + 3 The point is 1ha1 we can apply 1he comparison tcs1 srnrting with any term because the convergence of an infini1c series is dctennined by 1hc lnrgc-11 tail of 1he series. so the first lini1e number of lerms has no effect on the convergence or divergence of an infinite series.
Example 2: E.xamine lhe convergence of 00
I
I
I
11!
21
3!
S=E-=1 +-+ - + ··· n=I
SOLUTION:
~
I
= 11(11 -
1)(11 - 2) ...
3. 2 ::: 211 -
1
for
So
But
and so S
= L ~,, . converges. We'll see in Section 7 that 5 = e -
I.
m= I
Ano1hcr useful test for 1hc converge.nee of a posi1ivc scric..-. is the
L 11,, and L v,, art· hvo pMilive serit·s .rnch tlwr Lu,, and L v,, «:itlu:r boll, com·er~r. or horh di;:agt·. ,r I = 0 011d L v com·erRt•,·. rhe11 L u" couverges. If I = oo mrd L dirages. 1he11 L diverges. Limit Comparison Test: If
lim ~
14-,. .:x,.
I'
= / 11·i1J, 0 < I
I and r
= I are similar.
I""+' I= u,,
r < I. Choose,
~ 111,, 1,.1.- and appeal 10 lhe geomet.ric
82
ri es
12. Detennine whe1her I.he followin g series are absolutely convcrg~.nl. condi1ion:.illy convergent. or diver ent: 0
~
(a)
( - I )11
I
L - -. . ,r2 + I I II
(b)
L(-l)"+i _ _
(dl
(c)
~
,1 - I
,1.,,, I
-x,
11c-el
X·
= L (-1 )"+
13. lnvcs1jg:i1c 1hc convergence of the scric1- S
~.
+
I
2-4·6···(2n)
=
15. There is another test for convergence culled the Root Tt•sr: Let r
r