Student Solution Manual to accompany the 5th edition of Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach John Hamal Hubbard Barbara Burke Hubbard Cornell University Universite´ Aix-Marseille Matrix Editions Ithaca, NY 14850 MatrixEditions.com Copyright2015byMatrixEditions 214UniversityAve. Ithaca,NY14850 www.MatrixEditions.com All rights reserved. This book may not be translated, copied, or reproduced, in whole or in part, in any form or by any means, without prior written permission fromthepublisher,exceptforbriefexcerptsinconnectionwithreviewsorscholarly analysis. Printed in the UnitedStatesof America 10 9 8 7 6 5 4 3 2 ISBN: 978-0-9715766-9-8 Contents Note to students iv Chapter 0: Preliminaries 1 Chapter 1: Vectors, matrices, and derivatives 9 Chapter 1 review exercises 31 Chapter 2: Solving equations 40 Chapter 2 review exercises 74 Chapter 3: Higher partial derivatives, quadratic forms, and manifolds 85 Chapter 3 review exercises 122 Chapter 4: Integration 133 Chapter 4 review exercises 173 Chapter 5: Volumes of manifolds 182 Note on parametrizations 183 Chapter 5 review exercises 204 Chapter 6: Forms and vector calculus 207 Chapter 6 review exercises 266 Appendix A: Some harder proofs 277 Index 305 Note to Students This manual gives solutions to all odd-numbered exercises in the fifth edition of Vector Calculus, Linear Algebra, and Di↵erential Forms: A Unified Approach. You should not attempt to use Pleasethinkabouttheexercisesbeforelookingatthesolutions! Youwill this manual with a previous edi- get more out of them that way. Some exercises are straightforward; their tion of the textbook, as some sec- solutions are correspondingly straightforward. Others are “entertaining”. tions of the textbook have been You may be tempted go straight to the solution manual to look for renumbered, new sections have problems similar to those assigned as your homework, without reading the been added, some exercises have text. This is a mistake. You may learn some of the simpler material that been moved from one section to another, some earlier exercises way,butifyoudon’treadthetext,soonyouwillfindyoucannotunderstand have been discarded, and some the solutions. The textbook contains a lot of challenging material; you new ones have been added. really have to come to terms with the underlying ideas and definitions. Inthismanual,wedonotnum- Errata for this manual and for the textbook will be posted at ber every equation. When we do, www.matrixeditions.com/errata.html weuse(1),(2),... ,startingagain with (1) in a new solution. If you think you have found errors either in the text or in this manual, please email us at hubbard@matrixeditions.com or jhh8@cornell.edu so that we can make the errata lists as complete as possible. John H. Hubbard and Barbara Burke Hubbard iv Solutions for Chapter 0 0.2.1 a. The negation of the statement is: There exists a prime number such that if you divide it by 4 you have a remainder of 1, but which is not the sum of two squares. Solution0.2.1: Thenegationin b. The negation is: There exist x R and ✏>0 such that for all � >0 part a is false; the original state- there exists y R with y x <� an2d y2 x2 ✏. mentistrue: 5=4+1,13=9+4, 2 | � | | � |� c. Thenegationis: Thereexists✏>0suchthatforall� >0,thereexist 17 = 16 + 1, 29 = 25 + 4,..., 97=81+16,.... x,y R with x y <� and y2 x2 ✏. 2 | � | | � |� If you divide a whole number (prime or not) by 4 and get a 0.3.1 a. (A B) (A B) b. (A A) (B B) c. A A ⇤ ⇤ ⇤ ⇤ ⇤ ⇤ ⇤ remainderof3,thenitisneverthe sum of two squares: 3 is not, 7 0.4.1 a. No: Many people have more than one aunt, and some have none. is not, 11 is not, etc. You may b. No, 1 is not defined. 0 well be able to prove this; it isn’t c. No. all that hard. But the original statement about primes is pretty 0.4.3 Here are some examples: tricky. The negation in part b is false a. “DNAof”frompeople(excludingclonesandidenticaltwins)toDNA also, and the original statement patterns. is true: it is the definition of the b. f(x)=x. mapping x x2 being continu- 7! ous. Butinpartc,thenegationis 0.4.5 The following are well defined: g f : A C, h k : C C, true,andtheoriginalstatementis � ! � ! k g : B A, k h : A A, and f k : C B. The others are not, false. Indeed, if you take ✏ = 1 � ! � ! � ! unless some of the sets are subsets of the others. For example, f g is and any � > 0 and set x = 1, � y= 1 + �, then � not because the codomain of g is C, which is not the domain of f, unless � 2 C A. y2 x2 = y x y+x ⇢ | � | | � || | � 2 � 0.4.7 a. f g(h(3)) =f g( 1) =f( 3)=8 = + >1. � � 2 � 2 b. f g(h(1)) =f g( 2) =f( 5)=25 ✓ ◆ � � �� The original statement says that the function x x2 is uniformly 0.4.9I�fthearg�umen�tofthe�squarerootisnonnegative,thesquarerootcan 7! be evaluated, so the open first and the third quadrants are in the natural continuous, and it isn’t. domain. The x-axis is not (since y = 0 there), but the y-axis with the origin removed is in the natural domain, since x/y =0 there. 0.4.11 The function is defined for x R 10 and p( A) ( A)d+(A 1)Ad 1 <0. � � � � � �  � � By the intermediate value theorem, there must exist c with c < A such | | that p(c)=0. 0.5.3 Suppose f : [a,b] [a,b] is continuous. Then the function ! g(x) = x f(x) satisfies the hypotheses of the intermediate value theo- � rem (Theorem 0.5.9): g(a) 0 and g(b) 0. So there must exist x [a,b]  � 2 such that g(x)=0, i.e., f(x)=x. 0.5.5 First solution Let 1k=1bk be a rearrangement of 1k=1ak, which means that there exists aPbijective map ↵:N!N such thPat b↵(k) =ak. We give two solutions to Exer- Since the series 1k=1ak is absolutely convergent, we know it is conver- cise 0.5.5. gent. Let A= 1k=P1ak. We must show that P m ( ✏>0)( M) m M = b A <✏ . k 8 9 � ) � � � ! �kX=1 � � � First choose N such that 1N+1|ak|<✏��/2; in that��case, n P 1 1 (n>N) = a A a a <✏/2. k k k ) � � � | | | | �kX=1 � kX=n k=XN+1 � � Now set � � � � M =max ↵(1),...,↵(N) , { } i.e., M is so large that all of a ,...,a appear among b ,...,b . Then 1 N 1 M m N 1 m>M = b A a A + a <✏. k k k ) � � �� � � | | �kX=1 � �kX=1 � k=XN+1 � � � � Second solution � � � � � � � � The series 1k=1ak+|ak| is a convergent series of positive numbers, so its sum is the sup of all finite sums of terms. P If 1k=1bk is a rearrangement of the same series, then 1k=1bk+|bk| is also aPrearrangement of 1k=1ak+|ak|. P But the finite sums oPf terms of 1k=1ak +|ak| and 1k=1bk +|bk| are exactly the same set of numbers, so these two series converge to the same P P limit. The same argument says that 1 1 a = b . k k | | | | k=1 k=1 X X Solutionsfor Chapter0 3 Thus The first equality is justified becauseboth series 1 1 1 1 1 1 b = (b + b ) b = (a + a ) a = a . k k k k k k k k | | � | | | | � | | 1 1 k=1 k=1 k=1 k=1 k=1 k=1 (b + b ) and b X X X X X X k k k | | | | k=1 k=1 X X 0.6.1 a. Begin by listing the integers between 1 and 1, then list the converge, so their di↵erence does � also. numbersbetween 2and2thatcanbewrittenwithdenominators 2and �  whichhaven’talreadybeenlisted,thenlisttherationalsbetween 3and3 � Solution 0.6.1, part a: There thatcanbewrittenwithdenominators 3andhaven’talreadybeenlisted,  areinfinitelymanywaysofsolving etc. This will eventually list all rationals. Here is the beginning of the list: thisproblem,andmanyseemjust 3 1 1 3 8 5 7 5 4 2 1 1 2 as naturalas the one we propose. 1,0,1, 2, , , , ,2, 3, , , , , , , , , ,... � � �2 �2 2 2 � �3 �2 �3 �3 �3 �3 �3 3 3 b. Justasbefore,listfirstthefinitedecimalsin[ 1,1]withatmostone � digitafterthedecimalpoint(thereare21ofthem),thentheonesin[ 2,2] � with at most two digits after the decimal, and which haven’t been listed earlier (there are 380 of these), etc. 0.6.3 It is easy to write a bijective map ( 1,1) R. For instance, � ! Solution0.6.3: Thesameargu- x ⇡x f(x)= or g(x)=tan . mentholdsifoneusesthefunction 1 x2 2 � g, with derivative The derivative of the first mapping is ⇡ g0(x)= 2(1+cos2 ⇡x). (1 x2)+2x2 1+x2 2 f0(x)= � = >0 (1 x2)2 (1 x2)2 � � so the mappingis monotoneincreasingon ( 1,1), henceinjective. Sinceit � is continuous on ( 1,1) and � lim f(x)= and lim f(x)=+ , x 1 �1 x +1 1 &� % it is surjective by the intermediate value theorem. 0.6.5 a. To the right, if a A is an element of a chain, then it can be 2 continued: a,f(a),g f(a) ,f g(f(a)) ,..., and this is clearly the only pos�sible�con�tinuation�. Similarly, if b is an element of a chain, then the chain can be continued b,g(b),f g(b) ,g f(g(b)) ,..., and again this is the only poss�ible c�ont�inuation�. Let us set A = g(B) and B = f(A). Let f : A B be the unique 1 1 1 1 ! map such that f (g(b)) = b for all b B, and let g : B A be the 1 1 1 2 ! unique map such that g (f(a)) = a for all a A. To the left, if a A is 1 1 2 2 an element of a chain, we can extend to f (a),a. Then if f(a ) B , we 1 1 1 2 can extend one further, to g (f (a)), and if g (f (a)) A , we can extend 1 1 1 1 1 2 further to f g (f (a)) , g f (a) , f (a), a. 1 1 1 1 1 1 � � � � 4 Solutionsfor Chapter0 Eitherthiswillcontinueforeveroratsomepointwewillrunintoanelement of A that is not in A or into an element of B that is not in B ; the chain 1 1 necessarily ends there. b. As we saw, any element of A or B is part of a unique infinite chain to the right and of a unique finite or infinite chain to the left. Part c: c. Since every element of A and of B is an element of a unique chain, (1) can be uniquely continued and since this chain satisfies either (1), (2), or (3) and these are exclusive, forever the mapping h is well defined. (2) can be continued to an el- If h(a ) = h(a ) and a belongs to a chain of type 1 or 2, then so does 1 2 1 ement of A that is not in f(a ),hencesodoesf(a ),hencesodoesa ,andthenh(a )=h(a )implies 1 2 2 1 2 the imageof g a =a , since f is injective. 1 2 (3) can be continued to an el- Now suppose that a belongs to a chain of type 3. Then a is not the ement of B that is not in 1 1 first element of the list, so a A , and h(a ) = f (a ) is well defined. the imageof f 1 2 1 1 1 1 The element h(a ) is a part of the same chain, hence also of type 3, and 2 h(a ) = f (a ). But then a = g(f (a )) = g(f (a )) = a . So h is 2 1 2 1 1 1 1 2 2 injective. Nowanyelementb B belongstoamaximalchain. Ifthischain 2 is of type 1 or 2, then b is not the first element of the chain, so b B and 1 2 h(g (b)) = f(g (b)) = b, so b is in the image. If b is in a chain of type 3, 1 1 then b=f (g(b))=h(g(b)), so again b is in the image of h, proving that h 1 is surjective. d. This is the only entertaining part of the problem. In this case, there is only one chain of type 1, the chain of all 0’s. There are only two chains of type 2, which are f 1 g 1 f 1 g 1 f 1 +1 .... ! 2 ! 2 ! 4 ! 4 ! 8 f 1 g 1 f 1 g 1 f 1 1 .... � !�2 !�2 !�4 !�4 !�8 Allotherchainsareoftype3,andendtotheleftwithanumberin(1/2,1) orin( 1, 1/2), whicharethepointsinB f(A). Suchasequencemight � � � be 3 g 3 f 3 g 3 f 3 .... 4 ! 4 ! 8 ! 8 ! 16 Theexistenceofthisfunctionh Following our definition of h, we see that is Bernstein’s theorem. 0 if x=0 Solution0.6.7: TwosetsAand h(x)= x/2 if x= 1/2k for some k 0 8 B have the same cardinality if ± � >< x if x= 1/2k for some k 0. thereexistsaninvertiblemapping 6 ± � A ! B. Here, finding such an >: invertible map would be di�cult, 0.6.7 a. There is an obvious injective map [0,1) [0,1) [0,1), given ! ⇥ so instead we use Bernstein’s the- simply by g :x (x,0). orem, which says that if there is Weneedtoco7!nstructaninjectivemapintheoppositedirection;thatisa an injective map A B and an ! lotharder. Takeapointin(x,y) [0,1) [0,1),andwritebothcoordinates injective map B A, then there 2 ⇥ ! as decimals: is a invertiblemap A B. ! x=.a a a ... and y =.b b b ...; 1 2 3 1 2 3 Solutionsfor Chapter0 5 ifeithernumbercanbewrittenintwoways,oneendingin0’sandtheother in 9’s, use the one ending in 0’s. Remember that Now consider the number f(x,y) = .a1b1a2b2a3b3.... The mapping f injective = one to one is injective: using the even and the odd digits of f(x,y) allows you to surjective = onto reconstruct x and y. The only problem you might have is if f(x,y) could bijective = invertible. be written in two di↵erent ways, but as constructed f(x,y) will never end in all 9’s, so this doesn’t happen. Bernstein’s theorem is the ob- Note that this mapping is not surjective; for instance, .191919... is not ject of Exercise0.6.5. intheimage. ButBernstein’stheoremguaranteesthatsincethereareinjec- tive maps both ways, there is a bijection between [0,1) and [0,1) [0,1). ⇥ b. The proof of part a works also to construct a bijective mapping (0,1) (0,1) (0,1). But it is easy to construct a bijective mapping ! ⇥ (0,1) R, for instance x cot(⇡x). If we compose these mappings, we ! 7! find bijective maps R (0,1) (0,1) (0,1) R R. ! ! ⇥ ! ⇥ c. We can use part b repeatedly to construct bijective maps Part c: The map R f R R(f,id)(R R) R=R R2 (f,id)(R R) R2 (f,id):R R R ! ⇥ ! ⇥ ⇥ ⇥ ! ⇥ ⇥ mapsthefirstRt⇥oR⇥!Randthe =R⇥R3 (f!,id)(R⇥R)⇥R3 =R⇥R4··· . secondR to itself. Continuing this way, it is easy to get a bijective map R Rn. ! 0.6.9 It is not possible. For suppose it were, and consider as in equation 0.6.3 the decimal made up of the entries on the diagonal. If this number is rational, then the digits are eventually periodic, and if you do anything systematic to these digits, such as changing all digits that are not 7 to 7’s, and changing 7’s to 5’s, the sequence of digits obtained will still be Solution 0.7.3, part b: If you eventually periodic, so it will represent a rational number. This rational hadtroublewiththis,notethatit numbermustappearsomeplaceinthesequence,butitdoesn’t,sinceithas followsfromProposition0.7.5(ge- a di↵erent kth digit than the kth number for every k. ometrical representation of mul- The only weakness in this argument is that it might be a number that tiplication of complex numbers) can be written in two di↵erent ways, but it isn’t, since it has only 5’s and that 7’s as digits. 1 1 = , z z 0.7.1 “Modulus of z,” “absolute value of z,” and z are synonyms. “Real since zz�1 =�� 1��, wh|ic|h has length part of z” is the same as Rez = a. “Imaginary p|ar|t of z” is the same as 1. The modulus (absolute value) Imz =b. The “complex conjugate of z” is the same as z¯. of 3+4i is p25=5, so the modulus of (3+4i)�1 is 1/5. It fol- 0.7.3 a. The absolute value of 2+4i is |2+4i| = 2p5. The argument lowsfromthesecondstatementin (polarangle)of2+4iisarccos1/p5,whichyoucouldalsowriteasarctan2. Proposition 0.7.5 that the polar b. The absolute value of (3+4i) 1 is 1/5. The argument (polar angle) angleofz�1isminusthepolaran- is arccos(3/5). � gle of z, since the two angles sum � to 0, which is the polar angle of c. The absolute value of (1+i)5 is 4p2. The argument is 5⇡/4. (The the product zz�1 = 1. The polar complex number 1 + i has absolute value p2 and polar angle ⇡/4. De angleof 3+4i is arccos(3/5). Moivre’s formula says how to compute these for (1+i)5.) 6 Solutionsfor Chapter0 d. The absolute value of 1+4i is p17; the argument is arccos1/p17. 0.7.5 Parts 1–4 are immediate. For part 5, we find (z z )z = (x x y y )+i(y x +x y ) (x +iy ) 1 2 3 1 2 1 2 1 2 1 2 3 3 � =�(x1x2x3 y1y2x3 y1x2y3 x1�y2y3) � � � +i(x x y y y y +y x x +x y x ), 1 2 3 1 2 3 1 2 3 1 2 3 � which is equal to z (z z )=(x +iy )((x x y y )+i(y x +x y )) 1 2 3 1 1 2 3 2 3 2 3 2 3 � =(x x x x y y y y x y x y ) 1 2 3 1 2 3 1 2 3 1 2 3 � � � +i(y x x y y y +x y x +x x y ). 1 2 3 1 2 3 1 2 3 1 2 3 � Parts 6 and 7 are immediate. For part 8, multiply out: a b a2 b2 ab ab (a+ib) i = + +i a2+b2 � a2+b2 a2+b2 a2+b2 a2+b2 � a2+b2 ✓ ◆ ✓ ◆ =1+i0=1. Part 9 is also a matter of multiplying out: z (z +z )=(x +iy ) (x +iy )+(x +iy ) 1 2 3 1 1 2 2 3 3 =(x1+iy1)((x2+x3�)+i(y2+y3)) � =x (x +x ) y (y +y )+i y (x +x )+x (y +y ) 1 2 3 1 2 3 1 2 3 1 2 3 � =x1x2 y1y2+i(y1x2+x1y2)�+x1x3 y1y3+i(y1x3+x�1y3) � � =z z +z z . 1 2 1 3 0.7.7 a. The equation z u + z v = c represents an ellipse with foci | � | | � | Solution0.7.7: Rememberthat at u and v, at least if c> u v . If c= u v it is the degenerate ellipse | � | | � | thesetofpointssuchthatthesum consistingofthesegment[u,v],andifc< u v itisempty,bythetriangle of their distances to two points is inequality, which asserts that if there is a|z �sati|sfying the equality, then constant is an ellipse, with foci at thosepoints. c< u v u z + z v =c. | � || � | | � | b. Set z =x+iy; the inequality z <1 Rez becomes | | � x2+y2 <1 x, � corresponding to a region bopunded by the curve of equation x2+y2 =1 x. � We should worry that squar- If we square this equation, wpe will get the curve of equation ing the equation might have in- 1 troduced parasitic points, where x2+y2 =1 2x+x2, i.e., x= (1 y2), x2+y2 < 1 x. This is not � 2 � � � the case, since 1 x is positive which is a parabola lying on its side. The original inequality corresponds p � throughoutthe region. to the inside of the parabola. i p 1 8 0.7.9 a. The quadratic formula gives x= � ± � � , so the solutions 2 are x=i and x= 2i. �

Student Solution Manual to accompany the 5th edition of Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach

Similar Student Solution Manual to accompany the 5th edition of Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach