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Unformatted text preview: Instructor's Solution Manual for ADVANCED CALCULUS Gerald B. Folland Contents
1 Setting the Stage 1.1 Euclidean Spaces and Vectors 1.2 Subsets of Euclidean Space . . 1.3 Limits and Continuity . . . . . 1.4 Sequences . . . . . . . . . . . 1.5 Completeness . . . . . . . . . 1.6 Compactness . . . . . . . . . 1.7 Connectedness . . . . . . . . 1.8 Uniform Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1 1 2 3 4 5 5 7 8 8 9 10 10 10 11 12 14 15 17 19 19 20 21 22 24 25 25 26 27 28 30 31 32 2 Differential Calculus 2.1 Differentiability in One Variable . . . . . . . . . . . . . . 2.2 Differentiability in Several Variables . . . . . . . . . . . . 2.3 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . 2.4 The Mean Value Theorem . . . . . . . . . . . . . . . . . 2.5 Functional Relations and Implicit Functions: A First Look 2.6 HigherOrder Partial Derivatives . . . . . . . . . . . . . . 2.7 Taylor's Theorem . . . . . . . . . . . . . . . . . . . . . . 2.8 Critical Points . . . . . . . . . . . . . . . . . . . . . . . . 2.9 Extreme Value Problems . . . . . . . . . . . . . . . . . . 2.10 VectorValued Functions and Their Derivatives . . . . . . The Implicit Function Theorem and its Applications 3.1 The Implicit Function Theorem . . . . . . . . . . 3.2 Curves in the Plane . . . . . . . . . . . . . . . . 3.3 Surfaces and Curves in Space . . . . . . . . . . . 3.4 Transformations and Coordinate Systems . . . . 3.5 Functional Dependence . . . . . . . . . . . . . . Integral Calculus 4.1 Integration on the Line . . . . . . . . . . 4.2 Integration in Higher Dimensions . . . . 4.3 Multiple Integrals and Iterated Integrals . 4.4 Change of Variables for Multiple Integrals 4.5 Functions Defined by Integrals . . . . . . 4.6 Improper Integrals . . . . . . . . . . . . 4.7 Improper Multiple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 . . . . . . . iii . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv 5 Line and Surface Integrals; Vector Analysis 5.1 Arc Length and Line Integrals . . . . . 5.2 Green's Theorem . . . . . . . . . . . . 5.3 Surface Area and Surface Integrals . . . 5.4 Vector Derivatives . . . . . . . . . . . . 5.5 The Divergence Theorem . . . . . . . . 5.6 Some Applications to Physics . . . . . 5.7 Stokes's Theorem . . . . . . . . . . . . 5.8 Integrating Vector Derivatives . . . . . Infinite Series 6.1 Definitions and Examples . . . . . . . . 6.2 Series with Nonnegative Terms . . . . . 6.3 Absolute and Conditional Convergence 6.4 More Convergence Tests . . . . . . . . 6.5 Double Series; Products of Series . . . . Contents
34 34 35 36 38 38 39 40 41 43 43 44 45 46 47 49 49 50 51 53 54 56 57 59 59 60 61 62 63 65 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Functions Defined by Series and Integrals 7.1 Sequences and Series of Functions . . . . . . . . 7.2 Integrals and Derivatives of Sequences and Series 7.3 Power Series . . . . . . . . . . . . . . . . . . . 7.4 The Complex Exponential and Trig Functions . . 7.5 Functions Defined by Improper Integrals . . . . . 7.6 The Gamma Function . . . . . . . . . . . . . . . 7.7 Stirling's Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Fourier Series 8.1 Periodic Functions and Fourier Series . . . . . . . . 8.2 Convergence of Fourier Series . . . . . . . . . . . . 8.3 Derivatives, Integrals, and Uniform Convergence . . 8.4 Fourier Series on Intervals . . . . . . . . . . . . . . 8.5 Applications to Differential Equations . . . . . . . . 8.6 The InfiniteDimensional Geometry of Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . NOTE: Users of Advanced Calculus should be aware of the web site www.math.washington.edu/ folland/Homepage/index.html where a list of corrections to the book can be found. In particular, some errors in the exercises and in the answers in the back of the book were discovered in the course of preparing this solution manual. The solutions given here pertain to the corrected exercises. v g e g4 hg #9e yx#le ) 4 % ) e 4 ) e ) e ) yxw7v"e u tEsgfirq"pih gf H G C4 dV ba %V ' U SQQPI ' %4 V G U SQQPI c!`( Y6(T#R!XW (6$(16(T#R! 1FEDB!A % % )4 % % @9(18&765 &3#!2 (1 0 ) (&$#" ! ' % , , , . Setting the Stage Chapter 1 1.1 Euclidean Spaces and Vectors 2. 1. identities with the plus and minus signs gives (b). j i k g !fe g hfe di # ` A 8v##gqs % 4 444 3. 4. With as in the proof, equality holds precisely when the minimum value of is 0, that is, when for some . Thus equality holds in Cauchy's inequality precisely when and are linearly dependent. e i 5. The triangle inequality is an equality precisely when is 0, or when is a positive scalar multiple of or vice versa. z g y e y !@X5koke e g5y5e z!75ke g y g e p5pe z e y z !{ !77se z g z y g e ~k{xYs}o Cp4sg2fo { e { y g { y xkxxe {4 g {4 7kwfe zxg z!pe z y xg5ke g y hg #qe mwhk5 e g x "pe C g4 !xse hgwpe &v 9e twh g q hg7ue &q0 Bg tv9 e g q g g e hg #B7se r0h2s7ponm(9 e g e
then , so , so ; hence if also . Likewise, , that is, when the angle from then too. If not, then . 1.2 Subsets of Euclidean Space g 6. 8. This follows from the definitions by a simple calculation. 7. (a) If (b) If 1. (a)(d): See the answers in the back of the text. (e) and (f) , . segment (g) and . y AD0C sAD0C y Y vC svk C v&q 2 vBci RW0o C C pk s q q C vf2BWi . , and and then and or ; the latter possibility is excluded. (c) We always have . If and are proportional, then . nonzero vector perpendicular to , so 1 . The Pythagorean theorem follows immediately. . Taking the plus sign gives (a); adding these is the union of the unit circle and the line or . , so by (a), either to is a 2 Chapter 1. Setting the Stage 2. If , there is a ball contained in . is open, so every point of is an interior point of and hence of , so in fact and is an interior point of . Thus is open by Proposition 1.4a. Next, and are the complements of and , respectively, so they are closed by Proposition 1.4b. 6. One example (in 7. and ) is , for which . . 8. The sets in Exercise 1a and 1f are both examples. $# 1.3 Limits and Continuity
1. (a) (b) (c) 2. (a) Since (b) Since 5. The two formulas for agree along the curves the origin. It is discontinuous there since 6. Since and but , , so as is continuous except at . for all , we have as . Suppose . If is irrational, then , but there are points arbitrarily close to with . If is rational, then , . In both cases is discontinuous at . but there are points arbitrarily close to with 7. Clearly for all , so is continuous at 0. If is rational, then , but there are points arbitrarily close to with ; hence is discontinuous at . If is irrational and is the distance from to the nearest rational number with denominator , then for ; in any hence is continuous at . (There are only finitely many rational numbers with denominator bounded interval.)
H ! R D H I Sq C D "X8F D 4. and are continuous for for all , also continuous. since is continuous except at . Moreover, C Wci 3. D I V 6Q#i ! I Pq D C !Xi D D D C!0F D C !wD q s 8#i % D D #i D C "Xi D C D D !pG0F D C gfD C 4 Wo EC 0 i 0 C q X &0#i C 0& C 0 ` 9i C C C "`W0o C Ft &} C f 9 C C "80o W0o C C C "fD D F D 8Eci C g`0o C & tCci 0 0 C 0Ac C B s q8#89i 0 7 7 i80 7 w 7 v q 7 C Ac C 0 0 q q i 0 #89i i vr C @ 0 2 398v6 0 7 5 V XAi 2 31)'g`W9i 0 ( C C 4 0 C F`0o C & % C F`0o for and as as for . . . , we have , we have . as as as , so take $# 9. If then . Thus, if then . . R 5. This follows from the remarks preceding Proposition 1.4: , whereas and . is the disjoint union of z 9e 1 e v D0C j }R C ! & q e e 9e 1"v97kinm(u 4. The complements of and by Exercise 3 and Proposition 1.4b. are and , respectively, which are both open , , and 3. We use Proposition 1.4a. If , some ball centered at is contained in either hence in , so is an interior point of . If , there are balls and at and contained in and , respectively; the smaller of these balls is contained in is an interior point of . or and centered , so again R ss R k R i R 90C j 5s f ! i 96 k " ! s e q75 i " j . 1.4. Sequences 3 and , let . Then is open, and hence so is . 8. Given We have , so there exists such that . But this says that whenever , so is continuous at . One can replace "open" by "closed" in the hypothesis by the reasoning of the second paragraph of the proof of Theorem 1.13. , and let be small enough so that . Since is continuous, is a neighborhood of by Theorem 1.13 and the remarks following it. Hence it contains points in and points not in contains points in and points not in , and therefore . It follows that . . 1.4 Sequences 1. (a) Divide top and bottom by (b) to get .
I (c) Diverges since 2. 3. is , ! , and . 4. If and , then . By continuity of addition and multiplication (Theorem 1.10) and the sequential characterization of continuity (Theorem 1.15), the result follows.
H 5. If as , for any there exists such that whenever If , there exists such that whenever , and hence the other hand, if as , there exists such that for every there is an but . Let be such a point for . Then but I 6 6. If , , and , then the sequence must assume infinitely many distinct values, and for , all but finitely many of them are in ; thus is an accumulation point of . Conversely, if is an accumulation point of , for each positive integer the ball contains points of other than ; let be one. 7. If is an accumulation point of , then by Theorem 1.14 and Exercise 6. If and is not an accumulation point of , there is a neighborhood of that contains only finitely many points of . If is less than the minimum distance from to any of these points (which do not coincide with since ), is a neighborhood of that is disjoint from , and hence . &33ci 0 e 0 9 H C &% ! & 0uw9i e 9@u &! C . . On with . ! e e IV e 6Do Vi e Vi e I ! Q & (#9iv e V F ) % #I H e c C % e C H % whenever GD!4I a % ' H ! # D T%F e 9 9 i we e c 0 e e v(i # $0 ( D 0 4 C 0 I 2I I ##4 % 4 % BF$ 44 a VG uI 8(DF(73 C 0 " % % e ' C . for infinitely many . each. and points not in and hence g Conversely, suppose Since is continuous, Hence it contains points in not in . It follows that , and let ; let be small enough so that . is a neighborhood of by Theorem 1.13 again. , and so contains points in and points e v Suppose 3 v F 9. The fact that since is a onetoone correspondence between the points of the following consequences that we shall use: (i) If , . and the points of . (ii) If has , !C #ovh9i e i $9i2 l i i vg 5e e e e C % Xo g i g e i g Xg Xo g g X s g C % i g @ h9iv7 i v % % ' 0 e " I H I "% 1% ' $ e H C % e o I &2s9iv s9iv q ' 0 & 10 ) C % C e 0 H e I 6 ! e e C % e rp i u V q IV 86I 9eps ite j i wwe
0 H e ! e ! 9k &AC & 'C 0 e ! "
0 c 9iv U Vi e 4 Chapter 1. Setting the Stage 1.5 Completeness
1. (a) , so the inf and sup are and . (b) The supremum is the 0th element of the sequence; the infimum is the limit of the oddnumbered subsequence. , so the inf and sup are and . (c) 2. One example is 3. If Then , then 4. (a) If (b) The limit is zero if 5. We have . If , then . By induction, for all . This being the case, . Thus the sequence is increasing and bounded above by 2, so it converges to a limit . We have , hence , and hence or . The latter alternative is impossible since for all . 6. (a) Let obtain be the th term of the Fibonacci sequence, so . Since , we by dividing through by . Replacing by we get , and substituting in gives . is an increasing function of , and . (b) The function Hence, if then , and if then . Since and , it follows that for odd and for even. Next, , which by the hint is positive for odd and negative for even. (c) By (b), is an increasing sequence and is a decreasing sequence, bounded above and below, respectively, by . Their limits and both satisfy , and hence both are equal to . contains , we can pick ; then for all sufficiently large . Conversely, if , and then for , . 7. If converges to , and , then every ball about contains infinitely many so that we can pick 8. If is bounded and infinite, let be a sequence of distinct points of . By Theorem 1.19, this set has a convergent subsequence, and by Exercise 6 in 1.4 its limit is an accumulation point of . (At most one can be equal to ; throw it out if necessary.)
D$E0( q ) ' U fD q D q 3 1 r$Ex47I , then for all 9. If there are infinitely many for which . If there are only finitely many for which , then for sufficiently large, and hence . Since is arbitrary, we have and hence . ) ' 0(U ) 'U 0( A B@ A B@ ! $" $ &# C ! V v vr @ I ! ! ! #t9 1 9I x 10. We define a subsequence so that have . recursively. We take , and for , we choose . Then, with as in the definition of , we . It follows that . Similarly for e c e q e 9"c 0 ! e i ! P% I I # % 8&vRv e %c i $ s e ! #q e C g% e ! w"c ` # I I V tu60 B sm3 cv $ I ! v ! 4 ( I % 6 % % 3s % % 6 3q ! R 6 Rv ! 6 `s % V ki c"h5i0k`v% m`i " "10g % 6 V V "B v V I I V V Y 1( &h 6 6V w I I C % 9 Fc ! % % 2 1% ' 1 4 DY ' Av% ' E B ' % 1% ' % I % t! % %2% ' ! 1% ' % "! gA% ' c % ! D % le D ! D C! $ I V 6 lI $ FsD I 6V~lq C F}D D W D D D C 5 C i D0oru ( D D D C 7D C# (Vc0I ! %U"3 b 2 8cb V
(Exercise 1c in 1.4). and hence , 1 if or 1. , and nonexistent (or infinite) if
I % , let , considered as a fraction with denominator . is a subsequence of the given sequence that converges to . For , take ; for , take . . I 3 ) ' 05U D r$% D 3 1 $$% x42I E % " rI ) ' 0(U D q ) 'U `$h0( 2 I D $$% V cb 'V cv% w ! w" I ) ' 8(U 'V (% (c V C ! g $ (" ) ' U 05~% # ! D 1 )' $6h05U lD D ! 0"0 @ 3 and hence 1.6. Compactness 5 we have 12. With and as in the definition of and , the assertion that for is equivalent to the assertion that and for . If this holds, then for every , and hence . Conversely, if the latter condition holds, then for any there exists such that for , and so for ; hence . 1.6 Compactness 3. If is compact and is an infinite subset of , let be a sequence of distinct points of . This sequence has a convergent subsequence whose limit lies in , and is an accumulation point of (Exercise 6, 1.4). Conversely, suppose is not compact. If is not closed, there is a sequence in that converges to a point , and if is not bounded, there is a sequence in with . In either case, the set is an infinite subset of with no accumulation point in . (In the first case, the only accumulation point is ; in the second case, there is no accumulation point at all.) 5. By BolzanoWeierstrass: For , pick . Then for all , so some subsequence converges to a point . But since for , is actually in for all , i.e., By HeineBorel: Let be an open ball containing , and let . If the sets covered there would be a finite subcover; that is, . But this is false since Thus the sets do not cover , that is, . 6. (a) If , there is a sequence in that converges to , i.e., ; thus is compact, is closed, but . Then there exist , . Since is compact, by passing to a subsequence we may assume that But then also , so , contradicting . (c) One example is , . (b) Suppose 1.7 Connectedness
1. (a) The two branches ( and (b) One point in the set and the rest of the set. (c) The intersections with the halfspaces and ). . l h& i I ! ("q 1 wB89i vC fwhci q i t0 c) it 0A` C 0 0) CF if) C i0BX )$ C" q Wq) i 5 $ $ & I % i i Es such that . I i Ec C ! C 4. If not, there is a sequence but then in such that . Some subsequence , contrary to assumption. has a limit i q& (s ! "9 & I 6 V ! vci & D` i X& % C ! h& i 7I 1 C ! "X6 #i W9 C % X & 2 0 (b) bounded compact compact & Du t 2. (a) One example is (b) One example is , 1. (a) One example is , . . , . bounded. q) C " 11. If 1 I D D q A B@ q D $ A q !q D ) ' ) h0(U xD 0! 0 D 1 3 D lG1 D 0! D G $ A , then for any and for infinitely many . It follows that ; since can be arbitrarily small, the same is true with .
) ' 0(U I lD ! y! pD C % D Gq 6 Xi D0om V C "`i jg j " @D$qE0"0 D! C ~% Xi 1 2I A @ q )' ED 5h05U q vD Y D$16h)05U ' A D q D! 50"0 A @ 0 ! &" 1 3 ; . , . . 6 Chapter 1. Setting the Stage
are points in the unit sphere , the plane through , , and the origin (that is, the linear span of and ) intersects in a circle, and either of the two arcs between and provides a continuous path in from to . (If , any great circle through will do.) This argument works in any number of dimensions. 4. Suppose is disconnected, so where neither nor intersects the closure of the other one. Then is a disconnection of unless either or is empty, i.e., or . Likewise, we must have or . It cannot be that and are both contained in (resp. ), for then (resp. ) would be empty; so and or vice versa. Either alternative contradicts the assumption that . is connected when by Theorem 1.25, but not when . For example, take to be the unit sphere (Exercise 2) and to be a line through the origin; the intersection consists of two points. where and are open and disjoint. If , there is a ball centered at that 5. Suppose is contained in and hence is disjoint from ; hence . Likewise , so is a disconnection of . is a disconnection. If , there is a ball centered Conversely, suppose is open and at that is contained in (since is open) and a ball centered at that does not intersect (since ). The smaller of these two is a ball centered at that is contained in . Thus every point of is an interior point of , so is open; likewise, is open. 7. If is a disconnection of , define for and for . Each point of has a neighborhood that does not intersect , so that is constant on ; likewise with and switched. It follows that is continuous on . Conversely, if maps continuously onto , let and . If , then since is continuous, so . Thus , and likewise with and switched, so is a disconnection of . 8. Suppose is a disconnection of . Then or is empty. The latter alternatives are impossible: If since does not intersect the closure of , we would have definition of disconnection. $ G Vi ! im vv vvC p vD0C is a disconnection of unless , say, then ; but , contrary to the . C ) &Bqi 9. Pick or . If we are done. Otherwise, either or , in which case respectively; either way, the intermediate value theorem implies that C ! 9B t i Y9i C ! 9i i t C % 9i C 9i i 5e Vi e 5 C !9i p C "9i i ) C % "99B i 7 i 5e 6. If where and are closed and disjoint, it is immediate that is a disconnection of . Conversely, if is closed and is a disconnection of , suppose . Since is closed, we have ; since , we have . Hence , so is closed. Likewise, is closed. for some i q i i 89i 8i i Fci i @i i @i 1 8i i q i pe i i ! % 1 i v i 2 Vi k F i i 5 i ! G ) ! 3. If is neither strictly increasing nor strictly decreasing, one can find points such that (i) , and (ii) either and , or and ; we assume the former alternative. If or , then is not onetoone. Otherwise, the intervals and are nonempty, and one is contained in the other. Assuming is continuous, the intermediate value theorem implies that and , so there are points in and at which takes the same value, and again is not onetoone. i i c g e e e yg g e g 2. If g e g e e 1.8. Uniform Continuity 7 11. (a) The graph , , is arcwise connected almost by definition (it's an arc!), and is its closure. (Check that every point in has a neighborhood that does not intersect , and that every neighborhood of every point on the vertical line segment contains points of the graph .) So is connected by Exercise 8. (b) Suppose is continuous and satisfies and . The first component of is continuous, so by the intermediate value theroem, for each there exists so that and hence (= the only point in with coordinate ). As goes from to ( ,) must assume all values between and , and hence must assume all values between and (again because there is only one point in with a given coordinate in to ). By passing to a subsequence, by this range, and the coordinates of these points range from BolzanoWeierstrass we may assume that . Every neighborhood of contains points at which assumes any given value between and , so cannot be continuous at , contrary to assumption. 1.8 Uniform Continuity
1. Given 2. (a)
! ) Q#qisX9 ) q ) is 6#qi X9i &r#qiT`9iThn H ! )"2 i )X ! ) R#qis39iq H H C% D C !% ) ! #qi 9i wp &q t s }l &q } & q X ! c D # ! D T0$F # D $F ! ) C % V &wk , if then . , so . (b) For any hence 3. Given and we have , and likewise with and . , we can choose so that whenever and 5. If is unbounded, we can find a sequence in such that . If also is bounded, by passing to a subsequence we may assume that converges to some limit (which may not be in ). Then will be as small as we please provided and are sufficiently large, but for any we can find such that and hence . Thus is not uniformly continuous on . 0u ! % I w ) #si@9i 4. Suppose
) is uniformly continuous and whenever . It follows that counterexample, take and is Cauchy. Given , there exists so that , and there exists such that whenever whenever , so is Cauchy. For the or . H whenever . Let whenever and and %V I AD0C i # switched;
! . Then . 1 #iv q i % "D# iv &#(9i I Ws 2 0 #wci (s V b U S Q ) 6co6(Tgi V 6FXi I V 6 3 Rwciv % I w ! #iv h i H ) H ! v" ) H ! g2 C t% C % q ! ) hw5 i H) X H H H D ! ) hx i ) X C D0o 5 I %V C Wv% kWov C y AD fvC I %V 10. and , so there is a point such that , i.e., "f C "`ci i hci
! Q ) #qiv s9i C % V WI `wi 0 %qfAC 6co! "u ! V b U a !D8 % 1Fu(R % &I . I k I % V ` i 0AD0C  V b U 6co! gu v C 0 & C
0 C
,
0 b %V uc0I C 0 CD lD rFr5 D i D i 8F F y% V D D C C q C V "}6i Wo 2#i C ! V U S Q V U % C C C W6(Tu06! Yv`i " (Wo Wo
, and for , , so every interval about 0 contains small subintervals on which C 0 q V U s q#6! %Y C V U $6! @ 6v(Tk V U % VU SQ 6 ! @v!`i 6i 3Wo V C ## 0F i T4#4#v i vi wi ## 0F D 4 D # D #%rF DD9 C % # ! D %D i #
! D 0F # D T%F 2.1 Differentiability in One Variable Differential Calculus Chapter 2 1. Suppose Otherwise, let and . If be the points of on , then by the mean value theorem. at which vanishes, in increasing order. Then . Each of the differences on the right s## 2. We have have term has no limit. 3. is positive by the mean value theorem. since ; the first term on the right approaches 0 as . In particular, 5. For 4. 6. These formulas are obtained by applying l'H^ pital's rule 2 or 3 times. The general result is that o , where is the operator defined by . (Explicitly, .) D 0F F D 8F D 0 C 4 ( " 6Vs)76i ( V V 6i C$Wo C dxI 6V 2I V 6 ! I ! % " 6 6u'v( 6(% A 6 ' 6 V % V V % V " 6Vu%&i " ( 6(1 i V% I F $ 6# C " # V" # V ! 4 s 4 1 4 # D D V #uS 6## W"u S D $F A X8F D D D 8F " 0F V D 8. 7. 9. (a) (b) This is the case (c) For , . Assume ; then . The first term in brackets is a polynomial of degree in , and the second term is a polynomial of degree in . (d) The case is (b). Assuming by inductive hypothesis that , we have . By (c), where is a polynomial; hence the limit as is 0 by (a). C 3Wo , . of (a). I i since . By l'H^ pital, the latter quantity tends to o for some by Corollary 2.12. for all . 8 . The calculation for cross products is similar. . As . as . For we , but the second also, and so . e #v s q D D # E F s ) F D# F s ) F &v#o sq0Eo D D D D D D q e ) e t ! e v "h) C % i ke v q v q % ! # $ 8( C C C gH W0o H C @W0o 3W0o C C C C b C m H H H H $ 8% H H # ! U USQ V U USQ & C C ! (To U USQ (T"`6 ! 00 (T0i ' W0o C C 0 W0o @89i C 0 ci s 0#ci q C" u Rr " u " " ( r ) u ir " " u " ( r ) u ir ( ) u rB ` u y u % % v% $w i0V v! # vw ri0V BF ! v2 v% ! v g ! % # v2 % a )v v! . C 1Dv% D % C (R0o 2 2.2. Differentiability in Several Variables 10. is well defined since the two formulas agree at ; it is continuous on by inspection and continuous at and by definition of and . By the intermediate value theorem, for any between and there exists such that , and by the mean value theorem, for some (if ) or (if ). 9 7$ i 1 q % C (R0o i C Wo 2.2 Differentiability in Several Variables 1. (a) See the answer in the back of the text. 14 7 $5 4 f(1D 7 ( Dv5 f(1D 8v&x60(Dr7Twc8rkTw 89i % % V % % % % E%4 7 34 ( `(1D 7 ( &8 F(1D " # 7 v1c " # 7 mX89i 2. (a) See the answer in the back of the text. C 5 C d w(C CVi6Y% l a c(BC 9((C #(EC a g@ Ca2 C( 7 v "38% "Y v% ( ( r C "(% oC8 2 u ! C WC D % Xir uc vs` Ximci s` ; C WD 0oi$ C % C # C Wo (b) (c) (b) . ; ; ; ; ; ( 4. 8. Assume for simplicity; the general case follows by an elaboration of the argument as in the proof of Theorem 2.19. Suppose and on . Given , let be small enough so that . If , by the mean value theorem we have 5. 3. 7. (a) 6. Since (c) Taking follow that (b) With which implies the continuity of (b) The coefficient of (a) or , so , is largest. we see that for all ; but this is false. , we have and . at . as , so . ; the result follows. . If were differentiable it would and hence . . . . 9 C %4 89i @8 9B C C % C q C !`8i B 89i } `ci Ct! D0C y D7D % ! C"` D u# T{o Xo Xso e g z!Xo e i 5e i 5e g # T0 i 2.3 The Chain Rule 10 2. (a) and (c) See the answer in the back of the text. (b) , 6c9i 4## i vcci i 44 C 5 2v" w !`( 0 " i85 s(2s ir C % % w v1D 8 B(%RDv1 tEtE rv& 5 % % # g l S r"ci # D # D #%F #%rF # X0D"fi $#%DF s # #%F %F # D # D W V " V ( V W F V ! 9 g # vi p rp i E W vi vf vi @ g # C"`v2lv (d # % % d g i0V v% c # ) ( 7 V ( % ( v29 # )
( YF # 3. (a) (b) (c) 1. See the answers in the back of the text. . t` 2.4 The Mean Value Theorem 7. We have at 6. These are all similar; we just do (d). If and , so 4. 5. Both formulas for the tangent plane at where and . 1. (a) The conclusion is that the directional derivative vanishes at some point on the line segment, where . (Apply Rolle's theorem to the function on the interval .) (b) The conclusion is that there is a point such that . (Suppose on . If on then on . Otherwise, either the maximum or the minimum of on [which exist since is compact] is achieved at a point , and then .) g e @ki 2.5 Functional Relations and Implicit Functions: A First Look z e !so i 8$ 2. (a) If is convex, we can apply Theorem 2.39 to get (b) An example with : let be the square removed. Define on by if , and . Then on , but 1. (a) See the answer in the back of the text. (b) Differentiation in gives tiation in gives . V % V (% Ei0 pi w 1% (%2 i0 ir 1! vv%5 v% w X%! vv%5 v%! v7 ' 9ir e(!} V % V % sv% ' V9i } 8 e V v% ' e35 }v%mf svh ' e31 k % % v% rkw4 6 Vd 3 % 08vYF # 1 % "lw V Y % 0w FE C "l 2. (a) Elimination of gives . Therefore (b) Differentiating the original equations in , so and . , so , so , or . , so amount to , or . . with the segment on the axis if and , and if for . , we have , , . Hence the tangent plane is given by , these derivatives being evaluated . Chapter 2. Differential Calculus . and hence gives , and . and . Likewise, differen . Thus . . C % C % 24 z AD0C V g 9e g 8e wo # & #9 i V D 9i FD 9i 8i 8i li p# i pi 8i 8i fi p# i i ( ( ( ( % ) 7 iT w %2 v% !9 # ) G ( ( 7 i( w ( 7 i # ) ( vD ( ( % ( # G 7 i(( v% # # 7 ivm # ) G ( ( 3 ( % ( ( % # # ) 1 # isvy( # ) 7 i ( vm3 # ) W 87 iw( v% ( k isv%p # () # () YF # # ) 1 # USQ v(TYvd # v(Ts & v(To8 US v(TQ@v USQ U S Q ( U ( w # 8vY! oF # % # v! o %!u U ( ( % U ( w v1! o!q & U % vD" ( ( w 4 v! oq % H H # TsH u# ! H X U U Hr H # (S Q ! UX V V US H H (TS oQ v# T ! oU E# (ToQ U B H H Q H H Vv# H # (U TS XH V T (To ! oE USQ U
and # !% U o % V Fk #V V V # (! V% V V V F b # V W V V !@ V %V b% % 0V l b b% V V " VW b%p b" V % 0W W # V b 9i v" vp % d % vpv! ( i8u g % v~ vv% C % y$w g #  ir 7 va 2.6. HigherOrder Partial Derivatives 3. Differentiating the equations gives which are linear equations in and and F V V b%! V b! V # V ! b% # V 89i g
are IVs, 4. If to be solved simultaneously. . If are IVs, 7. Taking say 6. 5. For and the equation then becomes . Now take and as the independent variables. Differentiating (*) with respect to gives and , so and . Differentiating (*) with respect to gives , so . Substituting these into gives . % "&% 7 f C " C V 5 Th Th t C V y5 TB x V Ts % % Ch % t 2 C h % % @F% is the derivative when Likewise , is implicitly a function of and ; the equation , so . For are implicitly functions of and ; differentiating the given equations in gives and , whence . For , and are implicitly functions of and ; differentiating the given equations in gives and , whence . and , just use yields , and as independent variables means that is considered as a function of and ; by (2.44) it equals and . The product of these quantities is . For and are determined as functions of 2.6 HigherOrder Partial Derivatives 2. From Example 4, 4. We have 3. (a) . Hence . The derivatives are just like but with an extra subscript on the 's: , etc. Now it's just a matter of collecting terms. Similarly, , etc. (The multiplying in the answer in the book should be .) 1. These are routine exercises in elementary calculus. Half of the calculations were performed in Exercise 1, 2.2. 5. (b) ; hence . The result follows. and ; furthermore, . are both equal to . Hence and hence . Similarly, , which works out to be and . ; also . and . (*) 11 , . . , 1 3 b q 8C A A 8s C ~C V cb rd3 % 0V A s dq#i #iv q G C a q &rsh! %U v#i ! U a q V X Fr#i 7 (( f7 B! U 7 q % q 7 (a F( 7 (8Wu d (a r 0 ) (2s ir % m(# i 7 ( ( % % i81muiR ( vW f (2ui8% i ( si y i 7 %4 a q q 0 6@! V % 7 vmW#i ( ( ( ' ( iR ( )' ~hi hi h~h % q (w! V d G d G q 7 d ( 0W wv v ( #iR 7 ( ( ) vui ( % 'Byi !ui % 7 444 V444 V 0 4 4 4 V444 CA##"3 0A##g XT##t w 0A##g ui i 444 44 7 4 ##4t4 w ` A#4 #" )w T A#4 #4 " i A}A##p vs A}WT##pp `i 444 % 444 444 ##p A##p ( 1i51 gXi 4
; also , for ( rW (& & Q 44 ##4 I A r( G W kW ( ) W W W g ( W s ( X A W X % W X A W XA ( ` ( sA ( 7 7 W XA ( W XA v%
, and hence w W XA ) W X W X v W X A F V V g} W wqW XA
.
( ) BF li A pi q& & 1 % $& & # % v1h$ % & # # w % # # w # v! % % 12 6. , so . The result follows. . Also Chapter 2. Differential Calculus , so 7. 8. We have , so W sA and ## 9. are the same, with , so since replaced by and . Since . Adding these up gives 10. In one variable, the assertion is that binomial theorem. (Induction on , using the fact that .) The variable result follows by applying the onevariable result in each variable separately; the facts that for all and make everything turn out right. (This could be phrased as an induction on .) 5 2.7 Taylor's Theorem 11. This follows by applying the onedimensional binomial theorem in each variable as in the preceding problem. 1. (a) 2. (a) 3. By Lagrange's formula, since . In general, we have less than for provided ; so the 5th order polynomial suffices. (c) (b) (b) also , , , and , so . . , and for , and , and . , so , so for , so . , so , so , which is proved just like the , adding these gives ; also . . , and . This is ; I x ) s#s) v q q h) ) I 9 h ) s ) ) 3 x1 V ) $8#s) I q C " D 0F X $F D I C ! D 8F s $F D ! D C I D C 8F % 6V 8F 1 8F D h lF D C % 0FR D I I V 6 D q 0 F u# v v 6V D I D D 8F q8F Y EF D 8F C D !`0F 8F s F D D 0FR q8FR s F D D D 0 % s $F D 2 I I % I  r4$EI 4Y % I w E 1I u% E 9i & I u% E ( % 444 444 ##t ( $ I b 444 b I I ( o I 2 !`T##g ( I I I T 1 !cA##g I o !XI XS h 3I r(Tf h!Xci b USQ I sFs h!f % ##" 7 444 ( r ( d 7 D % ( $ % ( ad rw % lwp 444 ##g 7 v 7 ( $ v wl f ##" 7 % ( r % y 444 % d % % % % % ( 7 ( v2 w ( v2 v v u 4##4g 7 4 xli ( wli! wli! wiu y xBu y 4##4g 7 v% i % 4 % d % % % ( v i 8vfi v i!u # W ` i(TQ # U S U sli g`li! %Y 4 4#4#4 #7 44 7)i srW 7 ##4 i 4 84 4 #4 #4( r i 4 4 #4#g( l ( ) g##g88li !`"EI % 6V C I I V 6 dr1 I aCC (uI % & (wwC C i AD0C h BI 0V i0 q 0X i 5 " 0 V Xi I & 2.7. Taylor's Theorem 4. For the remainder is bounded by . This is less than for , so , so to three decimal . I 0 V 6. Setting 9. We have we have 7. With 5. (a) 8. A places all 4th order derivatives of (b) For by definition of (c) Thus for for small , and likewise if odd, the same reasoning shows that 10. By (2.70) we have is less than in absolute value, we have where fold application of l'H^ pital gives o , , . , we have , with no remainder. (The remainder is also known to vanish since vanish.) , we have , , and by Corollary 2.60, for sufficiently small even, if we have we have for small . changes sign along with at . . Since each component of . Hence for . 13 . . . for C C le 00o C C Wle 0o C g} 444 T##2 7h% A##4 phT(% v 44 6 (f v mWle 9i C e l % C C " % " # 6 v!uW0o F` C C Wle 0o 9i C C W0le & 2 30 C 0 9i C C C z W00o d~ C % ~9i C C le 0 0 o C C C C C C C Wle 0o W0le W00o e lf C "l C " 3# C leh le C C  v% C 2 %t v% Chg C # v C % v % v"% v% # C g v% vE qv%& v " v t Er % " " # ' W T e C ' Wv T C ' W8 YD % v1 vd d C e y ' r e ld % (2 i81 # " i8 C &D ## ! V C % D ## V ( 0D V ( D ## V Ff # ~ x # F$# x W1F D C "$# VD% # 06rf D ## r V % D G C " C le 0o C Wle C C C W0o W e C le 0o C C W0o le % v% C w sC 7v1 # % letg C t fC tv% 7# % t v % Cy7 m C " # " 6 v23vm # % % " # " 6 v% qv % % C 7 7 C t% 7 7 % }D % % D % ! D C ! C ! 8E0o D C D WCF % ! % % }1 5 % w C"s !f C C! 7 7 7 "f uv7 v!uv7 v % % C "3# C rE# C &f rx C 8v}v% l &E# % 8v}v r& ( C W ( C WC0o e Y8rx % v1~ % }vr % le d C W ( C C W0o erf r C C C p &p r x C 5 C F # % @vy # % 5v v ' W% Al e % ' 0o C d C vFx r % vD % (p i8! # k iqw W% ' Dle % ' D0o C % ' D e (% ' le (% ' e 0o C % C C C (YD0o W0o 2.8 Critical Points
14 1. We employ the notation , , as in Theorem 2.82. (a) and , so the critical points are , , and . Also , , , so these points are respectively a saddle, a minimum, and a minimum. (b) and , so the critical points are , , and . Also , , , so and are saddles, is a maximum, and are minima. (c) and . If then either or . In the first case, or ; in the second case, . So the critical points are , , and . Also, , , and , so and are saddles and is a minimum. (d) and . If either or then ; otherwise, only when , that is, . Thus the critical points are and all points on the and axes. Note that on the lines , , and ; when and when . Thus the points and are local elsewhere, , local (nonstrict) maxima when , and saddle points when . (nonstrict) minima when Also, , and is inside the triangle bounded by the lines on which , so it must be a maximum. (One could also check this by Theorem 2.82.) and . Thus (e) or . In the first case, or . In the second case, and hence , or . Thus the critical points are , , and . is obviously the global minimum. A straightforward but tedious application of Theorem 2.82 shows that are maxima and are saddles. (See also the solution to Exercise 1h below.) (f) and . If then ; substituting this into gives and . At this critical point, , , and , so the point and a maximum if . is a minimum if (g) , , and , so the critical points are those where , , and . The Hessian matrix is diagonal with diagonal entries , , . Thus is a minimum, is a maximum, and are saddles. (h) With we have , , . If , then either or . In the and first case the equations give or , and or ; ; , ; , . In the second case, the equations the solutions are give and hence . So the critical points are , , , and . One can analyze them without the tedium of computing all the second derivatives as follows. Since for , is obviously the global minimum. Since as , must have a global maximum; by examining the values at the critical points one sees that the maximum is at . Now consider for near . Since has a maximum at , has a maximum at . On the other hand, has a local minimum at . Thus is a saddle, and likewise so is . C D0o t}w vw " % % d C & % % % iT(sivDh}# v% Chapter 2. Differential Calculus 2.9. Extreme Value Problems 15 precisely on the planes , , , and that include (i) Note that the faces of the tetrahedron with vertices at , , , and . Since on the set where , all points on the lines where these planes intersect are degenerate critical points, and none of them are maxima or minima since changes sign whenever one crosses one of these planes. Since , , and , one sees that the , i.e., . This point only other critical point is where is inside the tetrahedron, there, and on the faces of the tetrahedron; hence this point is a local maximum. (b) For we have + higher order, and ; these all have local minima at the origin. However, is negative in the region between the two parabolas and and positive in the regions inside both or outside both. The origin is on the boundary of all these regions, so has neither a maximum or a minimum there.
# 2.9 Extreme Value Problems
1. 2. circle, the second case, 3. and , so the critical points are at , which are both outside the triangle. The sides of the triangle are segments in the lines , , and . We have , whose critical points are at , and . Also, , whose only critical point in the interval is at , where . Next, , whose only critical point in the interval is at , where . Finally, at all three and the vertices. Comparing all these values, we see that the minimum is maximum is . G C W 5 ! ' (% V 5 6 60 w5B % V ' %d 5C C % V w "l% ' C V C q q r&u E ' v2 w ( hv2@Rci % % % V v0 (w B ~ '%d 5C V 0 f ' q p2q a % % vl w 0( g`vc Rci uC '((1W8 le V% C ' W y7 'V e Y( 1W9i C % vg@%f % % C vE~p yp 'V cv le % pv% p# ! v H H H H H C e C US C !H ! s (Ts% UH a U S Q H !H H l& ! U H H W ! H %U ! H (TQ H (STst ! %U (TX# B ! (To V U Q % U S Q C U USQ ! st U % U a U U S Q ! syB ! (To C 0o C % Fkw g# vd w 0o H H b C H H H H ! H " H C U S USQ U U S Q % U USQ (TQ ~ (Ts% ! 2 (Ts&u ! % (To C C W % v! # % w F`W and , so the only critical point is (at and ). So the maximum is , and . On the unit circle, , whose maximum and minimum are and and the minimum is . . On the unit or . In and , so the only critical point is , and , so when or . In the first case, . So the minimum is and the maximum is . , so # & # # Eo 4 oXo e e e " 5. The second directional derivative of
" in the direction at is % vr 7 7 D% C D #6!`WAWF e 4. (a) The 2ndorder Taylor polynomial of at the origin is , so the origin is a degenerate critical point. ( 3. The origin is a global minimum for and a saddle point for righthand graph of Figure 2.5 (but upside down). . For it is a "shoulder point" as in the # D B # ! D B 2. We have a maximum if , , and , so the origin is a minimum if and , and , and a saddle if . (If , then [any combination of the two signs can occur], so the origin is still an extremum.) C pci 1 D C m C&% DD %3" $v%"E 3lv% % l7 7 r v% r ! # v% r ! C " # z sg5
1 D !B z C C C " % C q tt D p# # # # D X % F # e 8 ( ! D e # D ' 1 !B p# 6% D & # o e C C "l# 16 4. and . Next one analyzes . (ii) Chapter 2. Differential Calculus
; setting these simultaneously equal to 0 gives , on the four sides: (i) ; critical point ; critical point at , not in the range of ; critical point at , not in the range of interest. (iv) ; critical point at ; . Finally one checks the corners: , , and . The minimum is and the maximum and at , and interest. (iii) , is 56. gives . 7. As in Exercise 1e in 2.8, one finds that the critical points are and , and and . Since assumes both positive and negative values and vanishes at infinity, by Theorem 2.83b it has both an absolute maximum and an absolute minimum, which must occur at critical points; hence the maximum is and the minimum is . 8. If is in the first quadrant but outside the "triangle" bounded by the lines and and the hyperbola , then ; hence has a minimum but no maximum by Theorem , (see Exercise 1f in 2.8), and the value 2.83a. The only critical point is at of there is ; this has to be the minimum. , , and subject to 9. Lagrange's method works easily here: one has to solve the constraint . The first equation implies that or . If , the second and third equations imply that and hence . If , the second equation forces or . If , then , so ; if , then . So the constrained critical points are , , and . Clearly the first pair gives the minimum of and the last pair gives the maximum of .
#
D D I 10. With and the asserted result. , we have . Solving this pair of linear equations for 11. By Lagrange's method, we solve subject to . With , the first set of equations gives , , constraint equation then gives , and hence is a minimum because as on the constraint surface.) 12. We wish to minimize subject to the constraint (and method, we solve subject to the constraint , so . There is no maximum; no matter how large is. ). By Lagrange's ; the solution is obviously satisfies $ and yields , and the . (This 2 0 @ci 2 0 R l ' q# ' YD ' ' ' c# ' YD ' E YwE ' w # ' D ' % 1 V VD 986## 968F % % D % 1 R# 1 1$ V Fs `WC v v9i s r k % s e I # c# D !`## WD i 1F % % V f C F% c d C le 0o e lmf C r5 C d e C rx ~p % % 8&pw kv% 6. Clearly except when maximum by Theorem 2.83b, namely critical points.) , so is the absolute minimum. has an absolute . (See Exercise 1e in 2.8 for the analysis of the and 89i and 5. Clearly , so has a minimum on by Theorem 2.83a. We have . Setting these equal to 0 simultaneously gives . Substituting these values into and simplifying # 1~ % ys5 ( 5 1B(R8 % ( ( da C wrhW8 Dh(R0o &BW0o % C C C C ( (R % %% Dpvw v Y(Rci w % 1t w45 d d vD&swa s8 ( t % % v! v lWC 9i d % C D (R0o vdDy C G t}D w f0o d 5 D" v5 fv1 # % d 2 v5 fv C W e 5 ( ((D Vd ( V(G f ci Y % % C !` e 0o C"WC0o C C "sg} C 89i % # u0V # 0# s V V # 0## } % vFh # v%rA 7 # % w 1 # #cin1 #89i D00Co e C C C WD0o e W0 e C C ley} CFp % f C } C w ~ 2 j ( ( x lt r s s 8 p % % # D % !`# 4 i YF # D ## WD i `#%rF kr 1yule 0oC % ( % (D r f 7 2 l D V ci % Fx 2.10. VectorValued Functions and Their Derivatives 17 and the second one by , we wish to 13. Parametrizing the first line by minimize . We have and . so the critical point is , . The point on the first line closest to (and the minimum distance is ). the second one is 15. We wish to minimize subject to the constraints and , , and to obtain method, we solve simultaneously with the constraint equations gives , . , 16. (a) Lagrange's method gives the equations , . Eliminating and (and assuming or maximum is clearly positive) we obtain such that zero); either way, the critical points are those the constraints, the constant of proportionality is , and (b) Using the parametrization, maximum is obviously . . By Lagrange's ; solving this , , which is OK since the (whichever avoids division by is proportional to . By . , whose 17. The distance from to is , and the distance from to ). Thus the total travel time from to is is that the distances have to add up right: method gives the equations and whence . 19. Let be an orthonormal eigenbasis for , with eigenvalues . If , we have and ; thus we wish to maximize and minimize subject to the constraint . By an easy extension of the argument in Exercise 9, Lagrange's method shows that the critical points are the unit eigenvectors of . (Things are a little messier if the eigenvalues are not all distinct.) If is a unit eigenvector with eigenvalue , then , so the maximum and minimum are the largest and smallest of the 's. 2.10 VectorValued Functions and Their Derivatives 1. . VD 444 ##s t 444 R##9i #q 4 6p V 4#4#4t ciq ##9i 444 VD B v% tx 4 lvd v v5 V D F u # 444 76##cc 44 "0##4 c civ 18. Let be the product of the 's with the th term omitted. Lagrange's method gives the equations for all , from which it follows that the 's are all equal. Thus the maximum value of occurs at , and that value is . In other words, is at most when , that is, , with equality only when the 's are all equal. H H H Q U H ! H Q v V i I U H Q 0 ! U H I 6c" ! g$ ! I I H Q v V i H Q vA V 0i % h% U U H Q U B % ( is (remember that . The constraint . Thus, Lagrange's , H H s ! U # D q (TQ ! U US D V # G i 8D ## w B V B C ~ % % % X $ ivY 8` $ i Y v% ( WV vd&k % v% d !s5lv 14. We wish to minimize equation for gives dV C W ` ( d % ` @ % qv!s& Y vxBw % % % v ! pB g pB um #is c y`W AAi 3iq C W( W @ @v v % u v$ v% r V l V g} % v2 " v2 % V % V % 8v%&xv% 0rv%! t v% # 8vrwv% 08vg % V v!kv% 082 rd r C % "s !l } r % H ! U (To # G l i USQ D i D T# e V ci 6V 6V % ! i % V yv% H H Q r0 U V H ! !p V H U Q vA Rri U V mp v% w # ( D % 3 % !% isv% U C % ! % subject to the constraint . Solving the constraint . After a little calculation we find and . Setting these equal to 0 gives , so that ; hence and . % 3 ' Y#DvDn Yu A6D#ci 3 ' u #v s I 3 ' 00D s q I
3 ' w 9 n s I z r 3 ' W60 C C % I C r Y v s 3 I ' u q 3 v for all and , so e % 01 C # d o e ) e ) e ) e `ov qq0o 5 # C C C C 5 1 % % W00oTv % C # # # USQ (T5 i8% % uci , a Y C C % C1 Da WC0oC C % % C (RT % a 1 C 0V Y % % % ` a 1 % v% v v1 d v % `civ 18 3. (a) C % 2. (b) % 4. (a) (b) W 5. 7. 6. This is immediate from the definitions. , so . . ; this is the th component of 8. By (2.86), C C & & 7 ( & 9. (a) The unit sphere . point which gives the desired result. is compact, and is continuous on it, so for all . . achieves a maximum . Chapter 2. Differential Calculus . on q u 9u q ru V 9F u p e 9 z C rqz (b) Obviously , with equality if . . If . Thus , let and . Then is the smallest constant such that # # ' Y q j i } i 5e 10. (a) If so , we have . 1u (b) If and for for any ; in particular, if .
1 , the bound for , then in (a) is and . Also, for all , i.e., at some , so . , u V vi C % C vci V 1 2ci d V DD 1 % r "Y % 8Bq@ r8 v1r v$rim`ci u v v% % l l t C C FW0o6 C C C gW0o C C C C !uA"cW0o( W C 0o 8 ci 0ci 8 ci 6 $# 89i 89i C C uW0ov# j 2 0 ( ! U & V % V ! # US (TQ # # C C W0o 3ci C ! m& C % tR ci d % vx@v!$# % % vx@v! CghWCD0o C C C FuWD0o0# h9i % i C % v2 limci % C C W0oyEci % vxBvx g89i and (R vva !f % e % V 5 ww8$( Y t (RDv% B `w 3# Bv! % ui s V 6 v$w i % C D D0o % vp`w guci The Implicit Function Theorem and its Applications Chapter 3 3.1 The Implicit Function Theorem 2. With is nonzero when and the set where 1. With . Hence the equation and square root vanishes at and none for others. we have and , so at least one of and , which is the case when . If then is a single point, and if then the set where is empty. (Clearly when .) C & C &@ 3. With so and ; the tangent line is vertical at the former 4. The and intercepts of the graph are at point. Writing , we see that the graph is asymptotic to the curve as and asymptotic to the curve as ; hence maps onto . Also, , so is onetoone (in fact, strictly increasing). Hence exists. ( ~}
( jtGw 0 j j Fp 6( D0o C C W W % FXv $iT V 2 0 5id1i k and we have , , , so can be solved locally for or but not . Explicitly, ; but , and the , so there are two values of for some nearby values of and , we have . Hence the equation can be solved for and but not . 5. With 6. With At , then, we have . So the equations can be solved for , we have when and ; 19 , we have and or for . and . , , so , , . 20 7. With Chapter 3. The Implicit Function Theorem and its Applications
, we have , then, we have . The determinants of all the submatrices of this matrix are nonzero, so the equations can be solved for any pair of variables. which equals at . So the equations can be solved for . 3.2 Curves in the Plane , ; the endpoint is , and C C "uWo C Wov D 3. (a) (b) is the parabola is the halfline . . e e ls lf %t% W C C D0o 2 e 0 B x % dS W7B 2. (a) except at , so is locally a smooth curve. is also connected; see part (b). , is a straight line. If , is the unit circle. If is odd, is the graph of (b) If , a curve that is asymptotic to the line as but has a "bump" in the middle where it goes through the first quadrant between and . If is even, is a simple closed curve intermediate between the unit circle and the square with vertices at , . (c) If is even, the top and bottom halves of are graphs of , a continuous function on that is differentiable except at the endpoints; likewise with and switched. If is odd, is the graph of , a continuous function that is differentiable except at ; likewise with and switched. i C ' W0 R } i i FukY i C D0o C C W0o V i WC0o C C "s z m ks s VEW0o `ci i C C C C W0om@ 6iq`(f9i z! @i hcb C % q0o C C W0o 8i z lW0o C C i f i z C C RW0o r} F l C "} i C p C 7 e s z r % 1 % lgs imf i ( f WCv 'Ee @ p 8i 3 f C ' f W0 e 1. (a) is a smooth curve (an ellipse); only at ; near any point except ; near any point except (b) is the union of two smooth curves (a hyperbola); only at ; near any point except ; near any point. never vanishes; except near ; (c) is a smooth curve (one branch of a hyperbola); globally. (d) is the union of the three lines , , and ; at the three points where near all points of the lines and and these lines intersect; elsewhere, near all points of the lines and . (e) is the union of and the parabola ; at but not on the parabola; on the parabola; near any point of the parabola except . (f) is the parabola ; but is still a smooth curve there; globally; near any point except . ; at each point of . (g) is the discrete set c i  Vi C C W0o le 0o C z h C % WDvYD 8ci 5 AD 8i } 9. With , , and , we have % v% v% equals when all variables are set equal to 1. Hence the equations can be solved for and . , ci & ` q g `rg f " %1 8. With , we have , which % y xu% v% Y % v% t % C & u }( p 5( u 9i ( % v23 A( r C DDD0o 8 . At u } H H ! %v (Tv U USQ "U (TQ ! US , . r 0V T V i! i C ! C " C " e @ t u gw e F le i C C le v hW0o 2 C C W0o C C W0o C W C C W0o 2 D i 2 2 F} ci D g} 68im V ! i0 i V V V V V X 68i0w 68i X p i0 f i f C C e V W0om`89i lX 68t V w i0p iT i`89i ci t m i0r i V @ ptE u"!vc(bU Ts $e C fi r(Tvh! %s$e "o& V USQ U % ! U U SQ % i b f SQ XU(To& C C C V W D0o W D0o W C C i bRsb ( % b ( sb bsb (T! ! U USQ b 0C C "3 d C R0" ' C C W v0o Wv% to C D0o e W e C C D Wov
. b z i ` !`i F ( w ( C C "`Wo % lgs 3.3. Surfaces and Curves in Space (c) is the line ; but (d) is the astroid the points where (e) is a limacon; a reasonable sketch can be obtained by drawing a smooth curve from to to to to to (corresponding to for a smooth curve except at , where it has a selfintersection; never vanishes. V cb i C Wv% , a simple closed curve with cusps at , ). i 3.3 Surfaces and Curves in Space z ( C " ( ' C W @D0o W C C W C C W0o C W ( 1. (a) ! 7p "(2$i8 s fi 8 C % WC C C WD YW v1Dv (RD~WCv WCv 1Dv1D % ( c ~B lwpv% "` E8s(2 i8s ir C % % % % Dv1D D (1Dv%1D8 DD Dv DvDv%1D x ` C " z @ "E F z !3 # D 6. (a) This is obvious: 5. (a) One sees that 4. (a) The lefthand and righthand derivatives of . The curves described by for near or are both smooth; they are (c) We have tangent to the lines with slope at the origin. One can see this from the nonparametric representation too: if is small, then , so the equation gives , or . (b) For very large, goes from to . (b) Since , as traverses the intervals , traces out the line segments from to to to to and , when are the corners of the square. (b) is the plane ( on by substituting in and simplifying. Conversely, if satisfies this equation, then if ), because and . to if and only if either ; , so is asymptotic to , makes a loop through and (b) is the cone when ; (c) is the onesheeted hyperboloid everywhere independent. is still smooth at at at the origin (the vertex of the cone). . ; or are both 0. and (d) is the paraboloid ; 3. (a) Using polar coordinates in the 2. (a) With (b) With and is (b) Using polar coordinates in the we have . The cross product of the latter vectors is or . , we have . The cross product of the latter vectors is or . plane, plane, when , but the surface is nonsingular. . As goes from to to to and back to , and goes from . everywhere independent. , , , and . Since for , and , so the tangent plane is and . , , so the tangent plane ; the points , which are and where ). It is , 21 , 22 (c) two of the possibilities. Chapter 3. The Implicit Function Theorem and its Applications and 4. (a) The cross product of the normal vectors intersection is (set and solve for (b) The cross product of the normal vectors intersection is (set and solve for 5. (a) Substituting in the equation of the sphere gives Parametrize this ellipse in the usual way: , equals at 6. Perhaps the best way to nail this is to perform a rotation around the axis to make the plane horizontal. Namely, let and (The matrix is orthogonal since its columns are orthonormal, so this transformation preserves shapes.) In these coordinates the plane is and the cone is or . Thus the intersection is the curve in the plane with coordinates . This is clearly a circle if , an ellipse if , a parabola if , and a hyperbola if . Parametrizations may be obtained in the form , in the elliptic case and , in the hyperbolic case (for suitable ), and , in the parabolic case. 7. The statement is as follows, and the proof is as outlined in the text on p. 130. (a) Let and be realvalued functions of class on an open set in , and let . If and and are linearly independent, there is a neighborhood of in such that is the graph of a function from some interval in into ( , or similarly with the variables permuted). (b) Let be of class . If , there is an open interval containing such that the set is the graph of a function as in part (a). 3.4 Transformations and Coordinate Systems
1. (a) constant: circles centered at the origin. back of text for the inverse and Jacobian.) constant: halflines starting at the origin. (See answer in ( ( ( 2. (a) , (i.e., 7 B % F f % (c) parabola. . The range of is the parabola , and each line ). or } f (b) . constant: the lines . ( ). ! rQD F! C C uD V D D D m1 V B x g D constant: the curves ( ). maps onto this C i pe % D V C v9i ( U SQ q(TE j z mli iq ui j e 9i D # W1 D l e gs ! U p 9r(TQ US %RD @D i m V D D F C j ( % j D ( 8 h(i i j 0 T%Fqp # D D F V b FX (b) corresponds to ; so C D0 # D ! U U S b A(TQ Ah! U i 9i US 8(TQ ! U US 7 fwBFl (TQ f C $i " 51v% v% ( C DvRvd1T D0v `i d DvRv1 (1DD0o C C % WvR ( ( DD# `i DD8 DDv% % Dv1D `i h C " F} pcF C D0v (d) . and and ), so is , and one point in the works. and and ), so is , and one point in the works. . . or ; then ; we have works. (T5c U SQ ! U U ! r ' US (% wV v(TQ ! U "oh s % 0vu D9ihci !
D " are U s (Ts u v USQ e o
( ( ' 1 C ' ' ` % ( C D0 D , which e o h! U D B p D ` ' 8t5 ' u BFs lg hg5 w @ sr Fsr Fu}l C "s}l F} C Wq e q V V w# 1 D # %D Wv e C # D ## V " 8D V b f C" l! e C gu C ! U !r(TQ US ! U} (TQ 2 "Y ! ! U US ! U U 89i 1(TQYr(TQ ! U@h! 5 U S U S U $q S ! U U SQ US U(TQt! ! U@ "f (TY (TQt "Yh! %U 1U(STQYr(TQ ! U US F q V ! V US p(TQ D ! xu"U ! %U US fB (TQ ## V g 8D V # C D0ov Dv% 3.4. Transformations and Coordinate Systems , , and (b) Substituting the formulas for and into the equations for the lines gives ; the image is the triangle bounded by these lines. (c) , , and ; the region is the triangle with these vertices. 7 (R W0o uW0o % C C C C ! a
( ( 3. (a) If and , we have . If 4. (a) The lines and again in the third quadrant. (b) , which vanishes when and ; for these values, and . (c) The foci of the hyperbola are at where . When , the foci of the ellipse (a) we have ; with as in (a) we have . (c) The lines = constant are tangent to the hyperbolas = constant along the line (d) We have , so or ; hence minus sign is necessary to make when , and then (b) , . . meet the hyperbolas and and in the first quadrant and ; with are at , i.e., when . , where the . z &9iXi z RXimh9i z RXiq 9iv e # # # Xi q s 3q j 0 j g z g rBX oe Xoe ) ) qXiE XimhqXi C C C W0o @ b " C C " H i H ! U s ! % (ToTs ! 2 (To ! U @ U U S Q U U S Q C U US H H ! %s H (TQ USQ U U USQ U H (TE ! X H ! EB(TX H ! E ! %U ! E ! X U U USQ USQ (TEB(TX (TE ! %U U S QH H b f h C b hf hf C 0` D To f SQ " C k ( ( s V f V u6V f u V V r 6i 89i 68v1 w V % w 6V ( 68vY C Y % 1 d C Y h(1Dva s (1Dv% % % 1 5 1 X V ' F % 1 1 (d1Dva u 1 C " @ i , so and or ; in other words, , we have , we have , whence . Then and . . , . and hence . ). and . 5. Let 7. With (c) (e) (b) 6. (a) : the sphere of radius about . : the (half) cone or negative axis if or , the plane if (polar coordinates) in the plane. corresponding to the ray latter determinant is nonzero at , where , then is locally invertible there. Because of the form of , has the form where is . For near , let . Then , that is, , so solves the implicit function problem. Uniqueness follows from the uniqueness of . With so one easily computes that (the positive : the vertical halfplane , we have . Also . If the as in where 23 24 Chapter 3. The Implicit Function Theorem and its Applications 3.5 Functional Dependence 1. (a) (b) is nonsingular except on the plane . (c) is defined for and has rank 2 there (the second (d) (e) has rank 2 on the set where is defined (the first two rows are independent; the third is (f) times their sum). is nonsingular except on the plane . 2. (a) T HEOREM . Let be a map from a connected open set into . Suppose that has rank at every . Then every has a neighborhood such that and are functionally dependent on and is a smooth curve in . P ROOF : Let and . Since has rank 1, it has at least one nonzero entry, which we may take to be . By the implicit function theorem, the equation can be solved near and to yield as a function of and . Then becomes a function of and too. Implicit differentiation of the equations and with respect to (taking as the other independent variable) gives and . Solving the first of these for and substituting in the second gives , so is a (smooth) function of alone, say . Hence , and the range of near is the smooth curve . (b) T HEOREM . Let be a map from a connected open set into . Suppose that has rank at every . Then every in has a neighborhood such that , , and are functionally dependent on and is a smooth curve in . P ROOF : The proof is similar to part (a), , , . We may assume . Then so we just sketch it. Let we can solve the equation for as a function of and , making and into functions of and . One calculates that and , which both vanish since has rank 1. Thus and (which describes a smooth curve), and and . u @ ci m # V ci V Cr i ci ( j ( j j 89i u 189i C y # # rsv# T # # rs6# T ci 89i ci ci 8 9i y2 % j j C % 9 j C "f i %V rv are and times the first). and row is times the first). . has rank 1 everywhere (the second and third rows . C &% @ u # 89i 89i ci v i 9iv 3 D# kC i i i ci v & i 9iv x C C v1 v% @ % v% i C rw v% v % @ B B i8% $! % B B v ri8% i8% ri8% $ h! U vY % F C C C (STQ u"Yr(Tv1 U ! U U S Q % C u"U ) ! US 8(TQ % v% third is times the first plus times the second). 7 v%!E $ 7 % l % $v1 % %7 v% C @ v% % v2l v% has rank 2 everywhere (the first two rows are independent; the . . Chapter 4 Integral Calculus
4.1 Integration on the Line 3. We use Lemma 4.5. Given , let be a partition of such that . Let be the partition obtained by adding the subdivision points and to (if they are not already there). By Lemma 4.3, . Let be the partition of obtained from by omitting the subintervals belonging to or ; then . By Lemma 4.5, is integrable on . 4. If then and for any partition of ; it follows that 5. Given an interval , let and be the sup and inf of on , and let and be the sup and inf of on . If on then and , and if on then and ; in either case, . If changes sign on , then and , so . It follows that for any partition of we have , and it follows from Lemma 4.5 that is integrable. Moreover , , hence . so 6. Let . Given , the interval of length centered at contains for sufficiently . For , let be the interval of length centered at . Then every large, say is contained in , and the sum of the lengths of the 's is . for 8. (a) If , let ; then is a onetoone correspondence between partitions of and partitions of . The sup and inf of on are equal on , and the length of the former interval is times the length to the sup and inf of of the latter. It follows that and ; the result follows. 25 7. If and is continuous at , there is an interval . By Theorem 4.9c, containing such that . 2. If o s s 9 4 q 4 V A V v R # i T%v 8D V# V # R%rD % A% 0 V 6 866 V % 8 ~% C % D 1 i 1 1 i v x R%rD v # C !% D i %V 6 % C m% q v s X q e q e q v # R%D % 3 r 3 r ! B q C 1 s ! s 3 6 c s I 3 C 1 h C q 3 3 s R%#rD 3 X s q R%D # % q q q s s 6 # R% D ! % ( ! v % X X % ! R%rD # % C % 0" ` 0(U D ! ) ' C ! % ` ` (0( $ (0(U ) 'U ) ' C 1 m2 s q C " D u# # % D
, for any interval we have and , the orders are reversed ( and ; it follows that for any partition of and hence that . If , etc.) but the final result is the same. # R%D 0! $ 0" D D! C "X D # X 1. The sup and inf of of any interval on any nontrivial interval are 1 and 0, respectively. It follows that for any partition we have and , so and . % . 26 (b) and (c) are similar; one considers or Chapter 4. Integral Calculus 4.2 Integration in Higher Dimensions
and . 2. (a) Given a partition of , let and be the inf and sup of on . Then the graph of is contained in where , and the sum of the areas of these rectangles is . This can be made arbitrarily small (Lemma 4.5), so the graph has zero content. (b) The boundary of is the union of the graph of and three line segments, all of which have zero content, so is measurable. Let . Given a partition of , let be a partition of with the same 's, such that each and (as in part (a)) is among the 's. Then the upper and lower approximations to the area of corresponding to this partition are just and ; as these can be made arbitrarily close to , it follows that is the area of . 4. This really follows from Exercise 3 and the observation that if is a rectangle whose interior includes , then the outer area of plus the inner area of equals the area of , and likewise with replaced H Rs 1 1 B ' ( s ( A A A v A H H % %A % % v T V u84 8 % ( E % % 4 A V % A ! 3 H DE 3. Given , choose a partition of a rectangle containing such that . Let be the subrectangles of the partition that are contained in (so is the sum of their areas). For each , let be the rectangle with the same center as and side lengths times as big, where ; then , and . Let be the partition obtained by adding all the (resp. ) coordinates of the sides of the 's into the collection of 's (resp. 's) in . Then the 's are among the subrectangles of the partition , so . It follows that , and the reverse inequality is obvious.
A W( T ! &( A and ! A F # R%rD A A Fl 3 3 r% 3 ` # y I A 3 s I I 6 # R%rD V ! A A % 0C yqR%rD # s A % s A s A A C C % ! 1. (a) If (b) Given where , the same is true of , choose a finite collection of rectangles ( and for each . Then R 7 in# i i q . x iT i i " `R w i 8D h# 8% R w iT u " ! " 5 i i w iT u x w ic s I H } i q q H H ! H " !R " H C % % q 8D h# 8%V % s ( 9. and are uniformly continuous on , and is a compact set in the plane, so is uniformly continuous on it. Thus, given , we can choose such that whenever and are points in with and , and we can choose small enough so that and whenever and . Next, let . By Proposition 4.16 we can choose small enough so that and for any partition of with . Since is continuous, its infimum on is achieved at some point , so . Then for any we have # i # # R%D d% ! f i H C % i i x i R#%D i c ! # i Xi !Q # i si ) ! ! ! ! # # C % C)% iR%rD iR%rD # y # # R%rD % BDDrB % . 6 36v ) such that 4.3. Multiple Integrals and Iterated Integrals 27 is the same as the inner area of its interior , which is the same by . The inner area of as the inner area of since the boundary of has zero content; it follows that . Alternatively, one can redo the argument of Exercise 3 by considering a partition of and shrinking the rectangles that do not meet slightly to produce rectangles that do not meet . 5. Given any partition of a rectangle whose interior contains , the subrectangles of the partition fall into , (ii) those that intersect , and (iii) those that do not intersect three classes: (i) those contained in . (Any rectangle that intersects both and also intersects ; otherwise and would be a disconnection of .) The sum of the areas of the rectangles in class (i) (resp. class (ii), classes (i) and (ii)) approximates (resp. , ). It follows that , and hence by Exercises 3 and 4 that . 6. (a) is a union of open rectangles, so it is open (any point is an interior point). (b) Let be the approximation to obtained by stopping at the th stage, with the rectangles of length . is the union of 1 rectangle of area , two rectangles of area , . . . , rectangles of area . , and there is some overlap, so ; in fact The sum of these areas is with independent of . If is any partition of the unit square, the union of the (closed) subrectangles of that are contained in is a compact set, and the open rectangles out of which is built are an open cover of it, so by HeineBorel there is a finite subcover; in other words, all these subrectangles are contained in for some . It follows that , so . 7. Suppose . Moreover, for any constant and replaced by , . Finally, if , we apply this result with to get the desired conclusion. 4.3 Multiple Integrals and Iterated Integrals
1. (a)
7 7 ( ) ( ( w 8ps ( v%r fqv 2&prvd ( ( " # ' W% ( ' $a ( ' " # 7 7 ' u ( ( % ( v1( p # W 7B( f i # ( 7 8 i ( 7 ( i ( ( v ( "7i 7 ( v !i " . (b) . 2. . 3. (a) and (b): See answers in back of text. ( 5w ' to get for the case . With since , . Since , we can apply the mean value theorem , the claimed result si i c8F k i D C u## i & i i &i C ## (d) We have ## u # # C !T# i i 3 5 i 8Fu D ! @ D i i D i 8F 8F&# t i D hg #i&i  d1iu&i hp# # iFD C C i 8  i &i B% s (c) contains every the set of all such such that is dense. Hence and has a terminating base2 decimal expansion; by Exercise 4. . and ! 1R q s R` R 9 %! ! ` C "1F` X X F` ` ci ! ! &2AC hci i ` c R` R ! q 7 % r % 9 3 % H W H b ( (Tu USQ H W F k ! B } W # ' 5 ci y rEy y r # m 89i AD0C 8 i AD0C C AD0C ci C 0 2 0 % v% d1vRci % f 7 ) " 5 (5 pv% t60uRB pt ! ( i ( % ! ! fi %gp % ( ( ( 7 7 ( ( ( 5 p r 5 7 7 7 7 D # # D 5 u s # } # qA g # w g w # # D r i wfi ` i & i %V %V "$7$8 T8i 8 i t i # p 8i # ' # ' ~p ! v% WD # u v # # '# f # ( U % U S $S ! 3 ! oU ( # ( i! U p 2 ( i(TQ US U f 7 ( i(TQ " # 5 ( i 6(TQ 7 ( ( B ( # # u # i ( u # # ci # 89i # # C w# ' 11F} e % ' f e Chapter 4. Integral Calculus
and 4.4 Change of Variables for Multiple Integrals
13. (a) is not integrable on because it is unbounded ( as ). However, for fixed , is continuous on except at the three points , , and and is bounded in absolute value by ; hence it is integrable on . Likewise is integrable on . 28 9. (a) The region of integration is bounded below by the region in the first quadrant of the the parabola and above by the plane . (b) and (c): See answers in back of text. g 7Ff 89i # 7 " ) # ( } 89i " D G (Rv1
. and (c) The parabolas intersect at Thus 6. The region of integration is bounded above by 7. Reversing the order of integration gives 4. (a) and (b): See answers in back of text. (c) . 89i 10. The volume is ( 12. 11. The mass is 1. 8. See answer in back of text. 5. (a) of mass is the axis. Reversing the order of integration gives (b) (b) is (c) ; the moment is . So the center of mass is . . By symmetry, the . when and , so the net charge is coordinates are , and they are given by . . . and , below by . , so the coordinate of the center . The moment is , and likewise the moment , and on the left by . The moment is . . plane under . . , and . % % Y V 7 7 7 ( ( 7 7 % 7 ( % V 7 V % V ( V 7 V V ' 7 7 %V S ( 7 S % ' " 89i V r v FBci ' V
. , so ' qv% g v% ug 7 7 7 ( 7 ( 2i 7 8mi % 1 t89i V m 7 7 D % va @ci 0 % 7 7 7 7 p ( 1 } % 7 7  a C (DTW# % a D ( k q sC #q sC q C q r % %Y s 5 5 h89i V % fv! vp3g ( ( b
( 7 b b `7 h( T(% Tc% w ( U U Q U USQ 5 (Ts1 b w ! vs ( Q p5 ( Q U H W ( ( ( b% ( 7 ( H b ( 5% H H W H H H H W H H ( ( ( USQ U % ! U US ( ! U ! oH "5 H % ( (TQ (T` W % b q b USQ v% r q (Ts% ! ( ( ( 5 `b 7 ( `b 7 H ' % e E ' h2 7 C b W ( l gu H 7 7 b bV v f b A7 " 7 ( " H H H 6 @( 7 US ! %U (TQ H ( b ( W 3 y b b 4.4. Change of Variables for Multiple Integrals 5. In cylindrical coordinates with the origin at the center of the base, the mass is . s0w ' ( b b T T Tc% w 6. In cylindrical coordinates, ternatively, in spherical coordinates, the plane 8. By symmetry the coordinates of the centroid are all equal, so it suffices to calculate . The moment is , and the volume is , so . cb 7 H W 7 7 b A6 ! %U cb H W "U ! USQ q (TX ( "U ! W 1 ( 4. The equation of the cylinder is 3. The top and bottom hemispheres are given by . 2. The volume is The moment is , and . . , or . The moment vanishes by symmetry. , so ( . The moment is . Thus ). Hence . Al is given by , so b ( b H W ! U ( 7. 10. Let 11. Let in 9. Let , ; then , and , . The given parallelogram in the plane becomes the rectangle in the plane given by , . So the area of is ; the moment is , and the moment is , so and . and are their geometric centers, which correspond under the map (Shortcut: The centroids of . The center of is clearly , so the center of is .) , , space becomes the unit ball in . and ; then ; then , and the given ellipsoid space. Thus the volume of the ellipsoid is V b ' V u(c( TW 12. We have area is and , so , the moment is , and the moment is . ( s 7 } 7 ( ( (cb 'V r Thus and 7 7 . . Thus the 29 , . . b ( B 8i v% l 8i ) 8w i v5 ( g i % $ 8i x ir ) % i w i8% g wfi & %V f i v % v8x %V i & 7 " # vVp i & 7 " # 6 ( i wV 5i 7 " 6 i 7 " 6 ) & & # & # % 8 i X i F i s i 8xfi (% @ i wiT `i xi si ci i i3ci Xi &i ci `i % % S % U i X i p 8 i % U 8S wfi8 ! U % X i% wfi8( U TQ i 2pi8(STQ 2 i 2fi8(TQ i 2xi8B! U i i i qi 9i ci p 8 i # i ci wi i # i y i ~}"i i "i wi # W " ( % t (# # " ( g # ( 64 " ( v `i Q U (UU TSS 7 r(U TS oQ v% 8(STQ v% U ! U U v% A i6 (STQ v%! u ri! % " qv 4% iv(TQ 'g`i V 00# vS g # svS t sv0 u # 00V # `i V C 0o C C % #' " ) 67 v% v mX89i " vY `i " g #' " ( gp ci i Wo C % C Wci #
0 7 j and 8s i 2 t m V f6 C 2s B ! 4 7 V 82 i0! ci 7 ( 3 i 3 i81F # % # . 13. 30 , so 14. We have . Also, and ; hence the . and axes correspond to and . Therefore 15. In double polar coordinates, the unit sphere in . Thus the volume of the ball is . is given by Chapter 4. Integral Calculus , and the volume element is b c% w W H " " 4.5 Functions Defined by Integrals as a function of for each , and . For fixed , one studies the 1. (a) is obviously behavior of as as in Exercise 9, 2.1. First one verifies (by l'H^ pital) that o as for every and (by induction) that for where is a polynomial. By induction again, it follows that for all , as , and hence that exists and equals zero. 0 WC A s% C % g V 6ci C " C 0A6ci V 1 # C 2I # # ' " 6 ci09% uci C I WC 0 A ci WC 0 C 0o 7 b 7 7 ! $ b % H 6. For C 4. 2. (a) 5. 3. 7. (a) hand, (c) (b) (b) (b) For short, let we have . . . Then for , so . For , so , so , . . . . , so , so and , so , and so . . On the other , so . . But . 4.6. Improper Integrals
On the other hand, 31 8. We show that for each , exists and is given by (4.48), and hence is continuous by Corollary 4.53a. Let where and each is small enough so that the points are all in a ball contained in . By the mean value theorem, , and for each as . Hence, by the bounded convergence theorem, the integrals of these difference quotients, which are , converge to , which is therefore . 4.6 Improper Integrals
1. (a) Converges by comparison to (b) Diverges by comparison to . (for example), using the fact that and (c) The integrand equals (d) Converges since (e) For , which is roughly for for near 0. near 0, and , so the integrand is (f) For , , so converges; but for , , so F u 0V C ! u 0V C (e) verges. for near 0 and for near infinity; hence condiverges. (d) The integrand is for near 0; hence diverges. ( , d h! U (c) For near 0, less than ; hence , so converges. and converges. For and large, the integrand is all converge, though.) (b) The integrand is near 0 and near 1; hence converges but 3. (a) The integrand is comparable to
( q ! U V 6F t g near 0 and less than near infinity; hence converges. diverges. dV vDvF USQ ( v% v% ( ! U (T7 g !liW 6FY( i 6 V V 7 ' %V 7B ' $ (b) Diverges by comparision to near . since and for near 1; hence converges. ; hence diverges. 7kb 2. (a) Converges since for near 1. (e) Diverges by comparision to kb US ! U cb fi ! U&p(TQ w18% F wB ' V V c! I `WCo ! I C!`Wo9! I C (d) The integrand is bounded in absolute value by ; hence converges absolutely. , which is comparable to , so .( and for small .) % (k} i (c) Converges by comparison to large . for large 6 ! " ! B V t 0 ' w 7 ( ) . 9i 2 0 ) ) qXi $ ) ) ) 0 8V#qXi `s8 0}i Thus, in , the terms involving add up to the negative of this quantity. add up to , and the terms involving for 8i v% ( ) i v5 ( ) & d ( B 8i v 8i ) ( v5 w v% g & (%Y v%l i v2 i & % 8i 0 ) ) v qXi i qXi V ) , so C 0 i 9i X ) fi 4##4 4 V $ Yq)8 ) 7i q ) ) 3#s 7i ) f C C Wv0 ~ ) 9i i ( ) b % ! b ( T( % u u C Wo } % ( C % 4 C ( 4 C ( is % 4 Wo % 9Wo 4 % 9Wo } i ( 3 q8s q 0#is is Wo Wo Wo ui C C C % % 2 % uS r %2 S S %2 % %2 S S uYS S 2 2l % % % (2s % iq % S S #(% i0 3S (2uiq6 V % V
. On the other hand, since . Hence u8F i D 8F $## D i i ! i v #i p qr#i i gq C C # # ## C 0 ## 0 ## 2 0 # i i #i i F i i bV v B 1 F V 1 0V % ~ f b ! ' b " 0(%!5 Xh! U V b " % sh! %U 1 sh! %U 2 x ' i 21 i @i! ! q % i Ct% D ) 'U rP1 Wi 0( vRvRFuI 2I I x % C "`i % % i Xi Ff i % q % p i q &I q # I q Cr1 % % 444 % % 3 gW## t y i % ( f S S S i ( % S `S S 2% % S . C V "` 6uS 32 4. (a) This follows from Corollary 2.12 ( (b) , so converges if and only if } uS 11. We have 10. The antiderivative of 9. Integrating by parts, , so bounded and , and convergent. 6. (a) 8. Granted the hint, we have 5. (a) As in the preceding exercise, (b) 7. Let the other hand, can take . follows. On the one hand, The last integral is proper, and the first and third P.V. integrals exist (and are zero), but ; hence the original P.V. integral exists if and only if the coefficient of this term vanishes. 4.7 Improper Multiple Integrals
2 1. Spherical coordinates turn . Likewise, the integral over . all (b) One possibility: (which exists since . Since is increasing, we have . It follows that . for is . It follows that into becomes is bounded). Given , so ( for large. converges if and only if where , so since . Since , which converges precisely when , which converges precisely when ( b % . One way to carry out the hint is as . ) and , there exists and hence , for . As , Chapter 4. Integral Calculus for . elsewhere. for large ). for all , so , is absolutely such that for we have remains . On , we 7 & 89i C 4.7. Improper Multiple Integrals 4. (a) Since the first quadrant of the unit disc is contained in the square , using polar coordinates we have ; the integral diverges. (This could also be done by a .) calculation like that in part (b) below, using the fact that 7 i r`ci b 5 ci V US u0VF5 ci i0F (TQ U(Ts% SQ ci % % ! i vg i ci HW H H ! % (TQ 1 U US %! w) ( h ) b V c(3 i q H H US (TQ b ' kb ' " 6 " # " H W ( US b% ! E U S Q U (TX 7 ( (TQ r W H 7 V b u`b F 0W V @0( b ( 2. (a) In spherical coordinates, the integral is tends to 1 at infinity. 3. By the extreme value theorem, is bounded, say for . Then the integral defining (c) In spherical coordinates, the integral is . (e) In polar coordinates, the integral is 6 "# " 6 V b c% (d) (b) In polar coordinates, the integral is . For a given , let is dominated by ; the integral diverges. , which diverges since the integrand . be big enough so that (b) We have turns the indefinite integral into . Hence . Since , it follows that 1 ) . , so for a fixed u ci % US (TQ )B the substitution . and so . b ! %U B ! I 2 S ! C ) "sB`i 33 DD Yi p u ' by symme w
, and F` v f 4 i 3iq rfC Aid3iq q q ( b U S Q U A 8( ( u u ' % p w (T` ! ` #i 7 ' U % U (B! % % USQ F v1(T ! U` V(%B! &l% g U US U S Q U $! s% 1 (TQ (Tp ! u (Tu ' % USQ 4 4 % USQ % % U % (Ts FBB! s % ##4 (T9 ! ` USQ U % 44 F ##4 % ( %!3 "7$% % ! U (TQ US ! ` U h kb E % ! U % kb U % h! s(TQ % u US
b U % h! sU % U S Q ` (TQ (T ! %U S # D ' ! b% U % h! s% ' 7 w E q 3q b h! B k r(TQ 7 U US r si BX ! %vy2BvBD ` (T" ! A# D U #A# F v D USQ D U U D U S Q ! %uGu r(T# f d d d d %! u d X vw2cv w2cvw2 Yw ' #i Y2 F % v2 i w @ ` ' ` # i ( 7 " i ` w Bi #i D ' ` (T ! o D u #i # USQ U # 5.1 Arc Length and Line Integrals Line and Surface Integrals; Vector Analysis Chapter 5 1. (a) 2. (a) With the center at the origin and the major axis on the axis, the ellipse is described parametrically by , , and the whole length is 4 times the length in the first quadrant. Hence where . (b) The base of the cylinder is the circle in the plane; the semicircle where both , , . On the sphere we then coordinates are positive is given by have . Hence the arc length is V 0D ## F I I (b) . Let (d) (c) , so . Then , and , so 3. The element of arc length is curve is . . Thus the arc length or "mass" of the moment is U C S r (TQ 4. try. , and the . Thus , so 5. (a) Parametrize by . , ; then ; . and . 34 . . , so 7 v% p i F} w v1 5 v$Y % % % mx 2 b% 1mp % 4 % #i f f i i i to 5.2. Green's Theorem 8. As noted in the text, if is a partition of and is the partition obtained from by adding in the point if it is not already there, then , so in computing it is enough to consider partitions that contain . If is such a partition with , let and ; then and are partitions of and , respectively. Conversely, if and are partitions of and , we can concatenate them to obtain the partition of . In these cases we clearly have . It follows that , and taking the supremum over all gives . On the other hand, given , choose and so that for ; then . Since is arbitrary, , and we are done. " Y @ @ 1 E Y % s Y % cR1 @ 1 C % A1 $ 1 @ % RFl c Y 1 ! % @ 0A3 @ @ " q % @ vA1 @ q vA1 $ ` # R%D % % # R%v rD % % % Dr % % 3 # R%6 D v % % 0(U l @ ) ' 1 # R%D % % % 4 q i iq E g E u #i ##iq E ql #i iq E 4 4 u #i ## is E q #i #iq E # q q 5 !D iq F ( ( % % g 7 F YFk v% i (Y "} C F} u C C mp C k b F ! U 7 US 0(TQf0! U0p % U USQ ! U USQ U vr(STQ ` (T0u"f00(T0p0 ! %5 ` rp B @ i b b U bV % b U " cv(V % B! v v(% b US b &(TQY cvV(% " vB! o i F8 B! %U 6 i b # 7 ( ! w % u % C 4 ( d % 7 ` ( 4 7 ( 8t2l Y & vBm87i iq % q q % u Wvv8vDD w vR64 8i p %F&5%1 i @is q q b% USQ 1F (T7s ! k U ! kD (TQo i 4 U US U S ! U U S U (TQ wh"rA8(TQ h! o is b % q q U S Q ! U !Y"fC (T h"o~is ( % 4 ( ` v 8v2 i p 2 q q rgpC ( Aiis vR vRui % 7 iyhiq (b) Parametrize so (c) Parametrize by , ; then and ( 6. (a) (d) Parametrize the parabolic portion of , . Then by . . 9. By the mean value theorem, the displayed expression equals # i 7. (a) Parametrize (b) (b) With (c) On the segment from (0,0) to (1,0) we have and ; on the segment from (1,0) to (1,1) and , and on the segment from (1,1) to (0,0) we have and . Hence we have the integral is . points as in part (a), we have . by . By Exercise 9, 4.1, the sum of these quantities from E Y p i i . is , can be made as close to sufficiently fine. It follows that this integral equals 5.2 Green's Theorem 1. (a) Let be the unit disc; then circle has the wrong orientation). (b) ( , . , where the nexttolast step uses Cauchy's inequality. , by . Then we have and . . (remember that , (the minus sign is there because the is oriented clockwise!); then , so , and the straight portion by as we wish by taking for some , which 35 , . H H H U USQ U S ! vs (HTD 2# D ! E (UU TS vqU S (QU TuD 2 # D p (TE ((TTvD (qH D ! E # !%D u D H Q S Q U S Q U S Q U TQ ` H fys (Tvs (TuD # w H ! %vq (TuD # F USQ USQ U S Q H U U H ! UE ! UD d ! U D q "u ! vs (TP# c (Tvs (T$h# @ ` U USQ D USQ USQ D W H " " " A 8 B u b ' u w w u # ( D ( W H " " " h b ' s u # A ( D C % v @ e T ! ` H H H t i USQ % ! U (Tsw s v h"U v # 8(T vu r(TT( v% ! USQ d USQ % d 4 % C a C a C DF 63% } v$# Av$# rv$ va @r# i 36 (c) 2. Directly, by using the polar angle 5. The oriented boundary of consists of two vertical line segments, a segment of the axis, and the , traversed from right to left. The vertical segments contribute nothing to curve since on them, and the segment of the axis contributes nothing since on it. The integral over the curve is . C " i b% R0C h! vh! xsi U U C yp b% q q h2sC 3. If is the region inside , . The integrand is positive inside the unit disc and negative outside, so the integral is maximized by taking to be the unit disc and to be the unit circle. 8 (d) theorem vanishes. as the parameter for both circles: . By using Green's theorem: W ( pv F} ( H 7 b fb 4 ( H% !H H 6 H W H b USQ ! U (Ts% H % ! U 4. Take the arch given by of the axis (where : the region under it is bounded on the bottom by the segment ) and on top by the cycloid (traversed from right to left). Thus the area is or ; both integrals are equal to rU(STQwB } 6. We have where therefore follows from Corollary 5.17. is a connected subset of , by Theorem 1.26. It does not 7. (a) The image of under the map contain 0, hence must be contained in either or ; otherwise it would be disconnected. j v v e  C W"2 2 0o C f ! 4 2 g f i V i C"5 b s (b,c) Let , , and . Thus , where is given the positive orientation with respect to and is given the orientation induced from the one on by the change of variable . This may or may not be the positive orientation of with respect to , so in applying Green's theorem to the integral over there will be an ambiguity of sign. The result is . The easiest way to determine which sign is right is to observe that the area is positive; hence the sign must be if and if . . , . , so 0ci 5.3 Surface Area and Surface Integrals C ! e v # 3. With 2. 1. and , and Chapter 5. Line and Surface Integrals; Vector Analysis , so the integrand of the double integral in Green's , we have , so . . The result and hence . % 5 vs ( 5.3. Surface Area and Surface Integrals 7. By symmetry, the integrals of , , and over the unit sphere are equal, so the integral of vanishes. The integral is also easily computed in spherical coordinates: it is . C " ( E U(TQ7f (STQ S U Y2 5 rl B f 4 c r l6 ! U USQ % U b b7 ! %Uvq U S Q % b% (Ts7 !w "vs (Tsf ! o % % ( b ( H W ( USQ (US B(T7f E(TQ w! b% s !% U ! U ` Tq "` ( % D b" h ! Ub"w ! EB(TQ U US C by symmetry. b D ( # V # ## 2 D ' pFS D D % # USQ (T# 7 USQ D ! U H (T" " H # ! E ! D U U A2BvF!`vD # D 7 D # ! D ## V D # ' ! %#RI UQP 2 D 0@ # D D % # V b% b V U D BA# ' ! # SD A2hvD ' ! # # % # V I % D A#2BvF!` # D % b (TQ US ! b "U D % U(TE ! D SQ U # BF!`A# b D D V D ( V b 8D c( w BvD ' f F D D vD BA# ! w w b ( 0 4 A# D D 4. The integral can be set up in two ways, as in Example 1 (p. 232). First way: The upper half of the surface is , so the area is H W # D ( b s (TD X# D USQ H H USQ U U USQ US q (TuD 1# D us ! ! E (T (TQ The substitution simplifies this to tution turns this into coordinates to parametrize the ellipsoid. With the formula (5.20) for area yields the integral stitution turns this into w w (TuD s# USQ USQ USQ (To s (TD 1# D # obtained in the first way since ( if uate the integral as if .) ! D PA ! # "$ C y Since , is always positive, so finally for even functions. Finally, one uses a trig substitution , if ) or a table of integrals to eval, where if and . (Both expressions for the area have the limiting value as # " "# " % D V 0D ## l 5. Clearly The moment, in spherical coordinates, is b % b %
0 D # n 6. 8. (a) where and are the portions of the sphere and the paraboloid with , oriented with the normal pointing up and down, respectively. As in (b), the normal at a point on the sphere is , so . Also, the element of area in % ' E% ' V u yw 4 % ' 9 V 0wBx$% D C 4 5 p k1% T i % H H ( USQ (TE 7 ! U (b) Since lines through the center of a sphere are perpendicular to the sphere, the unit normal to the unit sphere at a point is simply the vector . Hence , which vanishes since is an odd function. Alternatively, in spherical coordinates one finds , so . C "w ( 4 g  y2 % !} v } 5l51ri % 4 9 % }  W EUvcp U(TQ H ! E ! 2v H H 42 ! " S U U ( i g
. , so ; taking (c) The triangle lies in the plane , so . (d) The normal is horizontal on the vertical side of the cylinder, so ( ) we have , ; on the bottom ( ), we have . , and the area is as parameters, we have ; hence , and the substi. Second way: Use modified spherical , , and , , and the sub. This is the same as the integral . . there. On the top , . Hence D " q g D # `" b 4 2 # " # ' B( i c U(STE Q (e)
! U o 37 . . 38 ! %s% U Chapter 5. Line and Surface Integrals; Vector Analysis 5.4 Vector Derivatives . 3. The first two formulas are most easily obtained just by writing out the indicated products and taking the indicated derivatives. The last one follows from the first one by using (5.29). 4. (Also 5, 6, 7.) These are straightforward calculations, but the verifications of (5.25), (5.27), and (5.33) require a little masochism. It is less frustrating to start with the complicated expressions (on the right in (5.25) and (5.27), on the left in (5.33)) and work toward the simpler expressions on the other side. 9. This follows immediately from (5.29) and (5.30). 5.5 The Divergence Theorem (b) , so polar coordinates yield . (e) , so the integral is . 3. , so volume of . 4. Let ; then divergence theorem. and , so the result follows from the (b) By (5.28) we have ; apply the divergence theorem to g 5. (a) . . i8" a 2. Directly: the integral is times the area of , i.e., so the integral is on (see the remark at the beginning of the exercises), so . By the divergence theorem: , . # F D v  B 4 ( 2 ( !t 4 H D b(w ! U 4 a ( D b D ( ( D }D V 2 i  4 ! DG % t % ( V # " # # `( # D b 7 D D b V ( (d) , and the volume of the ellipsoid is unit sphere by the change of variable , , . (reduce it to the volume of the ), so the integral is # D b 7 # V D V ~ # D 6 m X 4 7 (c) get , and the integrals of , , and . over the cube are equal by symmetry, so we H v $% ' 1. (a) , and the integral of over the unit ball vanishes by symmetry since 7 ( 7 ( bd A6 8 % ! v% v% # 0 ) D d D6 E2ui8F! % % b !g % v!t V 8. is skewsymmetric in and . is odd. 2. These are simple computations too. For (c), with 8 D and , one has 9( 9i 1. These are all simple computations. (% 1W 3FG D D u p ! D D EFG`W . Hence . " spherical coordinates is , so . On the paraboloid, with as parameters, we have that the component must be negative), so } % p i7 vx v% # c @ v2r v% y % b% v4 % ' 4 (remember 2 "s% ! U 7 $
" H W % `" b " 4 x 9 H W 4 2 "U ! b E % `b ( 2 y b % 5.6. Some Applications to Physics 39 , and likewise for we have . and , so , so ( 6. (a) (b) See Exercise 2c in 5.4. 5.6 Some Applications to Physics
(we take the density to be 1) at a particular 1. To evaluate the potential point , the key to proceeding efficiently is to rotate the coordinates so that is on the positive axis (or, equivalently, to observe that is spherically symmetric so that it suffices to take on the positive axis). For and we have , so in spherical coodinates we have is 2. Think of the ball as the union of thin spherical shells of radius ( ) and thickness . For a given , the shells with contribute nothing to the field, and the shells with contribute . Integrating from to gives the field as for and for , as claimed. (The potential can also be found by integrating in ; it is for and for .) 3. (a) We take . The field is ( ( ( u `b V u 89 ( b V u 89 ( b u 7 ! `i ( ! ! u u ! &G! ( D`vi 0 8 z b ( 7 C For mass for this is ; for it is located at the origin. The corresponding field . . The latter is the potential for a for and ( % USQ % f (T`B ( ! 9i u 8V ( p V ( b b % b (b) Since is continuous, is the limit of the mean value of on the sphere as . C 0 b ( (u 0 u u (u7f u } Q (WTSHU X E !vXi 5 "8 u 6l ! $ t % F b % ! ( U 1 h 9i @9i % C u u % b ( 7. (a) Since of the sphere we have on , by (5.39) we have . is the union (with the outward normal) and the sphere (with the inward normal). On and (as in Exercise 6c), so by Exercise 5a, , which is times the mean value of on the sphere . Likewise, the integral over (with the inward normal) is times the mean value of on this sphere. The sum is zero, so the mean values are equal. V (e) Choose so small that the ball is contained in the interior of on , so by the divergence theorem and part (c), . . Then C 4 ` C C q u b( z b ( u F r 4 & u V r u Cm V V u 8VD( b 7 u ! % u V u 89 ( b C C 00o x F`9i F ( ( $u % b ( q q 7 xC (d) is not of class on the region inside the sphere, so the divergence theorem doesn't apply. b b ( F D ( 4 (c) If is the sphere of radius hence D about , on D 1 ~E7 u 8V u E~ V u 89BF`9i 4 V u 89& ( ( V u 8BF ) z D F iqB V . ; % ' V D b1F(8FT(% ' 0F`b (Y % V D V D % %8D V % 'W8D V r % 1 F D V 8D D D i r1 fV v % c vD 4 c 8(k F 7$ 8Q % 4 P' b 4 % q " 8 Q % c @ uc 1 9 4 P' b% ) 9i !$ q i `9i ( ) 0C
0 ( 4 V ) qB ! ) ) s i sq}i ) ) iS ! S %2 @ E! T 3 @ 0 @ h E!`v T 3 ` @X0 turns this into 40 4. The argument is essentially the same as the proof of Theorem 5.46, using the facts that and for (Exercise 2d, 5.4). The twodimensional analogue of Green's formula (5.39) is easily obtained from Exercise 6, 5.2, and it yields the following analogue of (5.47): V ) ) h) 8qgfS iS S S % S 3 h X0 E ! `v S @ 3!`v `vD ' V 2 0 s 2 0 D ( ) D v f i 9i % i0V6 i8D ! V vD ' D V () F D (% The and cays like components of the integrand decay like as , while the component de, so the integral converges. The component vanishes since its integrand is odd. Since (via the substitution ), we obtain . (b) should be , but the integrand decays like as , so the integral diverges. is an antiderivative of , (c) Since H s ! I D k As , the first term approaches whose gradient is Multiplying top and bottom of the fraction by , so subtracting off the yields . (Of course the can be discarded too.) S where is the unit inward normal to the circle, namely, term on the right becomes Chapter 5. Line and Surface Integrals; Vector Analysis . The estimate (5.48) for the first as C
0 b % S u q 4 ) sqfi ) $S ) $S z y ) i0v i81 V % 5.7 Stokes's Theorem `9i C m) 3S 2 0 S S XvD ' S `9i 2. 1. by The integral is and from the second term, since , i.e., lies over the curve in the , or , an ellipse with semiaxes times the area of the region inside the ellipse, i.e., . on the circle, one obtains . The curve , so the integral is twice the area of the disc , plane given and . . s ! u U U @ rg` ! ` r gu 8ci (TD~ @ U S Q "U ! # s t ~ 8@ # t r E 8ci 8 3 # E 8iu # % g ~l 8i # t9i # 8i m m i g 8ci D Y S !v@ % (2 i ` i `ci V % V 3 Y S ui S "ui # "u9i # iu9 "k rv% u9i 87 i 1Fs ` i 3 (Thr USQ D `v 0 r(STu3g`c i U Q U @ v ! "ui (T` rh! @ gui # U S Q U ph! UYgf }9i # iu (Ts tF 8h! h} i }v9i USQ U `8i US vl8( TQ ` i S vp$U(TQ Xci # 8i vwYvp$! U k (wYvf3h"U % Xci a a ! rvd i v 7 v! 7 va vX8 7 v2 v 0# d a d a ( ( ( FB89i # i h( E r7 rv% Bci 8i ri y y 6 8Q P' ` 8Q P' 2 Y t D H H V U S Q v (ToF ! H W H H USQ C p 8Q " H ! s!s H (Tsf " "!U (TX# C C U U S Q % 4 p f 4 P ' P' C " 4 w 8Q P' w 8Q 4 P C"f 4w 8Q ' G F`(V i WV i H H H C " (S U ! U 4 2 8Q P' H H TQ "5 f 4 U C C ~ "!uDU (TS uQ D t 8Q 4 P' r D 4 P ' " Q r D
D 8P Q 4 x ' 5.8. Integrating Vector Derivatives . Thus 3. The equation of a nonvertical plane parallel to the axis has the form . The integral of this over the disc is twice the area of the disc, i.e., (the integral of vanishes by symmetry). # b % D % 4 % (@ # u c ` # (cc ` 0 B # l 4. is the circle of radius about the origin in the the disc of radius in the plane, so , by symmetry. Second method: parametrize plane. First method: is also the boundary of 7 ( is the ellipse 5. Like Exercise 4, this can be done two ways. is the region inside this ellipse, we have Alternatively, parametrize by , , . 6. (a) This is a simple calculation. U o y p !% H H H 4 f 2 H U S Q 6 (To!0 U uD 7 (TuD USQ (b) Parametrize by , , b% ! H W H ! U oED 7. Let (c) is not of class on any surface bounded by Stokes's theorem doesn't apply. be the annulus in between the circles , so and in the plane, oriented so that , the sign being if and if . It follows that . by ; then . (such a surface must intersect the axis), so ; then . , in the since , plane. If and . Then . Then . But ! P `b21"f 2 a 4 % t$ W b 4 4 P' & 8Q e hm , and the area of is 8. Use (5.26) and (5.30): 5.8 Integrating Vector Derivatives $ !f 4 4 2 8 P ' Q & 1. (a) ; (g) (f) (d) (c) (e) (b) . , so ; , so . , so . . , so ; , so . ; ; , so ; . Now apply Stokes's theorem. ; , so . ; , so , so . 41 ; . cV t b% C 4 "f 9 cV t b% b% f 4 f 2 5 4 A k 4 4 tC C !% l 4 C ! A 4 F 2 d F 4 ! ! C 8Q P' C ! s C " x " " " " 6 & F` " & " 6 v% " a d " 3 u89i@ # 3 f9ci " & " 6 v% " & " # ci3 f v a % kva vd " " 89i9$ vt " 6 ~ d " & C w B `89i@ # X89iu Y B "u89i@ Fu89i ci@ m 89iu l B # % l A # F h 8Q P' C C !srE v! v! 42 2. (a) 3. By the procedure outlined before Theorem 5.64 we can find such that , so by Theorem 5.63 there exists such that 4. (a) If (b) Let be the minimum distance from points in the compact set to the origin, and let Then the curve and the circle together bound a region , so by Green's theorem again, . (c) Proceeding as in (b), we have and , so (b) Take so by integration by parts, , let , so be the annulus between by Green's theorem. Thus , and , so . One solution is . , so . Hence, and . Then is independent of . . Then . Hence, and . (c) From Example 1 and part (b), for any closed curve in (oriented counterclockwise) we have and hence . By Proposition 5.60, is a gradient. Chapter 5. Line and Surface Integrals; Vector Analysis . This does not look hopeful, but , so we can take . . Then . , . , % S 0 A q% S 0 DS % D D $q% s% D F% Y% V q% q% $vD V 44 vD ##4 D $X% D C " %
. 2 0 I h0V s 5 0h0r1 q V 2r% 0u 0 s #iv 0}WsC q q C ! 2! i AD0C } 0 rW `i X xh fi0 C &}I 0tY#iv 0 V q 3 ! ! s C ! mPF! 2 0 I t q r I 0V Fs#iv q % C % D T~kI 0V ui S i h0V I mui 1" ' ! ( Wfs % " B 2 % V `S 10`S ! 2! !suS S uS %V vu X u80V} V C % Y0$ V % ( i0W#F` 'v10V # V C % C % g% % Y! ' ' ! s % C v% # % V T( gsi8@ g i8% ( 2! 2! ( % B % ( }si8% }si8%
, which does not tend to 0; the series diverges. and ratio terms is and ratio and ratio ; it converges for ; it converges for . ; it converges for . , so the full sum is 1. 6.1 Definitions and Examples Infinite Series Chapter 6 1. (a) This is a geometric series with initial term i.e., , to the sum 2. (a) The th term is (b) This is a telescoping series; the sum of the first (c) This is a geometric series with intial term 1 and ratio absolute value if and only if is closer to 1 than to , i.e., . (b) This is a geometric series with initial term or , to the sum (c) This is a telescoping series; the sum of the first (d) This is a geometric series with intial term , to the sum . terms is 4. (a) Note that if as . 3. If and vanishes as (d) The oddnumbered terms do not tend to zero; the series diverges. , so , so , then then all (b) Assume all Conversely, if
2 0 AI are positive. With , then
D where , and for , so Lagrange's formula reads is between 0 and . For we have , so this vanishes as . (For we still have .) The integral formula (2.56) gives , and the mean value theorem for integrals then gives for some . Now, for , we have , which are nonzero. Let as in (a), if . The ratio is less than 1 in . The sum is , then 0 @ D 43 q) S . ; then , so the series diverges. , i.e., , i.e., . , 44 Chapter 6. Infinite Series 6.2 Series with Nonnegative Terms 1. ; converges by comparison to 2. Practically any test you can think of will work on this one! (Ratio test, root test, integral test using Corollary 2.12, comparison to geometric series , ...) 3. ; diverges by comparison to D . 7. Diverges by ratio test: . 8. Diverges by comparison to 9. Converges by ratio test: , using (2.13). . 10. Converges by ratio test: . 13. By l'H^ pital's rule or Taylor's theorem, o . ; series converges by comparison to ; series diverges by comparison 18. Diverges by Raabe's test: 19. If , so . converges, then , so converges by comparison to for large . For such for we have and for , so , . 20. Use the integral test: and remains bounded as 21. Use the integral test: precisely when . by the substitution problem, the series converges if and only if . for and , so as in the preceding 17. Converges by Raabe's test: , so ( % `S S V uS S S wuS S TS q6 20VW S S `S S TS s6 V uS 2 0 % 30W 3 4 V `S S uS 6V V S 5 S s6 D D q D C 0 D D (2 % 0(2 % D V D V % % 02 % D V D V " % D ! D 0 V D C(2 % 0V D A FtB % 0V ! D A F V D 0 C 16. Converges by the extended root test (part (a) of Theorem 6.14): for large . b o q iW pb a V D 15. ; series diverges by comparison to 14. By rationalizing the numerator, to . . 12. Converges by root test: . (See Exercise 7 in 2.1, with .) 11. Diverges by ratio test: . (See Exercise 7 in 2.1, with The root test is a little easier if you know Stirling's formula.) 5 f ( 6. Converges by ratio test: %V 2 % VU SQ 0 6(Tt " 0V D V 0 % V r ( 0 ! 0V & D V D 7 0 % @( % T" % 0V " m D V D ( 0 V C(a 0( % D V D 2 30 #v" C V DV D % (2 % T! % 8V ( % D R D V 0 C ' ' w" ' 0F V V (2 0V D 5. Diverges by root test: 2 30 ! 0 " % V ( 8 4. Converges by ratio test:
D C 0 V " 0(%2 % ( ( ) V R D (2 0V "!U ( ) ( G ! A aV D D . . . . . . 6.3. Absolute and Conditional Convergence 45 . Also from Theorem 6.7, . (The error in adding or removing the initial term in the sum is negligible.) 23. The derivative of 6.7, so .08, gives 24.
( ! . , so is decreasing. 25. Let . If then , and Theorem 6.14a that converges. If then diverges. infinitely many ; it follows that for all but finitely many ; it follows from for infinitely many and so for 6.3 Absolute and Conditional Convergence , add up the positive terms until the sum exceeds 1; then put 2. To get a rearrangement whose sum is in one negative term; then add more positive terms until the sum exceeds 2; then put in another negative term, etc. Since the original series converges, only finitely many negative terms can be less than . it never drops below , so it tends to as After they are all used up, once the sum exceeds more terms are added. 3. For the series the th partial sum of , the th and th partial sums both coincide with ; hence they converge to . When this series is added to , the oddnumbered (positive) terms of the latter series survive unchanged, and for all , the th terms of the two series cancel and the th terms add up to give the negative , , , . . . . After omitting the resulting zero terms one obtains the stated rearranged series, terms whose sum is therefore . and 4. Let . Then if is odd, and if is even. Since D 5. (a) Suppose in which case . After throwing out finitely many terms we may assume for all , for all . By Taylor's theorem, where , so is the sum of the two absolutely convergent series and . Conversely, suppose where . We have , and where , so is the sum of Taylor's theorem again gives the two absolutely convergent series and . D Fs q D $ D D D # 8 # # u q # p 0# # q # # D #S # D 2 ! # Fs D D S D u q D p 0# Fs q D D 1S ! # D S1S D 2 ! , . 2 I D 4 v4`6@ D ) I %S p3 % H ~3 % I ) 3 C 0 D 6@ # D 3 @6 % S ( 7 (% @I % C ( 2I 1 % 444 S v##s ( 3 7 444 A## 444 ##2 w Y7 C C C 2 1. For (a) and (b), just use the fact that the convergent series and and for all . For (c), use the ratio test. to get a comparison with as in Theorem 6.7. Also, a (C D y% D m% D m% D ! D ! ! %V 1 k V V V C "" 0 V s" 0! 6 s " 0 V 0 S 0F C % 6 36 V IV S ( q y p 0V is , which is nonpositive for . By Theorem . The integral is , . Adding on the first two terms, namely .25 and % V p i8 1 'V , and is negligible, so ! A 5 5 V S G w 0 22. From Theorem 6.7, S 76C#S ( V w C #3S (h uS h uS s6 C C V V w0d a (hS pw5 5 0 ! S % %V G V 76C#vV G w5 5 0 hS S u#YS S q( S S S % C C V S 6 1 S V 1 . The integral is ! U ! ! 0V g 5( ! 5BC V a 0(2Y(C ! ( ! 0V 7 1 p i0 1 p 0V V i0 vfc 7 i0 V V
( q m H US (TQ s a(C 46 D Chapter 6. Infinite Series
(b) is convergent by the alternating series test, but not absolutely convergent since By Taylor's theorem, where converges as above; is absolutely convergent; but diverges.
2 6.4 More Convergence Tests
1. By the ratio test, the series converges absolutely for . At or the series becomes which are absolutely convergent by comparison to , i.e., or , and diverges for , both of . 4. By the ratio test, the series converges absolutely for , i.e., , and diverges for . At the series is , which diverges by comparison with . At it is , which converges (conditionally) by the alternating series test. 5. The series converges absolutely for the ratio test. (In detail: , i.e., 6. By the ratio test or the root test, the series converges absolutely or diverges according as is or , i.e., or . At the series is , which converges (conditionally) by the alternating series test. 7. By the ratio test, the series converges absolutely for , i.e., , and diverges for . The numerator of the coefficient of is clearly bigger than the denominator, so at or the terms of the series do not tend to zero and the series diverges. 8. By the ratio test, the series converges absolutely for . At both and the series is by the alternating series test. , i.e., , and diverges for , which converges (conditionally)
5 ! S 9. We have , so by the ratio test, the series converges absolutely for and diverges for . At , we have , , so the series diverges by Raabe's test. Howand ever, by the proof of Raabe's test, the th term is comparable to , and the terms decrease since . Hence, at , where there is an extra factor of , the series converges conditionally by the alternating series test. 10. By Taylor's theorem, converges conditionally, while conditionally. where . The series converges absolutely; hence the original series converges V #fXi0suin C W#F V d 0(! V G , and .) At the series is . At the series is diverges by comparison to (for example) which converges (conditionally) by the alternating series test. (f S (su % 0 v V % V (k S ( h % 0 % F S (r S V S ( % 0(! V 4f d ! , and diverges for , by . We have , which , % V % C d ! y! % F! 1 D V (%u 0V Wrf D q $ ' 0 V ! ( k d !f V % C V 0 !f% 4 " V 2 WC % 4 0 7 S iV 2 0V (V 5 S 0pf% 0 8p!l p(" % 0(! % S ( } % np tc D A D V ~! % % ! 4f ! ! 3. By the ratio test, the series converges absolutely for all : as 2 30 C 0 C a ! #1 ! a (2 % 0V "( 2. By the ratio test or the root test, the series converges absolutely for . At or the series becomes diverges for diverge since .
D , i.e., , and or , both of which . . D %V D ) l( D p c# Fs ( q D Fs D D ( ( ! "4C ! ! Wt5v % V 0 08 V V ! m! !&Y ! % p V aV D 0 C C V G d gRW#r( % d 0V Y W#r d 0(" ! s$ X ( ( ( s 0 (t 0Vs #(" % D V D a a F 0V w% V B p S V w C "} r} ( D C "} w C Fx a " 0V w% a d2 @ C ! D P Fs D S % YFf 1 w C ! WC#} d 0(s d V G % % C l( 0 Wgfv % 5 } } % &Qf % % y% p % QW5X a 1 } % aV d@ ! . 6.5. Double Series; Products of Series 47 . Hence, for is an alternating series (for any , since 15. The power series the powers of are all even), and the absolute value of the ratio of the th term to the th is . For this is less than 1 when , so the terms decrease after the first . one. Hence the error is smaller than the first neglected term, namely 16. Let . First Method: We have . Since decreases to 0 and the partial sums of converge to the full sum, Dirichlet's test gives the convergence of . Second Method: Let and . Then by Lemma 6.23, . We have , and the series is absolutely . convergent since 18. The series converges absolutely by comparison to the geometric series When it converges for by Corollary 6.27. When by Corollary 6.27, since . when it converges for 6.5 Double Series; Products of Series
1. (a) (b) A since H b I % H ! X A (% 3 T!@ "$ "##g2F`"3 8 444 % % (2$ T"3 3 8 A A 3 T8) d A "3 A 8 X A T w1 w q v D C X 0 D f 17. The convergence of follows from Dirichlet's test. (Take relabel as .) Absolute convergence is guaranteed for , since in Theorem 6.25, then and hence . (for any ). . . # d 0w% # # G V b q V G m D # # ##2 # s ` # W # v # q 0 # % # D A # 1A# ## 444 w## D D # D # qT0 # 1 % # h! U ' D D $ H H cb U(STQ" (TQ H US b % 0I ! # D % 00 v V b q !s D 6 ( % T(% % 0V comparison to is conditionally convergent, and , so the original series is conditionally convergent. % " Y where . Finally, is absolutely convergent by % % c ) 14. By Taylor's theorem, where , so S V ( ) q " q % Bc % F hS % E ) by comparison to . where 7 q 13. By Taylor's theorem, where , and hence . Hence the series converges absolutely m !% U S q q @ ! U %V 6 } # D q v " F} S # # # E# S D D 12. , so the series diverges. ! D D 11. Let convergent. On the other hand, , so
0 . Then , so the series is not absolutely is decreasing for , and hence so is , and converges by the alternating series test. CmS % V V % C D 6w fiS V V 6 wwiS D C"u 0 V w S D C ! D A % t% l A 3 8 A A 2 0 I 4 A Q A D A vq A A A % ! Q ! A A ) 60C ! A A A 3 D ! D ) q 3 3 3 A ) A A q I f w3 s D D A ) A ! $ D A A C % ! 2 9 D ) 2 % A 3 c s I A 3 2 A % D C m% 3 A 2 " 1 D 1 ) 60" 3 5 A % A A D A c s I C A % C t% 2 3 2 G! A A ) 1 A A A 1 0r 3 1 % D s I C A A A A % % D } A A A A q q C % 2 S! A A A D A A 3 D 2 30 C $g3 0"3 g3 t1 0 V ! T & I q w 0s q X u 0 X m( w Bn w A w ` A % f V tf % % 3 0V p3 % % 8 7 ! 0V % X % V % ` % (4##B 9V % `} % v## 44 4 % 444 4 % 444 4 w } % w## y wf T 4 4 4 wf ~i v## ( T i lA i V fi Xi i V 8}Bi h V 3 V A 8 A V A 8 A V A T V Xi i 3 3 48 2. 3. If 5. Let 4. . Clearly Case I: find such that for . Let that and hence Case II: for each , but We can then find such that then . Moreover, . It follows that the coefficient of in the Taylor series is , so that series converges for by the ratio test. The product of this series with itself is the series where . On the other hand, the sum of the latter series is , whose Taylor series is the geometric series . Equating coefficients of in these two series gives the asserted formula . (The justification for the last step is contained in Theorem 2.77: for , say, so is the th Taylor polynomial of . See also Corollary 7.22.) is convergent by the alternating series test, but not absolutely convergent. The Cauchy product of the series with itself is where . By the hint, . , which we think of as the limit of the square partial sums , and for each . The claim is that , whether these quantities are finite or not. , then . , so . We can then find ; then . Given , we can find for . . On the other hand, given such that as , so . But by the binomial theorem, ; hence Chapter 6. Infinite Series such that . Let Case III: 6. Since , it is approximated by its square partial sums, so given we have for sufficiently large. This is a double series (it's obtained from by replacing by 0 when ), so by Exercise 5, it can be summed as an iterated series first in , then in . Summing over only those such that gives , that is, the sum of the tail ends of the series over is less than . With notation as in Exercise 5, this implies that . But also , so . It follows that . 7. 8. The only nonzero terms in are at and at , so all . On the other hand, the only nonzero terms in are at and and the latter term is missing when . Hence the sum is 0 for and 1 for that the sum of the first iterated series is 0, whereas the sum of the second one is 1.
3 and only if . for some ; then . Given . It follows that . It follows that , we can find . . This has a finite limit as such that for at , . It follows 3 C 3 A C "% 3 l
3 A D ! C & 3 3 l A D A . It follows diverges. , we can . Let , let if . ; Chapter 7 Functions Defined by Series and Integrals
7.1 Sequences and Series of Functions
1. (a) if , . We have for , so the convergence is uniform there. , if . We have for , so (b) the convergence is uniform there. (c) if , . We have for or , so the convergence is uniform there. (d) for all , so uniformly on . (e) for all , but the maximum of on this interval is (at ), so the convergence is not uniform. However, for provided , and ; hence the convergence is uniform on . (f) for all , but the maximum of on this interval is (at ), so the convergence is not uniform. However, for , so the convergence is uniform on for any . for since for and for , and . (g) For any we have for and for , so the convergence is uniform on these intervals. 2. (a) The series is a geometric series, convergent for to the sum . The convergence is absolute and uniform on for any , by the Mtest with . The sum is continuous on . (b) The series is absolutely and uniformly convergent on by the Mtest with ( ); it diverges elsewhere since the th term . The sum is continuous on . (c) The series is absolutely and uniformly convergent on for any by the Mtest with ( converges by the ratio test). It diverges for since the th term . The sum is continuous on . (d) The series is absolutely and uniformly convergent on by the Mtest with ; the sum is everywhere continuous. ; the sum is (e) The series is absolutely and uniformly convergent on by the Mtest with everywhere continuous. 49
( V y j % % (Rv1 C 0 H B %H 1 0s H H C &% % % 31 AD C 0 C% V AD F C 2 0o H H 6 C "% 2 W Y0 V C "% H 2 H f i H H H q C D0 i q C C m% C " $d`r2iv 1 sr2i0 % ! ! r "`i C '2! i 2i ` # # R%0C # q I V q 6## rC `iv H C YiH I 7 2 0C i 2 C " 6 I H H H 1 q I rC hi0 19I 20 C @i I B C i j C 0 I q P0#iv H H D AY ub f }b 0C f i i H ! q C cb v qb 0C w i b i C "Ur7si0 H C C C 0 1 WWXiv q &f! C mi !Wo H q C q Fu ! 2sC "`iv C 0rkiv C 0 5b H i AD x H D0C p i V ~ C 0 j 50 Chapter 7. Functions Defined by Series and Integrals
(f) The series is absoutely and uniformly convergent on for any , and it diverges for (Theorem 6.9). The sum is continuous on
H as or 5. The series fails to converge absolutely by comparison to . However, decreases to 0 for each , so by the alternating series test, the series converges for each , and the absolute as difference between the th partial sum and the full sum is at most . The latter quantity is independent of and tends to zero as , so the convergence is uniform. 6. (a) Since converges we have , so for ( ) we have . Hence converges absolutely and uniformly on by the Mtest. (b) By the hint, . The first series on the right converges by the Mtext, since , and absolutely and uniformly for converges by comparison to . The second one is independent of , so it does not affect uniform convergence, but it decides the issue of absolute convergence. for 8. By Theorem 7.7, the point is to show that Cauchy on , so given there exists . Since and are continuous on for whenever is uniformly Cauchy on . But it is uniformly such that for whenever , we can take the limit as to conclude that , and hence is uniformly Cauchy on . 9. Since is continuous, the set in the hint is closed and hence compact. Moreover, since increases to , we have . By Exercise 5 in 1.6, if each is nonempty then so is . But if is in the latter set we have for all , which is false since . Hence some (and hence every with ) is empty, i.e., for all when . Thus uniformly on . 7.2 Integrals and Derivatives of Sequences and Series
!% 1. The series defining converges absolutely and uniformly on is continuous, and termwise integration is permissible: . Now, term of the last series is when
( ) 2. The series defining converges absolutely and uniformly on by the Mtest with Hence is continuous there, and is a telescoping series; the th partial sum is , so the full sum is 1. " t t p si p i 2 0C % vDDv5 8 C 0#DvRv! C d % ( ) % b (TQ US i j by the Mtest with is 0 when is odd and is odd, when when is even; hence the th , and when . . . This U 7. If then ) 0 9 1 2I ) i ! % & p Q#iW i I I 1 i Xi i Biv #4 #! 44 6 ) R%D # w % I w # i %D f q 0#iv ) si # %D 0 # R%D ) % I w #%F7 # D i ! C "% # D #%F )#i Xi 6 R%rD # W A f 2 30 I C 0 I i I A s q D c A s9v#i si A 8#i 0 i f q (iv uniformly on , we have for , and , where D (ED q c# $0V as as . But . . Hence 2 # V V l # 0 F5 # 0V@ q@#8)tl0V % # 1 X $0V 3 $0V V 0D 00 D V 0 w08) D WErD 2 i q 2 30 I C 0 H A H ~! D 2 0 AI 2 ! H I US r (TQ 4. Given , and for let . Then for all , and The Mtest therefore gives the uniform convergence of the series for
2 0 , let . For a given , let , so . . I I & I 0VElS i0 q I V i0 V V # ip ( p i 2 0 0 ! V Bg I pD ! uI ! #iv V ! H 3 %
H 1 @ I AD0C % $ I H H s $D B C r% C 09 H ! #i q #iv Y q H 2&q 3 q I ( ) 3. Let be the maximum value of on . Given , choose so that . Then if is large enough so that we have and for . That is, for all sufficiently large, so uniformly on . for for when is ! C AD0C i q c#i0 #i C y% 2 C y% H 2 C % H 1 by the Mtest with . AD0C #i q f 7.3. Power Series
3. (a) Just observe that (b) We have , and if , or discontinuous.
2 0 I 51 is if , if , and if . If the second term tends to zero like as it vanishes to begin with. Hence is , , or for , respectively. The convergence cannot be uniform near because the limit is 4. In each case it is a matter of using the Mtest to establish the uniform convergence of the derived series on compact subsets of the interval of convergence. In what follows we write down the derived series and the constants in the test. (b) (c) (e) (f) ; ; for (d) ; . ; for ; for 5. Using the observation that , we see that the derived series of is . This series converges uniformly on compact subsets of by the same argument as in Exercise 4, 7.1 (basically, a comparison to ), so the termwise differentiation is justified. 6. (a) For , as . (c) Given , pick so that when . Then is the sum of , , and . By part (b), the first term is within of provided is suffuciently large. The absolute value of the second term is at most . Finally, since integrable functions are bounded we have , so the third term is no bigger than , which is less than for large by part (a). In short, for sufficiently large, and we are done. 7.3 Power Series
. 1. (a) and diverges when (b) . 0 , so by the ratio test, converges when . 3. converges when and diverges when , so the radius of convergence is 2. If , then by the root test (Theorem 6.14a). . If then for large , so D ! &QX % , so by the root test, converges when and diverges when converges ! r)s @ I WoC i i0 (b) With , . V ! I Q#WoC D 5 i i q i D 8#WoC `i V y i 06V q C I WoC D V i i Wo i iv i C Wo H H qt qwC C C D6 iV ! # WoC 0 hii m % m% D Gu 5 v I I vI H s H q vI I vI PsC C % 1 q I q ! 2 s 3 0 I D C 0 I ! rX D s 3 0 H s s # ! D V A q } ) D X @ % s 3 ~W D D V % X @ D (a) ; for . for . . . . C % ~G I 6 V b C !f C ! C b i C C
! 3 % j v$e le fiF fi wi i w i0v% V H u!X 1 S S ) ) 7 V ! i0Vv1 % q 0 X % V U V ! o H H q $%vX % H ( V ) H q 0X " 0V ) C ! p C d Crt $0v$uvI ! RI Bi I V I I QP b C % b v9! RI u I I QP 1 . q 0 D C " D D r I G Tr rV for all D D D i D D %% W ww5 C " D C D B 3i Cd %% % vW ww5 EDvV b 5 wW ww5 GG %%
for % d C V %V wW G Y(G k6(a ! 8(TQ US ! dV V %V % a C WD 6fw(G p(a fr wW G ! d w5 W r % V ! "& r % V ! W %d 52 , 6. Each of the series for the three integrals in question, namely, , and , is an alternating series whose terms decrease monotonically to zero in absolute value, so the full sum lies in between any two successive partial sums. One simply computes the partial sums (with a calculator or otherwise) until the desired accuracy is attained. , so the (a) answer to three decimal places is .747. (b) , so the answer to three decimal places is .905. (c) This series, alas, converges much more slowly. One has to go to the 123rd partial sum to be sure that the answer to three decimal places is .822 and not .823. I used Maple to find that the 123rd partial sum is and the 124th partial sum is ; the full sum is in between. (It is actually .) % V V % 0 V q8 sv % vq s $ v%"S q % Bv! for 2 % V ` " & j} i D V B " & C 0 D
) ' 0(U ~% ) D r! Vs ! D 4. Let . If , let . Then so converges by the root test (Theorem 6.14a). If , then many , so and diverges. Chapter 7. Functions Defined by Series and Integrals for large , for infinitely % s %V s &t ! s D ) D V 5. (a) for . By Theorem 7.18, . (TQ US j i ws X } ! % & # v"S % V % `vlwS % " % 0V )v 7 % 0V i 8(TQ US j i 5f (b) for . By Theorem 7.18, j i }f (c) ing that for for (Exercise 3 in 6.1), with the understand. By Theorem 7.18, . However, the integral on the left is continuous for (it is improper but convergent at ), and the series on the right converges absolutely and uniformly for by comparison to . Thus the equality persists for . V v% 7. If hence then for odd. Likewise, if even. for . , so by Corollary 7.22, then for all and and hence 8. (a) The absolute value of the ratio of the th term to the th term is which vanishes as for all ; hence the series converges for all . 3 A A I I % "3 3 % l" " i0 A " A % I I % @ywI % TI % " I I " 3 i A A I " I$3 A 3 % " I" % " $ % " A v l" % I I " $ % " i " v W 2 30 r D B ity, (c) (d) (b) . (For the second equality, .) .) . (In the third equal, } # ! U # (TQ US ` # ! %W U S U k U U l # (ToQ T# $FD m `j # i(T#S Q l # #! 9D r B # v # ! %UW uR# U(STQo G # ! u # (To V V }@%rD D U USQ U U USQ USQ ! U U ! Yh! W(T@B(T"u "Y! % (T@(TQ USQ US U U S Q U S Q U U U S U S Q U h! @(T@r(TY! %"u ! @B(TQ @ (TY! %"` Ei! U U S X Ei9(TQ E US 8(TQ pW l ( ( V 444 d 4 a 4 4 Tm 6##frfr}% 7.4. The Complex Exponential and Trig Functions 9. The series converges for all % 7 % (2 %  " % m (2 % T" % % (2 % T" % 0V ` S % m} } b i AD p (% % T" % 0V i ! Q I I P e ld AD le AD i m m % 0V W &! RQ I I P I Q P I I Q P " iS p! Q Yg`! i0 #1! R0gX 1! RI I PI V U V h"!Y w(TQh (TQYiT V % 0V US U S V % 08 m % & 0V 8 V 0V V 0 0) r(T7 p3 % 0 A A U S Q (% % 0V (% % &4 % 0V $ x " 0V V p7 f 7 0Vv) 'u f ' V q h 0V V q 0V V s h 0V A i g A g ( ' V 7.4 The Complex Exponential and Trig Functions by the ratio test. Let , so i ( 10. (a) 11. (a) Integrate by parts: and 1. (a) Alternatively, one can examine the power series expansions. (b) First method: express the sinh and cosh of and in terms of sides of the asserted formulas, and simplify. Second method: set part (a). (b) From Example 2 and Abel's theorem, we have for , the series converging uniformly on that interval. (The convergence at comes from the alternating series test, and Abel's theorem then gives the uniformity on and the validity of the expansion at , since is continuous.) It then follows from Theorem 7.13a that for . (d) (b) (c) Setting in (b) and using (a) gives the observation that (c) . (see part (a)). . . Alternatively, since . . and and US (TQ U B! %Wp( u ! U 3. 2. If The asserted formulas follow by taking real and imaginary parts. (Of course some constants of integration are being suppressed here.) # USQ # # (T2p # ! " # vD U D uFr # ! # U R# (TF8 U S Q D E ## EDD 4 # ! # 3D U U S Q Wl # (Tor $D (c) with , . By Exercise 2, this is . . , and then yields the desired result. , multiply out the right , and use (7.34) and , by (7.30) and (7.32). ; then 53 . , and C C !`Wo 1 C V sb h i V U S Q t % U b hi 6rvB! %U h6 rr(Tvh! s% Bi % 2 0C @ 0C ! p2sC q q D i! hr ! U q C 1 6 r ! U i V H H 2 30 # # ( 6 V % V # H 1 s V US r6r(Tk u6 h! U H V U SQ V 6 (TQ C % H H ! q 1 !C i C t F ! V ! C ! 76 V b r x6gV rh"!UC b g v "U w6 rh"U V V % " FS " i0 h! U D V "X i i U 2 D 0U C S Q h ! ` h! X ~ W(T U r(TX iS B i0 V U S Q H& H C % i `i 2 h U(TQ & S 8(TQ & 6 US Ff (Tv 6 USQ V D ## 8FS } 2 0 oC 6 6 & & i % ( % 4##B %@% % 44 4  % % ( % #4#64h 4 b ( % v##B 4 b 4 4 444 4 444 ( H ) ## ( T H V H i `i C % 2 t ! i !  i V I QP 'b 6i9! RI '3  i h} H i 0 V H & Cd% 2 & 6 B & & & ci V i DF u ci Y i x i iD 9i i C q 2 30 0 i 8# 9i q #9i i 2 ! i 7.5 Functions Defined by Improper Integrals 54 1. For all wise, we have 2. Let 3. 5. 4. times gives for any other hand, . Formally differentiating . The latter integral is uniformly convergent on for any rem 7.38 with , so the differentiation is justified. On the other hand, ; the result follows. so by the fundamental theorem of calculus, by Theorem 7.38 with . Since is uniformly convergent on any (see Exercise 3), we can reverse the order of integration to get compact interval in . 6. convergent for for any of integration to get equality, use the result of Exercise 3, 7.4, or integrate by parts twice.) , and pick a point . The integral by Theorem 7.38 with as Chapter 7. Functions Defined by Series and Integrals , so the integral converges absolutely. Like, so the convergence is uniform. . Formally differentiating . The latter integral is uniformly convergent on , so the differentiation is justified. On the . Hence . By Theorem 7.39 (with replaced by ), . The last integral is a constant, . is uniformly , so we can reverse the order . (For the first 7. 8. If , set ; then , so . It follows that for since the sine function is odd, and of course the integrand vanishes if . The convergence cannot be uniform on if by Theorem 7.39, since the resulting function is discontinuous at 0. However, the convergence is uniform for or ( ). To see this, use integration by parts as in Example 3, 4.6, or Example 3, 7.5: we have , and for this is bounded in absolute value by , which vanishes as . convergent for by Theorem 7.38 with to get of Exercise 3, 7.4, or integrate by parts twice.) 9. Let for . Since , for the integrand is less than , whose integral is finite. Hence the convergence is uniform on for any , and so is continuous on . Next, formally . for . Hence , and By Exercise 8, the differentiation is justified and . . The integral is uniformly , so we can reverse the order of integration . (For the first equality, use the result times gives , by Theo 0 2 & C 0 i v Wo F5 i 6 i 65 i C H H 7Xr "0#i C % D q # q01 i ) V cb " ' ' ! ' V " " u " " 1 ' V & & & C b ' i b ' ` " & Wo "% @e Bi C i % !`i 6BF V C ! i 1 " " " `i H 6 BF % % V V % V 6! C t% !2 C C t% 0X 1 6 " & " " & vY i C $ V C F " 6 % " & V 6 " & " " & 6 v1 "& C C "Wo `i C 2% lb ' `i V 6cb H k ' " 6 ui V ' C g% 1 "& C % FP 3i i q C 1 rs#ci ci C ~ W9i C 1 y6 9i C1 9i ci Ei V 6#" & $ E9i " " C"Wo C 7 " ui 7 " & 7 7 i7 " Ei Xi USQ 3 Y rh! " & (T" & d@ (T" & Yi U USQ b ' ` " & 6 Wo C 7 " i 7 p 3S d#i @S V ! U i U S Q r r(T" & rh"" & U 1 rh!"" & Yi U " & Yi j ! " & 6 # %rD # WD `b @w# 6 D 1#%F # D # D # D #%rF C # 1 %rD ## 5DF` b # # D # D T%F }#%DF # D #2F`b ` u"!U ` b h! %U # D F`#%rF ' 3 ! # %rD ' # D ! q 6VXWD (TkE # (TQ USQ US v "U q ! ## F "Y ## YF ! s!}WD (Tf # (TQ D ! U D U % USQ US D V D U S USQ # F 6)(TQ # ( To 2 AD0C 10. (a) The integral over proper since use the identity to get integrable function 7.5. Functions Defined by Improper Integrals converges by comparison to , and the integral over is . One way to obtain the uniformity easily is to and the estimate , which for in a bounded set is less than a fixed . D " ! 13. Let 15. (a) Formal differentiation of the integral times yields the latter integral, for any , is uniform for ( so the differentiation is justified. (b) Integrate by parts: as assumed estimate on guarantees that for all , so the integral converges uniformly to a continuous function 14. (a) The integrand is at most of . Formal differentiation of the integral yields ; this integral is still convergent for because the factor kills the factor near , and the convergence is uniform for . Hence for . If , let , so , and . If , the substitution likewise yields . (b) The differential equations for give for . Since is continuous at 0, we have by Proposition 4.66, so . 12. As in Exercise 11, . Write this equation as to get . It follows that 11. (a) The differentiated integral is by Theorem 7.38 with . is continuous on the region , if we define is continuous in , and so is since the convergence is uniform ( ). Thus is continuous for . We have for , the differentiation being justified since the latter integral converges uniformly for . Hence, by the substitution and Proposition 4.66, . Thus for , and . (b) (c) By the substitution . (b) Solving the differential equation from (a), , and (c) Since is continuous in and by part (a), the formula . Also, clearly is even in both and , so . . and . Hence , , which converges absolutely and uniformly on . Hence, by integration by parts, . . By Exercise 8, this is and multiply through by the integrating factor , and . , so (Proposition 4.66). ) since . . The convergence of , persists for for any , hence 55 ; the  % v##6h 444 4 (% % v##B%@% 444 4
! % % ##B3% 444 4 b 4 % v##vh 444 4 ! " % ##B 444 4 % v##B%@% 444 4 ! U 444 %4 I v## % I ##B$% 444 4 % 4 b  I 6##6B b ' fb ' 4 444 4 4 lI v4 ( 4 I "I ! I " I 5 ! U V % % % (}v% T fv% }i 2fi F % ! b % `(fv% i fiWq fi i % b 4 i i % "`v% b % % uhi uhi `i (v% (Ehv% Tuhv% `v% hi i X `i V D g6# "$F # l # m tl # $ C 3F `ci ! 0 X V # # u # f g f" # ph g # fB # fB8 g g` Dci 7BF ( i 89i 7 7 7 ( b ' ( kb ' 4 ( 4 ` ( ) V %cb k b b '( ( 2 } w ' !` !f ' " kb ' 4 4 X } X7 % v##RsD(k % 6##B&!` v##4 ( 4 6##Rs 444 4 4 % 444 4 % 44 4 444 %4 7.6 The Gamma Function ` 6 S V V 6S % b % ` , so . 56 2. Let % 1. , , and D 0 F # 7 3. (a) By (7.52), ( ) ( ) (b) By (7.51), ( 7 " (c) Let , ; then 4. (a) The substitution (c) integrating from 0 to 1. (b) (d) The substitution in (7.53) turns turns (7.53) into 6. We have Hence, given that ' % 7. 5. Let (There is one more factor in the denominator than in the numerator, which absorbs the extra factor of 2. If , we simply have .) I ! ! U F` ( %V ! ! U sI If so the duplication formula is valid for . Thus, if the formula is valid for , and by induction, it is therefore valid for all . is odd, it is b ' I I ##6h 444 4 444 4 2sI v##B3% b ' 9 4 44 v##4 ( 4 4 % I ##B 444 %4 8. Since ; then , so ; use Theorem 7.55. . , Chapter 7. Functions Defined by Series and Integrals , we have into goes from by (7.51). . , by Exercise 7 we have . , and . If . to 0 as goes from 0 to 1. Thus is even, this is . ; the result follows by it is also valid for . . V a (q " F V V rq5D qpC 8Fs Ft` 8FT V D D V D C 1 C fis q q is s sS # # D F % !} %V 7 xc u V V V h % %  (% % 0V r`F ` % %2 % 2 % 1 B " % B %  m V b ! ( i @ v v i r 8 V ( ( ( ( 7 V V q q 3 i q q `i i 7 7 7 v Yi T 1i Xl1 v 1imY ##4 44 v ki0Vi ##4 & 44 ! v v 1i g7i & q }i 0 C 0 3 0C i i 5i 9 i ki V q i 6 i ` q Y`i @`i @`i v# ` i 2 q i @`i A `i S! v#i 3 0C E i 2pi V E i ! p i c i 7}i i fi hi V "% C b b 9 1 b % b  q b % " 0 6 V % v##rsv % 6##B04 % 444 4 V 444 % V '~ %V 'p % b % ! ! 7.7. Stirling's Formula Multiplying through by gives . Since is increasing, and we have just seen that it is . The sequence also converges . , the sequence bounded above by , so it converges to a limit to , but its terms are , so . In short, 9. (a) For since defining . Actually there is more to be said when , since then the integral is improper: the integrand blows up at . But . For in a finite interval, say , we have , so and . Hence uniformly for , so the differentiation in the limit as is justified. (b) . Reversing the order of integration (OK even if since the integral is absolutely convergent) turns into . Now, in the inner integral, make the substitution , so that goes , , and , from 0 to 1 when goes from to . Then so 10. (a) The series is series diverges. (b) The series is converges. , . The th term is asymptotic to . The th term is asymptotic to 11. The series is converges if and only if . The th term is asymptotic to . For Raabe's test, we have Since for small, this is approximately Hence Raabe's test gives convergence for and divergence for for large, and the limit is . but is indecisive for . ! % D D % 12. and . Hence the series converges if and only if for large, so the th term , i.e., . ! # uD # D 7.7 Stirling's Formula 1. Since the second derivative of gives where is asymptotic to is , Lagrange's form of the remainder immediately for . Thus, , and for this is, in absolute value, at most 3S D sF&fD . %V %V " D h V D lF @yS 0F g , so the series , so the series , so the 57 7 7 b b % ' 2 0 4 4 B7 7 7 " % 8% 4 6 % 7 % 4 " 7 % % % 7 7 V % % % % !X % V % ###B&V % `7 % ##Rs %444 4 % 444 4 % ##Rs ! A##B&% 444 %4 % %444 4 b ' F c% V % b V b c% ' 2 30 % % 0V % 58 2. , and 3. Note that Thus the numerator and denominator of Wallis's fraction are the whole fraction is approaches 1 as which tends to . as . Thus Chapter 7. Functions Defined by Series and Integrals , or and . , where means that the ratio . , so i % I ! D q 'tsI % bV U SQ D v0D (T718% H H H H H H (Tv D v(% y ! v fFT D v(% (Tv xF D v(% D C % USQ bV U D bV USQ D bV b%V b b s D C # U b V U b b V H H D ! 8D koD (% D ! 8D ko`v(% H H 8D ! v D v(% U bV US b b V US (T b V (TQ 0D ko`v(% Q 8D v(% D C % C # D b b s H  3 % ! ) d3 % cv(5 U ( bV  bV H % 3 & H H ( v(5 H H H t@ v( H (% Tcv(% H H W V b V H # USQ V (Tv ( (% y ! %v 2oT USQ b V ! %v U b 2o H H WU % b H V H (Tv o ! v U b ~o cv(% # bV C D B# `vcA# "o m` B# `i 6 X B# `V b V b ! U b % V b% H W b%V 6 c V V b H H H H HW H H 8 r u `b (% Tcv(V % D US V C % V b H US H H U V W! H ( (% pw V (TQ (% W (TQ H U(TQ cv(% D S bV C y # cv D bV H H 3 0 3 (To cv `cv(% (TQ D D V % USQ bV bV US 3 ! % b ! 0 V b ! 0i W$ 1 H h V H % H HW H H b V bV U US (TQ ! U cv(% bV H W f } p cv ! l p !"U cv H H H H C U(Tv cv D g% SQ bV bv V ! U cv(% @ D & # bV C H H ! %(T7 "U USQ H d3 % 0i 3 % ! U cv V bV V h cb (% C r D USQ V (Tvcb (% ~ 3 ! H H % ! " U bV cv(% H H V W b H ! v U
if bV cv # 8.1 Periodic Functions and Fourier Series Fourier Series Chapter 8 1. is odd, so and . This is 0 if H 4. 5. Here it is easier to use the exponential form of the series: 7. 6. 9. First method: Suppose . Then (the integrand is in , so adding it to the all integrals). By periodicity of , the second integral on the right equals 2. By the double angle formula, 3. is even, so struction of . For so the Fourier series is is is even, so is even, so . For is odd, so ; thus the Fourier series is and , integration by parts gives , . The constant term , . This is 0 when and is even and . . This is 0 if is even and . is the mean value of 8. is even, so , . The constant term . is the mean value of . 59 , and for , and the thing on the right is a Fourier series! . The constant term is is odd, say . is odd and when , . on Integration by parts gives , so is odd, say ; hence on ; thus the Fourier series , which is 0 by con , namely if . is even, say . . For , H Acv bV H H 7 ! U U7 H "3b ! $ b H pcb % ! U H % ! U V 1"v . Hence . . 60 first integral gives . Another application of periodicity shows that this is equal to D D 8F i D B8F method: Let points where . . is a continuous function of , and except at the finitely many is discontinuous we have . It follows that is constant, so C D "X0F D % F D `8F 8.2 Convergence of Fourier Series C Wo D 8F 1. Let be the sawtooth wave of Example 1. H V bV b bV cv(% m3 pct % cH v@ H X % }b ! V X % }o U b (a) The function depicted is (b) The function depicted is . H H ! b Ds i b b d q V ssw % q H ! H HC H H # H 0 5d H C ! ! ) ' U ! H V @#Y hs ! ! H #C V ! # @s C t% C % j i H 7 A %w( "18fg3 % 0V A V b ( ( bV b 8f"3 % 0V cv(5 b %V b SQ b V # 1 TQ "X b S b # 0## % c## V V # # @ T# o X # 0 # 0## % V C r% # V H C b 3 Y# 0 v## B! o # 1(T@ b USQ V b V b U H b %V b QUQ b V # 0 # % V # F# @ ## o 5 # 0 u## v## bB! %o 5 V bV U # 0V # % C ~% H # V C # 0 v## B! %o w V bV b U C h % b DvV g V 8 (V o EC V b H H C d V b (V o b V V b b V % b (ko ` 3 0V H A 3 0V A cbv ~cv(% V bV b H k3 % f Cpg 3 % V 3 b V b V C F V  3 0 r1 3 0 cv "`cv(% H ! U 0V % H U H ! W d h cH B! o ! U V bV b ! U V b % V b U H "W 0 vcB"o H 0 # b ! 6 H H U Q (STWD V U bV b%V H 8D ! o cv"qc H s H 8D fo sc% q b%V b b V b b%V b H 0D ko&X 8D fo h H ! U % USQ bV H c H 3 0HV 3 (To H c v(% b ! ! U~ "U ! ! H 3 % ! U H d3 % cv(% ! b VH 7 H 7 H H 7 7 U(T$b (T"`cb (STQ SQ USQ U cb (TQ V H c(V o US b % U bV 1! %v v(% 4. (a) Setting gives obtainable from the observation that scopes. Setting gives . H 5. Given 2. The function 3. (a) . Setting gives . (The function represented by the series is discontinuous at , so the sum of the series is the average of the left and right hand limits!) Again the term is , and for the sum of the th and th terms is , so and hence . (d) Setting (b) Setting (c) Setting gives the sum of the th and , or (b) (d) (c) Exercise 7, 8.1, so and and (8.23), there exists sum of the th and gives here is the function of Exercise 4, 8.1, shifted to the right by , and gives such that . , choose for or where th terms is for th terms is is the square wave of Exercise 1, . . small enough so that when . Let for , so by Exercise 5, or . or or . , so by Exercise 3, 8.1, we have . The ; thus 8.1, with where Chapter 8. Fourier Series ; setting , a result also , so that the series tele, or 8.1; hence term is when is the function of , , and for . Second when . By . . The gives . H H % 5b H H H b m! ! C V % U ! o xb H US (TQ V V H H Hq h 0V H H H 7 ! %o cv(% 2 cv(% `F (TQ 7 "o cv(% cv(% V %H U b V H b V H US V % ! U bV bV (TQ 8 US ! "U ! H b ! H AC H ! US (TQ s 0 ! %U cv(5 V % bV Hb ! H AC USQ (TH v H H H USQ (Tv v ! U ms H CGV b V D7 !`67 67 5 7 b F17 V b b " H W H H H ( 7 b F` b bV b% b%V USQ (T 7 V 5 b TcHv u 7 cm H % !7 b H H H H H s b b b U V % b }9( H H ! ( 8 H `DF }q( b q d USQ (T V `D t % b H V U U b V ! %uD ! %o 0D poD (% H C b Wso D H C b b q H &D ! H DF DD WsH H H 8D#fbo08D "Y V D q b b D V D csF0 EFrh D D ! q 1 H H W H H H H H H D C H $" (TQ T 7 "vH " ! v H " # U S H b b H H U b H H ! U C ! U T 2 (Tv b b USQ USQ (Tv b D # ! "! H q v H % H % sv Y Xv v % H v H #@ p q( 's 2 sw % (V % r x q % H H H x H % H " % } q % 9q4 X f q 's4 % sw ! s( % b oT V bd d V & q( % H q % sw 's 8.3. Derivatives, Integrals, and Uniform Convergence . We have Assume now that likewise . Moreover, , and this last integral is between 0 and H The first term on the right is at most at most . Similarly, abbreviating as in absolute value, and the second term is . Adding up these results, ( ! 8.3 Derivatives, Integrals, and Uniform Convergence H ! H H # 9 C % C % ! "! d d ! H H % q @ Xv H s( 'q9 % d ` V ! H s w % % h sW 's ! ! 6. Since for is uniformly continuous, given we can choose small enough so that for any when . The argument of the preceding exercise (slightly simplified, since ) shows that for all when . H 3. (a) From Exercise 4, 8.1, we have integration yields is 0.) 4. The function is continuous and piecewise smooth, and its derivative is where is the square wave (Exercise 1, 8.1). So by Corollary 8.27, we have , and in particular the latter series converges to for . On the other hand, termwise integration of the series for from 0 to gives, for , , or . Now, since , the equality of the two series for amounts for , and one verifies this by substituting to the assertion that for in Example 1, 8.2.
H ! H H H sY X ! V # 2. (a) H H 1. (c) Setting is (b) Integration of the result of (a) and multiplication by 4 gives , where (b) Termwise integration of the series in Exercise 7, 8.1 gives for , . The graph is the broken line joining the points , as desired. we have .( for , in (b) gives , or . Now, we have ( ), and termwise is odd, so its mean value on , and , , and . . . , and , and . for 61 . 62 5. Chapter 8. Fourier Series
is not continuous (it has jumps at the odd multiples of ), so Theorem 8.26 does not apply. (For those who like distributions: The derivative of is really , where is the periodic deltafunction with singularities at the odd multiples of . The Fourier series of is , so the correct conclusion is not but rather , which is true by Exercise 5, 8.1.) converges if and only if
I (c) The given series converges uniformly (Mtest with ), so its sum is continuous, but the differentiated series does not converge at most points (the terms do not tend to zero as ). 8.4 Fourier Series on Intervals
1. (a) The even periodic extension of is the constant function , which is its own Fourier series. The odd periodic extension is the square wave (Exercise 1, 8.1). (c) The even periodic extension of is the function of Exercise 4, 8.1. The odd periodic extension is given by for , so ; the Fourier series of the latter two functions are given by Example 1 and Exercise 6, 8.1. (d) Let be the triangle wave of Example 2, 8.1. Then the even periodic extension of is , and the odd periodic extension is . The Fourier series of these are, respectively, and (since ). 2. (a) The odd 2periodic extension of series is is where is the square wave of Exercise 1, 8.1; its Fourier . where is the function of . 6, 8.1; 3. We have and similarly for and . Replace by , then set odd and and use the facts that to deduce that for even. and , whence for C " # C y D # D d D V b ! U % # % r % `b V % USQ %V US % 8 } ! U V b 6( Th u 8 % `b v(TQ }0 % i # b 6(Tvi D %V USQ V (d) The 1periodic extension of is . Its Fourier series is . where is the function of Exercise 5, 8.1, with B# (c) The odd periodic extension of is given by where is the function of Exercise . on , that is, its Fourier series i 7 b (b) The even 4periodic extension of is a square wave, namely, Exercise 1, 8.1; since , its Fourier series is H V !h3 % 0 ! H % H 1 Y0W b% V c% 0V b bV b%V v ! U b cb 1% ( 8 3 % 0 b 2 3 % ! o 8( vV 5 V V U b o cvV V b b ( s B s wi % 7 7 3 % 0Vb 3 v(TQo A cv 7 US bV b d3 6(TQ A US 3 % (STQ A U 7 cb b T 3 % ! U cb Bb v 3 % 0 b 3 % ! o cv V U bV b ov H H 3 % "U A H `cb T d3 % v(TQ ! U S A cbvV 3 USQ bV V b !h3 % V H0 (% 3 6(To cv(% hco b cb % !"o U H H H H xo b b H H b ! (b) The even periodic extension of which is its own Fourier series. is (Exercise 3, 8.1). The odd periodic extension is ! " U % d ! " U H % B! U H H H $ (b) The series times. converges for all , so the given series can be differentiated any number of % V 6. (a) The series differentiated 6 times. , so the given series can be H b H b U cB! s% $ H d q hFI b H # 0V bvcB! o V b U H b%V 8c H 2 30 H , is 7 CC vw#D0C i " 3 i % R0C 8.5. Applications to Differential Equations 4. Extend as suggested in the hint and expand it in a Fourier sine series on : where . Then for even, by Exercise 3; and for odd, is symmetric about (as observed in the solution of is twice its integral over . Setting thus yields Exercise 3), so its integral over where . C # A A Vb 3 ! oTi (% m b U V V d3 ! %U `i % (0C % R0C ! %V ! U b "vi b " U # %V ! b ! vi # %V U V 8.5 Applications to Differential Equations 1. (a) 3. If is to satisfy where , we must have , assuming that termwise differentiation of the series is justified. To solve this ordinary differential equation, multiply through by the integrating factor to obtain , whence . For this to work, the following conditions are (more than) sufficient: (1) is of class on , and . (2) is as a function of for each , , and , , and are jointly continuous as functions of and . The boundary conditions on and guarantee that their odd periodic extensions are still at least , and that of is at least piecewise continuous. It follows that the Fourier sine coefficients of (namely, ) are absolutely summable, and those of (namely, ) are continuous in and satisfy for in any finite interval . We then have 2. One follows the separationof variables procedure as on p. 382 to find solutions of the form . The periodicity condition then forces , so the resulting analog of (8.35) is . To satisfy the initial condition one takes to be the Fourier series of . (The result looks a little neater in exponential form: where .) VcoI b #WCo # & " q " D#Wo # & " 0#i # C q & 0#i q 0C i WoC # C "1 (0C i 9i ci ci C s u0oC (0C i C l Wo C (0C 9i v " & " C Wo # & " i # i & " & " i # V & V I i i # cb 1 wi # Vb ! vi u9i U I & b c"vi # 9i V ! U H H " H s & 6 # U S D H H ! U H (TQ F H H ! U # U(TQ F S D H " E ' E H ' ! U US (TQ & 6 '
& 05 A(6V o " 7 ( " ( vWw( m 73 % 0 5 b b V C C V " ( " ( vWw( bVCC q C ra9i C (wC " C 3 % 0 ( " V bV CC q Ca CCd ( at r vWw( Yr"1ci ww 1 C ( } C# Cm C C V b U S Q C C V b U S Q C Ww#v (Tv0a % !Ww#v o((TvwG "la Ww#v v(ST" CCV b U Q V b U S Q bVCC Ca Cd RWCwC#vo6(T" vWw( !WR9i b 5 G (wD # % Cw( C ( ( 3 % b d3 % b A CC w( q Ww#voT2 3 % 6(TQ " " " w( C C V b US CC The Fourier on is , so the solution (8.35) of the heat equation is . , the error in discarding the terms after is cosine series for and 40 when . (c) For , . Almost good enough, but not quite! A slightly less crude estimate works:
" " " 3 % 3 Cd X % 0WCwC#vVb d3 % v(To & " " A vWwC ( !u9i V USQ bV C Ca V CC V U SQ bVCC C 3 % 0Ww#vb t3 % 6(To 8 vWw( ua To within this error, (b) When . , which is about 10 when , 12 when 63 , 64 Chapter 8. Fourier Series
This is enough to guarantee the absolute and uniform convergence of the series defining and , as well as the absolute and uniform convergence of the series defining and for and ( ), so that all formal calculations are justified. ci & ci for 4. (a) The odd periodic extension of the initial displacement is where is as in Exercise 2, 8.3, with , so its Fourier sine series can be read off from the answer to that exercise. The series for can then be read off from (8.37). we have , and , .147, , 0 when , so the first five coefficients (up to the overall factor of ) are , , 0. When we have , and , .147, .090, .059, .040 when , so the first five coefficients are ( times) .696, .331, .203, .133, .090. (Note: The norm of the initial displacement is , independent of , so the total energy of these waves is independent of and a direct comparision of the coefficients is appropriate.) (b) When , .803, .124, (Exercise 1b, 7.4), so ; hence any linear combination of and is also a linear combination of and (and vice versa). Now, for to satisfy Laplace's equation, we need , and the boundary conditions become . As in the text, this forces and . Hence , so (by the preceding remark) . Taking linear combinations, we arrive at the general solution . We then have , which must be the Fourier sine series of , and , which must be the Fourier series of . 7. (a) For to satisfy the polar Laplace equation, we need , or , a constant. This differential and equation for , together with the periodicity requirement, yields ( ). The general solution of the differential equation for is then a linear combination of and (or and if ). But and blow up at the origin and must be discarded. Hence we obtain the general solution , and the requirement that means that the 's are the Fourier coefficients of ; thus . (b) This follows immediately from part (a), (8.20), (8.22), and the observation that the Poisson kernel is an even function, so that one can replace by in the integral. (b) First we note that % H H H ` v H H CS 6 y CS C % C vRD0 H H X 3 H H V f @ wW i ( B( H V C p ( H H H @ W W 3 W @ W ` 6 b ! v b "U # V U V ! i 9i iD V U V ! b ! v b "U # C $W9i b ! U s 82`b "U # b "U $ci V # V ! V ! 8b "U !`b "U # V ! # V ! i@ i@ cb V 8i V b ! U Ei cb V C u Wou C V i i Fi@ i V i@ i 89i ! 8lsw "U A ! U U A (STQ A ! %U U S Q ! U A ! p TsD8 Ew "u #"sA (TQ QUQ US ! u (T&7A (Tu ! U Fw ! U U USQ USQ 6. (a) If solves the problem for and solves the problem in the general case. solves the problem for , then i V D b 5. For to be a solution of the modified wave equation, we must have , or , a constant. As and . Then on page 385, the boundary conditions force . The roots of are where , so . Taking linear combinations of these solutions for gives the desired result analogous to (8.37). (This is assuming . If have pure exponential decay with no oscillation.) not, the solutions for V 3 W 4 3 C aR8&vRvRd % G (C cb #! U %a ww% %t WCwh ## ` # bvV % bGV C% C # o p# Ca(C w(C aa 3 aR8 &vRvR % Ga w(C w(C ad a 0wG cb #! U 5 b( %VWCwC%rh## # bvV % C # o~3# ci b F5D `i C x i ! e H H vV q b % vRvRF H ! %U H (TQ # & i@ D b m H US V F C D b ~k r V % C 5i@ D b ~ui % V Vb ! %t}i U V cb H H 7i i fi@ i i3 FBi V V % i3 i xi i % i@ i ci o V b # 3 C W9i C C % w i H i (0C p ci 0C i (0C f i ## V # 9 H W H H H Ca G b W V ! V a b % D0V (0! D7 7 (b ` ( cv(% b bV H H H b bV V a b % 7 cv (5 (DvV 7 v8% H W H H CdGV b w "1  % 0 V a V b Dv7 " D7 ( % b H H H H o cv(% gpo cvy  % 0 cv(5 b bV b bV V bV H W H CG b V 7 " 7 V a b% V 7 7 cvm 7 D"1V 7 % bV V d G b H W H # !u b% b # D 2 D b D D C " # % C " x 8! # x D % % d # # t & D % 2 0 C 0 V I Pp #i Xi 5 q i i i R%#D xT D 2 0 C 0 V 5 " #i i #i i q % q mq#i i # i # i R%D 7 r#i q # R %rD # p q q q ! 2! i c i p2q D i i ! # ! C &D # ! ! &D # D i #%Fd # R%D " s Cg5 i i A H r i 5 v i i % A A 5 A sA A A A H V 6% ' i A A 'V 6% wi i V f 6% ' i 'V & 6% wi 8.6 The InfiniteDimensional Geometry of Fourier Series 8.6. The InfiniteDimensional Geometry of Fourier Series 2. If , let be its odd periodic extension and let be the Fourier series of , whose restriction to is the Fourier sine series of . Then as . Likewise for the cosine series. H H W ! U }3 "Y 7 ( ( 7 7 1 ( F # ( # v # 9# 5 # @ # 7 9# ( 5 ( C # iT i ( ` T F1 F D "` ` # D D 5R# i s 8D 5i s H H H C 0 ! 2 30 H "U # q ! U # !h b 0C H ! U # b % b C c0o i C USQ b ' U S (T"Y b " } (TQ C % %V cb "U (TQ ! US C " }3 ! @ ! U U }8(T@ (TQ 3 USQ US 3 %cb V 3 7 3 (Tgw t3 6(TQ USQ US A A T 6 }3 (TY (TQ USQ US r 3 v(Th& 3 6(TQ }8(TY (TQ USQ US 3 USQ US
!% U 1. Use the identities , or and , etc., to see (of ; the norm of C r% that course and when are 0 when ). The norm of or and is when for 3. is . and , so to make # # # 4. 5. so we must have we need and hence . Finally, and . . Then since the integrand is even; likewise, since the integrand is odd. ( A
H A 6. ( A
H A A 8. We have 7. Suppose, to begin with, that is continuous on except for a jump at . To simplify . For large enough so that and , define to be notation, we assume if or , and if . Then is continuous on . Moreover, if for all , the same is true of , since the values of on lie in between and ; so . Hence 9. (a) From Exercise 4, 8.1, of discontinuities at , on which that (c) From Exercise 3b, , so and this formula becomes . (b) From Exercise 6, 8.1, , or 8.3, , one likewise defines interpolates linearly between as . .) Hence , or . and . (When as to be and we have . ). , or . In the general case except on the intervals , and one finds and ). 65 . , , A b co ` b % H H W H C ~ b 3c C H t % H H !` W H b 1 H H H 3 D b b " D "lD C H U b % b bV b H % V b W 8 D 5oD V 8D ! o 8D poD (V 8D 5o D cv(% y 0D ko bV cv(% V 0D ! %o 01Xo D U D b V b ! $AC D ! D 66 10. First way: Second way: If formula is still valid when a function of . (d) From Exercise 7, 8.1, for since are the Fourier coefficients of , then the Fourier coefficients of are . Hence, by (8.46), . or we have , or . This (both sides vanish then), and the sum is clearly periodic as Chapter 8. Fourier Series . , and ...
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This note was uploaded on 10/21/2010 for the course MATHEMATIC MAT237Y1 taught by Professor Romauldstanczak during the Fall '09 term at University of Toronto Toronto.
 Fall '09
 RomauldStanczak
 Calculus, Continuity, Vectors, Limits, Sets, The Land

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