经济学中的数学Mathematics for economists课后答案answers2.pdf

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1、ANSWERS PAMPHLET51 b) x(s,t) 5 0 3 2 1 3 0 21 t 1 2 2 22 s, x 1 2y 1 3z 5 12. 10.37 a) Rewriting the symmetric equations, y 5 y02 (b6 a)x01 (b6 a)x and z 5 z02 (c6 a)x01 (c6 a)x. Then x(t) y(t) z(t) 5 x0 y0 z0 1 1 b6 a c6 a t or x(t) y(t) z(t) 5 x0 y0 z0 1 a b c t. b) The two planes are described by

2、 any two distinct equalities in system (20). For example, x 2 x0 a 5 y 2 y0 b and y 2 y0 b 5 z 2 z0 c . In other words, 2bx 1 ay 5 2bx01 ay0and cy 2 bz 5 cy02 bz0. c) i) x12 2 21 5 x22 3 4 5 x32 1 5 ,ii) x12 1 4 5 x22 2 5 5 x32 3 6 . d) i) 4x11 x25 11 and 5x22 4x35 11. ii) 5x12 4x25 23 and 6x22 5x35

3、 23. 10.38 a) Normals (1,2,23) and (1,3,22) do not line up, so planes intersect. b) Normals(1,2,23)and(22,24,6)dolineup,soplanesdonotintersect. 10.39 a) (x 2 1, y 2 2,z 2 3) ? (21,1,0) 5 0, so x 2 y 5 21. b) Thelinerunsthroughthepoints(4,2,6)and(1,3,11),sotheplanemust be orthogonal to the difference

4、 vector (23,1,5). Thus (x 2 1, y 2 1, z 1 1) ? (23,1,5) 5 0, or 23x 1 y 1 5z 5 27. c) The general equation for the plane isax1by1gz 5d. The equations to be satisfi ed are a 5d 6 a, b 5d 6 b, and c 5d 6 g. A solution isa5 16 a,b5 16 b.g5 16 c andd5 1, so (16 a)x 1(16 b)y 1(16 c)z 5 1. 10.40 Plug x 5

5、31t, y 5 127t, and z 5 323t into the equation x1y1z 5 1 of the plane and solve for t: (31t)1(127t)1(323t) 5 1 = t 5 26 3. The point is 116 3,2116 3,1 . 10.41 1 121 . . .4 121 . . .3 = 1 023 . . .5 012 . . .21 = x 55 1 3z y 5 21 2 2z. Taking z 5 t, write the line as x y z 5 5 21 0 1 t 3 22 1 . 52MATH

6、EMATICS FOR ECONOMISTS 10.42IS : 1 2 c1(1 2 t1) 2 a0Y 1 (a 1 c2)r 5 c02 c1t01 Ip1 G LM : mY 2 hr 5 Ms2 Mp. Iprises = IS moves up = Ypand rpincrease. Msrises = LM moves up = Ypdecreases and rpincreases. m rises = LM becomes steeper = Ypdecreases and rpincreases. h rises = LM fl atter = Ypincreases an

7、d rpdecreases. a0 rises = IS fl atter (with same r-intercept) = rpand Yprise. c0rises = IS moves up = Ypand rpincrease. t1ort2rises= IS steeper(withsamer-intercept)= rpandYpdecrease. Chapter 11 11.1 Suppose v15 r2v2. Then 1v12 r2v25 r2v22 r2v25 0. Suppose c1v11 c2v25 0. Suppose that ci? 0. Then civi

8、5 2cjvj, so vi5 (cj6 ci)vj. 11.2 a) Condition (3) gives the equation system 2c11c25 0 c11 2c25 0. The only solution is c15 c25 0, so these vectors are independent. b) Condition (3) gives the equation system 2c11c25 0 24c12 2c25 0. One solution is c15 22, c25 1, so these vectors are dependent. c) Con

9、dition (3) gives the equation system c15 0 c11 c25 0 c21 c35 0. Clearly c15 c25 0 is the unique solution. ANSWERS PAMPHLET53 d) Condition (3) gives the equation system c11 c35 0 c11 c25 0 c21 c35 0. Theonlysolutionisc15 c25 c35 0,sothesevectorsareindependent. 11.3 a) The coeffi cient matrix of the e

10、quation system of condition 3 is 110 000 101 011 .Therankofthismatrixis3,sothehomogeneousequation system has only one solution, c15 c25 c35 0. Thus these vectors are independent. b) The coeffi cient matrix of the equation system of condition 3 is 111 000 1210 000 . The rank of this matrix is 2, so t

11、he homogeneous equation system has an infi nite number of solutions aside from c15 c25 c35 0. Thus these vectors are dependent. 11.4 If c1v15 v2, then c1v11 (21)v25 0, and condition (4) fails to hold. If condition (4) fails to hold, then c1v11 c2v25 0 has a nonzero solution in which, say, c2? 0. The

12、n v25 (c16 c2)v1, and v2is a multiple of v1. 11.5 a) The negation of “c1v11c2v21c3v35 0 implies c15 c25 c35 0” is “Thereissomenonzerochoiceofc1,c2,andc3thatc1v11c2v21c3v35 0.” b) Suppose that condition (5) fails and that c1? 0. Then v15 2(c26 c1)v12 (c36 c1)v3. 11.6 Suppose hv1,.,vnj is a collection

13、 of vectors such that v15 0. Then 1v11 0v21 ? 1 0vn5 0 and the vectors are linearly dependent. 11.7 A? c1 . . . ck 5 c1v11?1ckvk,sothecolumnsofAarelinearlyindependent if and only if the equation system A ? c 5 0 has no nonzero solution. 54MATHEMATICS FOR ECONOMISTS 11.8 The condition of Theorem 2 is

14、 necessary and suffi cient for the equation system of Theorem 1 to have a unique solution, which must be the trivial solution.ThusAhasrankn,anditfollowsfromTheorem9.3thatdetA ? 0. 11.9 a) (2,2) 5 3(1,2) 2 1(1,4). b) Solvetheequationsystemwhoseaugmentedmatrixis 1101 1012 0113 . The solution is c15 0,

15、 c25 1, and c35 2. 11.10 No they do not. Checking the equation system Ax 5 b, we see that A has rank 2; this means that the equation system does not have a solution for general b. 11.11 For any column vector b, if the equation system Ax 5 b has a solution xp, then xp 1v11 ? 1 x p nvn5 b. Consequentl

16、y, if the equation system has a solution for every right-hand side, then every vector b can be written as a linear combination of the column vectors vi. Conversely, if the equation system fails to have a solution for some right-hand side b, then b is not a linear combination of the vi, and the vido

17、not span Rn. 11.12 The vectors in a are not independent. The vectors in b are a basis. The vectors in c are not independent. The vectors in d are a basis. 11.13 a) det ? 10 01 5 1 ? 0, so these vectors span R2and are independent. b) det ? 210 01 5 21 ? 0,sothesevectorsspanR2andareindependent. d) det

18、 ? 11 211 5 2 ? 0, so these vectors span R2and are independent. 11.14 Two vectors cannot span R3, so the vectors in a are not a basis. More than three vectors in R3cannot be independent, so the vectors in e fail to be a basis. The matrix of column vectors in b and c have rank less than 3, so neither

19、 of these collections of vectors is a basis. The 333 matrix A whose column vectors are the vectors of d has rank 3, so detA ? 0. Thus they are a basis according to Theorem 11.8. 11.15 Supposeaistrue.Thenthe n3nmatrixAwhosecolumnsarethevectorsvi has rank n, and therefore Ax 5 b has a solution for eve

20、ry right-hand side b. Thus the column vectors vispan Rn. Since they are linearly independent and span R3, they are a basis. Since the rank of A is n, detA ? 0. Thus a ANSWERS PAMPHLET55 implies b, c and d. Finally suppose d is true. Since detA ? 0, the solution x 5 0toAx 5 0isunique.Thusthecolumnsof

21、Aarelinearlyindependent; d implies a. Chapter 12 12.1 a) xn5 n.b) xn5 16 n.c) xn5 2(21) n21(n21). d) xn5 (21)(n 2 1)(n 2 1)6 n.e) xn5 (21)n.f) xn5 (n 1 1)6 n. g) xnis the truncation ofpto n decimal places. h) xnis the value of the nth decimal place inp. 12.2 a) 1/2 comes before 3/2 in original seque

22、nce. b) Not infi nite. c) 26 1 is not in original sequence. 12.3 If x and y are both positive, so is x 1 y, and x 1 y 5 |x| 1 |y| 5 |x 1 y|. If x and y are both negative, so is x 1 y, and 2(x 1 y) 5 2x 2 y; that is, |x 1 y| 5 |x| 1 |y|. If x and y are opposite signs, say x . 0 and y # 0, then x1y #

23、x 5 |x| # |x|1|y| and 2x 2 y # 2y 5 |y| # |x|1|y|. Since x 1 y # |x|1|y| and 2(x 1 y) # |x| 1 |y|, |x 1 y| # |x| 1 |y|. Itfollowsthat|x| 5 |y1(x2y)| # |y|1|x2y|;so|x|2|y| # |x2y|.Also, |y| 5 |x 1(y2 x)| # |x|1|y2 x| 5 |x|1|x 2 y|; so |y|2|x| # |x 2 y|. Therefore, fl fl |x| 2 |y| fl fl # |x 2 y|. 12.

24、4 If x and y are $ 0, so is xy and |x|y| 5 xy 5 |xy|. If x and y are # 0, xy 5 |xy|. If x $ 0 and y # 0, xy # 0 and |xy| 5 2xy. Then, |x|y| 5 x(2y) 5 2xy 5 |xy|. Similarly, for x # 0 and y $ 0. 12.5 |x 1 y 1 z| 5 |(x 1 y) 1 z| # |x 1 y| 1 |z| # |x| 1 |y| 1 |z|. 12.6 Follow the proof of Theorem 12.2,

25、 changing the last four lines to |(xn2 yn) 2 (x 2 y)| 5 |(xn2 x) 2 (yn2 y)| # |xn2 x| 1 |yn2 y| # 2 1 2 5. 56MATHEMATICS FOR ECONOMISTS 12.7 a) There exists N . 0 such that n $ N = |xn2 x0| , 1 2|x0|. Then, |x0| 2 |xn| # |x02 xn| , 1 2|x0|, or 1 2|x0| # |xn|, for all n $ N. Let B 5 min n 1 2|x0|,|x1

26、|,|x2|,.,|xN| o . For n # N, |xn| $ minh|xj| : 1 # j # Nj $ B. For n $ N, |xn| $ 1 2|x0| $ B. b) Let B be as in part a). Let. 0. Choose N such that for n $ N, |xn2 x0| #B|x0|. Then, for n $ N, fl fl fl fl 1 xn 2 1 x0 fl fl fl fl 5 |xn2 x0| |xn|x0| # |xn2 x0| B|x0| (since |xn| $ B for all n = 16 |xn|

27、 # 16 B for all n) # ? B|x0| B|x0| 5. Therefore, 16 xn 16 x0. 12.8 Suppose yn y with all yns and y nonzero. By the previous exercise, 16 yn 16 y. By Theorem 12.3 xn yn 5 xn? 1 yn x ? 1 y 5 x y . 12.9 Let. 0. Choose N such that for n $ N, |xn20| #?B where |yn| # B for all n. Then, for n $ N, |xn? yn2

28、 0| 5 |xn? yn| 5 |xn|yn| # B ? B 5. 12.10 Suppose that xn$ b for all n and that x , b. Choose. 0 such that 0 , b 2 x. So,1 x , b and I(x) 5 (x 2,x 1) lies to the left of b on the number line. There exists an N . 0 such that for all n $ N, xn I(x).Forthese xns, xn, x1, b,acontradictiontothehypothesis

29、 that xn$ b for all n. 12.11 TheproofofTheorem12.3carriesoverperfectly,usingthefactthat|x?y| # kxk ? kyk holds in Rn. 12.12 Suppose xn a and b ? a is an accumulation point. Choose5 ka 2 bk6 4. There exists an N such that n $ N = kxn2ak ,. Since b is an accumulation point, there exists an m $ N such

30、that kxm2 bk ,. ANSWERS PAMPHLET57 Then, ka 2 bk 5 ka 2 xm1 xm2 bk # kxm2 ak 1 kxm2 bk ,15 25 ka 2 bk 2 . Contradiction! 12.13 The union of the open ball of radius 2 around 0 with the open ball of radius 2 around 1 is open but not a ball. So is the intersection of these two sets. 12.14 Let x Rn1. Ch

32、 (x 2,x 1). 12.16 IfAisopen,forallx AthereisanopenballBxcontainingxandinA.Then xABx, A , xABx. Let A denote the interior of A. By defi nition, for all x A, Bx, A, so xABx, A, A , S 5 . (Otherwise it would be possible to construct a sequence in S converging to x.) Therefore x intT. 12.24 Any accumula

33、tion point of S is in clS. To see this, observe that if x is an accumulation point of S, then for all n . 0,B16 n(x) S ? . For each integer n, choose xnto be in this set. Then the sequence hxnj is in S and converges to x, so x clS. Conversely, if x is in clS, then there is a sequence hynj , S with l

34、imit x. Therefore, for all. 0 there is a y S such that ky 2 xk ,. Consequently, for all. 0,B(x) S ? , so x is an accumulation point of S. 12.25 There is no ball around b contained in (a,b, so this interval is not open. A sequence converging down to a from above and bounded above by b will be in the

35、interval, but its limit a is not. Hence the interval is not closed. The sequence converges to 0, but 0 is not in the set so the set is not closed. Noopenballofradiuslessthan1/2aroundthepoint1containsanyelements of the set, so the set is not open. A sequence in the line without the point converging t

36、o the point lies in the line without the point, but its limit does not. Hence the set is not closed. Any open ball around a point on the line will contain points not on the line, so the set is not closed. 12.26 Suppose that x is a boundary point. Then for all n there exist points xn S and ynin Sc, b

37、oth within distance 16 n of x. Thus x is the limit of both the hxnj and hynj sequences, so x clS and x clSc. Conversely, if x is in bothclS andclSc,thereexistsequenceshxnj , S andhynj , Scconverging to x. Thus every open ball around x contains elements of both sequences, so x is a boundary point. 12

38、.27 The whole space Rmcontains open balls around every one of its elements, so it is open. Any limit of a sequence in Rm is, by defi nition, in Rm, so Rm is closed. The empty set is the complement of Rm. It is the complement of a closed and open set, and consequently is both open and closed. Checkin

39、g directly, the empty set contains no points or sequences to falsify either defi niton, and so it satisfi es both. 12.28 The hint says it all. Let a be a member of S and b not. Let l 5 hx : x 5 x(t) ; ta 1 (1 2 t)b,0 # t , j. Let tp be the least upper bound of the set of all t such that x(t) S. Cons

40、ider the point x(tp). It must be in S because there exists a sequence htnj converging up to tpsuch that each 60MATHEMATICS FOR ECONOMISTS x(tn) S, x(tp) is the limit of the x(tn), and S is closed. Since S is open, there exists a ball of positive radius around x(tp) contained in S. Thus for . 0 suffi

41、 ciently small, x(t p1) S which contradicts the construction of tp. Chapter 13 13.1 See fi gures. 13.3 Intersect the graph with the plane z 5 k in R3, and project the resulting curve down to the z 5 0 plane. 13.4 A map that gives various depths in a lake. Close level curves imply a steep drop off (a

42、nd possibly good fi shing). 13.6 a) Spheres around the origin. b) Cylinders around the x3-axis. c) The intersection of the graph with planes parallel to the x1x3-plane are parabolas with a minimum at x 1 1 5 0. The level of the parabolas shrinks as y grows. d) Parallel planes. 13.9 See fi gures. 13.

43、10 The graph of f is G 5 h(x, y) : y 5 f(x)j. Consider the curve F(t) 5 (t, f(t). The point (x, y) is on the curve if and only if there is a t such that x 5 t and y 5 f(t); that is true if and only if y 5 f(x); and this is true if and only if (x, y) is in the graph of f. 13.11 a) (2235) x1 x2 x3 , b

44、) 223 124 10 ? x1 x2 , c) 1021 2326 021 x1 x2 x3 . 13.12 a) (x1x2) ? 121 211 ? x1 x2 , ANSWERS PAMPHLET61 b) (x1x2x3) ? 525 2521 x1 x2 x3 , c) (x1x2x3) 1223 224 2343 x1 x2 x3 . 13.13 f(x) 5 sinx, g(x) 5 ex, h(x) 5 logx. 13.14 Suppose f(x) is linear. Let ai5 f(ei), and suppose x 5 x1e11?1 xkek. Then

45、f(x) 5 f(x1e11 ? 1 xkek) 5 X i xif(ei) 5 X i aixi 5 (a1,.,ak)(x1,.,xk). 13.15 Consider the line x(t) 5 x 1 ty, and suppose f is linear. Then f(x(t) 5 f(x1ty) 5 f(x)1tf(y), so the image of the line will be a line if f(y) ? 0 and a point if f(y) 5 0. 13.16 Let hxnj n51 be a sequence with limit x. Sinc

46、e g is continuous, there exists an N such that for all n $ N, |g(xn) 2 g(x)| , g(x)6 2. Thus, for n $ N, g(xn) ? 0, since g(x) 2 g(xn) , g(x)6 2 implies that 0 , g(x)6 2 , g(xn). Sinceconvergencedependsonlyonthe“largeN”behaviorofthesequence, we can suppose without loss of generality that g(xn) is ne

47、ver 0. Now, f(xn) converges to f(x) and g(xn) converges to g(x), so the result follows from Exercise 12.8. 13.17 Suppose not; that is, suppose that for every n there exists an xn B16 n(xp) with f(xn) # 0. Since |xn2 xp| , 16 n, xn xp. Since f is continuous, f(xn) f(xp). Since each f(xn) # 0, f(xp) 5

48、 lim f(xn) # 0, but f(xp) . 0. (See Theorem 12.4). 13.18 From Theorem 12.5 conclude that a function f : Rk Rmis continuous if and only if its coordinate functions are continuous. From Theorems 12.2 and 12.3 and this observation, conclude that if f and g are continuous, then the coordinate functions

49、of f 1 g and f ? g are continuous. 62MATHEMATICS FOR ECONOMISTS 13.19 =: Suppose f 5 (f1,., fm) is continuous at xp. Let hxnj be a sequence converging to xp. Then, f(xn) f(xp). By Theorem 12.5, each fi(xn) fi(xp). So each fiis continuous. =: Reverse the above argument. 13.20 Suppose xn5 (xn1,xn2,.,xnk) x05 (x01,x02,., x0k). By Theorem 12.5, xni x0ifor each i. This says h(x1,.,xk) 5 xiis continuous. Any monomial g(x1,.,xk) 5 Cx

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