Solution of Principles of Mathematical Analysis(Rudin)

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魯丁 數學分析 答案 解答 Solution of Principles of Mathematical Analysis
A Chapter 1 The Real and Complex Number Systems Exercise 1.1 If r is rational (r+0)and a is irrational, prove that r+r and ra are irrational Solution. If r and r +3 were both rational, then =r-r would also be rational. Similarly if r were rational, then r would also be rational Exercise 1.2 Prove that there is no rational number whose square is First Solation. Since v12=2V3, we can invoke the previous problem and prove that v3 is irrational. If m. and n are integers having no common factor and such that m2= 3n then m is divisible by 3(since if m2 is divisible by 3, so is m). Let m=3k. Then m2=9k2, and we have 3k2=n. It then follows that n is also divisible by 3 contradicting the assumption that m and n have no common factor Second Solutio. Suppose m2= 12n2, where m and n have no common factor It follows that m must be even, and therefore n must be odd. Let m = 2r Then we have r2=3n2, so that r is also odd. Let r=2s+1 and n=2t+1 hen 4s2+4+1=3(42+4t+1)=12t2+12+3, so that 4(S2+-3t2-3t)= But this is absurd, since 2 cannot be a multiple of 4 CHAPTER I. THE REAL AND COMPLEX NUMBER SYSTEMS Exercise 1.3 Prove Proposition 1.15, i.e. prove the following, statements (a)#0 and ary =cz, then 1 =z (b)x≠0amdy=, then y=1 (c) Ifro and y=1, then y=l/c (d)x≠0,机en1/(1/x)=x Solution.(a) Suppose #0 and ry c2. By Axiom(M5)there exists an element 1/x such that 1/=1. By(M3)and (M4) we have(1/(ay) ((1/3)y= ly=y, and similarly(1/c(rz)=z. Hence y=z (b)Apply(a)with z=1 (c Apply(a)with z=1/c (d)Apply (a)with z replaced by 1/z,y=1/(1/c), and 2=c Exercise 1: 4 Let e be a nonempty subset of an ordered set; suppose a is a lower bound of E, and B is an upper bould of E. Prove that C<B Solution. Since E is nonempty, there exists z EE. Then by definition of lower and upper bounds we have <x < B, and hence by property ii in the definition ot an ordering: we have a< B unless a=x=6 Exercise 1. 5 Let A be a nonempty set of real numbers which is bounded below Lct -a be the set of all numbers -z, where 2 E A. Prove that inf A=-sup(A Solution: We need to prove that -sup (A) is the greatest lower bound of A For brevity, let o=sup(A). We need to show that a< s for all a E A and a2 B if B is any lower bound of A Suppose S E A. Then, -cE-A, and, hence - c S sup(-A) It follows that t2-sup(A), i.e., a <a. Thus a is a lower bound of A Now let B be any lower bound of A. This means b< a for all in A Hence-xs-6 for all r∈A, which says y≤-β for all y∈-A.This Imeans B is an upper bound of A. Hence -B2 sup(-A)by defin tion of sup, i. 6s-sup(-A), and so -sup(-A)is the greatest lower bound of A Exercise1,6Fiⅸxb>1. (a)If m, n, p,q are integers,n>0, q>0, ar /7.=p/a, prove that = q Hence it makes sense to define b=(6m)1/n (pRove that r and s are rational c)If r is real, define B(a to be the set of all numbers bt, where t is rational and t<r Prove that b=sup b(r) when r is rational. Hence it makes sense to define Sup Br for every real x (d) Prove that ba+y= bb3 for all real r and y Solution.(a) let k= mg=mp. Since there is only one positive real number c such that cng= b(Theorem 1.21), if we prove that both(6m)1/n and(bp)1/9 have this property, it will follow that they are equal. The proof is then a routine computation:((6 m)l/n)rq mq b and similarly for(bp), (6) Let r nd en r ane 6r+s=(6 (bm809 1/n y the laws of exponents for integer exponents. By the corollary to Theorem 1.21 we then haye br+s=(bmu)/nu( bnu)1/w=6 bs where the last equality follows from part (a) (c)It will simplify things later on if we amend the definition of b(a) slightly by defining it as (bt: t rational, t r]. It is then slightly more difficult to prove that b= sup b(r) if r is rational, but the technique of Problem 7 comes to our rescue. Here is how: It is obvious that b is an upper bound of b(r) We need to show that it is the least upper bound. The inequality b1/n t if m>(b-1)/(t-1)is proved just as in Problem 7 below. It follows that if 0<I<O, there exists an integer n with bi/n<br/, i.e., c <br-lIT E b(r) Hence a is not an upper bound of b(r), and so br is the least upper bound (d)By definition b*ty =sup b(a +y), where B(a+ y) is the set of all numbers bt with t rational and t a +y. Now any rational number t that is less than T+y can be written as r+s, where r and s are rational, r< s< y. To do this, let r be any rational number satisfying t-9<r<a, and lets=t-r. Conversely any pair of rational nurnbers r, s with r< r, s< y gives a rational sum t=T +s<a+. Hence B(a+y) can be described as the set of all numbers b,bs with r<a, s< y, and r and s rational, i.e. B(a +y) is the set of all products uv, where u E B(a)and v E B(g) Since any such product is less than sup b(a)sup b(9), we see that the num ber M= sup B(a sup B() is an upper bound for B(c+ g. On the other hand, suppose 0 <c< sup B(r)sup B(y). Then c/(sup B Bup b(y. Let m=(1/)(c/(sup B(a))+sup B(3). Then c/sup B()<m< sup b(y),and the ere exist u E B(a),vE B( such that c/m< u and m v, Hence we have CHAPTER 1. THE REAL AND COMPLEX NUMBER SYSTEMS c=(c/m m< uv E B(a+y), and so c is not an upper bound for B(+2).It follows thaT sup b(a sup b(y is the least upper bound of B(r +y),i.e bzty=6by as required Exercise 1.7 Fix b>1,y >0, and prove that there is a unique real r such that b y, by completing the following outline. (This a is called the logarithm of y to the base b. a) For any positive integer n, bn-1> (6) Hence 6-1>7(b/n-1) (c)Ift> I and n>(b-1)/(t-1), then 61/n< t (d)If w is such that bu< y, then bu+(1/n)<y for sufficiently large n: to see this apply part (c) with t=y6-w (e)If bu> y, then bu(i/n)>y for suficiently large n (f Let a be the set of all w such that b< 1, and show that sup A satisfies b=y (g)Prove that this is unique Solution.(a) The inequality bn-1 2 n(b-1)is equality if n= 1. Then, by induction b+1-1=b+1-b+(-1)=b(b0-1)+(b-1)≥bn(b-1)+(6b-1) (bn+1)(b-1)≥(n+1)(b-1) (6)Replace b by bi/m in part(a) (c) The inequality m>(b-1/( t-1)can be rewritten as n(t-1)>(6-1) and since 6-1>n(bi/m-1), we have n(t-1)>n(bi/m-1), which implies (d )The application of part (c with t=yb-w>I is immediate ( e) The application of part(c)with t=bw(1/)yields the result, as in part(d )above f) There are only three possibilities for the number r= sup A: 1)b <3; 2) b>3;3)b y. The first assumption, by part(d), implies that r+(1/n)EA for large n, contradicting the assumption that i is an upper bound for A. The second, by part(e), implies that c-(1/n)is an upper bound for A if n is large, contradicting the assumption that z is the smallest upper bound. Hence the only remaining possibility is that b=3 (9) Suppose z+a, say x >2.Then 62=b*+(2-x) =bb2->b2=y Lence a is unique. (It is easy to see that bw 1 if w>0. since there is a. positive rational number r=m with0 r< w, and br=(b m1/n. Then 6 >1 since b> 1, and(bm 17>1 since 1m=1<6m Exercise 1.8 Prove that no order can be defined in the complex field that turns it into an ordered field. Hint: -1 is a, square Solution. By Part(a) of Proposition 1. 18, either i or -i must be positive. Hence 1=2=(-i)2 must be positive. But then 1=(1)2, must also be positive, and this contradicts Part(a of Proposition 1.18, since 1 and -1 cannot both be positive Exercise 1.9 Suppose z =a+bi,w=c+di. Define z w if a< C, and also if a =c but b< d. Prove that this turns the set of all conplex numbers into an ordered set. (This type of order relation is called a dictionary order, or lexicographic order, for obvious reasons. ) Does this ordered set have the least ? upper bound property Solution. We need to show that either z w or =w, or w< 2. Now since the real numbers are ordered, we have a c or a=c, or c <a. In the first case z < w: in the third case 2< 2. Now consider the second case. We must have b<dor b=d or d 6. In the first of these cases x w. in the third case w<z and in the second case z= w We also need to show that if z w and w< u, then z <u. Let u=e+fi Since z w. we have either a< cor a=c and b< d. Since w< u we have either c< f or c=f and d g. Hence there are four possible cases Case 1: a< c and c< f. Then a< f and 5o z a, as required. Case 2: a< c and c=f and d<g. Again a< f, and z < u Case 3: a=c and b< d and c< f. Once again a f and so z< u Case 4: a=c and b< d and c=f, and d g. Then a=f and b< g, and oz<u Exercise 1.10 Suppose z=a-bi, w=u+2v, and 1/ 11-1/2 a 2 Prove that x2=w if v>0 and that(2)2=w if v<0.. Conclude that every complex number (with one exception) has two complex square roots Solution (a+b2 64)+abi. Now + and, since(eg/)1/2 x1/2y/2, ⊥ 1/2 2、1/2 10 CHAPTER 1. THE REAL AND COMPLEX NUMBER SYSTEMS Hence \21/2 2ab=2( Now(c2)1/2=a if c 20 and( 22)1/2=-c if r sO. We conclude that 2ab=v ifv≥0and2b=-i”≤0. Hence2=mv≥0. Replacing b by-b,we find that(2)2=uifv≤0. Hence every non-zero complex number has(at least) two complex square roots Exercise 1.11 If z is a complex number, prove that there exists an r>0 and a complex number w with w= l such that z= rw. Are w and r always uniquely determined by z? Solution. If z-0, we take r=0, w= 1. (In this case w is not unique. Otherwise we take r=z and w=x/z, and these choices are unique, since if = rU, we must have r=r=r=zh, z/r Exercise 1.12 If 21, -., in, are complex, prove that z1+22+…+zn|≤|z1|+|2 Solution. The case n=2 is Part(e of Theorem 1. 33. We can then apply this result and induction on n to get z1+22+…n}=|(xa1+22+…+n-1)+xn ≤|21+22+…+2n-1|+|z 1zn-1+ Exercise 1.13 If t, y are complex, prove that zl-l列lsx-yl Solution. Since a=m-y+y, the triangle inequality gives c≤-y+y1, so that |)-lyl lr-yl. Similarly Iyl-lal <ka-yI. Since | -lyl is a real number we have either ia-Igl=Ial-lyl or|-l3l=lyl-Iz. In either case, we havc shown that xl-13l s l c-yI Exercise I 14 If z is a complex number such that z= 1, that is, such that zZ=l, compute 1+22+1- S0lt0n.1+2|2=(1+2(1+2)=1+2+z+2=2-z+2, Similarly 2=(1-x)(1-2)=1-2-2+x2=2-z-z.Her nce 1+22+1 Exercise 1.15 Under what conditions does equality hold in the Schwarz in equality? Solution. The proof of Theorem 1. 35 shows that equality can hold if B=0 or if Ba;-C6i =0 for all j, i.e., the numbers a; are proportional to the numbers In terms of linear algebra this means the vectors a=(a1, a2,..., an)and 61, 62, .. bn)in complex n-dimensional space are linearly dependent. Co on- versely, if thesc vectors are linearly independent, then strict inequality holds. Exercise 1.16 Suppose k23, x,yE R, x-y=d>0, and r>0. Prove (a)If 2r> d, there are infinitely many z E RK such that If 2r= d, there is cxactly one such z (c If 2r< d, there is no such z How must these statements be modified if k is 2 or 1? Solution.(a) Let w be any vector satisfying the following two equations W·(x-y 4 From linear algebra it is known that all but one of the components of a solution w of the first equation can be arbitrary. The remaining coinponent is then uniquely determined. Also, if w is any non-zero solution of the first equation there is a unique positive number t such that tw satisfies both equations. (For example, if T1+91, the first equation is satisfied whenever 2(2-32)+…+2(2k-y y1 If(21, 22, .. Zk)satisfies this equation, so does(tz1, tu2,., tzk)for any real number t. Since at least two of these components can vary independently, we can find a solution with these components having any prescribed ratio. This 12 CHAPTER 1. THE REAL AND COMPLEX NUMBER SYSTEMS ratio does not change when we multiply by the positive number t to obtain a solution of both equations. Since there are infinitely many ratios, there are infinitely many distinct solutions. For each such soiution w the vector z x+y+w is a solution of the required cquation.For 2 +2w ⅹ-y +0+ 2 and a similar relation holds for z-yI2 (b) The proof of the triangle inequality shows that equality can hold in this inequality only if it holds in the schwarz inequality, i.e., one of the two vectors is a scalar multiple of the other. Further examination of the proof shows that the scalar must be nonnegative. Now the conditions of this part of the problem show that x-y=d=x-z+z-y Hence it follows that there is a nonnegative scalar t such that X- Z t(z-y) However, the hypothesis also shows immediately that t= 1, and so z is uniquely etermined as xTy (c If z were to satisfy this condition, the triangle inequality would be vio- lated. i.e. we would have x-y=d>2r z When k=2, there are precisely 2 solutions in case(a). When k=1, there are no solutions in case(a). The conclusions in cases()and (c) do not require Modification Exercise 1.17 Prove that x+y12+|x-y|2=2|x|2+2 if x E Rf and y e R. interpret this geometrically as a statement about parallelo grame Solution. The proof is a routine computation, using the relation x±y2=(x±y)·(x±y)=|x|2±2x·y+|y 【实例截图】
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