The Real And Complex Number Systems

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V fg , AV f , BV g , . (Theorem 6.10*) Letf be of bounded variation on R, and assume that f is bounded away from zero; that is, suppose that there exists a positive number m such that 0 m |fx| for all x R. Then g 1/f is also of bounded variation on R, and V g , V f, m 2 . Proof: Given any compacgt interval a, b, wehave which implies that V g a, b V fa, b m 2 V f, m 2 V g , V f, . m 2 (Theorem 6.11*) Letf be of bounded variation on R, and assume that c R. Then f is of bounded variation on , c and on c, and we have V f , V f , c V f c, . Proof: Given any a compact interval a, b such that c a, b. Then we have V f a, b V f a, c V f c, b. Since V f a, b V f , which implies that V f a, c V f , and V f c, b V f , , we know that the existence of V f , c and V f c, . Thatis,f is of bounded variation on , c and on c, . Since V f a, c V f c, b V f a, b V f , which implies that V f , c V f c, V f , , * and V f a, b V f a, c V f c, b V f , c V f c, which implies that V f , V f , c V f c, , ** we know that V f , V f , c V f c, . (Theorem 6.12*) Letf be of bounded variation on R. LetVx be defined on , x as follows: Vx V f , x if x R, andV 0. Then (i) V is an increasing function on , and (ii) V f is an increasing function on , . Proof: (i)Letx y, then we have Vy Vx V f x, y 0. So, we know that V is an increasing function on , . (ii) Let x y, then we have V fy V fx V f x, y fy fx 0. So,
we know that V f is an increasing function on , . (b) Show that Theorem 6.5 is true for , if ”monotonic” is replaced by ”bounded and monotonic.” State and prove a similar modefication of Theorem 6.13. (Theorem 6.5*) Iff is bounded and monotonic on , , then f is of bounded variation on , . Proof: Given any compact interval a, b, then we have V f a, b exists, and we have V f a, b |fb fa|, sincef is monotonic. In addition, since f is bounded on R, say |fx| M for all x, we know that 2M is a upper bounded of V f a, b for all a, b. Hence, V f , exists. That is, f is of bounded variation on R. (Theorem 6.13*) Letf be defined on , , then f is of bounded variation on , if, and only if, f can be expressed as the difference of two increasing and bounded functions. Proof: Suppose that f is of bounded variation on , , then by Theorem 6.12*, we know that f V V f, where V and V f are increasing on , . In addition, since f is of bounded variation on R, we know that V and f is bounded on R which implies that V f is bounded on R. So, we have proved that if f is of bounded variation on , then f can be expressed as the difference of two increasing and bounded functions. Suppose that f can be expressed as the difference of two increasing and bounded functions, say f f 1 f 2 , Then by Theorem 6.9*, andTheorem 6.5*, we know that f is of bounded variaton on R. Remark: The representation of a function of bounded variation as a difference of two increasing and bounded functions is by no mean unique. It is clear that Theorem 6.13* also holds if ”increasing” is replaced by ”strictly increasing.” For example, f f 1 g f 2 g, whereg is any strictly increasing and bounded function on R. One of such g is arctan x. 6.7 Assume that f is of bounded variation on a, b and let P x 0 , x 1 ,...,x n þa, b. As usual, write f k fx k fx k1 , k 1,2,...,n. Define AP k : f k 0, BP k : f k 0. The numbers and p f a, b sup f k : P þa, b kAP n f a, b sup |f k | : P þa, b kBP are called respectively, the positive and negative variations of f on a, b. For each x in a, b. LetVx V f a, x, px p f a, x, nx n f a, x, and let Va pa na 0. Show that we have: (a) Vx px nx.
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The Real And Complex Number Systems
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Proof: Consider 2n−2 ∑ (x + 1)
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Then S = N. Proof: (A ⇒ B): If S
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Note: There are many and many metho
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In addition, (√ a ) 2 − − b (
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Proof: Say {ar + b : a ∈ Z, b ∈
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(⇐)It is clear since every a n is
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Proof: Choose a 0 = [x], and thus c
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Proof: Use Cauchy-Schwarz inequalit
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In addition, ∑ a k b k + a j b j
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So, z 1 z 2 = ¯z 1¯z 2 (c) z¯z =
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so, |a − b| 2 ≤ ( 1 + |a| 2) (
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Proof: Note that in this text book,
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1.39 State and prove a theorem anal
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1.43 (a) Prove that Log (z w ) = wL
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For complex θ, we show that it hol
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So, ix−y = i (x + iy) = iz = log
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and the sum of their squares is giv
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Some Basic Notations Of Set Theory
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(i) If x ∈ R, it is clear that (x
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(a) A ∪ (B ∪ C) = (A ∪ B) ∪
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(e) A ∩ (B − C) = (A ∩ B) −
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Proof: Given x ∈ X, then f (x)
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By hyppothesis, we get T ⊆ f (S)
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Proof: Since S is an infinite set,
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Proof: Assume S˜R and let f be a o
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2.22 Let S denote the collection of
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Charpter 3 Elements of Point set To
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And thus by the remark in (e), it i
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3.7 Prove that a nonempty, bounded
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Bx, r x S Bx, r x T . So, at
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Claim that Bp, r S as follows. Let
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Covering theorems in R n 3.17 If S
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uncountable. Hence, by exercise 3.2
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Then we have (a) (b) (c) (d) x y
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(b) Give an example of a metric spa
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Hence from (1)-(4), we know that i
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3.44 intM A M A . Proof: Let B
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That is, intB which is absurb. He
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Limits And Continuity Limits of seq
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x n1 x n 1 1 x n x n 1 2 . Rema
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|a m a n| |a m a m1 a m1 a m2
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and use the fact if a n converges
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lim y0 fx, y 0ifx 0, 1ifx 0. and
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fx, y 1 r cos sinr2 cossin if x
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then we have lim fx, y L if x 0an
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this is because a can be an isolate
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irrational. Prove that: (a) ffx x
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In addition, since f1 f1 by f0 0,
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lim r n0 fx f lim n x n lim n fx
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if x Q, andgx 0ifx Q c . Remark:
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number f T supfx fy : x, y T is
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Proof: Since the hypothesis says th
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(f) S 0, 1 0, 1, T 0, 1 0, 1. S
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In Exercises 4.29 through 4.33, we
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function. Since f0, 1 0, 1 0, 1,
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4.40 If x is a point in a metric sp
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from b to A) are disjoint. So, if e
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x F k1 F k A B U V which im
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Proof: Assume that B C 1 is discon
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Proof: By Mean Value Theorem, wehav
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x r y r rz r1 x y ry r1 /2. So,
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hx gfx if x S. Iff is uniformly c
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dx, y , x, y A, wehave dfx, fy
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In addition, part (b) comes from in
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continuous at a, a. (2) (x y) Sinc
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Proof: Let D denote the set of dico
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(a) Provet that the formula df, g
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so, by induction, we find dp n1 , p
Page 139 and 140: Proof: If p and p are fixed points
Page 141 and 142: (b) Assume there is a subsequence p
Page 143 and 144: Hence, f is increasing on , a/3 a
Page 145 and 146: f k x f k 0 x 0 fk x x by induct
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Page 149 and 150: g x g x 3 2 a d f x a d f x 3 2
Page 151 and 152: which implies that F 0, where
Page 153 and 154: g fx gfx. Assume that there is a
Page 155 and 156: Remark: For x 0, we can show that
Page 157 and 158: Proof: Look at the Generalized Mean
Page 159 and 160: Proof: By Generalized Mean Value Th
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Page 163 and 164: Remark: If we can make sure that fx
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Page 167 and 168: fx fy x y f x /2 *’ Combi
Page 169 and 170: lim xa fx gx L. Remark: 1. The pr
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Page 175 and 176: f x 1 ...x n n fx 1 ...fx n n . C
Page 177 and 178: lx fc f cx c fx. (Exercise 2)
Page 179 and 180: So, we have Partial derivatives fb
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Page 185 and 186: Functions of Bounded Variation and
Page 187 and 188: For the part fx x fy 1 |fx fy| 1
Page 189: Claim that 1 sup A. Suppose NOT, t
Page 193 and 194: in Section 6.12. Proof: Sinceft : t
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Page 197 and 198: n k1 n ||fb k | |fa k || |fb k
Page 199 and 200: Supplement on lim sup and lim inf I
Page 201 and 202: (1) lim n→ inf a n − a n has
Page 203 and 204: Remark: The exercise is useful in t
Page 205 and 206: which implies that c ≤ a b since
Page 207 and 208: If lim sup n→ a n1 a n , then it
Page 209 and 210: We first prove lim n→ sup n ≤
Page 211 and 212: 0 ≤ d 1 * by Theorem 8.23 (i). N
Page 213 and 214: Consider which implies that which i
Page 215 and 216: So, L 1 5 since L ≥ 1. 2 Remark:
Page 217 and 218: For case (i), since 1 n log nlog lo
Page 219 and 220: ()Assume that ∑ n1 That is, p!
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Page 223 and 224: (b) Prove that ∑ n1 −1 n−1 /
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Page 229 and 230: Suppose NOT, it means that lim n→
Page 231 and 232: then since n lim n→ ∑ k1 n S n
Page 233 and 234: |sin k|r ∑ k So, ∑ |sink|r dive
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Page 237 and 238: lim n→ sin ... sin n 1 ... 1 n
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n c n ∑ a k b n−k k0 n ∑ k0
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So, n s n ∑ cos2k − 1x j1 sin
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n u n ∑−1 k a k k1 n ∑−1
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which implies that which implies th
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So, by above sayings, we have prove
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(a) If fn is multiplicative and if
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Sequences of Functions Uniform conv
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(b) Prove that h n (x) does not con
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Proof: Since g is continuous on a c
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Hence, {g (x) x n } converges unifo
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y continuity of f k(x0 ) (x) − f
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In addition, as c ≥ 1/2, if f n
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(Lemma) If {a n } and {b n } are tw
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9.16 Let {f n } be a sequence of re
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which implies that ∣ lim sup 1 k
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have ∫ x 0 ∞∑ t 2n − 1 2 n=
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[ ] π And as x ∈ , π , then n+p
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0.1 Supplement on some results on W
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which implies that h (x + t) − h
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we have ∫ d c |f n − g| 2 dx
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we complete it. 9.31 Given that two
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if x = λ (≠ 0) is a root of 1 +
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So, the series diverges. (ii) As
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9.38 For each real t, define f t (x
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So, B 0 = P 0 (0) = C 0 = 1, B 1 =
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Remark: (1) The reader can see the
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That is, lim n inf a n sup c n . n
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and lim n infa n b n lim n a n
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q! kq1 1 k! kq1 q! k! 1 q 1
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g x log 1 1 x x a x 2 x log
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The Real And Complex Number Systems

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