The Real And Complex Number Systems

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a − a n . So, as n ≥ maxN 1 , N 2 ,wehave a − a n a . That is, lim n→ a n a. (c) The sequence diverges to if and only if, lim inf n→ a n lim sup n→ a n . Proof: ()Given a sequence a n with lim n→ a n . So, given M 0, there is a positive integer N such that as n ≥ N, wehave M ≤ a n . * It implies that a n is not bounded above. So, lim sup n→ a n . In order to show that lim inf n→ a n . We first note that a n is bounded below. Hence, lim inf n→ a n ≠−. So, it suffices to consider that lim inf n→ a n is not finite. (So, we have lim inf n→ a n . ). Assume that lim inf n→ a n a, wherea is finite. Then given 1, and an integer m, there exists a positive Km m such that a Km a 1 which contradicts to (*) if we choose M a 1. So, lim inf n→ a n is not finite. (d) The sequence diverges to − if and only if, lim inf n→ a n lim sup n→ a n −. Proof: Note that, lim sup n→ −a n − lim inf n→ a n . So, by (c), we have proved it. (Theorem 8.4)Assume that a n ≤ b n for each n 1,2,.... Then we have: lim n→ inf a n ≤ lim n→ inf b n and lim n→ sup a n ≤ lim n→ sup b n . Proof: If lim inf n→ b n , there is nothing to prove it. So, we may assume that lim inf n→ b n . That is, lim inf n→ b n − or b, whereb is finite. For the case, lim inf n→ b n −, it means that the sequence a n is not bounded below. So, b n is also not bounded below. Hence, we also have lim inf n→ a n −. For the case, lim inf n→ b n b, whereb is finite. We consider three cases as follows. (i) if lim inf n→ a n −, then there is nothing to prove it. (ii) if lim inf n→ a n a, wherea is finite. Given 0, then there exists a positive integer N such that as n ≥ N a − /2 a n ≤ b n b /2 which implies that a ≤ b since is arbitrary. (iii) if lim inf n→ a n , then by Theorem 8.3 (a) and (c), we know that lim n→ a n which implies that lim n→ b n . Also, by Theorem 8.3 (c), wehave lim inf n→ b n which is absurb. So, by above results, we have proved that lim inf n→ a n ≤ lim inf n→ b n . Similarly, we have lim sup n→ a n ≤ lim sup n→ b n . 8.4 If each a n 0, prove that lim n→ inf a n1 a n ≤ lim n→ infa n 1/n ≤ lim n→ supa n 1/n ≤ lim n→ sup a n1 a n . Proof: ByTheorem 8.3 (a), it suffices to show that lim n→ inf a n1 a n We first prove ≤ lim n→ infa n 1/n and lim n→ supa n 1/n ≤ lim n→ sup a n1 a n . lim n→ supa n 1/n ≤ lim n→ sup a n1 a n .
If lim sup n→ a n1 a n , then it is clear. In addition, since a n1 a n is positive, a ≠−. So, we may assume that lim sup n1 n→ a n a, wherea is finite. Given 0, then there exists a positive integer N such that as n ≥ N, wehave a n1 a n a lim sup n→ a n1 a n which implies that So, which implies that a Nk a N a k ,wherek 1,2,.... a Nk 1 Nk a N 1 Nk a k Nk lim supa Nk 1 Nk k→ ≤ lim supa N 1 k Nk a Nk k→ a . So, lim supa Nk 1 Nk ≤ a k→ since is arbitrary. Note that the finitely many terms do NOT change the value of limit superiror of a given sequence. So, we finally have lim n→ supa n 1/n ≤ a lim n→ sup a n1 a n . Similarly for lim n→ inf a n1 a n ≤ lim n→ infa n 1/n . Remark: These ineqaulities is much important; we suggest that the reader keep it mind. At the same time, these inequalities tells us that the root test is more powerful than the ratio test. We give an example to say this point. Given a series 1 2 1 3 1 2 1 2 3 ... 1 2 2 n 3 1 n ... where n n a 2n−1 1 ,anda2n 1 , n 1, 2, . . . 2 3 with and lim n→ inf a n1 a n lim n→ supa n 1/n 1 2 1 0, lim n→ sup a n1 a n . 8.5 Let a n n n /n!. Show that lim n→ a n1 /a n e and use Exercise 8.4 to deduce that lim n n→ e. n! 1/n Proof: Since a n1 a n n 1n1 n! n 1!n n 1 1 n n → e, by Exercise 8.4, wehave lim n→ a n 1/n lim n n→ e. n! 1/n
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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
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Proof: If p and p are fixed points
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(b) Assume there is a subsequence p
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Hence, f is increasing on , a/3 a
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f k x f k 0 x 0 fk x x by induct
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h k1 h k k jk f j g kj x j0
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g x g x 3 2 a d f x a d f x 3 2
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which implies that F 0, where
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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 183 and 184: ux, y (1) arctany/x, ifx 0, y R
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 and 190: Claim that 1 sup A. Suppose NOT, t
Page 191 and 192: we know that V f is an increasing
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: which implies that c ≤ a b since
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→
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Page 233 and 234: |sin k|r ∑ k So, ∑ |sink|r dive
Page 235 and 236: n For the part 1 2 0 that x sint
Page 237 and 238: lim n→ sin ... sin n 1 ... 1 n
Page 239 and 240: Also, lim q→ fp, q lim q→ p si
Page 241 and 242: n c n ∑ a k b n−k k0 n ∑ k0
Page 243 and 244: So, n s n ∑ cos2k − 1x j1 sin
Page 245 and 246: n u n ∑−1 k a k k1 n ∑−1
Page 247 and 248: which implies that which implies th
Page 249 and 250: So, by above sayings, we have prove
Page 251 and 252: (a) If fn is multiplicative and if
Page 253 and 254: Sequences of Functions Uniform conv
Page 255 and 256: (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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