The Real And Complex Number Systems

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n ∑ k3 log k k which implies that 3 n log x x dx C O log n n 1 2 log2 n − 1 2 log2 3 C O n ∑ k1 log k k ,whereC is a constant log n n 1 2 log2 n A O ,whereC is a constant log n n , where A C log2 − 1 2 2 log2 3 is a constant. n 1 1 (b) ∑ k2 loglog n B O (B is constant) klogk n logn 1 Proof: Letfx defined on 2, , then 2 xlogx f′ 1 x − xlogx 1 log x 0on 2, . So, it is clear that fx is a positive and continuous function on 3, , with lim x→ fx lim 1 x→ x log x 0. So, by Theorem 8.23, wehave n ∑ k2 1 k log k n 2 dx x log x C O 1 n log n log log n B O 1 n log n where B C − log log 2 is a constant. 8.21 If 0 a ≤ 1, s 1, define s, a ∑ n0 n a −s . ,whereC is a constant ,whereC is a constant (a) Show that this series converges absolutely for s 1 and prove that k ∑ s, h k k s s if k 1, 2, . . . h1 where s s,1 is the Riemann zeta function. Proof: First, it is clear that s, a converges absolutely for s 1. Consider k ∑ s, h k h1 k ∑ h1 k ∑ h1 ∑ n0 ∑ n0 k ∑ ∑ n0 h1 k s ∑ n0 k s ∑ n0 k s s. k ∑ h1 1 n h k s k s kn h s k s kn h s 1 kn h s 1 n 1 s
(b) Prove that ∑ n1 −1 n−1 /n s 1 − 2 1−s s if s 1. Proof: Let n S n ∑ j1 −1 j−1 j s , and thus consider its subsequence S 2n as follows: 2n S 2n ∑ j1 2n ∑ j1 1 j s 1 j s n − 2 ∑ j1 n − 2 1−s ∑ j1 1 2j s which implies that lim n→ S 2n 1 − 2 1−s s. Since S n converges, we know that S 2n also converges and has the same value. Hence, ∑ n1 1 j s −1 n−1 /n s 1 − 2 1−s s. 8.22 Given a convergent series ∑ a n , where each a n ≥ 0. Prove that ∑ a n n −p converges if p 1/2. Give a counterexample for p 1/2. Proof: Since a n n −2p 2 ≥ a n n −2p a n n −p , we have ∑ a n n −p converges if p 1/2 since ∑ a n converges and ∑ n −2p converges if p 1/2. and For p 1/2, we consider a n 1 nlogn 2 , then ∑ a n converges by Cauchy Condensation Theorem ∑ a n n −1/2 ∑ 1 n log n diverges by Cauchy Condensation Theorem. 8.23 Given that ∑ a n diverges. Prove that ∑ na n also diverges. n Proof: Assume ∑ na n converges, then its partial sum ∑ k1 ka k is bounded. Then by Dirichlet Test, we would obtain ∑ka k 1 k ∑ a k converges which contradicts to ∑ a n diverges. Hence, ∑ na n diverges. 8.24 Given that ∑ a n converges, where each a n 0. Prove that ∑a n a n1 1/2 also converges. Show that the converse is also true if a n is monotonic. Proof: Since a n a n1 ≥ a 2 n a n1 1/2 , we know that ∑a n a n1 1/2
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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
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Remark: For x 0, we can show that
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Proof: Look at the Generalized Mean
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Proof: By Generalized Mean Value Th
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there is a point p such that fp 0.
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Remark: If we can make sure that fx
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which implies that Hence, g 1 h g
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fx fy x y f x /2 *’ Combi
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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 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
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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 225 and 226: then b k sin kx, a k1 − a k 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
Page 257 and 258: Proof: Since g is continuous on a c
Page 259 and 260: Hence, {g (x) x n } converges unifo
Page 261 and 262: y continuity of f k(x0 ) (x) − f
Page 263 and 264: In addition, as c ≥ 1/2, if f n
Page 265 and 266: (Lemma) If {a n } and {b n } are tw
Page 267 and 268: 9.16 Let {f n } be a sequence of re
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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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