The Real And Complex Number Systems

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Remark: In order to show the series ∑ sin 1 n diverges, we consider Cauchy Criterion as follows. n sin 1 ≤ sin 1 ... sin 1 2n n 1 n n and given x ∈ R, forn 0,1,2,..., we have |sin nx| ≤ n|sin x|. So, sin 1 2 ≤ sin 1 ... sin 1 n 1 n n for all n. Hence, ∑ sin 1 n diverges. Note: There are many methods to show the divergence of the series ∑ sin 1 n . We can use Cauchy Condensation Theorem to prove it. Besides, by (11), it also owrks. (9) O-Stolz’s Theorem. n 1 Proof: LetS n ∑ j1 j and X n log n. Then by O-Stolz’s Theorem, it is easy to see lim n→ S n . (10) Since n k1 1 1 k diverges, the series ∑ 1/k diverges by Theorem 8.52. (11) Lemma: If a n is a decreasing sequence and ∑ a n converges. Then lim n→ na n 0. Proof: Sincea n → 0anda n is a decreasing sequence, we conclude that a n ≥ 0. Since ∑ a n converges, given 0, there exists a positive integer N such that as n ≥ N, we have a n ..a nk /2 for all k ∈ N which implies that k 1a nk /2 since a n ↘. Let k n, then as n ≥ N, wehave n 1a 2n /2 which implies that as n ≥ N 2n 1a 2n which implies that lim n→ 2na 2n 0 since lim n→ a n 0. * Similarly, we can show that lim n→ 2n 1a 2n1 0. ** So,by(*)adn(**),wehaveprovedthatlim n→ na n 0. Remark: From this, it is clear that ∑ 1 n diverges. In addition, we have the convergence of ∑ na n − a n1 . We give it a proof as follows. Proof: Write
then since n lim n→ ∑ k1 n S n ∑ ka k − a k1 k1 n ∑ k1 a k − na n1 , lim n→ S n exists a k exists and lim n→ na n 0. B Prove that ∑ 1 p diverges, where p is a prime. Proof: GivenN, letp 1 ,...,p k be the primes that divide at least one integer≤ N. Then N ∑ n1 k 1 n ≤ j1 k j1 ≤ exp 1 1 p j 1 p j 2 ... 1 1 − p 1 j by 1 − x −1 ≤ e 2x if 0 ≤ x ≤ 1/2. Hence, ∑ 1 p diverges since ∑ 1 n diverges. Remark: There are many proofs about it. The reader can see the book, An Introduction To The Theory Of Numbers by Loo-Keng Hua, pp 91-93. (Chinese Version) C Discuss some series related with ∑ sink . k STUDY: (1) We have shown that the series ∑ sin 1 diverges. k (2) The series ∑ sinna b diverges where a ≠ n for all n ∈ Z and b ∈ R. Proof: Suppose that ∑ sinna b converges, then lim n→ sinna b 0. Hence, lim n→|sinn 1a b − sinna b| 0. Consider |sinn 1a b − sinna b| k ∑ j1 2 p j 2cos na b a 2 sin a 2 which implies that 2 cosna b cos a 2 − sinna b sin a 2 sin a 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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every g k is never zero in c , c
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f x 1 ...x n n f x 1 ...x n1 n x
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f x 1 ...x n n fx 1 ...fx n n . C
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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
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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!
Page 221 and 222: n S n ∑−1 k1 1 3k − 2 − 1
Page 223 and 224: (b) Prove that ∑ n1 −1 n−1 /
Page 225 and 226: then b k sin kx, a k1 − a k 1
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Page 229: Suppose NOT, it means that lim n→
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
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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
Page 269 and 270: which implies that ∣ lim sup 1 k
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Page 275 and 276: 0.1 Supplement on some results on W
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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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