The Real And Complex Number Systems

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(l) ∑ n2 n p 1 n−1 − 1 n Proof: Note that n p 1 n − 1 − 1 n 1 n 3 2 −p n n − 1 1 1 n−1 n So, as p 1/2, the series converges and as p ≥ 1/2, the series diverges by Limit Comparison Test. (m) ∑ n1 n 1/n − 1 n Proof: With help of Root Test, lim n→ sup n 1/n − 1 n 1/n 0 1, the series converges. (n) ∑ n1 n p n 1 − 2 n n − 1 Proof: Note that n p n 1 − 2 n n − 1 1 n 3 2 n 3 2 −p n n 1 n n − 1 n − 1 n 1 . So, as p 1/2, the series converges and as p ≥ 1/2, the series diverges by Limit Comparison Test. 8.16 Let S n 1 , n 2 ,... denote the collection of those positive integers that do not involve the digit 0 is their decimal representation. (For example, 7 ∈ S but 101 ∉ S.) Show that ∑ k1 1/n k converges and has a sum less than 90. Proof: DefineS j the j − digit number ⊆ S. Then#S j 9 j and S j1 S j . Note that ∑ 1/n k 9 j 10 . j−1 k∈S j So, ∑ k1 1/n k ≤ ∑ j1 9 j 10 j−1 90. In addition, it is easy to know that ∑ k1 1/n k ≠ 90. Hence, we have proved that ∑ k1 1/n k converges and has a sum less than 90. 8.17 Given integers a 1 , a 2 ,...such that 1 ≤ a n ≤ n − 1, n 2, 3, . . . Show that the sum of the series ∑ n1 a n /n! is rational if and only if there exists an integer N such that a n n − 1 for all n ≥ N. Hint: For sufficency, show that ∑ n2 n − 1/n! is a telescoping series with sum 1. Proof: ()Assume that there exists an integer N such that a n n − 1 for all n ≥ N. Then .
()Assume that ∑ n1 That is, p! ∑ np1 a n p! ∑ np1 ∑ n1 an n! N−1 ∑ n1 N−1 ∑ n1 N−1 ∑ n1 an n! ∑ nN an n! an n! ∑ n − 1 n! nN an n! ∑ nN 1 n − 1! − 1 n! N−1 ∑ an n! 1 N − 1! ∈ Q. n1 a n /n! is rational, say q p ,whereg. c. d. p, q 1. Then n! ∈ Z. Note that a n n! ≤ p! ∑ np1 p! ∑ n1 n − 1 n! an n! ∈ Z. p! p! 1since1 ≤ a n ≤ n − 1. So, a n n − 1 for all n ≥ p 1. That is, there exists an integer N such that a n n − 1for all n ≥ N. Remark: From this, we have proved that e is irrational. The reader should be noted that we can use Theorem 8.16 to show that e is irrational by considering e −1 . Since it is easy, we omit the proof. 8.18 Let p and q be fixed integers, p ≥ q ≥ 1, and let pn x n ∑ kqn1 n 1 k , s n ∑ k1 −1 k1 k (a) Use formula (8) to prove that lim n→ x n logp/q. Proof: Since we know that n ∑ k1 1 k log n r O 1 n , pn x n ∑ k1 qn 1 − k ∑ k1 logp/q O 1 n which implies that lim n→ x n logp/q. (b) When q 1, p 2, show that s 2n x n and deduce that ∑ n1 1 k −1 n1 n log 2. Proof: We prove it by Mathematical Induction as follows. As n 1, it holds trivially. Assume that n m holds, i.e., .
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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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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
Page 181 and 182: ux, y ey e y cosx 2 and vx, y e
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 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: For case (i), since 1 n log nlog lo
Page 221 and 222: n S n ∑−1 k1 1 3k − 2 − 1
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
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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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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