The Real And Complex Number Systems

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(b) Prove that {f n } converges pointwise but not uniformly on [0, π/2] . Proof: Note that each f n (x) is continuous on [0, π/2] , and the limit function { 0 if x ∈ (0, π/2] f = . 1 if x = 0 Hence, by <strong>The</strong>orem9.2, we know that {f n } converges pointwise but not uniformly on [0, π/2] . 9.29 Let f n (x) = 0 if 0 ≤ x ≤ 1/n or 2/n ≤ x ≤ 1, and let f n (x) = n if 1/n < x < 2/n. Prove that {f n } converges pointwise to 0 on [0, 1] but that l.i.m. n→∞ f n ≠ 0 on [0, 1] . Proof: It is clear that {f n } converges pointwise to 0 on [0, 1] . In order to show that l.i.m. n→∞ f n ≠ 0 on [0, 1] , it suffices to note that ∫ 1 Hence, l.i.m. n→∞ f n ≠ 0 on [0, 1] . Power series 0 f n (x) dx = 1 for all n. 9.30 If r is the radius of convergence if ∑ a n (z − z 0 ) n , where each a n ≠ 0, show that ∣ ∣ ∣ ∣∣∣ lim inf a n ∣∣∣ ≤ r ≤ lim sup a n ∣∣∣ n→∞ a n+1 n→∞ ∣ . a n+1 Proof: By Exercise 8.4, we have 1 ∣ lim n→∞ sup ∣ a n+1 ∣∣ ≤ r = a n 1 lim n→∞ sup |a n | 1 n ≤ 1 lim n→∞ inf ∣ ∣ a n+1 ∣∣ . a n Since and 1 ∣ lim n→∞ sup 1 lim n→∞ inf ∣ ∣ a n+1 a n ∣ a n+1 a n ∣ ∣∣ = lim n→∞ inf ∣ ∣∣ = lim n→∞ sup ∣ a n ∣∣∣ ∣a n+1 ∣ a n ∣∣∣ ∣ , a n+1 28
we complete it. 9.31 Given that two power series ∑ a n z n has radius of convergence 2. Find the radius convergence of each of the following series: In (a) and (b), k is a fixed positive integer. (a) ∑ ∞ n=0 ak nz n Proof: Since 2 = we know that the radius of ∑ ∞ n=0 ak nz n is 1 , (*) 1/n lim n→∞ sup |a n | 1 lim n→∞ sup |a k n| 1/n = 1 (lim n→∞ sup |a n | 1/n) k = 2k . (b) ∑ ∞ n=0 a nz kn Proof: Consider which implies that lim sup ∣ an z kn∣ ∣ 1/n = lim sup |a n | 1/n |z| k < 1 n→∞ n→∞ ( ) 1/k 1 |z| < = 2 1/k by (*). lim n→∞ sup |a n | 1/n So, the radius of ∑ ∞ n=0 a nz kn is 2 1/k . (c) ∑ ∞ n=0 a nz n2 Proof: Consider ∣ ∣ ∣∣ 1/n lim sup ∣a n z n2 = lim sup |a n | 1/n |z| n n→∞ and claim that the radius of ∑ ∞ n=0 a nz n2 is 1 as follows. If |z| < 1, it is clearly seen that the series converges. However, if |z| > 1, lim sup |a n| 1/n lim inf |z| n ≤ lim sup |a n | 1/n |z| n n→∞ n→∞ n→∞ 29
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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)
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So, we have Partial derivatives fb
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ux, y ey e y cosx 2 and vx, y e
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ux, y (1) arctany/x, ifx 0, y R
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Functions of Bounded Variation and
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For the part fx x fy 1 |fx fy| 1
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Claim that 1 sup A. Suppose NOT, t
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we know that V f is an increasing
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in Section 6.12. Proof: Sinceft : t
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2n h P 2 hx i hx i1 i1 n 2n
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n k1 n ||fb k | |fa k || |fb k
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Supplement on lim sup and lim inf I
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(1) lim n→ inf a n − a n has
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Remark: The exercise is useful in t
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which implies that c ≤ a b since
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If lim sup n→ a n1 a n , then it
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We first prove lim n→ sup n ≤
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0 ≤ d 1 * by Theorem 8.23 (i). N
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Consider which implies that which i
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So, L 1 5 since L ≥ 1. 2 Remark:
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For case (i), since 1 n log nlog lo
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()Assume that ∑ n1 That is, p!
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n S n ∑−1 k1 1 3k − 2 − 1
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(b) Prove that ∑ n1 −1 n−1 /
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then b k sin kx, a k1 − a k 1
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diverges, then ∑ where By (2) and
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Page 267 and 268: 9.16 Let {f n } be a sequence of re
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Page 275 and 276: 0.1 Supplement on some results on W
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Page 289 and 290: So, B 0 = P 0 (0) = C 0 = 1, B 1 =
Page 291 and 292: Remark: (1) The reader can see the
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Page 297 and 298: q! kq1 1 k! kq1 q! k! 1 q 1
Page 299: 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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