The Real And Complex Number Systems

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f fy c lim n fx n n y n x n . (ii) There is a good exercise; we write it as follows. Let f C 2 a, b, andc a, b. For small |h| such that c h a, b, write fc h fc f c hhh where 0 1. Show that if f c 0, then lim h0 h 1/2. Proof: Since f C 2 a, b, byTaylor Theorem with Remainder Term, we have fc h fc f ch f h 2! 2 ,where c, h or h, c f c hhh by hypothesis. So, f c hh f c h h f 2! and let h 0, we have c by continuity of f at c. Hence, limh 1/2 since f c 0. h0 Note: We can modify our statement as follows. Let f be defined on a, b, and c a, b. For small |h| such that c h a, b, write fc h fc f c hhh where 0 1. Show that if f c 0, and x x for x a h, a h, then lim h0 h 1/2. Proof: Use the exercise (c), we have f fc h 2fc fc h c lim h0 h 2 f lim c hh f c hh by hypothesis h0 h f lim c hh f c hh 2h since x x for x a h, a h. h0 2hh Since f c 0, we finally have lim h0 h 1/2. 5.25 Let f have a finite derivatiive in a, b and assume that c a, b. Consider the following condition: For every 0, there exists a 1 ball Bc; , whose radius depends only on and not on c, such that if x Bc; , andx c, then fx fc x c f c . Show that f is continuous on a, b if this condition holds throughout a, b. Proof: Given 0, we want to find a 0 such that as dx, y , x, y a, b, we have |f x f y| . Choose any point y a, b, and thus by hypothesis, given /2, there is a 1 ball By; , whose radius depends only on and not on y, such that if x By; , and x y, then, fx fy x y f y /2 . * Note that y Bx, , so, we also have ,
fx fy x y f x /2 *’ Combine (*) with (*’), we have |f x f y| . Hence, we have proved f is continuous on a, b. Remark: (i) The open interval can be changed into a closed interval; it just need to consider its endpoints. That is, f is continuous on a, b if this condition holds throughout a, b. The proof is similar, so we omit it. (ii) The converse of statement in the exercise is alos true. We write it as follows. Let f be continuous on a, b, and 0. Prove that there exists a 0 such that fx fc x c f c whenever 0 |x c| , a x, c, b. Proof: Given 0, we want to find a 0 such that fx fc x c f c whenever 0 |x c| , a x, c b. Sincef is continuous on a, b, we know that f is uniformly continuous on a, b. That is, given 0, there is a 0 such that as dx, y , wehave |f x f y| . * Consider dx, c , x a, b, thenby(*),wehave fx fc x c f c |f x f c| by Mean Value Theorem where dx , x . So, we complete it. Note: This could be expressed by saying that f is uniformly differentiable on a, b if f is continuous on a, b. 5.26 Assume f has a finite derivative in a, b and is continuous on a, b, with a fx b for all x in a, b and |f x| 1 for all x in a, b. Prove that f has a unique fixed point in a, b. Proof: Given any x, y a, b, thus, by Mean Value Theorem, wehave |fx fy| |f z||x y| |x y| by hypothesis. So, we know that f is a contraction on a complete metric space a, b. So,f has a unique fixed point in a, b. 5.27 Give an example of a pair of functions f and g having a finite derivatives in 0, 1, such that but such that lim x0 f x g x lim x0 fx gx 0, does not exist, choosing g so that g x is never zero. Proof: Letfx sin1/x and gx 1/x. Then it is trivial for that g x is never zero. In addition, we have fx f lim 0, and lim x does not exist. x0 gx x0 g x
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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
Page 115 and 116: from b to A) are disjoint. So, if e
Page 117 and 118: x F k1 F k A B U V which im
Page 119 and 120: Proof: Assume that B C 1 is discon
Page 121 and 122: Proof: By Mean Value Theorem, wehav
Page 123 and 124: x r y r rz r1 x y ry r1 /2. So,
Page 125 and 126: hx gfx if x S. Iff is uniformly c
Page 127 and 128: dx, y , x, y A, wehave dfx, fy
Page 129 and 130: In addition, part (b) comes from in
Page 131 and 132: continuous at a, a. (2) (x y) Sinc
Page 133 and 134: Proof: Let D denote the set of dico
Page 135 and 136: (a) Provet that the formula df, g
Page 137 and 138: so, by induction, we find dp n1 , p
Page 139 and 140: Proof: If p and p are fixed points
Page 141 and 142: (b) Assume there is a subsequence p
Page 143 and 144: Hence, f is increasing on , a/3 a
Page 145 and 146: f k x f k 0 x 0 fk x x by induct
Page 147 and 148: h k1 h k k jk f j g kj x j0
Page 149 and 150: g x g x 3 2 a d f x a d f x 3 2
Page 151 and 152: which implies that F 0, where
Page 153 and 154: 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
Page 161 and 162: there is a point p such that fp 0.
Page 163 and 164: Remark: If we can make sure that fx
Page 165: which implies that Hence, g 1 h g
Page 169 and 170: lim xa fx gx L. Remark: 1. The pr
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Page 173 and 174: f x 1 ...x n n f x 1 ...x n1 n x
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:
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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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Suppose NOT, it means that lim n→
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then since n lim n→ ∑ k1 n S n
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|sin k|r ∑ k So, ∑ |sink|r dive
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n For the part 1 2 0 that x sint
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lim n→ sin ... sin n 1 ... 1 n
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Also, lim q→ fp, q lim q→ p si
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n c n ∑ a k b n−k k0 n ∑ k0
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So, n s n ∑ cos2k − 1x j1 sin
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n u n ∑−1 k a k k1 n ∑−1
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which implies that which implies th
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So, by above sayings, we have prove
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(a) If fn is multiplicative and if
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Sequences of Functions Uniform conv
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(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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The Real And Complex Number Systems

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