The Real And Complex Number Systems

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W x f 1 f 2 ... f n ... ... ... ... f 1 n2 f 1 n f 2 n2 f 2 n ... f n n2 ... f n n . (b) Assuming the existence of n constants c 1 ,...,c n , not all zero, such that c 1 f 1 x ...c n f n x 0 for every x in a, b, show that Wx 0 for each x in a, b. Proof: Since c 1 f 1 x ...c n f n x 0 for every x in a, b, wherec 1 ,...,c n , not all zero. Without loss of generality, we may assume c 1 0, we know that c 1 f k 1 x ...c n f k n x 0 for every x in a, b, where 0 k n. Hence, we have Wx f 1 f 2 ... f n f 1 f 2 ... f n ... ... ... ... f 1 n1 f 2 n1 ... f n n1 0 since the first column is a linear combination of other columns. Note. A set of functions satisfing such a relation is said to be a linearly dependent set on a, b. (c) The vanishing of the Wronskian throughout a, b is necessary, but not sufficient. for linear dependence of f 1 ,...,f n . Show that in the case of two functions, if the Wronskian vanishes throughout a, b and if one of the functions does not vanish in a, b, then they form a linearly dependent set in a, b. Proof: Let f and g be continuous and differentiable on a, b. Suppose that fx 0for all x a, b. Since the Wronskian of f and g is 0, for all x a, b, wehave fg f g 0 for all x a, b. * Since fx 0 for all x a, b, wehaveby(*), fg f g 0 g 0 for all x a, b. f 2 f Hence, there is a constant c such that g cf for all x a, b. Hence, f, g forms a linearly dependent set. Remark: This exercise in (b) is a impotant theorem on O.D.E. We often write (b) in other form as follows. (Theorem) Letf 1 ,..,f n be continuous and differentiable on an interval I. If Wf 1 ,...,f n t 0 0forsomet 0 I, then f 1 ,..,f n is linearly independent on I Note: If f 1 ,..,f n is linearly independent on I, ItisNOT necessary that Wf 1 ,...,f n t 0 0forsomet 0 I. For example, ft t 2 |t|, andgt t 3 . It is easy to check f, g is linearly independent on 1, 1. AndWf, gt 0 for all t 1, 1. Supplement on Chain Rule and Inverse Function Theorem. The following theorem is called chain rule, it is well-known that let f be defined on an open interval S, letg be defined on fS, and consider the composite function g f defined on S by the equation
g fx gfx. Assume that there is a point c in S such that fc is an interior point of fS. Iff is differentiable at c and g is differentiable at fc, then g f is differentiable at c, andwe have g f c g fcf c. We do not give a proof, in fact, the proof can be found in this text book. We will give another Theorem called The Converse of Chain Rule as follows. (TheConverseofChainRule) Suppose that f, g and u are related so that fx gux. Ifux is continuous at x 0 , f x 0 exists, g ux 0 exists and not zero. Then u x 0 is defined and we have f x 0 g ux 0 u x 0 . Proof: Sincef x 0 exists, and g ux 0 exists, then fx fx 0 f x 0 x x 0 o|x x 0 | * and gux gux 0 g ux 0 ux ux 0 o|ux ux 0 |. ** Since fx gux, andfx 0 gux 0 , by (*) and (**), we know that f ux ux 0 x 0 g ux 0 x x 0 o|x x 0 | o|ux ux 0 |. *** Note that since ux is continuous at x 0 , we know that o|ux ux 0 | 0asx x 0 . So, (***) means that u x 0 is defined and we have f x 0 g ux 0 u x 0 . Remark: The condition that g ux 0 is not zero is essential, for example, gx 1on 1, 1 and ux |x|, wherex 0 0. (Inverse Function Theroem) Suppose that f is continuous, strictly monotonic function which has an open interval I for domain and has range J. (It implies that fgx x gfx on its corresponding domain.) Assume that x 0 is a point of J such that f gx 0 is defined and is different from zero. Then g x 0 exists, and we have g x 0 1 f gx 0 . Proof: Itisaresultofthe converse of chain rule note that fgx x. Mean Value Theorem 5.10 Given a function defined and having a finite derivative in a, b and such that lim xb fx . Prove that lim xb f x either fails to exist or is infinite. Proof: Suppose NOT, we have the existence of lim xb f x, denoted the limit by L. So, given 1, there is a 0 such that as x b , b we have |f x| |L| 1. * Consider x, a b , b with x a, then we have by (*) and Mean Value Theorem, |fx fa| |f x a| where a, x |L| 1|x a| which implies that
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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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Page 103 and 104: number f T supfx fy : x, y T is
Page 105 and 106: Proof: Since the hypothesis says th
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Page 109 and 110: In Exercises 4.29 through 4.33, we
Page 111 and 112: function. Since f0, 1 0, 1 0, 1,
Page 113 and 114: 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: which implies that F 0, where
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 and 166: which implies that Hence, g 1 h g
Page 167 and 168: fx fy x y f x /2 *’ Combi
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
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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
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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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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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