The Real And Complex Number Systems

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Proof: Since tan z = sin z sin (x + iy) sin x cosh y + i cos x sinh y = = cos z cos (x + iy) cos x cosh y − i sin x sinh y (sin x cosh y + i cos x sinh y) (cos x cosh y + i sin x sinh y) = (cos x cosh y − i sin x sinh y) (cos x cosh y + i sin x sinh y) ( sin x cos x cosh 2 y − sin x cos x sinh 2 y ) + i ( sin 2 x cosh y sinh y + cos 2 x cosh y sinh y ) = (cos x cosh y) 2 − (i sin x sinh y) 2 = sin x cos x ( cosh 2 y − sinh 2 y ) + i (cosh y sinh y) cos 2 x cosh 2 y + sin 2 x sinh 2 since sin 2 x + cos 2 x = 1 y (sin x cos x) + i (cosh y sinh y) = cos 2 x + sinh 2 since cosh 2 y = 1 + sinh 2 y y 1 2 = sin 2x + i sinh 2y 2 cos 2 x + sinh 2 since 2 cosh y sinh y = sinh 2y and 2 sin x cos x = sin 2x y sin 2x + i sinh 2y = 2 cos 2 x + 2 sinh 2 y sin 2x + i sinh 2y = 2 cos 2 x − 1 + 2 sinh 2 y + 1 sin 2x + i sinh 2y = cos 2x + cosh 2y since cos 2x = 2 cos2 x − 1 and 2 sinh 2 y + 1 = cosh 2y. 1.47 Let w be a given complex number. If w ≠ ±1, show that there exists two values of z = x + iy satisfying the conditions cos z = w and −π < x ≤ π. Find these values when w = i and when w = 2. Proof: Since cos z = eiz +e −iz , if we let e iz = u, then cos z = w implies 2 that w = u2 + 1 ⇒ u 2 − 2wu + 1 = 0 2u which implies that So, by <strong>The</strong>orem 1.51, (u − w) 2 = w 2 − 1 ≠ 0 since w ≠ ±1. e iz = u = w + ∣ w 2 − 1 ∣ 1/2 e iφ k , where φ k = arg (w2 − 1) 2 ( ) = w ± ∣ w 2 − 1 ∣ 1/2 e i arg(w 2 −1) 2 + 2πk , k = 0, 1. 2 32
So, ix−y = i (x + iy) = iz = log ∣ w ± ∣ w 2 − 1 ∣ ⎛ ( )⎞ 1/2 e i arg (w 2 −1) 2 ∣ +i arg ⎝w ± ∣ w 2 − 1 ∣ 1/2 e i arg ( w 2 −1 ) 2 ⎠ Hence, there exists two values of z = x+iy satisfying the conditions cos z = w and ⎛ ( )⎞ −π < x = arg ⎝w ± ∣ w 2 − 1 ∣ 1/2 e i arg(w 2 −1) 2 ⎠ ≤ π. For w = i, we have ( iz = log ∣ 1 ± √ ) 2 i∣ + i arg which implies that z = arg (( 1 ± √ ) ) 2 i For w = 2, we have ∣ iz = log ∣2 ± √ 3∣ + i arg (( 1 ± √ ) ) 2 i ( − i log ∣ 1 ± √ ) 2 i∣ . ( 2 ± √ ) 3 which implies that ( z = arg 2 ± √ ) 3 ∣ − i log ∣2 ± √ 3∣ . 1.48 Prove Lagrange’s identity for complex numbers: ∣ n∑ ∣∣∣∣ 2 n∑ n∑ a k b k = |a k | 2 |b k | 2 − ∑ ( ) 2 ak¯bj − ā j b k . ∣ k=1 k=1 k=1 1≤k
Page 1 and 2: The Real And Complex Number Systems
Page 3 and 4: Proof: Consider 2n−2 ∑ (x + 1)
Page 5 and 6: Then S = N. Proof: (A ⇒ B): If S
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Page 21 and 22: So, z 1 z 2 = ¯z 1¯z 2 (c) z¯z =
Page 23 and 24: so, |a − b| 2 ≤ ( 1 + |a| 2) (
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Page 37 and 38: Some Basic Notations Of Set Theory
Page 39 and 40: (i) If x ∈ R, it is clear that (x
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Page 43 and 44: (e) A ∩ (B − C) = (A ∩ B) −
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Page 55 and 56: Charpter 3 Elements of Point set To
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Page 61 and 62: Bx, r x S Bx, r x T . So, at
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Page 75 and 76: 3.44 intM A M A . Proof: Let B
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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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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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