The Real And Complex Number Systems

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lim n→ |sinn 1a b − sinna b| lim n→ sinn 1a b − sinna b lim n→ sup 2 cosna b cos a 2 − sinna b sin a 2 sin a 2 lim n→ sup 2 cosna b cos a sin a 2 2 |sin a| ≠ 0 which is impossible. So, ∑ sinna b diverges. Remark: (1) By the same method, we can show the divergence of ∑ cosna b if a ≠ n for all n ∈ Z and b ∈ R. (2) The reader may give it a try to show that, and p by considering ∑ n0 (**). (3) The series ∑ sink k p ∑ n0 cosna b Proof: First, it is clear that ∑ sink k |∑ sin k| ≤ 1 sin 1 2 partial sums as follows: Since sin p1 2 b sin b 2 sin a p 2 b * p p1 sin ∑ sinna b b 2 cos a p sin b 2 b ** n0 2 e inab . However, it is not easy to show the divergence by (*) and converges conditionally. converges by Dirichlet’s Test since , we consider its . In order to show that the divergence of ∑ sink k 3n3 n ∑ sin k ∑ sin 3k 1 sin 3k 2 sin 3k 3 k 3k 1 3k 2 3k 3 k1 k0 and note that there is one value is bigger than 1/2 among three values |sin 3k 1|, |sin 3k 2|, and|sin 3k 3|. So, 3n3 ∑ k1 sin k k n 1 2 ≥ ∑ 3k 3 k0 which implies the divergence of ∑ sink . k Remark: The series is like Dirichlet Integral sinx 0 x Integral converges conditionally. (4) The series ∑ |sink|r diverges for any r ∈ R. k Proof: We prove it by three cases as follows. (a) As r ≤ 0, we have So, ∑ |sink|r diverges in this case. k (b) As 0 r ≤ 1, we have ∑ |sin k|r k ≥ ∑ 1 k . dx. Also, we know that Dirichlet
|sin k|r ∑ k So, ∑ |sink|r diverges in this case by (3). k (c) As r 1, we have 3n3 ∑ |sin k| r k k1 n ∑ k0 n ≥ ∑ k0 |sin 3k 1| r 3k 1 1 2 r 3k 3 . ≥ ∑ |sin k| k . |sin 3k 2|r 3k 2 |sin 3k 3|r 3k 3 So, ∑ |sink|r diverges in this case. k (5) The series ∑ sin2p−1 k ,wherep ∈ N, converges. k Proof: We will prove that there is a positive integer Mp such that n ∑ k1 sin 2p−1 k ≤ Mp for all n. * So, if we can show (*), then by Dirichlet’s Test, we have proved it. In order to show (*), we claim that sin 2p−1 k can be written as a linear combination of sink, sin3k,..., sin2p − 1k. So, n ∑ k1 n sin 2p−1 k ∑ k1 n ≤ |a 1 | ∑ k1 |a 1 | a 1 sin k a 2 sin 3k ...a p sin2p − 1k n sin k ...|a p| ∑ sin2p − 1k k1 |a ≤ ... p| : Mp by Theorem 8.30. sin 1 2 sin 2p−1 2 We show the claim by Mathematical Induction as follows. As p 1, it trivially holds. Assume that as p s holds, i.e., then as p s 1, we have sin 2s−1 k ∑ j1 s a j sin2j − 1k
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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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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
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Page 193 and 194: in Section 6.12. Proof: Sinceft : t
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Page 209 and 210: We first prove lim n→ sup n ≤
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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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Page 219 and 220: ()Assume that ∑ n1 That is, p!
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Page 223 and 224: (b) Prove that ∑ n1 −1 n−1 /
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Page 239 and 240: Also, lim q→ fp, q lim q→ p si
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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
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Page 255 and 256: (b) Prove that h n (x) does not con
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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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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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