The Real And Complex Number Systems

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n ∑ kn a k ∑ kn a k b k b n b k ∑ C k d k − d k1 C d − C n−1 d n kn ∑ kn C k d k − d k1 − C n−1 d n by C lim k→ C k and lim k→ d k 0. In order to show the existence of lim n→ b n ∑ kn a k , it suffices to show the existence of lim n→ ∑ kn C k d k − d k1 . Since the series ∑ kn C k d k − d k1 exists, lim n→ ∑ kn C k d k − d k1 0. From above results, we have proved the convergence of lim n→ b n ∑ kn a k . Note: We also show that lim n→ b n ∑ kn a k 0 by preceding sayings. Supplement on the convergence of series. A Show the divergence of ∑ 1/k. We will give some methods listed below. If the proof is easy, we will omit the details. (1) Use Cauchy Criterion for series. Since it is easy, we omit the proof. (2) Just consider 1 1 2 1 3 1 4 ... 2 1 n ≥ 1 1 2 2 1 4 1 ...2n−1 2 n 1 n 2 → . Remark: We can consider 1 1 ... 1 1 ... 2 10 1 ...≥ 1 11 100 10 9 100 90 ... Note: The proof comes from Jing Yu. (3) Use Mathematical Induction to show that 1 k − 1 1 k 1 k 1 ≥ 3 if k ≥ 3. k Then 1 1 2 1 3 1 4 1 5 1 6 ....≥ 1 3 3 3 6 3 9 ... Remark: The proof comes from Bernoulli. (4) Use Integral Test. Since the proof is easy, we omit it. (5) Use Cauchy condensation Theorem. Since the proof is easy, we omit it. (6) Euler Summation Formula, the reader can give it a try. We omit the proof. (7) The reader can see the book, Princilpes of Mathematical Analysis by Walter Rudin, Exercise 11-(b) pp 79. Suppose a n 0, S n a 1 ...a n ,and∑ a n diverges. (a) Prove that ∑ an 1a n diverges. Proof: Ifa n → 0asn → , then by Limit Comparison Theorem, we know that diverges. If a n does not tend to zero. Claim that does not tend to zero. ∑ an 1a n a n 1a n
Suppose NOT, it means that lim n→ a n 1a n 0. That is, lim 1 n→ 0 lim 1 a 1 n→ a n 0 n which contradicts our assumption. So, ∑ an 1a n (b) Prove that and deduce that ∑ an S n Proof: Consider diverges. a N1 S N1 a N1 S N1 ... a Nk S Nk diverges by claim. ... a Nk S Nk ≥ 1 − S N S Nk then ∑ an S n diverges by Cauchy Criterion with (*). Remark: Leta n 1, then ∑ an S n (c) Prove that and deduce that ∑ an S2 n Proof: Consider and So, ∑ an S n 2 converges. ≥ a N1 ...a Nk S Nk 1 − S N S Nk , * ∑ 1/n diverges. a n S n 2 ≤ 1 S n−1 1 S n−1 − 1 S n − 1 S n a n S n−1 S n ≥ a n S n 2 , ∑ S 1 n−1 − S 1 n converges by telescoping series with S 1 n → 0. converges. (d) What can be said about ∑ Proof: For∑ the series ∑ For ∑ so ∑ an 1n 2 a n a n 1 na n and ∑ a n 1 n 2 a n an 1na n :asa n 1 for all n, theseries∑ an 1na n ∑ 1 1n an 1na n ∑ 1 an 1n 2 a n 1k 2 : Consider converges. (8) Consider ∑ sin 1 n diverges. Proof: Since a n converges. a n 0ifn ≠ k2 , 1ifn k 2 1 ≤ 1 1 n 2 a 1 n a n n 2 n , 2 lim sin 1 n n→ 1 n 1, the series ∑ 1 n diverges by Limit Comparison Theorem. diverges. As
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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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Page 179 and 180: So, we have Partial derivatives fb
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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:
Page 217 and 218: For case (i), since 1 n log nlog lo
Page 219 and 220: ()Assume that ∑ n1 That is, p!
Page 221 and 222: n S n ∑−1 k1 1 3k − 2 − 1
Page 223 and 224: (b) Prove that ∑ n1 −1 n−1 /
Page 225 and 226: then b k sin kx, a k1 − a k 1
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Page 231 and 232: then since n lim n→ ∑ k1 n S n
Page 233 and 234: |sin k|r ∑ k So, ∑ |sink|r dive
Page 235 and 236: n For the part 1 2 0 that x sint
Page 237 and 238: lim n→ sin ... sin n 1 ... 1 n
Page 239 and 240: Also, lim q→ fp, q lim q→ p si
Page 241 and 242: n c n ∑ a k b n−k k0 n ∑ k0
Page 243 and 244: So, n s n ∑ cos2k − 1x j1 sin
Page 245 and 246: n u n ∑−1 k a k k1 n ∑−1
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
Page 253 and 254: Sequences of Functions Uniform conv
Page 255 and 256: (b) Prove that h n (x) does not con
Page 257 and 258: Proof: Since g is continuous on a c
Page 259 and 260: Hence, {g (x) x n } converges unifo
Page 261 and 262: y continuity of f k(x0 ) (x) − f
Page 263 and 264: In addition, as c ≥ 1/2, if f n
Page 265 and 266: (Lemma) If {a n } and {b n } are tw
Page 267 and 268: 9.16 Let {f n } be a sequence of re
Page 269 and 270: which implies that ∣ lim sup 1 k
Page 271 and 272: have ∫ x 0 ∞∑ t 2n − 1 2 n=
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Page 275 and 276: 0.1 Supplement on some results on W
Page 277 and 278: 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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