The Real And Complex Number Systems

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Remark: There are many methods to show this. We do NOT give the detailed proof. But there are hints. (1) Taking log on n! n n 1/n , and thus consider 1 n log 1 n ... log n n → 0 1 log xdx −1. (2) Stirling’s Formula: n! n n e −n 2n e 12n ,where ∈ 0, 1. Note: In general, we have lim x→ Γx 1 x x e −x 2x 1, where Γx is the Gamma Function. The reader can see the book, Principles of Mathematical Analysis by Walter Rudin, pp 192-195. (3) Note that 1 1 x x ↗ e and 1 1 x x1 ↘ e on 0, . So, 1 1 n n1 n e 1 1 n which implies that en n e −n n! en n1 e −n . (4) Using O-Stolz’s Theorem: Let lim n→ y n and y n ↗.If lim x n1 − x n n→ y n1 − y n a, wherea is finite or , then lim x n n→ y n a. Let x n log 1 n ... log n n and y n n. Note: For the proof of O-Stolz’s Theorem, the reader can see the book, An Introduction to Mathematical Analysis by Loo-Keng Hua, pp 195. (Chinese Version) (5) Note that, if a n is a positive sequence with lim n→ a n a, then a 1 a n 1/n → a as n → . Taking a n 1 1 n n , then a 1 a n 1/n n n 1 n! 1 n → e. Note: For the proof, it is easy from the Exercise 8.6. We give it a proof as follows. Say lim n→ a n a. Ifa 0, then by A. P. ≥ G. P. , we have a 1 a n 1/n ≤ a 1 ...a n n → 0byExercise 8.6. So, we consider a ≠ 0 as follows. Note that log a n → log a. So, by Exercise 8.6, log a 1 ... log a n n → log a which implies that a 1 a n 1/n → a. 8.6 Let a n be real-valued sequence and let n a 1 ...a n /n. Show that lim n→ inf a n ≤ lim n→ inf n ≤ lim n→ sup n ≤ lim n→ sup a n . Proof: ByTheorem 8.3 (a), it suffices to show that lim n→ inf a n ≤ lim n→ inf n and lim n→ sup n ≤ lim n→ sup a n . 1/n
We first prove lim n→ sup n ≤ lim n→ sup a n . If lim sup n→ a n , there is nothing to prove it. We may assume that lim sup n→ a n − or a, wherea is finite. For the case, lim sup n→ a n −, then by Theorem 8.3 (d), wehave lim n→ a n −. So, given M 0, there exists a positive integer N such that as n ≥ N, wehave a n ≤−M. * Let n N, wehave n a 1 ...a N ..a n n a 1 ...a N n a N1 ...a n n ≤ a 1 ...a N n n − N n −M which implies that lim n→ sup n ≤−M. Since M is arbitrary, we finally have lim n→ sup n −. For the case, lim sup n→ a n a, wherea is finite. Given 0, there exists a positive integer N such that as n ≥ N, wehave a n a . Let n N, wehave n a 1 ...a N ..a n n a 1 ...a N n a N1 ...a n n ≤ a 1 ...a N n n − N n a which implies that lim n→ sup n ≤ a which implies that lim n→ sup n ≤ a since is arbitrary. Hence, from above results, we have proved that lim sup n→ n ≤ lim sup n→ a n . Similarly for lim inf n→ a n ≤ lim inf n→ n . Remark: We suggest that the reader keep it in mind since it is the fundamental and useful in the theory of Fourier Series. 8.7 Find lim sup n→ a n and lim inf n→ a n if a n is given by (a) cosn Proof: Note that, a b : a, b ∈ Z is dense in R. Bycosn cosn 2k, we know that lim n→ sup cos n 1 and lim n→ inf cos n −1.
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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
Page 157 and 158: Proof: Look at the Generalized Mean
Page 159 and 160: Proof: By Generalized Mean Value Th
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Page 163 and 164: Remark: If we can make sure that fx
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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
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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
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
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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:
Page 217 and 218: For case (i), since 1 n log nlog lo
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 229 and 230: Suppose NOT, it means that lim n→
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Page 233 and 234: |sin k|r ∑ k So, ∑ |sink|r dive
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Page 239 and 240: Also, lim q→ fp, q lim q→ p si
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Page 243 and 244: So, n s n ∑ cos2k − 1x j1 sin
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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
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Page 253 and 254: Sequences of Functions Uniform conv
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Page 257 and 258: 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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