The Real And Complex Number Systems

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Something around the number e 1. Show that the sequence 1 1 n n converges, and denote the limit by e. Proof: Since 1 1 n n n kn 1 n k k0 1 n 1 n nn 1 2! 1 n 1 1 2! 1 1 1 n ... n! 1 1 1 1 2 1 2 .. 1 ... 2 2 n1 3, 2 nn 1 1 .. 1 n! n 1 1 n 1 n n 1 and by (1), we know that the sequence is increasing. Hence, the sequence is convergent. We denote its limit e. Thatis, lim n 1 1 n n e. n 1 Remark: 1. <strong>The</strong> sequence and e first appear in the mail that Euler wrote to Goldbach. It is a beautiful formula involving e i 1 0. 1 2. Use the exercise, we can show that k0 e as follows. k! Proof: Let x n 1 1 n n , and let k n, wehave 1 1 2! 1 1 1 .. k n! 1 1 1 1 k n k 1 x k which implies that ( let k ) On the other hand, So,by(2)and(3),wefinallyhave n y n : i0 k0 1 i! e. 2 x n y n 3 1 k! e. 4 3. e is an irrational number. Proof: Assume that e is a rational number, say e p/q, where g.c.d. p, q 1. Note that q 1. Consider q!e q! q! k0 q k0 1 k! 1 k! q! kq1 and since q! k0 q 1 k! and q!e are integers, we have q! kq1 1 k! is also an integer. However, 1 k! ,
q! kq1 1 k! kq1 q! k! 1 q 1 1 q 1q 2 ... 1 q 1 1 q 1 2 ... 1 q 1, a contradiction. So, we know that e is not a rational number. n 1 4. Here is an estimate about e k0 k! , where 0 1. ( In fact, we know nn! that e 2. 71828 18284 59045 . . . . ) Proof: Since e k0 1 k! 0 e x n kn1 1 n 1! 1 n 1! 1 k! ,wherex n n k0 1 k! 1 1 n 2 1 n 2n 3 ... 1 1 n 2 1 n 2 2 1 n 1! n n 2 1 1 nn! since n 2 1 n 1 2 n . So, we finally have n e 1 k! , where 0 1. nn! k0 Note: We can use the estimate dorectly to show e is an irrational number. 2. For continuous variables, we have the samae result as follows. That is, x e. lim x 1 1 x Proof: (1) Since 1 1 n n e as n , we know that for any sequence a n N, with a n , wehave lim n 1 a 1 a n n e. 5 (2) Given a sequence x n with x n , and define a n x n , then a n x n a n 1, then we have a n x 1 1 1 1 n a n 1 xn 1 a 1 a n1 n . Since a n a 1 1 e and 1 1 n1 a n 1 an e as x by (5) we know 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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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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Page 247 and 248: which implies that which implies th
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Page 257 and 258: Proof: Since g is continuous on a c
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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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Page 281 and 282: we complete it. 9.31 Given that two
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Page 285 and 286: So, the series diverges. (ii) As
Page 287 and 288: 9.38 For each real t, define f t (x
Page 289 and 290: So, B 0 = P 0 (0) = C 0 = 1, B 1 =
Page 291 and 292: Remark: (1) The reader can see the
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The Real And Complex Number Systems

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