The Real And Complex Number Systems

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Proof: Consider f t+1 (x) − f t (x) = so as n = 1, 2, ..., we have ∞∑ n=0 [P n (t + 1) − P n (t)] xn n! = xe xt by f t (x) = xe xt / (e x − 1) ∞∑ = (n + 1) t n xn+1 (n + 1)! , n=0 P n (t + 1) − P n (t) = nt n−1 . by (b) (f) P n (1 − t) = (−1) n P n (t) Proof: Note that so we have n=0 Hence, P n (1 − t) = (−1) n P n (t) . (g) B 2n+1 = 0 if n = 1, 2, ... f t (−x) = f 1−t (x) , ∞∑ ∞ (−1) n P n (t) xn n! = ∑ P n (1 − t) xn n! . Proof: With help of (e) and (f), let t = 0 and n = 2k + 1, then it is clear that B 2k+1 = 0 if k = 1, 2, ... n=0 (h) 1 n + 2 n + ... + (k − 1) n = P n+1(k)−P n+1 (0) n+1 (n = 2, 3, ...) Proof: With help of (e), we know that P n+1 (t + 1) − P n+1 (t) n + 1 = t n which implies that 1 n + 2 n + ... + (k − 1) n = P n+1 (k) − P n+1 (0) n + 1 (n = 2, 3, ...) 38
Remark: (1) The reader can see the book, Infinite Series by Chao Wen-Min, pp 355-366. (Chinese Version) (2) There are some special polynomials worth studying, such as Legengre Polynomials. The reader can see the book, Essentials of Ordinary Differential Equations by Ravi P. Agarwal and Ramesh C. Gupta. pp 305-312. (3) The part (h) tells us one formula to calculte the value of the finite series ∑ m k=1 kn . There is an interesting story from the mail that Fermat, pierre de (1601-1665) sent to Blaise Pascal (1623-1662). Fermat used the Mathematical Induction to show that n∑ k (k + 1) · · · (k + p) = k=1 n (n + 1) · · · (n + p + 1) . (*) p + 2 In terms of (*), we can obtain another formula on ∑ m k=1 kn . 39
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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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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
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Page 257 and 258: Proof: Since g is continuous on a c
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Page 261 and 262: y continuity of f k(x0 ) (x) − f
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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
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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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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: So, B 0 = P 0 (0) = C 0 = 1, B 1 =
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Page 297 and 298: q! kq1 1 k! kq1 q! k! 1 q 1
Page 299: 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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