精華區beta Math 關於我們 聯絡資訊
※ 引述《LuisSantos ( )》之銘言: : 1. : x dy - y dx : Let ω = -------------- , (x , y) 屬於 R^2 - {0} . : x^2 + y^2 : Show that ω is a closed 1-form , but not exact on R^2 - {0} . 這個我不熟,但我猜只要套 Rudin 第十章的樣子就可以搞定的樣子。 : 2. Let S = {x屬於R^n | ||x|| ≦ 1} and f : S→R be a nonnegative continuous : function. : (a) Prove that f has the absolute maximum value on S. : (b) Let M be the absolute maximum value of f on S . : k 1/k : Show that lim (∫(f(x)) dx_1 dx_2 ... dx_n) = M . : k→∞ S 這一題跟某一題很像: 可參見:http://frankmath.cc/plover/Apostol.pdf (p.302 之後). 你自己依樣畫葫蘆就可以了。 : 3. Prove or disprove (by a counter example) the following statements: : If f : R→R is a continuous function , then f is an open mapping . 這不對! 因為定理是這個: Prove that every continuous open mapping of |R into |R is monotonic. 這也是 Rudin 的習題。 請至 bbs://bs2.to 版名:P_Apostol, 精華區-高等微積分-2 ◆ 高微書簡介-個人看法-6 ◆ [Rudin] 高微解答 Ch4. ex 15. : 4. Let f : [0,1]→R be defined by : 0 if x is irrational : f(x) = : 1 p : --- if x = --- : q q : where p , q ≧ 0 with no common factor . : Is f integrable on [0,1] ? 這就是鼎鼎大名的 爆米花函數 (或稱為 Riemamnn function). 跟上一題一樣, Ch4, ex. 18. : 5. Let C([0,1]) be the set of real continuous functions on [0,1] . : Show that the complement of the following set : 1 : A = { f屬於 C([0,1]) | 0 < ∫ f(x) dx < 3 } is closed : 0 這個我想應該也只是定義的使用。不行我們再討論討論。 : 6. Let (M,d) be a metric space . If A is compact in M and B is closed in M , : and A∩B = ψ . Show that there is an δ > 0 such that d(x,y) > δ for : all x 屬於 A and y 屬於 B . Rudin 習題,Ch4. ex. 21. : 7. Let k(x,y) be a continuous real-valued function on the square : S = [0,1] ╳ [0,1] . Assume that |k(x,y)| < 1 for each (x,y) 屬於 S . : Let A : [0,1]→R be continuous . Prove that there is a unique continuous : real-valued function f(x) on [0,1] such that : 1 : f(x) = A(x) + ∫ k(x,y)f(y) dy by contraction mapping theorem . : 0 這個只是用他給的 hint: Contraction Mapping Theorem :-) -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 122.116.231.200
math1209:第四題, http://frankmath.cc/plover/Apostol.pdf 02/24 06:39
math1209:ex 7.28 (p. 308~). 當然,如果你學過測度論那這會變得 02/24 06:40
math1209:很單純... 02/24 06:40
Potervens:the set of discontinuities has measure zero 02/24 10:59