Giáo trình giải tích 2 part 7 doc

u, v x, y x =1,y = −1,u=1,v = −1 x =0,y =1,u=0,v=0 ∂u ∂x x =1,y = −1 x =0,y =1 f(x, y)=x 2 + y 2 − 1=0 y = g(x) g F (x, y)=x 3 − y 3 =0 y = x ∂F ∂y (0, 0) = 0 n u =(u 0 , ··· ,u n−1 ) P u (x)=x n + u n−1 x n−1 + ···+ u 1 x + u 0 u = a x 0 P a P a (x 0 )=0,P  a (x 0 ) =0 U a V x 0 u ∈ U x(u) ∈ V P u (x)=0 C ∞ 3 x 3 + px + q =0 p, q ∆=4p 3 +27q 2 ∆ > 0 x ∗ (p, q) ∆ < 0 x − (p, q) <x 0 (p, q) <x + (p, q) ∆=0,q >0 x − (p, q) < 0 x 0+ (p, q) ∆=0,q <0 x 0− (p, q) < 0 x + > 0 (p, q)=(0, 0) x =0 x ∗ C ∞ x − ,x 0 ,x + C ∞ ✲ x ✻ y abS inf f sup f R n A =[a 1 ,b 1 ] ×···[a n ,b n ]. A v(A)=(b 1 − a 1 ) ···(b n − a n ) P A [a i ,b i ],i=1, ··· ,n a i = c i0 <c i1 < ···<c im i = b i m 1 m 2 ···m n A S =[c 1i 1 ,c 1i 1+1 ] ×···×[c ni n ,c ni n+1 ]. S ∈ P f : A → R P A L(f,P)=  S∈P inf x∈S f(x) v(S) U(f,P)=  S∈P sup x∈S f(x) v(S) L(f,P) ≤ U(f,P) P  P P  P P  P L(f,P) ≤ L(f,P  ) U(f, P  ) ≤ U(f,P). I (f)=sup P L(f,P) ≤ inf P U(f,P)=I(f) f A I(f)=I(f) f A  A f  A f(x)dx  b 1 a 1 ···  b n a n f(x 1 , ··· ,x n )dx 1 ···dx n f A >0 P A U(f,P) −L(f,P) <. f ≡ c U(f,P)=L(f,P)=cv(A) P f A  A f = cv(A) D(x)=  0 x 1 x [0, 1] P L(D,P)=0,U(D,P)=1. P A ξ P =(ξ S ,S ∈ P ) ξ S ∈ S S(f, P,ξ P )=  S∈P f(ξ S ) v(S) |P | S ∈ P f : A → R A ⊂ R n f A  A f = I lim |P |→0 S(f, P,ξ P )=I, ∀ξ P >0 δ>0 P A |P | <δ |S(f, P,ξ P ) −I| < ∀ ξ P P 0 A >0 δ>0 P A |P | <δ P P < n n =1 A =[a, b] P 0 N δ = /N P |P | <δ P P 0 ≤ × ≤ N × δ =  n>1 P 0 V 1 , ··· ,V k T δ = /T P A |P | <δ S ∈ P S ⊂ V i ,i=1, ··· ,k S P 0 v(S) ≤ δD D V 1 , ··· ,V k S  S∈P,S⊂V i ,∀i v(S) <δT =  ⇒ |f(x)| <M,∀x ∈ A P 0 U(f,P 0 ) −I</2 ,I− L(f,P 0 ) </2  := /2M δ>0 P |P | <δ P 1 P P 0 P 2 P P 0 ξ P  S∈P f(ξ S ) v(S) ≤  S∈P 1 f(ξ S ) v(S)+  S∈P 2 f(ξ S ) v(S) ≤ U(f,P 0 )+M/M < I +   S∈P f(ξ S ) v(S) ≥ L(f,P 0 ) −/2 >I− |S(f, P,ξ P ) −I| < ⇒ >0 δ>0 P N P S ∈ P ξ S ∈ S |f(ξ S ) −sup S f| </v(S)N |U(f,P) −I|≤|U(f,P) −  S∈P f(ξ S )v(S)| + |  S∈P f(ξ S )v(S) I | <  S∈P v(S)/v(S)N =  U(f,P) −I| < 2 |L(f,P) − I| < 2 |U(f,P) − L(f,P)| < 4 f A  f : A =[a 1 ,b 1 ] ×[a n ,b n ] → R P N [a i ,b i ],i =1, ··· ,n N +1 c ik = a i + k(b i − a i ) N ,k=0, ··· ,N N →∞ (b 1 − a 1 ) ···(b n − a n ) N n N  k 1 ,···,k n =1 f(c 1k 1 , ··· ,c nk n ) −→  A f.  1 0 xdx = lim N→∞ 1 N N  k=1 k N = lim N→∞ 1 N 2 N(N +1) 2 = 1 2 . C ⊂ R n C χ C (x)=  1 x ∈ C 0 x ∈ C A C C χ C A C v(C)=  A χ C . A C n =1,n=2 U(χ C ,P) P C L(χ C ,P) P C C A A C C C ⊂ R n f : A → R A C f C fχ C A f C  C f =  A fχ C . R(C) C B ⊂ R n µ(B)=0 µ n (B)=0 >0 S 1 ,S 2 , ··· B B ⊂∪ i S i  i v(S i ) <. N ⊂ R R R 2 µ(B)=0 f : B → R m µ(f(B)) = 0 µ(B i )=0,i=1, 2, ··· µ(∪ i B i )=0 >0 i S i1 ,S i2 , ··· B i  j v(S ij ) </2 i {S ij } ∪ i B i  ij v(S ij ) <  i 1 2 i <  R n f : A → R A ⊂ R n f f f S o(f, S)=sup x∈S f(x) − inf x∈S f(x) f a o(f,a) = lim r→0 + o(f,B(a, r)). r o(f,a)=0 f a B = {x : o(f,x) > 0} f (⇒) µ(B)=0 >0 B  = {x : o(f,x) ≥ } B  ⊂ B µ(B  )=0 B  S 1 , ··· ,S N B  N  i=1 v(S i ) < P A S ∈ P S ∩ B  = ∅ S ⊂ S i i ∈{1, ··· ,N} P 1 = {S ∈ P : S ∩B  = ∅} P 2 = {S ∈ P : ∃iS⊂ S i } S ∈ P 1 o(f,x) <,x∈ S S P S  ∈ P 1 sup S  f − inf S  f ≤ 2 U(f,P) −L(f,P)=(  S∈P 1 +  S∈P 2 )  (sup S f − inf S f)v(S)  ≤  S∈P 1 2v(S)+  S∈P 2 Mv(S), (M =sup A f − inf A f) ≤ 2v(A)+M N  i=1 v(S i ) < (2v(A)+M). f A (⇐) f A B =  k∈N B 1 k µ(B 1 k )=0, ∀k ∈ N k >0 P U(f,P) −L(f,P)=  S∈P (sup S f − inf S f)v(S) <  k .  S∩B 1 k =∅ 1 k v(S) ≤  S∩B 1 k =∅ (sup S f − inf S f)v(S) <  k {S ∈ P : S ∩B 1 k = ∅} B 1 k  S∩B 1 k =∅ v(S) < µ(B 1 k )=0  C ⊂ R n µ(∂C)=0. C ⊂ R n f : C −→ R f C f :[a, b] → R f C χ C χ C ∂C k ∈ N D k = {x ∈ [a, b]:o(f, x) ≥|f(a) − f(b)|/k} k f B =  k D k f  f(x)=sin 1 x x =0 f(0) = 0 [−1, 1] f(x, y)=x 2 +sin 1 y y =0 f(x, 0) = 0 A = {x 2 +y 2 ≤ 1} A R n f,g A α, β ∈ R αf + βg A  A (αf + βg)=α  A f + β  A g A 1 ,A 2 ⊂ A f A 1 ,A 2  A 1 ∪A 2 f =  A 1 f +  A 2 f −  A 1 ∩A 2 f. f ≤ g A  A f ≤  A g. |f| A |  A f|≤  A |f|. f A c ∈ A  A f = f(c)v(A). inf A fv(A) ≤  A f ≤ sup A fv(A)  A, B A ∪B v(A ∪B)=v(A)+v(B) −v(A ∩B). f [a, b] c ∈ [a, b] f(c)= 1 b −a  b a f(x)dx f :[a, b] −→ R d dx  x a f = f, x ∈ [a, b]. F f [a, b] F  (x)=f(x),x∈ [a, b]  b a f = F (b) −F(a). g :[a, b] −→ R f g([a, b])  g(b) g(a) f =  b a f ◦ gg  . u, v [a, b]  b a uv  = u(b)v(b) −u(a)v(a) −  b a u  v. • • ✲ x ✻ y g 2 g 1 x C a b C g 1 ,g 2 [a, b] g 1 ≤ g 2 C x ∈ [a, b] d(x) C x = x ×R ∩ C = x ×[g 1 (x),g 2 (x)] (C)=  C dxdy =  b a d(x)dx =  b a (  g 2 (x) g 1 (x) dy)dx f [a, b] ×[c, d] f [a, b] ×[c, d] V = {a ≤ x ≤ b, c ≤ y ≤ d, 0 ≤ z ≤ f(x, y)} x ∈ [a, b] S(x)= {(y, z):c ≤ y ≤ d, 0 ≤ z ≤ f(x, y)} =  d c f(x, y)dy (V )=  V dxdydz =  b a S(x)dx =  b a (  d c f(x, y)dy)dx C ⊂ R n × R m f : C → R Ω={x ∈ R n : ∃y ∈ R m , (x, y) ∈ C} C R n C x = {y ∈ R m :(x, y) ∈ C} C x  C x f(x, y)dy x ∈ Ω  C f(x, y)dxdy =  Ω (  C x f(x, y)dy)dx. C = A×B A, B R n , R m P, P  A, B P ×P  A ×B S × S  ,S ∈ P, S  ∈ P  L(f,P × P  )=  S×S  ∈P ×P  inf{f(x, y); x ∈ S, y ∈ S  }v(S × S  ) =  S∈P (  S  ∈P  inf{f(x, y); x ∈ S, y ∈ S  }v(S  ))v(S) ≤  S∈P inf{  S  ∈P  inf{f(x, y); y ∈ S  }v(S); x ∈ S}v(S) ≤  S∈P inf x∈S (  B f(x, y)dy)v(S) ≤ L(  B f(x, y)dy, P ). L(f,P × P  ) ≤ L(  B f(x, y)dy, P ) ≤ U(  B f(x, y)dy, P ) ≤ U(f,P × P  ).  A×B f =  A (  B f(x, y)dy)dx. C A, B C ⊂ A × B f A ×B 0 C  g 1 ,g 2 :[a, b] → R g 1 ≤ g 2 C = {(x, y):a ≤ x ≤ b, g 1 (x) ≤ y ≤ g 2 (x)} f C C  C f(x, y)dxdy =  b a (  g 2 (x) g 1 (x) f(x, y)dy)dx h 1 ,h 2 :Ω→ R Ω ⊂ R 2 h 1 ≤ h 2 C = {(x, y, z):(x, y) ∈ Ω,h 1 (x, y) ≤ z ≤ h 2 (x, y)} f C C  C f(x, y, z)dxdydz =  Ω (  h 2 (x,y) h 1 (x,y) f(x, y, z)dz)dxdy E = { x 2 a 2 + y 2 b 2 ≤ 1} E Ox [−a, a] C x = {y : − b a  a 2 − x 2 ≤ y ≤ b a  a 2 − x 2 } v(E)=  a −a (  b a √ a 2 −x 2 − b a √ a 2 −x 2 dy)dx =  a −a 2 b a  a 2 − x 2 dx =2ab(arcsin x a + x 2 a  a 2 − x 2 )| a −a = πab. . x 2 ≤ y ≤ b a  a 2 − x 2 } v(E)=  a −a (  b a √ a 2 −x 2 − b a √ a 2 −x 2 dy)dx =  a −a 2 b a  a 2 − x 2 dx =2ab(arcsin x a + x 2 a  a 2 − x 2 )| a −a = πab. . ≤ z ≤ h 2 (x, y)} f C C  C f(x, y, z)dxdydz =  Ω (  h 2 (x,y) h 1 (x,y) f(x, y, z)dz)dxdy E = { x 2 a 2 + y 2 b 2 ≤ 1} E Ox [−a, a] C x = {y : − b a  a 2 − x 2 ≤ y ≤ b a  a 2 − x 2 } v(E)=  a −a (  b a √ a 2 −x 2 − b a √ a 2 −x 2 dy)dx. 1] f(x, y)=x 2 +sin 1 y y =0 f(x, 0) = 0 A = {x 2 +y 2 ≤ 1} A R n f,g A α, β ∈ R αf + βg A  A (αf + βg)=α  A f + β  A g A 1 ,A 2 ⊂ A f A 1 ,A 2  A 1 ∪A 2 f =  A 1 f +  A 2 f −  A 1 ∩A 2 f. f