Giáo trình giải tích 2 part 8 ppt

A g Dg g : U −→ R n C 1 U ⊂ R n A A ⊂ U g det Dg =0 A f : g(A) −→ R f ◦ g|det Dg| A  g(A) f =  A f ◦ g|det Dg|. g ∈ C 1 A g(A) f ◦ g|det Dg| A g ∈ C 1 det Dg(a) =0 U a a g =(Φ n ◦ T n ) ◦···◦(Φ 1 ◦ T 1 ), T i (x)=a + σ i (x − a) σ i Φ i (x 1 , ··· ,x n )=(x 1 , ··· ,φ i (x), ··· ,x n )(i =1··· ,n) det Dg(a) =0 i, ∂g n ∂x i (a) =0. B(x)=a + σ(x −a) σ n i h = g ◦B ∂h n ∂x n (a)= ∂g n ∂x i (a) =0. Φ(x)=(x 1 , ··· ,x n−1 ,h n (x)) Φ ∈ C 1 det DΦ(a)= ∂h n ∂x n (a) =0 U a Φ Φ −1 ∈ C 1 g = h ◦ B −1 = G ◦ Φ ◦ T T = B −1 = a + σ −1 (x − a) G(x)=(h 1 (x), ··· ,h n−1 (x),x n ) G g(x):=T (x)=a + σ(x − a) σ |det T | =1 U a g(x):=Φ i (x)=(x 1 , ··· ,φ i (x), ··· ,x n ) i = n U a = S × [a n .b n ],S R n−1 Φ(U)=S × φ n (U) det DΦ= ∂φ n ∂x n .  Φ(U) f =  S (  φ n (x 1 ,···,x n−1 ,[a n ,b n ]) f(x)dx n )dx 1 ···dx n−1 . =  S (  b n a n f(x 1 , ··· ,φ n (x))| ∂φ n ∂x n |dx n )dx 1 ···dx n−1 =  U f ◦ Φ|det DΦ|. T Φ Φ ◦ T  Φ◦T (C) f =  T (C) f ◦ Φ|det DΦ| =  C f ◦ Φ ◦ T |det(DΦ) ◦ T ||det DT| =  C f ◦ (Φ ◦ T )|det D(Φ ◦ T )|. A A P S ∈ P S ∩ A = ∅ g S  g(A) f =  S∈P S∩A=∅  g(A∩S) f =  S∈P S∩A=∅  A∩S f ◦ g|det Dg| =  A f ◦ g|det Dg|.  g : U −→ R n ,U R n A A ⊂ U g A f g(A) f ◦ g|det Dg| A  g(A) f =  A f ◦ g|det g|. µ{x :detDg(x)=0} =0 A = A ∪ ∂A ∂g(A) ⊂ g(∂A) • g : R 2 −→ R 2 g(r, ϕ)=(r cos ϕ, r sin ϕ) | D(x, y) D(r, ϕ) | = |det Dg(r, ϕ)| =      cos ϕ −r sin ϕ sin ϕrcos ϕ      = r. ✲ ✻ 0 R 2π r ϕ ✲ g ✲ ✻ 0 R x y ϕ ✟ ✟ ✟ ✟ ✟✯ r A ⊂ R 2 g intA f A  g(A) f(x, y)dxdy =  A f(r cos ϕ, r sin ϕ)rdrdϕ.  x 2 +y 2 ≤R 2 e −x 2 −y 2 dxdy =  2π 0  R 0 e −r 2 rdrdϕ = π(1 − e −R 2 ). R → +∞  ∞ −∞ e −x 2 dx = √ π • g : R 3 −→ R 3 (r, ϕ, z) → (r cos ϕ, r sin ϕ, z). |det g(r, ϕ, z)| = r A ⊂ R 2 g intA f A  g(A) f(x, y, z)dxdydz =  A f(r cos ϕ, r sin ϕ, z)rdrdϕdz A = {x 2 + y 2 ≤ 1, 0 ≤ z ≤ 1}  A z 1+x 2 + y 2 dxdydz =  1 0  2π 0  1 0 z 1+r 2 rdrdϕdz =  1 0 zdz  2π 0 dϕ  1 0 r 1+r 2 dr = π 2 ln 2. • g : R 3 −→ R 3 , (ρ, ϕ, θ) → (ρ cos ϕ sin θ, ρsin ϕ sin θ, ρcos θ) |det Dg(ρ, ϕ, θ)| = ρ 2 sin θ s M 0 ✓ ✓ ✓ ✓✼ θ ϕ ρ ✁ ✁ ✁ ✁☛ x ✲ y ✻ s z  g(A) f(x, y, z)dxdydz =  A f(ρ cos ϕ sin θ, ρ sin ϕ sin θ, ρ cos θ)ρ 2 sin θdρdϕdθ, A g intA f  x 2 +y 2 +z 2 ≤R 2  x 2 + y 2 + z 2 dxdydz =  R 0  2π 0  π 0 ρ 3 sin θdρdϕdθ =  R 0 ρ 3 dρ  2π 0 dϕ  π 0 sin θdθ = R 4 4 2π.2. A R n λ ∈ R λ : R n → R n ,λ(x)=λx (y 1 , ··· ,y n )=(λx 1 , ··· ,λx n ) v(λ(A)) =  λ(A) dy =  A λ n dx = λ n v(A) v 1 , ··· ,v n ∈ R n A = {y ∈ R n : y = x 1 v 1 + ···+ x n v n , 0 ≤ x 1 , ··· ,x n ≤ 1} v(A)=|det(v 1 , ··· ,v n )| = |det(v ij ) 1≤i,j≤n | v i =(v i1 , ··· ,v in ) 1+x + x 2 + x 3 + ··· = 1 1 − x 1 x + 1 x 2 + 1 x 3 + ··· = 1 x(1 − 1/x) = 1 x − 1 ···+ 1 x 3 + 1 x 2 + 1 x +1+x + x 2 + x 3 + ··· =0 x =0, 1 ∞  k=1 x k k 2 ∞  k=0 k!x k ∞  k=0 k k +1 (x −1) k ∞  k=2 (x +2) k ln k ∞  k=1 1 2 k k x k ∞  k=1 (−1) k k x 2k+1 ∞  k=1 1 k  x 2  2k ∞  k=0 x k ∞  k=0 (k +1)x k ∞  k=0 x k+1 k +1 ∞  k=0 (k 2 +2k − 2)x k ∞  k=1 (−1) k x k k ∞  k=0 x 2k+1 2k +1 ∞  k=0 2k 2 +1 k! x k f (a, b) f c ∈ (a, b) f(x)=e − 1 x 2 (x =0),f(0) = 0 f f 0 f(x)=sin 3 x f(x)= x (1 − x)(1 − x 2 ) f(x)= 1 (1 − x) n f(x)= √ a + x f(x)=ln  1+x 1 − x (x)= 1 √ 2  x 0 e − t 2 2 dt (1) (x)=  x 0 sin t t dt (1) f T  a+T a f(x)dx =  T 0 f(x)dx, ∀a f [−π, π] a k ,b k a  k ,b  k f f  a  k = kb k ,b  k = −ka k f(x)=0 −π ≤ x ≤ 0 f (x)=1 0 <x≤ π f(x)=|x|, −π ≤ x ≤ π π 4 =1− 1 3 + 1 5 − 1 7 + ··· va π 2 8 = 1 1 2 + 1 3 2 + 1 5 2 + ··· x := 0, ±π f(x) t ∈ R \ Z f t (x)=costx, |x|≤π f t (x) tπ = 1 tπ + ∞  k=1 2t π(t 2 − k 2 ) π 2 sin t π = lim n→+∞ n  k=−n 1 (t − k) 2 sin tπ tπ = lim n→+∞ (1 − t 2 1 )(1 − t 2 2 ) ···(1 − t 2 n 2 ) f(x)=e x ,x∈ [0, 2π] f(x)=0 x ∈ [0,l] f(x)=1 x ∈ (l, 2l) f(x)=x, x ∈ (−2, 2) cos f(x)=1 x ∈ [0,π/2] f(x)=0 x ∈ (π/2,π] f(x)=x(π −x),x∈ [0,π] sin x =2 ∞  k=1 (−1) k+1 k sin kx, −π<x<pi x 2 ,x 3 ,x 4 −π<x<π R n |d(x, y) − d(y, z)|≤d(x, z). m, M, a, b > 0 x =(x 1 , ···,x n ) ∈ R n m max 1≤i≤n |x i |≤x≤M max 1≤i≤n |x i |. ax≤ n  i=1 |x i |≤bx. x, y ∈ R n | <x,y>| = xy, x + y = x + y. f,g [a, b]       b a fg      ≤   b a f 2  1/2   b a g 2  1/2 . T : R n → R m M T (h)≤Mh, ∀h ∈ R n . R n {x : x≤1}, {x : x =1}, {x =(x 1 , ···,x n ):x i ∈ Q,i=1, ···,n} a ∈ R x 1 = a, x k = x 2 k−1 − x k−1 +1 a (x k ) (x k ) (x k ) (x k ) lim k→∞ x k+1 x k = L>0 lim k→∞ k √ x k = L ∀>0, ∃A, B > 0,N ∈ N : k ≥ N ⇒ A(L − ) k <x k < B(L + ) k . x k = k k k! lim k→∞ k (k!) 1/k = e lim k→∞ x k = M (x k ) lim k→∞ 1 k (x 1 + ···+ x k )=M 0 <x k <y k x k+1 =(x k y k ) 1 2 y k+1 = 1 2 (x k +y k ) x k <y k (x k ), (y k ) (x k ) R n x k+1 −x k  < 1 k 2 + k , ∀k (x k ) x k =1+ 1 2 + ···+ 1 k (k ∈ N) R [a, b] (a, b) (a, b] B(a, r) R n A R [0, 1] A A [0, 1] U k (k ∈ N) R n  k∈N U k  k∈N U k  k∈N (R n \ U k )  k∈N (R n \ U k ) X ⊂ Y X ⊂ Y X ∪ Y = X ∪ Y ∂(X ∪Y ) ⊂ ∂X∪∂Y ∂(X ∩Y ) ⊂ ∂X∩∂Y ∂(X ×Y )=∂X× Y ∪X × ∂Y. X R n X Z R {x ∈ R n : x < 1} X x ∈ X d x > 0 d(x, y) ≥ d x , ∀y ∈ X. X K X ∩ K = ∅ d>0 d(x, y) ≥ d, ∀x ∈ K, ∀y ∈ X. X, K X ∩K = ∅ d>0 (F k ) R n k ∈ N F k F k ⊃ F k+1 (F k )=sup{d(x, y):x, y ∈ F k }→0 k →∞  k∈N F k I k =(0, 1 k ),k ∈ N  k I k = ∅ {(x, y) ∈ R 2 : |x|≤1} {(x, y) ∈ R 2 : x 4 + y 4 =1} {x ∈ R n : x≤r} {x ∈ R n :1≤x≤2} {x ∈ R n : x =1} Z [0, 1] K R n R n \ K K R n R n \ K L i ,i ∈ I, L i  L j = ∅ i, j  i∈I L i C C C = {(x, y) ∈ R 2 :0<y≤ x 2 ,x =0}∪{(0, 0)} C C (0, 0) C C = {(x, y) ∈ R 2 : y =sin 1 x ,x =0}∪{(0,y):|y|≤1} C C C C F 0 =[0, 1] F 1 =[0, 1 3 ] ∪ [ 2 3 , 1] F 0 F 2 =[0, 1 9 ] ∪ [= 2 9 , 1 3 ] ∪ [ 2 3 , 7 9 ] ∪ [ 8 9 , 1] F 1 F k F k−1 F k 2 k [ k 3 k , k +1 3 k ] C =  k F k x ∈ C x =  ∞ k=0 a k 3 k a k ∈{0, 2} C C (C)=∅ C C 1 3 +2. 1 9 + ···+2 k−1 /3 k + ···=1 (0, 0) f(x, y)= x 2 y x 4 + y 2 n → +∞ n 2 +n ∼ n 2 , √ n +1− √ n ∼ √ n 2 , (−1) n n 2 = O(n 2 ),n 2 +2 = o(n 3 ), sin n = O(1) x → 0 (1 + x) α =1+αx + o(x) (1 + x) α ∼ 1+αx e x =1+x + o(x) e x ∼ 1+x ln(1 + x)=x + o(x)ln(1+x) ∼ x sin x = x + o(x)sinx ∼ x cos x =1− 1 2 x 2 + o(x)cosx ∼ 1 − 1 2 x 2 x →∞ o, O x α ,x β ,a x ,b x , log c x, log d x (α, β, a, b, c, d > 0) f(x, y)= xy(x + y) x 2 + y 2 (x, y) =(0, 0) f(0, 0) = 0. f(x, y)= e xy − 1 2xy xy =0 f(x, y)=0 xy =0 f(x, y)= sin xy x x =0 f ((0,y)=y f(x, y)= xy x 2 + y 2 ,f(0, 0) = 0) f : R 2 → R (0, 0) f(x, y)= x 2 y x 4 + y 2 (x, y) =(0, 0) f(0, 0) = 0 f : R → R f(x + y)=f(x)+f(y), ∀x, y ∈ R f 0 f g : R → R g(x + y)=g(x)g(y), ∀x, y ∈ R g 0 g f :[0, 1] −→ [0, 1] f(x)= 1 q x = p q f(x)=0 x f f : R n → R m f R n f −1 (V ) V ⊂ R m f −1 (F ) F ⊂ R m U ⊂ R {(x, y) ∈ R 2 : x ∈ U} f : R n → R {x ∈ R n :0≤ f(x) ≤ 1} f : R → R U ⊂ R f(U) {A ∈ Mat(n, n) : det A =0} (n, n) n R f : R → R {x : f(x)=0}{x : f (x) > 1}{f(x):x ≥ 0}{f (x):0≤ x ≤ 1} f : R 2 → R {f(x, y):x 2 + y 2 =1} X ⊂ R n d(x, X)= inf y∈X d(x, y) R n  x → d(x, X) |d(x, X) −d(x  ,X)|≤d(x, x  ) . − t 2 1 )(1 − t 2 2 ) ···(1 − t 2 n 2 ) f(x)=e x ,x∈ [0, 2 ] f(x)=0 x ∈ [0,l] f(x)=1 x ∈ (l, 2l) f(x)=x, x ∈ ( 2, 2) cos f(x)=1 x ∈ [0,π /2] f(x)=0 x ∈ (π /2, π] f(x)=x(π −x),x∈ [0,π] sin x =2 ∞  k=1 (−1) k+1 k sin. ϕ, r sin ϕ, z)rdrdϕdz A = {x 2 + y 2 ≤ 1, 0 ≤ z ≤ 1}  A z 1+x 2 + y 2 dxdydz =  1 0  2 0  1 0 z 1+r 2 rdrdϕdz =  1 0 zdz  2 0 dϕ  1 0 r 1+r 2 dr = π 2 ln 2. • g : R 3 −→ R 3 , (ρ, ϕ,. −1) k ∞  k =2 (x +2) k ln k ∞  k=1 1 2 k k x k ∞  k=1 (−1) k k x 2k+1 ∞  k=1 1 k  x 2  2k ∞  k=0 x k ∞  k=0 (k +1)x k ∞  k=0 x k+1 k +1 ∞  k=0 (k 2 +2k − 2) x k ∞  k=1 (−1) k x k k ∞  k=0 x 2k+1 2k