Giáo trình giải tích 2 part 4 pdf

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Giáo trình giải tích 2 part 4 pdf

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lim (x,y)→(0,0) xy(x + y) x 2 + y 2 =0     xy(x + y) x 2 + y 2     ≤ 1 2 |x 2 + y 2 ||x + y| |x 2 + y 2 | ≤|x + y|→0 (x, y) → (0, 0). lim (x,y)→(0,0) sin xy x = lim (x,y)→(0,0) sin xy xy x =1.0=0. lim (x,y)→(0,0) x − y x + y (x k ,y k )=( 1 k , 1 k ) (x  k ,y  k )=( 1 k , 0) (0, 0) f(x k ,y k ) → 0 f(x  k ,y  k ) → 1. f(x, y) (x 0 ,y 0 ) f a 12 = lim y→y 0 lim x→x 0 f(x, y),a 21 = lim x→x 0 lim y→y 0 f(x, y),a= lim (x,y)→(x 0 ,y 0 ) f(x, y). x 0 =0,y 0 =0 f(x, y)=(x + y)sin 1 x sin 1 y . a 12 ,a 21 a =0 f(x, y)= x 2 − y 2 x 2 + y . a 12 =0,a 21 =1 a f(x, y)= xy x 2 + y 2 . a 12 = a 21 =0 a f(x, y)=x sin 1 y . a 12 =0 a 21 a =0 a = a 12 = a 21 f : X ×Y → R m x 0 ,y 0 X, Y lim y→y 0 f(x, y)=g(x), ∀x ∈ X. lim x→x 0 f(x, y)=h(y) y ∀>0, ∃δ>0:x ∈ X, d(x, x 0 ) <⇒ d(f(x, y),h(y)) <, ∀y ∈ Y. lim (x,y)→(x 0 ,y 0 ) f(x, y) = lim x→x 0 lim y→y 0 f(x, y) = lim y→y 0 lim x→x 0 f(x, y). x lim x→∞ f(x)=L, lim x→a f(x)=∞, lim x→∞ f(x)=∞. R n a ∈ R n a = ∞ F a (R n , R m ) a R n R m a f,ψ ∈ F a (R n , R m ) f = o(ψ) x → a ⇔ lim x→a f(x) ψ(x) =0. f,g,ψ ∈ F a (R n , R m ) f = o(ψ) g = o(ψ) x → a f + g = o(ψ) x → a f = o(ψ) x → a g <f,g>= o(ψ) x → a f,ψ ∈ F a (R n , R m ) f = O(ψ) x → a ⇔∃C>0,r >0:f (x) ≤Cψ(x), ∀x ∈ B(a, r). f,g,ψ ∈ F a (R n , R m ) f = O(ψ) g = O(ψ) x → a f + g = O(ψ) x → a f = O(ψ) x → a g <f,g>= O(ψ) x → a o(ψ),O(ψ) f = o(ψ) g = o(ψ) f = g f,g ∈ F a (R n , R) f ∼ g x → a ⇔ lim x→a f(x) g(x) =1. ∼ n →∞ P (n)=a p n p + a p−1 n p−1 + ···+ a 0 ∼ a p n p (a p =0) 1+2+···+ n = n(n +1) 2 = O(n 2 ) 1 2 +2 2 + ···+ n 2 = n(2n +1)(n +2) 6 = O(n 3 ) n! ∼  n e  n √ 2πn = O   n e  n+ 1 2  2 n ,n p , ln q n, n p ln q n n → +∞ p ∈ N 1 p +2 p + ···+ n p = O(n p+1 ) n →∞ f : X → R m X ⊂ R n a ∈ X lim x→a f(x)=f(a). (, δ) f a ∀>0, ∃δ>0: B(a, δ) ⊂ f −1 (B(f (a),) f : R n → R m f R n V ⊂ R m f −1 (V ) F ⊂ R m f −1 (F ) C(X, R m ) f : X → R m X f a f a m =1 f(x 1 , ··· ,x n )=x i (i =1, ··· ,n) e x ln x sin x arcsin x f(x 1 , ··· ,x n )=  0≤i 1 ···i n ≤N a i 1 ,···,i n x i 1 1 ···x i n n R n x → x i ,x→ a T : R n −→ R m T (αx + βy)=αT (x)+βT(y), ∀x, y ∈ R n ,α,β ∈ R. T m n (a ij ) m×n T (e j )= m  i=1 a ij e i ,j=1, ··· ,m. y = Tx    y 1 y m    =    a 11 a 12 ··· a 1n a m1 a m2 ··· a mn       x 1 x n    1 T ∃M>0: Tx≤Mx, ∀x ∈ R n . T  =max x=1 Tx , T f : K −→ R m f K f(K) f : K → R K ⊂ R n f K a,b ∈ K f(a)=sup x∈K f(x),f(b)= inf x∈K f(x) (y k ) f(K) x k ∈ K, y k = f(x k ) K x σ(k) ) x ∈ K f (y σ(k) = f(x σ(k) )) f(x) ∈ f(K) f(K) m =1 f(K) M =supf(K) m =inff(K) f(K) a, b ∈ K f(a)=M,f(b)=m  f : K −→ R m f K f K ∀>0, ∃δ>0:x, x  ∈ K, d(x, x  ) <δ =⇒ d(f(x),f(x  )) <. f ∃>0, ∀k ∈ N, ∃x k ,x  k ∈ K : d(x k ,x  k ) < 1 k , d(f(x k ),f(x  k )) ≥ . K (x σ(k) ) (x k ) x ∈ K d(x σ(k) ,x  σ(k) )) < 1 σ(k) (x  σ(k) ) (x  k ) x f d(f(x σ(k) ),f(x  σ(k) )) d(f(x),f(x)) = 0  f(x)= 1 x ,x∈ (0, +∞) E E E R N : E → R (N1)(N2)(N3) R n x → max 1≤i≤n |x i | x →  n i=1 |x i | x N 1 ,N 2 M,m mN 1 (x) ≤ N 2 (x) ≤ MN 1 (x), ∀x ∈ E. E f 1 , ··· ,f n E T : E → R n ,x 1 f 1 + ···+ x n f n → (x 1 , ··· ,x n ) N E = T −1 ◦ N E N R n E N R n  R n E S n−1 = {x ∈ R n : x =1} N S n−1 M =max x∈S n−1 N(x) m = min x∈S n−1 N(x) M,m > 0 x ∈ R n \{0} x x ∈ S n−1 (N2) mx≤N(x) ≤ Mx  f : C → R m f C f(C) f : C → R f C f(C) a, b ∈ C µ ∈ R f(a) <µ<f(b) c ∈ C : f (c)=µ f C f(C) f f f(C) A, B f(C) f U, V f −1 (A)=C ∩ U f −1 (B)=C ∩V U, V C C R 1  f :[a, b] → [a, b] x ∗ ∈ [a, b] f(x ∗ )=x ∗ f :[a, b] → R f(b),f(a) (x k ) f(x)=0 f : S n −→ R,n ≥ 1, x 0 ∈ S n f(x 0 )=f(−x 0 ) S n = {x ∈ R n+1 : x =1} g(x)=f(x) −f(−x) g S n g(S n ) R g(x)g(−x) ≤ 0 g(S n ) 0  M ⊂ R n f : M → M d ∃θ, 0 <θ<1: d(f ( x) ,f(y)) ≤ θd(x, y), ∀x, y ∈ M. f ∃!x ∗ ∈ M : f(x ∗ )=x ∗ x 0 ∈ M (x k ) x 1 = f(x 0 ),x k+1 = f(x k )(k =2, 3, ···) (x k ) x ∗ f (x k ) d(x k+1 ,x k )=d(f(x k ),f(x k−1 ) ≤ θd(x k ,x k−1 ) ≤···≤θ k d(x 1 ,x). m =1, 2, ··· d(x k+m ,x k ) ≤ d(x k+m ,x k+m− 1 )+···+ d(x k+1 ,x k ) ≤ (θ k+m + ···θ k )d(x 1 ,x 0 ) ≤ θ k 1 − θ d(x 1 ,x) → 0, k →∞. (x k ) lim x k = x ∗ M x ∗ ∈ M f f f(x ∗ )=x ∗ ¯x ∈ M f f(¯x)=¯x d(¯x, x ∗ )=d(f(¯x),f(x ∗ )) ≤ θd(¯x, x ∗ ). θ<1 d(¯x, x ∗ )=0 ¯x = x ∗  f : R → R 0 <θ<1 |f  (x)| <θ,∀x |f(x) − f(y)| = |f  (c)||x −y|≤θ|x − y|, ∀x, y ∈ R f f : M → M d(f(x),f(y)) <d(x, y), ∀x, y ∈ M,x = y f(x)=x + 1 x x ∈ M =[1, ∞) T : R n → R n (t ij ) T n  i,j=1 t 2 ij < 1hay n  i,j=1 |t ij | < 1 n max 1≤i,j≤n |t ij | < 1 (f k ) k∈N f k : X → R m ,X⊂ R n (f k ) f x ∈ X lim f k (x)=f(x) (f k ) X f ∀>0, ∃N(): k ≥ N ⇒ d(f k (x),f(x)) ≤ , ∀x ∈ X, M k =sup x∈X d(f k (x),f(x)) lim k→∞ M k =0 X ∞  k=0 f k = f 0 + f 1 + ···+ f k + ··· , f k : X → R m kS k = f 0 + f 1 + ···+ f k X (S k ) X R f k (x)=    1 − 1 k |x| |x|≤k, 0 |x| >k. (f k ) f(x) ≡ 1 sup x∈R |f k (x) − f(x)| =1 → 0, k →∞. ∞  k=0 x k f(x)= 1 1 − x [−1, 1) 0 ≤ r<1 [−r, r] S k (x)=1+x + ···+ x k = 1 − x k+1 1 − x [−r, r] sup |x|≤r |S k (x) − f(x)| =sup |x|≤r      x k+1 1 − x      = r k+1 1 − r → 0, k →∞. f (−1, 1) (f k ) X ∀>0, ∃N : k, l > N ⇒ d(f k (x),f l (x)) ≤ , ∀x ∈ X.  R [a, b] 1+ 1 1! x + ···+ 1 k! x k + ··· e x x − 1 3! x 3 + ···+ 1 (2k +1)! x 2k+1 + ··· sin x 1+ 1 2! x + ···+ 1 2k! x 2k + ··· cos x f k (x)=x k ,x ∈ [0, 1] (f k ) (f k ) f k k (f k ) f f x 0 >0 N k>N d(f(x),f k (x)) <  3 d(f(x 0 ),f k (x 0 )) <  3 k f k x 0 δ>0 d(f k (x),f k (x 0 )) <  3 d(x, x 0 ) <δ. d(f(x),f(x 0 )) ≤ d(f(x),f k (x)) + d(f k (x),f k (x 0 )) + d(f k (x 0 ),f(x 0 )) <, f x 0  R n X ⊂ R n BF(X,R n ) f : X → R n f ∈ BF(X, R m ) ⇔∃M>0:f(x)≤M,∀x ∈ X. f =sup x∈X f(x) BF(X,R m ) R n d(f,g)=f − g,f,g∈ BF(X, R m ) f,f k ∈ BF(X, R m ) (f k ) k∈N f f k → f BF(X,R m ) d(f k ,f) → 0 k →∞ X C(X, R m ) g :[a, b] → R a = a 0 <a 1 < ···<a k = b g g(x)=A i x + B i ,x∈ [a i−1 ,a i ],i=1, ···,k g A i ,B i g f :[a, b] → R f [a, b] f f f g : K → R K X 1 , ··· ,X k g f :[a, b] → R f [a, b] f max, min f f R n f [a, b] f f x = a + t(b − a) [a, b]=[0, 1] f B k (x)=B k f(x)= k  p=0 C p k f( p k )x p (1 − x) k−p . (x + y) k = k  p=0 C p k x p y k−p x x kx(x + y) k−1 = k  p=0 pC p k x p y k−p x 2 k(k −1)x 2 (x + y) k−2 = k  p=0 p(p − 1)C p k x p y k−p y =1−x r p (x)=C p k x p (1 −x) k−p k  p=0 r p (x)=1, k  p=0 pr p (x)=kx, k  p=0 p(p − 1)r p (x)=k(k −1)x 2 . k  p=0 (p − kx) 2 r p (x)=k 2 x 2  p=0 r p (x) − 2kx k  p=0 pr p (x)+ k  p=0 p 2 r p (x)=kx = k 2 x 2 − 2kx +(kx + k(k −1)x 2 )=kx(1 − x) M =max |x|≤1 |f(x)| >0 δ>0 |x − y| <δ |f(x) − f(y)| < f(x) − B k (x)=f(x) − k  p=0 C p k f( p k )x p (1 −x) k−p = k  p=0 (f(x) − f( p k ))r p (x).  1 p : | p k − x| <δ |f(x) −f( p k )| < r p (x) ≥ 0 |  1 |≤ k  p=0 r p (x)=. . a p n p (a p =0) 1 +2+ ···+ n = n(n +1) 2 = O(n 2 ) 1 2 +2 2 + ···+ n 2 = n(2n +1)(n +2) 6 = O(n 3 ) n! ∼  n e  n √ 2 n = O   n e  n+ 1 2  2 n ,n p , ln q n, n p ln q n n → +∞ p ∈ N 1 p +2 p + ···+. a 12 ,a 21 a =0 f(x, y)= x 2 − y 2 x 2 + y . a 12 =0,a 21 =1 a f(x, y)= xy x 2 + y 2 . a 12 = a 21 =0 a f(x, y)=x sin 1 y . a 12 =0 a 21 a =0 a = a 12 = a 21 f : X ×Y → R m x 0 ,y 0 X, Y lim y→y 0 f(x,. lim (x,y)→(0,0) xy(x + y) x 2 + y 2 =0     xy(x + y) x 2 + y 2     ≤ 1 2 |x 2 + y 2 ||x + y| |x 2 + y 2 | ≤|x + y|→0 (x, y) → (0, 0). lim (x,y)→(0,0) sin xy x =

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