Giáo trình giải tích 2 part 1 potx

TRƯỜNG ĐẠI HỌC ĐÀ LẠT KHOA TOÁN - TIN HỌC Y  Z TẠ LÊ LI GIẢI TÍCH 2 (Giáo Trình) Lưu hành nội bộ Y Đà Lạt 2008 Z R n R n 5 R n R n R n R n R n X f n : X → R n ∈ N (f n ) n∈N x ∈ X (f n (x)) n∈N D = {x ∈ X : (f n (x)) n∈N } (f n ) D  x → f(x) = lim n→∞ f n (x) (f n ) f D f n (x)=1− 1 n |x| (n ∈ N) R R f(x) = lim n→∞ (1 − 1 n |x|)=1, ∀x. f n (x)=x n (n ∈ N) R (−1, 1] f(x) = lim n→∞ x n =  0 |x| < 1 1 x =1 f n f (f n (x)) x ∈ D (f n ) f D >0 N n ≥ N ⇒|f n (x) − f(x)| <, ∀x ∈ D M n =sup x∈D |f n (x) − f(x)|→0 n →∞ M n =sup|f n (x) − f(x)| =1 (f n ) (g n ) f g D (f n + g n ) (cf n ) f + g cf D (f n ) D ∀>0, ∃N : n, m ≥ N ⇒ sup x∈D |f n (x) − f m (x)| < (f n ) f D ∀>0, ∃N : n ≥ N ⇒ sup x∈D |f n (x) − f(x)| </2 m, n ≥ N sup x∈D |f n (x) − f m (x)| < sup x∈D |f n (x) − f(x)|+sup x∈D |f m (x) − f(x)| <. (f n ) D x ∈ D (f n (x)) f(x) ∈ R m →∞  → 0 sup x∈D |f n (x) −f(x)|→0 n →∞ (f n ) f D  (f n ) f D f D lim lim n→∞ lim x→x 0 f n (x) = lim x→x 0 lim n→∞ f n (x) (f n ) [a, b] lim  lim n→∞  b a f n (x)dx =  b a lim n→∞ f n (x)dx (f n ) [a, b] (f  n ) [a, b] (f n (c)) c ∈ [a, b] (f n ) f [a, b] lim lim n→∞ f  n (x)=  lim n→∞ f n (x)   x 0 ∈ D >0 N |f N (x) − f(x)| </3, ∀x ∈ D. f N x 0 δ>0 |f N (x) −f N (x 0 )| </3, ∀x, |x −x 0 | <δ. |x − x 0 | <δ |f(x)−f(x 0 )|≤|f(x)−f N (x)|+|f N (x)−f N (x 0 )|+|f N (x 0 )−f(x 0 )| </3+/3+/3= f x 0 lim x→x 0 f(x) = lim x→x 0 lim n→∞ f n (x)=f(x 0 ) = lim n→∞ lim x→x 0 f n (x) f n f [a, b]       b a f n −  b a f      ≤|b −a| sup x∈[a,b] |f n (x) − f(x)|→0, n →∞ lim n→∞  b a f n =  b a f =  b a lim n→∞ f n F n (x)=  x c f  n (F n ) F [a, b] F (x)=  x c lim n→∞ f  n F n (x)=f n (x) − f n (c) f n = F n + f n (c) [a, b] f = F + lim n→∞ f n (c) f  (x)=F  (x)=  lim n→∞  x c f  n   = ( lim n→∞ f n )  (x)  X ∞  k=0 f k = f 0 + f 1 + ···+ f n + ··· f k X n S n = f 0 + ···+ f n D = {x ∈ X : (S n (x)) n∈N } S(x)= ∞  k=0 f k (x) D ∞  k=0 f k D (S n ) n∈N S D M n =sup x∈D |S n (x) − S(x)| =sup x∈D | ∞  k=n+1 f k (x)|→0, n →∞ ∞  k=0 x k =1+x + x 2 + ···+ x n + ··· D = {x ∈ R : |x| < 1} S(x)= 1 1 − x D r = {x : |x|≤r} 0 <r<1 S n (x)= 1 − x n+1 1 − x sup |xleqr |S n (x) − S(x)| =sup |x|≤r      x n+1 1 − x      ≤ r n+1 1 − r → 0, n →∞ D sup |x|≤1 |S n (x) − S(x)| =+∞ ∞  k=0 f k D ∀>0, ∃N : n, m ≥ N ⇒ sup x∈D | m  k=n f k (x)| < ∞  k=0 f k [a, b] f k [a, b] k ∈ N [a, b] lim  lim x→x 0 ∞  k=0 f k (x)= ∞  k=0 lim x→x 0 f k (x) f k [a, b]    b a  ∞  k=0 f k (x)  dx = ∞  k=0   b a f k (x)dx  f k [a, b] ∞  k=0 f  k [a, b] ∞  k=0 f k [a, b]   ∞  k=0 f k   (x)= ∞  k=0 f  k (x) |f k (x)|≤a k , ∀x ∈ D ∞  k=0 a k ∞  k=0 f k D (f k ) 0 ∞  k=0 ϕ k D ∞  k=0 f k ϕ k D (f n ) ∞  k=0 ϕ k D ∞  k=0 f k ϕ k |f k (x)|≤a k m  k=n |f(x)|≤ m  k=n a k ∞  k=0 f k  ∞  k=0 a k x k x 0 ∞  k=0 a k (x − x 0 ) k z = x − x 0 x 0 S(x)= ∞  k=0 a k (x −x 0 ) k R, 0 ≤ R ≤ +∞ R>0 S(x) |x − x 0 | <R |x − x 0 | >R S D r = {x : |x − x 0 |≤r} 0 <r<R R S 1 R = lim sup k→∞ k  |a k | x 0 0 z = x −x 0 |z|≤r<R ρ : r<ρ<R lim sup k 0 |a k | 1 k < 1 ρ , ∀k>k 0 . |a k z k | <  r ρ  k S(z) D r S(z) |z| <R |z| >R ρ : R<ρ<|z| lim sup k |a k | 1 k > 1 ρ |a k z k | >  |z| ρ  k k a k z k → 0 ∞  k=0 a k z k  1 R = lim k→∞ |a k+1 | |a k | ∞  k=0 k!x k R = lim k→∞ |a n | |a n+1 | = lim n→∞ k! (k +1)! =0 ∞  k=0 x k k! ∞ |x−x 0 | = R ∞  k=0 x k , ∞  k=1 x k k , ∞  k=1 x k k 2 1 |x| =1 ∞  k=0 x k x = ±1 ∞  k=1 x k k 2 |x| =1 ∞  k=1 x k k x =1 x = −1 ∞  k=0 a k (x − x 0 ) k R>0 S(x)= ∞  k=0 a k (x −x 0 ) k (x 0 −R, x 0 + R)  ∞  k=0 a k (x − x 0 ) k   = ∞  k=1 ka k (x − x 0 ) k−1   ∞  k=0 a k (x − x 0 ) k  dx = ∞  k=0 a k k +1 (x − x 0 ) k+1 + C  ∞  k=0 (−1) k x k = 1 1+x |x| < 1 ∞  k=1 (−1) k kx k−1 = − 1 (1 + x) 2 |x| < 1 ∞  k=0 (−1) k x k+1 k +1 =ln(1+x) |x| < 1 1 1+x 2 = 1 1 − (−x 2 ) =1−x 2 + x 4 − x 6 + ···= ∞  k=0 (−1) k x 2k , |x| < 1 arctan x = x − x 3 3 + x 5 5 − x 7 7 + ···= ∞  k=0 (−1) k x 2k+1 2k +1 , |x| < 1 f k (x)=x k ϕ k (x)=a k ∞  k=0 a k S S(x)= ∞  k=0 a k x k |x| < 1 lim x→1 − S(x)=S ln 2 = 1 − 1 2 + 1 3 − 1 4 + 1 5 −···+ (−1) n+1 n +1 + R n π 4 =1− 1 3 + 1 5 − 1 7 + 1 9 −···+ (−1) n 2n +1 + R n R n O( 1 n ) f x 0 f(x)= ∞  k=0 a k (x − x 0 ) k a k = f (k) (x 0 ) k! k =0, 1, 2, ··· n ∈ N x x 0  ∞  k=0 a k (x − x 0 ) k  (n) = ∞  k=n k(k −1) ···(k −n +1)a k (x − x 0 ) k−n x = x 0  f x 0 f x 0 Tf(x)= ∞  k=0 a k (x − x 0 ) k , a k = f (k) (x 0 ) k! Tf(x)=f(x) Tf(x) f(x)= ∞  k=0 sin 2 k x k! Tf(x) Tf(x) = f(x) f(x)=e − 1 x 2 x =0 f (0) = 0 f (k) (0) = 0, ∀k Tf(x) ≡ 0 = f(x) Tf(x)=f(x), |x − x 0 | <R f D = {x : |x − x 0 | <R} f C |f (k) (x)|≤C, ∀x ∈ (x 0 − R, x 0 + R) f x ∈ (x 0 −R, x 0 + R) θ ∈ (0, 1) |f(x) −T n (x)| = |R n (x)| =      f (n+1) (x 0 + θR) (n +1)! (x − x 0 ) n+1      ≤ CR n+1 (n +1)! . 1 ∞  k=0 ( 1) k x k +1 k +1 =ln (1+ x) |x| < 1 1 1+ x 2 = 1 1 − (−x 2 ) =1 x 2 + x 4 − x 6 + ···= ∞  k=0 ( 1) k x 2k , |x| < 1 arctan x = x − x 3 3 + x 5 5 − x 7 7 + ···= ∞  k=0 ( 1) k x 2k +1 2k +1 ,. < 1 f k (x)=x k ϕ k (x)=a k ∞  k=0 a k S S(x)= ∞  k=0 a k x k |x| < 1 lim x 1 − S(x)=S ln 2 = 1 − 1 2 + 1 3 − 1 4 + 1 5 −···+ ( 1) n +1 n +1 + R n π 4 =1 1 3 + 1 5 − 1 7 + 1 9 −···+ ( 1) n 2n. x 0 ) k   = ∞  k =1 ka k (x − x 0 ) k 1   ∞  k=0 a k (x − x 0 ) k  dx = ∞  k=0 a k k +1 (x − x 0 ) k +1 + C  ∞  k=0 ( 1) k x k = 1 1+x |x| < 1 ∞  k =1 ( 1) k kx k 1 = − 1 (1 + x) 2 |x| < 1 ∞  k=0 ( 1) k x k +1 k