n N Khi đó a b x n 2 3 n Đ nh lý 14 ị Đ nh lý v dãy trung gian ị ề Gi s ả ử i

N n khi đó a b x n 2 3 n đ nh lý 14 ị đ nh lý

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, ∀ n N Khi đó a b x n 2 3 n
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Đ nh lý 1.4 ( Đ nh lý v dãy trung gian) Gi s : ả ử i) lim a n = lim b n = n n ii) a n z n b n , ∀ n N Khi đó: lim z n =
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Đ nh lý 1.4 (Tiêu chu n Cauchy) Dãy s { x n } có gi i h n h u h n khi và ch khi nó là dãy Cauchy. T c là: > 0 , ∃ N 0 N : m , n > N 0 ta có: | x m x n | < Thí d 1. S d ng tiêu chu n Cauchy đ ch ng minh r ng dãy { a n } v i a n = 1 + 1 + ... + 1 n 2 , n N h i t Ch ng minh: Gi s ả ử > 0 cho tr c tuỳ ý. Khi đó: ướ a n + 1 1 p a n = ( n + 1 ) 2 1 1 + ... + ( n + p ) 2 < 1 < n ( n + 1 ) + ( n + 1 ) ( n + 2 ) + ... + ( n + p 1 ) ( n + p ) = = . 1 1 Σ + . 1 1 Σ + ... + . 1 1 Σ = n n + 1 n + 1 1 1 n + 2 1 n + p 1 1 n + p = n n + p < n < n > , ∀ p > 0 T đó theo tiêu chu n Cauchy thì dãy đã cho là m t dãy h i t . Thí d 2. Xét dãy { a n } v i: a n = 1 + 1 1 n ∈ N . 2 + ... + n , Ta nh n xét r ng n , ta l y s t nhiên ố ự p = n ta thu đ c: ượ 1 1 1 1 1 . a n + p a n . = | a 2 n a n | = n + 1 + n + 2 + ... + 2 n n 2 n = 2. Do đó n u ế l y = 1 thì không t n t i ch s N sao cho p > N p N thì: 2 n + p n . Đi u này ch ng t dãy { a n } không ph i là dãy c ơ b n nên nó là m t dãy phân kì. Đ nh nghĩa 1.8 (Ánh x co) * Đ nh nghĩa: Gi s ả ử f : R n R n , Ánh x f đ c g i là ánh x co n u t n t i ượ ế m t h ng s (0< <1) sao cho: x j , x jj ta có: " f ( x j ) − f ( x jj )" < " x j x jj " . Nh n xét: M i ánh x co đ u liên t c. * Nguyên lý ánh x co: M i ánh x co f : R n R n luôn t n t i m t đi m b t đ ng duy nh t. Ch ng minh: Gi s ả ử x o R n là m t đi m b t kỳ. Đ t: x 1 = f ( x 0 ), x 2 = f ( x 1 ) thì: 2 13 . . . . a a <
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f ( x 1 ) = f ( f ( x 0 )) = f 2 ( x 0 ) 14
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. . . x k = f ( x k 1 ) = f k ( x 0 ) . . . Ta ch ng minh dãy { x k } là dãy Cauchy. Th t v y: k, p là hai s nguyên d ng ươ tùy ý. Rõ ràng ta có: x k + p x k = f ( x ) − f ( x ) = p f ( x p ) − f ( x ) ≤ ... ≤ x p x k 0 M t khác: x p x = x p x p 1 + x p 1 x p 2 + ... + x 1 x x p x p 1 + x p 1 x p 2 + ... + x 1 x . Nh ng: ư x 1 x = x 1 x " x 2 x 1 " = f ( x 1 ) − f ( x 0 ) ≤ x 1 x 0 . . . x p x p 1 ≤ ... ≤ p 1 x 1 x 0 . C ng v ế v i v ế các b t đ ng th c trên ta đ c: ượ x p x x p x p 1 + x p 1 x p 2 + ... + x x 0 ≤ ( 1 + 2 p 1 0 k k + p k v i tùy ý, do đó dãy { x k } là dãy Cauchy. Do đó x R n sao cho: lim " x k x " = 0, do ánh x co là ánh x liên t c nên ta có: lim " f ( x ) − f ( x )" k 0 f ( x ) = x nên ta có: lim " f ( x ) − x " = 0 k k = . M t khác k k + 1 k k + Bây gi ta ch ng m nh s duy nh t: Gi s t n t i hai đi m b t đ ng x , x ∗∗ . Khi đó ta có: f ( x ) = x ; f ( x ∗∗ ) = x ∗∗ . T đó ta có: " x x
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