BOÄ MOÂN TOAÙN ÖÙNG DUÏNG ÑHBK

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TOAÙN 1 GIAÛI TÍCH HAØM MOÄT BIEÁN • BAØI 6: KHAI TRIEÅN TAYLOR •

TS . NGUYEÃN QUOÁC LAÂN (1 2 /2 0 0 7 )

CAÙC ÑÒNH LYÙ TRUNG BÌNH

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Cöïc trò taïi x0: ∃ ε > 0 : ∀ x ∈ (x0 – ε, x0 + ε) ⇒ f(x) ≤ f(x0) Fermat: f ñaït cöïc trò taïi x0 ∈ (a,b) & khaû vi taïi x0 ⇒ f’(x0) = 0 Minh hoaï hình hoïc:

ÑÒNH LYÙ ROLL

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Haøm f(x) lieân tuïc treân [a,b], khaû vi trong (a,

b),

f(a)

⇒ ∃ x0∈(a, b): f’(x0) = 0 Minh hoaï hình hoïc: VD: Chöùng minh phöông

trình

4ax3 + 3bx2 + 2cx – (a + b + c) = 0 coù ít nghieäm trong Giaûi:

nhaát

1

thöïc khoaûng Xeùt

=

f(b)

ÑÒNH LYÙ (SOÁ GIA) LAGRANGE

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Haøm f(x) lieân tuïc treân [a,b], khaû vi trong (a,b) ⇒ ∃ c ∈ (a, b): f(b) – f(a) = f’(c)(b – a) Aùp duïng: Khaûo saùt tính ñôn ñieäu cuûa haøm

y

=

f(x)

baèng ñaïo CMinh haøm VD: BÑThöùc arctgx − arctgy ≤ x − y

KHAI TRIEÅN TAYLOR

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Haøm y = f(x) coù ñaïo haøm taïi x0 ⇒ f(x) ≈ f(x0) + f’(x0)(xthöùc – x0) Taylor: f coù ñaïo haøm caáp n+1 Coâng treân (a,b); x0 , x∈(a, b) ( n) f ' ' ( x0 ) f ( x0 ) 2 f = f ( x0 ) + f ' ( x0 )( x − x0 ) + ( x − x0 ) + ... + ( x − x0 ) n + ? 2! n! n

⇒ f ( x) = ∑ k =0

( n +1) f ( k ) ( x0 ) f ( c) k ( x − x0 ) + ( x − x0 ) n+1 , c ∈ ( x0 , x ) k! (n + 1)!         Rn ( x ) : Phaàn dö

Lagrange CT Taylor (phaàn dö Peano): f coù ñhaøm ñeán caáp n treân (k) ( x ) n f(a,b) 0 ( x − x )k + o ( x − x )n , x → x f ( x) = ∑ 0 0 0 k ! k =0

(

)

KHAI TRIEÅN MAC – LAURINT

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x0 = 0: Khai trieån Mac – Laurint (phoå bieán) n f ( n ) ( 0) n f ( k ) ( 0) k f ( x) = f ( 0) + f ' ( 0) x +  + x + Rn ( x ) = ∑ x + Rn ( x) n! k! k =0 Phaàn

f ( n+1) ( c ) n+1 x , c = c( x ) ∈ ( 0, x ) dö Rn ( x) = (n + 1)!

(

)

Lagrange: n +1 , x→0 Phaàn dö Rn ( x) = o x

Peano: VD: Khai trieån Mac – Laurint cuûa haøm ex

a/

x2 xn b/ cosxx e = 1 + x + +  + + o( x n +1 ) , x → 0 2! n! Keát 2n x2 x4 x n cos x = 1 − + −  + ( − 1) + o( x 2 n +1 ) , x → 0 quaû: 2! 4! ( 2 n )!

MINH HOAÏ KHAI TRIEÅN MAC – LAURINT

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Minh hoaï hình hoïc khai trieån Mac - Laurint haøm p1 ( x) =f(x) x = sinx x3 p2 ( x ) = x − 6

x3 x5 p3 ( x) = x − + 6 120 Chuù yù: Ñoà thò

ña

thöùc

xaáp xæ tieán daàn veà ñoà thò

haøm

KHAI TRIEÅN MAC – LAURINT HAØM CÔ BAÛN

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Haøm löôïng giaùc: sinx, cosx. Haøm tgx (chæ ñeán caáp ba) 3 5 x x ( − 1) n−1 x 2n−1 sin x = x − + +  + + o x 2n , x → 0 3! 5! (2n − 1)!

( )

x2 x4 ( − 1) n x 2n cos x = 1 − + +  + + o x 2 n+1 , x → 0 2! 4! (2n)!

(

)

( )

x3 tgx = x + + o x 4 , x → 0 3 Khai trieån ex: taùch muõ chaün, leû & ñan daáu. cos chaün → muõ chaün; sin leû → muõ leû; tg leû → muõ leû. K0 ñan daáu → shx, chx

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