toan_1_Bai_07_Kynang_Taylor - bookbooming

BOÄ MOÂN TOAÙN ÖÙNG DUÏNG ÑHBK

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TOAÙN 1 GIAÛI TÍCH HAØM MOÄT BIEÁN • BAØI 7: KYÕ NAÊNG KHAI TRIEÅN TAYLOR •

TS . NGUYEÃN QUOÁC LAÂN (12/2007)

KHAI TRIEÅN CÔ BAÛN: MUÕ, LGIAÙC, HYPERBOLIC

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Töø khai trieån haøm y = e ⇒ Khai trieån sinx, chaüncoshx cosx,Muõ sinhx, n x2 x4 ( − 1) x 2 n cos x = 1 − + − ... + + o( x 2 n +1 ), x → 0 2 2! 4!

(2n)! x e x = 1 + x + + ...

2! x3 ( − 1) n x 2 n +1 ( 2 n + 2 ) sin x = x − + ... + +o x , x→0 3!

(2n + 1)!

Muõleû Töông töï nhö sin x, cos x nhöngkhoâng ñan daáu → shx, chx 2 2n x3 x 2 n +1 x x shx = x + + ... + + o( x 2 n + 2 ), chx = 1 + + ... + + o( x 2 n +1 ) 3! ( 2n + 1)!

2!

(2n)!

( )

x3 tgx = x + + o x 4 , x → 0 3

Chuù yù phaàn dö cosx, sinx, chx, shx: o nhoû cuûa

KHAI TRIEÅN CÔ BAÛN: LUYÕ THÖØA, 1/(1 ± x), LN(1 + x) -------------------------------------------------------------------------------------------------------------------------------------

Haøm nghòch ñaûo – inverse function (Toång

caáp soá nhaân): 1 1 n n n n 2 ( ) = 1+ x ++ x + o x , = 1 − x + x +  + ( − 1) x + o( x n ) 1− x 1+ x Toång quaùt: Haøm luyõ thöøa (1 + x)α → Nhò n thöùc Newton (1 + x) (1 + x ) α = 1 + αx + α ( α − 1) x 2 +  + α  ( α − n + 1) x n + o( x n ) 2! n!

3 VD: Khai trieån MacLaurint f ( x ) = 3 1 + x ñeáncaáp 2 3 1 x 1 1 x 1 1 1 x      3 haøm Gia (1 + x ) 3 = 1 + +  − 1 +  − 1 − 2  + o( x ), x → 0 3 3  3  2! 3  3  3  3! ûi: ln(1 + x): ∫1/ x 2 x3 (−1) n −1 n ln (1 + x ) = x − + +  + x + o( x n ) 2 3 n (1+x) → xn/n,

BAÛNG KHAI TRIEÅN CAÙC HAØM CÔ BAÛN: 7 HAØM ---------------------------------------------------------------------------------------------------------------------------

Haøm ex cos x sin x 1 1+ x 1 1− x

(1 + x )

α

ln (1 + x )

Khai trieån x2 x3 xn 1+ x + + + + 2! 3! n!

2n x2 x4 x n 1 − + −  + ( − 1) x 2n 2! 4! ( 2 n )!

2 n +1 x3 x5 x n x − + −  + ( − 1) x 2 n +1 3! 5! ( 2n + 1)!

1 − x + x − x +  + ( − 1) x n 2

3

n

Phaàn dö ec Lagrange x n +1 ( n + 1)! cos( sin ) c 2 n + 2 x ( 2 n + 2 )! cos( sin ) c 2 n +3 x ( 2n + 3)! ( − 1) n+1 1 n + 2 x n+1 (1 + c )

1 + x + x 2 + x3 +  + x n

α ( α − 1) 2 α  ( α − n + 1) n 1 + αx + x + + x 2! n! n x 2 x3 x 4 n +1 x x − + − +  + ( − 1) 2 3 4 n

PPHAÙP KHTRIEÅN MACLAURINT: TOÅNG, HIEÄU, TÍCH

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Ñöa haøm caàn khai trieån veà daïng toång, hieäu, tích (ñhaøm, tphaân) caùc haøm cô baûn. Aùp duïng kh/tr MacLaurint cô baûn 2 x − 5 ln(1 + x ) VD: Khai trieån ML ñeán f ( x ) = e + 1− x 2 caáp 3:     x2 x 2 3 ( ) Gia f ( x ) = 1 + x + + ...  + 2 1 + x + x + ... − 5 x − + ...  + o( x ) 2 2     ûi: VD: Khai trieån MacLaurint f ( x ) = cos x ⋅ cosh x x2 ñeán caáp 3: x 2 3  3  3 Gia f ( x ) = 1 − + o( x ) 1 + + o( x )  = 1 + o( x ), x → 0 2!  2!   ûi: Chuù yù: Coù theå söû duïng caû ñaïo haøm, tích

(

)

phaân (coi chöøng C!) VD: Khai trieån ML ñeán f ( x ) = ln x + 1 + x 2 + 1

KHTRIEÅN MACLAURINT HAØM THÖÔNG: DUØNG 1/(1 ± x) -----------------------------------------------------------------------------------------------------------------------------------------

1 Vôùi thöông (tyû soá, phaân soá) 2 1± x haøm soá: Duøng Chuù yù: ÔÛ maãu soá baét buoäc

phaûi xuaát hieän soá 1! ex 1 , caáp 2 b/ , caáp 3 VD: Khai trieån a / 2+ x cos x 2 2 MacLaurint    1 1 1 x x x x 2 2  = 1 + x + + o( x ) 1 − + + o( x )  Gia a / e ⋅ ⋅ 2 1+ x 2 2  2!  2 4  2 ûi: 2 2 x 1 1 3  x 3  b/ = = 1 +  + o( x )  +  + o( x )  + ...

2 3 cos x 1 − ( x 2! + o( x ) ) 2  2  1 VD: Khai trieån MacLaurint f ( x ) = 2 x − 4x + 3 ñeán caáp 2 1 1 1 1  1 1 1 1  =  − =  ⋅ − Gia f ( x ) =  ( x − 1)( x − 3) 2  x − 1 x − 3 2 3 1 − x 3 1 − x 

KHAI TRIEÅN MACLAURINT VÔÙI HAØM HÔÏP

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Haøm hôïp f(u(x)): Khai trieån laàn löôït töøng böôùc. Ñaàu tieân khai trieån MacLaurint u(x), sau ñoù khai trieån f(u) & caét ñeán luyõ thöøa ñöôïc yeâu caàu Luoân (Coù theå ñoåitra thöù Chuù yù quan troïng: kieåm töï). ñieàu kieän u(0) = 0!

2 4 VD: Khai trieån a / sin ( x ) b / cos x ñeáncaáp 3 u MacLaurint a / u = x 2 & u ( 0 ) = 0 ⇒ sin u = u − + ... = x 2 + o( x 4 ) Gia 3!

12   ûi: 2 4    x2 x4   x x 1 4 4  b / 1 −  − + o( x )  = 1 +  − + + o( x )  = 1 + u + ...

2  2 24     2  24        u VD (caûnh giaùc!): Khtrieån MacLaurint y = ln(2 +

KHAI TRIEÅN TAYLOR QUANH x – x0: ÑÖA VEÀ KTR ML

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Khai trieån Taylor f(x) quanh x = x0: Ñoåi bieán t = x – x0 vaø söû duïng khai trieån Mac Laurint Caùch 2: Bieán ñoåi ñeå (x – x0) xuaát hieän tröïc cho haøm f(t) tieáp trong haøm soá!

1 3 VD: Khai trieån Taylor f ( x ) = quanh x0 = 2 ñeáncaáp x 1 1 1 1 1 t haøm  f ( x ) = = = ⋅ = 1 + +  Giaûi: Caùch 1: t = x   x t + 2 2 1+ t 2 2  2  –2⇒ 1 1 1 = ⋅ Caùch 2: Taïo (x – 2) f ( x ) = ( x − 2) + 2 2 1 + ( x − 2) 2 trong haøm 2 VD: Khai trieån Taylor f ( x ) = 3 x quanh x0 = −8 ñeáncaáp 13 haøm3 ( x + 2 )  1  − ( x + 2)     ( x + 2) − 8 = −21 − = −21 + ⋅  Gia  + ...   8  8    3   ûi:

ÖÙNG DUÏNG KT TAYLOR. TÌM GIÔÙI HAÏN

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Tìm lim: Khai trieån ML vôùi phaàn dö Peano + Ngaét boû VCB VD: Tính VD: Tìm VD: Tìm

3  x3 4  x 4 x −  x − + o( x )  ( ) + o x 6 x − sin x   = lim 6 lim = lim 3 3 3 x →0 x →0 x →0 x x x

sin 3 x + 4 sin 3 x − 3 ln (1 + x ) sin 3 x − 3 ln(1 + x ) lim = lim x →0 x →0 x2 e x − 1 sin x

(

)

ln (1 + x )   1 lim  − 2  x →0 x( 1 + x ) x  

1   2  (SGK/8 lim  x − x ln1 +  x →∞   x  0)

x − (1 + x ) ln(1 + x ) x →0 x 2 (1 + x )  x − ln (1 + x ) x ln (1 + x )  = lim  2 − 2 x →0 x (1 + x )   x (1 + x )

= lim

ÖÙNG DUÏNG KT TAYLOR. TÍNH GAÀN ÑUÙNG --------------------------------------------------------------------------------------------------------------------------------

Tính gaàn ñuùng & öôùc löôïng sai soá: phaàn dö Lagrange ( n +1) (k) n f ( c) f ( x0 ) n +1 k f ( x) ≈ ∑ ( x − x0 ) , ∆ = Rn = x − x0 , c ∈ ( x0 , x ) k! (n + 1)! k =0 VD: Tính gaàn ñuùng giaù trò soá e vôùi ñoä chính c -4 1 1 1 e 3 xaùc 10 (SGK/79) , c ∈ ( 0,1) ⇒ e ≈ S , ∆ ≤ Gia e = 1 + + +  + + ( n + 1)!  1!  2!   n! ( n + 1)! ûi: S Töông töï: Caàn choïn bao nhieâu soá haïng trong khai trieån haøm y = ex ñeå coù theå -4 xaáp xæ e vôùi ñoä chính xaùc 10 VD: Goùc x naøo cho pheùp xaáp xæ sinx ≈ x

vôùi ñoä chính xaùc 10-4

VI PHAÂN

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Haøm khaû vi taïi x0 ⇔ ∆y = A∆x + o(∆x), ∆x → 0 : Soá gia haøm soá bieåu dieãn tuyeán tính theo ∆x y ( C ) : y = f ( x) vaø voâ cuøng beù baäc cao cuûa ∆x Vi phaân: dy = A∆x = f ( x0 + ∆x ) f’(x)dx Nhaän xeùt: Haøm coù ∆y ñaïo haøm ⇔ Coù vi f ( x0 ) ∆x f ' ( x ) ∆x phaân: Haøm khaû vi 1/ C: haèng soá ⇒ 0 dC = 0 & d(Cy) = Cdy Vi 2/

phaân

toång,

hieäu,

tích, thöông:

O

d ( u ± v ) = du ± dv d ( uv ) = vdu + udv

x0

x0 + ∆x

 u  vdu − udv d  = v2 v

x

VI PHAÂN HAØM HÔÏP

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 y = f ( x ) , x : bieánñoäclaäp  Vi phaân   ⇒ dy = y ' dx  y = f ( x ) , x = x( t ) : haømhôïp caáp 1: ⇒ Vi phaân caáp 1: baát bieán!

VD: Tính dy cuûa a/ y = sinx

b/ y =

sinx, b /=dycost = cos xdx = − cos x sin tdt hoaëcy = sin ( cos t ) ⇒ dy =  Gia x ûi: Vi

2 2 3 phaân x : Bieánñoäclaäp⇒ d y = f ' ' dx , d y = 

caáp y = f ( cao: x ) , x = x( t ) ⇒ d 2 y = f ' ' dx 2 + f ' d 2 x

(d

2

x = x' ' dt 2 )

VD: Tính d2y: a/ y = arctgx b/ y = arctgx, 2x sin t 2 2 2 2 2 x = sint dx b / d y = y ' ' dx − dt ÑS: a / d y = − 2 2 2 1+ x (1 + x )