1. 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)

2. KHAI TRIEÅN CÔ BAÛN: MUÕ, LGIAÙC, HYPERBOLIC
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Töø khai trieån haøm y = ex  Khai trieån sinx, cosx,
sinhx,Muõ
coshx
chaün
x2 x4
cos x  1    ... 
 1 x 2 n  ox 2 n1 , x  0
n
x2 2! 4! (2n)!
e x  1  x   ...
2!
x3
sin x  x   ... 
 1 x 2 n1  ox 2 n 2  , x  0
n
Muõ
leû 3! (2n  1)!

Töông töï nhö sin x, cos x nhöng khoâng daáu shx, chx
ñan
x 2 n 1
 ox 2 n  2 , chx  1   ...   ox 2 n 1 
x3 x2 x 2n
shx  x   ... 
3! 2n  1! 2! (2n)!
x3
 
tgx  x   o x 4 , x  0
Chuù yù phaàn dö cosx, sinx,
3 chx, shx: o nhoû cuûa soá

3. KHAI TRIEÅN CÔ BAÛN: LUYÕ THÖØA, 1/(1  x), LN(1 + x)
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Haøm nghòch ñaûo – inverse function (Toång caáp
soá nhaân):
 1  x    x n  ox n ,  1  x  x 2     1 x n  ox n 
1 1 n
1 x 1 x
Toång quaùt: Haøm luyõ thöøa (1 + x)  Nhò thöùc
Newton (1 + x)n
   1 2     n  1 n
1  x   1  x  x  x  ox n 
2! n!
VD: Khai trieån MacLaurint f  x   3 1  x ñeán 3 caáp
haøm
Giaûi 1  x 3  1     1    1  2   ox 3 , x  0
1 x 1  1  x 2 1  1  1  x 3
3 3  3  2! 3  3  3  3!
:
ln(1 + x): 1/(1+x) (1) n1 n
x  ox n 
x 2 x3
ln1  x   x     
 xn/n, ñan 2 3 n

4. BAÛNG KHAI TRIEÅN CAÙC HAØM CÔ BAÛN: 7 HAØM
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Haøm Khai trieån Phaàn dö
x 2 x3 xn Lagrange
ec
ex 1 x     x n 1
2! 3! n! n  1!
x2 x4 n x
2n
cossin c 2 n  2
cos x 1       1 x 2n x
2! 4! 2n ! 2n  2!
x3 x5 x 2 n 1 2 n 1 cossin c 2 n 3
x       1
n
sin x x x
3! 5! 2n  1! 2n  3!
1
1  x  x 2  x 3     1 x n
n
 1n1 1 n 2 x n1
1 x 1  c 
1
1  x  x 2  x3    x n
1 x
   1     n  1
1  x  
1  x  x  
2
xn
2! n!
x 2 x3 x 4 n 1 x
n
ln1  x  x        1
2 3 4 n

5. 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
VD: Khai trieån ML ñeán f  x   e x   5 ln1  x 
1 x
caáp 3:
Giaûi f  x   1  x   ...  21  x  x 2  ...  5 x   ...  ox 3 
 x2   x2

 2   2 
:
VD: Khai trieån MacLaurint ñeán f  x   cos x  cosh x
caáp 3:
Giaûi f  x   1   ox 1   ox 3   1  ox 3 , x  0
 x2 3  x2

 2!  2! 
:
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  

6. KHTRIEÅN MACLAURINT HAØM THÖÔNG: DUØNG 1/(1  x)
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Vôùi thöông (tyû soá, phaân soá) 2 1
1 x
haøm soá: Duøng
Chuù yù: ÔÛ maãu soá baét buoäc phaûi
xuaát hieän soá 1!
VD: Khai trieån a / ex 1
, caáp b/
2 , caáp 3
2 x cos x
MacLaurint 1  x x 2 2 
 1  x   ox 1    ox 
1
Giaûi a / e x   1 x2 2
2 1 x 2 2  2!  2 4 
: 2
x 3  x 3 
 1    ox     ox   ...
2 2
1 1

cos x 1  x 2!  ox 
b/ 2 3
2  2 
VD: Khai trieån MacLaurint ñeán f  x   2 1
x  4x  3
caáp 2
1 1 1 1  1 1 1 1 
Giaûi f  x     
 x  1 x  3 2  x  1 x  3   2 3  1  x 3  1  x 
:

7. 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
ñöôïcyù quan troïng:theå ñoåi thöù tra ñieàu
Chuù yeâu caàu (Coù Luoân kieåm töï).
kieän u(0) = 0!
VD: Khai trieån a / sin x 2  b / cos x ñeán 4
caáp
MacLaurint 2
Giaûi a / u  x & u 0  0  sin u  u   ...  x 2  ox 4 
u3
3! 12
:  
 x2 x4    x2 x4 4 

b / 1     ox   1      ox   1  u  ...
4 1
 2 24     24 
 
2
 
2
 u 
VD (caûnh giaùc!): Khtrieån MacLaurint y = ln(2 + x)

8. 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 cho haøm
f(t) 2: Bieán ñoåi ñeå (x – x0) xuaát hieän tröïc tieáp
Caùch
trong haøm soá!
VD: Khai trieån Taylor f  x   1 quanh x0  2 ñeán 3 caáp
x
haøm
Giaûi: Caùch 1: t = x – 2 f  x   1  1  1  1  1 1  t  
 
x t  2 2 1 t 2 2  2 

Caùch 2: Taïo (x – 2) trong f  x   1 1 1
 
 x  2  2 2 1   x  2 2
haøm
VD: Khai trieån Taylor f  x   3 x quanh x0  8 ñeán 2 caáp
haøm  x  21 3  21  1     x  2   ...
Giaûi 3  x  2  8  21     

 8    3  8  
:

9. ÖÙNG DUÏNG KT TAYLOR. TÌM GIÔÙI HAÏN
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Tìm lim: Khai trieån ML vôùi phaàn dö Peano +
Ngaét boû VCB
 4 
x   x   ox 
x3
 ox 4 
x3
VD: Tính lim x  sin x  lim    lim 6
6
x0 x 3 x 0 x3 x 0 x3
VD: Tìm lim sin 3x  4 sin 3 x  3 ln1  x  sin 3 x  3 ln1  x 
 lim
x0 e  1 sin x
x
  x 0 x2
VD: Tìm lim  1  ln1  x  x  1  x  ln1  x 
 lim
x 0  x 1  x  x2  x 2 1  x 
  x 0
 x  ln1  x  x ln1  x 
 lim  2  2
x 0
 x 1  x  x 1  x  

(SGK/80 lim  x  x 2 ln1  1 
 

x   x 
)

10. ÖÙNG DUÏNG KT TAYLOR. TÍNH GAÀN ÑUÙNG
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Tính gaàn ñuùng & öôùc löôïng sai soá: phaàn
dö Lagrange
n
f k 
 x0  f  n 1 c 
f ( x)    x  x0  ,   Rn  , c   x0 , x 
n 1
x  x0
k
k 0 k! (n  1)!
VD: Tính gaàn ñuùng giaù trò soá e vôùi ñoä chính xaùc
10-4 (SGK/79) 1 1 1 ec 3
Giaûi e  1       , c  0,1  e  S ,  
:  n  1!
1! 2! n! n  1!
S
Töông töï: Caàn choïn bao nhieâu soá haïng trong
khai trieån haøm y = ex ñeå coù theå xaáp xæ e
vôùi ñoä x naøo cho10-4
VD: Goùc chính xaùc pheùp xaáp xæ sinx  x vôùi ñoä
chính xaùc 10-4

11. 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 vaø voâ
cuøng beù baäc cao cuûa x y C  : y  f  x 
Vi phaân: dy = Ax =
f  x0  x 
f’(x)dx
Nhaän xeùt: Haøm coù
y
ñaïo haøm  Coù vi
f  x0 
phaân: Haøm khaû dC
1/ C: haèng soá  vi x f '  x0 x
= 0 & d(Cy) = Cdy
O x0 x0  x x
2/ Vi phaân
toång, hieäu, d u  v   du  dv  u  vdu  udv
d  
tích, thöông: d uv   vdu  udv v v2

12. VI PHAÂN HAØM HÔÏP
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Vi phaân  y  f  x , x : bieán laäp 
ñoäc
 y  f  x , x  xt  : haøm   dy  y ' dx
 
hôïp
caáp 1:
 Vi phaân caáp 1: baát
bieán!
VD: Tính dy cuûa a/ y = sinx b/ y = sinx, x
= cost b / dy  cos xdx   cos x sin tdt hoaëc  sin cos t   dy  
Giaûi y
:
Vi phaân caáp x : Bieán laäp d 2 y  f ' ' dx 2 , d 3 y  
ñoäc 
cao: f  x , x  xt   d 2 y  f ' ' dx 2  f ' d 2 x
y d 2
x  x' ' dt 2 
VD: Tính d2y: a/ y = arctgx b/ y = arctgx, x =
sint
ÑS: a / d 2 y   2x sin t 2
dx 2
b / d y  y ' ' dx 
2 2
dt
1  x 
2 2 1 x 2