Bài tập toán cao cấp part 7 pot

8.3. C´ac di . nh l´y co . ba ’ nvˆe ` h`am kha ’ vi 95 21. lim x→0+0  cotgx  tgx .(DS. 1) 22. lim x→0  5 2+ √ 9+x  1/ sin x .(DS. e −1/30 ) 23. lim x→0  cos x  cotg 2 x .(DS. e −1/2 ) 24. lim x→0+0  ln2x  1/lnx .(DS. 1) 25. lim x→0  1 + sin 2 x  1/tg 2 x .(DS. e) 26. lim x→0+0  cotgx  1/lnx .(DS. e −1 ) 27. lim x→π/2  sin x  tgx .(DS. 1) 28. lim x→0 e +x − e −x − 2x sin x −x .(DS. −2) 29. lim x→0 e −x − 1+x − x 2 2 e x 3 − 1 .(DS. − 1 6 ) 30. lim x→0 e −x − 1+x 4 sin 2x .(DS. − 1 2 ) 31. lim x→0 2 x − 1 − xln2 (1 − x) m − 1+mx .(DS. ln 2 2 m(m −1) ) 32. lim x→0  2 π arccosx  1/x .(DS. e − 2 π ) 33. lim x→∞ lnx x α , α>0. (DS. 0) 34. lim x→∞ x m a x ,0<a= 1. (DS. 0) 35. lim x→0+0 ln sin x ln(1 − cos x) .(DS. 1 2 ) 36. lim x→0  1 x 2 − cotg 2 x  .(DS. 2 3 ) 37. lim x→ π 4  tgx  tg2x .(DS. e −1 ) 38. lim x→ π 2 −0  tgx  cotgx .(DS. 1) 96 Chu . o . ng 8. Ph´ep t´ınh vi phˆan h`am mˆo . tbiˆe ´ n 8.3.3 Cˆong th´u . c Taylor Gia ’ su . ’ h`am f(x) x´ac di . nh trong lˆan cˆa . n n`ao d´ocu ’ adiˆe ’ m x 0 v`a n lˆa ` n kha ’ vi ta . idiˆe ’ m x 0 th`ı f(x)=f(x 0 )+ f  (x 0 ) 1! (x − x 0 )+ f  (x 0 ) 2! (x − x 0 ) 2 + ···+ + f (n) (x 0 ) n! (x − x 0 ) n + o((x − x 0 ) n ) khi x → x 0 hay: f(x)= n  k=0 f (k) (x 0 ) k! (x −x 0 ) k + o((x − x 0 ) n ),x→ x 0 . (8.15) Dath´u . c P n (x)= n  k=0 f (k) (x 0 ) k! (x −x 0 ) k (8.16) d u . o . . cgo . il`adath´u . c Taylor cu ’ a h`am f(x)ta . idiˆe ’ m x 0 , c`on h`am: R n (x)=f( x) − P n (x) d u . o . . cgo . il`asˆo ´ ha . ng du . hay phˆa ` ndu . th ´u . n cu ’ a cˆong th´u . c Taylor. Cˆong th ´u . c (8.15) du . o . . cgo . i l`a cˆong th´u . c Taylor cˆa ´ p n dˆo ´ iv´o . i h`am f(x)ta . i lˆan cˆa . ncu ’ ad iˆe ’ m x 0 v´o . i phˆa ` ndu . da . ng Peano (n´o c˜ung c`on d u . o . . cgo . i l`a cˆong th´u . c Taylor di . aphu . o . ng). Nˆe ´ u h`am f(x)c´oda . o h`am dˆe ´ ncˆa ´ p n th`ı n´o c´o thˆe ’ biˆe ’ udiˆe ˜ n duy nhˆa ´ tdu . ´o . ida . ng: f(x)= n  k=0 a k (x − x 0 ) k + o((x − x 0 ) n ),x→ x 0 v´o . ic´achˆe . sˆo ´ a k du . o . . c t´ınh theo cˆong th´u . c: a k = f (k) (x 0 ) k! ,k=0, 1, ,n. 8.3. C´ac di . nh l´y co . ba ’ nvˆe ` h`am kha ’ vi 97 Nˆe ´ u x 0 = 0 th`ı (8.15) c´o da . ng f(x)= n  k=0 f (k) (0) k! x k + o(x n ),x→ 0 (8.17) v`a go . i l`a cˆong th´u . c Macloranh (Maclaurin). Sau d ˆay l`a cˆong th´u . c Taylor ta . i lˆan cˆa . nd iˆe ’ m x 0 =0cu ’ amˆo . tsˆo ´ h`am so . cˆa ´ p I. e x = n  k=0 x k k! + o(x n ) II. sin x = x − x 3 3! + x 5 5! + ···+ (−1) n x 2n+1 (2n + 1)! + o(x 2n+2 ) = n  k=0 (−1) k x 2k+1 (2k + 1)! + o(x 2n+2 ) III. cos x = n  k=0 (−1) k x 2k (2k!) + o(x 2n+1 ) IV. (1 + x) α =1+ n  k=1 α(α −1) (α −k +1) k! x k + o(x n ) =1+ n  k=1  α k  x k + o(x n ) α(α −1) (α − k +1) k! =            α k   nˆe ´ u α ∈ R, C k α nˆe ´ u α ∈ N. Tru . `o . ng ho . . p riˆeng: IV 1 . 1 1+x = n  k=0 (−1) k x k + o(x n ), IV 2 . 1 1 − x = n  k=0 x k + o(x n ). 98 Chu . o . ng 8. Ph´ep t´ınh vi phˆan h`am mˆo . tbiˆe ´ n V. ln(1 + x)= n  k=1 (−1) k−1 k x k + o(x n ). ln(1 − x)=− n  k=1 x k k + o(x n ). Phu . o . ng ph´ap khai triˆe ’ n theo cˆong th´u . c Taylor Nhu . vˆa . y, d ˆe ’ khai triˆe ’ n h`am f(x) theo cˆong th´u . c Taylor ta pha ’ i´ap du . ng cˆong th´u . c f(x)=T n (x)+R n+1 (x), T n (x)= n  k=0 a k (x − x 0 ) k , a k = f (k) (x 0 ) k! · (8.18) 1) Phu . o . ng ph´ap tru . . ctiˆe ´ p: du . . a v`ao cˆong th´u . c (8.18). Viˆe . csu . ’ du . ng cˆong th´u . c (8.18) dˆa ˜ nd ˆe ´ nnh˜u . ng t´ınh to´an rˆa ´ tcˆo ` ng kˆe ` nh m˘a . c d`u n´o cho ta kha ’ n˘ang nguyˆen t˘a ´ cd ˆe ’ khai triˆe ’ n. 2) Phu . o . ng ph´ap gi´an tiˆe ´ p: du . . a v`ao c´ac khai triˆe ’ n c´o s˘a ˜ n I-V sau khi d˜abiˆe ´ ndˆo ’ iso . bˆo . h`am d˜a cho v`a lu . u´ydˆe ´ n c´ac quy t˘a ´ c thu . . chiˆe . n c´ac ph´ep to´an trˆen c´ac khai triˆe ’ n Taylor. Nˆe ´ u f(x)= n  k=0 a k (x − x 0 ) k + o((x − x 0 ) n ) g(x)= n  k=0 b k (x − x 0 ) k + o((x − x 0 ) n ) th`ı a) f(x)+g(x)= n  k=0 (a k + b k )(x − x 0 ) k + o((x − x 0 ) n ); 8.3. C´ac di . nh l´y co . ba ’ nvˆe ` h`am kha ’ vi 99 b) f(x)g(x)= n  k=0 c k (x − x 0 ) k + o((x − x 0 ) n ) c k = k  p=0 a p b k−p c) F (x)=f[g(x)] = n  j=0 a j  n  k=0 (b k (x −x 0 ) k − x 0  j +o   n  k=0 b k (x −x 0 ) k − x 0  n  3) D ˆe ’ khai triˆe ’ n c´ac phˆan th´u . ch˜u . uty ’ theo cˆong th´u . c Taylor thˆong thu . `o . ng ta biˆe ’ udiˆe ˜ n phˆan th´u . cd ´odu . ´o . ida . ng tˆo ’ ng cu ’ ad ath´u . c v`a c´ac phˆan th´u . cco . ba ’ n (tˆo ´ i gia ’ n !) rˆo ` i´apdu . ng VI 1 ,IV 2 . 4) Dˆe ’ khai triˆe ’ n t´ıch c´ac h`am lu . o . . ng gi´ac thˆong thu . `o . ng biˆe ´ ndˆo ’ i t´ıch th`anh tˆo ’ ng c´ac h`am. 5) Nˆe ´ u cho tru . ´o . c khai triˆe ’ nd a . o h`am f  (x) theo cˆong th´u . c Taylor th`ı viˆe . c t`ım khai triˆe ’ n Taylor cu ’ a h`am f(x)d u . o . . c thu . . chiˆe . nnhu . sau. Gia ’ su . ’ cho biˆe ´ t khai triˆe ’ n f  (x)= n  k=0 b k (x − x 0 ) k + o((x − x 0 ) n ), b k = f (k+1) (x 0 ) k! · Khi d´otˆo ` nta . i f (n+1) (x 0 )v`adod´o h`am f(x) c´o thˆe ’ biˆe ’ udiˆe ˜ ndu . ´o . i da . ng f(x)= n+1  k=0 a k (x − x 0 ) k + o((x − x 0 ) n+1 ) = f(x 0 )+ n  k=0 a k+1 (x − x 0 ) k+1 + o((x − x 0 ) n+1 ) trong d ´o a k+1 = f (k+1) (x 0 ) (k + 1)! = f (k+1) (x 0 ) k! · 1 k +1 = b k k +1 · 100 Chu . o . ng 8. Ph´ep t´ınh vi phˆan h`am mˆo . tbiˆe ´ n Do d´o f(x)=f(x 0 )+ n  k=0 b k k +1 (x − x 0 ) k+1 + o((x − x 0 ) n+1 ) (8.19) trong d ´o b k l`a hˆe . sˆo ´ cu ’ adath´u . c Taylor dˆo ´ iv´o . i h`am f  (x). C ´ AC V ´ IDU . V´ı du . 1. Khai triˆe ’ n h`am f(x) theo cˆong th´u . c Maclaurin dˆe ´ nsˆo ´ ha . ng o(x n ), nˆe ´ u 1) f(x)=(x +5)e 2x ;2)f(x)=ln 3+x 2 − x Gia ’ i 1) Ta c´o f(x)=xe 2x +5e 2x . ´ Ap du . ng I ta thu du . o . . c f(x)=x  n−1  k=0 2 k x k k! + o(x n−1 )  +5  n  k=0 2 k x k k! + o(x n )  = n−1  k=0 2 k k! x k+1 + n  k=0 5 · 2 k k! x k + o(x n ). V`ı n−1  k=0 2 k x k+1 k! = n  k=1 2 k−1 (k −1)! x k nˆen ta c´o f(x)=5+ n  k=1  2 k−1 (k −1)! + 5 · 2 k k!  x k + o(x n ) = n  k=0 2 k−1 k! (k + 10)x k + o(x n ). 2) T`u . d˘a ’ ng th´u . c f(x)=ln 3 2 +ln  1+ x 3  −ln  1 − x 2  v`a V ta thu d u . o . . c f(x)=ln 3 2 + n  k=1 1 k  1 2 k + (−1) k−1 3 k  x k + o(x n ).  8.3. C´ac di . nh l´y co . ba ’ nvˆe ` h`am kha ’ vi 101 V´ı du . 2. Khai triˆe ’ n h`am f(x) theo cˆong th´u . c Taylor ta . i lˆan cˆa . nd iˆe ’ m x 0 = −1dˆe ´ nsˆo ´ ha . ng o((x +1) 2n )nˆe ´ u f(x)= 3x +3 √ 3 − 2x − x 2 · Gia ’ i. Ta c´o f(x)= 3(x +1)  4 − (x +1) 2 = 3 2 (x +1)  1 − (x +1) 2 4  − 1 2 . ´ Ap du . ng cˆong th´u . c IV ta thu du . o . . c f(x)= 3 2 (x +1)+ 3 2 (x +1) n−1  k=1   − 1 2 k   (−1) k (x +1) 2k 4 k + o((x +1) 2n ) trong d ´o   − 1 2 k   (−1) k =(−1) k  − 1 2  − 1 2 − 1   − 1 2 −(k − 1)  k! = (2k −1)!! 2 k k! · Do d ´o f(x)= 3 2 (x +1)+ n−1  k=1 3(2k −1)!! 2 3k+1 k! (x +1) 2k+1 + o((x +1) 2n ).  V´ı du . 3. Khai triˆe ’ n h`am f(x) theo cˆong th´u . c Taylor ta . i lˆan cˆa . nd iˆe ’ m x 0 =2dˆe ´ nsˆo ´ ha . ng o((x −2) n ), nˆe ´ u f(x)=ln(2x −x 2 +3). Gia ’ i. Ta biˆe ’ udiˆe ˜ n 2x − x 2 +3=(3− x)(x + 1) = [1 − (x − 2)][3 + (x − 2)] = 3[1 − (x − 2)]  1+ x − 2 3  . 102 Chu . o . ng 8. Ph´ep t´ınh vi phˆan h`am mˆo . tbiˆe ´ n T`u . d ´o suy ra r˘a ` ng f(x) = ln3 + ln[1 −(x − 2)] + ln  1+ x − 2 3  v`a ´ap du . ng cˆong th´u . cVtathud u . o . . c f(x) = ln3 − n  k=1 1 k (x − 2) k + n  k=1 (−1) k−1 (x − 2) k k3 k + o((x − 2) n ) = ln3 + n  k=1  (−1) k−1 3 k − 1  (x − 2) k k + o((x − 2) n ).  V´ı du . 4. Khai triˆe ’ n h`am f(x) = ln cos x theo cˆong th ´u . c Maclaurin dˆe ´ nsˆo ´ ha . ng ch´u . a x 4 . Gia ’ i. ´ Ap du . ng III ta thu du . o . . c ln(cos x)=ln  1 − x 2 2 + x 4 24 + o(x 4 )  = ln(1 + t), trong d´otad˘a . t t = − x 2 2 + x 4 24 + o(x 4 ). Tiˆe ´ p theo ta ´ap du . ng khai triˆe ’ nV ln(cos x) = ln(1 + t)=t − t 2 2 + o(t 2 ) =  − x 2 2 + x 4 24 + o(x 4 ) − 1 2  − x 2 2 + x 4 4 + o(x 4 )  2  + o   − x 2 2 + x 4 24 + o(x 4 )  2  = − x 2 2 + x 4 24 − x 4 8 + o(x 4 )=− x 2 2 − x 4 12 + o(x 4 ).  V´ı du . 5. Khai triˆe ’ n h`am f(x)=e x cos x theo cˆong th´u . c Maclaurin d ˆe ´ nsˆo ´ ha . ng ch´u . a x 3 . Gia ’ i. Khai triˆe ’ ncˆa ` n t`ım pha ’ ic´oda . ng e x cos x = 3  k=0 a k x k + o(x 3 ). 8.3. C´ac di . nh l´y co . ba ’ nvˆe ` h`am kha ’ vi 103 V`ı x cos x = x +0(x), (x cos x) k = x k + o(x k ), k =1, 2, nˆen trong cˆong th´u . c e w = n  k=0 w k k! + o(w n ),w= x cos x ta cˆa ` nlˆa ´ y n = 3. Ta c´o w = x cos x = x − x 3 2! + o(x 4 ) w 2 = x 2 + o(x 3 ),w 3 = x 3 + o(x 3 ) v`a do d´o e x cos x = 3  k=0 w k k! + o(w 3 ) =1+x − x 3 2! + o(x 4 )+ 1 2  x 2 +0(x 3 )  + 1 3!  x 3 + o(x 3 )  +0(x 3 ) =1+x + 1 2 x 2 − 1 3 x 3 + o(x 3 ).  V´ı du . 6. Khai triˆe ’ n theo cˆong th´u . c Maclaurin d ˆe ´ n o(x 2n+1 )dˆo ´ iv´o . i c´ac h`am 1) arctgx, 2) arc sin x. Gia ’ i. 1) V`ı (arctgx)  = 1 1+x 2 = n  k=0 (−1) k x 2k + o(x 2n+1 ) nˆen theo cˆong th´u . c (8.19) ta c´o arctgx = n  k=0 (−1) k x 2k+1 (2k +1) + o(x 2n+2 ). V´o . i n =2tathud u . o . . c arctgx = x − x 3 3 + x 5 5 + o(x 6 ) 104 Chu . o . ng 8. Ph´ep t´ınh vi phˆan h`am mˆo . tbiˆe ´ n 2) Ta c´o (arcsinx)  = 1 √ 1 − x 2 =1+ n  k=1 (−1) k   − 1 2 k   x 2k + o(x 2n+1 ) =1+ n  k=1 (−1) k (2k − 1)!! 2 k k! x 2k + o(x 2n+1 ). T`u . d´o ´ap du . ng cˆong th´u . c (8.19) ta c´o arc sin x = x + n  k=0 (2k −1)!! 2 k k!(2k +1) x 2k+1 + o(x 2n+2 ). V´o . i n =2tathudu . o . . c arc sin x = x + 1 6 x 3 + 3 40 x 5 + o(x 6 ).  V´ı du . 7. Khai triˆe ’ n h`am f(x)=tgx theo cˆong th´u . c Maclaurin d ˆe ´ n o(x 5 ). Gia ’ i. Ta s˜e d`ung phu . o . ng ph´ap hˆe . sˆo ´ bˆa ´ tdi . nh m`a nˆo . i dung du . o . . c thˆe ’ hiˆe . n trong l`o . i gia ’ i sau dˆa y . V`ıtgx l`a h`am le ’ v`a tgx = x + o(x)nˆen tgx = x + a 3 x 3 + a s x 5 + o(x 6 ). Ta su . ’ du . ng cˆong th´u . c sin x =tgx·cos x v`a c´ac khai triˆe ’ nIIv`aIII ta c´o x − x 3 3 + x 5 5 + o(x 6 )=  x + a 3 x 3 + a 5 x 5 + o(x 6 )   1 − x 2 2! + x 4 4! +0(x 5 )  Cˆan b˘a ` ng c´ac hˆe . sˆo ´ cu ’ a x 3 v`a x 5 o . ’ hai vˆe ´ ta thu d u . o . . c      − 1 6 = − 1 2 + a 3 1 5! = 1 4! − a 3 2! + a 5 [...]... 9.2 - a ´ Dao h`m riˆng cˆp 1 110 e a - a ´ Dao h`m riˆng cˆp cao 113 e a ´ ` ’ Vi phˆn cua h`m nhiˆu biˆn 125 a a e e 9.2.1 9.2.2 ´ ’ Ap dung vi phˆn dˆ t´ gˆn d´ng 126 a e ınh ` a u 9.2.3 ´ C´c t´ chˆt cua vi phˆn 1 27 a ınh a ’ a 9.2.4 ´ Vi phˆn cˆp cao 1 27 a a 9.2.5 Cˆng th´.c Taylor 129 o u 9.2.6 9.3 ´ Vi phˆn cˆp... kha vi a y e 1 07 13 f (x) = sin x sin 3x n (−1)k 22k−1 (DS (1 − 22k )x2k + o(x2n+1 )) (2k)! k=0 ’ ’ ´ Khai triˆn h`m theo cˆng th´.c Taylor trong lˆn cˆn diˆm x0 dˆn e a o u e e a a o((x − x0)n ) (14-20)   1 n √  2  (x − 1)k + o((x − 1)n )) 14 f (x) = x, x0 = 1 (DS k=0 k 15 f (x) = (x2 − 1)e2x, x0 = −1 n e−2 2k−2 (k − 5) (DS (x + 1)k + o((x + 1)n )) (k − 1)! k=1 16 f (x) = ln(x2 − 7x + 12), x0 =... 3x 5 f (x) = ln(x2 + 3x + 2) n (−1)k−1 (1 + 2−k )xk + o(xn )) (DS ln2 + k k=1 (−1)k−1 − 2−k k x + o(xn )) k k=1 n 6 f (x) = ln(2 + x − x2) (DS ln2 + 1 − 2x2 2 + x − x2 n (−1)k+1 − 7 · 2−(k+1) 1 xk + o(xn )) (DS + 2 k=1 3 7 f (x) = n 5 3x2 + 5x − 5 (DS + (−1)k 2−(k+1) − 1 xk + o(xn )) 8 f (x) = 2 x +x−2 2 k=1 ’ ´ e Khai trˆn h`m theo cˆng th´.c Maclaurin dˆn 0(x2n+1 ) (9-13) e a o u 9 f (x) = sin2... lim (DS 2) x→0 x − sin x tgx + 2 sin x − 3x (DS 0) 22 lim x→0 x4 20 f (x) = ´ Chu.o.ng 8 Ph´p t´nh vi phˆn h`m mˆt biˆn e ı a a o e 108 ex − e−x − 2 x→0 x2 1 1 24 lim − x→0 x sin x 23 lim x2 25 26 27 28 (DS 1) (DS 0) cos x − e− 2 1 (DS − ) lim 4 x→0 x 12 √ 2 cos x 1 1− 1+x lim (DS ) 4 x→0 x 3 2 x ln cos x + 1 2 lim (DS − ) x→0 x(sin x − x) 4 √ 3 19 sin(sin x) − x 1 − x2 ) (DS lim 5 x→0 x 90 Chu.o.ng... o((x + 1)n )) (k − 1)! k=1 16 f (x) = ln(x2 − 7x + 12), x0 = 1 n 2−k + 3−k (DS ln6 − (x − 1)k + o(x − 1)n )) k k=1 (x − 1)x−2 , x0 = 2 3−x n 1 (−1)k (DS (x − 2) + + (x − 2)k + o((x − 2)n )) k k−1 k=2 17 f (x) = ln 18 f (x) = (x − 2)2 , x0 = 2 3−x n (DS (x − 2)k + o((x − 2)n )) k=2 2 19 f (x) = x − 3x + 3 , x = 3 x−2 n (DS 3 + (−1)k (x − 3)k + o((x − 3)n )) k=2 2 x + 4x + 4 , x0 = 2 x2 + 10x + 25 n (−1)k... vi phˆn h`m nhiˆu biˆn e ınh a a e e 110 9.3.1 9.3.2 9.3.3 9.1 9.1.1 Cu.c tri 145 Cu.c tri c´ diˆu kiˆn 146 e e o ` ´ ´ Gi´ tri l´.n nhˆt v` b´ nhˆt cua h`m 1 47 a o a a e a ’ a - Dao h`m riˆng a e - ´ Dao h`m riˆng cˆp 1 a e a ’ ’ ’ Gia su w = f (M), M = (x, y) x´c dinh trong lˆn cˆn n`o d´ cua diˆm a a a o ’ e a ’ ˜ ´ ´ e o u ´ a u M(x, y) Tai diˆm M ta . d ˆe ’ t´ınh gˆa ` nd´ung . . . . . 126 9.2.3 C´ac t´ınh chˆa ´ tcu ’ aviphˆan 1 27 9.2.4 Vi phˆan cˆa ´ pcao 1 27 9.2.5 Cˆong th´u . cTaylor 129 9.2.6 Vi phˆan cu ’ ah`amˆa ’ n 130 9.3 Cu . . c. ln(2 + x −x 2 ). (DS. ln2 + n  k=1 (−1) k−1 − 2 −k k x k + o(x n )) 7. f(x)= 1 − 2x 2 2+x − x 2 . (DS. 1 2 + n  k=1 (−1) k+1 − 7 · 2 −(k+1) 3 x k + o(x n )) 8. f(x)= 3x 2 +5x −5 x 2 + x − 2 .(D S. 5 2 + n  k=1  (−1) k 2 −(k+1) −1  x k +. −5) (k −1)! (x +1) k + o((x +1) n )) 16. f(x) = ln(x 2 −7x + 12), x 0 =1. (D S. ln6 − n  k=1 2 −k +3 −k k (x − 1) k + o(x − 1) n )) 17. f(x)=ln (x − 1) x−2 3 − x , x 0 =2. (D S. (x − 2) + n  k=2  1 k + (−1) k k