Bài tập toán cao cấp tập 3 part 7 docx

198 Chu . o . ng 13. L´y thuyˆe ´ tchuˆo ˜ i 23.  n1 (−1) n−1 (2n + 1)!! 2 · 5 ·8 ···(3n −1) .(D S. Hˆo . itu . tuyˆe . tdˆo ´ i) 24.  n1 (−1) n+1  1 −cos π √ n  .(D S. Hˆo . itu . c´o diˆe ` ukiˆe . n) 25.  n1 (−1) n sin π n n .(D S. Hˆo . itu . tuyˆe . tdˆo ´ i) 26.  n1 (−1) n √ n +2 .(D S. Hˆo . itu . c´o diˆe ` ukiˆe . n) 27.  n1 (−1) n n √ n .(D S. Phˆan k`y) 28.  n1 (−1) n+1 n −ln n .(D S. Hˆo . itu . c´o diˆe ` ukiˆe . n) 29.  n1 (−1) n−1 (n +1)a 2n . (D S. Hˆo . itu . tuyˆe . tdˆo ´ i khi |a| > 1, hˆo . itu . c´o diˆe ` ukiˆe . n khi |a| =1, phˆan k`y khi |a| < 1) 30.  n1 (−1) n (n + 1)( √ n +1− 1) .(D S. Hˆo . itu . tuyˆe . tdˆo ´ i) 31.  n1 (−1) n+1  2+ 1 n  n 5 n .(DS. Hˆo . itu . tuyˆe . tdˆo ´ i) 32.  n1 (−1) n tg π 3 n .(DS. Hˆo . itu . tuyˆe . tdˆo ´ i) Trong c´ac b`ai to´an sau d ˆay, h˜ay t`ım sˆo ´ sˆo ´ ha . ng cu ’ a chuˆo ˜ id˜acho cˆa ` nlˆa ´ yd ˆe ’ tˆo ’ ng cu ’ ach´ung v`a tˆo ’ ng cu ’ achuˆo ˜ itu . o . ng ´u . ng sai kh´ac nhau mˆo . td a . ilu . o . . ng khˆong vu . o . . t qu´a sˆo ´ δ cho tru . ´o . c 33.  n1 (−1) n−1 1 2n 2 , δ =0, 01. (DS. No =7) 13.3. Chuˆo ˜ il˜uy th`u . a 199 34.  n1 cos(nπ) n! , δ =0, 001. (D S. No =5) 35.  n1 (−1) n−1 √ n 2 +1 , δ =10 −6 .(DS. No =10 6 ) 36.  n1 cos nπ 2 n (n +1) , δ =10 −6 .(DS. No = 15) 37.  n1 (−1) n 2n (4n + 1)5 n , δ =0, 1?; δ =0, 01? (DS. No =2,No =3) 38.  n1 (−1) n n! , δ =0,1; δ =0, 001? (D S. No =4,No =6) 13.3 Chuˆo ˜ il˜uy th`u . a 13.3.1 C´ac di . nh ngh˜ıa co . ba ’ n Chuˆo ˜ il˜uy th`u . ad ˆo ´ iv´o . ibiˆe ´ n thu . . c x l`a chuˆo ˜ ida . ng  n0 a n x n = a 0 + a 1 x + a 2 x 2 + ···+ a n x n + (13.6) hay  n0 a n (x −a) n = a 0 + a 1 (x −a)+···+ a n (x −a) n + (13.7) trong d ´o c´ac hˆe . sˆo ´ a 0 ,a 1 , ,a n , l`a nh˜u . ng h˘a ` ng sˆo ´ .B˘a ` ng ph´ep d ˆo ’ i biˆe ´ n x bo . ’ i x − a t`u . (13.6) thu d u . o . . c (13.7). Do d ´odˆe ’ tiˆe . n tr`ınh b`ay ta chı ’ cˆa ` n x´et (13.6) l`a d u ’ (t ´u . c l`a xem a = 0). Chuˆo ˜ i (13.6) luˆon hˆo . itu . ta . id iˆe ’ m x = 0, c`on (13.7) hˆo . itu . ta . i x = a. Do d ´otˆa . pho . . pd iˆe ’ m m`a chuˆo ˜ il˜uy th`u . ahˆo . itu . luˆon luˆon = ∅. D ˆo ´ iv´o . ichuˆo ˜ il˜uy th `u . abˆa ´ t k `y (13.6) luˆon luˆon tˆo ` nta . isˆo ´ thu . . c R :0 R  +∞ sao cho chuˆo ˜ id ´ohˆo . itu . tuyˆe . tdˆo ´ i khi |x| <Rv`a phˆan k`y khi |x| >R.Sˆo ´ R d ´odu . o . . cgo . il`ab´an k´ınh hˆo . itu . cu ’ a chuˆo ˜ i 200 Chu . o . ng 13. L´y thuyˆe ´ tchuˆo ˜ i (13.6) v`a khoa ’ ng I(R)=(−R, R)du . o . . cgo . il`akhoa ’ ng hˆo . itu . cu ’ achuˆo ˜ i l˜uy th `u . a (13.6). B´an k´ınh hˆo . itu . R cu ’ a chuˆo ˜ il˜uy th`u . ac´othˆe ’ t´ınh thˆong qua c´ac hˆe . sˆo ´ cu ’ an´obo . ’ imˆo . t trong c´ac cˆong th´u . c R = lim n→∞ |a n | |a n+1 | , (13.8) ho˘a . c R = lim n→∞ 1 n  |a n | (13.9) nˆe ´ u gi´o . iha . no . ’ vˆe ´ pha ’ icu ’ a (13.8) v`a (13.9) tˆo ` nta . i. D - i . nh ngh˜ıa 13.3.1. Ngu . `o . i ta n´oi r˘a ` ng h`am f(x) khai triˆe ’ nd u . o . . c th`anh chuˆo ˜ il˜uy th`u . a  n0 a n x n trˆen khoa ’ ng ( −R, R)nˆe ´ u trˆen khoa ’ ng d ´o c h u ˆo ˜ id˜anˆeuhˆo . itu . v`a tˆo ’ ng cu ’ a n´o b˘a ` ng f(x), t´u . cl`a f(x)=  n0 a n x n ,x∈ (−R, R). D i . nh ngh˜ıa 13.3.2. 1 + Chuˆo ˜ il˜uy th`u . ada . ng f(x 0 )+ f  (x 0 ) 1! (x −x 0 )+···+ f (n) (x 0 ) n! (x − x 0 ) n + =  n0 f (n) (x 0 ) n! (x −x 0 ) n (13.10) d u . o . . cgo . i l`a chuˆo ˜ i Taylor cu ’ a h`am f(x)v´o . i tˆam ta . id iˆe ’ m x 0 (o . ’ d ˆay 0! = 1, f (0) (x 0 )=f(x 0 )). 2 + C´ac hˆe . sˆo ´ cu ’ a chuˆo ˜ i Taylor a 0 = f(x 0 ),a 1 = f  (x 0 ) 1! , ,a n = f (n) (x 0 ) n! (13.11) d u . o . . cgo . il`ac´achˆe . sˆo ´ Taylor cu ’ a h`am f(x). 13.3. Chuˆo ˜ il˜uy th`u . a 201 3 + Khi x 0 = 0, chuˆo ˜ i Taylor f(0) + f  (0) 1! x + ···+ f (n) (0) n! x n + ···=  n0 f (n) (0) n! x n (13.12) d u . o . . cgo . il`achuˆo ˜ i Maclaurin. 13.3.2 D - iˆe ` ukiˆe . n khai triˆe ’ nv`aphu . o . ng ph´ap khai triˆe ’ n D - i . nh l´y 13.3.1 (Tiˆeu chuˆa ’ n khai triˆe ’ n). H`am f(x) khai triˆe ’ ndu . o . . c th`anh chuˆo ˜ il˜uy th`u . a  n0 a n x n trˆen khoa ’ ng (−R, R) khi v`a chı ’ khi trˆen khoa ’ ng d´o h`am f(x) c´o da . o h`am mo . icˆa ´ p v`a trong cˆong th´u . c Taylor f(x)=f(0) + f  (0) 1! x + ···+ f (n) (0) n! x n + R n (x) phˆa ` ndu . R n (x) → 0 khi n →∞∀x ∈ (−R, R). Trong thu . . c h`anh ngu . `o . i ta thu . `o . ng su . ’ du . ng dˆa ´ uhiˆe . ud u ’ nhu . sau. D - i . nh l´y 13.3.2. D ˆe ’ h`am f(x) khai triˆe ’ ndu . o . . c th`anh chuˆo ˜ il˜uy th`u . a  n0 a n x n ,x∈ (−R, R) d iˆe ` ukiˆe . ndu ’ l`a trˆen khoa ’ ng d´o h`am f(x) c´o da . o h`am mo . icˆa ´ p v`a c´ac d a . oh`amd´obi . ch˘a . n, t´u . cl`a∃M>0:∀n =0, 1, 2, v`a ∀x ∈ (−R, R) th`ı |f (n) (x)|  M. Ta nˆeu ra d ˆay hai phu . o . ng ph´ap khai triˆe ’ n h`am th`anh chuˆo ˜ il˜uy th `u . a 202 Chu . o . ng 13. L´y thuyˆe ´ tchuˆo ˜ i 1. Phu . o . ng ph´ap I (phu . o . ng ph´ap tru . . ctiˆe ´ p) gˆo ` m c´ac bu . ´o . c sau: a) T´ınh c´ac hˆe . sˆo ´ theo cˆong th´u . c (13.11) b) Ch´u . ng to ’ r˘a ` ng lim n→∞ R n (x)=0. Nhu . o . . cd iˆe ’ mcu ’ aphu . o . ng ph´ap n`ay l`a t´ınh to´an qu´a cˆo ` ng kˆe ` nh v`a sau n˜u . a l`a viˆe . c kha ’ o s´at gi´o . iha . n R n (x) → 0(n →∞)la . i c`ang ph´u . c ta . pho . n. 2. Phu . o . ng ph´ap II (phu . o . ng ph´ap gi´an tiˆe ´ p) l`a phu . o . ng ph´ap du . . a trˆen ba ’ ng c´ac khai triˆe ’ n “c´o s˘a ˜ n” (hay Khai triˆe ’ nba ’ ng)c`ung v´o . i c´ac ph´ep t´ınh d ˆo ´ iv´o . i chuˆo ˜ il˜uy th`u . a. I. e x =1+x + x 2 2! + ···+ x n n! −···=  n0 x n n! , x ∈ R. II. sin x = x − x 3 3! + x 5 5! −···+(−1) n x 2n+1 (2n + 1)! + ···= =  n0 (−1) n x 2n+1 (2n + 1)! , x ∈ R. III. cosx =1− x 2 2! + x 4 4! −···+(−1) n x 2n (2n)! + ···= =  n0 (−1) n x 2n (2n)! , x ∈ R. IV. (1 + x) α =1+αx + α(α − 1) 2! x 2 + ···+ α(α −1)···(α −n +1) n! x n + =1+  n1  α n  x n , −1 <x<1,  α 0  =1,  α n  = α(α −1)···(α −n +1) n! ,  α n  = C α n nˆe ´ u α ∈ N. 13.3. Chuˆo ˜ il˜uy th`u . a 203 Khi α = −1 ta c´o 1 1+x =1−x + x 2 −···+(−1) n x n + =  n0 (−1) n x n , −1 <x<1. 1 1 −x =1+x + x 2 + ···+ x n + ···=  n0 x n , −1 <x<1. V. ln(1 + x)=x − x 2 2 + x 3 3 −···+(−1) n−1 x n n + ; −1 <x<1 ln(1 −x)=−x − x 2 2 −···− x n n − , −1 <x<1. C ´ AC V ´ IDU . V´ı du . 1. T`ım miˆe ` nhˆo . itu . cu ’ a chuˆo ˜ il˜uy th`u . a  n1 (−1) n−1 nx n . Gia ’ i. 1 + Tas˜e´apdu . ng cˆong th´u . c (13.8). V`ı a n =(−1) n−1 n v`a a n+1 =(−1) n (n + 1) nˆen ta c´o R = lim n→∞ |a n | |a n+1 | = lim n→∞ n n +1 =1. Nhu . vˆa . y chuˆo ˜ ihˆo . itu . b´o . i −1 <x<1. 2 + Ta c`on cˆa ` n kha ’ o s´at su . . hˆo . itu . cu ’ a chuˆo ˜ ita . i c´ac d ˆa ` um´ut cu ’ a khoa ’ ng hˆo . itu . . V´o . i x = −1 ta c´o  n1 (−1) n−1 n(−1) n =  n1 (−1) 2n−1 n =  n1 (−n). Do d ´o chuˆo ˜ id˜a cho phˆan k`y ta . idiˆe ’ m x = −1 (khˆong tho ’ a m˜an diˆe ` u kiˆe . ncˆa ` n!) 204 Chu . o . ng 13. L´y thuyˆe ´ tchuˆo ˜ i V´o . i x = 1 ta c´o  n1 (−1) n−1 n ⇒ lim n→∞ (−1) n−1 n khˆong tˆo ` nta . i Do d ´o chuˆo ˜ i phˆan k`yta . idiˆe ’ m x = 1. Vˆa . ymiˆe ` nhˆo . itu . cu ’ a chuˆo ˜ il`a (−1, 1).  V´ı d u . 2. T`ım khoa ’ ng hˆo . itu . cu ’ a chuˆo ˜ i  n1 (−1) n (x − 2) n n n · Gia ’ i. Trong tru . `o . ng ho . . p n`ay ta su . ’ du . ng cˆong th´u . c (13.9) v`a thu d u . o . . c R = lim n→∞ 1 n  |a n | = lim n→∞ 1 n     (−1) n n n    = lim n→∞ n =+∞. D iˆe ` ud´o c´o ngh˜ıa l`a chuˆo ˜ id˜achohˆo . itu . v´o . imo . i gi´a tri . x,t´u . cl`a I(R)=(−∞, +∞). V´ı d u . 3. T`ım khoa ’ ng hˆo . itu . cu ’ a chuˆo ˜ i  n0 n!x n , 0! ≡ 1. Gia ’ i. ´ Ap du . ng cˆong th´u . c (13.8) ta c´o R = lim n→∞ |a n | |a n+1 | = lim n→∞ n! n!(n +1) = lim n→∞ 1 n +1 =0. Vˆa . y R =0.D iˆe ` ud´o c´o ngh˜ıa r˘a ` ng chuˆo ˜ id˜a cho hˆo . itu . ta . idiˆe ’ m x =0.  V´ı du . 4. Khai triˆe ’ n h`am 1 4 −x th`anh chuˆo ˜ il˜uy th`u . ata . i lˆan cˆa . nd iˆe ’ m x 0 =2(c˜ung t´u . c l`a: theo c´ac l˜uy th`u . acu ’ ahiˆe . u x − 2 hay chuˆo ˜ il˜uy th `u . av´o . i tˆam ta . id iˆe ’ m x 0 = 2). 13.3. Chuˆo ˜ il˜uy th`u . a 205 Gia ’ i. Ta biˆe ´ ndˆo ’ i h`am d˜achodˆe ’ c´o thˆe ’ ´ap du . ng khai triˆe ’ nba ’ ng: 1 1 −t =  n0 t n , −1 <t<1. Ta c´o 1 4 −x = 1 2 · 1 1 − x −2 2 Xem t = x −2 2 ta c´o: 1 4 −x = 1 2  1+  x −2 2  +  x −2 2  2 + ···+  x − 2 2  n +  ⇒ 1 4 −x =  n0 1 2 n+1 (x −2) n . Khai triˆe ’ n n`ay chı ’ d ´ung khi    x −2 2    < 1 ⇔|x − 2| < 2 ⇔−2 <x− 2 < 2 ⇔ 0 <x<4.  Nhˆa . nx´et. Ba . nd o . cc˜ung dˆe ˜ d`ang thu du . o . . c khai triˆe ’ ntrˆend ˆay b˘a ` ng phu . o . ng ph´ap tru . . ctiˆe ´ p. V´ı d u . 5. Khai triˆe ’ n h`am f(x) = sin πx 4 th`anh chuˆo ˜ i Taylor v´o . i tˆam ta . id iˆe ’ m x 0 =2. Gia ’ i. Ta biˆe ´ nd ˆo ’ i h`am d˜a cho nhu . sau sin π 4 x = sin π 4 (x −2 + 2) = sin  π 4 (x − 2) + π 2  = cos π 4 (x −2). Xem π 4 (x −2) = t v`a ´ap du . ng khai triˆe ’ nba ’ ng I II ta c´o sin π 4 x =  n0 (−1) n (2n)!  π 4 (x −2)  2n =  n0 (−1) n π 2n 4 2n (2n)! (x −2) 2n ,x∈ R.  206 Chu . o . ng 13. L´y thuyˆe ´ tchuˆo ˜ i V´ı d u . 6. Khai triˆe ’ n h`am f(x)=ln 1+x 1 −x th`anh chuˆo ˜ i Maclaurin. Gia ’ i. Ta c´o ln 1+x 1 −x = ln(1 + x) − ln(1 −x). M˘a . t kh´ac ln(1 + x)=x − x 2 2 + ···+(−1) n−1 x n n + , −1 <x<1; ln(1 −x)=−x − x 2 2 −···− x n n − , −1 <x<1. T`u . d ´o ln(1 + x) − ln(1 − x)=2x + 2x 3 3 + 2x 5 5 + ···+ 2x 2n−1 2n −1 + − 1 <x<1.  B ` AI T ˆ A . P Su . ’ du . ng c´ac khai triˆe ’ nba ’ ng d ˆe ’ khai triˆe ’ n h`am th`anh chuˆo ˜ il˜uy th `u . av´o . i tˆam ta . id iˆe ’ m x 0 = 0 v`a chı ’ ra b´an k´ınh hˆo . itu . cu ’ a chuˆo ˜ i 1. f(x)=e −2x .(DS.  n0 (−1) n 2 n n! x n , R =+∞) 2. f(x)=ln 1 1 −2x .(D S.  n1 2 n n x n , R = 1 2 ) 3. f(x)=x ln  1+ x 3 3  .(D S.  n1 (−1) n x 3n+1 3 n n x n , R = 3 √ 3) 4. f(x)= 3 √ 1 −4x.(DS. 1 − 4 3 x +  n2 4 n · 2 · 5 ···(3n − 4) 3 n · n! x n ; R =1) 5. f(x) = sin 5x.(D S.  n0 (−1) n 5 2n+1 (2n + 1)! x 2n+1 ; R =+∞) 6. f(x) = cos x 3 3 .(D S.  n0 (−1) n x 6n 3 2n (2n)! ; R =+∞) 13.3. Chuˆo ˜ il˜uy th`u . a 207 B˘a ` ng c´ach biˆe ´ ndˆo ’ i (trong tru . `o . ng ho . . pcˆa ` n thiˆe ´ t) sao cho c´o thˆe ’ ´a p du . ng c´ac khai triˆe ’ nba ’ ng d ˆe ’ khai triˆe ’ n h`am f(x) th`anh chuˆo ˜ il˜uy th`u . a v´o . i tˆam ta . id iˆe ’ m x 0 . H˜ay chı ’ ra b´an k´ınh hˆo . itu . cu ’ a chuˆo ˜ i 7. f(x)=e − x 2 , x 0 = 10. (DS. e −5  n0 (−1) n n! (x −10) n 2 n , R =+∞) 8. f(x)=2 x , x 0 = a.(DS. 2 a  n0 ln n 2 n! (x −a) n , R =+∞) 9. f(x)=2 x 3 −x , x 0 = 0. (DS.  n0  ln 2 3  n n! x n , R =+∞) 10. f(x)=e 1−2x 3 , x 0 = 0. (DS.  n0 (−1) n 2 n e n! x 3n , R =+∞) 11. f(x)=(2+x)e x−1 , x 0 = −2. (DS. 1 e 3  n0 (x +2) n+1 n! , R =+∞) 12. f(x) = sin(a + x), x 0 =0. (D S. sin a  n0 (−1) n x 2n 2n! + cos a  n0 (−1) n x 2n+1 (2n + 1)! ,R=+∞) 13. f(x) = sin xcos 3x, x 0 =0. (D S. 1 2  n0 (−1) n (4x) 2n+1 (2n + 1)! − 1 2  n0 (−1) n (2x) 2n+1 (2n + 1)! ,R=+∞) 14. f(x)=    cos x − 1 x ,x=0 0,x=0 ; x 0 =0. (D S.  n0 (−1) n (x) 2n−1 (2n)! , R =+∞) 15. f(x) = cos x, x 0 = π 2 .(D S.  n1 (−1) n  x − π 2  2n−1 (2n −1)! , R =+∞) 16. f(x) = sin 2 x cos 2 x, x =0. [...]... ’ o 30 f (x) = ex−1 , x0 = 4 (DS e3 n (x − 4)n , R = +∞) n! 0 ˜ ´ y e o Chu.o.ng 13 L´ thuyˆt chuˆ i 210 n 3n − 2(−1)n n x , R = +∞) n! 0 , x0 = −1 (DS 32 23x−2 n (ln 8)n (x + 1)n , R = +∞) n! 0 31 f(x) = e3x − 2e−x , x0 = 0 (DS 32 f(x) = 1 33 f (x) = 2x ex−1, x0 = 1 (DS 2 n (ln 2 + 1)n (x − 1)n , R = +∞) n! 0 π 34 f (x) = sin 3x, x0 = 4 √ 2 (−1)n 32 n π 2n (DS x− 2 n 0 (2n)! 4 √ π 2 (−1)n 32 n+1... = +∞) π x 35 f (x) = cos , x0 = − 2 3 (−1)n x + √ (DS 3 n 0 (−1)n π π x+ x+ 2n+1 (2n + 1)!(2 3 3 36 f (x) = cos2 x, x0 = (DS 2n (2n)!22n+1 n 0 + π 3 1 + 2 n 2n+1 , R = +∞) π 4 (−1)n 4n−1 x − (2n − 1)! 1 π 4 2n−1 , R = +∞) 37 f (x) = sin x cos2 x, x0 = 0 (DS n 0 (−1)n (1 + 32 n+1 )x2n+1 , R = +∞) 4(2n)! 38 f (x) = cos x cos 2x, x0 = 0 (DS 1 2 (−1)n n 0 32 n 1 + x2n , R = +∞) (2n)! (2n)! ˜ 13. 4 Chuˆ i... Fourier o 211 39 f (x) = x ln x, x0 = 1 (DS n 1 (−1)n−1 (x − 1)n+1 + n n 1 (−1)n−1 (x − 1)n , R = 1) n 40 f (x) = ln(2x + 3) , x0 = 4 2 11 (−1)n ln 11 + (DS n 1 n+1 11 1 (x − 4)n , R = ) n+1 2 41 f (x) = ln (3 − 4x), x0 = −2 (DS ln 11 − n 1 42 f (x) = arctg 4 11 n (x + 2)n −19 3 , x∈ , ) n 4 4 x +3 , x0 = 0 x 3 (DS π − + 4 n ˜ ’ a Chı dˆ n f (x) = − 0 (−1)n+1 x2n+1 , R = 3) 32 n+1 2n + 1 2n 3 n+1 x = (−1)... k ∈ Z) 2 12 y − y = 2x − 3 (DS y = 1 − 2x + Cex) ˜ ’ ´ ’ a Chı dˆ n Dˆi biˆn z = y + 2x − 3 o e 1 1 + = C) 13 (x + 1 )3 dy − (y − 2)2 dx = 0 (DS − y − 2 2(x + 1)2 √ √ √ √ 14 ( xy + x)y − y = 0 (DS 2 y + ln |y| − 2 x = C) x+y 15 2 x−2y +3 2 3 x 18−y = C) y = 0 (DS − 2 ln 18 ln 3 ı a Chu.o.ng 14 Phu.o.ng tr`nh vi phˆn 230 16 (y + xy)dx + (x − xy)dy = 0 (DS x − y + ln |xy| = C) 17 yy + x = 1 (DS (x − 1)2... nπxdx −1 1 = 2 π 2 n2 x cos nπx 1 −1 − cos nπxdx = 4 π 2 n2 (−1)n , n = 1, 2, −1 1 (x3 + x2 − x − 1) sin nπxdx = bn = 12 (−1)n , n = 1, 2, 3 n3 π −1 ’ ’ V` h`m f (x) kha vi trˆn khoang (−1, 1) nˆn ta c´ ı a e e o 4 2 x +x −x−1=− + 2 3 π 3 2 n 1 3( −1)n (−1)n cos nπx + sin nπx , n2 πn3 ∀x ∈ (−1, 1) ’ V´ du 3 Khai triˆn h`m f (x) = x, x ∈ [0, ]: ı e a 1) theo c´c h`m cosin; a a 2) theo c´c h`m...˜ ´ y e o Chu.o.ng 13 L´ thuyˆt chuˆ i 208 (DS n (−1)n+1 · 24n 3 2n x , R = +∞) (2n)! 1 17 f(x) = ln(x2 + 3x + 2), x0 = 0 (DS (−1)n−1 [1 + 2−n ] ln 2 + n 1 xn , R = +∞) n 18 f(x) = ln(4 + 3x − x2 ), x0 = 2 DS ln 6 + n 1 19 f (x) = x2 1 (−1)n−1 3 n − 2−n (x − 2)n ) n 1 , x0 = 0; x0 = 4 − 2x − 3 (DS 1) n 0 2) n 0 1 1 (−1)n+1 − 4 12 3 n xn , |x| < 1; 1 (−1)n (−1)n − (x − 4)n ,... (−1)n−1 Thay x0 = 0, ta thu du.o.c 3 n2 n 1 n 1 π2 (−1)n−1 = Thay x = π thu du.o.c n2 12 n 1 5 f(x) = x2 , x ∈ (0, π), f(x) = f (x + pi) (DS π2 + 3 n 1 π 1 cos 2nx − sin 2nx ) 2 n n π2 1 = ) n2 6 ˜ ´ y e o Chu.o.ng 13 L´ thuyˆt chuˆ i 220 x 6 f(x) = cos , x ∈ (0, 2π], f (x) = f (x + 2π) 2 (DS 7 f (x) = (DS  0 cos 8 x = 2 π ´ nˆu − 3 < x e 2 x nˆu 0 < x < 3 ´ e 6 3 − 2 4 π n 1 n 1 n sin nx ) (2n... Chu.o.ng 13 L´ thuyˆt chuˆ i 212 - ˜ a Dinh ngh˜ 13. 4.1 1) Chuˆ i h`m dang ıa o a0 nπx nπx + + bn sin , an cos 2 n 1 ( 13. 14) ˜ ´ o a a ` a o a e o ’ o trong d´ > 0; an , bn l` c´c h˘ng sˆ (goi l` hˆ sˆ cua chuˆ i ( 13. 14)) ´ ˜ du.o.c goi l` chuˆ i lu.o.ng gi´c o a a o.ng gi´c ( 13. 14) du.o.c goi l` chuˆ i Fourier cua h`m ˜ ˜ ’ 2) Chuˆ i lu o a o a a ˜ ´ ’ f (x) theo hˆ lu.o.ng gi´c co so ( 13. 13) (goi... 0 v` chı ’ a ˜ u dˆ e e a o u o a a ’ ra b´n k´ hˆi tu a ınh o 21 f (x) = x2 22 f (x) = arcctgx (DS x3 (−1)n x2n−1 π x+ +···+ + , |x| 2 3 2n − 1 1) ˜ u 13. 3 Chuˆ i l˜y th`.a o u 209 π ˜ ’ ’ a e u Chı dˆ n Khai triˆn arctgx v` su dung hˆ th´.c arcctgx = − e a ’ 2 arctgx 23 f (x) = 1 (1 − x )3 (DS 24 f (x) = arc sin x 1 (n + 1)(n + 2)xn , R = 1) 2n 0 (2n − 1)!!x2n+1 , R = 1) 1 (2n)!!(2n + 1) (DS... Chu.o.ng 13 L´ thuyˆt chuˆ i 222 ` ´ 16 f(x) = x − [x] = {x} - phˆn thˆp phˆn cua sˆ x a a a ’ o (DS {x} =  −x 17 f(x) = x2  π (DS 5π + 12 n ´ nˆu − π e 1 1 − 2 π x ´ nˆu 0 < x e n 1 sin 2nπx ) n 0 π 4 3( −1)n − 1 cos nx − 2 sin(2n − 1)x ) 2 πn π (2n − 1 )3 1  0 khi − 2 x 0 18 f (x) = 1  x khi 0 < x 2 2 (DS 1 + 4 n − 1 nπx 2 nπx (−1)n+1 + sin ) cos 2 (2n − 1 )3 π 2 nπ 2 19 f (x) = x, x ∈ [3, 5] (DS . −2x .(D S.  n1 2 n n x n , R = 1 2 ) 3. f(x)=x ln  1+ x 3 3  .(D S.  n1 (−1) n x 3n+1 3 n n x n , R = 3 √ 3) 4. f(x)= 3 √ 1 −4x.(DS. 1 − 4 3 x +  n2 4 n · 2 · 5 ···(3n − 4) 3 n · n! x n ; R =1) 5. f(x). =10 −6 .(DS. No = 15) 37 .  n1 (−1) n 2n (4n + 1)5 n , δ =0, 1?; δ =0, 01? (DS. No =2,No =3) 38 .  n1 (−1) n n! , δ =0,1; δ =0, 001? (D S. No =4,No =6) 13. 3 Chuˆo ˜ il˜uy th`u . a 13. 3.1 C´ac di . nh. tru . ´o . c 33 .  n1 (−1) n−1 1 2n 2 , δ =0, 01. (DS. No =7) 13. 3. Chuˆo ˜ il˜uy th`u . a 199 34 .  n1 cos(nπ) n! , δ =0, 001. (D S. No =5) 35 .  n1 (−1) n−1 √ n 2 +1 , δ =10 −6 .(DS. No =10 6 ) 36 .  n1 cos