Công thức khai triển TAYLOR - GONTCHAROV vá áp dụng

8/2/2019 Cng thc khai trin TAYLOR - GONTCHAROV v p dng

1/63

I HC QUC GIA H NITRNG I HC KHOA HC T NHIN

TRN VN LONG

CNG THC KHAI TRINTAYLOR - GONTCHAROV

V P DNG

LUN VN THC S TON HC

H NI, NM 2009


2/63

I HC QUC GIA H NITRNG I HC KHOA HC T NHIN

TRN VN LONG

CNG THC KHAI TRINTAYLOR - GONTCHAROVV P DNG

Chuyn ngnh : GII TCH

M s : 60 46 01

LUN VN THC S TON HC

Ngi hng dn khoa hc:GS.TSKH. NGUYN VN MU

H NI - NM 2009


3/63

MC LC

M u 3

1 Khai trin Taylor 61.1 Mt s kin thc chun b . . . . . . . . . . . . . . . . . . . 6

1.1.1 Mt s tnh cht ca a thc . . . . . . . . . . . . . . 61.1.2 Mt s nh l c bn ca gii tch c in . . . . . . 71.2 Khai trin Taylor i vi a thc . . . . . . . . . . . . . . . . 81.3 Khai trin Taylor vi cc phn d khc nhau . . . . . . . . . 12

2 Cng thc khai trin Taylor - Gontcharov 182.1 Bi ton ni suy Newton v cng thc khai trin Taylor -

Gontcharov . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.1.1 Bi ton ni suy Newton . . . . . . . . . . . . . . . . 182.1.2 Cng thc khai trin Taylor - Gontcharov . . . . . . . 20

2.2 Khai trin Taylor - Gontcharov vi cc phn d khc nhau . . 242.2.1 Khai trin Taylor - Gontcharov vi phn d dng La-

grange . . . . . . . . . . . . . . . . . . . . . . . . . . 282.2.2 Khai trin Taylor - Gontcharov vi phn d dng Cauchy 29

2.3 S hi t trong khai trin Taylor v khai trin Taylor- Gontcharov 312.4 Bi ton ni suy Newton i vi hm a thc nhiu bin. . . 38

2.4.1 Bi ton ni suy Taylor i vi hm a thc nhiu bin 382.4.2 Bi ton ni suy Newton i vi hm a thc nhiu bin. 39

3 Mt s bi ton p dng 433.1 Khai trin Taylor ca mt s hm s cp v ng dng . . . . 433.1.1 c lng v nh gi sai s . . . . . . . . . . . . . . 433.1.2 Tnh gii hn hm s. . . . . . . . . . . . . . . . . . . 49

3.2 Khai trin Taylor- Gontcharov vi bi ton c lng hm s 54

Kt lun 61

Ti liu tham kho 62

2


4/63

M U

Khai trin a thc ni ring v khai trin hm s ni chung cng nhng

vn lin quan n n l mt phn quan trng ca i s v gii tch ton

hc. Cng vi cc bi ton ni suy, cc bi ton v khai trin hm s c v tr

c bit trong ton hc khng ch nh l nhng i tng nghin cu m

cn ng vai tr nh l mt trong nhng cng c c lc ca cc m hnh

lin tc cng nh cc m hnh ri rc ca gii tch trong l thuyt phng

trnh vi phn, l thuyt xp x, l thuyt biu din,....

L thuyt khai trin hm s cng cc bi ton ni suy lin quan ra i

rt sm vi cc cng trnh ca Taylor, Lagrange, Newton... Tuy nhin, vic

xy dng bi ton khai trin hm s tha mn nhng yu cu khc nhau

cng nh vic xy dng l thuyt hon thin v khai trin hm s ni chung

n nay vn ang c nhiu nh ton hc tip tc nghin cu v pht trin

theo nhiu hng.

L thuyt cc bi ton v khai trin hm s cng nh cc bi ton ni

suy c in c lin quan cht ch n cc c trng c bn ca hm s nh

tnh n iu, tnh li lm, tnh tun hon,... l nhng mng kin thc quan

trng trong chng trnh gii tch.

Trong cc gio trnh gii tch i hc ta bit bi ton ni suy Taylor

Gi s hm f xc nh trn tp hp R, trong l hp ca cckhong m trn trc thc. Gi s f kh vi cp n ti im a . Hy xcnh cc a thc Pn(x) c bc deg Pn(x)

n sao cho

P(k)n (a) = f(k)(a), k = 0, 1, . . . , n .

3


5/63

T ta c khai trin Taylor ca hm f(x) ti im a

f(x) = f(a)+1

1!f(a)(xa)+ 1

2!f(a)(xa)2+...+ 1

n!f(n)(a)(xa)n+Rn(f; x)

vi cc phn d dng Lagrange v Cauchy.

Trong khai trin Taylor, khi xt b im M(a, P(k)n (a)), k = 0, 1,...,n ta

thy chng cng nm trn ng thng x = a. Khi cho a thay i v nhn

gi tr ph thuc vo k th ta c mt b im mi dng

Mk(xk, P(k)n (xk)), k = 0, 1,...,n

Khi , ta thu c bi ton ni suy Newton v dn n khai trin Taylor-Gontcharov l mt m rng t nhin ca khai trin Taylor.

Lun vn tp trung i gii quyt vn xy dng cng thc nghim ca

bi ton ni suy Newton, a ra biu din hm s f(x) theo cng thc khai

trin Taylor- Gontcharov v c bit a ra cc nh gi phn d ca khai

trin Taylor - Gontcharov ca hm f(x) di hai dng Lagrange v Cauchy

cng nh m rng bi ton i vi hm a thc nhiu bin.

Lun vn gm phn m u v c chia thnh ba chng

Chng 1: Nhc li cc kin thc c bn v a thc v mt s nh l c

bn ca gii tch c in s dng trong lun vn. Tip theo tc gi trnh by

bi ton ni suy Taylor, khai trin Taylor v cc nh gi phn d ca khai

trin Taylor.

Chng 2: L phn chnh ca lun vn. Bt u bng vic kho st bi

ton ni suy Newton, a ra cng thc nghim ca bi ton ni suy New-

ton. T dn n khai trin Taylor- Gontcharov ca hm f(x) theo cc

mc ni suy x0, x1,...,xn v c bit a ra cc nh gi c lng phn d

ca khai trin Taylor- Gontcharov di dng Lagrange v Cauchy. Phn tip

theo, tc gi nh gi s hi t trong khai trin Taylor v khai trin Taylor- Gontcharov. Cui cng l m rng ca khai trin Taylor - Gontcharov cho

4


6/63

hm a thc nhiu bin.

Chng 3: cp n mt s ng dng ca khai trin Taylor v khai

trin Taylor - Gontcharov cng nh ca bi ton ni suy Newton trong clng v nh gi sai s, tm gii hn hm s.

Lun vn c hon thnh di s hng dn khoa hc ca NGND.GS.

TSKH Nguyn Vn Mu, trng i hc Khoa hc T nhin, i hc Quc

gia H ni, ngi Thy tn tnh hng dn, gip tc gi trong sut

qu trnh hon thnh bn lun vn ny. Tc gi xin by t lng bit n chn

thnh v knh trng su sc i vi Gio s.

Tc gi xin by t lng cm n ti cc thy c gio, cc thnh vin, cc

anh ch ng nghip trong Seminare Gii tch trng i hc Khoa hc T

nhin, i hc Quc gia H ni v nhng kin ng gp qu bu, s gip

tn tnh v s c v ht sc to ln trong sut thi gian qua.

Tc gi xin chn thnh cm n Ban gim hiu, phng sau i hc, khoaTon - C - Tin hc trng i hc Khoa hc T nhin, i hc Quc gia

H ni to iu kin thun li cho tc gi trong sut qu trnh hc tp

ti trng. Tc gi cng xin chn thnh cm n S Gio dc - o to Nam

nh, trng THPT M Tho v gia nh ng vin, to iu kin thun

li cho tc gi trong sut kha hc.

H ni, thng 12 nm 2009Tc gi

Trn Vn Long

5


7/63

CHNG 1

KHAI TRIN TAYLOR

1.1 Mt s kin thc chun b

1.1.1 Mt s tnh cht ca a thcCc nh ngha, tnh cht trong mc ny c trch t [2].

nh ngha 1.1. Cho vnh A l mt vnh giao hon c n v. Ta gi athc bc n bin x l biu thc c dng

Pn(x) = anxn + an1xn1 + ... + a1x + a0.(an = 0)

trong cc s ai A

c gi l cc h s, an gi l h s bc cao nht v

a0 gi l h s t do ca a thc.

Tp hp tt c cc a thc vi h s ly trong vnh A c k hiu lA[x]. KhiA l mt trng th vnhA[x] l mt vnh giao hon c n v.Cc vnh a thc thng gp lZ[x],Q[x],R[x],C[x].

nh ngha 1.2. Cho hai a thc

P(x) = anxn + an1xn1 + ... + a1x + a0

Q(x) = bnxn + bn1xn1 + ... + b1x + b0

Ta nh ngha cc php ton sau

P(x) + Q(x) = (an + bn)xn + (an1 + bn1)xn1 + ... + (a1 + b1)x + a0 + b0

P(x) Q(x) = (an bn)xn + (an1 bn1)xn1 + ... + (a1 b1)x + a0 b0P(x).Q(x) = c2nx

2n + c2n1x2n1 + ... + c1x + c0

trong ck = a0bk + a1bk1 + ... + akb0, k = 0, 1,...,n.

6


8/63

nh l 1.1. Gi sA l mt trng, f(x), g(x) l hai a thc khc 0 cavnh A[x]. Khi , tn ti duy nht cp a thc q(x), r(x) thuc A[x] saocho

f(x) = g(x).q(x) + r(x)vi deg(r(x)) < deg(g(x)). Khi r(x) = 0, ta nif(x) chia ht cho g(x).

Nuf(a) = 0 th ta ni a l nghim ca f(x). Bi ton tm cc nghim ca

f(x) trongA c gi l gii phng trnh i s bc n

anxn + an1xn1 + ... + a1x + a0 = 0(an = 0)

trong A.nh l 1.2. Gi sA l mt trng, a A v f(x) A[x]. Khi , dca php chia f(x) cho x a chnh l f(a).nh l 1.3. Mi a thc bc n u c khng qu n nghim thc.

nh l 1.4. a thc c v s nghim l a thc khng.

1.1.2 Mt s nh l c bn ca gii tch c in

Sau y, ta nhc li mt s nh l c bn s dng trong cc phn sau.

Cc nh l ny c trch t [3].

nh l 1.5 (Rolle). Gi s f : [a, b] R lin tc trn on [a; b] v co hm trn khong (a, b) . Nu f(a) = f(b) th tn ti t nht mt im

c (a; b) sao cho f(c) = 0.nh l 1.6 (Lagrange). Nu hm sf lin tc trn on[a; b] v c o hmtrn khong(a, b) th tn ti t nht mt imc (a; b) sao cho f(b)f(a) =f(c)(b a).nh l 1.7 (nh l v gi tr trung bnh ca tch phn). Nu hm s fkh tch trn on [a; b] v m f(x) M vix [a; b] th tn ti mt s [m; M] sao cho

ba

f(x)dx = (b a).

7


9/63

H qu 1.1. Nu f l hm s lin tc trn on [a; b] th tn ti t nhtmt im c (a; b) sao cho

b

a

f(x)dx = f(c)(b a).

nh l 1.8 (nh l v gi tr trung bnh m rng ca tch phn). Gi shai hm sf v g kh tch trn on [a; b] v tha mn:

a) m f(x) M vix [a; b]b) g(x) khng i du trn [a; b].

Khi , tn ti t nht mt s thc [m; M] sao chob

a

f(x).g(x)dx =

ba

g(x)dx.

H qu 1.2. Gi s f l hm s lin tc trn [a; b] v g l hm s kh tchtrn on [a; b]. Nu g(x) khng i du trn [a; b] th tn ti t nht mt s

thc c [a; b] sao chob

a

f(x).g(x)dx = f(c)

ba

g(x)dx.

nh l 1.9 (Bolzano - Cauchy). Gi sf l mt hm s lin tc trn on[a; b] v l mt s nm giaf(a) vf(b). Khi , tn ti t nht mt im

c [a, b] sao cho f(c) = . Ni mt cch khc, f ly mi gi tr trung giangia f(a) v f(b).

1.2 Khai trin Taylor i vi a thc

Cc nh l, nh ngha v bi ton trong mc ny ch yu c trch t

[1].

Ta thng thy trong cc sch gio khoa hin hnh, dng chnh tc ca mt

a thc i s P(x) bc n, n N, (thng c k hiu deg(P(x)) = n)c dng:

P(x) = p0xn + p1x

n1 + ... + pn, p0 = 0.8


10/63

a thc dng chnh tc l a thc c vit theo th t gim dn ca ly

tha.Tuy nhin, khi kho st cc a thc, ngi ta thng quan tm n c

mt lp cc a thc bc khng qu mt s nguyn dng n cho trc no .

V th, v sau, ngi ta thng s dng cch vit a thc P(x) di dngtng dn ca bc ly tha

P(x) = b0 + b1x + b2x2 + + bnxn. (1.1)

Nhn xt rng a thc (1.1) c tnh cht

P(k)(0) = k!bk, k = 0, 1, . . . , n

vP(k)(0) = 0, k = n + 1, n + 2, . . .

V th a thc (1.1) thng c vit di dng cng thc (ng nht thc)

Taylor

P(x) = a0 +a1

1!x +

a2

2!x2 + + an

n!xn. (1.2)

Vi cch vit (1.2) ta thu c cng thc tnh h s ak(k = 0, 1, . . . , n) ca

a thc P(x), chnh l gi tr ca o hm cp k ca a thc ti x = 0:

ak = P(k)(0), k = 0, 1, . . . , n .

T y ta thu c ng nht thc Taylor ti x = 0

P(x) = P(0)+P(0)

1!x +

P(2)(0)

2!x2 + + P

(n)(0)

n!xn. (1.3)

V d 1.1. Vit biu thc

Q(x) = (x2 2x 2)5 + (2x3 + 3x2 x 1)2.di dng (chnh tc) cng thc Taylor ti

Q(x) = a0 +a1

1!x +

a2

2!x2 + + a10

10!x10.

Tnh gi tr ca a8?

9


11/63

Theo cng thc (1.3) th ta c ngay h thc a8 = Q(8)(0)

Dng (1.2) cho ta mi lin h trc tip gia cc h s ca mt a thc chnh

tc vi cc gi tr o hm ca a thc ti x = 0. Trong trng hp tng

qut, cng thc Taylor ti x = x0 c dng:

P(x) = P(x0)+P(x0)

1!(xx0) + P

(2)(x0)

2!(xx0)2 + + P

(n)(x0)

n!(xx0)n.

(1.4)

V d 1.2. Vit biu thc:

Q(x) = (x

1)(x

2)...(x

9)

di dng (chnh tc) cng thc Taylor ti im x = 10:

Q(x) = a0 +a1

1!(x 10)x + a2

2!(x 10)2 + + a9

9!(x 10)9.

Tnh gi tr ca a7?

Cng thc (1.4) cho ta h thc a7 = Q(7)(10).

Trong trng hp a thc bc ty , ta c cc kt qu hon ton tng t.

V d 1.3. Chng minh rng nu a thc P(x) tha mn iu kin deg P(x) n v P(k)() = qk, k {0, 1, . . . , n}, trong , qk l cc s cho trc;P(0)(x) P(x) th P(x) c dng:

P(x) =n

k=0qk

k!(x )k.

ng thc trn c chng minh bng cch ly o hm lin tip hai v

v s dng gi thit v cc gi tr ban u P(k)() = qk, k {0, 1, . . . , n}..Vic chng minh tnh duy nht c suy ra t tnh cht ca cc a thc

(khc 0) bc khng vt qu n l n c khng qu n nghim (k c bi).

By gi, ta chuyn sang bi ton ni suy Taylor.

Bi ton 1.1 (Bi ton ni suy Taylor). Cho x0, akR vi k = 0, 1, . . . , N

1. Hy xc nh a thc T(x) c bc khng qu N 1 v tha mn cc iu

10


12/63

kin:

T(k)(x0) = ak, k = 0, 1, . . . , N 1. (1.5)

Gii. Trc ht, d thy rng a thc

T(x) =N1k=0

k(x x0)k.

c bc deg T(x) N 1. Tip theo, ta cn xc nh cc h s k R saocho T(x) tha mn iu kin

T(k)(x0) = ak,

k = 0, 1, . . . , N

1.

Ln lt ly o hm T(x) n cp th k, k = 0, 1, . . . , N 1 ti x = x0 vs dng gi thit

T(k)(x0) = ak, k = 0, 1, . . . , N 1.

Ta suy ra

k =ak

k!, k = 0, 1, . . . , N 1.

Thay gi tr ca k vo biu thc ca T(x), ta thu c

T(x) =N1k=0

ak

k!(x x0)k. (1.6)

Vi mi k = 0,1,...,N-1, ta c

T(k)(x) = ak +N1

j=k+1

aj

(j

k)!

(x x0)jk.

Do vy a thc T(x) tha mn iu kin

T(k)(x0) = ak, k = 0, 1, . . . , N 1.

Cui cng, ta chng minh rng a thc T(x) nhn c t (1.6) l a thc

duy nht tha mn iu kin ca bi ton ni suy Taylor .

Tht vy, nu c a thc T(x), c bc deg T(x) N 1 cng tha mn

iu kin ca bi ton (1.5) th khi , a thc

P(x) = T(x) T(x)11


13/63

cng c bc deg P(x) N 1 v ng thi tha mn iu kinP(k)(x0) = 0, k = 0, 1, . . . , N 1.

Tc l, a thc P(x) l a thc c bc khng qu N

1 m li nhn x0 lm

nghim vi bi khng nh thua N, nn P(x) 0, v do T(x) = T(x).nh ngha 1.3. a thc

T(x) =N1k=0

ak

k!(x x0)k

c gi l a thc ni suy Taylor.

Nhn xt 1.1. Ch rng a thc ni suy Taylor T(x) c xc nh t(1.6) chnh l khai trin Taylor n cp th N 1 ca hm s T(x) ti imx = x0.

1.3 Khai trin Taylor vi cc phn d khc nhau

Cc nh l, nh ngha v bi ton trong mc ny ch yu c trch t

[1].

Ta xt cng thc khai trin Taylor i vi a thc. Tip theo, trong mc

ny, ta s xc lp cng thc Taylor vi cc phn d khc nhau. Ta nhc li,

khi hm f kh vi ti im x = a th theo nh ngha, ta c

f(a + h) f(a) = f(a)h + o(h).Nu t a + h = x th h = x a v

f(x) f(a) = f(a)(x a) + o(x a).Ni mt cch khc, tn ti hm tuyn tnh

P1(x) = f(a) + f(a)(x a)

sao cho

f(x) = P1(x) + o(x a)trong

P1(a) = f(a), P1(a) = f(a).

Ta pht biu mt s bi ton sau y.

12


14/63

Bi ton 1.2. Gi s hm f xc nh trn tp hp R, trong lhp ca cc khong m trn trc thc. Gi s f kh vi cp n ti im a .Hy xc nh cc a thc Pn(x) c bc deg Pn(x) n sao cho

P(k)n (a) = f(k)(a), k = 0, 1, . . . , n .

Gi s P(x) l a thc (bc n) ty . Ta vit

P(x) =n

k=0

ak

k!(x a)k.

Khi

P()(a) =a

!

! = a.

Nu by gi ta t

a = f()(a), = 0, 1, . . . , n .

th

f()(a) = P()(a).

Nh vy bi ton c gii xong.

Tip theo, ta xt bi ton c lng hiu f(x)

Pn(x).

nh ngha 1.4. a thc

Tn(f; x) =n

k=0

f(k)(x0)

k!(x a)k

c gi l a thc Taylor bc n vi tm a ca hm f, kh vi cp n ti im

a.

Ta t

f(x) = Tn(f; x)+Rn(f; x) = f(a)+f(a)(xa)+...+ 1

n!f(n)(a)(xa)n+Rn(f; x).

(1.7)

Cng thc (1.7) c gi l cng thc Taylor (dng y ) ca hm f(x).

Nu a = 0 th (1.7) c gi l cng thc Maclaurin.

Biu thc Rn(f; x) c gi l phn d ca cng thc Taylor. Vi nhng

iu kin khc nhau t ra i vi hm f, phn d s c biu din bi cc

cng thc khc nhau. Li gii ca bi ton c lng hiu f(x)Pn(x) cngchnh l c lng cc biu thc phn d ny.

13


15/63

B 1.1. Nu hm c o hm n cp n ti im a v

(a) = (a) = ... = (n)(a) = 0,

th (x) = o((x a)n

) khi x a, tc l(x)

(x a)n 0(x a).

Chng minh. Ta chng minh bng phng php qui np. Vi n = 1 ta c(a) = (a) v

(x)

x

a

=(x) (a)

x

a

(a) = 0(x a)

tc l (x) = o((x a)). Gi s b ng vi n no , tc l vi iukin

(a) = (a) = ... = (n)(a) = 0,

th

(x) = o((x a)n).Ta cn chng minh rng vi (n+1)(a) = o th

(x) = o((x a)(n+1)), x a.

Ta xt hm (x) = (x). Ta c

(a) = (a) = ... = (n)(a) = 0.

v do (x) = o((x a)n), tc l(x)

(x a)n =(x)

(x a)n 0(x a).Theo nh l Lagrange, ta c

(x)

(x a)n+1 =(x) (a)(x a)n+1 =

()(x a)(x a)n+1 =

()( a)n

ax a

n.

trong , nm xen gia a v x. T ta thu c

()

( a)n 0(x

a), 0 0. (1.11)

15


17/63

Cng thc (1.10) c gi l cng thc Taylor i vi hm f vi phn d

Rn+1 di dng Schlomilch-Roche.

Chng minh. Khng gim tnh tng qut, ta xt x > x0. Xt hm s

h(t) = f(x) n

k=0

f(k)(t)k!

(x t)k (x t)pn!p

, x0 t x, (1.12)

trong p R, p > 0, l tham s.Hm h(t) lin tc trn on [x0, x], h(x) = 0 v o hm h(t) tn ti t (x0; x). Ta chn s sao cho

h(x0) = f(x)

n

k=0

f(k)(x0)

k!

(x

x0)

k

(x x0)p

n!p

= 0. (1.13)

Vi cch chn , hm h(t) tha mn mi iu kin ca nh l Rolle trn

on [x0, x]. Do , tn ti [x0, x], sao cho

h() = f(n+1)()

n!(x )n + (x )

p1

n! = 0. (1.14)

Tht vy, t h thc (1.11), ta c

h(t) = f(t) + f(t)1!

f(t)1!

(xt)+ f(t)2!

2(xt) ...+ f(n)(t)n!

(xt)n1

f(n+1)(t)

n!(x t)n + (x t)

p1

n!. (1.15)

D dng thy rng mi s hng v phi ca (1.15) tr hai s hng cui

cng u kh nhau ht. T bng cch thay t = ta thu c (1.14). T

(1.14) ta c

= f(n+1)()(x )np+1. (1.16)Thay t (1.16) vo (1.11) ta thu c iu phi chng minh.

Bng cch chn cc gi tr p > 0 hon ton xc nh, ta thu c nhng

trng hp ring i vi phn d Rn+1(f; x). Ta xt nhng trng hp quan

trng nht khi p = n + 1 v p = 1.

Khi p = n + 1 th t (1.11) ta thu c phn d ca cng thc Taylor di

dng Lagrange

Rn+1(f; x) =f(n+1)()

(n + 1)!(x x0)n+1, = x0 + (x x0), 0 < < 1. (1.17)

16


18/63

Khi p = 1 th t (1.11) ta thu c phn d ca cng thc Taylor di dng

Cauchy

Rn+1(f; x) =f(n+1)(x0 + (x x0))

n!(x

x0)

n+1(1

)n, 0 < < 1 (1.18)

trong = x0 + (x x0).

Nhn xt 1.2. Cng thc Maclaurin vi cc phn d (1.17) v (1.18) cdng tng ng

Rn+1(f; x) =f(n+1)(x)(n+1)!

xn+1, 0 < < 1 (dng Lagrange).

Rn+1(

f;

x) =

f(n+1)(x)

(n+1)! (1

)

nxn+1,0

< 1 v f(x) = x x

2

2+ + (1)n x

n1

n 1 + Rn(x). Trong trnghp ny ta vit hai cng tc phn d

Rn(x) =(1)n+1xnn(1 + x)n

, 0 < < 1

(dng Lagrange);

Rn(x) = (1)n+1 xn

(1 + x)

1

1 + x

n1, 0 < < 1

44


46/63

(dng Cauchy).

Gi s 0 x 1. Khi , p dng cng thc phn d dng Lagrange ta c

|Rn(x)| 1

n|x|n

0(n ).Trong trng hp 1 < x < 0, cng thc phn d dng Lagrange khng

cho ta kt lun v dng iu ca Rn (v y ch bit 0 < < 1 v

= (x, n)). S dng cng thc phn d dng Cauchy ta c

|Rn(x)| < |x|n

1 |x| 0(n ), 0 < |x| < 1,

v rng 0 1, Rn(x) 0 khi n . thy r iu ny ta t

S(x) = x x2

2+ (1)n x

n1

n 1.

Khi

Sn(x) + Rn(x) = Sn+1(x) + Rn+1(x)

vRn(x) Rn+1(x) = (1)n+1x

n

n.

Vi x > 1 v n v phi ca ng thc trn khng dn n 0. Do, Rn(x) khng dn n 0 khi n v khng tha mn tiu chun tnti gii hn Cauchy.

Nh vy phn d Rn(x) ca cng thc Taylor i vi hm ln(1 + x) ch

dn n 0 vi

1 < x < 1.

Xt hm f(x) = (1 + x)m. i vi hm ny ta ch cn xt m R\N. Tac

f(n)(x) = m(m 1)...(m n + 1)(1 + x)mn,f(n)(0) = m(m 1)...(m n + 1).

Cng thc Taylor theo cc ly tha ca x c dng

(1+x)m = 1+mx+m(m

1)

2!x2+

+

m(m

1)...(m

n + 1)

(n 1)!xn

1+R

n(x).

45


47/63

Khi , phn d di dng Lagrange c dng

Rn =m(m 1)...(m n + 1)

n!xn(1 + x)mn.

V di dng Cauchy

Rn =m(m 1)...(m n + 1)

(n 1)! xn(1 + x)m1

1

1 + x

n1.

Vi 0 x < 1, th theo trn ta c vi n > m

|Rn| |m(m 1)...(m n + 1)|n!

|x|n 0(n ).

Tht vy, ta tUn =

m(m 1)...(m n + 1)(n 1)! x

n.

Khi ,|Un+1||Un| =

|m n|n + 1

|x|.Suy ra

|Un+1

|=

|m n|n + 1 |

x

||Un

|.

Vi n ln, v phi nh hn 1 v|m n|

n + 1 1 khi n v 0 x < 1.

Do dy (Un) c gii hn u no . Chuyn qua gii hn ng thc trn

khi n , ta thu c |u| = |x|.|u| hay u = 0. Do Rn 0(n )vi 0 x < 1.Vi 1 < x < 0 th t cng thc phn d di dng Cauchy ta c

|Rn| C|m(m 1)...(m n + 1)|(n 1)! |x|n. (3.2)

Trong , C l s ph thuc x nhng khng ph thuc n.

Tht vy, c lng |Rn(x)|, ta xt1

1 + x

n1

1 1

n1= 1

v vi m 1 > 0 th (1 + x)m1 2m1.

46


48/63

Khi m 1 < 0(1 + x)m1 0, b > 0).

Gii. Gii hn cho c dng 0.. Ta a v dng 00

bng php

i bin.

t

x = t th t2 = x v khi x

0+ th t

0+. Ta c

limx0+

1

x

x

a arctan

x

a

b arctan

x

b

52


54/63

= limt0

a arctan

ta b arctan

tb

t3.

V mu s l a thc bc ba nn ta cn khai trin Taylor t s chnh xc

n 0(t3). Khi t 0+ th

a arctan

t

a=

a

ta

+ t33

a3

+ o(t3

3

a3 )

= t +

t3

3a+ o(t3), t 0+,

b arctan

t

b=

b

t

b+ t33

b3

+ o(t3

3

b3)

= t +

t3

3b+ o(t3), t 0+.

Khi

limx0+

1x

x

a arctanxa

b arctan

xb

= limt0

t + t

3

3a+ o(t3)

t + t3

3b+ o(t3)

t3

= limt0

ab3a

t3 + o(t3)

t3=

a b3ab

.

Vy

limx0+

1x

x

a arctanxa

b arctan

xb

= a b

3ab.

V d 3.14. Tnh gii hn

limx0

(cos(x.ex) ln(1 x) x)cotx3.

Gii. Gii hn cn tm c dng 1. Ta c

limx0(cos(x.e

x) ln(1 x) x)cotx3

= e limx0 cotx3

ln f(x)

vi

f(x) = cos(x.ex) ln(1 x) x.Ta cn tnh

limx0

cot x3 ln[cos(x.ex) ln(1 x) x]. rng

cot x3 = 1tan x3

= 1x3 + o(x3)

, x 0.

53


55/63

Do ta cn phi khai trin hm [f(x) 1] theo cng thc Taylor tngng vi o(x3) bng cch s dng cc khai trin sau

xex = x + x2 + o(x2), x 0.

cos t = 1 t2

2!+ o(t3), t 0.

Suy ra

cos xex = 1 x2

2 x3 + o(x3), x 0.

ln(1 x) = x + x2

2+

x3

3+ o(x3), x 0.

Ta thu c

f(x) 1 = 23

x3 + o(x3), x 0v

limx0

23x

3 + o(x3)

x3 + o(x3)= 2

3

Vy

limx0

(cos(x.ex) ln(1 x) x)cotx3 = e 23 .

3.2 Khai trin Taylor- Gontcharov vi bi ton clng hm s

Trong mc ny, ta s xt khai trin Taylor - Gontcharov ca mt s hm c

th cng nh nh gi c lng phn d ca khai trin Taylor - Gontcharov.

V d 3.15. Xc nh cc tam thc bc hai f(x) tha mn iu kin

f(0) = 1; f(3) = 0; f(5) = 5.Gii. Theo cng thc ni suy Newton trong trng hp n = 2 ta c

f(x) = f(0) + f(3)(x 0) + f(5)

(x 3)22

(0 3)2

2

= 1 + 5

(x 3)22

92

.

Hayf(x) =

5

2x2 15x 1.

54


56/63

V d 3.16. Xc nh a thc bc ba f(x) tha mn cc iu kin

f(n)(3n + 1) = n3 3n2 + n + 1, n = 0, 1, 2, 3.

Gii. a thc f(x) cn tm phi tha mn cc iu kin sauf(1) = 1; f(4) = 0; f(7) = 1; f(3)(10) = 4.

Theo cng thc ni suy Newton trong trng hp n = 3, ta nhn c a

thc cn tm c dng

f(x) =2

3x3 87

6x2 + 84x 1135

6.

V d 3.17. Vi hm f(x) = ex

, x R ta c f(n)

(x) = ex

, x R vi ccmc ni suy x0 = 0; x1 = 1; x2 = 2 ta c

f(0) = 1; f(0) = e; f(2) = e2.

Do , a thc ni suy Newton bc 2 ca f(x) c dng

P2(x) = 1 + e

x

0

dt1 + e2

x

0

t1

1

dt2dt1 = 1 + ex + e2

(x 1)22

12 .

T , khai trin Taylor-Gontcharov ca hm s f(x) = ex vi cc mc ni

suy trn c dng

ex = 1 + ex + e2

(x 1)22

12

+ R3(f; x).

Trong , phn d R3(f; x) c xc nh bi

R3(f; x) = f(3)()

3!(x x0)3 f(3)(1)

2!R1(x0, x)(x1 x0)2

f(3)(2)

1!R2(x0, x1, x)(x2 x0) = e

6x3 e

1

2x e2 (x 1)2 1 .

Trong , nm gia x v 0; 1 (0; 1), 2 (0;2).V d 3.18. Xt hm f(x) = sin x vi cc mc ni suy x0 =

6; x1 =

4; x2 =

3; x3 =

2. Ta c

f(n)(x) = sin(x + n

2), n = 0, 1, 2, . . . .

55


57/63

Mt khc

f(

6) =

1

2; f(

4) =

2

2; f(

3) =

3

2; f(

2) = 0.

Do , phn d trong khai trin Taylor-Gontcharov n bc 3 ca hm sf(x) = sin x l

R4(f; x) =f(4)()

4!(x

6)4 f

(4)(1)

3!

3

1728

x

6

dt f(4)(2)

2!.2

36

x

6

t1

4

dt2dt1

f(4)(3)

1!.

3

x

6

t1

4

t2

3

dt3

.dt2

.dt1

=sin

4!(x

6)4 sin 1

3!

3

1728(x

6) sin 2

2!.2

72

(x

4)2

2

144

sin 31!

.

3

1

6(x

3)3

2

288(x

4)

3

1296

.

Trong , nm gia x v

6; 1 (

6;

4), 2 (

6;

3); 3 (

6;

2).

V d 3.19. Cho N(x) l a thc c bc deg N(x) 3 v tho mn cciu kin

|N(k)(k)| 1 , k = 0, 1, 2, 3.Chng minh rng

|N(x)| 112

x [0, 3].

Gii. p dng cng thc ni suy Newton ti cc nt ni suy xi = i 1,i = 1, 2, 3, 4 , ta c

N(x) = N(0) + N(1)R(0, x) + N(2)R2(0, 1, x) + N(3)(3)R3(0, 1, 2, x),

trong

R(0, x) =

x0

dt = x,

R2(0, 1, x) =x0

t11

dtdt1 = x2!

(x 2),

56


58/63

R3(0, 1, 2, x) =

x0

t11

t22

dtdt2dt1 =x

3!(x 3)2.

Khi

N(x) = N(0) + N(1)x + N(2)x

2!(x 2) + N(3)(3)x

3!(x 3)2.T gi thit bi ton ta suy ra

|N(x)| 1 + |x| + x

2!(x 2)

+ x3!

(x 3)2

1 + x + x2!

(x 2) + x3!

(x 3)2.

t

f(x) = 1 + x + x2!(x 2) + x3!(x 3)2

=1

6x3 1

2x2 +

3

2x + 1.

Kho st hm s f(x) trn [0, 3] ta c

f(x) f(3) = 112

.

Vy

|N(x)| 112

.

V d 3.20. Cho N(x) l a thc c bc deg N(x) 5 c dngN(x) = a5x

5 + a4x4 + a3x

3 + a2x2 + a1x + a0

v tho mn cc iu kin

|N(4)(4)| 1 , |N(5)(5)| 1.Tm gi tr ln nht ca |N(4)(x)| trn [1, 6].

Gii. Ta nhn thy N(4)(x) l hm s bc nht nn gi tr ln nht caN(x) trn [1, 6] s t c ti x = 1 hoc x = 6.

Do vy |N(4)(x)| cng s t gi tr ln nht trn [1, 6] ti mt tronghai u mt x = 1 hoc x = 6.

p dng cng thc ni suy Newton ti cc nt ni suy xi = i 1 ,i = 1, 2, 3, 4, 5, 6 , ta c

N(x) = N(0) + N(1)R(0, x) + N(2)R2(0, 1, x) + N(3)(3)R3(0, 1, 2, x)

57


59/63

+N(4)(4)R4(0, 1, 2, 3, x) + N(5)(5)R5(0, 1, 2, 3, 4, x),

trong

R4(0, 1, 2, 3, x) =

x0

t11

t22

t33

dtdt3dt2dt1 = x4!

(x 4)3,

R5(0, 1, 2, 3, 4, x) =

x0

t11

t22

t33

t44

dtdt4dt3dt2dt1 =x

5!(x 5)4.

Suy ra

N(4)(x) = N(4)(4) + N(5)(5)(x

4).

Khi

N(4)(1) = N(4)(4) 5N(5)(5),N(4)(6) = N(4)(4) + 2N(5)(5).

T gi thit bi ton, ta suy ra

|N(4)(1)| = |N(4)(4) 5N(5)(5)| 6,

|N(4)(6)| = |N(4)(4) + 2N(5)(5)| 3.Vy gi tr ln nht ca |N(4)(x)| bng 6 ti x = 1.

Tng t nh vi khai trin Taylor, vi khai trin Taylor- Gontcharov ta

cng c nhn xt sau

Nhn xt 3.1. Vn mu cht trong vic tm cng thc nh gi phnd ca cng thc khai trin Taylor - Gontcharov i vi hm f cng vi cc

mc ni suy xi cho trc l vic tnh o hm cp cao ca hm f v biu

thc Rk(x0, x1, . . . , xk1, x). Tuy nhin, vic tnh o hm cp cao ca mthm s nhiu khi khng n gin. Bn cnh , khi mc ni suy tng ln th

vic tnh ton biu thc Rk(x0, x1, . . . , xk1, x) kh phc tp. Do , trongcc v d trn, v gi tr ca mc ni suy nh nn ta mi a ra c biu

thc nh gi phn d ca f(x) mt cch chnh xc. Cn trong trng hp

tng qut, ta ch c th a ra nh gi di dng Lagrange v Cauchy

nh trong (2.12) v (2.14).

58


60/63

Tuy nhin, trong trng hp hm f(x) c o hm gii ni

|f(n)(x)| M, x [a; b], n = 0, 1, 2, . . .

ta c kt qu sauBi ton 3.1. Gi s hm s f(x) lin tc trn [a; b], c o hm mi cptrn [a; b] v xi [a; b], i = 0, 1, 2, . . . , n. Khi , nu M > 0 sao cho

|f(n)(x)| M, x [a; b], n = 0, 1, 2, . . .

th Rn+1(f; x) 0 khi n .Gii. Theo cng thc xc nh ca R

n+1(f; x) ta c

Rn+1(f; x) =f(n+1)()

(n + 1)!(x x0)n+1

n

k=1

f(n+1)(k)

(n k + 1)!Rk(x0, x1, . . . , xk1, x)(x x0)nk+1.

Mt khc:

Rk(x0, x1, . . . , xk1, x) = x

x0

t1x1

...

tk1xk1

dtkdtk1...dt1

ba

t1a

...

tk1a

dtkdtk1...dt1 = 1

k!(b a)k.

Do

Rn+1(f; x) M(b a)(n+1)(n + 1)!

+n

k=1

M(n k + 1)! . 1k!(b a)

k(b a)nk+1

= M(b a)(n+1)

1

(n + 1)!+

nk=1

1

k!(n k + 1)!

M(b a)(n+1)

1

(n + 1)!+

nk=1

1

n2! n

n2 + 1!

= M(b a)(n+1)

1

(n + 1)!+

nn2

!

n n2

+ 1

!

.

59


61/63

V n Z nn n = 2p + i, (i = 0, 1). Khi ta cRn+1(f; x)

M(b a)2p+i+1

1(2p + i + 1)!

+2p + i

p!(p + i + 1)!

.

Trng hp 1. Nu i = 0, ta cRn+1(f; x) M(b a)2p+1(2p + 1)!

+(b a)p

p!

2.(b a)2p

(p + 1)

0(p ).

Trng hp 2. Nu i = 1, ta cRn+1(f; x)

M

(b a)2p+2

(2p + 2)!+

(b a)p

p! 2

.(b a)2(2p + 1)

(p + 1)(p + 2) 0(p ).

Vy Rn+1(f; x) 0 khi n .

60


62/63

KT LUN

Lun vn trnh by v thu c

- Mt s kt qu c bn v bi ton ni suy Taylor, khai trin Taylor,

nh gi cng thc phn d v s hi t ca khai trin Taylor.

- a ra cng thc nghim ca bi ton ni suy Newton, biu din hm

s f(x) theo cng thc khai trin Taylor- Gontcharov v c bit a ra cc

nh gi phn d ca khai trin Taylor - Gontcharov ca hm f(x) di hai

dng Lagrange v Cauchy. Bn cnh , lun vn nh gi s hi t ca

khai trin Taylor - Gontcharov v khi qut ha bi ton ni suy Newton

i vi hm a thc nhiu bin.

- Mt s ng dng ca khai trin Taylor v khai trin Taylor - Gontcharov

trong vic c lng v nh gi sai s, tnh gii hn hm s....

- Mt s hng nghin cu c th pht trin t ti ny l

1. Khai trin Taylor - Gontcharov i vi hm nhiu bin v nh gi phn

d ca n.

2. Cc ng dng ca khai trin Taylor - Gontcharov trong phng trnh vi

phn, trong l thuyt cc bi ton bin....

V thi gian v kin thc cn hn ch nn lun vn chc chn khng trnh

khi nhng thiu st. Tc gi rt mong nhn c s quan tm, ng gp

kin ca cc thy c v cc bn ng nghip bn lun vn c hon

thin hn.

Tc gi xin chn thnh cm n!

61


63/63

TI LIU THAM KHO

[1] Nguyn Vn Mu, Cc bi ton ni suy v p dng, NXBGD, 2007.

[2] Nguyn Vn Mu, a thc i s v phn thc hu t, NXBGD,2004.

[3] Nguyn Xun Lim, Gii tch, tp I, NXB Gio dc 1998.

[4] Nguyn Duy Tin, Trn c Long, Bi ging gii tch, NXB ihc Quc gia H ni 2004.

[5] Nguyen Van Mau, Algebraic Elements and Boundary Value Problemsin Linear Spaces, VNU Publishers, Hanoi 2005.

[6] Przeworska-Rolewicz, D. Equations with Transformed Argument. AnAlgebraic Approach, Amsterdam-Warsaw 1973.

[7] Przeworska-Rolewicz, D. Algebraic Analysis, PWN - Polish Scien-tific Publishers and D. Reidel Publishing Company, Warszawa - Dor-

drecht, 1988.

Công thức khai triển TAYLOR - GONTCHAROV vá áp dụng

Documents

Transcript of Công thức khai triển TAYLOR - GONTCHAROV vá áp dụng