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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
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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
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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
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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 .
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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
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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
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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.
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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).
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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
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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?
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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
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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
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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.
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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.
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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)
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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)
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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
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(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).
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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.
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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
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= 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.
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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.
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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, . . . .
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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),
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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)
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+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).
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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
!
.
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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 .
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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!
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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.