su_hoi_tu-libre

Chng 6 CHUI S V CHUI LY THA

Trong chng ny, chng ti trnh by nhng khi nim v tnh cht c bn thng c s dng v chui s. Mt s tnh cht c bn v chui s dng, chui an du nh tiu chun Leibnitz cng c gii thiu. Chng ti cng a ra nhng khi nim c bn mang tnh cht gii thiu v chui hm, phn quan trng m chng ti mun nhn mnh y l kho st s hi t cng nh khai trin mt s hm thng gp thnh chui ly tha. 6.1. Chui s 6.1.1. Cc khi nim c bn 1. nh ngha Cho dy s v hn , tng v hn +Znnu )( c gi l chui s, k hiu l: ......321 +++++ nuuuu =1n nu c gi l s hng th n. nu2. Dy tng ring t c gi l tng ring th n ca chui s nn uuuus ++++= ...321 =1n nu c gi l dy tng ring ca chui s . +Znns )( =1n nu3. Chui s hi t, phn k

Chui s c gi l hi t nu tn ti gii hn =1n nu ssLim nn = v c gi l tng ca n. Ta vit: .

s

sun

n ==1 Nu gii hn khng tn ti hay bng n

nsLim th chui s c gi l phn k

v khi chui s khng c tng. =1n nu

4. Phn d th n Trong trng hp chui s hi t c tng bng S th hiu S-S=1n nu n c gi l phn d th n ca chui s , k hiu l: r=1n nu n Vy, di dng ngn ng -N, ta c:

122

Chui s hi t =1n nu > nssNnN :,0 > nrNnN :,0 5. Cc v d 1) (tng cp s nhn v hn)

0

1 ... ...nn

q q q= = + + + + n

q Ta c tng ring . Xt cc trng hp sau 1 ... nn

S q= + + + a) q 1

Ta c11

1

n

n

qS

q

+= , suy ra

= 1,

1

1

1,

limq

q

q

Snn

b) q = 1

Ta c Do : li1 1 ... 1nS n= + + + = m .nn

S = + c) q = -1

Ta c 1, 2 1

1 1 1 ...0, 2n

n kS

n k

= += + = = . Do khng tn ti lim nn S Vy

0

1

1n

n

qq

= = , hi t, nu | | 1q < .

Chui s phn k nu | | th chui phn k 0

n

n

q= 1q

2) Cho chui s +=1 )1( 1n nn =+++++=+++++= )111(...)4131()3121()211()1( 1...4.313.212.11 nnnnsn

1

11 += n

Vy, chui s cho hi t v c tng bng 1. 1n

n

lims =6.1.2. Tiu chun hi t Cauchy 1. Tiu chun Cauchy Chui s hi t =1n nu >> qp ssNqpN :0,0 . 2. V d

123

Dng tiu chun Cauchy, chng t rng chui s =1 1n n phn k. Gii

=>==+++>+++++====>==

3

1

2

1

22

1...

2

1

2

1

2

1...

2

1

1

1

:2,:3

12

N

N

NNNNNN

ssssNNqNpN NNqp

6.1.3. iu kin cn chui s hi t 1. nh l Nu chui s hi t th li=1n nu m 0nn u = . Chng minh: Gi s l tng ca chui s hi t =1n nu n ns s Suy ra 1 0n n n

nu s s s s = =2. H qu Nu th chui s phn k. lim 0n

nu =1n nu

V d Chui s +=1 12n nn phn k v 12 += nnun 021 nkhi 3. Ch

ch l iu kin cn m khng chui s hi t. 0n nu =1n nu Chng hn, xt chui s =1 1n n n

n

n

nnnnnsn ==++++>++++= 1...1111...

3

1

2

1

1

1

M += nLimn += + nn sLim . Vy, chui s =1 1n n phn k. 6.1.4. Tnh cht cu chui s hi t 1.Tnh cht 1

124

Nu chui s hi t c tng l s, chui s hi t c tng l s th cc chui cng hi t v c tng l s s

=1n nu =1n nv)(

1n

nn vu =

Chng minh: Gi sn v sn ln lt l cc tng ring th n ca cc chui s v . =1n nu =1n nv Khi , v n

nlim s s = / /nnlim s s = /( )n nnlim s s s s + = /+ .p.c.m

V d Tnh tng ca chui s sau: +=1 12 43n n nn Gii Ta c

1

11 14( )

14 314

n

n

= = = v

1

11 13( )

13 213

n

n

= = = 1 1 13 4 1 1 1 112 4 3 3 2 6( ) ( )n n n nnn n n = = = 5= + =+ = +

2. Tnh cht 2 Nu chui s hi t c tng l s th chui s cng hi t v c tng l ks. =1n nu =1n nku Chng minh:

Gi sn ln lt l tng ring th n ca chui s: =1n nu kssLimkksLim n

nn

n== .p.c.m.

3. Tnh cht 3 Tnh hi t hay phn k ca 1 chui s khng thay i khi ta ngt b i khi chui s 1 s hu hn cc s hng u tin. Chng minh: Nu bt i t m s hng u tin, ta c chui s =1n nu += 1mn nu

125

Gi sn v sk ln lt l cc tng ring th n v th k ca cc chui s v

=1n nu += 1mn numkmk sss = +/

* Nu chui s hi t =1n nu m k m ks s+ + / mk ks s s chui s hi t. += 1mn nu * Nu chui s phn k khng c gii hn khi =1n nu kms + k v do smhu hn s k khng c gii hn khi k chui s phn k. += 1mn nu V d Xt s hi t ca chui s +=1 31n n Gii Chui ny suy t chui iu ho bng cch ngt b i 3 s hng u tin. M chui iu ho phn k nn chui +=1 31n n cng phn k. Bi tp Tnh tng ca cc chui sau

+=1 )4( 1)1 n nn =1 2 14 1)3 n n ++=1 22 )1( 12)4 n nn n ++=1 )2)(1( 1)2 n nnn +=1 10 52)5 n n nn =1 2 14 1)6 n n

6.2. Chui s dng 6.2.1. nh ngha

Chui s dng l chui s , m =1n nu 1,0 > nun V d

1

1

1.3nn n

= + l chui s dng.

6.2.2. nh l Chui s dng hi t khi v ch khi dy (sn) b chn trn. Chng minh:

126

V hi t nn dy (s=1n nu n) hi t. M v , suy ra dy (s0, 1nu n> n) tng, do (sn) b chn trn. Ngc li nu (sn) b chn trn, th tn ti di hn, v dy (sn) tng, do chui s hi t. =1n nu V d Xt s hi t ca cc chui s dng sau: 1) =1 21n n Ta c 2

12

)1(

1...

2.1

1

1

11...

2

1

1

1222

=++++++= nnnnS n Suy ra sn b chn. Vy chui trn hi t. 2) =1 1n n Ta c n

n

n

nnnnSn ==++++++= 1...111...

2

1

1

1

Suy ra sn khng b chn. Vy chui phn k. 6.2.3. Cc tiu chun hi t 1. Tiu chun so snh a. nh l

Gi s v l 2 chui dng tho =1n nu =1n nv 0nnvu nn , khi * Nu chui hi t th chui hi t. =1n nv =1n nu * Nu chui phn k th chui h phn k. =1n nu =1n nv Chng minh: Do tnh cht 3 ca chui s hi t, c th gi s 10 =n , ngha l nvu nn * Gi sn v sn ln lt l tng ring th n ca cc chui v =1n nu =1n nv s n sn (1) n

127

Nu chui hi t v c tng l s, ngha l =1n nv // ssLim nn = s n s (2) n T (1) v (2) Chui hi t. nssn + +

128

M chui 2

1

1n n

= + phn k nn chui 2 ln 1n nn= + phn k.

2. Tiu chun tng ng

Gi s v l 2 chui dng tho =1n nu =1n nv kvunnn =lim 1) Nu th hai chuis v, ng thi hi t hoc phn k. + kvunnn nn00 :0,0 .

Do nn

uk

v< + suy ra 0( ) ,n nu k v n n< + .

Nu hi t nn chui =1n nv = +1 )(n nvk hi t. Theo nh l trn ta suy ra chui hi t.

1n

n

u=

Nu phn k th ta cng lm tng t, tuy nhin ch t 1

n

n

v= lim nn nu kv = suy ra 1

lim nn

n

v

u k = . V nn 0 k< < + 10 k< < + . Do o nu chui hi t th t suy ra chui hi t. Vy phn k.

=1n nu1

n

n

v= 1 nn u=

Vy 2 chui , ng hi t hoc phn k. 1

nn

u= 1 nn v=

2) Gi s v hi t. 0k =1

n

n

v=

129

Khi t gi thit lim 0nn

n

u

v = ta c 0 00, 0 : ,nnun n nv > > < 0,n nu v n n < . V hi t, nn

1n

n

v= 1 nn v= hi t, do 1 nn u= hi t.

3) Chng minh hon ton tng t nh mc (2). Gi s k = + v 1

nn

v= phn k. T

lim nn

n

u

v = + suy ra lim 0nnn

v

u = . Do phn k, v nu hi t th theo (ii) suy ra =1n nu =1n nu 1 nn v= hi t mu thun. Ch

Thng ta so snh vi chui s quan trng chui cp s nhn v chui iu ho. V d Xt s hi t ca cc chui s sau: 1)

2

1

2 1

5 2 2

n

nn

n

n

=

+ ++ + Ta c

22 10

5 2 2

n

n n

nu

n

+ += + + > , vi mi . Ta s so snh vi chui s 1n 1 1 2( )5 nnn nv = == hi t. D thy rng 1lim =

n

n

n v

u, do chui s cho hi t.

2) 1

ln

n

n

n

=

Ta c ln 1

,n

nu

n n= vi mi . 3n

M chui 1

1

n n

= phn k ( v d trn), nn chui cho phn k.

3) 2

1

3 1

2n

n

n n n

=

++ + Ta c

2

3 10

2n

nu

n n n

+= >+ + , vi mi . 1n

130

Chn 1 0nvn n

= > . Ta c. Do 3lim =n

n

n v

u chui =1n nv hi t, nn = +++1 3 21n nn n hi

t. 3. Tiu chun /D Alemberta. nh l /D Alembert Nu chui s dng tho =1n nu DuuLim nnn =+ 1 th chui s s hi t khi =1n nu 1D Khi Chui s dng c th hi t hoc phn k. 1D = =1n nu Khi D =+ chui s dng phn k. =1n nu Chng minh: * 1D < 1 - D > 0 Chn D< 1 1 <

01 )( nnuDu nn >+ + 0 nn uLim 131

chui s dng phn k. =1n nu* Khi Vi M=1, :+=D 1: 1 >> +

n

n

u

uNnN

1`n nu u n+ > > N chui s dng phn k. 0 nnu =1n nu

b. V d Xt s hi t ca cc chui sau = ==+1n 1n nn u2 1n1 )!() 1

1

( 2) 2 2[ ]

2 ( 1)! 2

n

n

nn n n

n

u n nlim lim lim

u n

+ + + += =+ =

Chui s +=1 2 )!1(n nn phn k. == = 1n n1n n u5n2) 1

1

1 5 1 1.

5 51

n

n

nn n n

n

u n nlim lim lim

u n n

+ + + += =

Khi th cha c kt lun g, ngha l chui c th hi t, cng c th l phn k.

1=D Chng hn, xt chui =1 !n nnn ne Ta c 1

11

1 +=+

nn

n

n

e

u

u khi . V n en

n

1nu u = e em nu =1 !n nnn ne phn k. Vy chui cho hi t. 4. Tiu chun Cauchy Cho chui s dng . Gi s =1n nu Lun nn =lim . Khi 1) Nu L < 1 th hi t; =1n nu 2) Nu L > 1 th phn k. =1n nu Chng minh: Gi s: Lun n

n=lim .

- Khi L < 1. Ly r sao cho L < r < 1. Khi 00 ,:0 nnrun n n , ngha l 0, nnru

n

n < . V chui hi t nn chui = 0nn nr =1n nu hi t. - Khi L > 1. Ta c 00 ,1:0 nnun n n >> , tc l . Do khng dn v 0 khi . Vy chui phn k .

0,1 nnun > nun =1n nu Ch

Khi 1L = th cha c kt lun g, ngha l chui c th hi t, cng c th l phn k. V d Xt s hi t ca cc chui sau:

133

1) 1

2 1

3 2

n

n

n

n

=

+ + hi t, v 2 13l = < 2)

2

1

1n

n

n

n

=

+ phn k, 1>= el 5. Tiu chun tch phn Cauchy a. nh l Xt chui s dng . t hm s f(x) tha =1n nu 1,)( = nunf n Gi s hm f(x) lin tc, dng, gim trn );1[ + . Khi chui hi t hi t. =1n nu + 1 )( dxxf Chng minh: Theo gi thit, ta c vi mi k, hm f(x) gim trn on [k, k+1] nn ]1,[,)()()1(1 +=+=+ kkxukfxfkfu kk , theo nh l trung bnh tch phn ta c . Do vi mi k nn ta c

1

1 ( )k

k

k

u f x dx+

+ ku

1

1n

, , ..., , 1

2

12 )( udxxfu 23

23 )( udxxfu 1

1

)( nnnn udxxfu Suy ra:

2 3

2 31 2 1 1

... ( ) ( ) ... ( ) ( )n n

n

n

u u u f x dx f x dx f x dx f x dx+ + + + + + = 1 2 ... nu u u + + + Do :

n nn sdxxfus1

11 )(

t . Ta c, = nn dxxfI1

)( 1 ,n n nI Is u s (*) (? ) Gi s chui hi t. =1n nu Theo nh l mc 2, suy ra dy tng ring (sn-1) b chn. Do t bt ng thc (*) suy ra dy cng b chn. Hn na d thy dy { tng. Do vy tn ti, do li I}{I }In nmn n

134

hi t. 1

)( dxxf+

(? ) Gi s hi t. Khi b chn. T bt ng thc (*) suy ra b chn, cho nn chui hi t.

+1

)( dxxf }{ nI n{S }

1n

n

u=

b. V d 1) Xt s hi t ca chui

1

1,

n n = R (chui Riemann)

- Nu : t 0 > 1( )f xx= . Kim tra thy ( )f x tho tt c cc iu kin ca nh l.

Ta bit rng tch phn suy rng +1

1dx

x hi t khi v phn k khi 1 > 1 - Nu th 0 1lim lim 0

nu

n= Vy chui

1

1,

n n = R hi t khi v phn k khi 1 > 1

2) =1 3 2lnn nn Ta c

23 3

ln 1,

n

nu

n n=

2vi mi . M chui 3n =1

3

2

1

nn

phn k, nn chui cho

phn k. 3) = +1 3 4 11n nn n Ta c

6

1

2

1

3

43 4 1

.

~1

1

nnn

n

nn

n =+ . V chui =16

1

1

nn

phn k, nn chui = +1 3 4 11n nn n phn k. 4) =3lnn nn Gii Dng tiu chun tch phn, xt hm s

x

xxf

ln)( =

135

,),0( +=fD 2/ ln1)( x xxf = , exxf == 0)(/ Bng xt du o hm

x 0 e 3 + /f /f + 0 -

f

Hm lin tc, n iu gim, dng trong )(xf ),3[ + Mt khc,

2

3 3 3

ln lnln (ln ) ln (ln )

32b bbxdx x

xd x lim xd x limx

+ + ++ +

= = = += + 2 3lnln

22 bLimb

. Vy chui =3lnn nn phn k. 5) =2 ln1n nn Xt hm s

xxxf

ln

1)( = lin tc, dng trn ),2[ + v 2)( = nnfun

10ln

1ln)(

22

/ >

6) = +1 23n n n 6.3. Chui s an du - Chui s c du bt k 6.3.1. Chui an du 1. nh ngha Chui an du l chui s c dng hay ...321 + uuu ...321 ++ uuu , (1) Trong 0, 1

nu n>

V d ...

3

1

2

11 +

Ta quy c ch xt chui an du c dng . ( )= =+ 1 1321 1... n nn uuuu2. nh l Leibnitz a. nh l Nu dy { l mt dy gim v th chui hi t v

.

} nu 0nu khi n ( ) 11

1n

nn

u = ( ) 1 1

1

1n

nn

u u =

Chng minh: chng t dy tng ring (sn) hi t ta chng minh n c 2 dy con hi t (s2m) v (s2m+1)

Ta c s2(m+1) = s2m+2 =s2m + (u2m+1 - u2m+2 ) > s2m => (s2m) tng Mt khc, ta cng c [ ] 11227654321)1(2 )...()()()( uuuuuuuuuus mmm 0 m Ta li c: 12212 ++ += mmm uss Do 0nu 012 +mu sss m =+ + 012 112 :,0 mmmss m >>

2212 :,0 mmmss m >>+ * 122 mkn >= ssmk k21 * +>+= + ssmkmkn k 1222 1212 Vy > ssNnN n:,0 (.p.c.m) b. V d Xt s hi t cua chui an du = 1 1 1.)1(n n n Gii 0

1 = nn

nu v dy n iu gim hi t theo Leibnitz )( nu )( nuv tng 11 = usc. Ch

Nu chui (1) tho Leibnitz v hi t v s th chui hi t v -s ...)( 4321 ++ uuuu Nh vy nu cc gi thit ca nh l Leibnitz c tho th chui an du hi t v tng s ca n tho ...)( 4321 ++ uuuu 1us . d. Tnh gn ng tng ca chui an du hi t Nu chui an du ...)( 4321 ++ uuuu tho Leibnitz th chui phn d th n

cng hi t theo Leibnitz v theo ch trn ta c: ...21 ++ ++ nn uu 1+ nn ur Theo nh l Leibnitz, ta ch bit chui an du hi t nhng khng r bng bao nhiu nn ny sinh vn c lng tng .

ss

Ta xem s sn s vp phi sai s tuyt i l: 1+= nnn urss V d

Tr li chui = 1 1 1.)1(n n n , nu ta xem 78,02,025,033.05,0

5

1

4

1

3

1

2

115 ++++= ss

Vp phi sai s tuyt i l 167,06

165 = ur

Thng thng ta gp bi ton ngc li 138

Phi chn n ti thiu bng bao nhiu gi tr gn ng sn ca chui an du chnh xc n ( ngha l sai s tuyt i khng vt qu ). p dng vo v d trn, ta phi chn n sao cho: 65 ur Chng hn 001.0= , th th n phi tho

1

1+n 10001 99910001 + nn Vy, n ti thiu l 999. 6.3.2. Chui c du bt k 1. nh l

Nu chui s =1n nu hi t th hi t. nn u=1 Chng minh

Gi sn v sn ln lt l tng ring th n ca cc chui s v nn

u=1 =1n nu , ngha l nn uuuus ...321 +++= v nn uuuus ...321/ +++= Trong chui , k hiu

nn

u=1 l tng ca tt c cc s hng dng trong n s hng u tin +ns

ns l tng cc gi tr tuyt i ca tt c cc s hng m trong n s hng u tin. Ta

c

v + = nnn sss + += nnn sss /

R rng v l nhng dy tng v , (1) )( +ns )( ns /nn ss + /nn ss Theo gi thit, chui s =1n nu hi t v // ssn / /ns s n< (2) T (1) v (2) / /,n ns s n s s n+ < < Suy ra rng cc dy s v u hi t (v u tng v b chn trn.) )( +ns )( ns Do cng hi t. )( ns2. nh ngha

Chui s c gi l hi t tuyt i nu chui snn

u=1 =1n nu hi t. 3. V d

139

=1 3sinn nnx hi t tuyt i. Gii Ta c n

nn

nx

n

nx =333

1sinsin

m chui s =1 31n n hi t ( Chui Riemann vi )13 >= 4. Ch

iu kin =1n nu hi t ch l iu kin ch khng phi l iu kin cn chui s hi t. Ngha l c trng hp chui s hi t nhng chui s =1n nu =1n nu =1n nu phn

k, ta ni chui s bn hi t. =1n nu V d Chui s = 1 1 1.)1(n n n bn hi t v chui s = == 11 1 11)1( nn n nn l chui iu ho phn k. V d Xt tnh hi t ca cc chui s 1) 2

1

sin

n

n

n

=

Ta c |2

sin n

n|

2

1

n , do chui cho hi t

2) ( ) ++ nn nn 13 121 Ta c

2 1 2

13 1 3

| |n nn

nu

+ Chui cho hi t. Ch

Nu chui phn k th cha kt lun chui || nu nu hi t hay phn k. Tuy nhin, nu dng tiu chun DAlembert hay Cauchy m bit c || nu phn k th

cng phn k. nu Tht vy, t

140

0

111 | | | | 0,

n

n n n

n

uu u u n n

u

+ +> > > > >0 0 , do khng dn v 0, tc l khng tin v 0, suy ra chui phn k.

nu

nu

V d ( ) 21

!

nn e

n

Ta c ( )( )2 211 ! 1.1 ! 1nn nnu e n eu n ne++ = =+ + 2 1n+ + . Do chui cho phn k.

Trng hp phn k nhng || nu nu hi t th chui c gi l bn hi t.

nu V d ( ) 1

1

11

n

n n

= l bn hi t.

Bi tp 1) Chng t rng cc chui s sau bn hi t = +++1n 21n 1nn 1n1a )() = 1n 1n nn1b ln)() = ++1n 2n 1n 1n21c )() = ++1n 32n 3n 1n21d )() = 1n n 1n2 11e )() = +1n 2n 1n n1f )() 2) Cho chui s =1 !cosn nn a) Chng t rng chui s ny hi t theo Leibnitz, hn th na n cn hi t tuyt i. b) Phi chn n ti thiu l bao nhiu sn l tr gn ng ca tng ca chui vi chnh xc 001,0= 6.4. Chui lu tha 6.4.1. Chui hm 1. nh ngha Chui hm l chui , trong cc l cc hm ca x. ( )nu x ( )nu x Khi x = xo th chui hm tr thnh chui s )( 0xun . Nu chui s hi t th im xo gi l im hi t, nu n phn k th xo gi l im phn k. - Tp hp tt c cc im x m chui hm hi t c gi l min hi t ca chui hm.

141

- : gi l tng ring th ca chui hm. == nk kn xuxs 1 )()( n - Nu th S(x) gi l tng ca chui hm. Trong trng hp ny,

: gi l phn d th n ca chui hm. Do ta c

)()(lim xsxsn =)()()( xsxsxr nn =

...)()( 21 ++= ++ nnn uxuxr2. V d 1) =0n nx Chui ny hi t vi mi x tho |x| < 1 v c tng

xxS =1 1)( .

Vy min hi t ca chui trn l X = (-1; 1) 2)

1x

n c min hi t l (theo kt qu ca chui Riemann bit) );1( +=X

3) 3 2

cosnx

n x+ Ta c

3 2 3 2 3

sin 1 1,

nxx

n x n x n + + . M chui 31 1n n= hi t nn 3cos nxn x2+ hi t, x

Vy min hi t l . X = R6.4.2. Chui hm hi t u 1. nh ngha Chui hm c goi l hi t u ti hm S(x) trn X, nu )(xun

XxxrxSxSnnn nn >> ,)()()(:0,0 00 2. V d Chui ( ) + nx n2 1 hi t vi mi x (theo l Leibnitz) Ta c 1 2

1 1( ) ( ) ,

1 1n nr x u x x

x n n+ = < + + + R

Nh vy 1 1( ) , 11n

r x nn

< < > + Do 0, > ly 0 1 1n > . Khi 0 , ( ) ,nn n r x x < R Vy chui ( ) + nx n2 1 hi t u trn R . 3. Tiu chun v s hi t u

142

a. nh l (tiu chun Cauchy) Chui hm hi t u trn X khi v ch khi )(xun *0 00, : , ,n n p N n n > 1( ) ... ( ) ,n n pu x u x x X+ + + + < b. nh l (tiu chun Weierstrass) Cho chui hm . Nu c mt chui s dng hi t sao cho )(xun na

Xxnaxu nn ,1,)( th chui hm trn hi t tuyt i v u trn X. Chng minh. R rng chui )(xun , Xx hi t (theo tiu chun so snh) Do chui hi t tuyt i. )(xu

n V chui s hi t nn ta c na

Xxaa

xuxuxuxu

pnn

pnnpnn

Do 0sinlimsinlim2222=+=+ xn nxxn nx n xnx

b. Tnh cht 2 Cho chui hm hi t u v hm S(x) trn [a, b]. Nu cc s hng (x) u lin tc trn [a, b], th .

n

n xu )( nu

1n == ba )()()( n ba nnba dxxudxxudxxSc. Tnh cht 3 Cho chui hm

n

n xu )( hi t trn (a, b) ti S(x), cc s hng lin tc trn (a, b). Khi nu chui hi t u trn (a, b) th S(x) kh vi v S(x) =

.

)('),( xuxu nnn

n xu )('n

n xu )('

6.4.3. Chui ly tha 1. nh ngha Chui ly tha l chui hm c dng (1) ...10

0

++== xaaxan nn Ch

Nu chui lu tha c dng th bng cch t 00

( )nn

n

a x x= , 0xxX = ta a chui

v dng (1). V vy, ta quy c nghin cu chui ly tha c dng (1). V d 1)

0 2 1

n

n

x

n

= + trong 12 1na n= +

2) 2

0

2 1( 2)

7 1

n

n

n

nx

n

=

+ , trong 22 17 1 nn na n = + 2. nh l Abel Nu chui lu tha hi t ti =0n nn xa 00 = xx th n hi t tuyt i ti mi x tho

0xx < . Chng minh: Gi s chui lu tha hi t ti . Khi chui s =0n nn xa 0x =0 0n nn xa hi t,

khi . Do 0 0nna x n 144

00 : , 1n

nK a x K n >

Ta c 00 0

, 1

n n

n n

n n

x xa x a x K n

x x= .

Khi |0| | | x< th 0 0

n

n

xK

x

= hi t (v 10 . Chng minh: Tht vy nu c tho | m chui hi t ti . Khi theo nh l Abel n s hi t tuyt i ti

1x ||| 01 xx > 1x1| | | |x x < , m trong khong ny c cha im , iu ny

mu thun vi gi thit. 0x

4. Bn knh hi t ca chui ly tha a. nh ngha S c gi l bn knh hi t ca chui ly tha nu chui

hi t (tuyt i) vi mi | |

0r > =0n nn xa=0n nn xa x r< , v phn k vi mi | |x r> . Ch

Nu , th ch hi t ti 0r = =0n nn xa 0x = . b. nh l (Hadamard) (Cng thc tm bn knh hi t) Gi s 1|lim

| |n

nn

a

a += | (hoc lim | |)n na = . Khi bn knh hi t c tnh bng

cng thc:

1, 0

0,

, 0

r

< < += = ++ =

(*)

Chng minh:

145

Gi s 1|lim| |

n

nn

a

a += | . Ta c 1 1| ( ) | | |lim lim . | | . | || ( ) | | |n n

n n

u x ax x

u x a+ += =

* Nu , th chui hi t tuyt i khi + Do bn knh hi t 1r = * Nu = + th 0,x ta c 1| ( ) |lim

| ( ) |n

n

u x

u x

+ = + , do bn knh hi t . 0r = * Nu 0 = th ta c 1| ( ) |lim 0 1

| ( ) |n

n

u x

u x

+ = < , suy ra chui hi t tuyt i x R , do bn knh hi t r =+ . i vi trng hp lim | |n na = ta cng c chng minh tng t. 5. Bi ton tm min hi t ca chui ly tha - Bc 1. Tm bn knh hi t ca chui lu tha bng cng thc (*) r - Bc 2. Xt ti 2 im mt ,x r x r= = . - Bc 3. Kt lun min hi t. Ch

Nu chui ly tha c dng th bng cch t00

( )nn

n

a x x= , 0xxX = ta a v

dng trc khi p dng cng thc (*). 0

n

nn

a X=

V d Tm min hi t ca chui lu tha 1)

1

n

n

x

n

=

- p dng cng thc (*) trn, ta c 11

lim||

||lim 1 =+== + n naa nnnn => . 1r =

- Xt ti 1x = , ta c 0

1

n n

= phn k (chui iu ho).

- Ti 1x = : 0

( 1)n

n n

=

hi t theo tiu chun Leibnitz. Do min hi t ca chui l [ )1,1X =

146

2) 1

n n

n

n x=

Ta c lim limnn

n na n = = = + . Suy ra 0r = .

Vy min hi t ca chui l { }0X = . 3)

0 !

n

n

x

n

=

Ta c 1| | 1 !

lim lim 0| | ( 1)! 1

n

n nn

a n

a n + = = + = . Suy ra r = +

Vy min hi t ca chui l X = R . 4)

0

1 1

2 1 2

n n

n

n

n x

=

+ + + Ta t 1

2t

x= + , ta c chui ly tha 0 12 1

n

n

n

nt

n

=

+ + . Ta c

1 1lim lim

2 1 2n

nn n

na

n += = + = . Suy ra bn knh hi t 1 2r = = .

- Xt ti , ta c chui s 2t =0

2 2

2 1

n

n

n

n

=

+ + . Ch rng

2 1 12 122 2 11

2 1 2 10lim lim

n n

nn n nn

n ne

+ ++ ++ + = = . Do chui s l phn k. - Xt ti . Ta c chui s 2t =

0

2 2( 1)

2 1

n

n

n

n

n

=

+ + . Khi

2 1 2 1 122 2 1( 1) 1 0

2 1 2 1lim lim

nn n n

n

n n

ne

n n

+ +

+ = + = + + 2

. Do chui s phn k.

Vy tm hi t , hay 2 t < < 51 22 232

2

x

xx

< < < + > .

Do min hi t ca chui ly tha cho l 5 3( , ) ( , ).2 2

X +)

=6. Tnh cht c bn ca chui ly tha Gi s chui lu tha c khong hi t ( , nn xa r r . a. Tnh cht 1

147

Chui lu tha hi t u trn mi on );(];[ rrba Chng minh: Ly 00 x r< < , sao cho . Khi v [ ] [ 0 0, -x ,a b x ] 00 x r< < nn chui s hi t. Mt khc ta li c

00

n

nn

a x=[ ] nbaxxaxa nonnn ,,,

Do chui hi t u trn nn xa );(];[ rrba . b. Tnh cht 2 C th ly tch phn tng s hng ca chui trn );(];[ rrba . c. Tnh cht 3 Tng ca chui lu tha l 1 hm lin tc trong khong (-r; r). d. Tnh cht 4 C th ly o hm tng s hng ca chui. Chng minh: Suy ra t tnh cht 1. Bi tp Tm min hi t ca chui hm

1) nn

n

xn

)1(2

12

= 2) nn xn )2(11 2 += 3) ( )1 22 1nn x n= ++ 4) nn xxn n += 3211 2 5) ++=1 )2( )1(n nnnx 6) nn xn )4(11 2 += 7) =1 )2( 1n nx 8) nn n xn )2(31 2 = 6.5. Chui Taylor v chui Mac- Laurin 6.5.1. Khai trin 1 hm thnh chui lu tha 1. t vn Gi s hm c o hm mi cp trong mi ln cn no ca im x)(xf o v c th biu din di dng tng ca 1 chui lu tha trong ln cn y. 2. nh dng (0) ...)(...)()()()( 0

303

202010 ++++++= nn xxaxxaxxaxxaaxf

trong l cc hng s ,......,,,, 210 naaaa 3. Xc nh cc h s Theo tnh cht 3 ca chui lu tha, trong khong hi t, ta c: (1) ...)(...)(2)( 10021

/ ++++= nn xxnaxxaaxf (2) ...)()1(...2)( 202

// +++= nn xxannaxf 148

.............................................................

(n) ...!)( 0)( += anxf n

.

Th x = xo vo cc ng thc trn, ta c: nkk

xfa

k

k ,0!

)( 0)( ==

4. Kt qu Khi : ...)(

!

)(...)(

!2

)())(()()( 0

0)(

20

0//

00/

0 +++++= nn xxn

xfxx

xfxxxfxfxf

6.5.2. Chui Taylor 1. nh ngha Chui hm n

n

n

xxn

xf)(

!

)(0

0

0)( = c gi l chui Taylor ca hm trong ln cn

ca im x)(xf

o

Khi x = xo: Chui hm nn

n

xn

f=0 )( ! )0( c gi l chui Mac-Laurin ca hm )(xf Ch

Theo trn, nu hm s c o hm mi cp trong v c th biu din di dng tng ca 1 chui lu tha trong ln cn y th chui lu tha y phi l chui Taylor ca hm trong ln cn y.

)(xf0

Vx

2. iu kin hi t Ta xt xem nu chui Taylor ca hm no hi t th vi iu kin no tng ca n ng bng .

)(xf

)(xf

a) V d tng ca chui hi t khng bng hm s

Xt hm s ==

00

0)(2

1

xkhi

xkhiexfx

Hm kh vi v hn ln ti mi x v o hm mi cp ca ti )(xf )(xf 0=x Tht vy,

12

2

220 0

1( ) (0) 1

0 2

xt

tx x t t tt

txf x f e t

lim lim lim te lim limx x tee

== = = = = = 0 0)0(/ = f

149

/ /2

40

2

11 122 23

0 0

2( ) (0) 2

20

2 4 40

t

x t

t t tt t t

txx x

x x

ef x f exlim lim lim lim t e

x x x

t tlim lim lim

e e e

++ + +

== = == = = =

= 0)0(// = f

Vy, chui Mac-Lau rin ca hm l: f n hi t v c tng bng khng vi mi x ...0...0000 32 ++++++ nxxxx b) nh ngha hm khai trin c thnh chui Taylor: Hm s c gi l khai trin c thnh chui Taylor nu chui Taylor ca )(xfhm hi t v c tng ng bng )(xf c) Cc iu kin nh l 1 Gi s trong mt ln cn no ca im xo hm c o hm mi cp. )(xf Nu trong ( ) 0n

nlim R x = 10)1( )()!1( )()( ++ += nnn xxnfxR

vi l 1im no nm gia xo v x th c th khai trin hm thnh chui Taylor trong ln cn y.

)(xf

Chng minh: Tht vy, Khai trin Taylor ca n cp n l:)(xf )()()( xRxPxf nn += trong

10

)1(

)()!1(

)()( ++ += nnn xxnfxR v ( ) 0nnlim R x = nn ( ) ( )nnf x lim P x=

Mt khc, tng ring th n ca chui Tay lor ca hm , do )()( xsxP nn = f ...)(

!

)(...)(

!2

)()(

!1

)()()( 0

0)(

20

0//

00

/

0 +++++= nn xxn

xfxx

xfxx

xfxfxf (.p.c.m)

nh l 2 Nu trong ln cn no ca im xo hm c o hm mi cp v tr tuyt i ca mi o hm u b chn bi cng 1 s th c th khai trin hm thnh chui Taylor trong ln cn y.

)(xf

)(xf

Chng minh: Theo gi thit, Mxf n )()( trong ln cn

0Vx

150

nxxn

Mxx

n

fxR

n

n

n 0

1

0

)1(

)!1()!1(

)()( ++= +

+

Do chui s =1 0! )(n nnxx c min hi t l R s hng tng qut 0 0( )! n nx xn 0

1

01( )!

nnx x

n

+ + 0( ) nnR x Hm thnh chui Taylor trong ln cn y. )(xf6.5.3. Chui Mac-Laurin ca 1 s hm thng dng 1. ( ) xf x e= ...

!...

!3!211)0(

32)( ++++++=

n

xxxxnf

nn

N l 1 s dng c nh bt k, ta c ,1k ),,( NNx Meexf Nxk ==)()( khai trin hm thnh chui Mac-Laurin trong ln cn )(xf )(xf ),( NN + ca x = xo= 0

...!

...!3!2

132 ++++++=

n

xxxxe

nx

2. siny x= xkxxf k += 1)

2sin()()(

Hm khai trin c thnh chui Mac-Laurin xxf sin)( = ,...0)0(,1)0(,0)0(,1)0(,0)0( )4()3(/// ===== fffff Vy, ta c:

...)!12(

)1(...!7!5!3

sin12

1753 ++++= nxxxxxx nn

3. cosy x= Tng t nh trn, chui Mac-Laurin ca hm xxf cos)( = hi t v chnh n trn ton R:

...)!2(

)1(...!6!4!2

1cos2642 ++++=n

xxxxx

nn

4. (1 )y x = + (Chui nh thc) ...

!

)1)...(1(...

!2

)1(1)1( 2 ++++++=+ nx

n

nxxx

151

c bit * Khi :1= ...)1(...1

1

1 32 ++++=+ nn xxxxx 5. ln(1 )y x= + ...)1(...

432)1ln( 1

432 ++++=+ n

xxxxxx

nn

Bi tp 1) Khai trin hm 3xy = thnh chui Taylor ln cn im 1=x ( Vit 4 s hng u ca chui Taylor) 2) Khai trin hm

xy

1= thnh chui lu tha ca 3x 3) Khai trin thnh chui Mac-Laurin cc hm sau; a) )(

2

1 xx eey += b)

xexy 2= c) xy 2sin= 4) Khai trin hm s

2)( += x xxf thnh chui lu tha ca x v tm min hi

t ca chui va tm c.

152

Chng 6 CHUI S V CHUI LY THA V d Xt s hi t ca cc chui s dng sau: 3) M chui hi t nn chui hi t. Chng minh: Gi s: .

Chng minh: Theo gi thit, ta c vi mi k, hm f(x) gim trn on [k, k+1] nn

6.3. Chui s an du - Chui s c du bt k Chng minh:

6.4.1. Chui hm 1. nh ngha Chui hm l chui , trong cc l cc hm ca x.

6.4.2. Chui hm hi t u Chng minh. R rng chui , hi t (theo tiu chun so snh) 4. Tnh cht c bn ca chui hm hi t u

6.4.3. Chui ly tha Ta c . Khi th hi t (v ). Do vy chui lu tha hi t. b. nh l (Hadamard) (Cng thc tm bn knh hi t) Chng minh: Gi s . Ta c

Bi tp 6.5. Chui Taylor v chui Mac- Laurin

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