giao trinh - le minh luu

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TRNG I HC LT KHOA TON - TIN HC

Y Z

GII TCH S (Bai giang tom tat)

NGI BIN SON

L MINH LU

Y a Lat 2009 Z

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Mc lc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1Chng 1 L thuyt sai s. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .31.1 Cc loi sai s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Quy tc thu gn s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Ch s chc, khng chc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Hai bi ton v sai s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.5 Sai s cc php ton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5Chng 2 Xp x tt nht . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.1 Xp x tt nht trong khng gian nh chun . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.2 Xp x tt nht trong khng gian cc hm lin tc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3 Xp x tt nht trong khng gian Hilbert . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17Chng 3 Xp x hm bng a thc ni suy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.1 Bi ton ni suy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .193.2 Gii h i s tuyn tnh xc nh a thc ni suy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.3 Phng php ni suy Lagrange . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.4 Trng hp cc mc ni suy cch u . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .233.5 Sai s ca png php ni suy Lagrange . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.6 Chn mc ni suy ti u . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .26Chng 4 Tnh gn ng o hm v tch phn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314.1 Dng ni suy Lagrange tnh gn ng o hm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314.2 Tnh gn ng tch phn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32Chng 5 Gii phng trnh phi tuyn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375.1 Phng php th . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375.2 Phng php chia i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375.3 Phng php lp n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .385.4 Phng php dy cung . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405.5 Phng php tip tuyn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425.6 Gii a thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445.7 Gii h hai phng trnh phi tuyn bng phng php lp n . . . . . . . . . . . . . . . . . . . .46Chng 6 Gii h phng trnh i s tuyn tnh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476.1 Mt vi khi nim cn thit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

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6.2 Phng php Gauss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476.3 Phng php cn s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .516.4 Phng php lp n gii h i s tuyn tnh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 536.5 Phng php Jacobi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 536.6 Phng php Seidel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 556.7 Phng php Gauss-Seidel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57Chng 7 Gii gn ng phng trnh vi phn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 597.1 Phng php xp x lin tip . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .597.2 Phng php chui nguyn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 617.3 Phng php Euler . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .627.4 Phng php Euler ci tin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 657.5 Phng php Runge-Kutta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

Ti liu tham kho . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68


Chng 1L Thuyt Sai S

1.1 Cc loi sai sTrn thc t khi o mt i lng hoc xc nh mt i lng m ta k hiu l a,

thng thng khng xc nh c gi tr ng m ch bit c gi tr gn ng a. Vyta gp phi sai s. C nhiu loi sai s:

1. Sai s thc s: i lng4 := |a a|

gi l sai s thc s ca a.2. Sai s tuyt i: Nu bit 4a 0 sao cho

a4a a a+4ath 4a gi l sai s tuyt i ca a.

3. Sai s tng i: i lnga :=

4a|a|

gi l sai s tng i ca a.1.2 Qui tc thu gn sGi s ta c s gn ng a c vit di dng thp phn

a = (p10p + ...+ j10j + ...+ ps10ps)

trong j {0, 1, 2, ..., 9}, p > 0. Ta mun thu gn s a n hng th j. Gi s a ls thu gn n hng th j ca s a v phn vt b l . t:a = p10

p + ...+ j+110j+1 + j10

j.

Trong : j bng j + 1 nu 0, 5 10j < < 10j v bng j nu 0 < 0, 5 10j.Trng hp = 0.5 10j th j = j nu j chn v j = j + 1 nu j l. Nh vysai s thu gn l i lng a 0 tha|a a| a.


Doa = (p10p + ...+ j10j + ),

a = (p10p + ...+ j10j + j10j),ta c

|a a| = |(j j) 10j + | < 0.5 10j.V d: S pi ' 3, 1415 ' 3, 142 ' 3, 14.

Ch 1. Sai s tuyt i khng c trng cho chnh xc ca php o m ch cho php ta

hnh dung c gn nhau gia gi tr ng v gi tr gn ng.2. Sai s tuyt i cng th nguyn vi i lng o.3. Sai s tng i c trng cho chnh xc ca php o v khng c th nguyn.4. Sau khi thu gn s th sai s tuyt i tng ln.Gi a l gi tr ng, a l gi tr gn ng v gi a l s sau khi thu gn ca a th

|a a| |a a|+ |a a| 4a + a.

1.3 Ch s chc v khng chcGi s c s gn ng a vit dng

a = (p10p + ...+ j10j + ...+ ps10ps).

Vi sai s tuyt i ca a l 4a. Cho s 0 < 1. Nu 4a 10i th ch s igi l ch s chc, ngc li ch s i gi l ch s khng chc.Ch s chc vi = 1 gi l chc theo ngha rng. Ch s chc vi = 0, 56 gi lchc theo ngha hp.

Ch : Nu i l ch s chc th j,j i cng l ch s chc. Nu i khng chc th j,j i cng khng chc. Mt ch s l chc sau khi thu gn s c th n khng cn l chc. Trong k thut, ngi ta thng dng = 1 v nu ch s l chc th sau thu gn

n vn l chc (0, 56 1). Khi tnh ton, ta thng gi li cc ch s chc v ly ph thm t 1 n 2 ch s

khng chc v c k hiu ring ch cc ch s khng chc ny.


Sai s tng i ca mt s khng ph thuc vo v tr du phy ca n (du chmthp phn).

1.4 Hai bi ton v sai sXt s gn ng vit dng thp phn

a = (p10p + p110p1 + ...+ ps10ps).

C hai bi ton t ra:Bi ton 1:Bit s ch s chc ca a l a, tm sai s tng i a ca a. Gi a0 l s a m saukhi di du phy sao cho ch s chc cui hng n v v ton ch s chc. Ta c

p 10a1 a0 (p + 1) 10a1 10a .

Vy1

(p + 1) 10a1 a 1

p 1011 .

Nu khng bit p th ly1

10s a 1

10s1.

Bi ton 2:Bit sai s tng i l a, tm s ch s chc a. Gi s bit a > 0, ta vit

a = 10m vi 0.1 < < 1 v m l s nguyn. t am l s a nhng di du chmthp phn sao cho am c m+ 1 ch s trc du chm thp phn. Ta c:

am (p + 1) 10m,

suy ra4a = amam = ama = am10m (p + 1).

Bi v0, 2 < (p + 1) 10.

Vy c hai trng hp xy raa. Nu (p + 1) 1 th 4am 1 v am c m+ 1 ch s chc.b. Nu (p + 1) > 1 th 4am 10 v am c m ch s chc.


Cui cng ta c th kt lun, nu a = 10m , 0.1 < 1 v (p + 1) 1 th a cm+ 1 ch s chc, ngc li a c m ch s chc.

1.5 Sai s cc php tonGi s phi tm i lng y theo cng thc

y = f(x1, x2, ..., xn).

Gi xi , y, i = 1, 2, ..., n v xi, y, i = 1, 2, ..., n l cc gi tr ng v gn ng. Nu fkh vi lin tc ta c|y y| = |f(x1, ..., xn) f(x1, ..., xn)|

=ni=1

|f(x1, ..., i, ..., xn)|xi

|xi xi |,

y i [xi, xi ], i = 1, 2, ..., n. Ta c th coi (do f kh vi lin tc v xi kh gn xi),f(x1, ..., i, ..., xn)xi ' f(x1, ..., xn)xi

.Do

4y =ni=1

f(x1, ..., xn)xi4xi ,

vy =

4y|y| =

ni=1

ln f(x1, ..., xn)xi4xi .

a. Sai s php tnh cng, tr:y = f(x1, x2..., xn) =

ni=1

xi,

f(x1, x2..., xn)

xi= 14y =

ni=1

4xi.

Gi s 4xm = maxi=1,n{4xi}, v ch s chc cui ca xm hng th k, (4xm =10k). Ta c:

4y 4xm = 10k.Do khi lm php cng, tr nn qui trn cc xi n mc gi li 1 hoc 2 ch sbn phi hng th k.Ch : Khi tr hai s gn nhau cn ly cc s vi nhiu ch s chc v khi tr hais gn nhau kt qu mt chnh xc.


b. Sai s php ton nhn, chia:Gi s

y =x1, ..., xpxp+1, ..., xn

= f(x1, ..., xp, ..., xn).

Khi ln y =

pi=1

ln xi n

j=p+1

lnxj

suy ray =

ni=1

xi.

Nu xm = maxi=1,n{xi} v s ch s chc ca xm l k th y xm v s chs chc ca y khng vt qu k. V vy khi lm php ton nhn, chia ta ch cnly k + 1 hoc k + 2 ch s l .

c. Sai s php ly tha, khai cn v nghch o:Cho y = x, R,

y =

d ln ydx4x = ||x.

Nu > 1 th y > x tc l php ly tha lm gim chnh xc. Nu 0 < < 1 th y < x tc l php khai cn lm tng chnh xc. Nu = 1 th y = x v php nghch o c chnh xc khng i.


Chng 2Xp X Tt Nht

2.1. Xp x tt nht trong khng gian nh chunGi s X l khng gian tuyn tnh nh chun. L X l a tp tuyn tnh ng ca

X v f X. Bi ton t ra hy tm phn t f L sao cho: f f = inf

gL g f .

nh l 2.1.1 Nu L l khng gian con hu hn chiu ca X th vi mi f X lun tnti f L tha

f f = infgL

g f .

(Phn t f gi l phn t xp x tt nht f trn L).Chng minh: Xt

= {g L : g 2 f } L.D thy l tp ng, gii ni trong khng gian hu hn chiu nn l Compact.Xt hm (g) := f g .Ta c

|(g) (h)| = | f g f h | (f g) (f h) = h g .

Do l hm lin tc trn tp Compact nn hm t cc tiu trn . T f : (f ) = min

g(g).

Mt khc: Nu g L \ tc l g khng thuc th g f g f

> 2 f f = f = f ( y ch phn t khng ca khng gianX). Bi vy g L\, th gf > f ,tc l

infgL\

g h f .

Suy ra f f = min

g f g


f infgL\

g h .

Do f f = min

gL f g .

nh l c chng minh.Ch : Sinh vin c th tham kho mt chng minh khc sau y khi (trong nh l

trn) bit c s ca khng gian tuyn tnh L thy r hn ngha vn .Gi s {g1, g2, ..., gn} l cc phn t c lp tuyn tnh trong X . t (bao tuyn tnhca cc phn t {g1, g2, ..., gn} trong X)

{g1, g2, ..., gn} := {ni=1

aigi, ai R}.

D thy {g1, g2, ..., gn} l khng gian con tuyn tnh hu hn chiu trong X . tL = {g1, g2, ..., gn}. Xt phin hm

F0(c) :=ni=1

cigi , c = (c1, c2, ..., cn) Rn.

K hiu | c |= (ni=1 c2i ) 12 v t K = {c Rn, | c |= 1} th K Rn l tp compacttrong khng gian hu hn chiu. Do F0(c) l hm lin tc trn tp compact nn t cctiu ti c0 K tc l0 m := F0(c0) = min

cKF0(c).

Bi m khng th l 0 v m = 0 th F0(c0) = ni=1 c0igi = 0, tc l c0i = 0, i. iuny ko theo | c0 |= 0 l mu thun (v c0 K). Xt hmF (c) =

ni=1

cigi f .

Nu f L th ly f = f . Nu f khng thuc L thinfcRn

F (c) = > 0;

F (c) =ni=1

cigi f ni=1

cigi f

=| c |ni=1

ci| c |gi f .


t c = ( ci|c| , ci|c| , ..., ci|c|) th | c |= 1, tc l c K. T trn ta cF (c) =| c | F0(c) f m | c | f .

D thy rng m | c | f khi | c | . Theo nh ngha gii hn, tn tiM > 0, c Rn, | c |> M th

F (c) > + 1.

Bi qu cu ngB(0,M) := {c Rn, | c |M}

l tp compact trong Rn. Hn na F (c) l hm lin tc nn n t cc tr trn B(0,M).Tc l tn ti c B(0,M) sao cho F (c) = . Ly

f =ni=1

cigi,

d thy f l phn t xp x tt nht f trong L.V d: Xt X = L2[0, 1]. xt h hm {g0 = x0, g1 = x1, g2 = x2, ..., gn = xn}. t

L := {g1, g2, ..., gn} =ni=0

aixi, ai R}

l tp cc a thc thc bc khng qu n v L l khng gian con hu hn chiu caL2[0, 1]. Theo nh l 2.1.1 vi mi f L2[0, 1] lun tn ti a thc bc khng qu nl Qn sao cho

f Qn L2[0,1]= mingL

f g L2[0,1] .

Tc l ( 10

|f(x)Qn(x)|2dx) 1

2

= mingL

( 10

|f(x) g(x)|2dx) 1

2

.

nh ngha 2.1.2 Khng gian tuyn tnh nh chun X c gi l li cht (ngt)nu x, y X,

x = y = 1, x+ y = 2,th x = y.

nh l 2.1.3 Nu X l khng gian li cht v L l khng gian con hu hn chiuca X th phn t xp x tt nht f l duy nht.

Chng minh: t% = min

gL f g ,


c hai trng hp xy ra1) Trng hp % = 0 f L v f = f . Tc f l duy nht.2) Trng hp % 6= 0 th f / L v % > 0. Gi thit phn chng, tn ti f 1 v f 2 , f 1 6= f 2u l xp x tt nht f . Khi

f f 1 = %, f f 2 = %.

Ta c:% f f

1 + f

2

2 f f

1

2+ f f 2

2

=%

2+%

2= %.

Suy ra f f

1 + f

2

2= %.

T phn t (f1+f2 )2

cng l phn t xp x tt nht f trn L. By gi ta xt hai phnt trong X l ff1

2v ff2

2. D kim tra rng f f

1

% = f f

2

% = 1,

v hn na f f

1

%+f f 2

% = 2f (f

1 + f

2 )

%

= 2 f f1+f

1

2

% = 2

% f f

1 + f

2

2= 2.%

%= 2.

Bi X l li cht nnf f 1

% f f

2

%.

Vy f 1 = f 2 v nh l c chng minh.Ch :a. Nu X l li cht th vi hai im khc nhau trn mt cu n v, on thng ni

hai im khng c im chung no khc vi mt cu tr chnh hai im ny ( nghahnh hc ca khng gian li cht).

b. Khng gian hu hn chiu Rn v khng gian Hilbert l li cht.c. Khng gian C[0,1] (khng gian cc hm lin tc trn on [0, 1]) khng li cht. Thtvy, ch cn ly phn t y1(x) = 1, y2(x) = x, ta c y1, y2 C[0,1] v y1 = 1, y2 = 1.Hn na, d thy y1 + y2 = maxx[0,1]|1 + x| = 2 nhng y1 6= y2, vy khng gian


C[0,1]] khng li cht.d. Nu tn ti phn t xp x ti nht f ca f ta t f f := En(f).2.2 Xp x tt nht trong khng gian cc hm lin tc C[a,b]K hiu C[a,b] l khng gian cc hm lin tc trn [a, b] v L l tp mi a thc bckhng qu n.nh l 2.2.1 (Wallee' - Poussin) Gi s f C[a,b] v Qn L. Nu tn ti n + 2im phn bit

a x0 < x1 < ... < xn+1 b,sao cho

sign{(1)i(f(xi)Qn(xi))} = const, i = 0, 1, 2, ..., n+ 1,th

% := mini=0,n+1

|f(xi)Qn(xi)| En(f),

y En(f) := minQL f Q .Chng minh: Nu:

:= mini=0,n+1

|f(xi)Qn(xi)| = 0,

th nh l l hin nhin.Nu > 0 ta chng minh bng phn chng. Gi s ngc li

En(f) < = mini=0,n+1

|f(xi)Qn(xi)|.

Xt P L l xp x tt nht f trn L. Khi : f P = En(f) < min

i=0,n+1|f(xi)Qn(xi)|.

Ta c|P (xi) f(xi)| P f < min

i|f(xi)Qn(xi)|

|Q(xi) f(xi)|.Suy ra

sign[Q(xi) P (xi)]= sign{[Q(xi) f(xi)] + [f(xi) P (xi)]}


= sign[Q(xi) f(xi)], i = 0, 1, 2, ..., n+ 1.T a thc bc n l (Qn P ) i du n+ 2 ln trn [a, b] nn n c t nht (n+ 1)nghim, vy Qn(x) P (x).Xt

= mini=0,n+1

|f(xi)Qn(xi)| > maxx[a,b]

|f(x)Qn(x)|

mini=0,n+1

|f(xi)Qn(xi)| = .

iu ny l mu thun v nh l c chng minh.nh l sau y l kh quan trng v phn t xp x tt nht trong C[a,b] v rng ngoivic n ch ra c phn t xp x tt nht ca f lin tc m n cn cho ta cch xc nh

a thc xp x tt nht Qn(x).

nh l 2.2.2 (Chebyshev) iu kin cn v Qn l a thc bc khng qu nxp x tt nht ca f C[a,b] l tn ti (n+ 2) im phn bit,a x0 < x1 < ... < xn+1 b,

sao chof(xi)Qn(xi) = (1)i f Q ,

i = 0, 1, 2, ..., n+ 1, = 1.

(Vi = 1 hoc = 1 v khng ph thuc vo i. Dy im {xi}n+1i=0 c gi l dyim Chebyshev.)nh l ny c chng minh kh phc tp. Chng minh chi tit sinh vin c th tm

trong cc ti liu tham kho. Di y ch trnh by ngn gn ngi c hnh dung tng v k thut ca phng php chng minh.

a. iu kin : t = f Qn ,

= mini=0,n+1

|f(xi)Qn(xi)|.

T gi thit v nh l Wallee'-Poussin ta c = = min

i|f(xi)Qn(xi)|

En(f) f Qn = .Vy

En(f) = f Qn .


Tc l Qn l a thc xp x tt nht f .b. iu kin cn. Ta xy dng n+ 2 im Chebyshev nh sau

= f Qn = En(f).

t g = f Qn v lyy0 = a,

y1 = min{y : g(y) = }.Khng mt tng qut, xem

g(y1) = .

y2 = miny[y1,b]

{y : g(y) = };..................................

ym = miny[ym1,b]

{y : g(y) = (1)m}.

Nh vy ta xy dng c dy {yn}mn=0 bng quy np. Nu m < n+ 2 th bng cchxy dng cc dy ph hp ngi ta chng minh c rng trng hp ny khng xyra. Vy m n+ 2. Khi ta ch cn ly {y0, y1, ..., yn+1} lm dy im Chebyshev vnh l c chng minh.

Ta bit khng gian C[a,b] khng li cht nn vn t ra l liu nh l duy nhtv phn t xp x tt nht cn ng trong C[a,b] khng? Cu tr li l vn ng. iu c ch ra trong nh l sau:nh l 2.2.3 a thc xp x u tt nht ca f C[a,b] trn L l duy nht.Chng minh: Gi s Pn L,Qn L u l cc a thc xp x tt nht ca f v

Pn 6= Qn.Xt a thcPn +Qn

2 L,

ta cEn(f) Pn +Qn

2 f

12 f Pn +1

2 f Qn

=1

2En(f) +

1

2En(f) = En(f).

Vy Pn +Qn

2 f = En(f).


iu ny suy ra a thc Pn+Qn2

cng l a thc xp x tt nht f trn L. Gi s dy{xi}n+1i=0 l dy Chebyshev ng vi Pn+Qn2 thPn(xi) +Qn(xi)2 f(xi)

= En(f), i = 0, 1, 2, ..., n+ 1.Bi vy

2En(f) = |P (xi) f(xi) +Q(xi) f(xi)| |P (xi) f(xi)|+ |Q(xi) f(xi)| P f + Q f = 2En(f).

T ,|P (xi) f(xi)| = |Q(xi) f(xi)| = En(f), i.

Suy raP (xi) f(xi) = i(Q(xi) f(xi)), i = 1.

Ta c,2En(f) = |P (xi) f(xi) + i(P (xi) f(xi))|

= (1 + i)|P (xi) f(xi)|.Suy ra 1 + i = 2 tc l i = 1. Cui cng ta c:

Pn(xi) f(xi) = Qn(xi) f(xi), i Pn(xi) = Qn(xi),i.Bi Pn(x) v Qn(x) l cc a thc bc n trng nhau trn n+2 im nn Pn(x) = Qn(x)v nh l c chng minh.

nh l 2.2.4 Xp x tt nht ca mt hm chn (l) cng l mt hm chn (l).Chng minh: Gi s f l chn th khi thay x bi x ta nhn c

| f(x) f (x) |=| f(x) f (x) | En(f), x.T f (x) cng l xp x tt nht f . Bi phn t xp x tt nht l duy nht ta suy raf (x) = f (x), x. Tc l f l hm chn.

a. Xp x a thc bc khng Q0(x)Cho f C[a, b]. Hy tm a thc bc khng Q0(x) xp x tt nht hm lin tc ftrn on [a, b].

tm := min

x[a,b]f(x), M := max

x[a,b]f(x).

Khi m f(x) M, x [a, b].


Bi Q0(x) l a thc bc khng tc l hm hng nn ta ly Q0(x) = M+m2 v ch ra rnga thc ny chnh l a thc xp x tt nht f(x). Ta cM m

2 f(x)Q0(x) M m

2,

vy| f(x)Q0(x) | M m

2, x [a, b].

Gi s f(x0) = m, f(x1) = M, x0, x1 [a, b]. D thy rng x0 v x1 l dy imChebyshev bif(x0)Q0(x0) = M m

2,

f(x1)Q0(x1) = M m2

.

Theo nh l Chebyshev, Q0 l xp x tt nht ca f trn [a, b].b. Xp x tt nht a thc bc mt Q1(x)Xt hm f(x) li lin tc trn [a, b]. Nu f(x) l tuyn tnh th a thc xp x tt

nht cng l f(x). Gi s f(x) khng l hm tuyn tnh v Q1(x) = px + q l a thcxp x tt nht f(x). t U(x) := f(x) (px+ q) th U(x) cng l hm li nn t cctr ti im c [a, b] duy nht. Theo nh l Chebyshev th c ba im Chebyshev lunphin, vy hai im u v cui phi l a v b. im cn li l im c (a, b) m ti U(x) t cc tr.Ta c

U(a) = f(a) (pa+ q) = f Q1 ,U(c) = f(c) (pc+ q) = f Q1 ,

U(b) = f(b) (pb+ q)= f Q1 ; = 1.

TU(b) U(a)

= f(b) f(a) p(b a) = 0.Suy ra

p =f(b) f(a)

b a .

tnh q ta xt0 = U(a) + U(c)

= f(a) (pa+ q) + f(c) (pc+ q)


= f(a) + f(c) p(a+ c) 2q.Vy

2q = f(a) + f(c) f(b) f(a)b a (a+ c)

Suy raq =

f(a) + f(c)

2 (f(b) f(a))(a+ c)

2(b a) .

Cui cng ta d kim tra rng a thc Q1(x) = px + q tha cc iu kin ca nh lChebyshev.2.3 Xp x tt nht trong khng gian HilbertXt khng gian Hilbert H v {ei}i=1 l h trc chun y , tc l

< ei, ej >= ij, i, j Nv

Span(ei) = H.

Vi mi x H lp tng FourierSn :=

ni=1

ciei,

y ci :=< x, ei > l h s Fourier ca x. Vi mi n N,0 x Sn 2= x 2 2 < Sn, x > + Sn 2

= x 2 ni=1

c2i .

Vyni=1

c2i x 2, n N.

iu ny ch ra chui ni=1 c2i hi t v c bt ng thc Besseli=1

c2i x 2 .

Chng ta bit l chui Fourier i=1 ciei hi t v hn na x =i=1 ciei.By gi, gi s H0 l mt khng gian con ng ca khng gian Hilbert H v x H. Biton t ra l tm h0 H0 sao cho x h0 = inf

hH0 x h0 = d(x,H0).


Gi s h0 = argminhH0 x h0 v c nh phn t h H0 bt k. Vi R xthmF () := x h0 + h 2 .

o hm ca F lF () = 2 < x x0, h > +2 h 2 .

R rng rngF (0) = min

RF () = x h0 2 .

Bi vy F (0) = 0, tc l < x h0, h >= 0 vi mi h H0. iu ny ch ra phn tx h0 trc giao vi H0, (x h0)H0. Hn na,

h0 = arg minhH0

x h0 .

Thc vy vi mi h H0, ta c x h 2= (x h0) + (h0 h) 2

= x h0 2 + h0 h 2 x h0 2 .Tc l h0 = argminhH0 x h0 . Du bng ch xy ra khi v ch khi h = h0.

D thy rng nu khng gian H0 c s chiu hu hn th phn t xp x tt nhth0 = argminhH0 x h0 tn ti v duy nht.


Chng 3Xp X Hm Bng a Thc Ni Suy

Cho [a, b] R. Gi f(x), x [a, b] l hm cn xp x, 0(x), 1(x), ..., n(x) l hhm c lp tuyn tnh. t

Rn =

{(x) =

ni=0

aii(x), ai R}.

3.1 Bi ton ni suyCho f(x) R, xi [a, b], (i = 0, n) l cc im phn bit trn [a, b]. Tc l

a x0 < x1 < x2 < .... < xn1 < xn b.

Bi ton t ra, hy xc nh hm (x) Rn sao cho f(xi) = (xi) (cc im xigi l mc ni suy). gii bi ton trn ch cn tm cc gi tr a0, a1, ..., an sao chof(xi) =

nj=0 ajj(xi). Ta bit nu

rank

0(x0) 1(x0) ... n(x0)0(x1) 1(x1) ... n(x1)... ... ... ...0(xn) 1(xn) ... n(xn)

< n+ 1,th bi ton ni suy ni trn khng c li gii (v nh thc 4 = 0). Vn t ra ltm h hm {k(xi)}nk=0 nh th no vi mi mc ni suy xi th bi ton ca ta c ligii.

nh ngha 3.1.1 H hm {i(x)}ni=0 c gi l h Chebyshev, nu vi mi dyc0, c1, c2, ..., cn khng ng thi bng khng th hm xc nh bi P (x) :=ni=0 cii(x)c khng qu n nghim trn [a, b].

V d: Xt h hm {i(x)}ni=0, i(x) = xi th h {i(x)}ni=0 l Chebyshev v theonh l Taylor, cc a thc bc khng qu n c khng qu n nghim trn trng s thc.Ch : Nu h hm {i(x)}ni=0 l Chebyshev th n l h hm c lp tuyn tnh.(iu ngc li cha chc ng).nh l 3.1.2 Nu vi mi i = 0, 1, 2, ..., n cc hm i(x) l kh vi lin tc n cp


n+ 1 trn [a, b] v vi mi k = 0, 1, 2, ..., n c nh thc Wronskian:

W [0, 1, ..., k] =

0(x) 1(x) ... n(x)(1)0 (x)

(1)1 (x) ... (1)n (x)... ... ... ...

(n)0 (x)

(n)1 (x) ... (n)n (x)

6= 0,vi mi x [a, b]. Khi h hm {i}ni=0 l h hm Chebyshev.

nh l 3.1.3 Vi mi dy im {xi}ni=0 l cc mc ni suy, vi mi hm f(x) th,Tn ti a thc ni suy khi v ch khi h hm {i}ni=0 l h hm Chebyshev.Chng minh: T bi ton ni suy ta suy ra c a thc ni suy ta phi gii h i

s sau: { ni=0 cii(xj) = f(xj)

j = 0, n

h gii c (c nghim) th nh thc

4 =


6= 0,vi mi dy im {xi}ni=0 v xi 6= xj, i 6= j.

a. Gi s {i}ni=0 l h Chebyshev ta chng minh 4 6= 0. Gi s 4 = 0 khi tn ti j, j = 0, 1, 2, ..., n m nj=0 2j > 0 sao cho nj=0 jj(xi) = 0, nn hmP (x) =

ni=0 jj(xi) c (n + 1) nghim v iu ny l v l v h {i}ni=0 l hChebyshev.

b. Vi mi hm f , vi mi {xi}ni=0, xi 6= xj, i 6= j m tn ti a thc ni suy, tachng minh h {i}ni=0 l h Chebyshev. Gi s ngc li rng h {i}ni=0 khng l hChebyshev, t tn ti hm P (x) =ni=0 cii(x) c n+ 1 nghim trn on [a, b] mta c th sp xp cc nghim sao choa x0 < x1 < ... < xn b.

Ly cc nghim xi ny lm mc ni suy ta c{ ni=0 cii(xj) = 0,

j = 0, 1, 2, ..., n.

Binj=0 c2j > 0 nn 4 = 0, tc l h i s hoc v nghim hoc v nh l mu thun,v nh l c chng minh.


3.2 Gii h i s tuyn tnh xc nh a thc ni suyT h i s xc nh a thc ni suy, nu ma trn h s c nh thc

4 =


6= 0.t

4i =

0(x0) ... f(x0) ... n(x0)0(x1) ... f(x1) ... n(x1)... ... ... ... ...0(xn) ... f(xn) ... n(xn)

,v s dng phng php Cramer gii h th nghim s l ci = 4i4 . T ta c

Qn(x) =ni=0

cii(x) =ni=0

4i4 i(x).

Khai trin nh thc i trn theo ct th i ta nhn c4i =

nj=0

f(xj)4ij,

vi 4ij l phn b i s ca f(xj). Th nh thc 4i vo cng thc tnh Qn(x) ta nhncQn(x) =

i,j

4ji4 f(xj)i(x)

=nj=0

f(xj)

(ni=0

4ji4 i(x)

)=

nj=0

f(xj)j(x).

Bi vy, ta cQn(x) =

nj=0

f(xj)j(x).

Nhn xt: a. Vi mi hm f(x) do Qn(x) l a thc ni suy ca f(x) nn Qn(xj) =f(xj) =

nj=0 f(xj)j(x), suy ra

j(xi) = ij, i, j.b. Chn f(x) 1 th f(xj) = 1 vi mi j ta suy ra

ni=0

j(xi) = 1.


Ch : Cn quan tm n cc vn sau:1. Trong cc bi ton thc t khc nhau, cn chn cc h Chebyshev i th no choph hp?2. lch gia hm ni suy v a thc ni suy?3. Chn mc ni suy no c li nht?4. nh hng ca sai s v php o?3.3 a thc ni suy LagrangeTrong bi ton ni suy nu ly h Chebyshev l h hm {i(x)}ni=0 = {1, x, ..., xn}th a thc ni suy Pn(x) ca hm cn ni suy f(x) c dng:

Pn(x) =ni=0

cixi.

Bi ton ni suy quy v bi ton: Tm a thc Pn(x) sao cho:{Pn(xj) = yj,j = 0, 1, 2, ..., n.

xc nh a thc Pi(x) bc n tha Pi(xj) = ij . Khi Pi(x) c dngPi(x) = A(x x0)(x x1)...(x xi1)(x xi+1)...(x xn).

Do 1 = Pi(xi) = A(xi x0)(xi x1)...(xi xi1)(xi xi+1)...(xi xn), suy raPi(x) =

(x x0)(x x1)...(x xi1)(x xi+1)...(x xn)(xi x0)(xi x1)...(xi xi1)(xi xi+1)...(xi xn) .

VyPn(x) =

ni=0

yiPi(x).

a thc ni suy Pn(x) xc nh nh trn gi l a thc ni suy Lagrange.V d: Xt hm f cho di dng bng sau

x 0 2 3 5f(x) 1 3 2 5

Tm a thc ni suy Lagrange ca f .


Gii: Xc nh Pi(x), i = 0, 3 nh sau:P0(x) =

(x 2)(x 3)(x 5)(0 2)(0 3)(0 5) =

1

30(x 2)(x 3)(x 5),

P1(x) =x(x 3)(x 5)2(2 3)(2 5) =

1

6x(x 3)(x 5),

P2(x) =x(x 3)(x 5)3(3 2)(3 5) =

1

6x(x 2)(x 5),

P3(x) =x(x 2)(x 3)5(5 3)(5 2) =

1

30x(x 2)(x 3).

a thc ni suy l:Pm(x) = 1P0(x) + 3P1(x) + 2P2(x) + 5P3(x)

= 130(x2)(x3)(x5)+31

6x(x3)(x5)21

6x(x2)(x5)+5 1

30x(x2)(x3)

= 130(x2)(x3)(x5)+ 1

2x(x3)(x5) 1

3x(x2)(x5)+ 1

6x(x2)(x3).

3.4 Trng hp cc mc ni suy cch uTrong ni suy Lagrange nu cc mc ni suy cch u nhau, tc l

x1 x0 = x2 x1 = ... = xn xn1 = h.Khi x1 = x0 + h;x2 = x1 + h...xn = x0 + nh. t

t :=x x0h

.

Suy rai(x) =

(x x0)...(x xi1)(x xi+1)...(x xn)(xi x0)...(xi xi1)(xi xi+1)...(xi xn)

=ht(th h)...(th (i 1)h)(th (i+ 1)h)...(th nh)ih(ih h)...(ih (i 1)h)(ih (i+ 1)h)...(ih nh)

=(1)ni.t(t i)...(tm)

i!(n i)!(t i)

=(1)nt(t 1)...(t n)

n!

ni=0

(1)iCinf(xi)

t i .

Vy, ta c cng thcPn(x) =

(1)nt(t 1)...(t n)n!

ni=0

(1)iCinf(xi)

t i .


Nhn xt:a. H s (1)niCin

ti khng ph thuc vo hm y = f(x). Nn ta c th tnh sn v lpthnh bng tnh.b. Nu thm mc ni suy mi th phi tnh li t u.V d 1: Tm a thc ni suy trng vi y = 3x ti cc im x0 = 1;x1 = 0; x2 = 1.

x -1 0 1y 1

31 3 .

Q2(x) =1

3

x(x 1)1(1 1) + 1

(x+ 1)(x 1)1(0 1) + 3

(x+ 1)x

(1 + 1).1.

=1

3(x2 x) (x2 1) + 3

2(x2 + x)

=4

6x2 +

8

6x+ 1 =

2

3x2 +

4

3x+ 1.

T Q2(x) =

2

3x2 +

4

3x+ 1.

3.5 Sai s ca cng thc ni suy LagrangeCho f C(n+1)[a,b] . Xc nh sai s R(x) = f(x) Pn(x). Gi s

Pn(x) =ni=0

cixi,

l a thc ni suy Lagrange ca f(x) trn cc mc a x0 < x1 < ... < xn b, tc lP (xi) = yi, i = 0, 1, 2..., n.

Gi(x) = ni=0(x xi) = (x x0)(x x1)...(x xn).

By gi ly x [a, b] ty , ta c nh gi tr x ny v xem vi mi i th x 6= xi (v nux = xi th sai s l 0). Xt hm

(z) = f(z) Pn(z) k, k = Const.Ti im xi, c

(xi) = f(xi) P (xi) k(xi), i = 0, n.Chn k (x) = 0, khi ta c

k =Pn(x) + f(x)

(x).


D thy rng (z) = 0 ti (n+ 2) im l x0, x1, ..., xn, x. T suy ra:o hm bc 1, (1)(z) = 0 ti (n+ 1) im;o hm bc 2, (2)(z) = 0 ti n im;...........................o hm bc n, (n)(z) = 0 ti 2 im;

Cui cng s tn ti im = (x) [a, b] sao cho (n+1)() = 0. T o hm cpn+ 1 ca (z) suy ra

0 = (n+1)() = f (n+1)() k(n+ 1)!.

Vyk =

f (n+1)()

(n+ 1)!,

k =f(x) Pn(x)

(x).

Nh vyf (n+1)()

(n+ 1)!=f(x) Pn(x)

(x).

T ta cR(x) = f(x) Pn(x) = f

(n+1)()

(n+ 1)!ni=0(x xi).

Nu t:Mn+1 = sup

x[a,b]f (n+1)(x).

Suy ra:|f(x) Pn(x)| | Mn+1

(n+ 1)!ni=0(x xi)|.

V d 2: Trong V d 1, ta xc nh c a thc ni suy hm y = 3x ti cc mcni suy {1, 0, 1} l P2(x) = 23x2 + 43x + 1. By gi, ta xc nh sai s ti x. S dngcng thc sai s ta c

R2 = f(x) P2(x)

=f (3)()

3![(x x0)(x x1)(x x2)].

Nu tM = sup

x[1,1]|f (3)(x)| = 3(ln3)3.

Khi sai s cho bi|R2| 3

2(ln3)3|2i=0(x xi)|.


3.6 Chn mc ni suy ti uCho hm f(x) C(n+1)[a,b] vn t ra l chn cc mc ni suy nh th no sai strong cng thc ni suy l b nht?1. a thc ChebyshevXt a thc bc n xc nh bi

Tn(x) = cos[n arccos(x)],

trong n = 0, 1, 2, ..., v x [1, 1]. thy c dng hin ca a thc ta t = arccos(x) tng ng cos = x.

Thay vo biu thc ta c:Tn1(x) = cos[(n 1)]

= cosn cos sinn sin .Suy ra

Tn+1(x) + Tn1(x) = 2xTn(x).

Ta nhn c cng thc tnh nh sauTn+1(x) = 2xTn(x) Tn1(x).

Vi n = 0, 1, 2, ...,+, ta nhn cT0(x) = cos0 = 1;

T1(x) = x;

T2(x) = 2xT1(x) T0(x) = 2 x2 1,..................

Tn+1(x) = 2xTn(x) Tn1(x).Nhn xt: a thc Tn(x) l a thc bc n c h s u l 2n1.nh l 3.6.1 Trong tt c cc a thc bc n vi h s u l 1 th a thc Tn(x)

2n1 c lch so vi 0 nh nht trn [1, 1] (h s u hiu l h s ca s hng bc cao nhttrong a thc), tc l vi mi a thc

P (x) = xn + a1xn1 + ...+ an,

th P Tn

2n1= 1

2n1.


Chng minh: Theo cch xc nh ca a thc Chebyshev ta c|Tn(x)2n1

| = |cos[n arccosx]2n1

| 12n1

.

iu ny c|Tn(x)2n1

| = 12n1

,

suy ran = kpi,

tc l =

kpi

n, k = 0, 1, 2, ..., n.

Do arccos xk =

kpi

n xk = cos(kpi

n).

By gi xt a thcQ(x) =

Tn(xk)

2n1 P (xk) = (1)

k

2n1 P (xk).

T gi thit phn chng ta c:|P (xk)| P < 1

2n1.

iu ny dn n du ca biu thc Q(xk) ch ph thuc vo du ca biu thc (1)k2nk ,tc lSign(Q(xk)) = Sign[

(1)k2k1

], k = 0, 1, 2, ..., n.

Vy a thc Q(xk) i du n+ 1 ln nn Q(x) c t nht n nghim trn [a, b] l iu vl (bi Q(x) c bc khng qu n).2. Nghim ca a thc ChebyshevXt im xk l nghim ca Tn(x) ngha l Tn(xk) = 0. Bi Tn(x) = cos[n arccos(x)]

v theo cch t k = arccos(xk) nn xk = cos(k), suy ra k = (2k+1)pi2n . Vyxk = cos(

2k + 1

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

3. Nghim ca a thc Chebyshev trn on [a, b]


By gi xt x [a, b]. T cng thc nghim ca a thc Chebyshev xk trn ta suyra cng thc ca a thc Chebyshev trn on [a, b] bng php i bint =

2x a abb a , t [1, 1].

Tc lx =

1

2{(b a)t+ (a+ b)}.

Khi vi mi a thc Pn(x), x [a, b] ta nhn c (sau khi i bin) a thcP n(t) = Pn(

1

2[(b a)t+ (a+ b)]) = ((b a)

n

2ntn + ...),

suy ra P n 2

n

(b a)n 1

2n1,

vy P n (b a)

n

22n1.

T , ta c Pn = sup

x[a,b]|Pn(x)|

= supt[1,1]

|Pn(b a2

t+a+ b

2)|

= supt[1,1]

|(b a)b

2ntn + ...|

=(b a)n

2nsup

t[1,1]|tn + ...| (b a)

n

2n2n1=

(b a)n22n1

.

Nh vy vi mi a thc Pn xc nh trn [a, b] thPn (b b)

n

2n.

Nhng theo kt qu ca nh l trn th a thc Tn(t)2n1 l a thc c lch nh nht trnon [1, 1] nn

Tn(2xabba )

2n1.(b a)

n

2n=

(b a)n22n1

.

Gi s nghim ca Tn(t) trn [1, 1] l i = cos(2i+1n )pi, i = 1, 2, ..., n suy raxi =

1

2{(b a) cos 2i+ 1

2(n+ 1)pi + (a+ b)}, i = 0, 1, ..., n.


4. Chn mc ni suy ti uT cng thc sai s ca phng php ni suy

|f(x) Pn(x)| | Mn+1(n+ 1)!

ni=0

(x xi)|.

Ta c: (x) :=ni=0(x xi), l a thc c bc n+ 1. Theo kt qu va nhn c th (x) (b a)

n+1

22n+1.

Nu ta ly(x) =

Tn+1(2xabba )

2n.(b a)n+12n + 1

,

th(x) = Tn+1(

2xabba )

2n.(b a)

n+1

2n+1=

(b a)n+122n+1

.

Bi vy nu ta ly cc mc ni suy chnh l cc nghim xi, i = 0, 1, 2, ..., n th sai s|R(x)| = |f(x) Pn(x)|

| Mn+1(n+ 1)!

ni=0(x xi)| =Mn+1(n+ 1)!

(b a)n+122n+1

.

Vy cc nghim xi, i = 0, 1, 2, ..., n chnh l cc mc ni suy ti u.Kt lun: c mc ni suy ti u th cc mc ni suy xi l

xi =1

2{(b a) cos 2i+ 1

2(n+ 1)pi + (a+ b)}, i = 0, 1, ..., n.

Khi , sai s cho bi cng thc|R(x)| M

(n+ 1)!

(b a)n+122n+1

.

V d:Xt hm f(x) = x+ 1, Tm a thc ni suy Q3 ca f trn [0, 1] vi cc mc nisuy ti u. Nh vy, theo kt qu trn ta c 4 mc ni suy ti u l:

xi =1

2{cos(2i+ 1

4pi) + 1}, i = 0, 1, 2, 3.


a thc ni suyQ3(x) =

(x x1)(x x2)(x x3)(x0 x1)(x0 x2)(x0 x3) +

x1 + 1

(x x0)(x x2)(x x3)(x1 x0)(x1 x2)(x1 x3)

+x2 + 1

(x x0)(x x1)(x x3)(x2 x0)(x2 x1)(x2 x3) + +

x3 + 1

(x x0)(x x1)(x x2)(x3 x0)(x3 x1)(x3 x2) .

c lng tt nht ca php ni suy trong trng hp ny l:|x+ 1Q3(x)| 1

4!27.


Chng 4Tnh Gn ng o Hm V Tch Phn

4.1 Dng ni suy Lagrange tnh gn ng o hmTa phi tnh o hm ca mt hm dng bng hoc mt hm dng gii tch phc

tp th thng thng ta dng phng php tnh gn ng. Chng hn ta c th thay hmf(x) bng a thc ni suy no ca f(x) l P (x) vi phn d R(x):

f(x) = P (x) +R(x),

R(x) =f (n+1)()

(n+ 1)!

ni=0

(x xi), = (x) (x0, x).

Khi f (x) = P (x) +R(x).

Phn ny ta s dng ni suy Lagrange tnh o hm. Thay gn ng f(x) bng athc ni suy Lagrange cp n l Qn, khi

f(x) = Qn(x) +R(x).

Trong Qn(x) =

nk=0

ykPk(x),

Pk(x) =(x x0)(x x1)...(x xk1)(x xk+1)...(x xn)

(xk x0)(xk x1)...(xk xk1)(xk xk+1)...(xk xn) .

Theo cng thc sai s ca php ni suy Lagrange ta c:R(x) =

f (n+1)()

(n+ 1)!

ni=0

(x xi), = (x) (x0, x).

Suy raR(x) =

d

dx{f

(n+1)()

(n+ 1)!

ni=0

(x xi)}

=f (n+1)()

(n+ 1)!.d

dx{

ni=0

(x xi)}.

T , ta cR(xk) =

f (n+1)()

(n+ 1)!

d

dx{i6=k

(xk xi)}.


V d trong trng hp s dng ni suy Lagrange cp 1, Q1(x) ta c:Q1(x) = y0

x x1x0 x1 + y1

x x0x1 x0 ,

R(x) =f()

2!(x x0)(x x1).

VyQ(x0) =

y1 y0x1 x0 , R

(x0) =f()

2!(x1 x0).

Khi c cng thc tnh o hm ca f lf (x0) =

f(x1) f(x0)x1 x0

f()

2!(x1 x0).

D thy rng y li l khai trin Taylor ca f ti x0 bit.4.2 Tnh gn ng tch phn.Gi s cn tnh tch phn:

I :=

ba

f(x)dx.

tnh gn ng tch phn I , thng thng ta thay hm f(x) bng a thc ni suy Q(x)ri ly tch phn a thc ni suy ny lm gi tr gn ng ca tch phn I , tc l

I :=

ba

f(x)dx ' ba

Q(x)dx.

4.2.1 Phng php hnh thang.Cho hm f(x) xc nh trn [a, b]. Chia on [a, b] thnh n on con bng nhau bi

cc im chia xi, vi i := 0, , n sao cho:a = x0 < x1 < < xn = b.

Khi xi = a+ ih, h = ban , i = 0, 1, ..., n.

-x

6y

a bxi1 xi


Thay din tch hnh thang cong bng din tch hnh thang trn on [xi1, xi], ta c xixi1

f(x)dx =yi1 + yi

2h.

Tng t vi cc on cn li v ly tng trn tt c on con, ta c: ba

f(x)dx =n1i=0

xi+1xi

f(x)dx 'n1i=0

h.yi + yi+1

2.

Cui cng ta c cng thc hnh thang sau: ba

f(x)dx ' b a2n

(y0 + 2y1 + ...+ 2yn1 + yn).

Cng thc sai sa. Sai s a phng: Do ta s dng a thc ni suy Lagrange Q1(x) xp x f(x)trn [xi1, xi], i = 1, 2, ..., n. Cng thc c th :

f(x) = yi1x xi

xi1 xi + yix xi1xi xi1 +R1(x), xi

xi1f(x)dx =

yi1 yi2

.h+

xixi1

R1(x)dx.

Trong |R1(x)| = |f()

2(x xi1)(x xi)| M

2(x xi1)(xi x)

vi M = supx[a,b] |f(x)|. T c| xixi1

f(x)dx yi1 yi2

.h| xixi1

|R1(x)|dx

M2

xixi1

(x xi1)(xi x)dx = Mh3

12.

b. Sai s ton phn: Sai s ton phn l sai s trn on [a, b]. K hiu R l sai ston phn, ta c

R = n.Mh3

12=M(b a)

12h2.

V d: Tnh tch phn I = 10ex

2dx bng phng php hnh thang vi n = 10. Ta c:

M := maxx[0,1]

{|y(2)(x)|} = maxx[0,1]

{|4(4x+ 1)ex2|} = 20.e.


Cng thc tnh nh sau:I =

10

ex2

dx

' 120{1+e+2(e(0,1)2+e(0,2)2+e(0,3)2+e(0,4)2+e(0,5)2+e(0,6)2+e(0,7)2+e(0,8)2+e(0,9)2)}.

Trong sai s ton phn cho bi:|R| (b a)Mh

2

12=

(0, 1)2.20.e

12.

4.2.2 Phng php ParabolCho hm f(x) xc nh trn [a, b]. Chia on [a, b] thnh 2n on con bng nhau bi

cc im chia xi, vi i := 0, , 2n sao cho:a = x0 < x1 < < x2n = b.

Khi xi = a+ ih, h = ba2n , i = 0, 1, ..., 2n.

-x

6y

a bx2i2

M2i2

x2i1

M2i1

x2i

M2i

Trn mi on [x2i2, x2i], i = 1, 2, ..., n, ta thay f(x) bng a thc ni suy bc haiQ2(x) vi cc mc ni suy x2i2, x2i1, x2i Cng thc c th :

Q2(x) = y2i2(x x2i1)(x x2i)

(x2i2 x2i1)(x2i2 x2i)

+y2i1(x x2i2)(x x2i)

(x2i1 x2i2)(x2i1 x2i)

+y2i(x x2i2)(x x2i1)

(x2i x2i2)(x2i x2i1) .

Theo cng thc ni suyf(x) = Q2(x) +R2(x).


T x2ix2i2

f(x)dx =

x2ix2i2

Q2(x)dx+

x2ix2i2

R2(x)dx.

D thy x2ix2i2

Q2(x)dx =h

3(y2i2 + 4y2i1 + y2i).

Vy x2ix2i2

f(x)dx =h

3(y2i2 + 4y2i1 + y2i) +

x2ix2i2

R2(x)dx.

Tng t vi cc on cn li v ly tng trn tt c on con, ta c: ba

f(x)dx =ni=1

x2ix2i2

f(x)dx 'ni=1

h

3(y2i2 + 4y2i1 + y2i).

Cui cng ta c cng thc tnh tch phn nh sau: ba

f(x)dx ' b a6n

(y0 + 4y1 + 2y2 + ...+ 4y2n1 + y2n).

Cng thc sai sa. Sai s a phng: t

R =

xi+hxih

f(x)dx h3[f(xi h) + 4f(xi) + f(xi + h)].

Xt hm(t) =

xi+txit

f(x)dx t3[f(xi t) + 4f(xi) + f(xi + t)].

tF (t) := (t) ( t

h)5(h), 0 t h.

D th li rng F (0) = F (h) = 0; F (0) = F(0) = 0,F (3)(t) = t

3[f (3)(xi + t) f (3)(xi t)] 60t

2

h5(h).

S dng nh l gi tr trung bnh cho f (3) nn tn ti [xi + t, xi t] sao chof (3)(xi + t) f (3)(xi t) = f (4)()(xi + t xi + t).

Suy raF (3)(t) = 2t

2

3[f (4)() +

90

h5(h)].


Bi F (0) = F (h) nn tn ti t1 (0, h),F (t1) = 0.

Bi F (0) = F (t1) nn tn ti t2 (0, t1),F(t2) = 0.

Bi F(0) = F(t2) nn tn ti t3 (0, t2),F (3)(t3) = 0.

Suy ra(h) = h

5

90f (4)(), xi t3 xi + t3.

Vy xi+hxih

f(x)dx =h

3[f(xi h) + 4f(xi) + f(xi + h)] h

5

90f (4)().

t M := maxx[a,b] |f (4)(x)|, ta nhn c

|R| Mh5

90.

b. Sai s ton phn lRTp (b a)

180Mh4.

V d: Tnh tch phn I = 10ex

2dx bng phng php Parabol vi n = 5. Ta c:

M := maxx[0,1]

{|y(4)(x)|} = maxx[0,1]

{|4(4x4 + 12x3 + 3)ex2|} = 76.e.

Cng thc tnh nh sau:I =

10

ex2

dx

' 130{1+e+2(e(0,2)2+e(0,4)2+e(0,6)2+e(0,8)2)+4(e(0,1)2+e(0,3)2+e(0,5)2+e(0,7)2+e(0,9)2)}.

Trong sai s ton phn cho bi:|R| (b a)Mh

4

180=

(0, 1)4.76.e

180.


Chng 5Gii Phng Trnh Phi Tuyn

Chng ny trnh by mt s phng php gii phng trnh f(x) = 0 trong f : R R.

5.1 Phng php thu tin ta tm cch a phng trnh f(x) = 0 v dng tng ng

g(x) = h(x),

Tip theo v th ca cc hm y = g(x) v y = h(x) tm giao im ca cc thny. Hanh giao im chnh l nghim cn tm.

-x

6y

0

y = g(x)

y = h(x)

a b

5.2 Phng php chia i.Gi s f(x) lin tc trn (a, b) v f(a).f(b) < 0 th f(x) c t nht mt nghim trn

(a, b). Ta dng phng php chia i lin tip on [a, b] tm gi tr gn ng canghim nh sau. Khng mt tng qut xem f(a) < 0 < f(b). Chia i [a, b] bi imc =

a+ b

2. Nu f(c) = 0, th c = . Nu f(c) 6= 0, xy ra hai trng hp:

a. Nu f(c)f(a) < 0, th chn on [a1, b1], a1 = a, b1 = c.b. Nu f(c)f(b) < 0, th chn on [a1, b1], a1 = c, b1 = b.Khi f(a1) v f(b1) tri du. Tip tc qu trnh nu trn, cui cng hoc c c

f(c) = 0, hoc ta xy dng c dy on tht li [an, bn], n N, m bn an = b a2ntha f(an) < 0 < f(bn). Theo nguyn l dy on tht li s tn ti , an <


bn,n N. Ta chng minh f() = 0. Tht vy. Do bn an = b a2n

0, khi n,nn c

limn

an = limn

bn = .

Bi f lin tc suy ra, f() = limn

f(an) 0 v f() = limn

f(an) 0. Vy f() = 0.

-x

6y

0a1 = a a2 a3

b4

a4

b1 = bb2b3

f(a)

f(b)

Thut tan thc hin nh sau:Bc 1. Ly c = a+b

2.

Nu f(c) = 0. Ta c c l nghim v dng thut ton.Nu f(c).f(a) < 0 th b := c.Nu f(c).f(a) > 0 th a := c.Bc 2. Nu |b a| < th nghim gn ng l c. Nu khng quay li bc 1.V d: Dng phng php chia i gii gn ng f(x) = x4+2x3x 1 trn on

[0, 1] vi s bc n = 5. Do f(0) = 1, f(1) = 1 nn nghim x [0, 1].Chia i [0, 1], c1 = 0.5, f(0.5) = 1.19. Ta chn [a1, b1] = [0.5, 1]. Tip tc chia ita c:

F (0.75) = 0.59, f(0.875) = 0.05,f(0.8125) = 0.304, f(0.8438) = 0.135, f(0.8594) = 0.043.

Vy nghim gn ngx ' 1

2(0.859 + 0.875) = 0.867.


5.3 Phng php lp n. gii c bng phng php lp ta a phng trnh f(x) = 0 v dng

x = (x).

Nu (x) [a, b] ,(x) q < 1, x [a, b] th vi mi x0 [a, b] dy lpxn+1 = xn , n = 0, 1, 2, ...,+,

hi t ti nghim x ca phng trnh f(x) = 0. Nu cho trc > 0, khi php lp thacng thc

| xn+1 xn |< (1 q)q

th dng v ly xn+1 l nghim gn ng (cng thc ny l iu kin dng). Tht vy:Dy {xn} l dy Cauchy v vi mi n, cn [a, b] sao cho| xn+1 xn |=| (xn) (xn1) |

=| (cn) | | xn+1 xn | q | xn+1 xn | .T , ta c

| xn+1 xn | q | xn xn1 |,| xn xn1 | q | xn1 xn2 |,| x2 x1 | q | x1 x0 | .

Vy| xn+1 xn | qn | x1 x0 | .

Vi mi p N ta c:| xn+p xn |

=| xn+p xn+p1 + xn+p1 ...+ xn+1 xn | qn+p+1 | x1 x0 | +...+ qn | x1 x0 |= qn(1 + q + q2 + ...+ qn1) | x1 x0 | .

T | xn+p xn | q

n

1 q | x1 x0 | .

Bi 0 q < 1 dy {xn} hi t ti x. Qua gii hn ta climn

xn = limn

(xn),

suy rax = (x).


Cho p + trong biu thc| xn+p xn | q

n

1 q | x1 x0 | .

Ta c| xn x | q

n

1 q | x1 x0 | .

C nhiu cch a phng trnh f(x) = 0 v dng gii c bng phng php lp n,chng hn xt hm no (x) 6= 0,x [a, b], t

(x) = x+ (x)f(x).

Hm (x) c chn sao cho x [a, b], | (x) < 1. V d, nu f (x) 6= 0 ta c thly

(x) = 1f (x)

.

Khi (x) = x f(x)

f (x).

Hn na o hm cp 1 ca l =

f.f

(f )2.

Bi (x) = 0 v lin tc nn tn ti > 0 sao chox (x , x + ) , | (x) | q < 1.

V d: Gii phng trnh f(x) = x3 + x 1000 = 0 trn [9, 10].C nhiu cch a f(x) = 0 v dng x = (x). C th ta dng dng:

x = 31000 x.

Doq := max

x[9,10]|(x)| = max

x[9,10]|13(1000 x) 23 | 1

3 (999) 23 ' 1

300.

Xp x ban u x0 = 10, ta c cng thc lp:xn+1 =

31000 xn, n = 0, 1, 2, 3....

Vi n = 1 c x1 = 3990, n = 2 c x2 = 31000 x1 ' 9.96666. Vi n = 3 nghimgn ng x3 ' 9.96667. Sai s cho bi| x3 x | 1

300 0.0001.


5.4 Phng php dy cungGi s phng trnh f(x) = 0 c nghim duy nht trn [a, b], f C2[a,b], v f , fkhng i du trn [a, b]. im x [a, b] c gi l im Fourier, nu f(x)f(x) > 0.a) Trng hp f < 0, f > 0, x [a, b]. D thy trong trng hp ny a l im

Fourier v f(a) > 0.Gi xk l xp x th k ca nghim, x0 = b l xp x ban u

-x

6y

0

y = f(x)

a b x1x2

M(a, f(a))

N0(b, f(b))N1(x1, f(x1))

Honh giao im ca dy cung MNk v trc honh trong M(a, f(a)) vNk(xk, f(xk)) l xp x xk+1.Phng trnh ng thng qua M v Nk nh sau:

y = f(a) +f(xk) f(a)

xk a (x a).Cho y = 0 ta c:

xk+1 = xk f(xk)(xk a)f(xk) f(a) .

Suy raxk+1 xk = f(xk)

f(xk) f(a)(xk a).

Cui cng ta nhn c cng thc dy cungxk+1 = xk f(xk)

f(xk) f(a)(xk a), k = 1, 2, ...,+.

Vy trong trng hp ny dy {xk} l dy n iu gim hi t ti nghim.


b) Trng hp f > 0, f > 0. Cng tng t nh trng hp a) trnh by trnnhng vi im b l Fourier v xp x ban u l a ta c cng thc lp sau:

xk+1 = xk f(xk)(xk b)f(xk) f(b) .

Cng thc sai s1. Cng thc sai s th nht: Gi s |f (x)| m > 0, x [a, b]. Khi c

|f(xk)| = |f(xk) f(x)|

= |f (uk)(xk x)| m|xk x|.Vy ta c cng thc sai s th nht:

|xk+1 x| |f(xk)|m

.

2. Cng thc sai s th hai: Gi sx [a, b], 0 < m |f (x)| M.

T cng thc dy cung ta c:xk+1 = xk f(xk)(xk a)

f(xk) f(a) .

Suy raf(xk) = f(xk) f(a)

xk a (xk+1 xk).

Xt v tri, do f(x) = 0 v t cng thc s gia, tn ti uk (x, xk) sao chof(xk) = f(x) f(xk) = (x xk)f (uk).

Xt v phi, t cng thc s gia, tn ti xk (xk+1, xk) sao chof(xk) f(a)

xk a .(xk+1 xk) = f(xk)(xk+1 xk).

Vy(x xk)f (uk) = f (xk)(xk+1 xk).

T (x xk+1 + xk+1 xk)f (uk) = f (xk)(xk+1 xk).


Hay|x xk+1| = |f

(xk) f (uk)||f (uk)| .|xk+1 xk|.

Hn na do|f (xk f (uk)| M m,

suy ra c cng thc sai s th 2,|xk+1 x| M m

m|xk+1 xk|.

V d: Gii phng trnh f(x) = x3 0.02x2 0.2x 1.2 = 0 trn [0, 1.5] vi = 0.002.D thy im Fourier l b = 1.5 v m = 3.49. t

fn = x3n 0.2x2n 0.2 1.2

ta cxn+1 = xn fn

1.425 fn (1.5 xn).

Bi x3 = 1.198, f(x3) ' 0.0072. Sai s cho bi| x3 x |< 0.0072

3.49' 0.002.

5.5 Phng php tip tuynXt hm f(x) c nghim duy nht, c cc o hm cp 1 v 2 lin tc v gi nguyn

du (f , f hoc dng hoc m) trn on [a, b]. Chn im x0 sao cho f(x0)f(x0) > 0,im ny c gi l im Fourier.Phng trnh tip tuyn vi ng cong y = f(x) ti im M0(x0, f(x0)) c dng

y = f (x0)(x x0) + f(x0).

Tip tuyn ct trc hanh ti im x1,0 = f (x0)(x1 x0) + f(x0).

T , ta cx1 = x0 f(x0)

f (x0).

Vi vai tr x0 l xn ta c dng tng qutxn+1 = xn f(xn)

f (xn), n = 1, 2, ...,+.


Dy lp ny ta gi l dy lp Newton.nh l 5.5.1 Vi cc gi thit trn, dy lp Newton hi t ti nghim ca phng

trnh f(x) = 0.Chng minh: Khng gim tng qut coi f > 0, f < 0. Cc trng hp khc chng

minh tng t. Trc tin, ta khai trin Taylor hm f(xn) ti xn1 c dngf(xn) = f(xn1) + f (xn1)(xn xn1) + f(cn1)

2(xn xn1)2.

Txn = xn1 f(xn1)

f (xn1),

suy raxn xn1 = f(xn1)

f (xn1).

Thay vo biu thc trn, ta cf(xn) = f(xn1) f (xn1). f(xn1)

f (xn1)

+f(cn1)

2(xn xn1)2

=f(cn1)

2(xn xn1)2 0.

Mt khcxn+1 xn = f(xn)

f (xn)

= f(cn1)(xn xn1)2

2f (xn) 0

v dy {xn} l n iu khng gim. Nu xn > x th do f < 0 nnf(xn) < f(x

) = 0.

Mu thun vi f(xn) 0. T dy {xn} c gii hn l .K hiu

M := supx[a,b]

{|f (x)|}.

Ta c|f(xn)| = |f (xn)|.|xn+1 xn| M |xn+1 xn|.


m t v mt hnh hc ca thut ton nu trong chng minh ca nh l trn, taxem hnh sau:

-x

6

y

0

y = f(x)

a b

x

x1x2

M(b, f(b))

N1(x1, f(x1))

N(a, f(a))Cng thc sai sGi s |f(x)| M1 v |f (x)| M2 vi mi x [a, b]. Mt mt ta c

f(xn+1) = f(xn+1) f(x) = f (xn+1)(xn+1) x).

Suy ra|xn+1 x| f(xn+1)

M2|xn+1 xn|2.

Mt khc t cng thc Taylor, ta cf(xn+1) = f(xn) + f

(xn)(xn+1 xn) + f(n)2

(xn+1 xn)2

=f(n)

2(xn+1 xn)2.

T , suy ra|f(xn+1)| M1

2|xn+1 xn|2.

Vy ta c cng thc sai s|xn+1 x| M1

2M2|xn+1 xn|2.

Tc hi t nhanh hn phng php dy cung, lp trnh tng i n gin.V d: Dng phng php tip tuyn gii phng trnh tg(x)

x 1 = 0 trn (pi, 3.pi

2) sai

s = 0.0001.


Vit li phng trnh trn nh sau:f(x) = sin(x) x sin(x) = 0.

D thy f (x) < 0, f(x) < 0,x (pi, 3.pi2). Dng cng thc

xn+1 = xn f(xn)f (xn)

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

Ly x0 = 3.pi2 , ta nhn c x1 = 9, x2 = 4.50004, x3 = 4.49343. Sai s cho bi cngthc:|x3 x| 3.pi

4|x3 x2|2

=3.pi

4|4.49343 4.50005|2

=3.pi

4|0.00662|2 ' 0.0001.

5.6 Gii a thcPhn ny xt phng php gii phng trnh f(x) = 0 vi f(x) l mt a thc bc n

no . Gi s f(x) =ni=0 aixi ta c th vit li n dng Hornerf(x) = a0 + x(a1 + ...+ x(an1 + xan))...).

tbn1 = an

bn2 = an1 + xbn1

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

bi1 = ai + xbi

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

b0 = a1 + xb1

f(x) = a0 + xb0.

tnh gi tr ca a thc m ta dng cch t trn, ta gi l phng php Horner.Tnh ton c thc hin theo s sau:

Gi s cn chia a thc f(x) cho (x x0). Ta vit nh sauf(x) = (x x0)Qn1(x) + f(x0).

Trong Qn1(x) = 0 + 1x+ ...+ n1xn1.


T cf(x) = n1xn + (n2 x0n1)xn1 + ...+ (0 x01)x+ f(x0) x00.

So snh cc h s, ta can = n1,

an1 = n2 x0n1,.................

a1 = 0 x01,a0 = f(x0) x00.

So snh vi phng php Horner trnh by trn vi f(x0) ta suy ra bi = i, i :=0, 1...n. S dng lin tip phng php, ta c

f(x) = (x x0)Qn1(x) +R0(x0),Qn1(x) = (x x0)Qn2(x) +R1(x0),

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

Q1(x) = (x x0)Q0(x) +Rn1(x0),Q0(x) = Rn(x).

Vyf(x) = R0(x0) +R1(x0)(x x0) + ...+Rn(x0)(x x0)n.

Theo cng thc Taylor thf(x) =

ni=0

f (i)(x0)

i!(x x0)i.

So snh hai biu thc cng ca f(x) suy ra:f (i)(x0) = i!Ri(x0), i := 0, 1, ..., n.

trong Ri(x0) c tnh t phng php Horner.5.7 Gii h hai phng trnh phi tuyn bng phng php lp nXt hai phng trnh vi hai n:{

F1(x, y) = 0,F2(x, y) = 0.

Gi s h c nghim c lp. s dng phng php lp ta a h trn v dng:{x = 1(x, y),y = 2(x, y).


Thut ton thc hin nh sau:{xn+1 = 1(xn, yn),yn+1 = 2(xn, yn).

Trong (x0, y0) l gi tr ban u.Gi s h ch c nghim duy nht (x, y) trong U = [a, b] [a, b] R2. Cc hm

1(x, y),2(x, y) kh vi lin tc trong U .|1x

|+ |2y

| q1 < 1,

|1x

|+ |2y

| q2 < 1.

Hoc|1x

|+ |1y

| q1 < 1,

|2x

|+ |2y

| q2 < 1.

Cc gi tr (xn, yn) U, n = 0, 1, 2, ...,+, th dy im xp x (xn, yn) xc nh trnhi t ti nghim (x, y) ca h. Sai s cho bi:|xn x|+ |yn y| max{q1, q2}

1max{q1, q2}(|xn xn1|+ |yn yn1)|.


Chng 6Gii H Phng Trnh i S Tuyn Tnh

6.1 Mt vi khi nim cn thitXt h phng trnh i s tuyn tnh

a11x1 + a12x2 + ...+ a1nxn = b1,a21x1 + a22x2 + ...+ a2nxn = b2,......................................an1x1 + an2x2 + ...+ annxn = bn.

Ma trn h s ca h

A :=

a11 a12 ... a1na21 a22 ... a2n.....................an1 a12 ... ann

Vit di dng ton t tuyn tnh

Ax = b.

Trong b = (b1, b2, ..., bn)

> Rn x = (x1, x2, ..., xn) Rn.

6.2 Phng php Gauss6.2.1 Phng php Gauss gii h i s tuyn tnh thun tin cho thut ton ta xt h i s tuyn tnh trn vi cch t (cc s hng

t do) ai,n+1 := bi. Khi h i s tuyn tnh tr thnha11x1 + a12x2 + ...+ a1nxn = a1,n+1,a21x1 + a22x2 + ...+ a2nxn = a2,n+1,......................................................an1x1 + an2x2 + ...+ annxn = an,n+1.

Phng php c chia thnh hai bc


a. Bc thun: Dng php bin i tng ng a h v dng tam gic trn (D)sau:

x1 + b(0)12 x2 + b

(0)13 x3 + ...+ b

(0)1,n1xn1 + b

(0)1,nxn = b

(0)1,n+1

0.x1 + x2 + b(1)23 x3 + ...+ b

(1)2,n1xn1 + b

(1)2nxn = b

(0)2,n+1

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

0.x1 + 0.x2 + 0.x3 + ...+ xn1 + b(n2)n1,nxn = b

(n2)n1,n+1

0.x1 + 0.x2 + 0.x3 + ... + 0.xn1 + xn = b(n1)n,n+1

Phng php c th nh sau: Gi s a11 6= 0 ta chia phng trnh u cho phn tdn a11 c:x1 + b

(0)12 x2 + b

(0)13 x3 + ...+ b

(0)1,n1xn1 + b

(0)1,nxn = b

(0)1,n+1.

Vib(0)1j =

a1ja11

.

kh x1 trong cc phng trnh cn li ta ln lt nhn phng trnh trn choa21, a31, ..., an1 v ly cc phng trnh th i, i = 2, 3, ..., n tr tng ng cho cc phngtrnh va nhn c. Kt qu ta nhn c h mi:

a11x1 + a12x2 + ...+ a1nxn = a1,n+1,

0x1 + a(1)22 x2 + ...+ a

(1)2nxn = a

(1)2,n+1,

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

0x1 + a(1)n2x2 + ...+ a

(1)nnxn = a

(1)n,n+1.

Trong a(1)ij = aij aikb(0)kj , i = 2, 3, ..., n, j = 2, 3, ..., n.

Tip tc ta chia phng trnh th hai cho phn t dn a(1)22 cx2 + b

(1)23 x3 + ...+ b

(1)1nxn = b

(1)2,n+1.

Vib(1)2j =

a(1)2j

a(1)22

, j = 3, 4, ..., n.

Tng t nh trn cho ti khi nhn c h tam gic trn (T ) sau:a11x1 + a12x2 + ...+ a1nxn = a1,n+1,

0x1 + a(1)22 x2 + ...+ a

(1)2nxn = a

(1)2,n+1,

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

0x1 + 0.x2 + ...+ a(n1)nn xn = a

(n1)n,n+1.


Chia phng trnh th i trong h ny cho phn t dn tng ng a(i1)ii , i = 1, 2, ..., nta nhn c h c dng tam gic trn (D).b. Bc ngc: Dng php th lin tip bt u t phng trnh cui cng tr ln

xc nh nghim xn, xn1, ..., x2, x1. Cng thc tnh nh sau:a(k)ij = a

(k1)ij a(k1)ik b(k1)kj , i, j = k, k + 1, ..., n,

b(k1)kj =

a(k1)kj

a(k1)kk

, j = k + 1, ..., n.

Phng php trn thc hin c nu a(k1)kk 6= 0. Nu khng th bng cch chuyn dng(ct) tng ng c c trng hp trn.V d: Phng php c trnh by c th (cho h 4 phng trnh 4 n) nh sau

a11x1 + a12x2 + a13x3 + a14x4 = a15,a21x1 + a22x2 + a23x3 + a24x4 = a25,a31x1 + a32x2 + a33x3 + a34x4 = a35,a41x1 + a42x2 + a43x3 + a44x4 = a45.

a. Bc thun: Gi s a11 6= 0, ta chia phng trnh th nht cho phn t dn a11 thuc:x1 + b

(0)12 x2 + b

(0)13 x3 + b

(0)14 x4 = b

(0)15

vi b(0)1j = a1ja11 . kh x1 trong ba phng trnh cn li ta ln lt nhn phng trnh trn cho

a21, a31, a41 v ly cc phng trnh th i, i = 2, 3, 4 tr tng ng cho cc phng trnhva nhn c. Kt qu ta nhn c h mi:

a(1)22 x2 + a

(1)23 x3 + a

(1)24 x4 = a

(1)25 ,

a(1)32 x2 + a

(1)33 x3 + a

(1)34 x4 = a

(1)35

a(1)42 x2 + a

(1)43 x3 + a

(1)44 x4 = a

(1)45 .

Trong a(1)ij = aij aikb(0)kj , i = 2, 3, 4, j = 2, 3, 4, 5.

Tip tc ta chia phng trnh th nht cho phn t dn a(1)22 cx2 + b

(1)23 x3 + b

(1)14 x4 = b

(1)25 .


Vib(1)2j =

a(1)2j

a(1)22

, j = 3, 4, 5.

Tng t cch trn ta kh x2 trong h ta nhn c h{a(2)33 x3 + a

(2)34 x4 = a

(2)35 ,

a(2)43 x3 + a

(2)44 x4 = a

(2)45 ,

via(2)ij = a

(1)ij a(1)i2 b(1)2j , i = 3, 4, j = 3, 4, 5.

Chia phng trnh u ca h cho a(2)33 ca(3)44 x4 = a

(3)45 .

Trong a(3)ij = a

(2)ij a(2)i3 b(2)3j , i = 4, j = 4, 5.

Chia phng trnh nhn c cho a(3)44 c

x4 =a(3)45

a(3)44

= b(3)45 .

Gp cc phng trnh nhn c, ta c hx1 + b

(0)12 x2 + b

(0)13 x3 + b

(0)14 x4 = b

(0)15 ,

0.x1 + x2 + b(1)23 x3 + b

(1)24 x4 = b

(1)25 ,

0.x1 + 0.x2 + x3 + b(2)34 x4 = b

(2)35 ,

0.x1 + 0.x2 + 0.x3 + x4 = b(3)45 .

Bc thun kt thc.b. Bc ngc: T phng trnh cui cng, ta c

x4 = b(3)45 .

Th x4 vo phng trnh th ba cx3 = b

(2)35 b(2)34 x4.

Th x3, x4 vo phng trnh th hai c:x2 = b

(1)25 b(1)24 x4 b(1)23 x3.


Th x2, x3, x4 vo phng trnh cn li ta nhn c nghimx1 = b

(0)15 b(0)14 x4 b(0)13 x3 b(0)12 x2.

n y kt thc bc nghch v tm c tt c cc nghim ca h. Thut ton kt thc.6.2.2 Dng phng php Gauss tnh nh thc tnh nh thc ca ma trn A ta thc hin ch bc thun tng t nh khi gii h

i s tuyn tnh vi v phi l 0 cho ti khi nhn c h tam gic trn (T ),a11x1 + a12x2 + ...+ a1nxn = 0,

0x1 + a(1)22 x2 + ...+ a

(1)2nxn = 0,

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

0x1 + 0.x2 + ...+ a(n1)nn xn = 0.

Ma trn h s:

T =

a11 a12 ... a1n0 a

(1)22 ... a

(1)2n

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

0 0 ... a(n1)nn

By gi ta chia dng (i) ca ma trn T cho phn t dn a(i1)ii s nhn c ma trn

B :=

1 b

(0)12 b

(0)13 ... b

(0)1n

0 1 b(1)22 ... b

(1)2n

..........................0 0 0 ... 1

D thy detB = 1. Vy

detA = a(0)11 .a

(1)22 ...a

(n1)nn .

6.2.3 Tm ma trn nghch o bng phng php Gauss tm ma trn nghch o A = (xij)n ca ma trn khng suy bin A = (aij)n bngphng php Gauss ta gii n h phng trnh tuyn tnh c cng ma trn h s A tc l

nk=1

aikxkj = ij, i, j = 1, 2, ..., n.

Trong ij =

{0, if i 6= j,1, if i = j.


6.3 Phng php cn s (phng php Cholesky)Xt h phng trnh tuyn tnh vi ma trn h s l ma trn i xng tc l aij =

aji, i, j = 1, 2, ..., n. Ta biu din ma trn A thnh tch hai ma trn:A = STS,

trong S = (sij l ma trn tam gic trn tc l (sij = 0,i > j v ST l ma trnchuyn v ca ma trn S. phn tch ma trn A thnh tch ca STS, ta s dng cngthc nhn hai ma trn:

aij =n

k=1

sikskj, i j.

Hays11 = (aii

i1k=1

s2ki)12 , i = 2, 3, ..., n, ;

sij =aij

i1k=1 skiskjsii

,i < j,sij = 0, i > j.

Nu sii 6= 0, i = 1, 2, ..., n, th detA = s11.s22...snn 6= 0 khi h c nghim duy nht.C th ta tm c:

S :=

s11 s12 ... s1n0 s22 ... s2n.....................0 0 ... tnn

By gi gii

Ax = b

ta vitSTS = b.

t Sx = y c STy = b. u tin, ta gii h tam gic diSTy = b,

xc nh y. Sau khi gii h ny, ta nhn c nghimy1 =

b1s11

;


yi =bi

i1k=1 skiyksii

,i > 1.

Sau gii h Sx = y tm nghim x, c th nghim s c tnh nh sau:xn =

ynsnn

;

xi =yi

nk=i+1 sikxk

sii, i < n.

6.4 Phng php lp n gii h i s tuyn tnhTrong khng gian Rn, xt cc chun sau:

x = max1in

|xi| :

x1 =ni=1

|xi| :

x2 = n

i=1

x2i .

Khi ta c cc chun tng thch ca ma trn A l

A = max1in

nj=1

|ai,j| :

A1 = max1jn

ni=1

|ai,j| :

A2 =

max1in

i(ATA),

trong i(ATA) l cc gi tr ring ca ma trn ATA. gii bng phng php lp n u tin ta a phng trnh Ax = b v dng

x = Bx+ g.

T nguyn l nh x co ta c kt qu sau: Nu B < 1 th dy lp:xk+1 = Bxk + g, k = 1, 2, ...,+,


trong x0 Rnchn bt k l hi t ti nghim x duy nht.Sai s c cho bi:xk x B

k

1 Bx1 x0,

xk x B1 Bxk xk1.

6.5 Phng php JacobiMa trn A = (aij)n gi l c ng cho tri nu mt trong hai iu kin sau tha

c1.j 6=i

|aij| < |aii|, i = 1, 2, ..., n,

c2.i6=j

|aij| < |ajj|, j = 1, 2, ..., n.

nh l Nu A c ng cho tri th c th a phng trnh Ax = b v dngx = Bx+ g vi ma trn B c chun nh hn 1.

Chng minh: Xt trng hp c1, ta11x1 + a12x2 + ...+ a1nxn = b1,a21x1 + a22x2 + ...+ a2nxn = b2,..............................................an1x1 + an2x2 + ...+ annxn = bn,

ta chia phng trnh th i cho aii v chuyn cc s hng j 6= i sang v phi ta cxi = a11

aiix1 a12

aiix2 ... a1n

aiixn +

biaii.

Phng trnh cho tng ng vi phng trnhx1 = 0.x1 a12a11x2 ... a1na11 xn + b1aii ,x2 = a21a22x1 + 0.x2 ... a2na22 xn + b2a22 ,..............................................................xn = an1annx1 an2annx2 ...+ 0.xn + bnann .

t ma trn B = (bij)n vibij =

{0, j = i,aij

aii, j 6= i

vgi =

biaii.


Khi phng trnh c dngx = Bx+ g.

Hn naB = max

1in

ni=1

|bij| < 1.

Trng hp c2:T

a11x1 + a12x2 + ...+ a1nxn = b1,a21x1 + a22x2 + ...+ a2nxn = b2,.................................................an1x1 + an2x2 + ...+ annxn = bn.

t: zi = aiixi v thay vo phng trnh v chuyn v, ta cz1 = 0.z1 a12a22 z2 ... a1nann zn + b1,z2 = a21a11x1 + 0.z2 ... a2nannxn + bn,....................................................zn = an1a11 x1 an2a22 x2 ...+ 0.xn + bn.


{0, j = i, aij

ajj, j 6= i,

vgi = bi.


Hn naB1 = max

1jn

ni=1

|bij| < 1.

V d: Dng phng php lp n gii h phng trnh10x1 + 2x2 + x3 = 10,x1 + 10x2 + 2x3 = 12,x1 + x2 + 10x3 = 8.

Ma trn h sA :=

10 2 11 10 21 1 10

D thy A l ma trn ng cho tri.

x1 = 0x1 0.2x2 0.1x3 + 1,x2 = 0.1x1 + 0x2 0.2x3 + 1.2,x3 = 0.1x1 0.1x2 + 0x3 + 0.8.


Vit li dngx = Bx+ g,

trong B :=

0 0.2 0.10.1 0 0.20.1 0.1 0

v

g =

11.20.8

D kim li

B = max1i3

3j=1

|bij| = max{0.3, 0.3, 0.2} = 0.3 < 1.

Dng phng php Jacobi, ta c dy lp xc nh bix(n+1)1 = 0x

(n)1 0.2x(n)2 0.1x(n)3 + 1,

x(n+1)2 = 0.1x(n)1 + 0x(n)2 0.2x(n)3 + 1.2,x(n+1)3 = 0.1x(n)1 0.1x(n)2 + 0x(n)3 + 0.8,

trong n = 0, 1, 2, ...,+ v xp x ban u

x(0) =

11.20.8

Vi n = 1,

x(1)1 = 0 1 0.2 1.2 0.1 0.8 + 1 = 0.68,x(1)2 = 0.1 1 + 0 1.2 0.2 0.8 + 1.2 = 0.94,x(1)3 = 0.1 1 0.1 1.2 + 0 0.8 + 0.8 = 0.58.

Vi n = 2, x(2)1 = 0 0.68 0.2 0.94 0.1 0.58 + 1,x(2)2 = 0.1 0.68 + 0 0.94 0.2 0.58 + 1.2,x(2)3 = 0.1 0.68 0.1 0.94.2 + 0 0.58 + 0.8.

Sai s cho bi cng thcx(2) x B

k

1 Bx1 x0 =(0.3)2

0.7 (0.22).


6.6 Phng php SeidelT phng trnh

x = Bx+ g,

ta phn tch ma trn B thnh tng ca hai ma trn tam gic trn v di. Tc lB = B1 +B2

trong

B1 :=

0 0 ... 0b21 0 ... 0................bn1 bn2 ... 0

v

B2 :=

b11 b12 ... b1n0 b22 ... b2n

.................0 0 ... bnn

Phng trnh vit li

x = B1x+B2x+ g.

Php lp xc nh nh sau:x(k+1) = B1x

(k+1) +B2x(k) + g, k := 1, 2, ...,+

C th cng thc lp xc nh theo

x(k+1)i =

i1j=1

bijx(k+1)j +

nj=i

bijx(k)j + gi.

Cc thnh phn x(k+1)j va tnh c vi j = 1, 2..., i 1 s dng ngay vo tnh thnhphn x(k+1)i .

nh l 6.6.1 Nu B < 1 th phng php Seidel hi t.Chng minh: D thy t nguyn l im bt ng th phng trnh trn c nghim

duy nht x. xt s hi t ca dy lp theo phng php Seidel ta xt

x(k+1)i xi =

i1j=1

bijx(k+1)j +

nj=i

bijx(k)j + gi xi


Phng php Seidel hi t tt hn phng php lp n. Tit kim b nh v hp lhn ch cc thnh phn x(k+1)i va tnh c huy ng tnh cc thnh phn tiptheo. C th cho cc v d ch ra rng hai phng php khng ng thi hi t hoc phnk.

6.7 Phng php Gauss-SeidelGi s ma trn A c ng cho tri. Nh phng php Jacobi ta a phng trnh

v dng gii bng phng php lp. Ta11x1 + a12x2 + ...+ a1nxn = b1a21x1 + a22x2 + ...+ a2nxn = b2.............................................an1x1 + an2x2 + ...+ annxn = bn

Ta chia phng trnh th i cho aii v chuyn cc s hng j 6= i sang v phi ta cxi = a11

aiix1 a12

aiix2 ... a1n

aiixn +

biaii.

Phng trnh cho tng ng vi phng trnhx1 = 0.x1 a12a11x2 ... a1na11 xn + b1aiix2 = a21a22x1 + 0.x2 ... a2na22 xn + b2a22..........................................................xn = an1annx1 an2annx2 ...+ 0.xn + bnann


{0, j = i,aij

aii, j 6= i,

vgi =

biaii.


Hn naB = max

1in

ni=1

|bij| < 1.

By gi ta s dng phng php Seidel p dng cho h trn nh sau:

x(k+1)i =

i1j=1

aijaii

x(k+1)j

nj=i

aijaii

x(k)j + gi.

Phng php ny gi l phng php Gauss-Seidel.


Chng 7Gii Gn ng Phng Trnh Vi Phn

Chng ny trnh by mt s phng php gii bi ton phng trnh vi phn cp 1. gii bi ton ny, ngi ta phn bit hai loi phng php l phng php gii tch vphng php s. Cc phng php gii tch cho php tm nghim gn ng dng biuthc gii tch cn trong cc phng php s th tm gn ng hm nghim y(x) ti mts im x0, x1, ..., xn trn [a, b].

7.1 Phng php xp x lin tipXt bi ton Cauchy {

y = f(x, y),y(0) = y0, 0 x 1.

Trong f(x, y) tha iu kin Lipschitz theo y trn min m D trong khng gian R2,tc l

|f(x, y1) f(x, y2)| L|y1 y2|, (x, y1), (x, y2) D, L = const.Ta ly L.

y = max0x1

ex|y(x)|.

y(n+1) = y0 +

x0

f(t, y(n)(t))dt, n = 0, 1, 2, ...

Ta ch ra dy lp hi t ti nghim duy nht y(x). t rn+1 = y(n+1) y ,y = y0 +

x0

f(t, y(t))dt

Ta c vi mi x [0, 1] xt|y(n+1)(x) y(x)| =

x0

[f(t, y(n)(t)) f(t, y(t))]dt|

L x0

|y(n)(t) y(t)|dt

= L

x0

etet|y(n)(t) y(t)|dt

L x0

ety(n) ydt


= Lrn

x0

etdt = Lrn(ex 1)

.

T suy raex|y(n)(x) y(x)| Lrn (e

x 1)

ex

= Lrn1 ex

Lrn1 e

.

Nh vyex|y(n)(x) y(x)| Lrn1 e

.

Ly max hai v ca bt ng thc trn, ta cmaxx[0,1]

ex|y(n)(x) y(x)| L(1 e)

rn,

tc l rn+1 qrn, vi q = 1e < 1. Suy rarn qnr0.

By gi, ta c lng r0. Ta cr0 = y0 y

= sup0x1

ex|y(x) y(x0)| = sup0x1

ex|(y)()||x x0|

sup0x1

ex||f(, y())|

sup(x,y)D

|f(x, y)|.

t M := sup(x,y)D |f(x, y)| th r0 M . Vyrn Mqn.

V d: Gii bi ton Cauchy{y = x y,y(0) = 1, 0 x 1.

C f(x, y) = x y, |f(x,y)y

| = 1, v do l = 1 v = L = 1. Chun c xc nhbi

y = maxx[0,1]

ex|y(x)|.

iu kin Lipshitz l|f(x, y1) f(x, y2)| |y1 y2|.


Min D m cha {(x, y) : x [0, 1], |y| b}, khi |f(x, y)| |x|+ |y| 1 + b =:M.

rn Mqn, q = (1 e1) < 1.Dy lp xc nh theo

y(0)(x) = 1,

y(1)(x) = 1 +

x0

(t y(t))dt = 1 + x2

2 x,

y(2)(x) = 1 +

x0

(t y(1)(t))dt = 1 x+ x2 x3

6,

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

Cng thc tng quty(n+1)(x) = 1 +

x0

(t y(n)(t))dt.

7.2 Phng php chui nguynXt bi ton Cauchy {

y = f(x, y),y(0) = y0, 0 x 1.

Gi s f(x, y) l hm gii tch trong ln cn (x0, y0), tc l f(x, y) khai trin thnh chuily tha trong ln cn

f(x, y) =

m,n=0

m+nf(x0, y0)

xmyn(x x0)m(y y0)n

m!n!.

Khi nghim y(x) cng c th khai trin thnh chui ly tha

y(x) =i=0

y(i)(x0)i!

(x x0)i.

Nh vy hm nghim y s c xc nh nu ta tnh c cc o hm ca y ti x0,v cc o hm ny d dng tnh c. C th nh sau:y(x0) = y0,

y(x) = f(x, y) y(x0) = f(x0, y0),

y(x) =f(x, y)

x+f(x, y)

yy(x, y)


y(x0) = f(x0, y0)x

+f(x0, y0)

yf(x0, y0),

Tip tc tnh cc o hm y(k), k = 3, 4, ... ta xc nh c hm nghim ca phngtrnh y(x)

y(x) =i=0

y(i)(x0)

i!(x x0)i.

Nh vy, phng php ny n gin nhng tnh ton phc tp v bn knh hi t cachui nghim kh xc nh.

V d: Gii bi ton Cauchy{y = x y,y(0) = 1, 0 x 1.

Nh vy f(x, y) = x y l gii tch trong ln cn (0, 1). Ta tnh cc o hmy(0) = 1,

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

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

y(k)(x) = y(k1)(x) y(k)(0) = (1)k2, k 2.Vy

y(x) = 1 x+ 2k=2

(1)kxkk!

= 2ex + x 1.

= 1 x+ 2(k=0

(x)kk!

1 + x)

Vy nghim ca bi ton l:y(x) = 2ex + x 1.

7.3 Phng php Euler7.3.1 Cng thc Euler


Xt bi ton Cauchy {y = f(x, y),y(0) = y0, a x b.

t vn l tm gn ng hm nghim y(x) ti mt s im a x0 < x1 < ... < xn b. Tc l tm nghim yi gn ng vi y(xi) vi mi i ( y y ch nghim ng). Nucc im chia xi cng nhiu th ta cng c hnh nh gn ng ca hm nghim y(x).Xt trng hp cc bc cch u, tc l xi+1 xi = h, i = 0, 1, 2, ..., n. T khai trinTaylor, ta c

y(xi+1) = y(xi) +y(xi)1!

(xi+1 xi) + y()2!

(xi1 xi)2.

Suy ray(xi+1) = y(xi) +

hf(xi, y(xi))

1!+ o(h2).

Vy ta c th coiy(xi+1) ' y(xi) + hf(xi, y(xi))

1!.

Ta c cng thc Euler nh sauyi+1 = yi + hf(xi, yi).

7.3.2 Cng thc sai s ca phng php Eulera. Sai s a phng: Xt sai s mc phi trn mt bc vi gi thit bc trc

tnh ng. Ti bc th i ta xt hm y(x) l nghim ca bi ton{y = f(x, y(x)),y(xi) = yi.

Nghim ng ca bi ton nyy(xi+1) = y(xi) +

hf(xi, y(xi))

1!+ o(h2).

Bi y(xi) = yi v yi+1 = yi + hf(xi, yi) nny(xi+1) = yi + hf(xi, yi) + o(h

2) = yi+1 + o(h2).

T suy ray(xi+1) yi+1 = o(h2).

b. Sai s ton phn: Xt sai s trn ton on [x0, xn]. Gi s hm f(x, y) tha iukin Lipshitz theo y|f(x, y1) f(x, y2)| L|y1 y2|,


v c M sao chof(x, y(x))

xM, x [x0, xn].

T cng thc Euler suy ra:1. Gi tr gn ng theo Euler l

yn = yn1 + hf(xn1, yn1).

2. Gi tr ng ca nghim ti xn ly(xn) = y0 +

xnx0

f(t, y(t))dt.

By gi, ta tn = yn y(xn),

i = yi y(xi), i = 0, 1, 2, ..., n.K hiu

4i = i+1 i= yi+1 yi (y(xi+1) y(xi))

= hf(xi, yi) xi+1xi

(f(t, y(t))dt

= hf(xi, yi) f(xi, y(xi)) xi+1xi

(f(t, y(t)) f(xi, yi))dt.

T suy ra|4i = |i+1 i|

Lh|yi y(xi)|+M xi+1xi

|t xi|dt

Lh|i|+ Mh2

2.

Vy|i+1| (1 + Lh)|i|+ Mh

2

2, i = 0, 1, 2, ..., n.

By gi xt|n| (1 + Lh)|n1|+ Mh

2

2

(1 + Lh){(1 + Lh)|n2|+ Mh2

2}+ Mh

2

2

= (1 + Lh)2|n2|+ Mh2

2(1 + (1 + Lh))

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


7.4 Phng php Euler ci tinThay v dng cng thc Euler ta s dng cng thc sau

yi+1 = yi +h

2(f(xi, yi) f(xi+1, yi+1)).

Phng php tin hnh nh sau: Lyy(0)i+1 = yi + hf(xi, yi),

Lp theo k 1,y(k)i+1 = yi +

h

2(f(xi, yi) + f(xi+1, y

(k1)i+1 )).

Khi thy|y(k)i+1 y(k1)i+1 |

th dng li v ly y(k)i+1 ' yi+1.7.5 Phng php Runge-Kuttat y1 = y0 +y0 , trong

y0 = pr1k1(h) + ...+ prrkr(h),

ki(h) = hf(i, i); i = x0 + ih, 1 = 0, i = 1, 2, ..., r,

i = y0 + i1k1(h) + ...+ i,i1ki1(h).

GiR(h) := y(x0 + h) y1 = y(x0 + h) y(x0)y0 ,

Nu (s+1)r (0) 6= 0, thr(h) =

ri=0

(i)r (0)

i!hi + 0(hs+1).

Khi ta chn cc h s i, ij, prj t iu kin (i)r (0) = 0 vi mi i = 0, 1, 2, ..., s v(s+1)r (0) 6= 0 vi s cng ln cng tt. Vy

(i)r (0) = y(i)0 [pr1k1(h) + ...+ prrkr(h)], i = 0, 1, ..., s.

Tc l, xc nh cc h s ta cn gii h phi tuyn saupr1k1(h) + ...+ prrkr(h) = y

(i)0 , i = 0, 1, 2, ..., s.

Cc trng hp ca phng php Runge-Kutta


a. Trng hp r = 1 ta nhn c cng thc Eulery1 = y0 + hf(x0, y0).

b. Trng hp r = 2 ta nhn c cng thc Euler ci tiny0 := y0 + hf(x0, y0),

y1 = y0 + hf(x0, y0) + f(x1, y0)

2.

c. Trng hp r = 4 c cng thc RK4 sauy0 =

1

6(k1 + 2k2 + 2k3 + k4),

k1 = hf(x0, y0),

k2 = hf(x0 +h

2, y0 +

k12),

k3 = hf(x0,+h

2, y0 +

k22),

k4 = hf(fx0 + h, y0 + k3).

Nhn xt:1. Sai s a phng ca cng thc (RK4) l

R = h5(5)4 ()

120, [a, b].

2. Phng php RK4 d lp trnh, chnh xc cao, tng i n gin.V d: Dng phng php Runge-Kutta (RK4) gii gn ng phng trnh vi phn

sau vi n = 5. {y = x y,y(0) = 1, 0 x 1.

Chia [0, 1] thnh 5 on bi xi = 0.2 i, i = 0, 1, 2, 3, 4, 5. Ta c: y0 = 1, y0 =k1 + 2k2 + 2k3 + k4, trong :

k1 = 0.2 (0 1) = 0.2,k2 = 0.2 (0 + 0.1 (1 + 0.1)) = 0.2,

k3 = 0.2 (0 + 0.1 (1 + 0.2/2)),k4 = 0.2 (0 + 0.2 (1 + k3)).


Ta tnh c y1 = y0 +0. Tng t y1 = k1 + 2k2 + 2k3 + k4, trong :k1 = 0.2 (x1 + 0.1 y1),

k2 = 0.2 (x1 + 0.1 (y1 + k1/2)),k3 = 0.2 (x1 + 0.1 (y1 + k2/2)),k4 = 0.2 (x1 + 0.2 (y1 + k3)).

Vy ta tnh c y2 = y1 +y1 . Tip tc qu trnh cho ti khi tnh c y5 th dngthut ton v vit kt qu thnh bng.


Ti liu tham kho1. P. K. Anh, Gii tch s, Nh xut bn i hc Quc gia H Ni, 1996.2. T. T. Ai, Phng php s, Nh xut bn i hc Quc gia H Ni, 2001.3. N. Bacvalop, Methodes Numeriques, 1976.4. I. C Berezin, H. P. Jicop, Phng php tnh, Ting Nga, NXB. Mockva, 1959.5. B. F. Demiovich, Computational Mathematics, Mir Publishers, Moscow, 1973.6. T. V. nh, Phng php tnh, Nh xut bn Gio dc, 1998.7. P. V. Hp, L. . Thnh, Phng Php Tnh, nh xut bn i hc v Trung hc

chuyn nghip, H Ni, 1977.8. H. X. Hun, Gio trnh cc phng php s, Nh xut bn i hc Quc gia H

Ni, 2004.9. Gio trnh gii tch s, i hc KHTN-TP. H Ch Minh.

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