Giao Trinh Pptinh

7/31/2019 Giao Trinh Pptinh

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I HC NNGTRNG I HC BCH KHOA

KHOA CNG NGH THNG TIN^[]\\][^

Bin son: GV. Th Tuyt Hoa

BI GING MN

PHNG PHP TNH

(Dnh cho sinh vin khoa Cng ngh thng tin)

( TI LIU LU HNH NI B )

NNG, NM 2007


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MC LC

CHNG I NHP MN.................................................................................. 5

1.1. Gii thiu mn phng php tnh .............................................................. 5

1.2. Nhim v mn hc ..................................................................................... 5

1.3. Trnh t gii bi ton trong phng php tnh........................................... 5

CHNG II SAI S ...................................................................................... 7

2.1. Khi nim ................................................................................................... 7

2.2. Cc loi sai s............................................................................................. 7

2.3. Sai s tnh ton ........................................................................................... 7

CHNG III TNH GI TR HM ..............................................................93.1. Tnh gi tra thc. S Hoocner........................................................... 9

3.1.1. t vn ............................................................................................ 9

3.1.2. Phng php........................................................................................ 9

3.1.3. Thut ton............................................................................................ 9

3.1.4. Chng trnh ..................................................................................... 10

3.2. S Hoocner tng qut.......................................................................... 10

3.2.1. t vn .......................................................................................... 10

3.2.2. Phng php...................................................................................... 10

3.2.3. Thut ton.......................................................................................... 12

3.3. Khai trin hm qua chui Taylo............................................................... 12

CHNG IV GII GN NG PHNG TRNH........................... 14

4.1. Gii thiu.................................................................................................. 14

4.2. Tch nghim............................................................................................. 14

3.3. Tch nghim cho phng trnh i s...................................................... 16

4.4. Chnh xc ho nghim.............................................................................. 174.4.1. Phng php chia i........................................................................ 17

4.4.2. Phng php lp................................................................................ 19

4.4.3. Phng php tip tuyn..................................................................... 21

4.4.4. Phng php dy cung...................................................................... 22


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CHNG V GII H PHNG TRNH

I S TUYN TNH .................................................. 26

5.1. Gii thiu.................................................................................................. 26

5.2. Phng php Krame................................................................................. 26

5.3. Phng php Gauss.................................................................................. 27

5.3.1. Ni dung phng php...................................................................... 27

5.3.2. Thut ton.......................................................................................... 27

5.4. Phng php lp Gauss - Siedel (t sa sai) ...........................................28

5.4.1. Ni dung phng php...................................................................... 28

5.4.2. Thut ton.......................................................................................... 30

5.5. Phng php gim d .............................................................................. 31

5.5.1. Ni dung phng php...................................................................... 31

5.5.2. Thut ton.......................................................................................... 32

CHNG VI TM GI TR RING - VECT RING........................... 34

6.1. Gii thiu.................................................................................................. 34

6.2. Ma trn ng ng.................................................................................... 34

6.3. Tm gi tr ring bng phng php anhilepski....................................35

6.3.1. Ni dung phng php...................................................................... 35

6.3.2. Thut ton.......................................................................................... 37

6.4. Tm vectring bng phng php anhilepski..................................... 386.4.1. Xy dng cng thc .......................................................................... 38

6.4.2. Thut ton.......................................................................................... 39

CHNG VII NI SUY V PHNG PHP

BNH PHNG B NHT........................................... 41

7.1. Gii thiu.................................................................................................. 41

7.2. a thc ni suy Lagrange ........................................................................ 42

7.3. a thc ni suy Lagrange vi cc mi cch u ..................................... 43

7.4. Bng ni suy Ayken................................................................................. 44

7.4.1. Xy dng bng ni suy Ayken.......................................................... 45

7.4.2. Thut ton.......................................................................................... 46

7.5. Bng Ni suy Ayken (dng 2).................................................................. 46

7.6. Ni suy Newton........................................................................................ 48

7.6.1. Sai phn............................................................................................. 48


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7.6.2. Cng thc ni suy Newton................................................................ 49

7.7. Ni suy tng qut (Ni suy Hecmit) ........................................................51

7.8. Phng php bnh phng b nht .......................................................... 53

CHNG VIII TNH GN NG TCH PHN XC NH.................. 57

8.1. Gii thiu.................................................................................................. 578.2. Cng thc hnh thang ............................................................................... 57

8.3. Cng thc Parabol.................................................................................... 58

8.4. Cng thc Newton-Cotet ......................................................................... 59

MT S CHNG TRNH THAM KHO..................................................... 62

TI LI U THAM KHO.................................................................................. 68


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CHNG I NHP MN

1.1. Gii thiu mn phng php tnh

Phng php tnh l b mn ton hc c nhim v gii n kt qu bng s

cho cc bi ton, n cung cp cc phng php gii cho nhng bi tontrong thc t m khng c li gii chnh xc. Mn hc ny l cu ni gia

ton hc l thuyt v cc ng dng ca n trong thc t.

Trong thi i tin hc hin nay th vic p dng cc phng php tnh cng

trnn ph bin nhm tng tc tnh ton.

1.2. Nhim v mn hc

- Tm ra cc phng php gii cho cc bi ton gm: phng php (PP)

ng v phng php gn ng.+ Phng php: ch ra kt qu di dng mt biu thc gii tch c th.

+ Phng php gn ng: thng cho kt qu sau mt qu trnh tnh

lp theo mt quy lut no , n c p dng trong trng hp bi

ton khng c li gii ng hoc nu c th qu phc tp.

- Xc nh tnh cht nghim

- Gii cc bi ton v cc tr

- Xp x hm: khi kho st, tnh ton trn mt hm f(x) kh phc tp, ta cth thay hm f(x) bi hm g(x) n gin hn sao cho g(x) f(x). Vic lachn g(x) c gi l php xp x hm

- nh gi sai s : khi gii bi ton bng phng php gn ng th sai s

xut hin do s sai lch gia gi tr nhn c vi nghim thc ca bi

ton. V vy ta phi nh gi sai s t chn ra c phng php ti

u nht

1.3. Trnh tgii bi ton trong phng php tnh- Kho st, phn tch bi ton

- La chn phng php da vo cc tiu ch sau:

+ Khi lng tnh ton t

+ n gin khi xy dng thut ton

+ Sai s b


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+ Kh thi

- Xy dng thut ton: s dng ngn ng gi hoc s khi (cng mn

cng tt)

- Vit chng trnh: s dng ngn ng lp trnh (C, C++, Pascal,

Matlab,)- Thc hin chng trnh, th nghim, sa i v hon chnh.


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CHNG II SAI S

2.1. Khi nim

Gi s x l s gn ng ca x* (x* : sng),

Khi = xx gi l sai s thc s ca x

V khng xc nh c nn ta xt n 2 loi sai s sau:

- Sai s tuyt i: Gi s xxxchosaobedu0x * >

Khi x gi l sai s tuyt i ca x

- Sai s tng i :x

xx

=

2.2. Cc loi sai sDa vo nguyn nhn gy sai s, ta c cc loi sau:

-Sai s gi thit: xut hin do vic gi thit bi ton t c mt siukin l tng nhm lm gim phc tp ca bi ton.

-Sai s do s liu ban u: xut hin do vic o c v cung cp gi truvo khng chnh xc.

-Sai s phng php : xut hin do vic gii bi ton bng phng phpgn ng.

-Sai s tnh ton : xut hin do lm trn s trong qu trnh tnh ton, qutrnh tnh cng nhiu th sai s tch lu cng ln.

2.3. Sai s tnh ton

Gi s dng n s gn ng )n,1i(xi = tnh i lng y,

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

Trong : f l hm kh vi lin tc theo cc i s xi

Khi sai s ca y c xc nh theo cng thc sau:

Sai s tuyt i: =

=n

1ii

i

xx

fy

Sai s tng i: =

=

n

1ii

i

xx

flny

- Trng hp f c dng tng: n21i x......xx)x(fy ==


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i1x

f

i

=

suy ra =

=n

1iixy

- Trng hp f c dng tch:

nx*...*1kx

kx*...*2x*1x)ix(fy

+

==

)xln...x(ln)xln...xlnx(lnx......x

x...x.xlnfln n1mm21

n1m

m21 +++++== ++

ix

1

x

fln

ii

=

=>

==

=

=n

1ii

n

1i i

iy xx

x

Vy =

=n

1iiy x

- Trng hp f dng lu tha: y = f(x) = )0(x >

xlnflnyln ==

xx

fln =

Suy ra xx

x.y =

=

V d. Cho 13.12c;324.0b;25.10a

Tnh sai s ca:

cb

ay

3

1 = ; cbay3

2 =

GiI c2

1ba3)cb()a(y 31 ++=+=

=cc

21

bb

aa3 ++

)cb(cb)a(a)cb()a(y 3332 +=+=

)c

c

2

1

b

b(cb

a

aa3y 32

+

+

=


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CHNG III TNH GI TR HM

3.1. Tnh gi tra thc. S Hoocner

3.1.1.t vn

Cho a thc bc n c dng tng qut :

p(x) = a0xn + a1x

n-1 + ... + an-1x+ an (a#0)

Tnh gi tra thc p(x) khi x = c (c: gi tr cho trc)

3.1.2. Phng php

p dng s Hoocner nhm lm gim i s php tnh nhn (ch thchin n php nhn), phng php ny c phn tch nh sau:

p(x) = (...((a0x + a1)x +a2)x+ ... +an-1 )x + an p(c) = (...((a0c + a1)c +a2)c+ ... +an-1 )c + an t p0 = a0p1 = a0c + a1 = p0c + a1

p2 = p1c+ a2

. . . . . . . .

pn = pn-1c + an = p(c)

S Hoocnera0 a1 a2 .... an-1 an

p0*c p1*c .... pn-2*c pn-1*c

p0 p1 p2 ... pn-1 pn= p(c)

Vd: Cho p(x) = x6 + 5x4 + x3 - x - 1 Tnh p(-2)

p dng s Hoocner:

1 0 -5 2 0 -1 -1

-2 4 2 -8 16 -30

1 -2 -1 4 -8 15 -31

Vy p(-2) = -31

3.1.3. Thut ton

+ Nhp vo: n, c, cc h s ai ( n,0i = )


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+ X l: t p = a0

Lp i = 1 n : p = p * c + ai

+ Xut kt qu: p

3.1.4. Chng trnh

#include

#include

main ( )

{ int i, n; float c, p, a [10];

clrsr ();

printf (Nhap gia tri can tinh : ); scanf (%f,&c);

printf (Nhap bac da thuc : ); scanf (%d,&n);printf (Nhap cc h s: \n);

for (i = 0, i


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Xc nh bnXt y=0, t (2) => p(c) = bn

Xc nh bn-1p(x) = (x-c) p1 (x) + p(c) (1

)

Trong p1(x) : a thc bc n-1

n1n2n2n

11n

0 b)byb...ybyb(y)cy(p +++++=+

t x=y+c ta c:

n1n2n2n

11n

0 b)byb...ybyb)(cx()x(p +++++= (2)

ng nht (1) & (2) suy ra:

p1(x) = b0yn-1 + b1y

n-2 + ...+ bn-2y + bn - 1

Xt y = 0, p1(c) = bn-1Tng t ta c: bn-2 = p2(c), , b1 = pn-1(c)

Vy bn-i = pi(c) (i = 0-->n) , b0 =a0

Vi pi(c) l gi tra thc bc n-i ti c

S Hoocner tng qut:

a0 a1 a2 .... an-1 an

p0*c p1*c .... pn-2*c pn-1*c

p0 p1 p2 ... pn-1 pn= p(c)=bn

p0*c p1

*c .... pn-2

*c

p0 p1 p2

... pn-1 = p1(c)=bn-1

...

V d: Cho p(x) = 2x6 + 4x5 - x2 + x + 2. Xc nh p(y-1)


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p dng s Hoocner tng qut :

\p(x) 2 4 0 0 -1 1 2

-2 -2 2 -2 3 -4

p1(x) 2 2 -2 2 -3 4 -2

-2 0 2 -4 7p2(x) 2 0 -2 4 -7 11

-2 2 0 -4

p3(x) 2 -2 0 4 -11

-2 4 -4

p4(x) 2 -4 4 0

-2 6

p5(x) 2 -6 10

-22 -8

Vy p(y-1) = 2y6 - 8y5 + 10y4 - 11y2+11y- 2

3.2.3. Thut ton

- Nhp n, c, a [i] (i = n,0 )

- Lp k = n 1

Lp i = 1 k : ai = ai-1 * c + ai

- Xut ai (i = n,0 )

3.3. Khai trin hm qua chui Taylo

Hm f(x) lin tc, kh tch ti x0 nu ta c th khai trin c hm f(x) qua

chui Taylor nh sau:( )

!n

)xx)(x(f...

!2

)xx)(x(f

!1

)xx)(x(f)x(f)x(f

n00

n20000

0

++

+

+

khi x0 = 0, ta c khai trin Macloranh:

!n

x)0(f...

!2

x)0(f...

!1

x)0(f)0(f)x(f

n)n(2

++

++

++

V d: ...!6

x

!4

x

!2

x1Cosx

642

++


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BI TP

1. Cho a thc p(x) = 3x5 + 8x4 2x2 + x 5a.

Tnh p(3)b. Xc nh a thc p(y-2)

2. Khai bo (nh ngha) hm trong C tnh gi tra thc p(x) bc ntng qut theo s Hoocner

3. Vit chng trnh (c s dng hm cu 1) nhp vo 2 gi tr a, b.Tnh p(a) + p(b)

4. Vit chng trnh nhp vo 2 a thc pn(x) bc n, pm(x) bc m v gi trc. Tnh pn(c) + pm(c)

5. Vit chng trnh xc nh cc h s ca a thc p(y+c) theo s Hoocner tng qut

6. Khai bo hm trong C tnh gi tr cc hm ex, sinx, cosx theo khaitrin Macloranh.


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CHNG IV GII GN NG PHNG TRNH

4.1. Gii thiu

tm nghim gn ng ca phng trnh f(x) = 0 ta tin hnh qua 2 bc:

- Tch nghim: xt tnh cht nghim ca phng trnh, phng trnh c

nghim hay khng, c bao nhiu nghim, cc khong cha nghim nu c.

i vi bc ny, ta c th dng phng php th, kt hp vi cc nh

l m ton hc h tr.

- Chnh xc ho nghim: thu hp dn khong cha nghim hi tc

n gi tr nghim gn ng vi chnh xc cho php. Trong bc ny ta

c th p dng mt trong cc phng php:

+ Phng php chia i+ Phng php lp

+ Phng php tip tuyn

+ Phng php dy cung

4.2. Tch nghim

* Phng php th:

Trng hp hm f(x) n gin

- V th f(x)

- Nghim phng trnh l honh giao im ca f(x) vi trc x, t suy

ra s nghim, khong nghim.

Trng hp f(x) phc tp

- Bin i tng ng f(x)=0 g(x) = h(x)

- V th ca g(x), h(x)

- Honh giao im ca g(x) v h(x) l nghim phng trnh, t suyra s nghim, khong nghim.

*nh l 1:

Gi s f(x) lin tc trn (a,b) v c f(a)*f(b)


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V d 1. Tch nghim cho phng trnh: x3 - x + 5 = 0

Gii: f(x) = x3 - x + 5

f(x) = 3x2 - 1 , f(x) = 0 x = 3/1

Bng bin thin:

x - 3/1 3/1 +

f(x) + 0 - 0 +

f(x)yC phng trnh co 1 nghim x (1, 2)

4

421

1

y = 2x

y = -x + 42


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* nh ly 2: (Sai s)

Gia s la nghim ung va x la nghim gn ung cua phng trnhf(x)=0, cung nm trong khoang nghim [ a,b] va f '(x) = m 0 khi a x

b. Khi o

m

)x(fx

V du 3. Cho nghim gn ung cuaphng trnh x4 - x - 1 = 0 la 1.22.Hay c lng sai s tuyt i la bao nhiu?

Gii: f (x) = f (1.22) = 1.224 - 1.22 - 1 = - 0,0047 < 0

f(1.23) = 0.588 > 0

nghim phng trnh x (1.22 , 1.23)

f '(x) = 4 x3 -1 > 4*1.223 - 1 = 6.624 = m x (1.22 , 1.23)

Theo nh ly 2 : x = 0.0047/6.624 = 0.0008 (v |x - | < 0.008)

3.3. Tch nghim cho phng trnh i s

Xt phng trnh i s: f(x) = a0xn + a1x

n-1 + + an-1x + an = 0 (1)

nh l 3:

Cho phng trnh (1) c m1 = max {ai} i = n,1

m2 = max {ai} i = 1n,0

Khi mi nghim x ca phng trnh u tho mn:

20

1

n2

n1 xa

m1xam

ax =++

=

nh l 4:

Cho phng trnh (1) c a0 > 0, am l h s m u tin. Khi mi nghim

dng ca phng trnh u m 0a/a1N += ,

vi a = max {ai} n,0i = sao cho ai < 0.

V d 4. Cho phng trnh: 5x5 - 8x3 + 2x2 - x + 6 = 0

Tm cn trn nghim dng ca phng trnh trn

Gii: Ta c a2 = -8 l h s m u tin, nn m = 2

a = max( 8, 1) = 8

Vy cn trn ca nghim dng: 5/81N +=

*nh ly 5:


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Cho phng trnh (1), xet cac a thc:

1(x) = xn f (1/x) = a0 + a1x + ... + anxn

2(x) = f(-x) = (-1)n (a0x

n - a1xn-1 + a2x

n-2 - ... + (-1)nan)

3(x) = xn f(-1/x) = (-1)n (anx

n - an-1xn-1 + an-2x

n-2 - ... + (-1)na0)

Gia s N0, N1, N2, N3 la cn trn cac nghim dng cua cac a thc f(x),1(x), 2(x), 3(x). Khi o moi nghim dng cua phtrnh (1) u nmtrong khoang [1/N1, N0] va moi nghim m nm trong khoang [-N2,-1/N3]

V du 5. Xt phng trnh

3x2 + 2x - 5 = 0 N0 = 1 + 3/5 (nh ly 4)

1(x) = 3 + 2x - 5x2 N1 khng tn tai (a0 < 0)

2(x) = 3x2 - 2x - 5 N2 = 1 + 5/3 (nh ly 4)

3(x) = 3 - 2x - 5x2 N3 khng tn tai (a0 < 0)

Vy: moi nghim dng x < 1 + 3/5

moi nghim m x > - (1 +5/3) = - 8/3

4.4. Chnh xc ho nghim

4.4.1. Phng php chia i

a. tngCho phng trnh f(x) = 0, f(x) lin tc v tri du ti 2 u [a,b]. Gi s

f(a) < 0, f(b) < 0 (nu ngc li th xt f(x)=0 ). Theo nh l 1, trn [a,b]

phng trnh c t nht 1 nghim .

Cch tm nghim :

t [a0, b0] = [a, b] v lp cc khong lng nhau [ai , bi ] (i=1, 2, 3, )

[ai, (ai-1+ bi-1)/2] nu f((ai-1+ bi-1)/2) >0

[ai, bi] =[(ai-1+ bi-1)/2,bi] nu f((ai-1+ bi-1)/2) < 0

Nh vy:

- Hoc nhn c nghim ng mt bc no :

= (ai-1+ bi-1)/2 nu f((ai-1+ bi-1)/2) = 0

- Hoc nhn c 2 dy {an} v {bn}, trong :


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{an}: l dy n iu tng v b chn trn

{bn}: l dy n iu gim v b chn di

nn == nnn

blimalim l nghim phng trnh

V d 6. Tm nghim phng trnh: 2x + x - 4 = 0 bng pphp chia i

Gii:

- Tch nghim: phng trnh c 1 nghim x (1,2)

- Chnh xc ho nghim: p dng phng php chia i ( f(1) < 0)

Bng kt qu:

an bn )2

ba(f nn

+

1 2 +

1.5 -

1.25 -

1.375 +

1.438 +

1.406 +

1.391 -

1.383 +

1.387 -1.385 -

1.386 1.387

386.1blimalim n11n

nn

==

Kt lun: Nghim ca phng trnh: x 1.386

b. Thut ton

- Khai bo hm f(x) (hm a thc, hm siu vit)

- Nhp a, b sao cho f(a)0

- Lp

c = (a+b)/2

nu f(c) > 0 b = c

ngc li a = c

trong khi (f(c)> ) /* a - b > v f(c) != 0 */


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- Xut nghim: c

4.4.2. Phng php lp

a. tngBin i tng ng: f(x) = 0 x = g(x)

Chn gi tr ban u x0 khong nghim (a,b),tnh x1 = g(x0), x2 = g(x1), , xk= g(xk-1)Nh vy ta nhn c dy {xn}, nu dy ny hi t th tn ti gii hn

= nn xlim (l nghim phng trnh )

b. ngha hnh hcHonh giao im ca 2 th y=x v y=g(x) l nghim phng trnh

Trng hp hnh a: hi tn nghim Trng hp hnh a: khng hi tn nghim (phn ly nghim)

Sau y ta xt nh l viu kin hi tn nghim sau mt qu trnh lp

nh l (iu kin )

Gi s hm g(x) xc nh, kh vi trn khong nghim [a,b] v mi gi tr g(x)

u thuc [a,b]. Khi nu q > 0 sao cho g(x)q


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- Trong trng hp tng qut, nhn c xp x xn vI chnhxc cho trc, ta tin hnh php lp cho n khi 2 xp x lin tiptho mn:

+ qq1

xx n1n

V d 7. Tm nghim: x3 - x - 1 = 0 bng phng php lp

Gii: - Tch nghim: phng trnh c mt nghim (1,2)

- Chnh xc ho nghim:

32

33 1xx;x

1xx;1xx01xx +=

+===

Chn g(x) = 3 1x +

1)1x(

131)x('g3

2 p dng phng php lp (chn x0 = 1)

x g(x) =3 1x +

1 1.260

1.260 1.312

1.312 1.322

1.322 1.3241.324 1.325

1.325 1.325

x4 - x5 < = 10-3

Nghim phng trnh x 1.325

c. Thut ton

- Khai bo hm g(x)

- Nhp x- Lp: y= x

x = g(x)

trong khi x - y>

- Xut nghim: x (hoc y)


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4.4.3. Phng php tip tuyn

a. tngChn x0 khong nghim (a, b)

Tip tuyn ti A0 (x0, f(x0)) ct trc x ti im c honh x1,

Tip tuyn ti A1 (x1, f(x1)) ct trc x ti im c honh x2, ,Tip tuyn ti Ak(xk, f(xk)) ct trc x ti im c honh xk,

C tip tc qu trnh trn ta c th tin dn n nghim ca phng trnh.

* Xy dng cng thc lp:

Phng trnh tip tuyn ti Ak(xk, f(xk))

y - f(xk) = f(xk)*(x - xk)

Tip tuyn ct trc x ti im c to (xk+1, 0)

Do vy: 0 f(xk) = f(xk)*(xk+1 - xk)

)x('f

)x(fxx

k

kk1k =+

b. ngha hnh hc

nh l (iu kin hi t theo Furi_iu kin )

Gi s [a,b] l khong nghim ca phng trnh f(x)=0. o hm f(x),

f(x) lin tc, khng i du, khng tiu dit trn [a,b]. Khi ta chn xp

x nghim ban u x0[a,b] sao cho f(x0)*f(x0) > 0 th qu trnh lp s hitn nghim.

V d 8. Gii phng trnh: x3 + x - 5 = 0 bng phng php tip tuyn

Gii: - Tch nghim:

f(x) = x3 + x - 5

a x2 x1 x0 b

x[ ]

A1

f(x)

tip tuynA0


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22

f(x) = 3x2 + 1 > 0 x

= )x(flimn , +=+ )x(flimn

Phng trnh trn c 1 nghim duy nht

f(1)* f(2) = (-3)*5 < 0

Vy phng trnh c 1 nghim duy nht x (1, 2)

- Chnh xc ho nghim:

f(x) = 6x > 0 x (1, 2)

f(x) > 0 x

Tho mn iu kin hi t Furi, p dng phng php tip tuyn

Chn vi x0 = 2 ( v f(2). f(2) > 0)

x f(x)/f(x)2 0.3851.615 0.0941.521 0.0051.516 0.0001.516

Vy nghim x 1.516

c. Thut ton

- Khai bo hm f(x), fdh(x)- Nhp x

- Lp y= x

x = y f(y)/fdh(y)

trong khi x - y>

- Xut nghim: x (hoc y)

4.4.4. Phng php dy cunga. tng

Gi s [a, b] l khong nghim phng trnh f(x)=0. Gi A, B l 2 im

trn th f(x) c honh tng ng l a, b. Phng trnh ng thng

qua 2 im A(a,f(a)), B(b, f(b)) c dng:

ab

ax

)a(f)b(f

)a(fy

=


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23

Dy cung AB ct trc x ti im c to (x1, 0)

Do :ab

ax

)a(f)b(f

)a(f0 1

=

)a(f)b(f

)a(f)ab(ax1

=

Nu f(a)*f(x1)


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24

Bng kt qu:

a b x f(x)

1

1.333

1.379

1.385

1.386

2 1.333

1.379

1.385

1.386

1.386

-0.447

-0.020

-0.003

-0.000

Vy nghim phng trnh: x 1.386

c. Thut ton

- Khai bo hm f(x)

- Nhp a, b

- Tnh x = a (b-a)f(a) / (f(b)-f(a))

- Nu f(x)*f(a) Ngc li

Lp a = x

x = a (b-a)f(a) / (f(b)-f(a))

trong khi x - a>

- Xut nghim: x


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BI TP

1. Tm nghim gn ng cc phng trnh:a. x

3

x + 5 = 0 b. x

3

x 1 = 0c. sinx x + 1/4 = 0 d. x4 4x 1= 0

bng phng php chia i vi sai s khng qu 10-3

2. Tm nghim gn ng cc phng trnh:a. x3 x + 5 = 0 b. x4 4x 1 = 0

bng phng php dy cung vi sai s khng qu 10-2

3. Tm nghim gn ng cc phng trnh:a. ex 10x + 7 = 0 b. x3 + x 5 = 0

bng phng php tip tuyn vi sai s khng qu 10-34. Dng phng php lp tm nghim dng cho phng trnh

x3 x 1000 = 0 vi sai s khng qu 10-3

5. Tm nghim dng cho phng trnh: x3 + x2 2x 2 = 06. Tm nghim m cho phng trnh: x4 - 3x2 + 75x 1000 = 07. Dng cc phng php c th tm nghim gn ng cho phng trnh

sau: cos2x + x 5 = 0

8. Vit chng trnh tm nghim cho c dng tng qut:f(x) = a0xn + a1xn-1 + + an-1x + an = 0a. p dng phng php chia ib. p dng phng php dy cung

9. Vit chng trnh tm nghim cho phng trnh ex 10x + 7 = 0 bngphng php tip tuyn.

10.Vit chng trnh xc nh gi tr x1, x2 theo nh l 3.11.Vit chng trnh tm cn trn ca nghim dng phng trnh i s

theo nh l 4.


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26

CHNG V GII H PHNG TRNH

I S TUYN TNH

5.1. Gii thiu

Cho h phng trnh tuyn tnh:

a11x1 + a12x2 + ... + a1nxn = a1n+1

a21x1 + a22x2 + ... + a2nxn = a2n+1

an1x1 + an2x2 + ... + annxn = ann+1

H phng trnh trn c thc cho bi ma trn:

a11 a12 ... a1n a1n+1

a21 a22 ... a2n a2n+1

....Ann+1 =

an1 an2 ... ann ann+1

Vn : Tm vectnghim )x,...,x,x(x n21=

* Phng php:

- Phng php ng (Krame, Gauss, khai cn): c im ca cc phng

php ny l sau mt s hu hn cc bc tnh, ta nhn c nghim ng

nu trong qu trnh tnh ton khng lm trn s

- Phng php gn ng (Gauss Siedel, gim d): Thng thng ta cho

n s mt gi tr ban u, t gi tr ny tnh gi tr nghim gn ng tt hn

theo mt qui tc no . Qu trnh ny c lp li nhiu ln v vi mt s

iu kin nht nh, ta nhn c nghim gn ng.

5.2. Phng php Krame

- Khai bo hm Dt tnh nh thc ma trn vung cp n

- Nhp n, aij (i = 1n,1j;n,1 += )

- d = Dt (A)

- Xt + d = 0

+ d # 0 {di = Dt(Ai) ; xi = di/d }


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5.3. Phng php Gauss

5.3.1. Ni dung phng php

- Bin i Ma trn A v ma trn tam gic trn

a11 a12 ... a1n a1n+1

a21 a22 ... a2n a2n+1

........A =

an1 an2 ... ann ann+1

a11 a12 ... a1n a1n+1

0 a'22 ... a'2n a'2n+1

...... A=

0 0 ... a'nn a'nn+1Cch bin i A A: Thc hin n-1 ln bin i

Ln bin i i (lm cho aji = 0; j = i + 1 n) bng cch:

dng j = dng j + dng i * m (m = -aji / aij )

- Tm nghim theo qu trnh ngc: xn nn-1 ... x1

V d 1. Gii h phng trnh

1 2 -1 3 5 1 2 -1 3 5

-2 X 2 1 0 -1 2 0 -3 2 -7 -8

1 X -1 3 2 4 8 5/3 0 5 1 7 13

1 X -2 0 5 1 4 4/3 0 4 3 7 14

1 2 -1 3 5 1 2 -1 3 5

0 -3 2 -7 -8 0 -3 2 -7 -8

0 0 13/3 -14/3 -1/3 0 0 13/3 -14/3 -1/313

17

0 0 17/3 -7/3 10/3

0 0 0 49/13 49/13

x4 = 1; x3 = 1; x2 = 1; x1 = 1

Vy nghim h phng trnh )1,1,1,1(x =

5.3.2. Thut ton

- Nhp n, aij ( 1n,1j,n,1i +== ) (nhp trc tip hoc t file)


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- Bin i A A (ma trn tam gic trn)

Lp i = 1 n -1Tm j sao cho aji # 0

+ Xt aij = 0 Hon i dng i v dng j cho nhau

+ Lp j = i + 1 n m = -aij/aii

Lpk = i n +1 ajk= ajk+ aik* m- Tm nghim

iij

n

1ijij1ini a/xaax

=

+=+ ( i =n 1)

Lp i = n 1s = 0

lp j = i + 1 n S = S + aij * xjxi = (ain+1 - s)/aii

- Xut xi (i=1n)

5.4. Phng php lp Gauss - Siedel (tsa sai)


Bin i h phng trnh v dng:

+= gxBx)x,......,x,x(x n21=

; )g,......,g,g(g n21=

; B = {bij}n

Cch bin i:

a11x1 +a12x2 + ....+ a1nxn = a1n+1

a21x1 +a22x2 + ....+ a2nxn = a2n+1

.......

an1x1 +an2x2 + ....+ annxn = ann+1

)1j(a/)xaa(x 11jn

1jj11n1 = =+

....

)nj(a/)xaa(x nnjn

1jnj1nnn =

=+

Tng qut:


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29

)ij(a/)xaa(x iijn

1jij1ini =

=+ (*)

Cho h phng trnh xp x nghim ban u: )x,...,x,x(x 0n02

000 =

Thay 0x

vo (*) tnh: )x,...,x,x(x 1n12

101 =

)ij(a/)xaa(x ii0j

n

1jij1in

1i =

=+

Tng t, tnh 2x

, 3x

,

Tng qut: )ij(a/)xaa(x iikj

n

1jij1in

1ki =

=+

+

Qu trnh lp s dng khi tho mn tiu chun hi t tuyt i:

)n,1i(xxk

i

ik

i =


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30

0 -0,2 -0,1

-0,1 0 -0,2B =-0,1 -0,1 0

)8.0,2.1,1(g = Do 13.0bmax1r

3

1jij

i


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31

xi = yi }

trong khi (t)

- Xut xi (i =1n)

5.5. Phng php gim d


Bin i h phng trnh v dng:

a1n + 1 - a11x1 - a12x2 - ... - a1nxn = 0

a2n +1 - a21x1 - a22x2 - ... - a2nxn = 0 (1)

.......

ann + 1 - an1x2 - an2x2 - ... - annxn = 0

Chia dng i cho aii # 0b1n + 1 - b12x2 - b13x2 - ... - x1 = 0

b2n + 1 - b21x1 b23x3 - ... - x2 = 0 (2)

.......

bnn + 1 - bn1x1 - bn2x2 - ... - xn = 0

Cho vectnghim ban u )x,...,x,x(x 0n02

010 =

V 0x

khng phi l nghim nn:

b1n+1 - b12x20 - b13x3

0 - ... - x10 = R1

0

b2n+1 - b21x10 - b23x3

0 - ... - x20 = R2

0

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

bnn+1 - bn1x10 - bn2x2

0 - ... - xn0 = Rn

0

0n

02

01 R,.......,R,R l cc s d do s sai khc gia 0x

vi nghim thc ca

h phng trnh

Tm Rs0 = max {|R1

0|, | R20|, ... | Rn

0|} va lam trit tiu phn t o bng

cach cho xs mt s gia xs = Rs0, ngha la xs1 = xs0 + Rs0

Tnh li cc s d :

Rs1 = 0

Ri1 = Ri

0 - bis * xs = Ri0 - bis * Rs0 (i = 1 n)

C tip tuc qua trnh lp trn cho n khi : Rik< (i = 1n) th Xk =(x1

k, x2k,... xn

k) la nghim cua h phtrnh.


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V d 3. Gii h phng trnh:

10 -2 -2 6

-2 10 -1 7

1 1 -10 8

Gii: Bin i v h phng trnh tng ng

0,6 + 0,2 x2 + 0,2x3 - x1 = 0

0,3 + 0,2 x1 + 0,2x3 - x2 = 0

0,8 + 0,1 x1 + 0,1x2 - x3 = 0

Cho )8.0,7.0,6.0(R)0,0,0(x 00 ==

}Rmax{R 0i03 = 3,1i =

x31 = 8.0Rx

0

3

0

3 =+ R2 = 78.08.01.07.0R.bR

0323

02 =+=+

76.08.02.06.0R.bRR 031301

11 =+=+=

)0,78.0,76.0(R1 =

Tng t ta c bng kt qu:

x1 x2 x3 R1 R2 R30 0 0 0.6 0.7 0.8

0.8 0.76 0.78 00.78 0.92 0 0.08

0.92 0 0.18 0.170.96 0.04 0 0.19

0.99 0.07 0.02 00.99 0 0.03 0.01

0.99 0.01 0 0.011 0.01 0 0

1 0 0.01 0

1 0 0 0Vy nghim h phng trnh x = (1, 1, 1)

5.5.2. Thut ton

- Nhp n, aij, xi

- Bin i h phng trnh (1) v dng (2)


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for (i=1, i


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34

CHNG VI TM GI TR RING - VECTRING

6.1. Gii thiu

Cho ma trn vung cp n

a11 a12 ... a1na21 a22 ... a2n

....... A =

an1 an2 ... ann

Tm gi tr ring, Vectring

x ca ma trn A

Ngha l: tm v

x sao cho :

det (A - E) = 0 ( E : Ma trn n v)

(A - E) x = 0

trnh vic khai trin nh thc (i hi s php tnh ln) khi tm ta cth p dng phng php anhilepski. phng php ny ta ch cn tm

ma trn B sao cho B ng dng vi ma trn A v B c dng ma trn

Phrbemit.

p1 p2 ... pn-1 pn

1 0 ... 0 00 1 ... 0 0

....

P =

0 0 ... 1 0

Khi gi tr ring ca ma trn A cng l gi tr ring ca ma trn B.

6.2. Ma trn ng ng

6.2.1.nh ngha

Ma trn B gi l ng dng vi ma trn A (B A) nu tn ti ma trnkhng suy bin M (det(M) 0) sao cho B = M-1A M

6.2.2. Tnh cht:

A B B A

A B, B C A C

A B gi tr ring ca A v B trng nhau.


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6.3. Tm gi tr ring bng phng php anhilepski


Thc hin n-1 ln bin i:

* Ln bin i 1: Tm M-1 , M sao cho A1 = M-1 A M A

v dng n ca A1 c dng: 0 0 0 ... 1 0

1 0 ... 00 1 ... 0

an1 an2 ... annM-1 =

0 0 ... 1

M-1n-1j = anj

1 0 ... 0 00 1 ... 0 0

1nn

1n

a

a

1nn

2n

a

a

1nna

1

1nn

nn

a

a

M =

0 0 ... 0 1

1nna

1

nu j = n -1

Mn-1j =

1nn

nj

a

a

nu j # n - 1

A1 = M-1 A M A

* Ln bin i 2: Chn M-1, M sao cho A2 = M-1 A1 M A1

v dng n-1 ca A2 c dng: 0 0 0 ... 1 0 0

A2 A1 , A1 A => A2 A (tnh cht)

. * Ln bin i th n-1

Ta nhn c ma trn An-1 A v An-1 c dng ca P.

Khi nh thc

det (P-E) = (-1)n (n - p1n-1 - - pn-1 - pn)

det (p-E) = 0 n - p1n-1 - - pn-1 - pn = 0


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Gii phng trnh, suy ra

V d 1. Tm gi tr ring ca ma trn:

2 1 0

1 3 1A =0 1 2

n = 3

ta tm:

p1 p2 P31 0 0P =

0 1 0

Ln 1: Chn

2 1 -2

1 5 -5A1 = M-1A M =

0 1 0

Ln 2: Chn

7 -14 81 0 0A2 = M

-1A1M=

0 1 0

=P

Gi tr ring l nghim phng trnh: 3 - 72 + 14 - 8 = 0

(-2) (-1) (-4) = 0 = 2; =1; =4

1 0 00 1 2M-1 =

0 1 0

1 0 00 1 -2M =

0 0 1

1 5 -50 1 0M-1 =

0 0 1

1 -5 5

0 1 0M =

0 0 1


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6.3.2. Thut ton

- Nhp n, aij ( i,j = 1n)

- Khai bo hm nhn 2 ma trn vung cp n

(C = A x B => kjikn

1kij bac = = )

- Lp k = n -1 1 (phn t bin i : ak+1 k)

/* Tnh 2 ma trn M, M1 (M1 la ma tran nghich dao cua M) */

for i = 1 n

for j = 1 n

if i k

if i = j {M[i,j] = 1; M1[i,j] = 1 }else {M[i,j] = 0; M1[i,j] = 0 }

else { M1[i,j] = a[k+1,j]

if (j = k) M[i,j] = 1/a[k+1,k]

else M[i,j] = - a[k+1,j]/a[k+1,k] }

/* Gi hm nhn 2 ln */

Ln 1 : vo A, M; ra B

Ln 2 : vo M1; B; ra A- Xut aij ( i,j = 1n)

Thut ton nhn 2 ma trnfor (i=1, i < = n; i++)

for (j=1; j< = n; j++) {

c[i] [j] = 0

for (k=1; k < = n; k++) c[i] [j] + = a [i] [k] * b [k] [j]

}


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6.4. Tm vectring bng phng php anhilepski

6.4.1. Xy dng cng thc

Gi

y l vectring ca ma trn P A

Ta c: (P - E)

y = 0

P y = E y

M-1. A. M .

y = E

y

Nhn 2 v cho M:

M M-1. A M

y = M E

y

A M

y = E M

y

t

x = M

y

A

x= E

x

(A - E)

x = 0

Vy

x = M

y l vectring ca A

1n211

11

2n1

1n M.M.M.A.M...M.MP

=

Mi: Ma trn M xc nh c ln bin i th i

v M = M1 M2 ... Mn-1

Xc nh

y

(P-E)

y = 0

p1 - p2 ... pn-1 pn y1

1 ... 0 0 y2...... ...

0 0 ... 1 - yn

= 0

(p1 - )y1 + p2y2 + ... + pn-1yn-1 + pnyn = 0y1 - y2 = 0

.....

yn-1 - yn = 0

cho: yn = 1 yn-1 = ,

yn-2 = yn-1 = 2 , ... , y1 = n-1


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Vy

y = (n-1, n-2, ... , 2, , 1)

V d 2. Tm vectring ca A

2 1 0

1 3 1A =

0 1 2Gii: Gi

y l vectring ca ma trn P A

v d 1 ta c:

1 = 2

y 1 = (4, 2, 1)

2 = 1

y 2 = (1, 1, 1)

3 = 4

y 3 = (16, 4, 1)

Tm M:

1 0 0 1 -5 -5 1 -5 50 1 -2 0 1 0 0 1 -2M = 12

11 M.M =

0 1 0 0 0 1

=

0 0 1

x = M

y

1 -5 5 4 -1

0 1 -2 2 0

x1 =

0 0 1 1

=

1

1 -5 5 1 1

0 1 -2 1 -1

x2 =

0 0 1 1

=

1

1 -5 5 16 1

0 1 -2 4 2

x3 =

0 0 1 1

=

1

Vy vectring ca A:

x1 = (-1, 0, 1)

x2 = (1, -1, 1)

x3 = (1, 2, 1)

6.4.2. Thut ton

B sung thm lnh trong thut ton tm tr ring nh sau:


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- Khi to B1 = E

- Lp k = n-1 1

/* Tnh 2 ma trn M, M1 */

/* Gi hm nhn 3 ln */

Ln 1: vo A, M; ra B

Ln 2: vo M1, B; ra A

Ln 3: vo B1, M; ra B

/* Gn li ma trn B1=B */

- Xut aij, bij


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41

CHNG VII NI SUY V PHNG PHP

BNH PHNG B NHT

7.1. Gii thiu

Trong ton hc ta thng gp cc bi ton lin quan n kho st v tnh

gi tr cc hm y = f(x) no . Tuy nhin trong thc t c trng hp ta

khng xc nh c biu thc ca hm f(x) m ch nhn c cc gi tr

ri rc: y0, y1, ..., yn ti cc im tng ng x0, x1, ..., xn.

Vn t ra l lm sao xc nh gi tr ca hm ti cc im cn li.

Ta phi xy dng hm (x) sao cho:

(xi) = yi = f (xi) vi n,0i =

(x) f (x) x thuc [a, b] v x xi

- Bi ton xy dng hm (x) gi l bi ton ni suy

- Hm (x) gi l hm ni suy ca f(x) trn [a, b]

- Cc im xi ( n,0i = ) gi l cc mc ni suy

Hm ni suy cng c p dng trong trng hp xc nh c biu

thc ca f(x) nhng n qu phc tp trong vic kho st, tnh ton. Khi

ta tm hm ni suy xp x vi n n gin phn tch v kho st hn.

Trong trng hp ta chn n+1 im bt k lm mc ni suy v tnh gitr ti cc im , t xy dng c hm ni suy (bng cng thc

Lagrange, cng thc Newton,).

Trng hp tng qut: hm ni suy (x) khng ch tho mn gi tr hm timc ni suy m cn tho mn gi tro hm cc cp ti mc .

(x0) = f(x0); (x1) = f(x1);

(x0) = f(x0); (x1) = f(x1);

Ngha l ta tm hm ni suy ca f(x) tha mn bng gi tr sau:


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42

xi x0 x1 ... xn

yi =f(xi) y0 y1 ... yn

y'i=f(xi) y'0 y'1 ... y'n

y'i=f(xi) y'0 y'1 ... y'n

7.2. a thc ni suy Lagrange

Gi s f(x) nhn gi tr yi ti cc im tng ng xi ( n,0i = ), khi a thc

ni suy Lagrange ca f(x) l a thc bc n v c xc nh theo cng thc sau:

=

=n

0i

inin )x(py)x(L

MS

)x(TS

)xx)...(xx)(xx)...(xx)(xx(

)xx)...(xx)(xx)...(xx)(xx()x(p

ni1ii1ii1i0i

n1i1i10in =

=

+

+

t W(x) = (x - x0)(x - x1)... (x - xn)

Suy ra: TS(x) =ix-x

W(x); )(xW'MS i=

Ln(x) = W(x) =

n

0i ii

i

)(xW')x-(x

y

V d 1. Cho hm f(x) tho mn:

xi 0 1 2 4

f(xi) 2 3 -1 0

Tm hm ni suy ca f(x), tnh f(5)

Gii:

Cch 1: W(x) = x (x - 1) (x - 2) (x - 4)

W(0) = (-1) (-2)(-4) = -8

W(1) = 1 (-1) (-3) = 3

W(2) = 2 (1) (-2) = -4

W(4) = 4 (3) (2) = 24

L3(x) = ))2x(4

1

)1x(3

3

)8(x

2)(4x)(2x)(1x(x

+

+


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43

= ))4x)(1x(x)4x)(2x(x4)4x)(2x)(1x((4

1++

= ))1x(x)2x(x4)2x)(1x()(4x(4

1++

= )2x6x4)(4x(

4

1 2

Cch 2:

L3(x) =)2)(1(2

)4x)(1x(x1

)3)(1(1

)4x)(2x(x3

)4)(2)(1(

)4x)(2x)(1x(2

+

= )2x6x4)(4x(4

1 2

7.3. a thc ni suy Lagrange vi cc mi cch u

Gi s hm f(x) nhn gi tr yi ti cc im tng ng xi ( n,0i = ) cch umt khong h.

th

xxt 0

= , khi :

x - x0 = h*t xi - x0 = h *i

x- x1 = h(t - 1) xi = x1 = h(i-1)

... ...

x - xi-1 = h(t- (i-1)) xi - xi-1 = h

x - xi+1 = h(t -(i+1)) xi - xi+1 = -h

... ...

x - xn = h(t - n) xi - xn = -h(n - i)

)in(*...*2*1*)1(1*...*)1i(i

)nt(*...*))1i(t)(1i(t(*...*)1t(t)htx(p

in0'n

+=+

=in)1)!*(in(!i*)it(

)nt(*...*)1t(t

Ln(x0 + ht) = t(t -1) ... (t - n)=

n

0i

ini

)!in(!i)it(

)1(y

Ln(x0 + ht) = =

n

0i

in

iin

it

cy.)1(

!n

)nt)...(1t(t

V d 2. Tm hm ni suy ca f(x) tho mn:


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44

xi 0 2 4

f(x0) 5 -2 1

Gii:

Cch 1:

W(x) = x (x - 2) (x - 4)

W(0) = (0 - 2) (0 - 4) = -8

W(2) = (2 - 0) (2 - 4) = -4

W(4) = (4 - 0) (4 - 2) = 8

L2(x) = )8).4x(

1

)4)(2x(

2

)0x(8

5)(4x)(2x(x

+

= ))4x(4

1

)2x(

2

x4

5()4x)(2x(x8

1

++

= ))2x(x)4x(x4)4x)(2x(5(8

1++

= )20x24x5(4

1)40x48x10(

8

1 22 +=+

Cch 2:

)2t

C.1

1t

C2

0t

C5

(!2

)2t)(1t(t

)t2(L

22

12

02

2 +

=

= )2t

1

1t

4

t

5(

2

)2t)(1t(t

+

+

= )1t(t)2t(t4)2t)(1t(5(2

1 2 ++

= 5t12t5)10t24t10(2

1 22 +=+

Vy 5x6x45)x(L 22 +=

7.4. Bng ni suy Ayken


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45

Khi tnh gi tr ca hm ti mt im x=c no bt k m khng cn phi

xc nh biu thc ca f(x). Khi ta c th p dng bng ni suy Ayken

nh sau

7.4.1. Xy dng bng ni suy Ayken

c-x0 x0-x1 x0-x2 x0-xn d1

x1-x0 c-x1 x1-x2 x1-xn d2

x2-x0 x2-x1 c-x2 x2-xn d3

xn-x0 xn-x1 xn-x2 c-xn dn

W(c) = (c- x0)( c- x1)( c- xn) : Tch cc phn t trn ng choW(xi) = (xi - x0)( xi x1) (xi - xi-1) (xi - xi+1) ... (xi - xn)

(c- xi) W(xi) = (xi - x0)( xi x1) (xi - xi-1) (c- xi)(xi - xi+1) ... (xi - xn)

di = (c-xi) W(xi) : Tch cc phn t trn dng i (i=0,1, ,n)

f(c) Ln(c) = W(c).=

n

0i ii

i

)(xW')xc(

y

f(c) W(c) =

n

0i i

i

d

y

V d 3. Tnh f (3. 5) khi bit f(x) tho mn

xi 1 2 3 4 5

yi 3 2 7 -1 0

Gii Xy dng bng ni suy Ayken

2.5 -1 -2 -3 -4 60

1 1.5 -1 -2 -3 -9

2 1 0.5 -1 -2 2

3 2 1 -0.5 -1 3

4 3 2 1 -1.5 -36

W(3.5) = 1.40625


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46

f(3.5) L4 (3.5) =3

1

2

7

9

2

20

1+

7.4.2. Thut ton

- Nhp: n, xi, yi (i = 0, n), c

- w = 1; s = 0;- Lp i = 0 n

{ w = w*(c - xi)

d = c - xi

Lp j = 0 n

Nu j != i th d = d * (xi - xj)

s = s + yi/d }

- Xut kt qu: w * s

7.5. Bng Ni suy Ayken (dng 2)

Xt hm ni suy ca 2 im: x0, x1

L01 =01

01

10

10 xx

xxy

xx

xxy

+

=01

0110

xx

)xx(y)xx(y

=

Hm ni suy ca hai im x0, xi

Xt hm p(x) c dng:

y0 x0-x

y1 x1-x

x1-x0

y0 x0-x

yi xi-xL0i(x) =

xi-x0

L01(x) x1-x

L0i(x) xi-xp(x) =

xi - x1


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47

L01(x0) (xi x0) - L0i(x0) (x1 x0) y0(xi - x1)p(x0) =

xi - x1=

xi - x1= y0

y1 (xi - x1)P(x1) =

xi - x1= y1

-y1 (x1 - xi)P(xi) = xi - x1= yi

Vy p(x) l hm ni suy ca 3 im x0, x1, xi

Tng qut: Hm ni suy ca n+1 im x0, x1,... xn

L012...n-2 n-1(x) xn-1-x

L012...n-2 n(x) xn-xL012...n(x) =

xn - xn-1

Bng Ni suy Ayken (dng 2)

xi yi Loi(x) Lo1i(x) Lo12i(x) ... Lo12...n(x) xi - x

x0 y0 x0 - x

x1 y1 Lo1(x) x1 - x

x2 y2 Lo2(x) Lo12(x) x2 - x

x3 y3 Lo3(x) Lo13(x) Lo123(x)

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

xn yn Lon(x) Lo1n(x) Lo12n(x) ... Lo12...n(x) xn - x

V d 4. Cho f(x) tho mn:

xi 1 2 3 4 5

yi

2 4 5 7 8

Tnh f (2.5)


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48

Gii: p dng bng Ayken (dng 2)

xi yi Loi(x) Lo1i(x) Lo12ix Lo123ix xi - x

1 2 -1.5

2 4 5 -0.5

3 5 4.25 4.625 0.5

4 7 4.5 4.875 4.5 1.5

5 8 4.25 4.875 4.562 4.407 2.5

Vy f(2.5) 4.407

Ch thch : L01(-2.5) = (2(-0.5) - 4(-1.5)) / (2-1) = 5

7.6. Ni suy Newton

7.6.1. Sai phn

Cho hm f(x) v h l hng s, khi :

f(x) = f (x + h) - f(x) c gI l sai phn cp 1 I vI bc h.

2f(x) = [f(x)] : sai phn cp 2

Tng qut: kf(x) = [k-1 f(x)] : sai phn cp k

Cch lp bng sai phn:

xi f(xi) f(xi) 2f(xi) 3f(xi) nf(xi)

x0 y0

x1 y1 f(x0)

x2 y2 f(x1) 2f(x0)

x3 y3 f(x2) 2f(x1) f3(x0)

.... .... ...

xn yn f(xn-1) nf(x0)


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50

V Ln(x) f(x) nn:

Ln(x0) f(x0) ; Ln(x0) f(x0) ;

2Ln(x0) 2f(x0) ; ; nLn(x0) nf(x0)

Vy :

!nh

)xx)...(xx)(xx()x(f...

!2h

)xx)(xx()x(f

h

xx)x(f)x(f)x(L

n1n10

0n

210

020

00n

++

+

+

V d 5. Xy dng hm ni suy Newton tho mn:

xi 1 2 3 4 5yi 2 4 5 7 8

Gii

Lp bng sai phn:

xi f(xi) f(xi) 2f(xi) 3f(xi) 4f(xi)

1 2

2 4 2

3 5 1 -1

4 7 2 1 2

5 8 1 -1 -2 -4

Hm ni suy Newton:

!4

)xx)(xx)(xx)(xx(4

!3

)xx)(xx)(xx(

2!2

)xx)(xx(

1

xx

22)x(L3210

210100

n

+

+


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51

7.7. Ni suy tng qut (Ni suy Hecmit)

Xy dng hm ni suy ca f(x) tho mn gi tr hm v gi tro hm cc

cp theo bng gi tr sau:

xi x0 x1 ... xn

yi =f(xi) y0 y1 ... yn

y'i=f(xi) y'0 y'1 ... y'n

yi'= f(xi) y''0 y1 ... yn

...

yi(k) =f(k)(xi)

y1(k) y2

(k) yn(k)

Gi s hm ni suy cn tm l a thc bc m: Hm(x)

m = n + =

k

1iis (Si : s gi thit c cho o hm cp i )

Hm(x) = Ln(x) + W(x) Hp(x)

( V Hm(xi) = Ln(xi) + W(xi) Hp(xi) = yi )

Vi: W(x) = (x-x0) * (x-x1)*....*(x-xn)

p= m - (n + 1)

o hm cp 1:

Hm(x) = Ln(x) + W(x) Hp(x) + W(x)Hp(x)

Xt ti cc im xi:

Hm(xi) = Ln(xi) + 2W(xi) Hp(xi) + W(xi)Hp(xi) = yi

=> Hp(xi)

o hm cp 2:

Hm(x) = Ln(x) + 2W(x) Hp(x) + W(x) Hp(x) + W(x)Hp(x)

0


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52

Xt ti cc im xi:

Hm(xi) = Ln(xi) + 2W(xi) Hp(xi) + W(xi) Hp(xi) + W(xi)Hp(xi) =yi

=> Hp(xi)

Tng t: o hm n cp k suy ra Hp(k-1)

(xi)

Ta xc nh hm Hp(x) tho mn:

xi x0 x1 ... xn

Hp(xi) h0 h1 ... hn

Hp(xi) h'0 h'1 ... h'n

...

Hp(k-1)(xi)

h0(k-1) h1

(k-1) ... hn(k-1)

V bn cht, bi ton tm hm Hp(x) hon ton ging bi ton tm hm

Hm(x). Tuy nhin y bc ca n gim i (n+1) v gi thit vo hm

gim i mt cp.

Tip tc gii tng t nh trn, cui cng a v bi ton tm hm nI suyLagrange (khng cn o hm). Sau thay ngc kt qu ta c hm ni

suy Hecmit cn tm Hm(x).

V d 6. Tm hm ni suy ca hm f(x) tho mn:

xi 0 1 3

f(xi) 4 2 0

f(xi) 5 -3

Gii: Hm ni suy cn tm l a thc H4(x)

H4(x) = L2(x) + W(x) H1(x)

0


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54

nhng cha xc nh c gi tr ca cc tham s a, b, c. xc nh

c cc tham s ny, ta tm cch tnh mt s cp gi tr tng ng (xi,

yi), i=1, 2, ,n bng thc nghim, sau p dng phng php bnh

phng b nht.

* Trng hp: y = ax + b

Gi i sai s ti cc im xi

i = yi - a - bxi

Khi tng bnh phng cc sai s: =

=n

1i

2iS

Mc ch ca phng php ny l xc nh a, b sao cho S l b nht. Nh

vy a, b l nghim h phng trnh:

0a

S=

0b

S=

Ta c: S = (yi2 + a2 + b2xi2 - 2ayi - 2bxiyi + 2abxi)

=

+= n

1iii )bx2y2a2(a

S

=

+= n

1iiii

2i )ax2yx2bx2(b

S

==

=+n

1ii

n

1ii yxbna

===

=+n

1i

ii

n

1i

2i

n

1i

i yxxbxa

Gii h phng trnh ta c: a, b

* Trng hp y = a + bx + cx2

Gi i sai s ti cc im xi

i = yi - a - bxi - cxi2

1

1


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55

Khi tng bnh phng cc sai s: =

=n

1i

2iS

Cc h s a, b xc nh sao cho S l b nht. Nh vy a, b, c l nghim

ca h phng trnh:

0a

S=

=== =++

n

1ii

n

1i

2

i

n

1ii yxcxbna

0a

S=

=====++

n

1iii

n

1i

3i

n

1i

2i

n

1ii yxxcxbxa

0c

S=

=====++

n

1ii

2i

n

1ii

n

1i

3i

n

1i

2i yx4xcxbxa

Gii h phng trnh ta c a, b, c

* Trng hp: y = aebx

Ly Logarit cs e hai v: Lny = lna + bx

t Y = lny; A = lna; B = b; X = x

Ta a v dng: Y = A + BX

Gii h phng trnh ta c A, B => a = eA, b=B

* Trng hp y = axb

Ly Logarit cs 10 hai v: Lgy = lga + blgx

t Y = lgy; A = lga; B = b; X = lgx


Gii h phng trnh ta c A, B => a = 10A, b=B

V d 7. Cho bit cc cp gi tr ca x v y theo bng sau:

xi 0.65 0.75 0.85 0.95 1.15

yi 0.96 1.06 1.17 1.29 1.58

Lp cng thc thc nghim ca y dng aebx


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56

Gii

Ta c: y = aebx

Ly Logarit cs e hai v: Lny = lna + bx

t Y = lny; A = lna; B = b; X = x


Xi = xi 0.65 0.75 0.85 0.95 1.15

Yi = lnyi -0.04 0.06 0.18 0.25 0.46

Xi Xi2 XiYi Yi

4.35 3.93 0.92 0.89

Phng php bnh phng b nht: A, B l nghim h phng trnh

==

=+n

1ii

n

1ii YXBnA

===

=+n

1iii

n

1i

2i

n

1ii YXXBXA

5A + 4.35B =0.89

4.35A + 3.93B = 0.92

Gii h phng trnh ta c: A = -.069, B = 1

Suy ra: a = eA = , b = B =1

Vy f(x) = xe21


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57

CHNG VIII TNH GN NG TCH PHN XC NH

8.1. Gii thiu

Xt hm s f(x) lin tc trn [a,b], nu xc nh c nguyn hm F(x) ta

c cng thc tnh tch phn:

=b

a

)a(F)b(Fdx)x(f

Nhng trong a s cc trng hp ta khng xc nh c nguyn hm ca,

hoc khng xc nh c biu thc ca f(x) m ch nhn c cc gi tr

ca n tI nhng im ri rc. Trong trng hp nh vy ta c th s dng

cc cng thc gn ng sau tnh tch phn:

- Cng thc hnh thang.

- Cng thc Parabol

- Cng thc Newton _Cotet

8.2. Cng thc hnh thang

Chia [a, b] thnh n on bng nhau vi khong cch h = (b - a)/n theo cc

im chia: x0=a, x1=a+h, ..., xn = b

=+++==

b

a

2x

x

x

x

x

ax 1

n

1n

1

0

Sdx)x(f...dx)x(fdx)x(fdx)x(f

S l din tch gii hn bi ng cong f(x), x=a, x=b, v trc x

Xt trn [x0, x1], ta xem ng cong f(x) l ng thng

Sf(x)

x0 =a

S1

Sn

x1 xn-1 xn = b


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58

)yy(h2

1SS 10hthang1 +=

Tng t:

)yy(h2

1

S 212 + ...

)yy(h2

1S n1nn +

Vy: +++++ b

an1n210 )yy2...y2y2y(2

hdx)x(f

8.3. Cng thc Parabol

Chia [a, b] thnh 2n on bng nhau vi khong cch h = (b - a)/2n theo cc

im chia: x0=a, x1=a+h, ..., x2n = b

+++=b

a

x

x

x

x

x

x

n2

2n2

4

2

2

0

dx)x(f...dx)x(fdx)x(fdx)x(f

Xt trn [x0, x2] xem ng cong f(x) l Parabol (ni suy bc 2 ca 3 im

x0, x1, x2)

)xx)(xx(

)xx)(xx(y

)xx)(xx(

)xx)(xx(y

)xx)(xx(

)xx)(xx(y)x(L)x(f

1202

102

2101

201

2010

2102

+

+

+

=

2

0

2

0

x

x

x

x2 dx)x(Ldx)x(f

Thay x0 = a, x1 = a + h , x2 = a+2h vo, ta c:

++2

0

x

x210 )yy4y(3

hdx)x(f

Tng t:


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59

++4

2

x

x432 )yy4y(3

hdx)x(f

++ n2

2n2

x

x21n22n2 )yy4y(3

hdx)x(f

Vy: ++++++ b

an21n22n2210 )yy4y2...y2y4y(3

hdx)x(f

V d. Tnh J = +5

12x1

dx theo 3 cch

Gii

Cch 1: 4/5arctgarctgxJ 51 == 0.588

Cch 2: chia [1, 5] thnh 4 on bng nhau (h=1) vi cc im chiaxi 1 2 3 4 5

yi 1/2 1/5 1/10 1/17 1/26

Cng thc hnh thang:

J (1/2 + 2/5 +2/10 +2/17 + 1/26) /2 0.628

Cch 3: Cng thc Parabol:

J (1/2 + 4/5 +2/10 +4/17 + 1/26) /3 0.591

8.4. Cng thc Newton-Cotet

Chia [a, b] thnh n on bng nhau vi khong cch h = (b - a)/n vi x0=a;

x1 = a + h , ...., xn = b.

t x = a + (b - a)t => dx = (b - a) dt

xi a a+h a + 2h ... b

ti 0 1/n 2/n ... 1

Khi :

=+=b

a

1

0

1

0

dt)t()ab(dt)t)ab(a(f)ab(dx)x(f

Vi (t)= f(a + (b - a)t

Xem (t) l hm ni suy Lagrange ca n + 1 im: t0, t1, ..., tn


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60

)n

1n1)...(

n

11)(01(

)

n

1nt)...(

n

1t)(0t(

y

...)1

n

1)...(

n

2

n

1)(0

n

1(

)1t)...(n

2t)(0t(

y)1)...(

n

2)(

n

1(

)1t)...(n

2t)(

n

1t(

y)t(L)t(

n

10n

+

+

+

=

Khi : 1

0

1

0n dt)t(Ldt)t(

t

+

+

=

1

0

in dt

)1n

i(...)

n

1i

n

i)(

n

1i

n

i(...)

n

1

n

i)(0

n

i(

)1t...()n

1it)(

n

1it(...)

n

1t)(0t(

P

Vy: =

b

a

n

0i

inipy)ab(dx)x(f

Xt n = 1 ( h = b-a )

=

=1

0

01 2

1dt

10

1tP ; =

=

1

0

11 2

1dt

01

0tP

+=+=b

a

1010 )yy(

2

h)

2

y

2

y)(ab(dx)x(f Cng thc hnh thang

Lu : Gi tr ca inP c th tra trong bng sau:

n inP

1 1/2 1/2

2 1/6 4/6 1/6

3 1/8 3/8 3/8 1/8

4 9/71 16/45 2/15 16/45 9/70

5 19/288 25/95 25/144 25/144 25/95 19/288


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61

BI TP

1. Khai bo (nh ngha) hm trong C tnh gn ng tch phn xc nhca f(x) tr n [a, b] (i kiu con tr hm)

a. Dng cng thc hnh thangb. Dng cng thc Parabolc. Dng cng thc Newton-cotet

2. Vit chng trnh tnh gn ng tch phn xc nh trn [a, b] ca 1 hm

f(x) c th (s dng cc hm khai bo trong cu 1). So snh kt qu,

nhn xt.


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62

MT S CHNG TRNH THAM KHO

1. Tnh gn ng tch phn xc nh

# include # include "conio.h"# include "math.h"# define PI 3.14159float d[10];int n;double g(double x){

return 1/(1+x*x);}double tp(double (*f)(double),float a,float b){

int n=100,i;float s,h=(b-a)/n;s=(f(a)+f(b))/2;for (i=1; i


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63

while (1) {printf("\n Nhap can de tinh tich phan: "); scanf("%f%f",&a,&b);

/*printf("a= "); scanf("%f",&a);printf("b= "); scanf("%f",&b);*/printf("\nS1=%.3f",tp(sin,0,PI));

printf("\nS2=%.3f",tp(cos,0,PI/2));printf("\nS3=%.3f",tp(g,a,b));nhap(d,&n);printf("\nS4=%.3f",tp(f,a,b));

printf("\n\n Ban tiep tuc ko(c/k)?");tt=getch();if (tt!='c') break;

}}

2. Tim nghiem gan dung cua phtrinh da thuc bac n bang PP chia doi

# include # include "conio.h"# include "math.h"# define eps 1e-3float f(float);

void nhap(float *, int );float d[10]; int n;

void main(){ float a,b,c; char tt;

while (1) {printf("\n Nhap bac phuong trinh: ");scanf("%d",&n);nhap(d,n);printf("\n Nhap khoang nghiem: "); scanf("%f%f",&a,&b);

/* printf("a= "); scanf("%f",&a);

printf("b= "); scanf("%f",&b);*/if (f(a)*f(b)= 1e-3 && f(c)!=0) {

printf("\n%.3f %.3f %.3f",a,b,f(c));if (f(b)*f(c)>0)b=c;else a=c;c=(a+b)/2;


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64

}printf("\n\n Nghiem phtrinh: %.3f",c);

}else

if (f(a)*f(b)>0) printf(" ( %f, %f) khong phai la khoangnghiem",a,b);

else if (f(a)==0) printf(" \n Nghiem phtrinh: %.3f",a);else printf(" \n Nghiem phtrinh: %.3f",b);

printf("\n\n Ban tiep tuc ko(c/k)?");tt=getch();if (tt!='c') break;}

}

void nhap(float *a, int n)

{ int i;printf("\n Nhap he so cua phuong trinh:\n");for (i=0;i


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65

printf("\n%.3f %.3f %f",a,-f(a)/fdh(a),b);*/do {b=a;a=b-f(b)/fdh(b);printf("\n%.3f %.3f %f",b,-f(b)/fdh(b),a);}

while (fabs(a-b) >= 1e-3 );printf("\nNghiem phtrinh: %.3f",a);printf("\nTiep tuc ko(c/k)?");tt=getch();if (tt=='k' || tt=='K') break;}

}float f(float x){

return exp(x)-10*x+7;}

float fdh(float x){return exp(x)-10;

}

4. Gii h phtrnh i s tuyn tnh bng PP Gauss

# include # include "conio.h"

# include "math.h"void nhap(float *a, int n,int m);void xuatmt(float *a, int n,int m);

main(){ float a[10][10];

float x[10],m,s;char tt;int n,i,j,k;while (1) {

printf("\n Nhap n= "); scanf("%d",&n);

printf("\n Nhap he so cua he phuong trinh:\n");for (i=1;i


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for (i=1;i


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67

for (i=1;i


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TI LI U THAM KHO

[1] ng Quc Lng, Phng php tnh trong k thut, Nh xut bn xydng H ni, 2001

[2] Phan Vn Hp, Gio trnh C s phng php tnh tp I,II. Trng HTng hp H ni, 1990

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