Me1bb99t Se1bb91 Hc6b0e1bb9bng De1baabn e1baa5n Mc3a1y Tc3adnh Ce1baa7m Tay

Bi ging Phng php s dnh cho Khoa C-H Thy Li-V Mnh Ti 2012-2013

MT S HNG DN N MY TNH CM TAY TNH NHANH CC KT QU S:

I. Hng dn n my tnh 500 MS, 570 MS, 500 ES, 570 ES hoc cc my khc tng ng, gii gn ng nghim ca phng trnh phi tuyn bng cng thc lp:

() = Vi

() = ( + ) Ta gi s vi cng thc lp c thit lp l

{ =

+ = () = ( + )

+) Bc 1: Nhp gi tr ban u: Nhp gi tr v lu vo mt b nh A bng cch n cc phm 2 Shift Sto A.

+) Bc 2: Nhp biu thc lp: Thao tc trn biu thc

() = ( + )

u c th ta n Alpha A. Sau khi nhp xong th lu vo chnh A (Shift Sto A) ta c gi tr 1 = 2.63905733 +) Bc 3: Lp: n phm = lin tip th ta c cc kt qu 2; 3; v cho bi bng sau.

n

0 2

1 2.63905733

2 2.83113021

3 2.891221301

4 2.910128666

5 2.916085467

6 2.917962846

7 2.918554596

8 2.918741122

9 2.918799918

10 2.918818451

+) Ch :

- Mi ln n = th nh ghi kt qu lun.

- Nu mun tnh gi tr ca () = (

2 + 10) th ta lm nh sau:


Nhp biu thc (): u c ta n Alpha X sau n CALC my hi X? Ta nhp gi tr cn tnh vo, v d 10 = 2.918818451 ta nhp 2.918818451 ri n = ta c kt qu (10) = 1.081925472 10

4.

II. Hng dn n my tnh 500MS; 500ES; 570MS; 570ES, gii h phng trnh tuyn tnh bng phng php lp Gauss-Seidel.

V d vi h 3 phng trnh 3 n sau:

{71 22 + 33 = 321 + 92 + 3 = 4

1 + 42 + 83 = 4

Dy lp Gauss-Seidel vi gi tr ban u (0) (0) (0) (0)0 1 2 3( ; ; ) (0;0;0)x x x x

( 1) ( ) ( )

1 2 3

( 1) ( 1) ( )

2 1 3

( 1) ( 1) ( 1)

3 1 2

2 3 3 (1)

7 7 7

2 1 4 (2)

9 9 9

1 1 1 (3)

8 2 2

k k k

k k k

k k k

x x x

x x x

x x x

+)Bc 1: Nhp v lu gi tr ban u: (0) 0 0 0

0 1 2 3( ; ; ) (0;0;0)x x x x

0 Shift Sto A

0 Shift Sto B

0 Shift Sto C

+) Bc 2: Nhp cc biu thc ca dy lp:

Nhp biu thc 1 :

0 x Alpha A + 2/7 x Alpha B - 3/7 X Alpha C + 3/7

Shift Sto A

Nhp biu thc 2:

(-) 2/9 X Alpha A + 0 x Alpha B - 1/9 x Alpha C + 4/9

Shift Sto B

(1)

2 0.349206x

Nhp biu thc 3

1/8 x Alpha A - 1/2 x Alpha B + 0 X Alpha C +

(1)

1 0.428571x


Shift Sto C

(1)

3 0.378968x

+) Bc 3:

Ri n phm = (2)1 0.365929x

Ri n phm = (2)2 0.321019x

Ri n phm = (2)

3 0.385232x

(*)Tip tc nh bc 3 ta c 2x , V ta c bng kt qu

k ( )

1

kx ( )2kx ( )3

kx

0

1

2

3

4

5

0

0.428571

0.365929

0.355192

0.356612

0.356500

0

0.349206

0.321019

0.322709

0.322636

0.322638

0

0.378968

0.385232

0.383044

0.383258

0.383243

Nghim (5) = (0.356500; 0.322638; 0.383243)

Ch : +) c kt qu no th phi ghi lun vo bng kt qu.

+)Nu h n phng trnh n n th ta lm tng t nh trn nhng phi dng mt b n bin nh, v phi nhp n biu thc ng thi bc 3 phi n n-1 ln quay tr li biu thc ban u

+)Lp n th phc tp hn v phi dng n 2 b nh.

Hng dn n my tnh 500MS; 500ES; 570MS; 570ES, gii h phng trnh tuyn tnh bng phng php lp n.

V d vi h 3 phng trnh 3 n:

{71 22 + 33 = 321 + 92 + 3 = 4

1 + 42 + 83 = 4

n 2

ln

n 2

ln

n 2

ln

n n-1

ln


Dy lp n vi gi tr ban u (0) 0 0 00 1 2 3( ; ; ) (0;0;0)x x x x

( 1) ( ) ( )

1 2 3

( 1) ( ) ( )

2 1 3

( 1) ( ) ( )

3 1 2

2 3 3 (1)

7 7 7

2 1 4 (2)

9 9 9

1 1 1 (3)

8 2 2

k k k

k k k

k k k

x x x

x x x

x x x

+)Bc 1: Nhp v lu gi tr ban u: (0) 0 0 0

0 1 2 3( ; ; ) (0;0;0)x x x x

0 Shift Sto A

0 Shift Sto B

0 Shift Sto C

+) Bc 2: Nhp cc biu thc ca dy lp:

Nhp biu thc th nht :

0 x Alpha A + 2/7 x Alpha B - 3/7 X Alpha C + 3/7

Shift Sto D

Nhp biu thc th 2:

(-) 2/9 x Alpha A + 0 x Alpha B - 1/9 x Alpha C + 4/9

Shift Sto X

(1)

2 0.444444x

Nhp biu thc th 3

1/8 x Alpha A - 1/2 x Alpha B + 0 x Alpha C + 1/2

Shift Sto Y

(1)

3 0.500000x

Bc tip theo ta lm li bc 2 nhng phi dng trn b nh D, X, Y.

Nhp biu thc th nht :

0 x Alpha D + 2/7 x Alpha X - 3/7 X Alpha Y + 3/7

Shift Sto A

Nhp biu thc th 2:

(1)

1 0.428571x

(2)

1 0.341270x


(-) 2/9 x Alpha D + 0 x Alpha X - 1/9 x Alpha Y + 4/9

Shift Sto B

(2)

2 0.293651x

Nhp biu thc th 3

1/8 x Alpha D - 1/2 x Alpha X + 0 x Alpha Y + 1/2

Shift Sto C

(2)

3 0.331349x

+) i vi my 500 ES hoc 570 ES, tnh tip () ta lm nh sau: n bn phm REPLAY ngc ln 5 ln, sau n phim

REPLAY sang tri 1 ln v n DEL ri n = th ta c 1(3) =0.370465

Tng t li N bn phm REPLAY ngc ln 5 ln, sau n phim REPLAY sang tri 1 ln v n DEL ri n = th ta c

2(3) =0.331790

C th ta c bng sau:

K ( )

1

kx ( )2kx ( )3

kx

0

1

2

3

4

5

0

0.428571

0.341270

0.370465

0.353725

0.356434

0

0.444444

0.293651

0.331790

0.318137

0.323571

0

0.500000

0.331349

0.395833

0.380413

0.385147

6 0.355957 0.322443 0.382768

7 0.356654 0.322813 0.383088

Nghim () = (0.356654; 0.322813; 0.383088)

Ch : MY ES th c b s trn my nn c lm phng php lp n nh hng dn trn. Cn my MS khng b s nn tnh vi k=3 th ta li quay li bc 2.

III. Gii gn ng phng trnh vi phn thng trn my tnh in t f(x) 570 ES. Bi ton Cauchy

{ = 2 + 2 0 < 1

(0) = 0

Vi = 0.1 a. Cng thc Euler:

{+1 = (, ) +

0 = (0) = 0

Tc l


{+1 = 0.1(

2 + 2) +

0 = 1

+) Bc 1: Khai bo cng thc +1 = 0.1(2 +

2) + : u c ta n Alpha X; u c ta n Alpha Y. (Trong qui trnh ny ta dng nh X cha gi tr v nh Y cha gi tr ca .) +) Bc 2: Tnh ton ln 1: n CALC, my hi: X?

Khai bo 0 = 0: Bm phm 0 = My hi tip: Y?

Khai bo: 0 = 1: Bm 1 = s cho kt qu 1 = 1.1 a kt qu vo b nh Y: Shift Sto Y

Sau tr v cng thc nhp: Bm phm REPLAY ngc ln mt ln.

+) Bc 3: Qui trnh: n CALC my hi: X?

Khai bo 1 = 0.1: Bm phm 0.1 = My hi tip: Y?

Bm = (1 = 1.1 c sn trong nh Y nn khng cn khai bo li) s cho kt qu 2 = 1.222 a kt qu vo b nh Y: Shift Sto Y

Sau tr v cng thc nhp: Bm phm REPLAY ngc ln mt ln. +) Bc 4: Lp li Qui trnh vi thay i duy nht l khi my hi X? th ta khai bo cc gi tr tip theo: 0.2; 0.3; 0.4; . ; 1. Ta c bng gi tr sau

n

0 0 1

1 0.1 1.1

2 0.2 1.222

3 0.3 1.3803284

4 0.4 1.586859049

5 0.5 1.863671213

6 0.6 2.246998253

7 0.7 2.800898367

8 0.8 3.649401534

9 0.9 5.062214689

10 1.0 7.724816445


b. Cng thc Euler ci tin ta cng lm tng t nhng ta dng them b nh th 3. Cng thc Euler ci tin

{

+1 = + (, )

+1 = +

2((, ) + (+1, +1))

0 = (0) = 1

Tc l ta c c

+1 = 0.05 (2 +

2 + +12 + ( + 0.1(

2 + 2))

2) + ()

+) Bc 1: Khai bo cng thc (*): u c th ta n Alpha X; au c ta n Alpha Y; u c +1 ta n Alpha A. (Tc l ta dng nh X lu ; nh Y lu v nh A lu +1). +) Bc 2: Tnh ton ln 1: Bm CALC, my hi: X?

Khai bo 0 = 0: Bm phm 0 = My hi tip: Y?

Khai bo: 0 = 1: Bm 1 = My hi tip: A?

Khai bo 1 = 0.1: Bm 0.1 = s cho kt qu 1 = 1.1 a kt qu vo b nh Y: Shift Sto Y

Sau tr v cng thc nhp: Bm phm REPLAY ngc ln mt ln.

+) Bc 3: Qui trnh: n CALC my hi: X?

Khai bo 1 = 0.1: Bm phm 0.1 = My hi tip: Y?

Bm = (1 = 1.1 c sn trong nh Y nn khng cn khai bo li) My hi tip: A? Khai bo 2 = 0.2: Bm 0.2 = s cho kt qu 2 = 1.222 a kt qu vo b nh Y: Shift Sto Y

Sau tr v cng thc nhp: Bm phm REPLAY ngc ln mt ln. +) Bc 4: Lp li Qui trnh vi thay i duy nht l khi my hi X? A? th ta khai bo cc gi tr tip theo: 0.2 (0.3); 0.3 (0.4); 0.4 (0.5); . ; 0.9 (1.0) Ta c bng gi tr sau

n

0 0 1

1 0.1 1.051

2 0.2 1.109467678

3 0.3 1.178909589

4 0.4 1.26373277

5 0.5 1.369718589


6 0.6 1.504725574

7 0.7 1.679805854

8 0.8 1.911109639

9 0.9 2.22338971

10 1.0 2.657022002

IV. Tnh gi tr ca hm trn mt on [; ] vi cc bc chia u =

nhm tnh

gn ng tch phn

= ()

Thao tc trn my tnh c chc nng Table, v d my 500ES hoc 570ES.

+) Bc 1: Vo ch Table: Mode 7 (7. TABLE) mn hnh hin

f(X)=

+) Bc 2: Nhp biu thc (), v d () = 3 5

2+6 trn [; ] = [1; 3] vi = 0.2.

Ta nhp biu thc bng cch u c ta n Alpha X. +) n = min hnh hin Start? Ta nhp gi tr = 1 n 1 = my hi End? Ta nhp = 3 n 3 =, my hi Step? Ta nhp = 0.2: n 0.2 = my s cho ra bng kt qu vi 11 gi tr ti cc im = 1; 1.2; ; 3.0 nh sau

i ()

1 1 2.2857

2 1.2 3.0651

3 1.4 4.0273

4 1.6 5.2154

5 1.8 6.6835

6 2.0 8.5

7 2.2 10.75

8 2.4 13.541

9 2.6 17.006


10 2.8 21.312

11 3.0 26.666

y hn ch ch l cho t ch s sau du phy. Ta c th tnh gi tr bng cch nhp

biu thc v dng phm CALC tnh gi tr th s cho kq vi 10 ch s.

+) Bc 1: Nhp biu thc () = 3 5

2+6 bng cch u c ta n Alpha X.

+) Bc 2: n CALC my hi X? ta nhp 0 = 1 sau n = ta c (1) = 2.285714286 +) n phm REPLAY ngc ln mt ln nhn thy biu thc ta va nhp, sau li thao

tc li bc 2 v khai bo 1 = 1.2 ta c (1.2) = 3.065149808

Ngoi ra ta c th tham kho cc phn mm gii gn ng khc nh Maple, Mathematica, Excel.

Thao tc trn Maple 13

Gii gn ng phng trnh sau bng phng php Newton trn Maple

3 + 42 15 = 0

>


Bng 1. Nghim ca phng trnh () = + = qua pp Newton v gi tr ca () tng ng.


NI DUNG N THI KIM TRA GIA K

Cu 1 (3,5 im)

+ Sai s tuyt i, sai s tng i, quy trn s v sai s mc phi

+ Tm ch s chc, bi ton ngc ca sai s.

+ Tnh gn ng mt biu thc vi sai s cho trc.

Cu 2 (3,5 im) v Cu 3 (3,0 im) Bao gm ton b chng 2: Gii gn ng mt phng trnh phi

tuyn:

+ Phng php chia i gii gn ng nghim, nh gi sai s mc phi v ngc li i tm gn ng

nghim khi bit sai s.

+ Phng php lp n gii gn ng nghim, nh gi sai s mc phi v ngc li i tm gn ng

nghim khi bit sai s.

+ Phng php Newton (tip tuyn) gii gn ng nghim, nh gi sai s mc phi v ngc li i tm

gn ng nghim khi bit sai s.

+ Phng php dy cung gii gn ng nghim, nh gi sai s mc phi v ngc li i tm gn ng

nghim khi bit sai s

NI DUNG V CU TRC THI

KT THC MN HC

Hnh thc thi: T lun

Thi gian: 60 pht

Cu 1 (3.5 im)

Gii gn ng phng trnh

Gii gn ng h phng trnh i s tuyn tnh

T chn mt v d gii s gn ng mt phng trnh siu vit hoc a thc bng t nht hai phng php khc nhau v so snh tc hi t nghim.

Cu 2 (3 im)

Tnh gn ng o hm

Tnh gn ng tch phn xc nh

T chn v d tnh gn ng tch phn theo 2 phng php v so snh kt qu.

Cu 3 (3.5 im)

Gii gn ng phng trnh vi phn

Gii gn ng phng trnh o hm ring

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Me1bb99t Se1bb91 Hc6b0e1bb9bng De1baabn e1baa5n Mc3a1y Tc3adnh Ce1baa7m Tay