giai on cuoi ky 1 in
DESCRIPTION
sssadTRANSCRIPT
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N TP CUI K MN PHNG PHP TNHBi ging in t
TS. L Xun i
Trng i hc Bch Khoa TP HCMKhoa Khoa hc ng dng, b mn Ton ng dng
TP. HCM 2013.
TS. L Xun i (BK TPHCM) N TP CUI K MN PHNG PHP TNH TP. HCM 2013. 1 / 1
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Cho M = 3.3. Shift-STO-MCu 1. Cho phng trnh ex + 2x2 8M = 0 trong khong cch lynghim [1, 2]. Dng phng php Newton, chn x0 ti bin theo iu kinFourier, tnh nghim gn ng x2 v nh gi sai s x2Gii.Ta c f (1) < 0, f (2) > 0, f (x) = ex + 4x > 0,x [1, 2] vf (x) = ex + 4 > 0, x [1, 2] nn chn x0 = 2. Ta xy dng dy (xn)theo cng thc
xn = xn1 f (xn1)f (xn1)
= xn1 exn1 + 2x2n1 8M
ex + 4x
Ta c |f (x)| > min{|f (1)|, |f (2)|} = |f (1)| = m. Shift-STO-A. Do nghim gn ng xn c nh gi sai s so vi nghim chnh xc x nhsau
|x xn| 6 |f (xn)|m
=|exn + 2x2n 8M|
m= xn
TS. L Xun i (BK TPHCM) N TP CUI K MN PHNG PHP TNH TP. HCM 2013. 2 / 1
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n xn xn0 2
1 1.73428805
2 1.70410221 0.000651577
Bm my. Tnh xn
X eX + 2X 2 8M
eX + 4X
CALC x = 2 x1CALC Ans x2Sai s
abs(eX + 2X 2 8M)A
CALC Ans x2Kt qu. x2 1.7041; x2 0.0007
TS. L Xun i (BK TPHCM) N TP CUI K MN PHNG PHP TNH TP. HCM 2013. 3 / 1
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Cu 2. Cho h phng trnh
2.1Mx1 +0.25x2 +0.53x3 = 3.410.32x1 +3.2Mx2 0.13x3 = 4.720.24x1 0.35x2 +4.4Mx3 = 5.12
S dng phng php Jacobi vi x (0) = (0.3, 2.5,1.9)T , tm vc t lpx (3)
Gii.
x1 =1
2.1M (3.41 0.25x2 0.53x3)= 3.412.1M 0.252.1M x2 0.532.1M x3
x2 =1
3.2M (4.72 0.32x1 + 0.13x3)= 4.723.2M 0.323.2M x1 + 0.133.2M x3
x3 =1
4.4M (5.12 0.24x1 + 0.35x2)= 5.124.4M 0.244.4M x1 + 0.354.4M x2
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x1x2x3
= 3.412.1M4.72
3.2M5.124.4M
+ 0 0.252.1M 0.532.1M 0.323.2M 0 0.133.2M 0.244.4M 0.354.4M 0
x1x2x3
Khi cng thc lp c dng
X (m) = TjX(m1) + Cj , m = 1, 2, . . .
Chn X (0) =
0.32.51.9
tnh X (1),X (2),X (3)TS. L Xun i (BK TPHCM) N TP CUI K MN PHNG PHP TNH TP. HCM 2013. 5 / 1
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MatA =
0 0.25
(2.1M) 0.53(2.1M) 0.32(3.2M) 0 0.13(3.2M) 0.24(4.4M) 0.35(4.4M) 0
, MatB =
3.41(2.1M)4.72
(3.2M)5.12
(4.4M)
,MatC =
0.32.51.9
Bm my. Mode - 6 -Matrix.Dim - MatA - 3 3- ACShift 4 - Dim - MatB - 3 1 - ACShift 4 - Dim - MatC - 3 1 - ACShift 4 - MatB+MatA*MatC= x (1) - ACShift 4 - MatB+MatA*MatAns= x (2) - ACShift 4 - MatB+MatA*MatAns= x (3) - AC
TS. L Xun i (BK TPHCM) N TP CUI K MN PHNG PHP TNH TP. HCM 2013. 6 / 1
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m x(m)1 x
(m)2 x
(m)3
0 0.3 2.5 -1.9
1 0.5472 0.4144 0.4079
2 0.4459 0.4354 0.3535
3 0.4493 0.4378 0.3557
Kt qu. x(3)1 0.4493; x (3)2 0.4378; x (3)3 0.3557
Cu 3. Cho h phng trnh
2.5Mx1 +0.45x2 +0.54x3 = 4.410.32x1 +2.5Mx2 0.33x3 = 1.720.31x1 0.35x2 +4.1Mx3 = 3.12
S dng phng php Gauss-Seidel vi x (0) = [1.3, 0.5, 0.1]T , tm vctlp x (3)
Kt qu. x(3)1 ; x (3)2 ; x (3)3
TS. L Xun i (BK TPHCM) N TP CUI K MN PHNG PHP TNH TP. HCM 2013. 7 / 1
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Bm my. 1.3 Shift-STO-A, 0.5 Shift-STO-B, 0.1 Shift-STO-C, M = 3.3Shift-STO-M.4.41 0.45B 0.54C
2.5MShift-STO-A.
1.72 0.32A + 0.33C2.5M
Shift-STO-B.
3.12 0.31A + 0.35B4.1M
Shift-STO-C.
Thc hin lin tip thm 2 ln na c x (3).
Kt qu. x(3)1 = 0.5091; x
(3)2 = 0.1977; x
(3)3 = 0.2240
Cu 4. Cho bng sx 2.2 2.5 2.7 3.0
y 2.34 M 1.5M 2.1Dng a thc ni suy
Newton, xp x gi tr ca o hm ti x = 2.4Kt qu. y (2.4)
TS. L Xun i (BK TPHCM) N TP CUI K MN PHNG PHP TNH TP. HCM 2013. 8 / 1
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xk f (xk) T sai phn I T sai phn II T sai phn III
2.2 2.34M2.342.52.2
2.5 M 5M+67.86(2.72.2)1.5MM2.72.5
40M31.63(3.02.2)
2.7 1.5M 77.5M3.02.52.11.5M32.7
3.0 2.1
Nh vy cng thc ni suy Newton tin l
N (1)4 (x) = 2.34 +M 2.342.5 2.2 .(x 2.2) +
5M + 6 7.86 (2.7 2.2) (x 2.2)(x 2.5)+
+40M 3 1.63 (3.0 2.2) (x 2.2)(x 2.5)(x 2.7)
y M2.342.52.2 + 5M+67.86(2.72.2) (2x 4.7) + 40M31.63(3.02.2) (3x2 14.8x + 18.19).CALC M=3.3, X=2.4. Kt qu. y (2.4) 7.0600TS. L Xun i (BK TPHCM) N TP CUI K MN PHNG PHP TNH TP. HCM 2013. 9 / 1
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Cu 5. Cho bng sx 1.2 1.3 1.4 1.5 1.7
y M 2.5 2M 4.5 5.5S dng phng
php bnh phng b nht, tm hm dng f (x) = A + Bx + Cx2 xp x ttnht bng s trnKt qu. A = , B = , C =Bm my. 3.3 Shift STO M. Bm Mode 3 - STAT. Chn 3- +cx2. Nhpd liu ca 2 ct x , y . AC - Thot ra. Chn Shift 1 - chn 7 - Reg - chn1- A =. Chn Shift 1 - chn 7 - Reg - chn 2- B =. Chn Shift 1 - chn 7- Reg - chn 3- C =.Kt qu. A = 33.9839, B = 48.7636, C = 15.0442Cu 6. Cho tch phn I =
2.51.3
lnx + 2Mdx . Hy xp x tch phn I bng
cng thc hnh thang m rng vi n = 8Kt qu. I Gii.
h =b an
=2.5 1.3
8=
3
20, x0 = 1.3, xk = 1.3 + k .
3
20,
yk = lnxk + 2M = ln
1.3 + k .
3
20+ 2M
TS. L Xun i (BK TPHCM) N TP CUI K MN PHNG PHP TNH TP. HCM 2013. 10 / 1
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Vy I h
2
7k=0
(yk + yk+1) =
3
40
7k=0
(ln
1.3 + k .
3
20+ 2M + ln
1.3 + (k + 1).
3
20+ 2M
)=
Kt qu. I 1.2835Cu 7. Cho bng s
x 1.0 1.2 1.4 1.6 1.8 2.0 2.2
f (x) 2 3.3M 2.4 4.3 5.1 6.2 M.S
dng cng thc Simpson m rng tnh tch phn I =2.21.0
[Mxf 2(x) + x2]dx
Kt qu. I Gii.
h =b an
=2.2 1.0
n= 0.4 n = 3, x0 = 1.0, xk = 1.0 + 0.4k ,
x k =xk1 + xk
2= 1.0 + (2k 1) 0.2, yk = Mxk f 2(xk) + x2k ,
y k = Mxk f
2(x k) + x2k
TS. L Xun i (BK TPHCM) N TP CUI K MN PHNG PHP TNH TP. HCM 2013. 11 / 1
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I h
6
2k=0
(yk + 4yk+1 + yk+1) =
0.4
6(
2k=0
(yk) + 42
k=0
(y k+1) +2
k=0
(yk+1)) =
0.4
6((y0 + y1 + y2) + 4(y
1 + y
2 + y
3) + (y1 + y2 + y3)) =
0.4
6((y0 + 2y1 + 2y2 + y3) + 4(y
1 + y
2 + y
3))
Bm my.M 1.0 22 + 1.02 shift-STO-AM 1.4 2.42 + 1.42 shift-STO-BM 1.8 5.12 + 1.82 shift-STO-CM 2.2 M2 + 2.22 shift-STO-D0.4
6 (A + 2B + 2C + D) shift-STO-X
M 1.2 (3.3M)2 + 1.22 shift-STO-AM 1.6 4.32 + 1.62 shift-STO-BM 2.0 6.22 + 2.02 shift-STO-C4 0.4
6 (A + B + C ) shift-STO-Y
Kt qu: X + YKt qu. I 252.4364TS. L Xun i (BK TPHCM) N TP CUI K MN PHNG PHP TNH TP. HCM 2013. 12 / 1
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Cu 8. Xt bi ton Cauchy{y (x) = My2 + x .y(x) + x2, x > 1.1,y(1.1) = 1.6
S dng cng thc Runge-Kutta cp 4, hy xp x gi tr ca hm tix = 1.3 vi bc h = 0.2.Kt qu. y(1.3) Gii.Vi h = 0.2, x1 = x0 + 0.2 = 1.3, y0 = 1.6. Ta c
K 01 = hf (x0, y0) = 0.2(My2 + x .y(x) + x2), K 02 = hf (x0 +
h2 , y0 +
K012 ),
K 03 = hf (x0 +h2 , y0 +
K022 ), K
04 = hf (x0 + h, y0 + K
03 ).
Cng thc tnh nghim gn ng l
y(1.3) y1 = y0 + 16
(K 01 + 2K02 + 2K
03 + K
04 ) =
TS. L Xun i (BK TPHCM) N TP CUI K MN PHNG PHP TNH TP. HCM 2013. 13 / 1
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Bm my. 3.3 Shift-STO-M. 0.2(My2 + x .y + x2).Tnh K 01 . CALC X = 1.1,Y = 1.6. K 01 Shift-STO-ATnh K 02 . CALC X = 1.1 +
0.2
2,Y = 1.6 +
A
2. K 02 Shift-STO-B
Tnh K 03 . CALC X = 1.1 +0.2
2,Y = 1.6 +
B
2. K 03 Shift-STO-C
Tnh K 04 . CALC X = 1.1 + 0.2,Y = 1.6 + C . K 04 Shift-STO-D
y(1.3) y1 = y0 + 16
(K 01 + 2K02 + 2K
03 + K
04 ) =
= 1.6 +1
6(A + 2B + 2C + D)
Kt qu. y(1.2) 40.3086
TS. L Xun i (BK TPHCM) N TP CUI K MN PHNG PHP TNH TP. HCM 2013. 14 / 1
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Cu 9. Cho hm s f (x) = ex . ln(x4 + 1)Mx . S dng sai phn hngtm, xp x gi tr ca f (0.7) v f (0.7) vi bc h = 0.15Kt qu. f (0.7) ; f (0.7) Gii. Cng thc sai phn hng tm
f (x0) f (x0 + h) f (x0 h)
2h f (0.7) f (0.7 + 0.15) f (0.7 0.15)
2 0.15
f (x0) f (x0 + h) 2f (x0) + f (x0 h)
h2
f (0.7) f (0.7 + 0.15) 2f (0.7) + f (0.7 0.15)0.152
Bm my. eX . ln(X 4 + 1)MXCALC X=0.7 + 0.15, M = 3.3 = shift-STO-ACALC X=0.7, M = 3.3 = shift-STO-BCALC X=0.7 0.15, M = 3.3 = shift-STO-Cf (0.7)
A C2 0.15 , f
(0.7) A 2B + C
0.152
Kt qu. f (0.7) 0.5301 ; f (0.7) 11.9020TS. L Xun i (BK TPHCM) N TP CUI K MN PHNG PHP TNH TP. HCM 2013. 15 / 1
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Cu 10. Xt bi ton bin{(x + M)y (x) + x3y (x) 15My(x) = x(x + 1), 0 6 x 6 1
y(0) = 0.5M, y(1) = 1.2
S dng phng php sai phn hu hn, xp x gi tr ca hm trong[0, 1] vi bc h = 0.25Kt qu. y(0.25) = , y(0.50) = , y(0.75) =Gii.x0 = 0, x1 = 0.25, x2 = 0.5, x3 = 0.75, x4 = 1.p(x) = x + M, q(x) = x3, r(x) = 15M, f (x) = x(x + 1);p1 = x1 + M, p2 = x2 + M, p3 = x3 + M; q1 = x
31 , q2 = x
32 , q3 = x
33 ;
r1 = r2 = r3 = 15M;f1 = x1(x1 + 1), f2 = x2(x2 + 1), f3 = x3(x3 + 1)
y0 = 0.5M, y4 = 1.2
(p1h2 q12h )y0 + (r1 2p1h2 )y1 + (p1h2 + q12h )y2 = f1
(p2h2 q22h )y1 + (r2 2p2h2 )y2 + (p2h2 + q22h )y3 = f2
(p3h2 q32h )y2 + (r3 2p3h2 )y3 + (p3h2 + q32h )y4 = f3
TS. L Xun i (BK TPHCM) N TP CUI K MN PHNG PHP TNH TP. HCM 2013. 16 / 1
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y0 = 0.5M, y4 = 1.2
(r1 2p1h2 )y1 + (p1h2 + q12h )y2 + 0y3 = f1 (p1h2 q12h )y0(p2h2 q22h )y1 + (r2 2p2h2 )y2 + (p2h2 + q22h )y3 = f2
0y1 + (p3h2 q32h )y2 + (r3 2p3h2 )y3 = f3 (p3h2 + q32h )y4
Bm my. Mode-5 - EQN.
r1 2p1h2
= 15M 2.(0.25 + M)(0.25)2
.
p1h2
+q12h
=0.25 + M
0.252+
(0.25)3
2(0.25)0
f1 (p1h2 q1
2h)y0 = 0.25(0.25 + 1) (0.25 + M
0.252 (0.25)
3
2(0.25))0.5M
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p2h2 q2
2h=
0.5 + M
0.252 (0.5)
3
2(0.25)
r2 2p2h2
= 15M 2.(0.5 + M)(0.25)2
.
p2h2
+q22h
=0.5 + M
0.252+
(0.5)3
2(0.25)f2 = 16x22 = 0.5(0.5 + 1)
0p3h2 q3
2h=
0.75 + M
0.252 (0.75)
3
2(0.25)
r3 2p3h2
= 15M 2(0.75 + M)0.252
f3 (p3h2
+q32h
)y4 = 0.75(0.75 + 1) (0.75 + M0.252
+(0.75)3
2(0.25))1.2
Kt qu. y(0.25) = 0.7473, y(0.50) = 0.4909, y(0.75) = 0.6225
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