APPLIED NUMERICAL METHODS USING MATLAB phần 2 potx

... r (1) 1 →     a (2) 11 a (2) 12 a (2) 13 b (2) 1 a (2) 21 a (2) 22 a (2) 23 b (2) 2 a (2) 31 a (2) 32 a (2) 33 b (2) 3     =    2 −1 −10 01 12 03 /2 −1 /21    : r (2) 1 : r (2) 2 : r (2) 3 (2. 2.10b) Here, ... (2. 2.1b)×a (1) m2 /a (1) 22 (m = 3) from it to get a (0) 11 x 1 + a (0) 12 x 2 + a (0) 13 x 3 = b (0) 1 (2. 2.3a) a (1...

Ngày tải lên: 09/08/2014, 12:22

... yields u 11 = √ a 11 ,u 12 = a 12 /u 11 ,u 13 = a 13 /u 11 ,u 14 = a 14 /u 11 (P2.7 .2. 1) u 22 =  a 22 − u 2 12 ,u 23 = (a 23 − u 13 u 12 )/u 22 ,u 24 = (a 24 − u 14 u 12 )/u 22 (P2.7 .2. 2) u 33 =  a 33 − u 2 23 − ... u 22 u 23 u 24 00u 33 u 34 000u 44     =     u 2 11 u 11 u 12 u 11 u 13 u 11 u 14 u 12 u 11 u 2 12 + u 2 22 u 12 u 1...

Ngày tải lên: 09/08/2014, 12:22

... = solve(’x1 ^2 + 4*x2 ^2- 5=0’, 2* x 1 ^2 - 2* x1 - 3*x2 -2. 5 = 0’) x1 = [ 2. ] x2 = [ 0.500000] [ -1 .20 6459] [ 0.941336] [0.60 322 9 -0.3 926 30*i] [-1.095668 -0.540415e-1*i] [0.60 322 9 +0.3 926 30*i] [-1.095668 ... 0 f 2 (x 1 ,x 2 ) = 2x 2 1 − 2x 1 − 3x 2 − 2. 5 = 0 (4.6.6) SECANT METHOD 189 2 1 0 −1 2 1.8 2 2 .2 (a) f 42 ( x ) = tan (p − x ) − x 10 0 20 −10 5...

Ngày tải lên: 09/08/2014, 12:22

... (5.9.3c), t 2 1 + (−t 1 ) 2 = 2 3 ,t 1 =−t 2 =− 1 √ 3 PROBLEMS 20 7 for five different sets of data rates a = [ 32 32 32 32] , [64 323 2 32] , [ 128 32 32 32] , [25 6 32 32 32] , and [5 12 32 32 32] and plots ... 1.0 022 e-13 (1.0437e-13) Flops 8055 61 02 (e) x 2 + y 2 + z 2 = 14 x 2 + 2y 2 − z = 6 x − 3y 2 + z 2 = 2 (P4.8.5) (f) x 3 − 12y + z 2 = 5 3x 2...

Ngày tải lên: 09/08/2014, 12:22

... quadl Gauss (P5.11.1 ,2) 0.001 4. 621 2e -2 2.9 822 e -2 8.4103e -2 0.0001 9. 427 8e-3 9. 427 7e-3 0.00001 2. 1853e-1 2. 9858e-3 8.4937e -2 (P5. 12. 1 ,2) 0.001 1 .23 93e-5 1.3545e-5 0.0001 8.3 626 e-3 5.0315e-6 6.4849e-6 0.00001 ... quad8) (P5.8.1) |error| 1.3771e -2 0 4.9967e-7 flops 5 024 7757 131 28 369 75 822 29 4 ORDINARY DIFFERENTIAL EQUATIONS 0.5 0 −0.5 2 1 0 −1 2 −1 0 1...

Ngày tải lên: 09/08/2014, 12:22

... OPTIMIZATION f (x) = x 1 + x 2 = 2 f (x) = x 1 + x 2 = 2 0 0 2 2 h (x) = x 1 + x 2 2 = 0 22 2 2 f (x) = x 1 + x 2 h (x) = x 1 + x 2 2 = 0 5 0 0 0 2 2 2 2 22 −5 Figure 7.11 The ... 7 .2. ∂ ∂x 1 l(x,λ) (7 .2. 3a) = 1 + 2 x 1 = 0,x 1 =−1 /2 (E7 .2. 4a) ∂ ∂x 2 l(x,λ) (7 .2. 3a) = 1 + 2 x 2 = 0,x 2 =−1 /2 (E7 .2. 4b) ∂ ∂λ l(x,λ) (7 .2. 3b) = x 2...

Ngày tải lên: 09/08/2014, 12:22

... 2. 5)*sin(x (2) - 2. 9);%(P7 .2. 2) d2f(1) = 12* x(1) ^2 - 24 + 20 *cos(x(1) - 2. 5)*cos(x (2) - 2. 9); %(P7 .2. 3) d2f (2) = 12* x (2) ^2 - 32 + 20 *cos(x(1) - 2. 5)*cos(x (2) - 2. 9); %(P7 .2. 3) ∂ 2 f/∂x 2 1 = 12x 2 1 − 24 ... [−1.6 926 −0.1183] (2) [2. 5463 −0.1896] (7) [ 2. 6573 2. 821 9] +, + m (3) [2. 520 9 2. 9 027 ] +, + G (8) [−0. 322 7 2. 425 7] (4) [−0.3865 2...

Ngày tải lên: 09/08/2014, 12:22

... conditions u(0,y) = 2e y ,u(1,y)= 2e 2x+y , u(x, 0) = 2e 2x ,u(x ,2) = 2e 2x +2 (P9.1.10) Divide the solution region into M x × M y = 20 × 40 sections. (e) ∂ 2 u(x, y) ∂x 2 + ∂ 2 u(x, y) ∂y 2 = 0for0≤ x ... 0 007/30 7 /24 25 / 42/ 35/6      , p(:, :, 2) =      −1 /20 0 −5/8 1 /21 5/16 0 0 05/16 1 /20 00−1 /2 −5 /24 0 −5/40 5/6      , (9.4.17) p(:, :, 3) =    ...

Ngày tải lên: 09/08/2014, 12:22

... 3matrix,the principal minor matrices are a 11 ,a 22 ,a 33 ,  a 11 a 12 a 21 a 22  ,  a 22 a 23 a 32 a 33  ,  a 11 a 13 a 31 a 33  ,   a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33   among which the leading ... j sin θ (F.40) a 2 = b 2 + c 2 − 2bc cos A(F.39a) sin θ = 1 j2 (e jθ − e −jθ ) (F.41a) b 2 = c 2 + a 2 − 2ca cos B(F.39b) cos θ = 1 2 (e j...

Ngày tải lên: 09/08/2014, 12:22