Surface creation Interpolation methods using Avenue

... with mathematics if the right tools for understanding mathematics are provided. A visual approach based on Windows in Computing for Numerical Methods Using Visual C++ is our step in achieving this ... Simulation 1.1 Numerical Approximations 1.2 C++ for Numerical Modeling 1.3 Mathematical Modeling 1.4 Simulation and Its Visualization 1.5 Numerical Methods 1.6 Num...

Ngày tải lên: 10/12/2013, 14:15

... alternatives to the 341 Independent Component Analysis. Aapo Hyv ă arinen, Juha Karhunen, Erkki Oja Copyright  2001 John Wiley & Sons, Inc. ISBNs: 0-4 7 1-4 0540-X (Hardback); 0-4 7 1-2 213 1-7 (Electronic) ... basic ICA model in Part II. 18 Methods using Time Structure The model of independent component analysis (ICA) that we have considered so far consist...

Ngày tải lên: 20/01/2014, 11:20

... the Lexicon,. Cam- bridge University Press, Cambridge, OK. 270 WORD-SENSE DISAMBIGUATION USING STATISTICAL METHODS Peter F. Brown, Stephen A. Della Pietra, Vincent J. Della Pietra, and ... pages 52-6,t, Wa- terloo, Canada. [Lesk, 1986] Lesk, M. E. (1986). Auto- mated sense disambiguation using machine- readable dictionaries: How to tell a pine cone from an ice cream con...

Ngày tải lên: 08/03/2014, 07:20

... important for the user to know the underlying concepts in solving the problem. Computing for Numerical Methods Using Visual C++ is one effort to integrate C++ with the visual solution to problems using ... 11, 2007 8:36 Char Count= 0 PREFACE Computing for Numerical Methods Using Visual C++ has been written to promote the use of Visual C++ in scientific...

Ngày tải lên: 19/03/2014, 14:13

... (2.2.12) APPLIED NUMERICAL METHODS USING MATLAB  Won Young Yang Chung- Ang University, Korea Wenwu Cao Pennsylvania State University Tae-Sang Chung Chung-Ang University, Korea John Morris The ... = -1 end Example 3. An if-elseif-else-end Block %nm119_3: example of if-elseif-else-end block ift>0,sgnt = 1 elseif t<0, sgnt = -1 else sgnt = 0 end Example 4. An i...

Ngày tải lên: 08/04/2014, 09:58

Ngày tải lên: 08/04/2014, 10:26

... 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, instead ... (2. 2.1b)ìa (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) 22 x...

Ngày tải lên: 09/08/2014, 12:22

... dy 1 h 1 3 (S 2,2 + 2S 1,2 ) = 3 1 3 (3 +2 3) = 2 (E3 .3. 7b) S 2,1 = dy 2 h 2 3 (S 3, 2 + 2S 2,2 ) = 1 1 3 (3 + 2 ì (3) ) = 2 (E3 .3. 7c) S 0 ,3 = S 1,2 − S 0,2 3h 0 = 3 −( 3) 3 = 2 (E3 .3. 8a) S 1 ,3 = S 2,2 − ... a 32 − l 31 u 12 a (1) 33 = a 33 − l 31 u 13    (2.4.6a) step 2: →    u 11 u 12 u 13 l 21 u 22 = a (1) 22 u 23 = a (1) 23 l 31 l...

Ngày tải lên: 09/08/2014, 12:22

... around the solution (e.g., Fig. 4. 4c or 4. 4d), whereas it con- verges to the solution quickly when f(x) has a steady slope as illustrated in Figs. 4. 4a and 4. 4b. 4. 5 SECANT METHOD The secant method ... “ roots()”. 1.55 1.5 1 .45 1.0000 1.5000 1.3750 1 .42 97 1 .40 77 1 .4 1.35 1 1.2 1 .4 x 2 y = xy = x y = g b ( x ) 1.0000 1.5000 1 .41 67 1 .41 42 1 .41 42 y = g c ( x )...

Ngày tải lên: 09/08/2014, 12:22

... −0.03 650 380828 h 2 = 0.0100000000 0.70 355 94917 0.032 956 5188 −0.00 354 728 950 h 3 = 0.0010000000 0.706 753 1100 0.0031936183 −0.000 353 67121 h 4 = 0.0001000000 0.7070714247 0.0003183147 −0.0000 353 5 652 h 5 = ... 12896 x 0 = [1 0 .5] x (P4.7.7) ||f(x)|| Flops 21 25 2378 x 0 = [1 0 .5] x [0.2321 1 .50 67] (P4.7.8) ||f(x)|| 1.07 (1.07) Flops 651 6 6492 x 0 = [1 0 .5] x (P4.7.9) ||f...

Ngày tải lên: 09/08/2014, 12:22

... solving ODEs, it gives us a Applied Numerical Methods Using MATLAB  , by Yang, Cao, Chung, and Morris Copyr ight  2005 John Wiley & Sons, I nc., ISBN 0-471 -69 833-4 263 292 ORDINARY DIFFERENTIAL ... Gauss (P5.11.1,2) 0.001 4 .62 12e-2 2.9822e-2 8.4103e-2 0.0001 9.4278e-3 9.4277e-3 0.00001 2.1853e-1 2.9858e-3 8.4937e-2 (P5.12.1,2) 0.001 1.2393e-5 1.3545e-5 0.0001 8. 362 6e...

Ngày tải lên: 09/08/2014, 12:22

... 0,x 2 =−1/2λ(E7.2.4b) ∂ ∂λ l(x,λ) (7. 2.3b) = x 2 1 + x 2 2 − 2 = 0 (E7.2.4c) x 2 1 + x 2 2 (E7.2.4c) = 2 (E7.2.4a,b) → (−1/2λ) 2 + (−1/2λ) 2 = 2,λ=±1/2 (E7.2.5) x 1 (E7.2.4a) =−1/2λ =∓1,x 2 (E7.2.4b) =−1/2λ =∓1 (E7.2.6) Now, ... (f o =−156.66) [−0.5,−1.0] [2.9035, −2 .74 68] [−2 .74 68, −2 .74 68] [−2 .74 68, −2 .74 68] [2.9029, 2.9028] (f o =−128.39) (f o =−100.12) (f o =−100.12) (f o =−15...

Ngày tải lên: 09/08/2014, 12:22

... 0.53 1. 08 0.46 — 1 −1.21 −1.21 −1.12 −1. 18 −1.21 −1.21 −1.15 1.26 1 −0. 58 −0. 58 −0.76 −0.64 −0. 58 −0. 58 −0.71 −1.70 c o 3 −1.76 −1.76 −1.44 −1.65 — −1.76 −1.76 −1.54 −1 .84 — −0.00 −0.00 −3 .82 1 −0.00 ... −1.33 −1 .84 −0.65 1 −0.00 0.00 0.29 −3 .82 −1.41 −0.00 −0.00 −0.00 −0.00 0.00 −0.00 −0.00 −0.00 −22.1 −16.4 1.21 1.21 1.12 1. 18 — 1.21 1.21 1.15 −1.26 — x o 1 0. 58 0. 58...

Ngày tải lên: 09/08/2014, 12:22

Ngày tải lên: 21/10/2014, 10:09