By C. Woodford, C. Phillips

ISBN-10: 9400713657

ISBN-13: 9789400713659

This publication is for college students following an introductory direction in numerical equipment, numerical suggestions or numerical research. It introduces MATLAB as a computing setting for experimenting with numerical tools. It methods the topic from a realistic perspective; concept is stored at a minimal commensurate with finished assurance of the topic and it comprises plentiful labored examples which offer effortless realizing via a transparent and concise theoretical therapy. This version areas even larger emphasis on ‘learning by means of doing’ than the former variation. totally documented MATLAB code for the numerical tools defined within the e-book may be on hand as supplementary fabric to the ebook on http://extras.springer.com

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**Sample text**

Conversely, if the function is very flat in the region of a root the range of possible x values would be comparatively large. 2. 1 Finding an Interval Containing a Root In the case of the function already considered we were fortunate to have available a graph which could be used to determine an initial interval containing a root. However if such a graph is not readily available it would be advisable to generate a series of x and f values to gain a feel for the function and the presence of roots and enclosing intervals.

8 Gauss–Seidel Iteration 41 grams respectively, and for large pies 60, 50 and 90 grams. The baker wishes to use up all the ingredients. Formulate a linear system to decide if this is possible. Let the number of small, medium and large pies to be made be x, y and z. Taking into account that 1 kilo = 1000 grams we have 3x + 2y + 4z = 250 (flour to be used) 4x + 3y + 6z = 350 (sugar to be used) 6x + 5y + 9z = 530 (fruit to be used). If the solution to these equations turns out to consist of whole numbers then the baker can use up all the ingredients.

The extension of Gaussian elimination to larger systems 24 2 Linear Equations is straightforward. The aim, as before, is to transform the original system to upper triangular form which is solved by backward substitution. Discussion The process we have described may be summarised in matrix form as premultiplying the coefficient matrix A by a series of lower-triangular matrices to form an upper-triangular matrix. We have ⎞ ⎞⎛ ⎞⎛ ⎞⎛ ⎛ 1 0 0 0 1 0 0 0 2 3 4 −2 1 0 0 0 ⎟ ⎟⎜ ⎟⎜ 1 ⎟⎜ ⎜ ⎜ ⎜ ⎟⎜ ⎜0 1 1 0 0⎟ 4 −3 ⎟ 0 0⎟ ⎟ ⎟⎜ 0 ⎟⎜ − 2 1 0 0 ⎟⎜ 1 −2 ⎜ ⎟ ⎟⎜ 0 −3/( 7 ) 1 0 ⎟⎜ −2 0 1 0 ⎟⎜ 4 ⎜0 0 3 −1 1 1 0 ⎠ ⎠⎝ ⎠⎝ ⎠⎝ ⎝ 2 17 3 3 −4 2 −2 0 0 − 62 − 0 − 1 0 1 0 0 1 75 7 2 ⎛ ⎞ 2 3 4 −2 ⎜ ⎟ 7 ⎜ −2 4 −2 ⎟ ⎜ ⎟ =⎜ 47 ⎟ − 75 ⎝ 7 7 ⎠ 161 525 where it can be seen that sub-diagonal entries correspond to the multipliers in the elimination process.

### Numerical Methods with Worked Examples: Matlab Edition (2nd Edition) by C. Woodford, C. Phillips

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