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Iterativos Methods


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Iterativos Methods

  1. 1. SOLUTION OF EQUATIONS FOR ITERATIVOS METHODS<br />Jeannie Castaño 2053298<br />
  2. 2. JACOBI<br />A iterative method is a method that progressively it calculates approaches to the solution of a problem. To difference of the direct methods, in which the process should be finished to have the answer, in the iterative methods you can suspend the process to the i finish of an iteration and an approach is obtained to the solution.<br />For example the method of Newton-Raphson<br />
  3. 3.
  4. 4. The Jacobi method de is the iterative method to solve system of equations simple ma and it is applied alone to square systems, that is to say to systems with as many equations as incognito.<br />The following steps should be continued: <br />First the equation de recurrence is determined. Of the equation i and the incognito iclears. In notation matricial is written: <br /> where x is the vector of incognito <br />It takes an approach for the solutions and to this it is designated for <br />3. You itera in the cycle that changes the approach<br />
  5. 5. Leaving of x=1, y=2 applies two iterations of the Jacobi method to solve the system:<br />
  6. 6. This Di it is used as unemployment approach in the iterations until Di it is smaller than certain given value.<br />Being Di:<br />
  7. 7. GAUSS-SEIDEL<br />Given a square system of n linear equations with unknown x:<br />Where:<br />
  8. 8. GAUSS-SEIDEL<br />Then A can be decomposed into a lower triangular component L*, and a strictly upper triangular component U:<br />The system of linear equations may be rewritten as:<br />
  9. 9. GAUSS-SEIDEL<br />The Gauss–Seidel method is an iterative technique that solves the left hand side of this expression for x, using previous value for x on the right hand side. Analytically, this may be written as:<br />However, by taking advantage of the triangular form of L*, the elements of x(k+1) can be computed sequentially using forward substitution:<br />
  10. 10. To solve the following exercise for the method of Gauss-Seidel with an initial value of (0,0,0). Use three iterations<br />1ra iteration<br />
  11. 11. 2da iteration<br />3ra iteration<br />
  12. 12. GAUSS-SEIDEL RELAXATION<br />To solve the previous exercise for the gauss method for relaxation. .Use two iterations<br />Replacing with the initial values (0,0,0)<br />
  13. 13. 1ra iteration<br />
  14. 14. 2da iteration<br />
  15. 15. BIBLIOGRAFIA<br /><br /><br /><br />