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An example of using Newton–Raphson method to solve numerically the equation f(x) = 0. In mathematics, to solve an equation is to find its solutions, which are the values (numbers, functions, sets, etc.) that fulfill the condition stated by the equation, consisting generally of two expressions related by an equals sign.
Once y is also eliminated from the third row, the result is a system of linear equations in triangular form, and so the first part of the algorithm is complete. From a computational point of view, it is faster to solve the variables in reverse order, a process known as back-substitution. One sees the solution is z = −1, y = 3, and x = 2. So ...
At any step in a Gauss-Seidel iteration, solve the first equation for in terms of , …,; then solve the second equation for in terms of just found and the remaining , …,; and continue to . Then, repeat iterations until convergence is achieved, or break if the divergence in the solutions start to diverge beyond a predefined level.
These equations describe boundary-value problems, in which the solution-function's values are specified on boundary of a domain; the problem is to compute a solution also on its interior. Relaxation methods are used to solve the linear equations resulting from a discretization of the differential equation, for example by finite differences.
The storage and computation overhead is such that the standard simplex method is a prohibitively expensive approach to solving large linear programming problems. In each simplex iteration, the only data required are the first row of the tableau, the (pivotal) column of the tableau corresponding to the entering variable and the right-hand-side.
To solve the equations, we choose a relaxation factor = and an initial guess vector = (,,,). According to the successive over-relaxation algorithm, the following table is obtained, representing an exemplary iteration with approximations, which ideally, but not necessarily, finds the exact solution, (3, −2, 2, 1) , in 38 steps.
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