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Root-finding by iteration |
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| Jun16-12, 08:42 AM | #1 |
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Root-finding by iteration
Dear all,
I have a question about root-finding. In fact my problem is about solving system of nonlinear equations but I have simplified my question as follows: Suppose I would like to find the root of the following function by iteration: [itex]y=f(x,p(x))[/itex] If I can calculate the total derivative with respect to x, I can use Newton-Raphson method effectively. However, for some reason, I can't calculate the total derivative. I can calculate the partial derivative though. I am using the naive method of finding the root by "partial derivative". To this end, in each iteration I replace p(x) by its value evaluated from the previous value of x, treating it as a constant, and apply Newton-Raphson method. Convergence is achieved in most cases but I'm not sure if this is the right way. My question is: 1. Is there a theory behind this method? Is it related to fixed-point iteration? 2. Does the convergence depend on p(x)? Your help would be appreciated, Hassan |
| Jun16-12, 09:06 AM | #2 |
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Is this function just a 'complicated' (because of p(x)) one-dimensional function? This is what you seem to imply since y is only a function of x. If this is the case, then in terms of the theory and application, the one-dimensional root finding methods like Newton-Rhapson and other similar ones should suffice provided that the function has the desired properties (like continuity, differentiability over a given interval). Hopefully this answers your first question. For the second question, it depends on whether the function given p(x) has differentiability and continuity requirements fulfilled. In terms of whether a root actually exists, then we can use the first derivative (might need the second for points of inflection) and the mean value theorem (or something similar) to show that a root exists given that the function is continuous (and differentiable). If your function has the above properties, you will be able to use any root-finding algorithm for your function y (assuming only depends on x) and the algorithm should be able to tell you whether or not a real root even exists (although most algorithms generate complex roots as well). |
| Jun16-12, 09:52 AM | #3 |
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Thanks a lot,
I think I should explains the original problem. I have the following system of "nonlinear" equations in matrix form: [itex]Ax=b[/itex] A=A(x,p(x)) is a symmetric sparse matrix and the number of unknowns could be as many as one million. The matrix elements depend on p(x), with x being the vector of unknowns. Since p depends on all unknowns, the matrix of derivatives becomes non-sparse and , I guess, non-symmetric. That's why I can't use the derivatives. Beside that, the dependence of p(x) on x is complicated and is not in the form of a functional but it is evaluated by numerical methods. The only method I know for solving system of nonlinear equations is the Newton-Raphson method. The method explained in my problem seems to be different and I have no material to support it. I am looking for a material which discusses the method. Back to one dimensional problem, can I solve the equation [itex]x^{3}-6=0[/itex] by writing it as [itex]x^{2}x-6=0[/itex] and iterate as [itex]x_{k}^{2}x_{k+1}-6=0[/itex]? This example doesn't converge in my code. Thanks again. |
| Jun18-12, 12:26 AM | #4 |
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Root-finding by iteration
Sorry for the confusing setup of the problem.
After some research, I realized that the method is in fact the fixed point iteration method. In each iteration, the Newton-Raphson method first solve the following equation for [itex]x_{k+1}[/itex] . [itex]f(x_{k+1},p(x_{k}))=0[/itex] and yields the following fixed-point iteration: [itex]x_{k+1}=g(x_{k})[/itex] In the fixed-point method, the convergence depends both on g(x) and the starting point [itex]x_{0}[/itex] . |
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