Solving Nonlinear Equations using Newton-Raphson Method

I have solved the following by hand but am having difficulties implementing the code. If anyone is able to assist me I would great appreciate it.
I would like to use Newton-Raphson to solve:
[ exp( X1*X2 ) = [ 1.2 cos( X1 + X2 ) ] 0.5 ]
Starting at X1(0) = 1 and X2(0) = .5. My tolerance is 0.0005.
Thank you in advance.

3 comentarios

If you regard both x1 and x2 as your unknowns, that is not a well-defined problem. You have two unknowns and only one equation, for which there is an infinite continuum of solutions. The fact that you have found a solution does not alter this fact.
If you tried to apply the Newton-Raphson method, you would discover that the necessary Jacobian matrix which is to be used in the recursion is not square and therefore has no inverse, and that means you could not make the recursion work.
Alex
Alex el 27 de Mzo. de 2014
Editada: Alex el 27 de Mzo. de 2014
The question posted funny. It should be exp(X1*X2) and cos(X1+X2) in the first matrix and 1.2 and 0.5 in the next
Jimmy
Jimmy el 12 de Nov. de 2014
Did anyone figure this problem out? I'm having the same problem, I can get the correct answer by hand but not with my code.

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Respuestas (1)

Roger Stafford
Roger Stafford el 27 de Mzo. de 2014
Editada: Roger Stafford el 27 de Mzo. de 2014
The recursion works very much the way it would in one dimension except that instead of dividing by the function's derivative, you multiply by the two functions' inverse Jacobian. Read about it at:
http://en.wikipedia.org/wiki/Newton's_method#k_variables.2C_k_functions

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el 27 de Mzo. de 2014

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el 12 de Nov. de 2014

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