Solving Nonlinear Equations using Newton-Raphson Method

4 vues (au cours des 30 derniers jours)
Alex
Alex le 27 Mar 2014
Commenté : Jimmy le 12 Nov 2014
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 commentaires
Alex
Alex le 27 Mar 2014
Modifié(e) : Alex le 27 Mar 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 le 12 Nov 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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Réponses (1)

Roger Stafford
Roger Stafford le 27 Mar 2014
Modifié(e) : Roger Stafford le 27 Mar 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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