Solving Nonlinear Equations using Newton-Raphson Method

4 visualizaciones (últimos 30 días)
Alex
Alex el 27 de Mzo. de 2014
Comentada: Jimmy el 12 de Nov. de 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 comentarios
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.

Iniciar sesión para comentar.

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

Etiquetas

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by