Solving Nonlinear Equations using Newton-Raphson Method

4 次查看（过去 30 天）
Alex
Alex 2014-3-27
评论： Jimmy 2014-11-12
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 个评论
显示 1更早的评论
Alex
Alex 2014-3-27
编辑：Alex 2014-3-27
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 2014-11-12
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.

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Roger Stafford
Roger Stafford 2014-3-27
编辑：Roger Stafford 2014-3-27
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

类别

在 Help Center 和 File Exchange 中查找有关 Newton-Raphson Method 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by