Solving Nonlinear Equations using Newton-Raphson Method

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Alex
Alex 2014 年 3 月 27 日
コメント済み: Jimmy 2014 年 11 月 12 日
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 件のコメント
Alex
Alex 2014 年 3 月 27 日
編集済み: Alex 2014 年 3 月 27 日
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 年 11 月 12 日
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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回答 (1 件)

Roger Stafford
Roger Stafford 2014 年 3 月 27 日
編集済み: Roger Stafford 2014 年 3 月 27 日
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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