PICTUREI want to do a Newton's iteration using Jacobian and I did not understand how to do that. Can someone explain the steps to me please?


  • $\begingroup$ Can you post more details and where you stuck? $\endgroup$ – user251257 Dec 10 '16 at 11:15
  • $\begingroup$ @user251257 I added a picture. I do not know how to find J. I have the solutions but i didn't understand then. So i wanted to know if someone can tell me the steps.. $\endgroup$ – JaneDoee Dec 10 '16 at 11:18
  • $\begingroup$ do you know how to compute derivatives? $\endgroup$ – user251257 Dec 10 '16 at 11:19
  • $\begingroup$ @user251257 yes $\endgroup$ – JaneDoee Dec 10 '16 at 11:23
  • $\begingroup$ just compute the 4 partial derivatives and write them down as the Jacobian $J$. $\endgroup$ – user251257 Dec 10 '16 at 11:24

Okay, you have a system of two equations, namely \begin{split} f_1(x_1,x_1) = 0\\ f_2(x_1,x_2) = 0 \end{split}

The Newton method reads $$\left[\begin{array}{c} x_1\\ x_2\end{array}\right]^{n+1} = \left[\begin{array}{c} x_1\\ x_2\end{array}\right]^{n} - \left(J^{-1}\right)^{n}\left[\begin{array}{c} f_1(x_1,x_2)\\ f_2(x_1,x_2)\end{array}\right]^n \tag1$$ Where $J^{-1}$ is the inverse of the Jacobian, that is $$J=\left[\begin{array}{cc} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2}\end{array}\right] $$ Use the equation $(1)$ recursively, starting with $n=0$ with your initial guess $$\left[\begin{array}{c} x_1\\ x_2\end{array}\right]^{0}=\left[\begin{array}{c} 1\\ 1\end{array}\right]$$

  • $\begingroup$ I just posted the solutions i have. I don't understand why they do not use J inverse for the first step $\endgroup$ – JaneDoee Dec 10 '16 at 11:46
  • $\begingroup$ Okay, it is the same. For sure you calculate the "inverse" of $J$, solving $\vec{v}$ from $$J^{n}\vec{v}=\vec{f}^{n}\tag2$$ and later you update your solution with $$\vec{v} = \vec{x}^{n+1}-\vec{x}^{n}\tag3$$ You will also notice that the equation $(1)$ I wrote is formally the same as $(2)$ with $(3)$ $\endgroup$ – HBR Dec 10 '16 at 11:52
  • $\begingroup$ I don't understand what v is . $\endgroup$ – JaneDoee Dec 10 '16 at 11:55
  • $\begingroup$ $\vec{v}$ is defined on your homework. The procedure you follow in your solution is first solve $\vec{v}$ from the system in $(2)$ and update your solution by using $(3)$. Do you understand it know? $\endgroup$ – HBR Dec 10 '16 at 11:57
  • $\begingroup$ no .. :/ i still dont know how to replace the values or understand what is going on $\endgroup$ – JaneDoee Dec 10 '16 at 12:13

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