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To solve a system of nonlinear equations which are equal to zero, ${f_i}(X) = 0\forall i \in 1:m,X = \left[ {{x_1},{x_2}, \cdots ,{x_n}} \right]$, we use Jacobian matrix and Newton algorithm. Hence, at each iteration, updated $X$ can be find by the following equation

${X^{(k + 1)}} = {X^{(k)}} + {J^{ - 1}}\left[ {\begin{array}{*{20}{c}} {{f_i}({X^{(k)}})} \\ {{f_2}({X^{(k)}})} \\ \vdots \\ {{f_m}({X^{(k)}})} \end{array}} \right]k = 0:j$

Where $J = \left[ {\begin{array}{*{20}{c}} {\frac{{\partial {f_1}({X^k})}}{{\partial {x_1}}}}&{\frac{{\partial {f_1}({X^k})}}{{\partial {x_2}}}}& \cdots &{\frac{{\partial {f_1}({X^k})}}{{\partial {x_n}}}}\\ {\frac{{\partial {f_2}({X^k})}}{{\partial {x_1}}}}&{\frac{{\partial {f_2}({X^k})}}{{\partial {x_2}}}}& \cdots &{\frac{{\partial {f_2}({X^k})}}{{\partial {x_n}}}}\\ \vdots & \vdots & \vdots & \vdots \\ {\frac{{\partial {f_m}({X^k})}}{{\partial {x_1}}}}&{\frac{{\partial {f_m}({X^k})}}{{\partial {x_2}}}}& \cdots &{\frac{{\partial {f_m}({X^k})}}{{\partial {x_n}}}} \end{array}} \right]$

The question is how to represent the above in terms of Jacobian matrix and first differential equations of ${f_i}(X)$ or $X$.

  • I don't understand, you already wrote the desired answer. Do you mean how to analytically write the inverse? It's useless (except possibly in 1D and 2D), don't bother. – Ian Nov 08 '17 at 01:56
  • It is true that the desired answer has written above but my question relates to the previous step, explicitly before identifying optimal $X(k)$. –  Nov 09 '17 at 18:04
  • I still don't understand. $J$ is the Jacobian matrix, and it is by definition given by first partial derivatives of the $f_i$. So what is your question? – Ian Nov 09 '17 at 18:11
  • Let's forget the above. One way to express ${f_i}(x)=0$$\forall {f_i}$ is $\Delta F(X) = J\Delta X $. Now, I want to write the latest in terms of first order differential equation. –  Nov 09 '17 at 18:22
  • Why? Do you want to devise some kind of "continuous flow" method for solving nonlinear systems? Newton's method is not of this character at all, so you'll want to think differently to find such a thing. – Ian Nov 09 '17 at 18:22
  • Thank you. I think you restate my problem in a few words. However, could you explain the assumptions that we could derive $\dot X = f(X)$ from a Newton's method? –  Nov 10 '17 at 14:57

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