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I just read here that Jacobian matrices can be used to solve a system of nonlinear equations, and I am wondering exactly how this might work. I have the following system to solve:

$$\frac{(1-x-2y-3z)(3-3x-2y-z)^3}{(4x)^4}-1=0$$ $$\frac{(1-x-2y-3z)(3-3x-2y-z)}{(2y)^2}-1=0$$ $$\frac{(1-x-2y-3z)^3(3-3x-2y-z)}{(4z)^4}-1=0$$ On the article that I linked, it states that a quick way to solve a non-linear system is by solving $$J_F(x_{n+1}-x_n)=-F(x_n)$$ I'm really not too sure what any of this means. I am familiar with the Jacobain matrix, as well as the standard Newton's method using $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$. What does $F(x)$ indicate in the above equation? Can someone provide an example of a simpler problem that uses this method?

高田航
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    https://math.stackexchange.com/questions/1179036/solving-a-non-linear-multivariable-system-of-equations – Moo Nov 15 '17 at 23:41
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    There are many solutions (also complex ones): $$x = 5.04835,y= 8.11751,z= -3.26314, \x= -3.07816,y= 8.1143,z= 5.3475, \ x= 0.554965,y= -0.892357,z= 0.358717,\ x = -0.594701,y= 0.828277,z= -0.288399,\x= -0.421321,y= 0.431613,z= 0.110539,\x= 0.430132,y= 1.13386,z= 0.747241,\x = 0.271627,y= 0.378312,z= -0.131725,\x= 0.228651,y= 0.234236,z= 0.0599894$$ – Moo Nov 16 '17 at 15:53
  • @Moo Thank you for your help. I see that there is only one answer in your list of solutions that satisfies $0<x,y,z<1$. Is there a way to show that that is the unique solution in $[0,1]^3$? – 高田航 Nov 16 '17 at 21:51
  • I am not sure, in 1 dimensional systems it relatively easy. I haven't looked at this in a very long time and I am not sure if the same is true in multidimensional systems. – Moo Nov 16 '17 at 23:02

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