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I want to use Newton's method to find two solutions to the system of equations using $L_\infty$ norm and an f-test with $tol_f = 10^{-6}.$

$$\frac{1}{2}x_1+x_2+\frac{1}{2}x_3 - \frac{x_6}{x_7}=0$$

$$x_3+x_4+2x_5-\frac{2}{x_7}=0$$

$$x_1+x_2+x_5- \frac{1}{x_7} = 0$$

$$-28837 x_1 - 139009 x_2 - 78213 x_3 + 18927 x_4 + 8427 x_5 + \frac{13492}{x_7} -10690 \frac{x_6}{x_7} =0$$

$$x_1 + x_2 +x_3 + x_4 + x_5 -1 =0$$

$$400x_1 x^3_4 - 178370 x_3 x_5 = 0$$

$$x_1 x_3 - 2.6058 x_2 x_4 = 0$$

So, how do I need to re-write these equations to be able to use Newton's method? I am not sure what to do about all the multiplications/divisions (like $x_6/x_7$). I am not sure how to approach this problem since all the Newton examples in my textbook deal with pairs of equations in two unknowns. What would be an efficient way of doing this?

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    $\begingroup$ You just need the Jacobian matrix, en.wikipedia.org/wiki/… $\endgroup$ – David Apr 14 '16 at 0:29
  • $\begingroup$ @David Once I worked out the partial derivatives the Jacobian becomes: $$\begin{bmatrix} 1/2 & 1 & 1/2 & 0 & 0 & -1/x_7 & x_6/x_7^2\\ 0 & 0 & 1 & 1 & 2 & 0 & 2/x_7^2\\ 1 & 1 & 0 & 0 & 1 & 0 & 1/x_7^2\\ -28837 & -139009 & -78213 & 18927 & 8427 & \frac{-10690}{x_7} & \frac{-13492}{x_7^2} + \frac{10690x_6}{x_7^2}\\ 1 & 1 & 1 & 1 & 1 & 0 & 0\\ 400 x_4^3 & 0 & -178370x_5 & 1200x_1 x_4^2 & -178370x_3 & 0 & 0\\ x_3 & -2.6058x_4 & x_1 & -2.6058 x_2 & 0 & 0 &0 \end{bmatrix}$$ So, how do we proceed? And how does $L_\infty$ factor into this? $\endgroup$ – Merin Apr 14 '16 at 2:06
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    $\begingroup$ Starting at $\{3.,1.,0.,1.,2.,3.,4.\} $, I get a result of $\{0.456931, -0.000407198, -0.0021252, 0.915172, -0.36957, 2.61055, 11.5004\}$ (plug these numbers into each equation to verify them) using Newton's method. I cannot figure out what you mean by that norm and f-test - are those some stopping criteria methods? $\endgroup$ – Moo Apr 14 '16 at 2:36
  • $\begingroup$ How did you choose $\{ 3, 1, 0, 1, 2, 3, 4 \}$ for the starting points? So the norm $L_\infty$ is defined as the maximum row sum ($max_{1 \leq i \leq n} \{ \sum^n_{j=1} |a_{ij}| \}$). And the f-test is: $|f(x^{(k)})| \leq tol_f.$ $\endgroup$ – Merin Apr 14 '16 at 3:25
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    $\begingroup$ The starting point shouldn't be very important, you'll want to avoid having $x_7=0$ though. $\endgroup$ – David Apr 14 '16 at 3:31
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If you want to go back to a situation where you would have only two equations for two variables, you could use equations $(1)$ to $(5)$ to express $x_1$ to $x_5$ as functions of $x_6$ and $x_7$.

If you do it, equation $(6)$ will stay quite long but workable. However, equation $(7)$ will become quite nice since (hoping no error in my calculations) it will reduce to $$\frac{(4034991986 x_6-1103097977 x_7+707920326) (231034 x_6-55381 x_7+33462)}{849165120000 x_7^2}=0$$ which gives two possible relations between $x_6$ and $x_7$.

For each of the possible relations (they are nicely linear), you need to solve equation $(6)$ which only contains $x_6$ as variable. It is a quartic in $x_6$.

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  • $\begingroup$ Is there some simple algorithm where you could directly use Newton's method? $\endgroup$ – Merin Apr 14 '16 at 15:04
  • $\begingroup$ Let me return the question : do you know a good starting point ? This problem may have a lot of solutions. $\endgroup$ – Claude Leibovici Apr 14 '16 at 18:10
  • $\begingroup$ No, what would be a good starting point? We randomly chose $\{ 3,1,0,1,2,3,4 \}.$ $\endgroup$ – Merin Apr 14 '16 at 19:37
  • $\begingroup$ Do we not need the Jacobian matrix at all? $\endgroup$ – Merin Apr 14 '16 at 20:21

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