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I have to solve a system of nonlinear equations using jacobian but I'm not sure how to solve for the solutions. I remember one of my friends doing $Ax = B$; where jacobian matrix was $A$, but im not sure what was $B$.

$10\cos(t_3)-9\cos(t_4) = 8 - 3.5\cos(15)$
$10\sin(t_3)-9\sin(t_4) = -3.5\sin(15)$

I have to solve for $t_3$ and $t_4$. And it tells me to set it in terms of $t_2$ (which is $15$ in the problem). Answers were $t_3 = 53$ degrees and $t_4 = 81$ degrees

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Hint

Let me start very basic : you want to solve $F(x,y)=0$ and $G(x,y)=0$ starting from an initial guess $(x_0,y_)$.

Expand each equation as a first order Taylor series. So $$F(x_0,y_0)+F'_x (x_0,y_0)(x-x_0)+F'_y (x_0,y_0)(y-y_0)=0$$ $$G(x_0,y_0)+G'_x (x_0,y_0)(x-x_0)+G'_y (x_0,y_0)(y-y_0)=0$$ or $$F'_x (x_0,y_0) \Delta X+F'_y (x_0,y_0)\Delta Y=-F(x_0,y_0)$$ $$G'_x (x_0,y_0)\Delta X+G'_y (x_0,y_0)\Delta Y=-G(x_0,y_0)$$ I suppose you see $A$ and $B$.

But since the problem is not linear, you must restart the calculation using $x_0=x_0+\Delta X$, $y_0=y_0+\Delta Y$ untill they do not change for the required accuracy.

I am sure that you can take from here and solve your specific problem.

Applying the method, you should arrive to $t_3=52.9807$ and $t_4=81.0408$ (I suggest you do all the work using radians to make you life simpler). There is another solution.

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  • $\begingroup$ Your comment has made it click for me! I understand it now! Thanks! I'm confused though. What are x0 and y0? Are they random guesses? $\endgroup$ – user3509716 Aug 30 '14 at 3:40
  • $\begingroup$ Don't be confuse, please. You are very welcome ! No, as for any Newton based method, they must be "reasonable" guesse" of the solution. $\endgroup$ – Claude Leibovici Aug 30 '14 at 4:14
  • $\begingroup$ $t3_0=1$ and $t4_0=3/2$ seem to be good guesses. Try and report ! $\endgroup$ – Claude Leibovici Aug 30 '14 at 4:27
  • $\begingroup$ thank you thank you. I have a question, For delta x and delta y i keep getting positive numbers so they keep going away from the answer. $\endgroup$ – user3509716 Aug 30 '14 at 15:01
  • $\begingroup$ I agree but I must go now. For the time being, set limit the $\Delta$'s between say $-0.05$ and $+0.05$ if they try to be larger in absolute value. I think I know what is the problem. $\endgroup$ – Claude Leibovici Aug 30 '14 at 15:59

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