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I have a system of equations:

  1. $x_1^3 + x_1^2x_2 -x_1x_3 + 6 = 0$
  2. $e^{x_1} + e^{x_2} - x_3=0$
  3. $x_2^2 - 2x_1x_2 = 4$

and the question asks me to evaluate the left hand side of the equations at my estimate for the solution to ensure they agree with the right hand side to within $0.001$.

First I used Newton's method to find a solution to the system, which is $[-1.45604279595534, -1.66423046608154, .422493404446532]$ on Maple.

However, I do not exactly understand the question.

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up vote 1 down vote accepted

I had a look to your problem since it is an interesting system of equations for which there is no way to know a priori the number of possible solutions.

From the second equation, you can eliminate x3 and express is as a function of x1 and x2. From the third equation, you can eliminate x1 as a function of x2. So, what is left is the first equation in which only appears x2. This equation has four roots which are -3.48958, -1.00851, 0.701462, 2.52722.

So, depending where you start your iterations, if the system solves, it could correspond to any of these solutions.

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The question is asking for you to find an approximate solution such that when you plug your solution into the left-hand-side of each equation, the resulting values are within $0.001$ of the right hand side of each solution.

For this specific question you need to get $0 \pm 0.001$ when you plug in to the first equation, $0 \pm 0.001$ for the second, and $4 \pm 0.001$ for the third.

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So for the first equation, I use my estimate and get $-0.615168477883073508385505820046007451335328$ which is within $[-.001,0.001]$. Is that it? – user87274 Dec 12 '13 at 2:46
I'm not checking the arithmetic (I don't have a CAS, and there's no way I'm doing that by hand or with a hand-calculator), but if you get $-0.6\ldots$, that's certainly outside of $[-0.001, 0.001]$. So, your estimate must be made tighter. – apnorton Dec 12 '13 at 3:03
I cannot just come up with an estimate! This is what I got by using Newton's method. – user87274 Dec 12 '13 at 3:06
@AJR Newton's Method is an estimate. The more times you perform the iterations in Newton's Method, the more accurate your estimate becomes (assuming it is a "well behaved" function to use for Newton's Method--some functions don't converge well). – apnorton Dec 12 '13 at 3:07
Yes, but after 8 iterations it stops because to Maple it is the best approximate to the solution. – user87274 Dec 12 '13 at 3:09

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