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April 2nd, 2017, 01:20 PM   #1
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Thumbs down System of non-linear equations!

Hello guys. I need help for one math project.

It's about a system of non-inear equations. The system go like this:


a) Calculate all the zeros in the system (manually)
b) Find the Jacobian matrix, J(x). (Notice that J(x) is singular for x(3) = 0)
c) Consider independent two starting solutions:
1) x(0) = {-0.01, -0.01, -0.01}T
2) x(0) = {-0.1, -0.1, -0.1}T

I have tried something but I don't know whether I am right...

Please help me. Thank you!

Last edited by skipjack; April 3rd, 2017 at 06:17 AM.
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April 2nd, 2017, 02:49 PM   #2
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Well, what did you try?

A couple of obvious points- from $\displaystyle x_3^4- 1= 0$ we have $\displaystyle x_3^4= 1$ and so $\displaystyle x_3= 1$, $\displaystyle x_3= -1$, $\displaystyle x_3= i$, or $\displaystyle x_3= -i$ (the last two if you are looking for solutions in the complex numbers).

And, of course, since $\displaystyle x_1x_2x_3= 0$ must have $\displaystyle x_1= 0$ or $\displaystyle x_2= 0$ (since we have already determined that $\displaystyle x_3$ cannot be 0).
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Last edited by skipjack; April 3rd, 2017 at 06:21 AM.
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April 2nd, 2017, 03:56 PM   #3
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Rather than use subscripts, I'm going to use x for x1, y for x2, and z for x3. If somebody would like to redo my post using subscripts, that's fine with me. Assuming we're only working with real numbers, substituting Country Boy's value of z in the first equation makes it an equation with an x term, a y term (with a coefficient of 1 that drops out), and a constant. Make the right side have the constant. As Country Boy said, x or y must be 0, and you can substitute in 0 for x and solve for y and sub in 0 for y and solve for x.
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April 3rd, 2017, 06:04 AM   #4
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Quote:
Originally Posted by EvanJ View Post
Rather than use subscripts, I'm going to use x for x1, y for x2, and z for x3. If somebody would like to redo my post using subscripts, that's fine with me. Assuming we're only working with real numbers, substituting Country Boy's value of z in the first equation makes it an equation with an x term, a y term (with a coefficient of 1 that drops out), and a constant. Make the right side have the constant. As Country Boy said, x or y must be 0, and you can substitute in 0 for x and solve for y and sub in 0 for y and solve for x.
First of all, Thank you!

I tried to make it like you said, and for y=0 in the end i am getting x=1, y=0 and z= +-1, and when i put x=0 i am getting y=+-1, x=0, z=+-1
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April 3rd, 2017, 06:06 AM   #5
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Quote:
Originally Posted by Country Boy View Post
Well, what did you try?

A couple of obvious points- from $\displaystyle x_3^4- 1= 0$ we have $\displaystyle x_3^4= 1$ and so $\displaystyle x_3= 1$, $\displaystyle x_3= -1$, $\displaystyle x_3= i$, or $\displaystyle x_3= -i$ (the last two if you are looking for solutions in the complex numbers).

And, of course, since $\displaystyle x_1x_2x_3= 0$ must have $\displaystyle x_1= 0$ or $\displaystyle x_2= 0$ (since we have already determined that $\displaystyle x_3$ cannot be 0).
First of all, Thank you!

I tried to make it like you said, and for x2 = 0 in the end I am getting x1 = 1, x2 = 0 and x3 = ±1, and when I put x1 = 0 I am getting x2 = ±1, x1 = 0, x3 = ±1

Last edited by skipjack; April 3rd, 2017 at 06:23 AM.
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