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laniik
science forum beginner

Joined: 19 Jun 2006
Posts: 4

Posted: Fri Jul 14, 2006 6:38 pm    Post subject: Iterative solution to non-linear equations

Hi, I have three equations with three unknowns. All of the equations
are non-linear. I would like to try to solve for the three unknowns
via iterative solution. What are some methods for this?

Thanks.
Lynn Kurtz
science forum Guru

Joined: 02 May 2005
Posts: 603

Posted: Fri Jul 14, 2006 6:53 pm    Post subject: Re: Iterative solution to non-linear equations

On 14 Jul 2006 11:38:39 -0700, "laniik" <laniik@yahoo.com> wrote:

 Quote: Hi, I have three equations with three unknowns. All of the equations are non-linear. I would like to try to solve for the three unknowns via iterative solution. What are some methods for this? Thanks.
laniik
science forum beginner

Joined: 19 Jun 2006
Posts: 4

Posted: Fri Jul 14, 2006 7:01 pm    Post subject: Re: Iterative solution to non-linear equations

[Mr.] Lynn Kurtz wrote:
 Quote: On 14 Jul 2006 11:38:39 -0700, "laniik" wrote: Hi, I have three equations with three unknowns. All of the equations are non-linear. I would like to try to solve for the three unknowns via iterative solution. What are some methods for this? Thanks. http://www.google.com/search?q=iterative+methods+nonlinear+equations&start=0&ie=utf-8&oe=utf-8&client=firefox-a&rls=org.mozilla:en-US:official --Lynn

I've looked at those. They seem to be targeted towards solving one
non-linear equation via somthing like Newtons method, which I
understand. I just dont see how to apply that to a system of equations.
C6L1V@shaw.ca
science forum Guru

Joined: 23 May 2005
Posts: 628

Posted: Fri Jul 14, 2006 7:30 pm    Post subject: Re: Iterative solution to non-linear equations

laniik wrote:
 Quote: [Mr.] Lynn Kurtz wrote: On 14 Jul 2006 11:38:39 -0700, "laniik" wrote: Hi, I have three equations with three unknowns. All of the equations are non-linear. I would like to try to solve for the three unknowns via iterative solution. What are some methods for this? Thanks. http://www.google.com/search?q=iterative+methods+nonlinear+equations&start=0&ie=utf-8&oe=utf-8&client=firefox-a&rls=org.mozilla:en-US:official --Lynn I've looked at those. They seem to be targeted towards solving one non-linear equation via somthing like Newtons method, which I understand. I just dont see how to apply that to a system of equations.

The Newton-Raphson method applies to general systems of equations. If
the system is f(x) = 0, where x \in R^r and f = (f_1, ..., f_n), and if
x_0 \in R^n is a starting point, the N-R method just takes a linear
approximation and solves that: f(x) =approx= f(x_0) + H(x_0) (x - x_0),
where H(x) = the Hessian matrix, H_{ij}(x) = d f_i(x)/ dx_j , and all
vectors are regarded as column vectors. If H = H(x_0) is invertible,
the next approximation is x = x_0 - H^(-1) f(x_0). Of course, just as
the solution.

There are many improvements possible. Do a Google search on
Newton-Raphson method.

R.G. Vickson
Jim Rockford
science forum beginner

Joined: 30 Jun 2006
Posts: 3

Posted: Fri Jul 14, 2006 10:08 pm    Post subject: Re: Iterative solution to non-linear equations

C6L1V@shaw.ca wrote:
 Quote: The Newton-Raphson method applies to general systems of equations. If the system is f(x) = 0, where x \in R^r and f = (f_1, ..., f_n), and if x_0 \in R^n is a starting point, the N-R method just takes a linear approximation and solves that: f(x) =approx= f(x_0) + H(x_0) (x - x_0), where H(x) = the Hessian matrix, H_{ij}(x) = d f_i(x)/ dx_j , and all vectors are regarded as column vectors. If H = H(x_0) is invertible, the next approximation is x = x_0 - H^(-1) f(x_0). Of course, just as in 1 dimension, you should start with a "reasonable" approximation to the solution.

Actually, H is the Jacobian matrix, not the Hessian.

J
C6L1V@shaw.ca
science forum Guru

Joined: 23 May 2005
Posts: 628

Posted: Sat Jul 15, 2006 1:05 am    Post subject: Re: Iterative solution to non-linear equations

Jim Rockford wrote:
 Quote: C6L1V@shaw.ca wrote: The Newton-Raphson method applies to general systems of equations. If the system is f(x) = 0, where x \in R^r and f = (f_1, ..., f_n), and if x_0 \in R^n is a starting point, the N-R method just takes a linear approximation and solves that: f(x) =approx= f(x_0) + H(x_0) (x - x_0), where H(x) = the Hessian matrix, H_{ij}(x) = d f_i(x)/ dx_j , and all vectors are regarded as column vectors. If H = H(x_0) is invertible, the next approximation is x = x_0 - H^(-1) f(x_0). Of course, just as in 1 dimension, you should start with a "reasonable" approximation to the solution. Actually, H is the Jacobian matrix, not the Hessian.

You're right, and I remembered that immediately after pressing the
"post" button.

RGV

 Quote: J

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