Convergence of the Newton-Raphson method applied to a nonlinear system
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I'm working on a nonlinear system of 2 equations and try to solve it with the Newton-Raphson method.
I have a function $f(x,y)$. In order to use the method, I have to find a starting vector $(x,y)$. But the method will only converge for specific starting vectors.
Is there a formula that gives the convergence of the Newton-Raphson method on a system of equations ?
Thank you all.
convergence nonlinear-system newton-raphson
asked Dec 5 '18 at 16:28
LucLuc
1
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add a comment |
$begingroup$
I'm working on a nonlinear system of 2 equations and try to solve it with the Newton-Raphson method.
I have a function $f(x,y)$. In order to use the method, I have to find a starting vector $(x,y)$. But the method will only converge for specific starting vectors.
Is there a formula that gives the convergence of the Newton-Raphson method on a system of equations ?
Thank you all.
convergence nonlinear-system newton-raphson
asked Dec 5 '18 at 16:28
LucLuc
1
$endgroup$
add a comment |
$begingroup$
I'm working on a nonlinear system of 2 equations and try to solve it with the Newton-Raphson method.
I have a function $f(x,y)$. In order to use the method, I have to find a starting vector $(x,y)$. But the method will only converge for specific starting vectors.
Is there a formula that gives the convergence of the Newton-Raphson method on a system of equations ?
Thank you all.
convergence nonlinear-system newton-raphson
asked Dec 5 '18 at 16:28
LucLuc
1
$endgroup$
I'm working on a nonlinear system of 2 equations and try to solve it with the Newton-Raphson method.
I have a function $f(x,y)$. In order to use the method, I have to find a starting vector $(x,y)$. But the method will only converge for specific starting vectors.
Is there a formula that gives the convergence of the Newton-Raphson method on a system of equations ?
Thank you all.
convergence nonlinear-system newton-raphson
convergence nonlinear-system newton-raphson
asked Dec 5 '18 at 16:28
LucLuc
1
asked Dec 5 '18 at 16:28
LucLuc
1
asked Dec 5 '18 at 16:28
LucLuc
1
asked Dec 5 '18 at 16:28
LucLuc
1
asked Dec 5 '18 at 16:28
LucLuc
1
1
add a comment |
add a comment |
2 Answers
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As long as you pick even a somewhat reasonable guess, the method should find the closest root. If the equations were linear, and you never encounter a situation where you divide by zero, you won't even have to use a reasonable guess.
$$
{displaystyle x_{n+1}=x_{n}-m{frac {f(x_{n})}{f'(x_{n})}}.}
$$
answered Dec 5 '18 at 16:36
nessness
375
$endgroup$
$begingroup$
The problem is that I use the N-R method in a program and I don't know some parameters of my function $f(x,y)$. That means I cannot make a "reasonable guess". What I do know is that $f(x,y)$ is nonlinear. I would need a formula that determines wether the method converges in terms of the function $f(x,y)$.
$endgroup$
– Luc
Dec 5 '18 at 16:51
add a comment |
$begingroup$
No, in general there is no easy way to determine whether or not Newton-Raphson will converge for a given starting value, other than to try it. The region of convergence can be very complicated, and may have a fractal boundary.
answered Dec 5 '18 at 16:59
Robert IsraelRobert Israel
321k23210463
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
As long as you pick even a somewhat reasonable guess, the method should find the closest root. If the equations were linear, and you never encounter a situation where you divide by zero, you won't even have to use a reasonable guess.
$$
{displaystyle x_{n+1}=x_{n}-m{frac {f(x_{n})}{f'(x_{n})}}.}
$$
answered Dec 5 '18 at 16:36
nessness
375
$endgroup$
$begingroup$
The problem is that I use the N-R method in a program and I don't know some parameters of my function $f(x,y)$. That means I cannot make a "reasonable guess". What I do know is that $f(x,y)$ is nonlinear. I would need a formula that determines wether the method converges in terms of the function $f(x,y)$.
$endgroup$
– Luc
Dec 5 '18 at 16:51
add a comment |
$begingroup$
As long as you pick even a somewhat reasonable guess, the method should find the closest root. If the equations were linear, and you never encounter a situation where you divide by zero, you won't even have to use a reasonable guess.
$$
{displaystyle x_{n+1}=x_{n}-m{frac {f(x_{n})}{f'(x_{n})}}.}
$$
answered Dec 5 '18 at 16:36
nessness
375
$endgroup$
$begingroup$
The problem is that I use the N-R method in a program and I don't know some parameters of my function $f(x,y)$. That means I cannot make a "reasonable guess". What I do know is that $f(x,y)$ is nonlinear. I would need a formula that determines wether the method converges in terms of the function $f(x,y)$.
$endgroup$
– Luc
Dec 5 '18 at 16:51
add a comment |
$begingroup$
As long as you pick even a somewhat reasonable guess, the method should find the closest root. If the equations were linear, and you never encounter a situation where you divide by zero, you won't even have to use a reasonable guess.
$$
{displaystyle x_{n+1}=x_{n}-m{frac {f(x_{n})}{f'(x_{n})}}.}
$$
answered Dec 5 '18 at 16:36
nessness
375
$endgroup$
As long as you pick even a somewhat reasonable guess, the method should find the closest root. If the equations were linear, and you never encounter a situation where you divide by zero, you won't even have to use a reasonable guess.
$$
{displaystyle x_{n+1}=x_{n}-m{frac {f(x_{n})}{f'(x_{n})}}.}
$$
answered Dec 5 '18 at 16:36
nessness
375
answered Dec 5 '18 at 16:36
nessness
375
answered Dec 5 '18 at 16:36
nessness
375
answered Dec 5 '18 at 16:36
nessness
375
375
$begingroup$
The problem is that I use the N-R method in a program and I don't know some parameters of my function $f(x,y)$. That means I cannot make a "reasonable guess". What I do know is that $f(x,y)$ is nonlinear. I would need a formula that determines wether the method converges in terms of the function $f(x,y)$.
$endgroup$
– Luc
Dec 5 '18 at 16:51
add a comment |
$begingroup$
The problem is that I use the N-R method in a program and I don't know some parameters of my function $f(x,y)$. That means I cannot make a "reasonable guess". What I do know is that $f(x,y)$ is nonlinear. I would need a formula that determines wether the method converges in terms of the function $f(x,y)$.
$endgroup$
– Luc
Dec 5 '18 at 16:51
$begingroup$
The problem is that I use the N-R method in a program and I don't know some parameters of my function $f(x,y)$. That means I cannot make a "reasonable guess". What I do know is that $f(x,y)$ is nonlinear. I would need a formula that determines wether the method converges in terms of the function $f(x,y)$.
$endgroup$
– Luc
Dec 5 '18 at 16:51
$begingroup$
The problem is that I use the N-R method in a program and I don't know some parameters of my function $f(x,y)$. That means I cannot make a "reasonable guess". What I do know is that $f(x,y)$ is nonlinear. I would need a formula that determines wether the method converges in terms of the function $f(x,y)$.
$endgroup$
– Luc
Dec 5 '18 at 16:51
add a comment |
$begingroup$
No, in general there is no easy way to determine whether or not Newton-Raphson will converge for a given starting value, other than to try it. The region of convergence can be very complicated, and may have a fractal boundary.
answered Dec 5 '18 at 16:59
Robert IsraelRobert Israel
321k23210463
$endgroup$
add a comment |
$begingroup$
No, in general there is no easy way to determine whether or not Newton-Raphson will converge for a given starting value, other than to try it. The region of convergence can be very complicated, and may have a fractal boundary.
answered Dec 5 '18 at 16:59
Robert IsraelRobert Israel
321k23210463
$endgroup$
add a comment |
$begingroup$
No, in general there is no easy way to determine whether or not Newton-Raphson will converge for a given starting value, other than to try it. The region of convergence can be very complicated, and may have a fractal boundary.
answered Dec 5 '18 at 16:59
Robert IsraelRobert Israel
321k23210463
$endgroup$
No, in general there is no easy way to determine whether or not Newton-Raphson will converge for a given starting value, other than to try it. The region of convergence can be very complicated, and may have a fractal boundary.
answered Dec 5 '18 at 16:59
Robert IsraelRobert Israel
321k23210463
answered Dec 5 '18 at 16:59
Robert IsraelRobert Israel
321k23210463
answered Dec 5 '18 at 16:59
Robert IsraelRobert Israel
321k23210463
answered Dec 5 '18 at 16:59
Robert IsraelRobert Israel
321k23210463
321k23210463
add a comment |
add a comment |
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