points at which newtons method fail
$begingroup$
Consider the system
$$ begin{align*} x-1 &= 0\ xy-1 &= 0\ end{align*}$$
For what starting point will Newton's method fail for solving this
system? Explain
Try
I know for multidimensional problems, the newton's algorithm is similar to the single variable case: we have
where
$Df(x_k)$ is the jacobian matrix and $x_0$ is the starting point. In our case
$$ f(x,y) = (x-1, xy-1)$$ so
the jacobian is
$$ Df(x,y) = begin{bmatrix} 1 & 0 \ y & x end{bmatrix} $$
Notice that the determinant of jacobian is $x$ and this better not be $0$. Therefore, any $x_0 = (x,y)$ will do unless $x=0$. In other words, the points $(0,alpha)$ will fail if used as starting point for newton method.
Is this correct?
numerical-methods nonlinear-system
asked Jan 21 at 7:56
Jimmy SabaterJimmy Sabater
2,714323
$endgroup$
add a comment |
$begingroup$
Consider the system
$$ begin{align*} x-1 &= 0\ xy-1 &= 0\ end{align*}$$
For what starting point will Newton's method fail for solving this
system? Explain
Try
I know for multidimensional problems, the newton's algorithm is similar to the single variable case: we have
where
$Df(x_k)$ is the jacobian matrix and $x_0$ is the starting point. In our case
$$ f(x,y) = (x-1, xy-1)$$ so
the jacobian is
$$ Df(x,y) = begin{bmatrix} 1 & 0 \ y & x end{bmatrix} $$
Notice that the determinant of jacobian is $x$ and this better not be $0$. Therefore, any $x_0 = (x,y)$ will do unless $x=0$. In other words, the points $(0,alpha)$ will fail if used as starting point for newton method.
Is this correct?
numerical-methods nonlinear-system
asked Jan 21 at 7:56
Jimmy SabaterJimmy Sabater
2,714323
$endgroup$
1
$begingroup$
Looks good. Depending on how much work you want to do for the problem, you may consider this: for what starting points will Newton's method work for one step, and then on the second step something bad happens?
$endgroup$
– obscurans
Jan 21 at 8:00
add a comment |
$begingroup$
Consider the system
$$ begin{align*} x-1 &= 0\ xy-1 &= 0\ end{align*}$$
For what starting point will Newton's method fail for solving this
system? Explain
Try
I know for multidimensional problems, the newton's algorithm is similar to the single variable case: we have
where
$Df(x_k)$ is the jacobian matrix and $x_0$ is the starting point. In our case
$$ f(x,y) = (x-1, xy-1)$$ so
the jacobian is
$$ Df(x,y) = begin{bmatrix} 1 & 0 \ y & x end{bmatrix} $$
Notice that the determinant of jacobian is $x$ and this better not be $0$. Therefore, any $x_0 = (x,y)$ will do unless $x=0$. In other words, the points $(0,alpha)$ will fail if used as starting point for newton method.
Is this correct?
numerical-methods nonlinear-system
asked Jan 21 at 7:56
Jimmy SabaterJimmy Sabater
2,714323
$endgroup$
Consider the system
$$ begin{align*} x-1 &= 0\ xy-1 &= 0\ end{align*}$$
For what starting point will Newton's method fail for solving this
system? Explain
Try
I know for multidimensional problems, the newton's algorithm is similar to the single variable case: we have
where
$Df(x_k)$ is the jacobian matrix and $x_0$ is the starting point. In our case
$$ f(x,y) = (x-1, xy-1)$$ so
the jacobian is
$$ Df(x,y) = begin{bmatrix} 1 & 0 \ y & x end{bmatrix} $$
Notice that the determinant of jacobian is $x$ and this better not be $0$. Therefore, any $x_0 = (x,y)$ will do unless $x=0$. In other words, the points $(0,alpha)$ will fail if used as starting point for newton method.
Is this correct?
numerical-methods nonlinear-system
numerical-methods nonlinear-system
asked Jan 21 at 7:56
Jimmy SabaterJimmy Sabater
2,714323
asked Jan 21 at 7:56
Jimmy SabaterJimmy Sabater
2,714323
asked Jan 21 at 7:56
Jimmy SabaterJimmy Sabater
2,714323
asked Jan 21 at 7:56
Jimmy SabaterJimmy Sabater
2,714323
asked Jan 21 at 7:56
Jimmy SabaterJimmy Sabater
2,714323
2,714323
1
$begingroup$
Looks good. Depending on how much work you want to do for the problem, you may consider this: for what starting points will Newton's method work for one step, and then on the second step something bad happens?
$endgroup$
– obscurans
Jan 21 at 8:00
add a comment |
1
$begingroup$
Looks good. Depending on how much work you want to do for the problem, you may consider this: for what starting points will Newton's method work for one step, and then on the second step something bad happens?
$endgroup$
– obscurans
Jan 21 at 8:00
1
1
$begingroup$
Looks good. Depending on how much work you want to do for the problem, you may consider this: for what starting points will Newton's method work for one step, and then on the second step something bad happens?
$endgroup$
– obscurans
Jan 21 at 8:00
$begingroup$
Looks good. Depending on how much work you want to do for the problem, you may consider this: for what starting points will Newton's method work for one step, and then on the second step something bad happens?
$endgroup$
– obscurans
Jan 21 at 8:00
add a comment |
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$begingroup$
Looks good. Depending on how much work you want to do for the problem, you may consider this: for what starting points will Newton's method work for one step, and then on the second step something bad happens?
$endgroup$
– obscurans
Jan 21 at 8:00