Information about Newton's method at a certain stepNewton's Method for Roots of PolynomialsProve or disprove - Newton's method convergence in higher dimensionsHow to solve for the interval of convergence in Newton's Method?In practice, what does it mean for the Newton's method to converge quadratically (when it converges)?Newton's method for nth roots of complex numbersUsing of Newton's method to find a complex root of a polynomialUse Newton's method on a constant?Convergence of a variant of Newton's MethodConditions for Newton's method to converge.Global convergence with Newton method for nonlinear systems
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Information about Newton's method at a certain step
Newton's Method for Roots of PolynomialsProve or disprove - Newton's method convergence in higher dimensionsHow to solve for the interval of convergence in Newton's Method?In practice, what does it mean for the Newton's method to converge quadratically (when it converges)?Newton's method for nth roots of complex numbersUsing of Newton's method to find a complex root of a polynomialUse Newton's method on a constant?Convergence of a variant of Newton's MethodConditions for Newton's method to converge.Global convergence with Newton method for nonlinear systems
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We are preforming Newton's method with step $F'(x_n)Delta x=-F(x_n)$, $x_n+1=Delta x +x_n$, where $F:mathbbR^n rightarrow mathbbR^n$. We can assume that it will converge. At a give time we know the values of $x_0$ up to $x_k$.
Can we use $||F'(x_0)-F'(x_k)||$ to say anything about the convergence?
newton-raphson
asked Mar 14 at 16:49
MathiasMathias
64
$endgroup$
add a comment |
$begingroup$
We are preforming Newton's method with step $F'(x_n)Delta x=-F(x_n)$, $x_n+1=Delta x +x_n$, where $F:mathbbR^n rightarrow mathbbR^n$. We can assume that it will converge. At a give time we know the values of $x_0$ up to $x_k$.
Can we use $||F'(x_0)-F'(x_k)||$ to say anything about the convergence?
newton-raphson
asked Mar 14 at 16:49
MathiasMathias
64
$endgroup$
$begingroup$
I think one can construct a parameterised example where the norm of the difference is arbitrary and the sequence of iterates is the same, so no. The reason is that the norm is too 'coarse'. In practice, it gives an estimate of how 'linear' the problem is.
$endgroup$
– copper.hat
Mar 14 at 16:57
add a comment |
$begingroup$
We are preforming Newton's method with step $F'(x_n)Delta x=-F(x_n)$, $x_n+1=Delta x +x_n$, where $F:mathbbR^n rightarrow mathbbR^n$. We can assume that it will converge. At a give time we know the values of $x_0$ up to $x_k$.
Can we use $||F'(x_0)-F'(x_k)||$ to say anything about the convergence?
newton-raphson
asked Mar 14 at 16:49
MathiasMathias
64
$endgroup$
We are preforming Newton's method with step $F'(x_n)Delta x=-F(x_n)$, $x_n+1=Delta x +x_n$, where $F:mathbbR^n rightarrow mathbbR^n$. We can assume that it will converge. At a give time we know the values of $x_0$ up to $x_k$.
Can we use $||F'(x_0)-F'(x_k)||$ to say anything about the convergence?
newton-raphson
newton-raphson
asked Mar 14 at 16:49
MathiasMathias
64
asked Mar 14 at 16:49
MathiasMathias
64
asked Mar 14 at 16:49
MathiasMathias
64
asked Mar 14 at 16:49
MathiasMathias
64
asked Mar 14 at 16:49
MathiasMathias
64
64
$begingroup$
I think one can construct a parameterised example where the norm of the difference is arbitrary and the sequence of iterates is the same, so no. The reason is that the norm is too 'coarse'. In practice, it gives an estimate of how 'linear' the problem is.
$endgroup$
– copper.hat
Mar 14 at 16:57
add a comment |
$begingroup$
I think one can construct a parameterised example where the norm of the difference is arbitrary and the sequence of iterates is the same, so no. The reason is that the norm is too 'coarse'. In practice, it gives an estimate of how 'linear' the problem is.
$endgroup$
– copper.hat
Mar 14 at 16:57
$begingroup$
I think one can construct a parameterised example where the norm of the difference is arbitrary and the sequence of iterates is the same, so no. The reason is that the norm is too 'coarse'. In practice, it gives an estimate of how 'linear' the problem is.
$endgroup$
– copper.hat
Mar 14 at 16:57
$begingroup$
I think one can construct a parameterised example where the norm of the difference is arbitrary and the sequence of iterates is the same, so no. The reason is that the norm is too 'coarse'. In practice, it gives an estimate of how 'linear' the problem is.
$endgroup$
– copper.hat
Mar 14 at 16:57
add a comment |
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$begingroup$
I think one can construct a parameterised example where the norm of the difference is arbitrary and the sequence of iterates is the same, so no. The reason is that the norm is too 'coarse'. In practice, it gives an estimate of how 'linear' the problem is.
$endgroup$
– copper.hat
Mar 14 at 16:57