Mixing
Newton's method for a vector field
Clash Royale CLAN TAG#URR8PPP
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Let $f : mathbbR^n to mathbbR^n$ be $C^2$ and let $f(x^*)=0$. Since
$$f(x^*) approx f(x) + Df(x) (x^* - x)$$
we can have the iterative procedure
$$x_k+1 = x_k - Df(x_k)^-1 f(x_k)$$
Is $G(x): = x - Df(x)^-1 f(x)$ invertible near $x=x_0$? Are there any results on the convergence of this procedure?
I tried to use the inverse function theorem. However, I do not know how to prove that $$DG(x_0) = I - D(Df(x)^-1 f(x)) bigg |_x_0$$ is non-singular.
optimization numerical-methods dynamical-systems nonlinear-optimization newton-raphson
edited Jul 19 at 10:29
Rodrigo de Azevedo
12.6k41751
asked Jul 18 at 17:50
Arthur
19812
 |Â
show 3 more comments
up vote
5
down vote
favorite
Let $f : mathbbR^n to mathbbR^n$ be $C^2$ and let $f(x^*)=0$. Since
$$f(x^*) approx f(x) + Df(x) (x^* - x)$$
we can have the iterative procedure
$$x_k+1 = x_k - Df(x_k)^-1 f(x_k)$$
Is $G(x): = x - Df(x)^-1 f(x)$ invertible near $x=x_0$? Are there any results on the convergence of this procedure?
I tried to use the inverse function theorem. However, I do not know how to prove that $$DG(x_0) = I - D(Df(x)^-1 f(x)) bigg |_x_0$$ is non-singular.
optimization numerical-methods dynamical-systems nonlinear-optimization newton-raphson
edited Jul 19 at 10:29
Rodrigo de Azevedo
12.6k41751
asked Jul 18 at 17:50
Arthur
19812
Do you have an expression for its determinant?
– Emil
Jul 18 at 18:29
Perhaps the method needs to be contractive? I think Lyapunov stability en.m.wikipedia.org/wiki/Lyapunov_stability means it does not blow up. Or perhaps it was the energy method one used. I think Taylor expansions can be used to find bounds of the error ?
– Emil
Jul 18 at 18:39
I was likely thinking about the fixed point theorem when talking about the mapping being contractive: en.m.wikipedia.org/wiki/Banach_fixed-point_theorem. I have some faint memory about negative eigenvalues being good as well, but can't remember the exact motivation, probably something about matrix powers.
– Emil
Jul 18 at 18:51
Asking for $G$ to be invertible is a strange question: ideally $G(x)=x_0$ and $G$ is not invertible! This will happen when $f$ is linear. The general situation seems quite complex and I'm not sure I expect $G$ invertible for "typical" $f$.
– user7530
Jul 18 at 18:58
We know that if $B:= I + A$ and $||A|| < epsilon$, then $det(B) not = 0$. I tried to use this fact. However, it is not clear to me why the norm of jacobian of $Df(x)^-1 f(x)$ at $x_0$ is small?
– Arthur
Jul 18 at 19:08
 |Â
show 3 more comments
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Let $f : mathbbR^n to mathbbR^n$ be $C^2$ and let $f(x^*)=0$. Since
$$f(x^*) approx f(x) + Df(x) (x^* - x)$$
we can have the iterative procedure
$$x_k+1 = x_k - Df(x_k)^-1 f(x_k)$$
Is $G(x): = x - Df(x)^-1 f(x)$ invertible near $x=x_0$? Are there any results on the convergence of this procedure?
I tried to use the inverse function theorem. However, I do not know how to prove that $$DG(x_0) = I - D(Df(x)^-1 f(x)) bigg |_x_0$$ is non-singular.
optimization numerical-methods dynamical-systems nonlinear-optimization newton-raphson
edited Jul 19 at 10:29
Rodrigo de Azevedo
12.6k41751
asked Jul 18 at 17:50
Arthur
19812
Let $f : mathbbR^n to mathbbR^n$ be $C^2$ and let $f(x^*)=0$. Since
$$f(x^*) approx f(x) + Df(x) (x^* - x)$$
we can have the iterative procedure
$$x_k+1 = x_k - Df(x_k)^-1 f(x_k)$$
Is $G(x): = x - Df(x)^-1 f(x)$ invertible near $x=x_0$? Are there any results on the convergence of this procedure?
I tried to use the inverse function theorem. However, I do not know how to prove that $$DG(x_0) = I - D(Df(x)^-1 f(x)) bigg |_x_0$$ is non-singular.
optimization numerical-methods dynamical-systems nonlinear-optimization newton-raphson
edited Jul 19 at 10:29
Rodrigo de Azevedo
12.6k41751
asked Jul 18 at 17:50
Arthur
19812
edited Jul 19 at 10:29
Rodrigo de Azevedo
12.6k41751
edited Jul 19 at 10:29
Rodrigo de Azevedo
12.6k41751
edited Jul 19 at 10:29
Rodrigo de Azevedo
12.6k41751
12.6k41751
asked Jul 18 at 17:50
Arthur
19812
asked Jul 18 at 17:50
Arthur
19812
asked Jul 18 at 17:50
Arthur
19812
19812
Do you have an expression for its determinant?
– Emil
Jul 18 at 18:29
Perhaps the method needs to be contractive? I think Lyapunov stability en.m.wikipedia.org/wiki/Lyapunov_stability means it does not blow up. Or perhaps it was the energy method one used. I think Taylor expansions can be used to find bounds of the error ?
– Emil
Jul 18 at 18:39
I was likely thinking about the fixed point theorem when talking about the mapping being contractive: en.m.wikipedia.org/wiki/Banach_fixed-point_theorem. I have some faint memory about negative eigenvalues being good as well, but can't remember the exact motivation, probably something about matrix powers.
– Emil
Jul 18 at 18:51
Asking for $G$ to be invertible is a strange question: ideally $G(x)=x_0$ and $G$ is not invertible! This will happen when $f$ is linear. The general situation seems quite complex and I'm not sure I expect $G$ invertible for "typical" $f$.
– user7530
Jul 18 at 18:58
We know that if $B:= I + A$ and $||A|| < epsilon$, then $det(B) not = 0$. I tried to use this fact. However, it is not clear to me why the norm of jacobian of $Df(x)^-1 f(x)$ at $x_0$ is small?
– Arthur
Jul 18 at 19:08
 |Â
show 3 more comments
Do you have an expression for its determinant?
– Emil
Jul 18 at 18:29
Perhaps the method needs to be contractive? I think Lyapunov stability en.m.wikipedia.org/wiki/Lyapunov_stability means it does not blow up. Or perhaps it was the energy method one used. I think Taylor expansions can be used to find bounds of the error ?
– Emil
Jul 18 at 18:39
I was likely thinking about the fixed point theorem when talking about the mapping being contractive: en.m.wikipedia.org/wiki/Banach_fixed-point_theorem. I have some faint memory about negative eigenvalues being good as well, but can't remember the exact motivation, probably something about matrix powers.
– Emil
Jul 18 at 18:51
Asking for $G$ to be invertible is a strange question: ideally $G(x)=x_0$ and $G$ is not invertible! This will happen when $f$ is linear. The general situation seems quite complex and I'm not sure I expect $G$ invertible for "typical" $f$.
– user7530
Jul 18 at 18:58
We know that if $B:= I + A$ and $||A|| < epsilon$, then $det(B) not = 0$. I tried to use this fact. However, it is not clear to me why the norm of jacobian of $Df(x)^-1 f(x)$ at $x_0$ is small?
– Arthur
Jul 18 at 19:08
Do you have an expression for its determinant?
– Emil
Jul 18 at 18:29
Do you have an expression for its determinant?
– Emil
Jul 18 at 18:29
Perhaps the method needs to be contractive? I think Lyapunov stability en.m.wikipedia.org/wiki/Lyapunov_stability means it does not blow up. Or perhaps it was the energy method one used. I think Taylor expansions can be used to find bounds of the error ?
– Emil
Jul 18 at 18:39
Perhaps the method needs to be contractive? I think Lyapunov stability en.m.wikipedia.org/wiki/Lyapunov_stability means it does not blow up. Or perhaps it was the energy method one used. I think Taylor expansions can be used to find bounds of the error ?
– Emil
Jul 18 at 18:39
I was likely thinking about the fixed point theorem when talking about the mapping being contractive: en.m.wikipedia.org/wiki/Banach_fixed-point_theorem. I have some faint memory about negative eigenvalues being good as well, but can't remember the exact motivation, probably something about matrix powers.
– Emil
Jul 18 at 18:51
I was likely thinking about the fixed point theorem when talking about the mapping being contractive: en.m.wikipedia.org/wiki/Banach_fixed-point_theorem. I have some faint memory about negative eigenvalues being good as well, but can't remember the exact motivation, probably something about matrix powers.
– Emil
Jul 18 at 18:51
Asking for $G$ to be invertible is a strange question: ideally $G(x)=x_0$ and $G$ is not invertible! This will happen when $f$ is linear. The general situation seems quite complex and I'm not sure I expect $G$ invertible for "typical" $f$.
– user7530
Jul 18 at 18:58
Asking for $G$ to be invertible is a strange question: ideally $G(x)=x_0$ and $G$ is not invertible! This will happen when $f$ is linear. The general situation seems quite complex and I'm not sure I expect $G$ invertible for "typical" $f$.
– user7530
Jul 18 at 18:58
We know that if $B:= I + A$ and $||A|| < epsilon$, then $det(B) not = 0$. I tried to use this fact. However, it is not clear to me why the norm of jacobian of $Df(x)^-1 f(x)$ at $x_0$ is small?
– Arthur
Jul 18 at 19:08
We know that if $B:= I + A$ and $||A|| < epsilon$, then $det(B) not = 0$. I tried to use this fact. However, it is not clear to me why the norm of jacobian of $Df(x)^-1 f(x)$ at $x_0$ is small?
– Arthur
Jul 18 at 19:08
 |Â
show 3 more comments
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
This even doesn't hold for 1-D function since $$DG(x_0)=1-left(dfracf(x)f'(x)right)'|_x_0=1-dfracf'^2(x_0)-f(x_0)f''(x_0)f'^2(x_0)=0$$which is singular. To show that for higher dimensions let's define $$Df^-1(x)=[a_ij(x)]\Df^-1(x)f(x)=[c_i(x)]$$and $$D(Df^-1(x)f(x))=[b_ij(x)]$$therefore $$c_i(x)=sum_k=1^na_ik(x)f_k(x)$$where $$f(x)=beginbmatrixf_1(x)\f_2(x)\.\.\.\f_n(x)endbmatrix$$also$$b_ij(x)=dfracpartial c_ipartial x_j=sum_k=1^ndfracpartial a_ik(x)partial x_jf_k(x)+sum_k=1^na_ik(x)dfracpartial f_k(x)partial x_j$$when substituting $x=x_0$, $f_k(x_0)$ becomes zero since $f(x_0)=0$ therefore$$b_ij(x_0)=sum_k=1^na_ik(x_0)dfracpartial f_k(x)partial x_j|_x_0$$but $dfracpartial f_k(x)partial x_j|_x_0$ is the (k,j)th entry of $Df(x_0)$ which by substitution leads to $$D(Df^-1(x)f(x))|_x_0=Df^-1(x_0)Df(x_0)=I$$or $$Large DG(x_0)=0$$
answered Jul 22 at 20:26
Mostafa Ayaz
8,6023630
add a comment |Â
up vote
2
down vote
Hint.
The process
$$
x_k = x_k-1-Df^-1(x_k-1)f(x_k-1)
$$
can be written as
$$
x_k = phi(x_k-1)\
x_k-1 = phi(x_k-2)
$$
and then
$$
|x_k-x_k-1| = |phi(x_k-1)-phi(x_k-2)| = |phi'(xi)||x_k-1-x_k-2|
$$
for a suitable $xiin (x_k-1-x_k-2)$ hence is sufficient that $|phi'(xi)| < 1$ for convergence
answered Jul 18 at 19:06
Cesareo
5,7722412
Thank you for the comment on the convergence. I guess $f(x_k)$ is a typo. It should be $f(x_k-1)$.
– Arthur
Jul 18 at 19:13
@Arthur Yes. It was a typo already corrected. Thanks.
– Cesareo
Jul 18 at 19:15
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
This even doesn't hold for 1-D function since $$DG(x_0)=1-left(dfracf(x)f'(x)right)'|_x_0=1-dfracf'^2(x_0)-f(x_0)f''(x_0)f'^2(x_0)=0$$which is singular. To show that for higher dimensions let's define $$Df^-1(x)=[a_ij(x)]\Df^-1(x)f(x)=[c_i(x)]$$and $$D(Df^-1(x)f(x))=[b_ij(x)]$$therefore $$c_i(x)=sum_k=1^na_ik(x)f_k(x)$$where $$f(x)=beginbmatrixf_1(x)\f_2(x)\.\.\.\f_n(x)endbmatrix$$also$$b_ij(x)=dfracpartial c_ipartial x_j=sum_k=1^ndfracpartial a_ik(x)partial x_jf_k(x)+sum_k=1^na_ik(x)dfracpartial f_k(x)partial x_j$$when substituting $x=x_0$, $f_k(x_0)$ becomes zero since $f(x_0)=0$ therefore$$b_ij(x_0)=sum_k=1^na_ik(x_0)dfracpartial f_k(x)partial x_j|_x_0$$but $dfracpartial f_k(x)partial x_j|_x_0$ is the (k,j)th entry of $Df(x_0)$ which by substitution leads to $$D(Df^-1(x)f(x))|_x_0=Df^-1(x_0)Df(x_0)=I$$or $$Large DG(x_0)=0$$
answered Jul 22 at 20:26
Mostafa Ayaz
8,6023630
add a comment |Â
up vote
1
down vote
accepted
This even doesn't hold for 1-D function since $$DG(x_0)=1-left(dfracf(x)f'(x)right)'|_x_0=1-dfracf'^2(x_0)-f(x_0)f''(x_0)f'^2(x_0)=0$$which is singular. To show that for higher dimensions let's define $$Df^-1(x)=[a_ij(x)]\Df^-1(x)f(x)=[c_i(x)]$$and $$D(Df^-1(x)f(x))=[b_ij(x)]$$therefore $$c_i(x)=sum_k=1^na_ik(x)f_k(x)$$where $$f(x)=beginbmatrixf_1(x)\f_2(x)\.\.\.\f_n(x)endbmatrix$$also$$b_ij(x)=dfracpartial c_ipartial x_j=sum_k=1^ndfracpartial a_ik(x)partial x_jf_k(x)+sum_k=1^na_ik(x)dfracpartial f_k(x)partial x_j$$when substituting $x=x_0$, $f_k(x_0)$ becomes zero since $f(x_0)=0$ therefore$$b_ij(x_0)=sum_k=1^na_ik(x_0)dfracpartial f_k(x)partial x_j|_x_0$$but $dfracpartial f_k(x)partial x_j|_x_0$ is the (k,j)th entry of $Df(x_0)$ which by substitution leads to $$D(Df^-1(x)f(x))|_x_0=Df^-1(x_0)Df(x_0)=I$$or $$Large DG(x_0)=0$$
answered Jul 22 at 20:26
Mostafa Ayaz
8,6023630
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
This even doesn't hold for 1-D function since $$DG(x_0)=1-left(dfracf(x)f'(x)right)'|_x_0=1-dfracf'^2(x_0)-f(x_0)f''(x_0)f'^2(x_0)=0$$which is singular. To show that for higher dimensions let's define $$Df^-1(x)=[a_ij(x)]\Df^-1(x)f(x)=[c_i(x)]$$and $$D(Df^-1(x)f(x))=[b_ij(x)]$$therefore $$c_i(x)=sum_k=1^na_ik(x)f_k(x)$$where $$f(x)=beginbmatrixf_1(x)\f_2(x)\.\.\.\f_n(x)endbmatrix$$also$$b_ij(x)=dfracpartial c_ipartial x_j=sum_k=1^ndfracpartial a_ik(x)partial x_jf_k(x)+sum_k=1^na_ik(x)dfracpartial f_k(x)partial x_j$$when substituting $x=x_0$, $f_k(x_0)$ becomes zero since $f(x_0)=0$ therefore$$b_ij(x_0)=sum_k=1^na_ik(x_0)dfracpartial f_k(x)partial x_j|_x_0$$but $dfracpartial f_k(x)partial x_j|_x_0$ is the (k,j)th entry of $Df(x_0)$ which by substitution leads to $$D(Df^-1(x)f(x))|_x_0=Df^-1(x_0)Df(x_0)=I$$or $$Large DG(x_0)=0$$
answered Jul 22 at 20:26
Mostafa Ayaz
8,6023630
This even doesn't hold for 1-D function since $$DG(x_0)=1-left(dfracf(x)f'(x)right)'|_x_0=1-dfracf'^2(x_0)-f(x_0)f''(x_0)f'^2(x_0)=0$$which is singular. To show that for higher dimensions let's define $$Df^-1(x)=[a_ij(x)]\Df^-1(x)f(x)=[c_i(x)]$$and $$D(Df^-1(x)f(x))=[b_ij(x)]$$therefore $$c_i(x)=sum_k=1^na_ik(x)f_k(x)$$where $$f(x)=beginbmatrixf_1(x)\f_2(x)\.\.\.\f_n(x)endbmatrix$$also$$b_ij(x)=dfracpartial c_ipartial x_j=sum_k=1^ndfracpartial a_ik(x)partial x_jf_k(x)+sum_k=1^na_ik(x)dfracpartial f_k(x)partial x_j$$when substituting $x=x_0$, $f_k(x_0)$ becomes zero since $f(x_0)=0$ therefore$$b_ij(x_0)=sum_k=1^na_ik(x_0)dfracpartial f_k(x)partial x_j|_x_0$$but $dfracpartial f_k(x)partial x_j|_x_0$ is the (k,j)th entry of $Df(x_0)$ which by substitution leads to $$D(Df^-1(x)f(x))|_x_0=Df^-1(x_0)Df(x_0)=I$$or $$Large DG(x_0)=0$$
answered Jul 22 at 20:26
Mostafa Ayaz
8,6023630
answered Jul 22 at 20:26
Mostafa Ayaz
8,6023630
answered Jul 22 at 20:26
Mostafa Ayaz
8,6023630
answered Jul 22 at 20:26
Mostafa Ayaz
8,6023630
8,6023630
add a comment |Â
add a comment |Â
up vote
2
down vote
Hint.
The process
$$
x_k = x_k-1-Df^-1(x_k-1)f(x_k-1)
$$
can be written as
$$
x_k = phi(x_k-1)\
x_k-1 = phi(x_k-2)
$$
and then
$$
|x_k-x_k-1| = |phi(x_k-1)-phi(x_k-2)| = |phi'(xi)||x_k-1-x_k-2|
$$
for a suitable $xiin (x_k-1-x_k-2)$ hence is sufficient that $|phi'(xi)| < 1$ for convergence
answered Jul 18 at 19:06
Cesareo
5,7722412
Thank you for the comment on the convergence. I guess $f(x_k)$ is a typo. It should be $f(x_k-1)$.
– Arthur
Jul 18 at 19:13
@Arthur Yes. It was a typo already corrected. Thanks.
– Cesareo
Jul 18 at 19:15
add a comment |Â
up vote
2
down vote
Hint.
The process
$$
x_k = x_k-1-Df^-1(x_k-1)f(x_k-1)
$$
can be written as
$$
x_k = phi(x_k-1)\
x_k-1 = phi(x_k-2)
$$
and then
$$
|x_k-x_k-1| = |phi(x_k-1)-phi(x_k-2)| = |phi'(xi)||x_k-1-x_k-2|
$$
for a suitable $xiin (x_k-1-x_k-2)$ hence is sufficient that $|phi'(xi)| < 1$ for convergence
answered Jul 18 at 19:06
Cesareo
5,7722412
Thank you for the comment on the convergence. I guess $f(x_k)$ is a typo. It should be $f(x_k-1)$.
– Arthur
Jul 18 at 19:13
@Arthur Yes. It was a typo already corrected. Thanks.
– Cesareo
Jul 18 at 19:15
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint.
The process
$$
x_k = x_k-1-Df^-1(x_k-1)f(x_k-1)
$$
can be written as
$$
x_k = phi(x_k-1)\
x_k-1 = phi(x_k-2)
$$
and then
$$
|x_k-x_k-1| = |phi(x_k-1)-phi(x_k-2)| = |phi'(xi)||x_k-1-x_k-2|
$$
for a suitable $xiin (x_k-1-x_k-2)$ hence is sufficient that $|phi'(xi)| < 1$ for convergence
answered Jul 18 at 19:06
Cesareo
5,7722412
Hint.
The process
$$
x_k = x_k-1-Df^-1(x_k-1)f(x_k-1)
$$
can be written as
$$
x_k = phi(x_k-1)\
x_k-1 = phi(x_k-2)
$$
and then
$$
|x_k-x_k-1| = |phi(x_k-1)-phi(x_k-2)| = |phi'(xi)||x_k-1-x_k-2|
$$
for a suitable $xiin (x_k-1-x_k-2)$ hence is sufficient that $|phi'(xi)| < 1$ for convergence
answered Jul 18 at 19:06
Cesareo
5,7722412
edited Jul 18 at 19:14
answered Jul 18 at 19:06
Cesareo
5,7722412
answered Jul 18 at 19:06
Cesareo
5,7722412
answered Jul 18 at 19:06
Cesareo
5,7722412
5,7722412
Thank you for the comment on the convergence. I guess $f(x_k)$ is a typo. It should be $f(x_k-1)$.
– Arthur
Jul 18 at 19:13
@Arthur Yes. It was a typo already corrected. Thanks.
– Cesareo
Jul 18 at 19:15
add a comment |Â
Thank you for the comment on the convergence. I guess $f(x_k)$ is a typo. It should be $f(x_k-1)$.
– Arthur
Jul 18 at 19:13
@Arthur Yes. It was a typo already corrected. Thanks.
– Cesareo
Jul 18 at 19:15
Thank you for the comment on the convergence. I guess $f(x_k)$ is a typo. It should be $f(x_k-1)$.
– Arthur
Jul 18 at 19:13
Thank you for the comment on the convergence. I guess $f(x_k)$ is a typo. It should be $f(x_k-1)$.
– Arthur
Jul 18 at 19:13
@Arthur Yes. It was a typo already corrected. Thanks.
– Cesareo
Jul 18 at 19:15
@Arthur Yes. It was a typo already corrected. Thanks.
– Cesareo
Jul 18 at 19:15
add a comment |Â
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Do you have an expression for its determinant?
– Emil
Jul 18 at 18:29
Perhaps the method needs to be contractive? I think Lyapunov stability en.m.wikipedia.org/wiki/Lyapunov_stability means it does not blow up. Or perhaps it was the energy method one used. I think Taylor expansions can be used to find bounds of the error ?
– Emil
Jul 18 at 18:39
I was likely thinking about the fixed point theorem when talking about the mapping being contractive: en.m.wikipedia.org/wiki/Banach_fixed-point_theorem. I have some faint memory about negative eigenvalues being good as well, but can't remember the exact motivation, probably something about matrix powers.
– Emil
Jul 18 at 18:51
Asking for $G$ to be invertible is a strange question: ideally $G(x)=x_0$ and $G$ is not invertible! This will happen when $f$ is linear. The general situation seems quite complex and I'm not sure I expect $G$ invertible for "typical" $f$.
– user7530
Jul 18 at 18:58
We know that if $B:= I + A$ and $||A|| < epsilon$, then $det(B) not = 0$. I tried to use this fact. However, it is not clear to me why the norm of jacobian of $Df(x)^-1 f(x)$ at $x_0$ is small?
– Arthur
Jul 18 at 19:08