Mixing
Newton's method - optimal step size
I am wondering how to prove the following affirmation:
We have : Min $f(x) = frac{1}{2}x^TDx - c^Tx$
with $f: R^n → R^1$ and $D$ a symetric positive matrix of size $nxn$.
Suppose $d_k = -nabla f(x_k)$ is a descent direction in $x_k$.
Show that the optimal solution to the problem $minlimits_{σ_k > 0} f(x_k + σ_kd_k)$ is:
$σ_k = -frac{∇f(x_k)^Td_k}{d_k^TDd_k}$.
Thanks for your help!
Louis
optimization proof-writing nonlinear-optimization
asked Nov 20 '18 at 21:50
Louis-Philippe Noël
123
add a comment |
I am wondering how to prove the following affirmation:
We have : Min $f(x) = frac{1}{2}x^TDx - c^Tx$
with $f: R^n → R^1$ and $D$ a symetric positive matrix of size $nxn$.
Suppose $d_k = -nabla f(x_k)$ is a descent direction in $x_k$.
Show that the optimal solution to the problem $minlimits_{σ_k > 0} f(x_k + σ_kd_k)$ is:
$σ_k = -frac{∇f(x_k)^Td_k}{d_k^TDd_k}$.
Thanks for your help!
Louis
optimization proof-writing nonlinear-optimization
asked Nov 20 '18 at 21:50
Louis-Philippe Noël
123
add a comment |
I am wondering how to prove the following affirmation:
We have : Min $f(x) = frac{1}{2}x^TDx - c^Tx$
with $f: R^n → R^1$ and $D$ a symetric positive matrix of size $nxn$.
Suppose $d_k = -nabla f(x_k)$ is a descent direction in $x_k$.
Show that the optimal solution to the problem $minlimits_{σ_k > 0} f(x_k + σ_kd_k)$ is:
$σ_k = -frac{∇f(x_k)^Td_k}{d_k^TDd_k}$.
Thanks for your help!
Louis
optimization proof-writing nonlinear-optimization
asked Nov 20 '18 at 21:50
Louis-Philippe Noël
123
I am wondering how to prove the following affirmation:
We have : Min $f(x) = frac{1}{2}x^TDx - c^Tx$
with $f: R^n → R^1$ and $D$ a symetric positive matrix of size $nxn$.
Suppose $d_k = -nabla f(x_k)$ is a descent direction in $x_k$.
Show that the optimal solution to the problem $minlimits_{σ_k > 0} f(x_k + σ_kd_k)$ is:
$σ_k = -frac{∇f(x_k)^Td_k}{d_k^TDd_k}$.
Thanks for your help!
Louis
optimization proof-writing nonlinear-optimization
optimization proof-writing nonlinear-optimization
asked Nov 20 '18 at 21:50
Louis-Philippe Noël
123
asked Nov 20 '18 at 21:50
Louis-Philippe Noël
123
asked Nov 20 '18 at 21:50
Louis-Philippe Noël
123
asked Nov 20 '18 at 21:50
Louis-Philippe Noël
123
asked Nov 20 '18 at 21:50
Louis-Philippe Noël
123
123
add a comment |
add a comment |
1 Answer
1
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oldest
votes
As this is a smooth convex ($D$ is symetric positive) problem, to minimize $sigma_kmapsto f(x_k+sigma_kd_k)$ you can search for $sigma_k$ such that
$$
frac{d}{dsigma_k}(sigma_kmapsto f(x_k+sigma_kd_k))=0
$$
hint:
$$
frac{d}{dsigma_k}(sigma_kmapsto f(x_k+sigma_kd_k))=langle d_k,nabla f(x_k+sigma_kd_k)rangle
$$
where $langle .,.rangle$ is the usual scalar product and $nabla f(x_k+sigma_kd_k)=D(x_k+sigma_kd_k)-c=nabla f(x_k)+sigma_kDd_k$
note:
This method is the gradient descent method with Cauchy step, not the Newton method. The Newton method includes second order correction
$$
x_{k+1}=x_k-H_f^{-1}(x_k)nabla f(x_k)
$$
where $H_f$ is the Hessian. With your $f$, $H_f(x_k)=D$ (constant) and $nabla f(x_k)=Dx_k-c$, thus:
$$
x_{k+1}=x_k-H_f^{-1}(x_k)nabla f(x_k)=x_k-D^{-1}D(x_k-c)=D^{-1}c
$$
and the method converge in one iteration to the solution $x^*=D^{-1}c$
In practice, if we do not want to solve $Dx=c$, we still use first order methods but not the steepest descent one. Its convergence is very slow when $D$ is badly-conditioned. The conjugate gradients method is generally a better choice.
answered Nov 20 '18 at 22:31
Picaud Vincent
1,21838
Thanks @Picaud Vincent! Exactly what I needed.
– Louis-Philippe Noël
Nov 21 '18 at 19:08
@Louis-PhilippeNoël so maybe you can voteup? :)
– Picaud Vincent
Nov 21 '18 at 19:15
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
As this is a smooth convex ($D$ is symetric positive) problem, to minimize $sigma_kmapsto f(x_k+sigma_kd_k)$ you can search for $sigma_k$ such that
$$
frac{d}{dsigma_k}(sigma_kmapsto f(x_k+sigma_kd_k))=0
$$
hint:
$$
frac{d}{dsigma_k}(sigma_kmapsto f(x_k+sigma_kd_k))=langle d_k,nabla f(x_k+sigma_kd_k)rangle
$$
where $langle .,.rangle$ is the usual scalar product and $nabla f(x_k+sigma_kd_k)=D(x_k+sigma_kd_k)-c=nabla f(x_k)+sigma_kDd_k$
note:
This method is the gradient descent method with Cauchy step, not the Newton method. The Newton method includes second order correction
$$
x_{k+1}=x_k-H_f^{-1}(x_k)nabla f(x_k)
$$
where $H_f$ is the Hessian. With your $f$, $H_f(x_k)=D$ (constant) and $nabla f(x_k)=Dx_k-c$, thus:
$$
x_{k+1}=x_k-H_f^{-1}(x_k)nabla f(x_k)=x_k-D^{-1}D(x_k-c)=D^{-1}c
$$
and the method converge in one iteration to the solution $x^*=D^{-1}c$
In practice, if we do not want to solve $Dx=c$, we still use first order methods but not the steepest descent one. Its convergence is very slow when $D$ is badly-conditioned. The conjugate gradients method is generally a better choice.
answered Nov 20 '18 at 22:31
Picaud Vincent
1,21838
Thanks @Picaud Vincent! Exactly what I needed.
– Louis-Philippe Noël
Nov 21 '18 at 19:08
@Louis-PhilippeNoël so maybe you can voteup? :)
– Picaud Vincent
Nov 21 '18 at 19:15
add a comment |
As this is a smooth convex ($D$ is symetric positive) problem, to minimize $sigma_kmapsto f(x_k+sigma_kd_k)$ you can search for $sigma_k$ such that
$$
frac{d}{dsigma_k}(sigma_kmapsto f(x_k+sigma_kd_k))=0
$$
hint:
$$
frac{d}{dsigma_k}(sigma_kmapsto f(x_k+sigma_kd_k))=langle d_k,nabla f(x_k+sigma_kd_k)rangle
$$
where $langle .,.rangle$ is the usual scalar product and $nabla f(x_k+sigma_kd_k)=D(x_k+sigma_kd_k)-c=nabla f(x_k)+sigma_kDd_k$
note:
This method is the gradient descent method with Cauchy step, not the Newton method. The Newton method includes second order correction
$$
x_{k+1}=x_k-H_f^{-1}(x_k)nabla f(x_k)
$$
where $H_f$ is the Hessian. With your $f$, $H_f(x_k)=D$ (constant) and $nabla f(x_k)=Dx_k-c$, thus:
$$
x_{k+1}=x_k-H_f^{-1}(x_k)nabla f(x_k)=x_k-D^{-1}D(x_k-c)=D^{-1}c
$$
and the method converge in one iteration to the solution $x^*=D^{-1}c$
In practice, if we do not want to solve $Dx=c$, we still use first order methods but not the steepest descent one. Its convergence is very slow when $D$ is badly-conditioned. The conjugate gradients method is generally a better choice.
answered Nov 20 '18 at 22:31
Picaud Vincent
1,21838
Thanks @Picaud Vincent! Exactly what I needed.
– Louis-Philippe Noël
Nov 21 '18 at 19:08
@Louis-PhilippeNoël so maybe you can voteup? :)
– Picaud Vincent
Nov 21 '18 at 19:15
add a comment |
As this is a smooth convex ($D$ is symetric positive) problem, to minimize $sigma_kmapsto f(x_k+sigma_kd_k)$ you can search for $sigma_k$ such that
$$
frac{d}{dsigma_k}(sigma_kmapsto f(x_k+sigma_kd_k))=0
$$
hint:
$$
frac{d}{dsigma_k}(sigma_kmapsto f(x_k+sigma_kd_k))=langle d_k,nabla f(x_k+sigma_kd_k)rangle
$$
where $langle .,.rangle$ is the usual scalar product and $nabla f(x_k+sigma_kd_k)=D(x_k+sigma_kd_k)-c=nabla f(x_k)+sigma_kDd_k$
note:
This method is the gradient descent method with Cauchy step, not the Newton method. The Newton method includes second order correction
$$
x_{k+1}=x_k-H_f^{-1}(x_k)nabla f(x_k)
$$
where $H_f$ is the Hessian. With your $f$, $H_f(x_k)=D$ (constant) and $nabla f(x_k)=Dx_k-c$, thus:
$$
x_{k+1}=x_k-H_f^{-1}(x_k)nabla f(x_k)=x_k-D^{-1}D(x_k-c)=D^{-1}c
$$
and the method converge in one iteration to the solution $x^*=D^{-1}c$
In practice, if we do not want to solve $Dx=c$, we still use first order methods but not the steepest descent one. Its convergence is very slow when $D$ is badly-conditioned. The conjugate gradients method is generally a better choice.
answered Nov 20 '18 at 22:31
Picaud Vincent
1,21838
As this is a smooth convex ($D$ is symetric positive) problem, to minimize $sigma_kmapsto f(x_k+sigma_kd_k)$ you can search for $sigma_k$ such that
$$
frac{d}{dsigma_k}(sigma_kmapsto f(x_k+sigma_kd_k))=0
$$
hint:
$$
frac{d}{dsigma_k}(sigma_kmapsto f(x_k+sigma_kd_k))=langle d_k,nabla f(x_k+sigma_kd_k)rangle
$$
where $langle .,.rangle$ is the usual scalar product and $nabla f(x_k+sigma_kd_k)=D(x_k+sigma_kd_k)-c=nabla f(x_k)+sigma_kDd_k$
note:
This method is the gradient descent method with Cauchy step, not the Newton method. The Newton method includes second order correction
$$
x_{k+1}=x_k-H_f^{-1}(x_k)nabla f(x_k)
$$
where $H_f$ is the Hessian. With your $f$, $H_f(x_k)=D$ (constant) and $nabla f(x_k)=Dx_k-c$, thus:
$$
x_{k+1}=x_k-H_f^{-1}(x_k)nabla f(x_k)=x_k-D^{-1}D(x_k-c)=D^{-1}c
$$
and the method converge in one iteration to the solution $x^*=D^{-1}c$
In practice, if we do not want to solve $Dx=c$, we still use first order methods but not the steepest descent one. Its convergence is very slow when $D$ is badly-conditioned. The conjugate gradients method is generally a better choice.
answered Nov 20 '18 at 22:31
Picaud Vincent
1,21838
edited Nov 20 '18 at 22:46
answered Nov 20 '18 at 22:31
Picaud Vincent
1,21838
answered Nov 20 '18 at 22:31
Picaud Vincent
1,21838
answered Nov 20 '18 at 22:31
Picaud Vincent
1,21838
1,21838
Thanks @Picaud Vincent! Exactly what I needed.
– Louis-Philippe Noël
Nov 21 '18 at 19:08
@Louis-PhilippeNoël so maybe you can voteup? :)
– Picaud Vincent
Nov 21 '18 at 19:15
add a comment |
Thanks @Picaud Vincent! Exactly what I needed.
– Louis-Philippe Noël
Nov 21 '18 at 19:08
@Louis-PhilippeNoël so maybe you can voteup? :)
– Picaud Vincent
Nov 21 '18 at 19:15
Thanks @Picaud Vincent! Exactly what I needed.
– Louis-Philippe Noël
Nov 21 '18 at 19:08
Thanks @Picaud Vincent! Exactly what I needed.
– Louis-Philippe Noël
Nov 21 '18 at 19:08
@Louis-PhilippeNoël so maybe you can voteup? :)
– Picaud Vincent
Nov 21 '18 at 19:15
@Louis-PhilippeNoël so maybe you can voteup? :)
– Picaud Vincent
Nov 21 '18 at 19:15
add a comment |
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