Mixing
In-exact line search
$begingroup$
In my class notes, the author says:
"If $f:mathbb{R}^n to mathbb{R}$ is bounded below and $p_k$ is a descent direction and the $alpha-beta$ also known as Armijo-Goldstein condition is met then either $nabla f(x_k) to 0$ or the angle between $nabla f(x_k)$ and $p_k$ approaches 90 degrees."
I think the author meant $| nabla f(x_k) | to 0$ as this would imply that we are approaching the stationary point of $f$??
or in the worst case scenario $nabla f(x_k)$ and $p_k$ becomes orthogonal.
It means that $p_k$ still decreases the $f$ but the rate of decrease decreases?
optimization convex-optimization numerical-optimization
asked Jan 18 at 14:37
Dreamer123Dreamer123
32729
$endgroup$
add a comment |
$begingroup$
In my class notes, the author says:
"If $f:mathbb{R}^n to mathbb{R}$ is bounded below and $p_k$ is a descent direction and the $alpha-beta$ also known as Armijo-Goldstein condition is met then either $nabla f(x_k) to 0$ or the angle between $nabla f(x_k)$ and $p_k$ approaches 90 degrees."
I think the author meant $| nabla f(x_k) | to 0$ as this would imply that we are approaching the stationary point of $f$??
or in the worst case scenario $nabla f(x_k)$ and $p_k$ becomes orthogonal.
It means that $p_k$ still decreases the $f$ but the rate of decrease decreases?
optimization convex-optimization numerical-optimization
asked Jan 18 at 14:37
Dreamer123Dreamer123
32729
$endgroup$
1
$begingroup$
$nabla f(x_k) to 0$ is is equivalent to $|nabla f(x_k) |to0$
$endgroup$
– daw
Jan 18 at 17:27
$begingroup$
Ah yes from the properties of the norm
$endgroup$
– Dreamer123
Jan 18 at 18:08
add a comment |
$begingroup$
In my class notes, the author says:
"If $f:mathbb{R}^n to mathbb{R}$ is bounded below and $p_k$ is a descent direction and the $alpha-beta$ also known as Armijo-Goldstein condition is met then either $nabla f(x_k) to 0$ or the angle between $nabla f(x_k)$ and $p_k$ approaches 90 degrees."
I think the author meant $| nabla f(x_k) | to 0$ as this would imply that we are approaching the stationary point of $f$??
or in the worst case scenario $nabla f(x_k)$ and $p_k$ becomes orthogonal.
It means that $p_k$ still decreases the $f$ but the rate of decrease decreases?
optimization convex-optimization numerical-optimization
asked Jan 18 at 14:37
Dreamer123Dreamer123
32729
$endgroup$
In my class notes, the author says:
"If $f:mathbb{R}^n to mathbb{R}$ is bounded below and $p_k$ is a descent direction and the $alpha-beta$ also known as Armijo-Goldstein condition is met then either $nabla f(x_k) to 0$ or the angle between $nabla f(x_k)$ and $p_k$ approaches 90 degrees."
I think the author meant $| nabla f(x_k) | to 0$ as this would imply that we are approaching the stationary point of $f$??
or in the worst case scenario $nabla f(x_k)$ and $p_k$ becomes orthogonal.
It means that $p_k$ still decreases the $f$ but the rate of decrease decreases?
optimization convex-optimization numerical-optimization
optimization convex-optimization numerical-optimization
asked Jan 18 at 14:37
Dreamer123Dreamer123
32729
asked Jan 18 at 14:37
Dreamer123Dreamer123
32729
asked Jan 18 at 14:37
Dreamer123Dreamer123
32729
asked Jan 18 at 14:37
Dreamer123Dreamer123
32729
asked Jan 18 at 14:37
Dreamer123Dreamer123
32729
32729
1
$begingroup$
$nabla f(x_k) to 0$ is is equivalent to $|nabla f(x_k) |to0$
$endgroup$
– daw
Jan 18 at 17:27
$begingroup$
Ah yes from the properties of the norm
$endgroup$
– Dreamer123
Jan 18 at 18:08
add a comment |
1
$begingroup$
$nabla f(x_k) to 0$ is is equivalent to $|nabla f(x_k) |to0$
$endgroup$
– daw
Jan 18 at 17:27
$begingroup$
Ah yes from the properties of the norm
$endgroup$
– Dreamer123
Jan 18 at 18:08
1
1
$begingroup$
$nabla f(x_k) to 0$ is is equivalent to $|nabla f(x_k) |to0$
$endgroup$
– daw
Jan 18 at 17:27
$begingroup$
$nabla f(x_k) to 0$ is is equivalent to $|nabla f(x_k) |to0$
$endgroup$
– daw
Jan 18 at 17:27
$begingroup$
Ah yes from the properties of the norm
$endgroup$
– Dreamer123
Jan 18 at 18:08
$begingroup$
Ah yes from the properties of the norm
$endgroup$
– Dreamer123
Jan 18 at 18:08
add a comment |
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1
$begingroup$
$nabla f(x_k) to 0$ is is equivalent to $|nabla f(x_k) |to0$
$endgroup$
– daw
Jan 18 at 17:27
$begingroup$
Ah yes from the properties of the norm
$endgroup$
– Dreamer123
Jan 18 at 18:08