Is $S = {alpha in mathbf{R}^3 mid alpha_1 + alpha_2e^{-t} + alpha_3 e^{-2t} leq 1.1 mbox{ for } tgeq 1}$...
$$S = { alpha in mathbf{R}^3 mid alpha_1 + alpha_2e^{-t} +
alpha_3 e^{-2t} leq 1.1 mbox{ for } tgeq 1}$$
Is $S$ affine, and is it a polyhedron?
I thought it's basically a linear function in $alpha$, thus affine. Since the linear function is characterized by $t$, as $t$ changes, there are many linear functions, thus a polyhedron? Is my reasoning correct, please?
Supplement:
Plotted @A.Γ's equation
convex-analysis convex-optimization
asked 2 days ago
zyxue
1819
add a comment |
$$S = { alpha in mathbf{R}^3 mid alpha_1 + alpha_2e^{-t} +
alpha_3 e^{-2t} leq 1.1 mbox{ for } tgeq 1}$$
Is $S$ affine, and is it a polyhedron?
I thought it's basically a linear function in $alpha$, thus affine. Since the linear function is characterized by $t$, as $t$ changes, there are many linear functions, thus a polyhedron? Is my reasoning correct, please?
Supplement:
Plotted @A.Γ's equation
convex-analysis convex-optimization
asked 2 days ago
zyxue
1819
1
A polyhedron is an intersection of a finite number of half-spaces. Compare with: $alpha_1cos t+alpha_2sin tle 1$, $tin[0,2pi]$ is a unit disc (not a polyhedron), though for each $t$ it is a half-plane.
– A.Γ.
yesterday
I plotted your equation. It's not affine, then why is it not a polyhedron? Is it because the number of half spaces not finite?
– zyxue
yesterday
It is not likely to be a polyhedron because the normal vector $(1,e^{-t}, e^{-2t})$ moves along a space non-linear curve. Yes, finite number is important. Any convex closed set is an intersection of half-spaces.
– A.Γ.
yesterday
For "Any convex closed set (doesn't have to be a polyhedron) is an intersection of half-spaces.", it doesn't matter whether the number is finite or not, right?
– zyxue
yesterday
1
The number of half-spaces is determined by the number of exposed faces, it maybe whatever, but for polyhedrons they are finitely many. The set $S$ here is convex and closed.
– A.Γ.
yesterday
add a comment |
$$S = { alpha in mathbf{R}^3 mid alpha_1 + alpha_2e^{-t} +
alpha_3 e^{-2t} leq 1.1 mbox{ for } tgeq 1}$$
Is $S$ affine, and is it a polyhedron?
I thought it's basically a linear function in $alpha$, thus affine. Since the linear function is characterized by $t$, as $t$ changes, there are many linear functions, thus a polyhedron? Is my reasoning correct, please?
Supplement:
Plotted @A.Γ's equation
convex-analysis convex-optimization
asked 2 days ago
zyxue
1819
$$S = { alpha in mathbf{R}^3 mid alpha_1 + alpha_2e^{-t} +
alpha_3 e^{-2t} leq 1.1 mbox{ for } tgeq 1}$$
Is $S$ affine, and is it a polyhedron?
I thought it's basically a linear function in $alpha$, thus affine. Since the linear function is characterized by $t$, as $t$ changes, there are many linear functions, thus a polyhedron? Is my reasoning correct, please?
Supplement:
Plotted @A.Γ's equation
convex-analysis convex-optimization
convex-analysis convex-optimization
asked 2 days ago
zyxue
1819
asked 2 days ago
zyxue
1819
edited yesterday
asked 2 days ago
zyxue
1819
asked 2 days ago
zyxue
1819
asked 2 days ago
zyxue
1819
1819
1
A polyhedron is an intersection of a finite number of half-spaces. Compare with: $alpha_1cos t+alpha_2sin tle 1$, $tin[0,2pi]$ is a unit disc (not a polyhedron), though for each $t$ it is a half-plane.
– A.Γ.
yesterday
I plotted your equation. It's not affine, then why is it not a polyhedron? Is it because the number of half spaces not finite?
– zyxue
yesterday
It is not likely to be a polyhedron because the normal vector $(1,e^{-t}, e^{-2t})$ moves along a space non-linear curve. Yes, finite number is important. Any convex closed set is an intersection of half-spaces.
– A.Γ.
yesterday
For "Any convex closed set (doesn't have to be a polyhedron) is an intersection of half-spaces.", it doesn't matter whether the number is finite or not, right?
– zyxue
yesterday
1
The number of half-spaces is determined by the number of exposed faces, it maybe whatever, but for polyhedrons they are finitely many. The set $S$ here is convex and closed.
– A.Γ.
yesterday
add a comment |
1
A polyhedron is an intersection of a finite number of half-spaces. Compare with: $alpha_1cos t+alpha_2sin tle 1$, $tin[0,2pi]$ is a unit disc (not a polyhedron), though for each $t$ it is a half-plane.
– A.Γ.
yesterday
I plotted your equation. It's not affine, then why is it not a polyhedron? Is it because the number of half spaces not finite?
– zyxue
yesterday
It is not likely to be a polyhedron because the normal vector $(1,e^{-t}, e^{-2t})$ moves along a space non-linear curve. Yes, finite number is important. Any convex closed set is an intersection of half-spaces.
– A.Γ.
yesterday
For "Any convex closed set (doesn't have to be a polyhedron) is an intersection of half-spaces.", it doesn't matter whether the number is finite or not, right?
– zyxue
yesterday
1
The number of half-spaces is determined by the number of exposed faces, it maybe whatever, but for polyhedrons they are finitely many. The set $S$ here is convex and closed.
– A.Γ.
yesterday
1
1
A polyhedron is an intersection of a finite number of half-spaces. Compare with: $alpha_1cos t+alpha_2sin tle 1$, $tin[0,2pi]$ is a unit disc (not a polyhedron), though for each $t$ it is a half-plane.
– A.Γ.
yesterday
A polyhedron is an intersection of a finite number of half-spaces. Compare with: $alpha_1cos t+alpha_2sin tle 1$, $tin[0,2pi]$ is a unit disc (not a polyhedron), though for each $t$ it is a half-plane.
– A.Γ.
yesterday
I plotted your equation. It's not affine, then why is it not a polyhedron? Is it because the number of half spaces not finite?
– zyxue
yesterday
I plotted your equation. It's not affine, then why is it not a polyhedron? Is it because the number of half spaces not finite?
– zyxue
yesterday
It is not likely to be a polyhedron because the normal vector $(1,e^{-t}, e^{-2t})$ moves along a space non-linear curve. Yes, finite number is important. Any convex closed set is an intersection of half-spaces.
– A.Γ.
yesterday
It is not likely to be a polyhedron because the normal vector $(1,e^{-t}, e^{-2t})$ moves along a space non-linear curve. Yes, finite number is important. Any convex closed set is an intersection of half-spaces.
– A.Γ.
yesterday
For "Any convex closed set (doesn't have to be a polyhedron) is an intersection of half-spaces.", it doesn't matter whether the number is finite or not, right?
– zyxue
yesterday
For "Any convex closed set (doesn't have to be a polyhedron) is an intersection of half-spaces.", it doesn't matter whether the number is finite or not, right?
– zyxue
yesterday
1
1
The number of half-spaces is determined by the number of exposed faces, it maybe whatever, but for polyhedrons they are finitely many. The set $S$ here is convex and closed.
– A.Γ.
yesterday
The number of half-spaces is determined by the number of exposed faces, it maybe whatever, but for polyhedrons they are finitely many. The set $S$ here is convex and closed.
– A.Γ.
yesterday
add a comment |
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1
A polyhedron is an intersection of a finite number of half-spaces. Compare with: $alpha_1cos t+alpha_2sin tle 1$, $tin[0,2pi]$ is a unit disc (not a polyhedron), though for each $t$ it is a half-plane.
– A.Γ.
yesterday
I plotted your equation. It's not affine, then why is it not a polyhedron? Is it because the number of half spaces not finite?
– zyxue
yesterday
It is not likely to be a polyhedron because the normal vector $(1,e^{-t}, e^{-2t})$ moves along a space non-linear curve. Yes, finite number is important. Any convex closed set is an intersection of half-spaces.
– A.Γ.
yesterday
For "Any convex closed set (doesn't have to be a polyhedron) is an intersection of half-spaces.", it doesn't matter whether the number is finite or not, right?
– zyxue
yesterday
1
The number of half-spaces is determined by the number of exposed faces, it maybe whatever, but for polyhedrons they are finitely many. The set $S$ here is convex and closed.
– A.Γ.
yesterday