Find tangent cone to closed convex set
Let $X = {xinmathbb{R}^5: x_i geq 0, x_1+2x_2-x_3 = 6,x_1 + x_2 + 2x_4 -x_5 = 4}$, and I am looking to find the tangent cone to $X$ at the point $(0,4,2,0,0)$. How would I go about this?
I know that $X$ is closed and convex and therefore the tangent cone is just ${{a(y-(0,4,2,0,0)): y in X, a geq 0}}$, but I think because it is in five dimensions I am having trouble picturing it. One idea I had was to find the normal cone, i.e. ${v in mathbb{R}^5: langle v, x-(0,4,2,0,0)rangle leq 0 forall x in X}$, but I can't figure out how to do this either
calculus optimization convex-optimization
asked Dec 9 '18 at 20:55
John Smith
1448
add a comment |
Let $X = {xinmathbb{R}^5: x_i geq 0, x_1+2x_2-x_3 = 6,x_1 + x_2 + 2x_4 -x_5 = 4}$, and I am looking to find the tangent cone to $X$ at the point $(0,4,2,0,0)$. How would I go about this?
I know that $X$ is closed and convex and therefore the tangent cone is just ${{a(y-(0,4,2,0,0)): y in X, a geq 0}}$, but I think because it is in five dimensions I am having trouble picturing it. One idea I had was to find the normal cone, i.e. ${v in mathbb{R}^5: langle v, x-(0,4,2,0,0)rangle leq 0 forall x in X}$, but I can't figure out how to do this either
calculus optimization convex-optimization
asked Dec 9 '18 at 20:55
John Smith
1448
add a comment |
Let $X = {xinmathbb{R}^5: x_i geq 0, x_1+2x_2-x_3 = 6,x_1 + x_2 + 2x_4 -x_5 = 4}$, and I am looking to find the tangent cone to $X$ at the point $(0,4,2,0,0)$. How would I go about this?
I know that $X$ is closed and convex and therefore the tangent cone is just ${{a(y-(0,4,2,0,0)): y in X, a geq 0}}$, but I think because it is in five dimensions I am having trouble picturing it. One idea I had was to find the normal cone, i.e. ${v in mathbb{R}^5: langle v, x-(0,4,2,0,0)rangle leq 0 forall x in X}$, but I can't figure out how to do this either
calculus optimization convex-optimization
asked Dec 9 '18 at 20:55
John Smith
1448
Let $X = {xinmathbb{R}^5: x_i geq 0, x_1+2x_2-x_3 = 6,x_1 + x_2 + 2x_4 -x_5 = 4}$, and I am looking to find the tangent cone to $X$ at the point $(0,4,2,0,0)$. How would I go about this?
I know that $X$ is closed and convex and therefore the tangent cone is just ${{a(y-(0,4,2,0,0)): y in X, a geq 0}}$, but I think because it is in five dimensions I am having trouble picturing it. One idea I had was to find the normal cone, i.e. ${v in mathbb{R}^5: langle v, x-(0,4,2,0,0)rangle leq 0 forall x in X}$, but I can't figure out how to do this either
calculus optimization convex-optimization
calculus optimization convex-optimization
asked Dec 9 '18 at 20:55
John Smith
1448
asked Dec 9 '18 at 20:55
John Smith
1448
edited Dec 10 '18 at 18:43
asked Dec 9 '18 at 20:55
John Smith
1448
asked Dec 9 '18 at 20:55
John Smith
1448
asked Dec 9 '18 at 20:55
John Smith
1448
1448
add a comment |
add a comment |
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