Characterizing the frontier between upward sloping and downward sloping parts of a curve
$begingroup$
Let me start with an example, let
$$mathbf{T}=begin{bmatrix}2&1\1&3end{bmatrix}$$
and $α=(2,2)$. Abusing notation, write $mathbf{λ}^{α}=(λ^2,(1-λ)^2)$ for any $0≤λ≤1$. If we compute $T^{top}mathbf{λ}^{α}$ for every value of $λ$, we obtain a series of points $(x,y)$ that connect $(1,2)$ and $(3,1)$ through a curve. As long as for all $i$, $α_i>1,$ this curve goes through a point in which $x$ is minimum (call it $minX$) and another in which $y$ is minimum ($minY$). In the example, $minX=(frac{3}{4},frac{19}{16})$ and $minY=(frac{13}{9},frac{2}{3}),$ that correspond to weights $(frac{3}{4},frac{1}{4})$ and $(frac{1}{3},frac{2}{3}),$ respectively.
I'm only interested in those $mathbf{λ}$ that yield points between $minX$ and $minY,$ i.e., points that are the result of $frac{1}{3}leqλleqfrac{3}{4}$. In these points, the slope of the curve (represented in the $x,y$ plane) is negative and $minX$ and $minY$ are the "frontier" points that separate the part in which the curve is negatively sloped from those where it is not. In fact, the slope of the curve at $minX$ is zero and it approaches $infty$ as we move closer to $minY.$
Similarly, for $n>2$, I'd be interested in those points sitting in the part of curved surface that is "negatively sloped". In $Bbb R^3$, I want to exclude those portions of the surface for which $x,y$ and $z$ increase simultaneously. Of course, I can still find values for $minX,$ $minY$ and $minZ,$ but I don't know how to determine the rest of the "frontier" points.
Additional information: If we define $mathbf{k}=mathbf{T}^{top}mathbf{barλ}^{α}$ for any $mathbf{barλ}=(barλ_1,barλ_2,...,barλ_n))$, with $sum_{i=1}^n barλ_i=1$ and $0leq barλ_i leq 1$, the $mathbf{λ}$ yielding points between $minX$ and $minY$ have an additional property: they are the ones for which the solution to the problem
$$max sum_{i=1}^n lambda_i$$
$$s.t. T^{top}λ^{α}=mathbf{k},$$
denoted $mathbf{λ}^*$, is such that $sum_i lambda_i^*=1.$ Observe that $mathbf{barλ}$ is part of the feasible set, and therefore $sum_i lambda_i^*geq1.$ What I would like is to characterize the $mathbf{λ}$ that are in the frontier between those resulting in $sum_i lambda_i^*=1$ and in $sum_i lambda_i^*>1.$
(I'm not sure about the tag, so I apologize if I've misled you and I'd be happy to change it if you can suggest a better one).
algebraic-geometry
asked Nov 6 '18 at 14:09
PatricioPatricio
3377
$endgroup$
|
show 4 more comments
$begingroup$
Let me start with an example, let
$$mathbf{T}=begin{bmatrix}2&1\1&3end{bmatrix}$$
and $α=(2,2)$. Abusing notation, write $mathbf{λ}^{α}=(λ^2,(1-λ)^2)$ for any $0≤λ≤1$. If we compute $T^{top}mathbf{λ}^{α}$ for every value of $λ$, we obtain a series of points $(x,y)$ that connect $(1,2)$ and $(3,1)$ through a curve. As long as for all $i$, $α_i>1,$ this curve goes through a point in which $x$ is minimum (call it $minX$) and another in which $y$ is minimum ($minY$). In the example, $minX=(frac{3}{4},frac{19}{16})$ and $minY=(frac{13}{9},frac{2}{3}),$ that correspond to weights $(frac{3}{4},frac{1}{4})$ and $(frac{1}{3},frac{2}{3}),$ respectively.
I'm only interested in those $mathbf{λ}$ that yield points between $minX$ and $minY,$ i.e., points that are the result of $frac{1}{3}leqλleqfrac{3}{4}$. In these points, the slope of the curve (represented in the $x,y$ plane) is negative and $minX$ and $minY$ are the "frontier" points that separate the part in which the curve is negatively sloped from those where it is not. In fact, the slope of the curve at $minX$ is zero and it approaches $infty$ as we move closer to $minY.$
Similarly, for $n>2$, I'd be interested in those points sitting in the part of curved surface that is "negatively sloped". In $Bbb R^3$, I want to exclude those portions of the surface for which $x,y$ and $z$ increase simultaneously. Of course, I can still find values for $minX,$ $minY$ and $minZ,$ but I don't know how to determine the rest of the "frontier" points.
Additional information: If we define $mathbf{k}=mathbf{T}^{top}mathbf{barλ}^{α}$ for any $mathbf{barλ}=(barλ_1,barλ_2,...,barλ_n))$, with $sum_{i=1}^n barλ_i=1$ and $0leq barλ_i leq 1$, the $mathbf{λ}$ yielding points between $minX$ and $minY$ have an additional property: they are the ones for which the solution to the problem
$$max sum_{i=1}^n lambda_i$$
$$s.t. T^{top}λ^{α}=mathbf{k},$$
denoted $mathbf{λ}^*$, is such that $sum_i lambda_i^*=1.$ Observe that $mathbf{barλ}$ is part of the feasible set, and therefore $sum_i lambda_i^*geq1.$ What I would like is to characterize the $mathbf{λ}$ that are in the frontier between those resulting in $sum_i lambda_i^*=1$ and in $sum_i lambda_i^*>1.$
(I'm not sure about the tag, so I apologize if I've misled you and I'd be happy to change it if you can suggest a better one).
algebraic-geometry
asked Nov 6 '18 at 14:09
PatricioPatricio
3377
$endgroup$
$begingroup$
There are some things that need to be cleaned up I think. If $alpha$ is an ordered pair, then what do you mean by "As long as $alpha > 1$? Why are you taking the transpose of the symmetric matrix $T$? And lastly, it seems to me that for the example given, the curve is a straight line segment between $(1,3)$ and $(2,1)$.
$endgroup$
– dbx
Nov 6 '18 at 14:33
$begingroup$
@dbx, I meant that each component of $α$ needed to be larger than $1$. I've edited my question and hope it is clearer now. I need to use the transpose because I want to combine columns rather than rows, is just an unhappy coincidence that I've chosen a symmetric matrix. I've corrected also the definition of $λ^{α},$ so it is now the curve it was meant to be. Thank you very much.
$endgroup$
– Patricio
Nov 6 '18 at 15:10
$begingroup$
I've been playing around with this very interesting problem, and I don't have a solution -- but I will note that Mathematica also doesn't know how to solve it in general.
$endgroup$
– dbx
Nov 6 '18 at 16:10
$begingroup$
@dbx, how did you ask Mathematica?
$endgroup$
– Patricio
Nov 7 '18 at 7:46
$begingroup$
I didn't save my code, but it's straightforward to write $Tcdot lambda^alpha$ as a system of two nonlinear equations in $lambda$. Their derivatives are easy enough to find, and finding your frontier points corresponds to finding their critical points. I only tried the Solve[ ] function, which failed.
$endgroup$
– dbx
Nov 7 '18 at 18:45
|
show 4 more comments
$begingroup$
Let me start with an example, let
$$mathbf{T}=begin{bmatrix}2&1\1&3end{bmatrix}$$
and $α=(2,2)$. Abusing notation, write $mathbf{λ}^{α}=(λ^2,(1-λ)^2)$ for any $0≤λ≤1$. If we compute $T^{top}mathbf{λ}^{α}$ for every value of $λ$, we obtain a series of points $(x,y)$ that connect $(1,2)$ and $(3,1)$ through a curve. As long as for all $i$, $α_i>1,$ this curve goes through a point in which $x$ is minimum (call it $minX$) and another in which $y$ is minimum ($minY$). In the example, $minX=(frac{3}{4},frac{19}{16})$ and $minY=(frac{13}{9},frac{2}{3}),$ that correspond to weights $(frac{3}{4},frac{1}{4})$ and $(frac{1}{3},frac{2}{3}),$ respectively.
I'm only interested in those $mathbf{λ}$ that yield points between $minX$ and $minY,$ i.e., points that are the result of $frac{1}{3}leqλleqfrac{3}{4}$. In these points, the slope of the curve (represented in the $x,y$ plane) is negative and $minX$ and $minY$ are the "frontier" points that separate the part in which the curve is negatively sloped from those where it is not. In fact, the slope of the curve at $minX$ is zero and it approaches $infty$ as we move closer to $minY.$
Similarly, for $n>2$, I'd be interested in those points sitting in the part of curved surface that is "negatively sloped". In $Bbb R^3$, I want to exclude those portions of the surface for which $x,y$ and $z$ increase simultaneously. Of course, I can still find values for $minX,$ $minY$ and $minZ,$ but I don't know how to determine the rest of the "frontier" points.
Additional information: If we define $mathbf{k}=mathbf{T}^{top}mathbf{barλ}^{α}$ for any $mathbf{barλ}=(barλ_1,barλ_2,...,barλ_n))$, with $sum_{i=1}^n barλ_i=1$ and $0leq barλ_i leq 1$, the $mathbf{λ}$ yielding points between $minX$ and $minY$ have an additional property: they are the ones for which the solution to the problem
$$max sum_{i=1}^n lambda_i$$
$$s.t. T^{top}λ^{α}=mathbf{k},$$
denoted $mathbf{λ}^*$, is such that $sum_i lambda_i^*=1.$ Observe that $mathbf{barλ}$ is part of the feasible set, and therefore $sum_i lambda_i^*geq1.$ What I would like is to characterize the $mathbf{λ}$ that are in the frontier between those resulting in $sum_i lambda_i^*=1$ and in $sum_i lambda_i^*>1.$
(I'm not sure about the tag, so I apologize if I've misled you and I'd be happy to change it if you can suggest a better one).
algebraic-geometry
asked Nov 6 '18 at 14:09
PatricioPatricio
3377
$endgroup$
Let me start with an example, let
$$mathbf{T}=begin{bmatrix}2&1\1&3end{bmatrix}$$
and $α=(2,2)$. Abusing notation, write $mathbf{λ}^{α}=(λ^2,(1-λ)^2)$ for any $0≤λ≤1$. If we compute $T^{top}mathbf{λ}^{α}$ for every value of $λ$, we obtain a series of points $(x,y)$ that connect $(1,2)$ and $(3,1)$ through a curve. As long as for all $i$, $α_i>1,$ this curve goes through a point in which $x$ is minimum (call it $minX$) and another in which $y$ is minimum ($minY$). In the example, $minX=(frac{3}{4},frac{19}{16})$ and $minY=(frac{13}{9},frac{2}{3}),$ that correspond to weights $(frac{3}{4},frac{1}{4})$ and $(frac{1}{3},frac{2}{3}),$ respectively.
I'm only interested in those $mathbf{λ}$ that yield points between $minX$ and $minY,$ i.e., points that are the result of $frac{1}{3}leqλleqfrac{3}{4}$. In these points, the slope of the curve (represented in the $x,y$ plane) is negative and $minX$ and $minY$ are the "frontier" points that separate the part in which the curve is negatively sloped from those where it is not. In fact, the slope of the curve at $minX$ is zero and it approaches $infty$ as we move closer to $minY.$
Similarly, for $n>2$, I'd be interested in those points sitting in the part of curved surface that is "negatively sloped". In $Bbb R^3$, I want to exclude those portions of the surface for which $x,y$ and $z$ increase simultaneously. Of course, I can still find values for $minX,$ $minY$ and $minZ,$ but I don't know how to determine the rest of the "frontier" points.
Additional information: If we define $mathbf{k}=mathbf{T}^{top}mathbf{barλ}^{α}$ for any $mathbf{barλ}=(barλ_1,barλ_2,...,barλ_n))$, with $sum_{i=1}^n barλ_i=1$ and $0leq barλ_i leq 1$, the $mathbf{λ}$ yielding points between $minX$ and $minY$ have an additional property: they are the ones for which the solution to the problem
$$max sum_{i=1}^n lambda_i$$
$$s.t. T^{top}λ^{α}=mathbf{k},$$
denoted $mathbf{λ}^*$, is such that $sum_i lambda_i^*=1.$ Observe that $mathbf{barλ}$ is part of the feasible set, and therefore $sum_i lambda_i^*geq1.$ What I would like is to characterize the $mathbf{λ}$ that are in the frontier between those resulting in $sum_i lambda_i^*=1$ and in $sum_i lambda_i^*>1.$
(I'm not sure about the tag, so I apologize if I've misled you and I'd be happy to change it if you can suggest a better one).
algebraic-geometry
algebraic-geometry
asked Nov 6 '18 at 14:09
PatricioPatricio
3377
asked Nov 6 '18 at 14:09
PatricioPatricio
3377
edited Dec 31 '18 at 13:53
Patricio
asked Nov 6 '18 at 14:09
PatricioPatricio
3377
asked Nov 6 '18 at 14:09
PatricioPatricio
3377
asked Nov 6 '18 at 14:09
PatricioPatricio
3377
3377
$begingroup$
There are some things that need to be cleaned up I think. If $alpha$ is an ordered pair, then what do you mean by "As long as $alpha > 1$? Why are you taking the transpose of the symmetric matrix $T$? And lastly, it seems to me that for the example given, the curve is a straight line segment between $(1,3)$ and $(2,1)$.
$endgroup$
– dbx
Nov 6 '18 at 14:33
$begingroup$
@dbx, I meant that each component of $α$ needed to be larger than $1$. I've edited my question and hope it is clearer now. I need to use the transpose because I want to combine columns rather than rows, is just an unhappy coincidence that I've chosen a symmetric matrix. I've corrected also the definition of $λ^{α},$ so it is now the curve it was meant to be. Thank you very much.
$endgroup$
– Patricio
Nov 6 '18 at 15:10
$begingroup$
I've been playing around with this very interesting problem, and I don't have a solution -- but I will note that Mathematica also doesn't know how to solve it in general.
$endgroup$
– dbx
Nov 6 '18 at 16:10
$begingroup$
@dbx, how did you ask Mathematica?
$endgroup$
– Patricio
Nov 7 '18 at 7:46
$begingroup$
I didn't save my code, but it's straightforward to write $Tcdot lambda^alpha$ as a system of two nonlinear equations in $lambda$. Their derivatives are easy enough to find, and finding your frontier points corresponds to finding their critical points. I only tried the Solve[ ] function, which failed.
$endgroup$
– dbx
Nov 7 '18 at 18:45
|
show 4 more comments
$begingroup$
There are some things that need to be cleaned up I think. If $alpha$ is an ordered pair, then what do you mean by "As long as $alpha > 1$? Why are you taking the transpose of the symmetric matrix $T$? And lastly, it seems to me that for the example given, the curve is a straight line segment between $(1,3)$ and $(2,1)$.
$endgroup$
– dbx
Nov 6 '18 at 14:33
$begingroup$
@dbx, I meant that each component of $α$ needed to be larger than $1$. I've edited my question and hope it is clearer now. I need to use the transpose because I want to combine columns rather than rows, is just an unhappy coincidence that I've chosen a symmetric matrix. I've corrected also the definition of $λ^{α},$ so it is now the curve it was meant to be. Thank you very much.
$endgroup$
– Patricio
Nov 6 '18 at 15:10
$begingroup$
I've been playing around with this very interesting problem, and I don't have a solution -- but I will note that Mathematica also doesn't know how to solve it in general.
$endgroup$
– dbx
Nov 6 '18 at 16:10
$begingroup$
@dbx, how did you ask Mathematica?
$endgroup$
– Patricio
Nov 7 '18 at 7:46
$begingroup$
I didn't save my code, but it's straightforward to write $Tcdot lambda^alpha$ as a system of two nonlinear equations in $lambda$. Their derivatives are easy enough to find, and finding your frontier points corresponds to finding their critical points. I only tried the Solve[ ] function, which failed.
$endgroup$
– dbx
Nov 7 '18 at 18:45
$begingroup$
There are some things that need to be cleaned up I think. If $alpha$ is an ordered pair, then what do you mean by "As long as $alpha > 1$? Why are you taking the transpose of the symmetric matrix $T$? And lastly, it seems to me that for the example given, the curve is a straight line segment between $(1,3)$ and $(2,1)$.
$endgroup$
– dbx
Nov 6 '18 at 14:33
$begingroup$
There are some things that need to be cleaned up I think. If $alpha$ is an ordered pair, then what do you mean by "As long as $alpha > 1$? Why are you taking the transpose of the symmetric matrix $T$? And lastly, it seems to me that for the example given, the curve is a straight line segment between $(1,3)$ and $(2,1)$.
$endgroup$
– dbx
Nov 6 '18 at 14:33
$begingroup$
@dbx, I meant that each component of $α$ needed to be larger than $1$. I've edited my question and hope it is clearer now. I need to use the transpose because I want to combine columns rather than rows, is just an unhappy coincidence that I've chosen a symmetric matrix. I've corrected also the definition of $λ^{α},$ so it is now the curve it was meant to be. Thank you very much.
$endgroup$
– Patricio
Nov 6 '18 at 15:10
$begingroup$
@dbx, I meant that each component of $α$ needed to be larger than $1$. I've edited my question and hope it is clearer now. I need to use the transpose because I want to combine columns rather than rows, is just an unhappy coincidence that I've chosen a symmetric matrix. I've corrected also the definition of $λ^{α},$ so it is now the curve it was meant to be. Thank you very much.
$endgroup$
– Patricio
Nov 6 '18 at 15:10
$begingroup$
I've been playing around with this very interesting problem, and I don't have a solution -- but I will note that Mathematica also doesn't know how to solve it in general.
$endgroup$
– dbx
Nov 6 '18 at 16:10
$begingroup$
I've been playing around with this very interesting problem, and I don't have a solution -- but I will note that Mathematica also doesn't know how to solve it in general.
$endgroup$
– dbx
Nov 6 '18 at 16:10
$begingroup$
@dbx, how did you ask Mathematica?
$endgroup$
– Patricio
Nov 7 '18 at 7:46
$begingroup$
@dbx, how did you ask Mathematica?
$endgroup$
– Patricio
Nov 7 '18 at 7:46
$begingroup$
I didn't save my code, but it's straightforward to write $Tcdot lambda^alpha$ as a system of two nonlinear equations in $lambda$. Their derivatives are easy enough to find, and finding your frontier points corresponds to finding their critical points. I only tried the Solve[ ] function, which failed.
$endgroup$
– dbx
Nov 7 '18 at 18:45
$begingroup$
I didn't save my code, but it's straightforward to write $Tcdot lambda^alpha$ as a system of two nonlinear equations in $lambda$. Their derivatives are easy enough to find, and finding your frontier points corresponds to finding their critical points. I only tried the Solve[ ] function, which failed.
$endgroup$
– dbx
Nov 7 '18 at 18:45
|
show 4 more comments
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$begingroup$
There are some things that need to be cleaned up I think. If $alpha$ is an ordered pair, then what do you mean by "As long as $alpha > 1$? Why are you taking the transpose of the symmetric matrix $T$? And lastly, it seems to me that for the example given, the curve is a straight line segment between $(1,3)$ and $(2,1)$.
$endgroup$
– dbx
Nov 6 '18 at 14:33
$begingroup$
@dbx, I meant that each component of $α$ needed to be larger than $1$. I've edited my question and hope it is clearer now. I need to use the transpose because I want to combine columns rather than rows, is just an unhappy coincidence that I've chosen a symmetric matrix. I've corrected also the definition of $λ^{α},$ so it is now the curve it was meant to be. Thank you very much.
$endgroup$
– Patricio
Nov 6 '18 at 15:10
$begingroup$
I've been playing around with this very interesting problem, and I don't have a solution -- but I will note that Mathematica also doesn't know how to solve it in general.
$endgroup$
– dbx
Nov 6 '18 at 16:10
$begingroup$
@dbx, how did you ask Mathematica?
$endgroup$
– Patricio
Nov 7 '18 at 7:46
$begingroup$
I didn't save my code, but it's straightforward to write $Tcdot lambda^alpha$ as a system of two nonlinear equations in $lambda$. Their derivatives are easy enough to find, and finding your frontier points corresponds to finding their critical points. I only tried the Solve[ ] function, which failed.
$endgroup$
– dbx
Nov 7 '18 at 18:45