Is there a systematic way to construct functions with prescribed local extrema?
I'm teaching multivariable calculus and having a hard time coming up with optimization problems.
Suppose I have three lists of points ${a_1, dotsc, a_r}$, ${b_1, dotsc, b_s}$, and ${c_1, dotsc, c_t}$ in $Bbb R^n$. Is it possible to construct a function $f:Bbb R^ntoBbb R$ that has a local max at every $a_i$, a local min at every $b_i$, a saddle point at every $c_i$, and no other local extrema? Furthermore, is it possible to accomplish this with $finBbb R[x_1, dotsc, x_n]$?
A weaker version of this question is: given points ${p_1,dotsc,p_k}$ in $Bbb R^n$, is it possible to construct a function $f:Bbb R^nto Bbb R$ whose set of critical points is exactly ${p_1,dotsc,p_k}$?
So, for instance, is there a systematic way to construct $finBbb R[x, y, z]$ with a local maximum at $(1, -3, 2)$ and a saddle point at $(2, -8, 1)$?
multivariable-calculus optimization hessian-matrix
asked Nov 25 at 6:42
Brian Fitzpatrick
21k42958
add a comment |
I'm teaching multivariable calculus and having a hard time coming up with optimization problems.
Suppose I have three lists of points ${a_1, dotsc, a_r}$, ${b_1, dotsc, b_s}$, and ${c_1, dotsc, c_t}$ in $Bbb R^n$. Is it possible to construct a function $f:Bbb R^ntoBbb R$ that has a local max at every $a_i$, a local min at every $b_i$, a saddle point at every $c_i$, and no other local extrema? Furthermore, is it possible to accomplish this with $finBbb R[x_1, dotsc, x_n]$?
A weaker version of this question is: given points ${p_1,dotsc,p_k}$ in $Bbb R^n$, is it possible to construct a function $f:Bbb R^nto Bbb R$ whose set of critical points is exactly ${p_1,dotsc,p_k}$?
So, for instance, is there a systematic way to construct $finBbb R[x, y, z]$ with a local maximum at $(1, -3, 2)$ and a saddle point at $(2, -8, 1)$?
multivariable-calculus optimization hessian-matrix
asked Nov 25 at 6:42
Brian Fitzpatrick
21k42958
2
This could be tricky. Suppose you want two local maxima, no local minima, and no saddle points. Is there such a function? Surprisingly, the answer is yes, but I would be hard pressed to construct it systematically.
– Rahul
Nov 25 at 8:44
@Rahul Good point. I've updated the question with a weaker version which may have an answer.
– Brian Fitzpatrick
Nov 25 at 21:05
add a comment |
I'm teaching multivariable calculus and having a hard time coming up with optimization problems.
Suppose I have three lists of points ${a_1, dotsc, a_r}$, ${b_1, dotsc, b_s}$, and ${c_1, dotsc, c_t}$ in $Bbb R^n$. Is it possible to construct a function $f:Bbb R^ntoBbb R$ that has a local max at every $a_i$, a local min at every $b_i$, a saddle point at every $c_i$, and no other local extrema? Furthermore, is it possible to accomplish this with $finBbb R[x_1, dotsc, x_n]$?
A weaker version of this question is: given points ${p_1,dotsc,p_k}$ in $Bbb R^n$, is it possible to construct a function $f:Bbb R^nto Bbb R$ whose set of critical points is exactly ${p_1,dotsc,p_k}$?
So, for instance, is there a systematic way to construct $finBbb R[x, y, z]$ with a local maximum at $(1, -3, 2)$ and a saddle point at $(2, -8, 1)$?
multivariable-calculus optimization hessian-matrix
asked Nov 25 at 6:42
Brian Fitzpatrick
21k42958
I'm teaching multivariable calculus and having a hard time coming up with optimization problems.
Suppose I have three lists of points ${a_1, dotsc, a_r}$, ${b_1, dotsc, b_s}$, and ${c_1, dotsc, c_t}$ in $Bbb R^n$. Is it possible to construct a function $f:Bbb R^ntoBbb R$ that has a local max at every $a_i$, a local min at every $b_i$, a saddle point at every $c_i$, and no other local extrema? Furthermore, is it possible to accomplish this with $finBbb R[x_1, dotsc, x_n]$?
A weaker version of this question is: given points ${p_1,dotsc,p_k}$ in $Bbb R^n$, is it possible to construct a function $f:Bbb R^nto Bbb R$ whose set of critical points is exactly ${p_1,dotsc,p_k}$?
So, for instance, is there a systematic way to construct $finBbb R[x, y, z]$ with a local maximum at $(1, -3, 2)$ and a saddle point at $(2, -8, 1)$?
multivariable-calculus optimization hessian-matrix
multivariable-calculus optimization hessian-matrix
asked Nov 25 at 6:42
Brian Fitzpatrick
21k42958
asked Nov 25 at 6:42
Brian Fitzpatrick
21k42958
edited Nov 25 at 21:04
asked Nov 25 at 6:42
Brian Fitzpatrick
21k42958
asked Nov 25 at 6:42
Brian Fitzpatrick
21k42958
asked Nov 25 at 6:42
Brian Fitzpatrick
21k42958
21k42958
2
This could be tricky. Suppose you want two local maxima, no local minima, and no saddle points. Is there such a function? Surprisingly, the answer is yes, but I would be hard pressed to construct it systematically.
– Rahul
Nov 25 at 8:44
@Rahul Good point. I've updated the question with a weaker version which may have an answer.
– Brian Fitzpatrick
Nov 25 at 21:05
add a comment |
2
This could be tricky. Suppose you want two local maxima, no local minima, and no saddle points. Is there such a function? Surprisingly, the answer is yes, but I would be hard pressed to construct it systematically.
– Rahul
Nov 25 at 8:44
@Rahul Good point. I've updated the question with a weaker version which may have an answer.
– Brian Fitzpatrick
Nov 25 at 21:05
2
2
This could be tricky. Suppose you want two local maxima, no local minima, and no saddle points. Is there such a function? Surprisingly, the answer is yes, but I would be hard pressed to construct it systematically.
– Rahul
Nov 25 at 8:44
This could be tricky. Suppose you want two local maxima, no local minima, and no saddle points. Is there such a function? Surprisingly, the answer is yes, but I would be hard pressed to construct it systematically.
– Rahul
Nov 25 at 8:44
@Rahul Good point. I've updated the question with a weaker version which may have an answer.
– Brian Fitzpatrick
Nov 25 at 21:05
@Rahul Good point. I've updated the question with a weaker version which may have an answer.
– Brian Fitzpatrick
Nov 25 at 21:05
add a comment |
1 Answer
1
active
oldest
votes
In the simpler case of $f:mathbb{R}^2tomathbb{R}$
(i) minima if both $f_{xx}$ and $f_{yy}$ are positive and $f^2_{xy}< fxxfyy$,
(ii) maxima if both $f_{xx}$ and $f_{yy}$ are negative and $f^2_{xy} <f_{xx}f_{yy}$,
(iii)saddle points if $f_{xx}$ and $f_{yy}$ have opposite signs or $f^2_{xy} >f_{xx}f_{yy}$.
Take local maximum at $(1,-3)$ and saddle point at $(2,-8)$ from example, the $f$ satisfying the following sufficient but not necessary condition will meet our requirement:
$f_x(1)=f_x(2)=f_y(-3)=f_y(-8)=0$
$f_{xx}(1)<0, f_{yy}(-3)<0, f_{xx}(2)>0 , f_{yy}(-8)<0, f_{xy}=f_{yx}=0$
apparently
$f_x=k(x-1)(x-2), f_y=-m(y+8)(y+5)(y+3)$ satisfies all above condition with $k, m>0$ (draw a graph of $f_x, f_y$ will help in drawing the conclusion).
Solve the PDE and choose $k,m$, integration constants freely will arrive at a function to our satisfaction. The case with $f:mathbb{R}^3tomathbb{R}$ can be generated similarly. Only problem the function thus created may be too simple to your like as $x,y,z$ are always decoupled. May be you can start with some 2 variable function with desired complexity and extrema and saddle points, then add 3rd,4th ... decoupled dimension...
answered Nov 25 at 22:02
Lance
54229
Would that it were so simple. The goal is to have no other critical points than the ones specified, but your function has critical points at all six points in ${1,2}times{-3,-5,-8}$.
– Rahul
Nov 26 at 6:41
@Rahul I see. Thanks.
– Lance
Nov 27 at 6:08
add a comment |
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In the simpler case of $f:mathbb{R}^2tomathbb{R}$
(i) minima if both $f_{xx}$ and $f_{yy}$ are positive and $f^2_{xy}< fxxfyy$,
(ii) maxima if both $f_{xx}$ and $f_{yy}$ are negative and $f^2_{xy} <f_{xx}f_{yy}$,
(iii)saddle points if $f_{xx}$ and $f_{yy}$ have opposite signs or $f^2_{xy} >f_{xx}f_{yy}$.
Take local maximum at $(1,-3)$ and saddle point at $(2,-8)$ from example, the $f$ satisfying the following sufficient but not necessary condition will meet our requirement:
$f_x(1)=f_x(2)=f_y(-3)=f_y(-8)=0$
$f_{xx}(1)<0, f_{yy}(-3)<0, f_{xx}(2)>0 , f_{yy}(-8)<0, f_{xy}=f_{yx}=0$
apparently
$f_x=k(x-1)(x-2), f_y=-m(y+8)(y+5)(y+3)$ satisfies all above condition with $k, m>0$ (draw a graph of $f_x, f_y$ will help in drawing the conclusion).
Solve the PDE and choose $k,m$, integration constants freely will arrive at a function to our satisfaction. The case with $f:mathbb{R}^3tomathbb{R}$ can be generated similarly. Only problem the function thus created may be too simple to your like as $x,y,z$ are always decoupled. May be you can start with some 2 variable function with desired complexity and extrema and saddle points, then add 3rd,4th ... decoupled dimension...
answered Nov 25 at 22:02
Lance
54229
Would that it were so simple. The goal is to have no other critical points than the ones specified, but your function has critical points at all six points in ${1,2}times{-3,-5,-8}$.
– Rahul
Nov 26 at 6:41
@Rahul I see. Thanks.
– Lance
Nov 27 at 6:08
add a comment |
In the simpler case of $f:mathbb{R}^2tomathbb{R}$
(i) minima if both $f_{xx}$ and $f_{yy}$ are positive and $f^2_{xy}< fxxfyy$,
(ii) maxima if both $f_{xx}$ and $f_{yy}$ are negative and $f^2_{xy} <f_{xx}f_{yy}$,
(iii)saddle points if $f_{xx}$ and $f_{yy}$ have opposite signs or $f^2_{xy} >f_{xx}f_{yy}$.
Take local maximum at $(1,-3)$ and saddle point at $(2,-8)$ from example, the $f$ satisfying the following sufficient but not necessary condition will meet our requirement:
$f_x(1)=f_x(2)=f_y(-3)=f_y(-8)=0$
$f_{xx}(1)<0, f_{yy}(-3)<0, f_{xx}(2)>0 , f_{yy}(-8)<0, f_{xy}=f_{yx}=0$
apparently
$f_x=k(x-1)(x-2), f_y=-m(y+8)(y+5)(y+3)$ satisfies all above condition with $k, m>0$ (draw a graph of $f_x, f_y$ will help in drawing the conclusion).
Solve the PDE and choose $k,m$, integration constants freely will arrive at a function to our satisfaction. The case with $f:mathbb{R}^3tomathbb{R}$ can be generated similarly. Only problem the function thus created may be too simple to your like as $x,y,z$ are always decoupled. May be you can start with some 2 variable function with desired complexity and extrema and saddle points, then add 3rd,4th ... decoupled dimension...
answered Nov 25 at 22:02
Lance
54229
Would that it were so simple. The goal is to have no other critical points than the ones specified, but your function has critical points at all six points in ${1,2}times{-3,-5,-8}$.
– Rahul
Nov 26 at 6:41
@Rahul I see. Thanks.
– Lance
Nov 27 at 6:08
add a comment |
In the simpler case of $f:mathbb{R}^2tomathbb{R}$
(i) minima if both $f_{xx}$ and $f_{yy}$ are positive and $f^2_{xy}< fxxfyy$,
(ii) maxima if both $f_{xx}$ and $f_{yy}$ are negative and $f^2_{xy} <f_{xx}f_{yy}$,
(iii)saddle points if $f_{xx}$ and $f_{yy}$ have opposite signs or $f^2_{xy} >f_{xx}f_{yy}$.
Take local maximum at $(1,-3)$ and saddle point at $(2,-8)$ from example, the $f$ satisfying the following sufficient but not necessary condition will meet our requirement:
$f_x(1)=f_x(2)=f_y(-3)=f_y(-8)=0$
$f_{xx}(1)<0, f_{yy}(-3)<0, f_{xx}(2)>0 , f_{yy}(-8)<0, f_{xy}=f_{yx}=0$
apparently
$f_x=k(x-1)(x-2), f_y=-m(y+8)(y+5)(y+3)$ satisfies all above condition with $k, m>0$ (draw a graph of $f_x, f_y$ will help in drawing the conclusion).
Solve the PDE and choose $k,m$, integration constants freely will arrive at a function to our satisfaction. The case with $f:mathbb{R}^3tomathbb{R}$ can be generated similarly. Only problem the function thus created may be too simple to your like as $x,y,z$ are always decoupled. May be you can start with some 2 variable function with desired complexity and extrema and saddle points, then add 3rd,4th ... decoupled dimension...
answered Nov 25 at 22:02
Lance
54229
In the simpler case of $f:mathbb{R}^2tomathbb{R}$
(i) minima if both $f_{xx}$ and $f_{yy}$ are positive and $f^2_{xy}< fxxfyy$,
(ii) maxima if both $f_{xx}$ and $f_{yy}$ are negative and $f^2_{xy} <f_{xx}f_{yy}$,
(iii)saddle points if $f_{xx}$ and $f_{yy}$ have opposite signs or $f^2_{xy} >f_{xx}f_{yy}$.
Take local maximum at $(1,-3)$ and saddle point at $(2,-8)$ from example, the $f$ satisfying the following sufficient but not necessary condition will meet our requirement:
$f_x(1)=f_x(2)=f_y(-3)=f_y(-8)=0$
$f_{xx}(1)<0, f_{yy}(-3)<0, f_{xx}(2)>0 , f_{yy}(-8)<0, f_{xy}=f_{yx}=0$
apparently
$f_x=k(x-1)(x-2), f_y=-m(y+8)(y+5)(y+3)$ satisfies all above condition with $k, m>0$ (draw a graph of $f_x, f_y$ will help in drawing the conclusion).
Solve the PDE and choose $k,m$, integration constants freely will arrive at a function to our satisfaction. The case with $f:mathbb{R}^3tomathbb{R}$ can be generated similarly. Only problem the function thus created may be too simple to your like as $x,y,z$ are always decoupled. May be you can start with some 2 variable function with desired complexity and extrema and saddle points, then add 3rd,4th ... decoupled dimension...
answered Nov 25 at 22:02
Lance
54229
edited Nov 25 at 22:12
answered Nov 25 at 22:02
Lance
54229
answered Nov 25 at 22:02
Lance
54229
answered Nov 25 at 22:02
Lance
54229
54229
Would that it were so simple. The goal is to have no other critical points than the ones specified, but your function has critical points at all six points in ${1,2}times{-3,-5,-8}$.
– Rahul
Nov 26 at 6:41
@Rahul I see. Thanks.
– Lance
Nov 27 at 6:08
add a comment |
Would that it were so simple. The goal is to have no other critical points than the ones specified, but your function has critical points at all six points in ${1,2}times{-3,-5,-8}$.
– Rahul
Nov 26 at 6:41
@Rahul I see. Thanks.
– Lance
Nov 27 at 6:08
Would that it were so simple. The goal is to have no other critical points than the ones specified, but your function has critical points at all six points in ${1,2}times{-3,-5,-8}$.
– Rahul
Nov 26 at 6:41
Would that it were so simple. The goal is to have no other critical points than the ones specified, but your function has critical points at all six points in ${1,2}times{-3,-5,-8}$.
– Rahul
Nov 26 at 6:41
@Rahul I see. Thanks.
– Lance
Nov 27 at 6:08
@Rahul I see. Thanks.
– Lance
Nov 27 at 6:08
add a comment |
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2
This could be tricky. Suppose you want two local maxima, no local minima, and no saddle points. Is there such a function? Surprisingly, the answer is yes, but I would be hard pressed to construct it systematically.
– Rahul
Nov 25 at 8:44
@Rahul Good point. I've updated the question with a weaker version which may have an answer.
– Brian Fitzpatrick
Nov 25 at 21:05