Is there a systematic way to construct functions with prescribed local extrema?












3














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







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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
















3














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







share|cite|improve this question




















  • 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














3












3








3


1





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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 at 21:04

























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














  • 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










1 Answer
1






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oldest

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0














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...






share|cite|improve this answer























  • 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











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1 Answer
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active

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0














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...






share|cite|improve this answer























  • 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
















0














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...






share|cite|improve this answer























  • 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














0












0








0






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...






share|cite|improve this answer














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...







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 25 at 22:12

























answered Nov 25 at 22:02









Lance

61229




61229












  • 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












  • @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


















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