Ygtjki

Dual Citizen. Exited the US on Italian passport recently

Realistic Alternatives to Dust: What Else Could Feed a Plankton Bloom?

How come people say “Would of”?

Should I use my personal or workplace e-mail when registering to external websites for work purpose?

Why could you hear an Amstrad CPC working?

What is the meaning of Triage in Cybersec world?

What is the best strategy for white in this position?

Is domain driven design an anti-SQL pattern?

I see my dog run

What is the steepest angle that a canal can be traversable without locks?

Are there any other methods to apply to solving simultaneous equations?

CiviEvent: Public link for events of a specific type

How to answer pointed "are you quitting" questioning when I don't want them to suspect

How long do I have to send payment?

Limit the amount of RAM Mathematica may access?

What do the Banks children have against barley water?

Idiomatic way to prevent slicing?

What is the motivation for a law requiring 2 parties to consent for recording a conversation

Why do UK politicians seemingly ignore opinion polls on Brexit?

If the Wish spell is used to duplicate the effect of Simulacrum, are existing duplicates destroyed?

Should I write numbers in words or as numerals when there are multiple next to each other?

Is it possible for the two major parties in the UK to form a coalition with each other instead of a much smaller party?

Carnot-Caratheodory metric

What are my rights when I have a Sparpreis ticket but can't board an overcrowded train?

Classifying the stationary points of $f(x, y) = 4xy-x^4-y^4 $

2

$begingroup$

$f(x, y) = 4xy-x^4-y^4 $

The gradient of this function is $0$ in $(-1, -1), (0, 0),(1, 1)$

I tried to compute the determinant of the Hessian Matrix, but it's 0 for every point, I always get a null eigenvalue for each point, so no matter what I can't find out what kind of points these are.

For $(0, 0)$, since $f(0, 0) = 0$, $f(x, 0) = -x^4$ and $f(0, y) = -y^4$. Since both will always be negative, isn't that supposed to be a maximum or a minimum point? According to Wolfram it's a saddle point and I can't understand why.

multivariable-calculus maxima-minima hessian-matrix

share|cite|improve this question

edited Apr 6 at 12:19

YuiTo Cheng

2,3144937

asked Apr 6 at 9:59

Frost832Frost832

325

$endgroup$

add a comment | 

2

$begingroup$

$f(x, y) = 4xy-x^4-y^4 $

The gradient of this function is $0$ in $(-1, -1), (0, 0),(1, 1)$

I tried to compute the determinant of the Hessian Matrix, but it's 0 for every point, I always get a null eigenvalue for each point, so no matter what I can't find out what kind of points these are.

For $(0, 0)$, since $f(0, 0) = 0$, $f(x, 0) = -x^4$ and $f(0, y) = -y^4$. Since both will always be negative, isn't that supposed to be a maximum or a minimum point? According to Wolfram it's a saddle point and I can't understand why.

multivariable-calculus maxima-minima hessian-matrix

share|cite|improve this question

edited Apr 6 at 12:19

YuiTo Cheng

2,3144937

asked Apr 6 at 9:59

Frost832Frost832

325

$endgroup$

add a comment | 

$begingroup$

$f(x, y) = 4xy-x^4-y^4 $

The gradient of this function is $0$ in $(-1, -1), (0, 0),(1, 1)$

I tried to compute the determinant of the Hessian Matrix, but it's 0 for every point, I always get a null eigenvalue for each point, so no matter what I can't find out what kind of points these are.

For $(0, 0)$, since $f(0, 0) = 0$, $f(x, 0) = -x^4$ and $f(0, y) = -y^4$. Since both will always be negative, isn't that supposed to be a maximum or a minimum point? According to Wolfram it's a saddle point and I can't understand why.

multivariable-calculus maxima-minima hessian-matrix

share|cite|improve this question

edited Apr 6 at 12:19

YuiTo Cheng

2,3144937

asked Apr 6 at 9:59

Frost832Frost832

325

$endgroup$

share|cite|improve this question

edited Apr 6 at 12:19

YuiTo Cheng

2,3144937

asked Apr 6 at 9:59

Frost832Frost832

325

share|cite|improve this question

edited Apr 6 at 12:19

YuiTo Cheng

2,3144937

asked Apr 6 at 9:59

Frost832Frost832

325

edited Apr 6 at 12:19

YuiTo Cheng

2,3144937

edited Apr 6 at 12:19

YuiTo Cheng

2,3144937

asked Apr 6 at 9:59

Frost832Frost832

325

asked Apr 6 at 9:59

Frost832Frost832

325

2 Answers
2

active

oldest

votes

2

$begingroup$

Note that:

• 
  $f(x,x)=4x^2-2x^4=2x^2(2-x^2)$ and therefore $f(x,x)>0$ of $xinleft(-sqrt 2,sqrt2right)setminus0$;


• 
  $f(x,0)=-x^4<0$ (unless $x=0$).

Therefore, yes, $(0,0)$ is a saddle point.

share|cite|improve this answer

answered Apr 6 at 10:20

José Carlos SantosJosé Carlos Santos

173k23133242

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $(1, 1)$, I computed the determinant $Delta f(x, y) = f(x, y) - f(1, 1) = 4xy-x^4-y^4-2)$, now to study its sign, can I consider $f(1, y) > 0$ and $f(x, 1) > 0$ ?
  $endgroup$
  – Frost832
  Apr 6 at 13:34
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  At tht point, all eigenvalues of the Hessean are smaller than $0$. Therefore, $f$ has a local maximum there.
  $endgroup$
  – José Carlos Santos
  Apr 6 at 13:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much.
  $endgroup$
  – Frost832
  Apr 6 at 13:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad I could help.
  $endgroup$
  – José Carlos Santos
  Apr 6 at 13:57
  

  
  
  
  

add a comment | 

2

$begingroup$

Note that the gradient of $f$ is given by $nabla f(x,y)=(4y-4x^3,4x-4y^3)$ and thus the Hessian of $f$ is given by

$$H_f(x,y)=beginpmatrix-12x^2 &4\4 &-12y^2endpmatrix$$

Now, we look at the Hessian for $(0,0),(1,1),(-1,-1)$:

$$H_f(0,0)=beginpmatrix0 &4\4 &0endpmatrix, H_f(1,1)=beginpmatrix-12 &4\4 &-12endpmatrix,H_f(-1,-1)=beginpmatrix-12 &4\4 &-12endpmatrix$$

Now, by checking the principal minors of the matrices, you can note that $H_f(1,1)$ and $H_f(-1,-1)$ are negative definite.

Thus, by the sufficient criteria for local extrema, we have that $(1,1)$ and $(-1,-1)$ are local maxima.

In a similar manner, $H_f(0,0)$ is indefinite, and thus $(0,0)$ is indeed a saddle point.

share|cite|improve this answer

edited Apr 6 at 10:27

answered Apr 6 at 10:20

blubblub

3,299929

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3176830%2fclassifying-the-stationary-points-of-fx-y-4xy-x4-y4%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

Note that:

• 
  $f(x,x)=4x^2-2x^4=2x^2(2-x^2)$ and therefore $f(x,x)>0$ of $xinleft(-sqrt 2,sqrt2right)setminus0$;


• 
  $f(x,0)=-x^4<0$ (unless $x=0$).

Therefore, yes, $(0,0)$ is a saddle point.

share|cite|improve this answer

answered Apr 6 at 10:20

José Carlos SantosJosé Carlos Santos

173k23133242

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $(1, 1)$, I computed the determinant $Delta f(x, y) = f(x, y) - f(1, 1) = 4xy-x^4-y^4-2)$, now to study its sign, can I consider $f(1, y) > 0$ and $f(x, 1) > 0$ ?
  $endgroup$
  – Frost832
  Apr 6 at 13:34
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  At tht point, all eigenvalues of the Hessean are smaller than $0$. Therefore, $f$ has a local maximum there.
  $endgroup$
  – José Carlos Santos
  Apr 6 at 13:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much.
  $endgroup$
  – Frost832
  Apr 6 at 13:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad I could help.
  $endgroup$
  – José Carlos Santos
  Apr 6 at 13:57
  

  
  
  
  

add a comment | 

2

$begingroup$

Note that:

• 
  $f(x,x)=4x^2-2x^4=2x^2(2-x^2)$ and therefore $f(x,x)>0$ of $xinleft(-sqrt 2,sqrt2right)setminus0$;


• 
  $f(x,0)=-x^4<0$ (unless $x=0$).

Therefore, yes, $(0,0)$ is a saddle point.

share|cite|improve this answer

answered Apr 6 at 10:20

José Carlos SantosJosé Carlos Santos

173k23133242

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $(1, 1)$, I computed the determinant $Delta f(x, y) = f(x, y) - f(1, 1) = 4xy-x^4-y^4-2)$, now to study its sign, can I consider $f(1, y) > 0$ and $f(x, 1) > 0$ ?
  $endgroup$
  – Frost832
  Apr 6 at 13:34
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  At tht point, all eigenvalues of the Hessean are smaller than $0$. Therefore, $f$ has a local maximum there.
  $endgroup$
  – José Carlos Santos
  Apr 6 at 13:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much.
  $endgroup$
  – Frost832
  Apr 6 at 13:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad I could help.
  $endgroup$
  – José Carlos Santos
  Apr 6 at 13:57
  

  
  
  
  

add a comment | 

$begingroup$

Note that:

• 
  $f(x,x)=4x^2-2x^4=2x^2(2-x^2)$ and therefore $f(x,x)>0$ of $xinleft(-sqrt 2,sqrt2right)setminus0$;


• 
  $f(x,0)=-x^4<0$ (unless $x=0$).

Therefore, yes, $(0,0)$ is a saddle point.

share|cite|improve this answer

answered Apr 6 at 10:20

José Carlos SantosJosé Carlos Santos

173k23133242

$endgroup$

share|cite|improve this answer

answered Apr 6 at 10:20

José Carlos SantosJosé Carlos Santos

173k23133242

answered Apr 6 at 10:20

José Carlos SantosJosé Carlos Santos

173k23133242

answered Apr 6 at 10:20

José Carlos SantosJosé Carlos Santos

173k23133242

2

$begingroup$

Note that the gradient of $f$ is given by $nabla f(x,y)=(4y-4x^3,4x-4y^3)$ and thus the Hessian of $f$ is given by

$$H_f(x,y)=beginpmatrix-12x^2 &4\4 &-12y^2endpmatrix$$

Now, we look at the Hessian for $(0,0),(1,1),(-1,-1)$:

$$H_f(0,0)=beginpmatrix0 &4\4 &0endpmatrix, H_f(1,1)=beginpmatrix-12 &4\4 &-12endpmatrix,H_f(-1,-1)=beginpmatrix-12 &4\4 &-12endpmatrix$$

Now, by checking the principal minors of the matrices, you can note that $H_f(1,1)$ and $H_f(-1,-1)$ are negative definite.

Thus, by the sufficient criteria for local extrema, we have that $(1,1)$ and $(-1,-1)$ are local maxima.

In a similar manner, $H_f(0,0)$ is indefinite, and thus $(0,0)$ is indeed a saddle point.

share|cite|improve this answer

edited Apr 6 at 10:27

answered Apr 6 at 10:20

blubblub

3,299929

$endgroup$

add a comment | 

2

$begingroup$

Note that the gradient of $f$ is given by $nabla f(x,y)=(4y-4x^3,4x-4y^3)$ and thus the Hessian of $f$ is given by

$$H_f(x,y)=beginpmatrix-12x^2 &4\4 &-12y^2endpmatrix$$

Now, we look at the Hessian for $(0,0),(1,1),(-1,-1)$:

$$H_f(0,0)=beginpmatrix0 &4\4 &0endpmatrix, H_f(1,1)=beginpmatrix-12 &4\4 &-12endpmatrix,H_f(-1,-1)=beginpmatrix-12 &4\4 &-12endpmatrix$$

Now, by checking the principal minors of the matrices, you can note that $H_f(1,1)$ and $H_f(-1,-1)$ are negative definite.

Thus, by the sufficient criteria for local extrema, we have that $(1,1)$ and $(-1,-1)$ are local maxima.

In a similar manner, $H_f(0,0)$ is indefinite, and thus $(0,0)$ is indeed a saddle point.

share|cite|improve this answer

edited Apr 6 at 10:27

answered Apr 6 at 10:20

blubblub

3,299929

$endgroup$

add a comment | 

$begingroup$

Note that the gradient of $f$ is given by $nabla f(x,y)=(4y-4x^3,4x-4y^3)$ and thus the Hessian of $f$ is given by

$$H_f(x,y)=beginpmatrix-12x^2 &4\4 &-12y^2endpmatrix$$

Now, we look at the Hessian for $(0,0),(1,1),(-1,-1)$:

$$H_f(0,0)=beginpmatrix0 &4\4 &0endpmatrix, H_f(1,1)=beginpmatrix-12 &4\4 &-12endpmatrix,H_f(-1,-1)=beginpmatrix-12 &4\4 &-12endpmatrix$$

Now, by checking the principal minors of the matrices, you can note that $H_f(1,1)$ and $H_f(-1,-1)$ are negative definite.

Thus, by the sufficient criteria for local extrema, we have that $(1,1)$ and $(-1,-1)$ are local maxima.

In a similar manner, $H_f(0,0)$ is indefinite, and thus $(0,0)$ is indeed a saddle point.

share|cite|improve this answer

edited Apr 6 at 10:27

answered Apr 6 at 10:20

blubblub

3,299929

$endgroup$

share|cite|improve this answer

edited Apr 6 at 10:27

answered Apr 6 at 10:20

blubblub

3,299929

answered Apr 6 at 10:20

blubblub

3,299929

answered Apr 6 at 10:20

blubblub

3,299929