Need help in finding extreme values

[aruwin]

My teacher has written this problem.

Find the extreme values of the function f(x,y) = x^4 + y^4 - 2(x - y)^2 ≤ 0.

But what does the "≤ 0" mean? I know literally it means less than 0 but it seems like an unusual function.

In addition, he gave us some hints which I don't understand at all. Here it is,



from the partial derivatives,

f_x = 4x^3 - 4(x-y) = 0

f_y = 4y^3 + 4(x-y) = 0



we get (x,y) = (0,0),(√2,-√2),(-√2,√2) for O,A,B.



(I think A is is f_xx(x,y) and B is f_xy(x,y) = 2 , but I am not sure because he didn't say anything)



At the origin O, = (0,0),

f_xx(0,0)x^2 + z.f_xy(0,0)xy + f_yy(0,0)y^2 = -4x^2 + 8xy - 4y^2

= -4(x^2 - 2xy +y^2)

= -4(x-y)^2 ≤ 0



So that's the hint that he gave to the class. Can you explain this because I don't understand anything from it. It's different than how I would usually find extreme values.

 

[tiny-tim]

hi aruwin!

(try using the X² and X₂ buttons just above the Reply box )

My teacher has written this problem.

Find the extreme values of the function f(x,y) = x^4 + y^4 - 2(x - y)^2 ≤ 0.

i think you must have copied that wrong, i think the ≤ 0 is a comment the teacher made when telling you how to check that it's not a saddle point, and has nothing to do with the definition of f

from the partial derivatives,

f_x = 4x^3 - 4(x-y) = 0

f_y = 4y^3 + 4(x-y) = 0

we get (x,y) = (0,0),(√2,-√2),(-√2,√2) for O,A,B.

if you add f_x + f_y, you get 4x³ = -4y³,

so x = -y, and you can easily prove that x = 0 √2 or -√2

(and he's arbitrarily called those solutions O A and B)

At the origin O, = (0,0),

f_xx(0,0)x^2 + z.f_xy(0,0)xy + f_yy(0,0)y^2 = -4x^2 + 8xy - 4y^2

= -4(x^2 - 2xy +y^2)

= -4(x-y)^2 ≤ 0

the formula f_xx(0,0)x² + 2.f_xy(0,0)xy + fyyy(0,0)y² is the formula for checking whether you have a maximum minimum or saddle point

(there should be a proof of that somewhere in your notes)

he's evaluated everything at x = y = 0 (ie, point O), eg f_xx = 12x² - 4 at x = 0, ie f_xx|_x=0 = -4

 

[aruwin]

hi aruwin!

(try using the X² and X₂ buttons just above the Reply box )


i think you must have copied that wrong, i think the ≤ 0 is a comment the teacher made when telling you how to check that it's not a saddle point, and has nothing to do with the definition of f

No, I didn't copy it wrong. I have asked him again what it means and he said that it's the region where z = f(x,y) ≤ 0.
Without "f ≤ 0", f -> infinity as x,y -> infinity, thus no finite upper bound.
And this part:
At the origin O = (0,0),
f_xx(0,0)x^2 + z.f_xy(0,0)xy + f_yy(0,0)y^2 = -4x^2 + 8xy - 4y^2
= -4(x^2 - 2xy +y^2)
= -4(x-y)^2 ≤ 0

is a taylor expansion.

This whole thing is confusing, I need help to understand this. I couldn't ask him further more because it was time :(

 

[chiro]

No, I didn't copy it wrong. I have asked him again what it means and he said that it's the region where z = f(x,y) ≤ 0.
Without "f ≤ 0", f -> infinity as x,y -> infinity, thus no finite upper bound.
And this part:
At the origin O = (0,0),
f_xx(0,0)x^2 + z.f_xy(0,0)xy + f_yy(0,0)y^2 = -4x^2 + 8xy - 4y^2
= -4(x^2 - 2xy +y^2)
= -4(x-y)^2 ≤ 0

is a taylor expansion.

This whole thing is confusing, I need help to understand this. I couldn't ask him further more because it was time :(

At the point where you get a maximum (even one in a multi-dimensional setting), all second derivatives should be negatives. It's a similar interpretation to the normal univariate case where you do a second-derivative test to check whether you have a local minimum or maximum.

Before going further though, what kinds of concepts are you using in this course? Are you using Lagrange multipliers? Hessian matrices?

Maybe you could give the readers a bit of pretext for this problem by mentioning the specific concepts that are tied to the lecture and the associated problem.

 

[aruwin]

At the point where you get a maximum (even one in a multi-dimensional setting), all second derivatives should be negatives. It's a similar interpretation to the normal univariate case where you do a second-derivative test to check whether you have a local minimum or maximum.

Before going further though, what kinds of concepts are you using in this course? Are you using Lagrange multipliers? Hessian matrices?

Maybe you could give the readers a bit of pretext for this problem by mentioning the specific concepts that are tied to the lecture and the associated problem.

The concepts in this lecture are about extreme values and applying taylor expansion.Does this give you any idea on how to solve this? And why is taylor expansion used in this problem?

 

[chiro]

The concepts in this lecture are about extreme values and applying taylor expansion.Does this give you any idea on how to solve this? And why is taylor expansion used in this problem?

A taylor expansion is basically an expansion of a function about a point. In short, it takes a function that has certain properties: differentiability being the key one, but also having a radius of convergence around the point of expansion.

The idea is that you can represent any function by choosing some point and 'expanding' around that point so that you get a function in terms of f(x+a) instead of f(x).

The reasons for doing this are varied, but one reason has to do with getting good approximations for when you need to get an approximation of a function at a point that is close to some other point (in this case the 'a' above), and you know the derivatives at the point a.

Now in your expansion, we know that the first partial derivatives are 0 since they are extreme points, which means in the expansion itself, the df/dx and df/dy are 0 and we also know that f(0,0) = 0. This means that this expansion takes into account the first and second partial derivatives for the expansion of f(x,y) around the point (0,0).

Does this help you?

 

[aruwin]

A taylor expansion is basically an expansion of a function about a point. In short, it takes a function that has certain properties: differentiability being the key one, but also having a radius of convergence around the point of expansion.

The idea is that you can represent any function by choosing some point and 'expanding' around that point so that you get a function in terms of f(x+a) instead of f(x).

The reasons for doing this are varied, but one reason has to do with getting good approximations for when you need to get an approximation of a function at a point that is close to some other point (in this case the 'a' above), and you know the derivatives at the point a.

Now in your expansion, we know that the first partial derivatives are 0 since they are extreme points, which means in the expansion itself, the df/dx and df/dy are 0 and we also know that f(0,0) = 0. This means that this expansion takes into account the first and second partial derivatives for the expansion of f(x,y) around the point (0,0).

Does this help you?

But why do we need to expand the function?And secondly,how to expand it?Is there any formula involved in that case?

 

[chiro]

But why do we need to expand the function?And secondly,how to expand it?Is there any formula involved in that case?

There is a formula for the general multivariable case and it's basically summing all the derivatives at some centre of expansion while postmultiplying each term by a centred polynomial with respect to its coeffecient.

This is probably a mouthful, so I'll just link you to the wiki site (which is also a mouthful).

http://en.wikipedia.org/wiki/Taylor_series#Taylor_series_in_several_variables

There is no one use for Taylor series: it's merely a way to represent a function (including multivariable function) in a different way, kind of like representing 2 + 2 = 1 + 3.

The reason its useful is because it's used in many approximation applications where we typically centre around a point and ignore a majority of the terms (this is called a Taylor polynomial), and again it's not just used for one particular thing, but for mathematics results in general (usually again with approximation).

The formula is located on the wiki, but I'm afraid you might get a little lost by reading the general definition.

For the f(x,y) case, let's say we centre it around a and b (a is x co-ordinate, b is y co-ordinate). We then calculate all the partial derivatives at the (a,b) point. For the nth derivative we have N+1 terms, corresponding to all the partial derivatives that are possible for 2 variables (x,y). For other numbers of variables we basically look at the number of individual terms for the expansion of an n-nomial. So for five variables, we look at the number of individual terms for (x + y + z + w + u)^n and then each individual simplified term in terms of the multinomial expansion will correspond to a derivative.

Lets stick to the case of only x and y. This means for every nth order term, we will have N partial derivatives. Let's look at the first couple:

df/dx, df/dy (N = 1)
d^2f/dx^2, d^2f/dxdy, d^2f/dy^2 (N=2)
d^3f/dx^3, d^3f/dx^2dy, d^3f/dxdy^2, d^3f/dy^3 (N=3)

and so on.

These are evaluated at your centre of expansion (which in the above is (a,b) for 2D function f(x,y)). You then add the centred expansion corresponding to the derivative: again a few examples,

df/dx*(x-a), df/dy(y-b), df^3/dx^2dy*(x-a)^2*(y-b)/3!, df^2/dy^2*(y-b)^2/2! (remember the division of n! for an n-th order derivative).

Once you match the derivative with its proper expansion, you take all these terms and add them together and that gives the expansion around some centre (a,b) for the 2D case.

Again for clarification look at the wiki or another source for definitions and more examples.

 

[aruwin]

There is a formula for the general multivariable case and it's basically summing all the derivatives at some centre of expansion while postmultiplying each term by a centred polynomial with respect to its coeffecient.

This is probably a mouthful, so I'll just link you to the wiki site (which is also a mouthful).

http://en.wikipedia.org/wiki/Taylor_series#Taylor_series_in_several_variables

There is no one use for Taylor series: it's merely a way to represent a function (including multivariable function) in a different way, kind of like representing 2 + 2 = 1 + 3.

The reason its useful is because it's used in many approximation applications where we typically centre around a point and ignore a majority of the terms (this is called a Taylor polynomial), and again it's not just used for one particular thing, but for mathematics results in general (usually again with approximation).

The formula is located on the wiki, but I'm afraid you might get a little lost by reading the general definition.

For the f(x,y) case, let's say we centre it around a and b (a is x co-ordinate, b is y co-ordinate). We then calculate all the partial derivatives at the (a,b) point. For the nth derivative we have N+1 terms, corresponding to all the partial derivatives that are possible for 2 variables (x,y). For other numbers of variables we basically look at the number of individual terms for the expansion of an n-nomial. So for five variables, we look at the number of individual terms for (x + y + z + w + u)^n and then each individual simplified term in terms of the multinomial expansion will correspond to a derivative.

Lets stick to the case of only x and y. This means for every nth order term, we will have N partial derivatives. Let's look at the first couple:

df/dx, df/dy (N = 1)
d^2f/dx^2, d^2f/dxdy, d^2f/dy^2 (N=2)
d^3f/dx^3, d^3f/dx^2dy, d^3f/dxdy^2, d^3f/dy^3 (N=3)

and so on.

These are evaluated at your centre of expansion (which in the above is (a,b) for 2D function f(x,y)). You then add the centred expansion corresponding to the derivative: again a few examples,

df/dx*(x-a), df/dy(y-b), df^3/dx^2dy*(x-a)^2*(y-b)/3!, df^2/dy^2*(y-b)^2/2! (remember the division of n! for an n-th order derivative).

Once you match the derivative with its proper expansion, you take all these terms and add them together and that gives the expansion around some centre (a,b) for the 2D case.

Again for clarification look at the wiki or another source for definitions and more examples.

Thanks for the explanation. OK, I think I know the purpose of taylor expansion now but what I don't get is how do I determine the maximum value and minum value of the function after expanding it?

 

[chiro]

Thanks for the explanation. OK, I think I know the purpose of taylor expansion now but what I don't get is how do I determine the maximum value and minum value of the function after expanding it?

Well if the function is continuous and differentiable, then the same idea is used for the multi-variable case as it is for the univariate (single-variable) case.

What happens is that you find stationary points within some region and then test the nature of those points with information on the second derivative.

Although this sounds simple, depending on your constraints and function, it can be pretty easy or pretty hard.

For your example, you don't actually have hard constraints and because of this you can do first and second derivative tests to first find the stationary points (which you have done) and then find out what kind of point they are (minimum, maximum, point of inflection).

If you need to introduce constraints, the first level used is what are known as Lagrange multipliers. From this you start to build up more complicated techniques which are the subject of a field known as Mathematical Optimization which considers the problems of whether global minima or global maxima even exist, and when they exist for more abstract problems.

Are you or have you covered a topic known as Lagrange multipliers and its use for optimization?

 

[aruwin]

Well if the function is continuous and differentiable, then the same idea is used for the multi-variable case as it is for the univariate (single-variable) case.

What happens is that you find stationary points within some region and then test the nature of those points with information on the second derivative.

Although this sounds simple, depending on your constraints and function, it can be pretty easy or pretty hard.

For your example, you don't actually have hard constraints and because of this you can do first and second derivative tests to first find the stationary points (which you have done) and then find out what kind of point they are (minimum, maximum, point of inflection).

If you need to introduce constraints, the first level used is what are known as Lagrange multipliers. From this you start to build up more complicated techniques which are the subject of a field known as Mathematical Optimization which considers the problems of whether global minima or global maxima even exist, and when they exist for more abstract problems.

Are you or have you covered a topic known as Lagrange multipliers and its use for optimization?

Yes, I need to introduce constraints in this problem and I have to find the global max and global minimum. But I haven't covered Lagrange multipliers yet. Note that sometimes the teacher gives us questions that are not fully taught yet.

 

[chiro]

Yes, I need to introduce constraints in this problem and I have to find the global max and global minimum. But I haven't covered Lagrange multipliers yet. Note that sometimes the teacher gives us questions that are not fully taught yet.

Well you should probably look up Lagrange multipliers as this probably the most important foundational concept in multi-variable optimization.

If it's not in your notes or your own resources, do a quick google search and access many of the notes on introductory multivariable calculus and optimization that talk about these.

 

[aruwin]

Well you should probably look up Lagrange multipliers as this probably the most important foundational concept in multi-variable optimization.

If it's not in your notes or your own resources, do a quick google search and access many of the notes on introductory multivariable calculus and optimization that talk about these.

I am so sorry.I have made a mistake. I don't need to find the global maximum/minimum but the local maximum/minimum.
OK, so after I have expanded the function at point (0,0),(√2,-√2) and (-√2,√2), how do I determine the local max/min values?
At (0,0), using the taylor expansion formula :
f_xx(0,0)x^2 + 2.f_xy(0,0)xy + f_yy(0,0)y^2 = -4x^2 + 8xy - 4y^2

So, what do I do next?How do I determine the conditions for it to become local maximum and minimum?
Also, I think another important point to take note is that in the beginning of the question,we are given f(x,y) ≤ 0 and therefore, the result of the expansion -4x^2 + 8xy - 4y^2 ≤ 0.

Now I am stucked =( Help,please...

 

[tiny-tim]

hi aruwin!

ah, a local maximum/minimum!

i think you're missing the overall picture …
​

you have a roughly circular region on whose boundary f(x,y) = 0, and inside it f(x,y) is negative

you've found, by differentiating once, that there can only be three extreme points, O A and B, and you know that A and B are outside the region, so that leaves only O

so you differentiate again, and apply the formula, and you get a quadratic, in this case 4(x-y)²

…we are given f(x,y) ≤ 0 and therefore, the result of the expansion -4x^2 + 8xy - 4y^2 ≤ 0

no, f(x,y) ≤ 0 is irrelevant, that quadratic would be negative anyway

because it is negative, the "acceleration" in any direction (not just x and y) near O is negative, and so O is a maximum

 

[aruwin]

hi aruwin!

ah, a local maximum/minimum!

i think you're missing the overall picture …
​

you have a roughly circular region on whose boundary f(x,y) = 0, and inside it f(x,y) is negative

you've found, by differentiating once, that there can only be three extreme points, O A and B, and you know that A and B are outside the region, so that leaves only O

so you differentiate again, and apply the formula, and you get a quadratic, in this case 4(x-y)²


no, f(x,y) ≤ 0 is irrelevant, that quadratic would be negative anyway

because it is negative, the "acceleration" in any direction (not just x and y) near O is negative, and so O is a maximum

I am so sorry if I am not getting this as quickly as I should,so please be patient with me =)
Ok, why are points A and B outside of the region?When I substitue the values of (√2,-√2)and (-√2,√2) into the original function f(x,y), the result is negative. So doesn't that mean they are inside the region?

 

[tiny-tim]

sorry, yes, they are in the region …

ok, so now you need to check A and B also! ​

 

[aruwin]

hi aruwin!

so you differentiate again, and apply the formula, and you get a quadratic, in this case 4(x-y)²

Oh no, now I realize another mistake that I made. There was a mistake in the taylor expansion. I forgot to divide by the factorial nth term which is 2!OK, now the result of the expansion should be : -2(x-y)^2 (I think it can't be expanded more because after the second partial differentiation, there's no more x and y left)

Here,I understand that the quadratic will always be a negative because of the "-2" at the front.

OK, and my teacher had just given the answer for point O (0,0). Here it is and please explain to me why is the answer so?I don't get it, I can't even imagine how the shape of the graph would turn out.

Answer for point (0,0):
For all values of (x,y) near (0,0),
f(x,y)≥f(0,0) → At (0,0) f(x,y) has a local minimum
f(x,y)≤f(0,0) → At (0,0) f(x,y) has a local maximum*And for the other two points, (√2,-√2)and (-√2,√2) I still haven't got the answers.

 

[tiny-tim]

hi aruwin!

Answer for point (0,0):
For all values of (x,y) near (0,0),
f(x,y)≥f(0,0) → At (0,0) f(x,y) has a local minimum
f(x,y)≤f(0,0) → At (0,0) f(x,y) has a local maximum

that's not the answer for this f,

that's the definition of "local minimum" and "local maximum" for any f …

eg, if you can find a neighbourhood of (0,0) such that f(x,y) is always higher than f(0,0), then we say that there is a local minimum at (0,0)​

the reason why the f_xx etc fomula works is that you need to check the sign of f_kk for any direction k (not just x and y)

and f_kk is, approximately, a linear combination of f_xx f_xy and f_yy

 

[aruwin]

hi aruwin! that's not the answer for this f,

that's the definition of "local minimum" and "local maximum" for any f …

eg, if you can find a neighbourhood of (0,0) such that f(x,y) is always higher than f(0,0), then we say that there is a local minimum at (0,0)​

the reason why the f_xx etc fomula works is that you need to check the sign of f_kk for any direction k (not just x and y)

and f_kk is, approximately, a linear combination of f_xx f_xy and f_yy

Oh, I thought he was giving the answer lol. So he was just giving the definition.
OK, so how do I know if f(x,y) is greater or lesser than f(0,0)?? What I understand is that from the taylor expansion, we know that f(0,0) is always negative (or 0).In other words, f(x,y) is always negative(or 0) at (0,0). Right?
And since at the beginnging of the problem, the constraint f(x,y) ≤ 0 is given, it is also negative or 0 but how to know that it's bigger or than f(0,0)? If I can understand this part, I can conclude that O has a local maximum or minimum value of 0. Although I am pretty sure that it has a maximum value at O because the rest of the values of x and y are below zero.

 

[tiny-tim]

… how do I know if f(x,y) is greater or lesser than f(0,0)?? What I understand is that from the taylor expansion, we know that f(0,0) is always negative (or 0).In other words, f(x,y) is always negative(or 0) at (0,0). Right?

if z = f(x,y) is a map of the ground, then we know that if there's a cross-roads at O, with the roads going north-south and east-west,

then from f_x and f_y, we know both roads are flat at O

and from f_xx and f_yy, we also know that both roads have a maximum at O​

but we don't know whether a road going from P = (-x,-y) to Q= (x,y) has a maximum at O

so we use that quadratic formula to give us the test for that road, POQ

if that test shows us that POQ always has a maximum, wherever P is, then obviously there is a local maximum at O, ie a little hilltop at O

 

[aruwin]

then from f_x and f_y, we know both roads are flat at O

and from f_xx and f_yy, we also know that both roads have a maximum at O​

Oh,it means there's a hill at O,right? f_x and f_y are flat because we let f_x=0 and f_y=0,correct?
And from f_xx and f_yy, we get -4 and since they're negative values, we know that the roads(curves) are like concave downwards and that means there's flat peak at O making it maximum. Is my imagination correct?

 

[aruwin]

but we don't know whether a road going from P = (-x,-y) to Q= (x,y) has a maximum at O

so we use that quadratic formula to give us the test for that road, POQ

if that test shows us that POQ always has a maximum, wherever P is, then obviously there is a local maximum at O, ie a little hilltop at O

So let's use that quadratic that we got which is -2(x-y)^2. So how to use this to test for POQ? Do I just substitute the x and y values of P and Q into the quadratic function?OK,so let's say P=(-1,-2) and Q = (1,2). Subtituting these I get -2 and no matter what values I put,they're always going to be negative and that means they're always going to be maximum because it can never go beyond 0,right?
Now what makes POQ lower than the north-south and west-east roads? In other words, what makes f(x,y) smaller than f(0,0)?
I am going to apply what I think and please correct me if I am wrong.

The function at O is -2(x-y)^2.
Substituting values of x and y with 0,
so f(0,0) = 0 and
substituting values of x and y with 1 and 2,
so f(1,2) = -2 and
thus, f(1,2) is lower than f(0,0). Generalizing it, we have f(x,y) lower than or equals to(when we substitute with 0) f(0,0) and so, f(x,y) has a local maximum value of 0 at (0,0).

 

[haruspex]

the formula f_xx(0,0)x² + 2f_xy(0,0)xy + f_yy(0,0)y² is the formula for checking whether you have a maximum minimum or saddle point

Would you mind explaining how you use that? I would have used f_xx*f_yy - f_xy² to check for a saddle point (negative), then, if positive, the sign of f_xx (say) to distinguish max from min. That suggests the test using the OP formula is to see if the resulting quadratic has a real root (true for a saddle). Is that it?

 

[tiny-tim]

hi aruwin!

(try using the X² button just above the Reply box )

So let's use that quadratic that we got which is -2(x-y)^2. So how to use this to test for POQ? Do I just substitute the x and y values of P and Q into the quadratic function?OK,so let's say P=(-1,-2) and Q = (1,2). Subtituting these I get -2 and no matter what values I put,they're always going to be negative and that means they're always going to be maximum because it can never go beyond 0,right?

right!

in fact you don't even need to substitute, since you can see that -2(x-y)² will always be negative …

the road will always have a hump at O, whatever its direction is!​

(btw, f(x,y) at P (or Q) might be larger … all the test tells you is the shape of f(x,y) very close to O)

Now what makes POQ lower than the north-south and west-east roads? In other words, what makes f(x,y) smaller than f(0,0)?
I am going to apply what I think and please correct me if I am wrong.

The function at O is -2(x-y)^2.
Substituting values of x and y with 0,
so f(0,0) = 0 and
substituting values of x and y with 1 and 2,
so f(1,2) = -2 and
thus, f(1,2) is lower than f(0,0). Generalizing it, we have f(x,y) lower than or equals to(when we substitute with 0) f(0,0) and so, f(x,y) has a local maximum value of 0 at (0,0).

sorry, i don't understand this

our test simply gives you f_kk(0,0), and the approximately quadratic shape of POQ very close to O

Would you mind explaining how you use that? I would have used f_xx*f_yy - f_xy² to check for a saddle point (negative), then, if positive, the sign of f_xx (say) to distinguish max from min. That suggests the test using the OP formula is to see if the resulting quadratic has a real root (true for a saddle). Is that it?

yes, that's the test in eg http://www.ucl.ac.uk/Mathematics/geomath/level2/pdiff/pd9.html

aruwin's professor is using a slightly different test

 

[aruwin]

hi aruwin!


sorry, i don't understand this

I mean, in the definition,
if f(x,y) ≤ f(0,0), then f(x,y) has a maximum value at O,right? I was trying to show that f(x,y)≤f(0,0) because I'm still not sure about how to prove that.

 

[tiny-tim]

I mean, in the definition,
if f(x,y) ≤ f(0,0), then f(x,y) has a maximum value at O,right?

yes

I was trying to show that f(x,y)≤f(0,0) because I'm still not sure about how to prove that.

in a neighbourhood of (0,0), f(x,y) is approximately f(0,0) + xf_x(0,0) + yf_y(0,0) … you're probably familiar with this

similarly, in a neighbourhood of (0,0), f(x,y) is approximately f(0,0) + xf_x(0,0) + x²f_xx(0,0)/2 + yf_y(0,0) + y²f_yy(0,0)/2 + xyf_xy(0,0)

(didn't your professor prove that for you?)

 

[chiro]

Just a heads up for aruwin: remember that the taylor series represents the function if the derivatives exist and you have convergence within a given region. Since your function is a smooth function, the derivatives will exist everywhere and it should converge for all real numbers.

Remember this in the context of tiny-tim's advice in how the taylor series represents a function in the neighbourhood of the centred point, and how you can use some of the terms to get an approximation of the function 'around that centred point' as tiny-tim has pointed out.

In short:

taylor series = function, and taylor polynomial (the approximation) approximates the function around the centre used (the polynomial is the one that you are using to show that around (0,0), everything is negative approximately).

 

[aruwin]

yes


in a neighbourhood of (0,0), f(x,y) is approximately f(0,0) + xf_x(0,0) + yf_y(0,0) … you're probably familiar with this

similarly, in a neighbourhood of (0,0), f(x,y) is approximately f(0,0) + xf_x(0,0) + x²f_xx(0,0)/2 + yf_y(0,0) + y²f_yy(0,0)/2 + xyf_xy(0,0)

(didn't your professor prove that for you?)

Oh, I think I get it. f(0,0) = 0, right? and f(x,y) is always negative,correct? So that means f(x,y) ≤ f(0,0). And like chiro has pointed out above, the polynomial that I'm, using is -2(x-y)² and it shows that everything around zero is negative,right? Hence, the local maximum is (0,0).
So, next up is to check the other 2 critical points,A and B.

 

[haruspex]

-2(x-y)² shows that everything around zero is negative,right? Hence, the local maximum is (0,0).

Not so fast. -(x-y)² ≤ 0. The possibility of 0 means we have not yet established whether O is a maximum or a saddle. To resolve this, you have to look further down the Taylor expansion.
In fact, just think what happens to f along the line x = y. It reduces to 2x⁴. So clearly O is a saddle.
It does bother me that the OP specifies f = etc. ≤ 0. I'm not sure what that ≤ 0 is there for. Is it saying f is only defined where the result ≤ 0? If so, we have the awkward situation that at O it is only defined in the second and fourth quadrants. In that sense, O is a local maximum.

 

[aruwin]

Not so fast. -(x-y)² ≤ 0. The possibility of 0 means we have not yet established whether O is a maximum or a saddle. To resolve this, you have to look further down the Taylor expansion.
In fact, just think what happens to f along the line x = y. It reduces to 2x⁴. So clearly O is a saddle.
It does bother me that the OP specifies f = etc. ≤ 0. I'm not sure what that ≤ 0 is there for. Is it saying f is only defined where the result ≤ 0? If so, we have the awkward situation that at O it is only defined in the second and fourth quadrants. In that sense, O is a local maximum.

Yes, without the ≤0, O is a saddle. The ≤0 is there to state the region of f(x,y).

Now, I am going to try for points A and B.

The taylor expansion at A(√2,-√2) is
f(x,y)≈ -8 + 1/2![20(x-√2)²+8(x-√2)(y+√2)+20(y+√2)²]+...

The taylor polynomial until the 2nd term is
f(x,y) = -8+10(x-√2)²+4(x-√2)(y+√2)+10(y+√2)²

I'm stucked here, from the equation, it seems that at (√2,-√2), f(x,y) is always going to be negative but if I put other values it can be either positive or negative. Help please,I am stucked here.

 

[tiny-tim]

Not so fast. -(x-y)² ≤ 0. The possibility of 0 means we have not yet established whether O is a maximum or a saddle. To resolve this, you have to look further down the Taylor expansion.

not exactly … our quadratic formula gives all the information we need except in one direction (which we can check "by hand"):

0 is a possibility only along x = y

so, near (0,0), our quadratic formula, -4(x-y)², tells us that there is a maximum along any direction other than x = y

however, if gives us no information about x = y, so that might be a maximum also, or it might not … we need a cubic or quartic formula to check, or we can check "by hand"

(we could have got all this by simply putting all the lines x = constant*y …
obviously the x⁴ + y⁴ is insignificant, and the -2(x-y)² dominates (unless the constant = 0), so clearly there is a maximum in every direction except that there is a minimum along x = y

the surface starts to go down everywhere, except that it goes slowly up (~ r⁴) along x = y )​

… we have the awkward situation that at O it is only defined in the second and fourth quadrants …

no, it's defined everywhere except between a couple of curves that approach O tangentially to x = y

 

[aruwin]

tiny-tim, can you help me with points A and B? I have expanded the function and I am stucked =(

 

[tiny-tim]

… I have expanded the function and I am stucked =(

(you mean "stuck" )

but doesn't your professor want you to use the f_xx etc formula?

 

[aruwin]

(you mean "stuck" )

but doesn't your professor want you to use the f_xx etc formula?

Isn't the f_xx formula is derived from the taylor expansion??I'll copy it again here =)

The taylor expansion at A(√2,-√2) is
f(x,y)≈ -8 + 1/2![20(x-√2)2+8(x-√2)(y+√2)+20(y+√2)2]+...

The taylor polynomial until the 2nd term is
f(x,y) = -8+10(x-√2)2+4(x-√2)(y+√2)+10(y+√2)2

I think I have figured something out.From the quadratic equation that we got from the taylor expansion,-8 is the constant and we have a square and thus, f(x,y)is always going to be bigger or equals to -8 and so we have a minimum value of -8.Am I right?

 

[haruspex]

not exactly … our quadratic formula gives all the information we need except in one direction (which we can check "by hand"):

I'd understood from earlier posts that the purpose of the exercise was to practise use of Taylor expansions. You can resolve the ambiguity by going further down the expansion, but in the present case, as you and I have both indicated, it's easier 'by hand'.

no, it's defined everywhere except between a couple of curves that approach O tangentially to x = y

ah yes... it's defined in a figure-of-8, which does spill into the first and third quadrants.

Yes, without the ≤0, O is a saddle. The ≤0 is there to state the region of f(x,y).

OK, but on that argument the whole of the curve along which f = 0 constitutes a local maximum.

I'm stuck here, from the equation, it seems that at (√2,-√2), f(x,y) is always going to be -8 but if I put other values it is always positive or can I just ignore the "-8"?

For the purpose of asking whether it's a max etc., the -8 is irrelevant. You don't care what the value of f is there, you only care how the value changes as you move from that point in different directions. If it increases no matter which way you move it's a minimum.