Second partial test

1. Oct 30, 2012

Zondrina

1. The problem statement, all variables and given/known data

1. What happens to D = fxxfyy - (fxy)2 at (0,0) for f(x,y) = 9x4 - 6x2y2 + y4? Classify the critical point at (0,0).

2. How about if f(x,y) = (y - x2)(y - x4) ?

2. Relevant equations

See above ^.

3. The attempt at a solution

1. Okay, so after taking all the partial derivatives and computing D, we get that D = 0 at (0,0) so that the second partial test fails.

So we have to consider some other method to see what happens at the origin.

Suppose we set y=0, so that f(x,0) = 9x4. Then fx(x,0) = 36x3.

Now if x < 0, fx < 0 and if x > 0, fx > 0. So it is clear that (0,0) acts like an inflection point along the x-axis and we can conclude that (0,0) is a minimum?

I'm not sure about my logic here since it's my first time trying one of these. Anyone have any pointers?

2. Same process as before, I take all my derivatives and compute D to get D = 0 so that at (0,0) my test fails.

Again I'll suppose that y = 0, so that f(x,0) = x6 and fx(x,0) = 6x5.

Now if x < 0, fx < 0 and if x > 0, fx > 0. So it is clear that (0,0) acts like an inflection point along the x-axis once again and we can conclude that (0,0) is a minimum?

I'm hoping my reasoning is correct.

2. Oct 30, 2012

Zondrina

So for question (1), the first three partials of x are all equal to zero at (0,0) as are the first three partials of y. The fourth partials of both though are both non zero at (0,0) so I have that :

n+1 = 4 which means n = 3 which is odd.

So we have n is odd and the fourth partials of x and y are both greater than zero so that (0,0) is a minimum.

Would this be more appropriate than what I did above?

3. Oct 30, 2012

Dick

Those don't sound like descriptions of inflection points. In both cases the curve y=0 has a minimum at y=0. For 2, can you think of a curve which has negative values near (0,0)? For 1, try 'completing the square' in y or factoring and thinking some more about it.

Last edited: Oct 30, 2012
4. Oct 30, 2012

Ray Vickson

The functions f1 (question 1) and f2 (question 2) behave very differently. For f2 the point (0,0) is neither a max nor a min; that is, in any two-dimensional neighbourhood of (0,0) there are points giving f2(x,y) > f2(0,0) and other points (x,y) giving f2(x,y) < f2(0,0).

RGV

5. Oct 30, 2012

Zondrina

Ahhh I see what you're getting at. So I'll formalize these thoughts a bit here.

For (1), the test fails so we turn to other methods.

If we set y = 0, then fx has a critical point at 0 and fxx has an inflection point at 0. If x < 0, then fx < 0 and fxx < 0. If x > 0, fx > 0 and fxx > 0. So for y=0, the point (0,0) acts like an inflection point along the x-axis and we can clearly see that (0,0) is a min for f ( Same process for (0,y) and the same result ).

Now for (2), the test fails again so we turn to other methods.

Now, for any neighborhood around (0,0), $\exists Q, P \in N_δ((0,0)) | f(Q) > f(0,0) \wedge f(P) < f(0,0)$.

So for example, take any x>1 and y>x4 as to make f(Q) > f(0,0). Take any x > 1 and x2 < y < x4 ( Say y = x3 ) as to make f(P) < f(0,0). Thus f has a saddle point at (0,0) and we are done.

I never considered that I could argue using neighborhoods around the point. Very informative.

6. Oct 30, 2012

Dick

y=x^3 is a good choice for the second one. For the first one showing it's a minimum along x=0 and y=0 does not suffice to show it's a minimum. The second one satisfies that as well. Write the first one as an expression involving squares.

7. Oct 30, 2012

Ray Vickson

Your analysis of (1) is incorrect. When y = 0 you have f1 = 9x^4, and this has a global min at x = 0---just look at the graph of x^4. It is true that f1' = 0 and f1'' = 0 at x = 0, but that does not mean that you have a saddle point.

Similarly, when x = 0 you have f1 = y^4, which is minimized at y = 0.

That takes care of the two axes. However, you have a 2-dimensional problem, so you need to worry about points off the two axes as well.

Personally, if I were doing it I would first write X = x^2 and Y = y^2, to get f1 = 9X^2 - 6XY + Y^2, which is much easier to work with. If you stare at it for a while you may see something rather nice happening.

RGV

8. Oct 30, 2012

Ray Vickson

Your analysis of (1) is incorrect. When y = 0 you have f1 = 9x^4, and this has a global min at x = 0---just look at the graph of x^4. It is true that f1' = 0 and f1'' = 0 at x = 0, but that does not mean that you have a saddle point.

Similarly, when x = 0 you have f1 = y^4, which is minimized at y = 0.

That takes care of the two axes. However, you have a 2-dimensional problem, so you need to worry about points off the two axes as well.

Personally, if I were doing it I would first write X = x^2 and Y = y^2, to get f1 = 9X^2 - 6XY + Y^2, which is much easier to work with. If you stare at it for a while you may see something rather nice happening.

The function f2 of question 2 has the interesting, but counter-intuitive property, that if you go away from (0,0) in any direction, the function goes up strictly (that is, for *any* direction p = (p_x,p_y) in R^2 we have f(t*p) > f(0,0) for small scalar t > 0); nevertheless, (0,0) is NOT a local minimum! Don't worry if you don't "see it"; I saw a similar example a few years ago in an optimization seminar. Later, after you have done with this exercise I can reveal the secret.

RGV

9. Oct 30, 2012

Zondrina

Okay so at least (2) looks good :). I'm still a bit shaky as to why the argument implies that there is a saddle point though.

For (1) I'm pretty sure I could re-write it as (y2-3x2)2, but I'm terrible at completing squares ( A skill I never mastered ).

EDIT : Oh wait, let me try your trick by subbing in for x^2 and y^2 first.

10. Oct 30, 2012

Dick

In this case just factoring will do it. You did that. Now classify the critical point.

11. Oct 30, 2012

Zondrina

EDIT : That was a dumb idea I had since applying the test would give the same result.

Last edited: Oct 30, 2012
12. Oct 30, 2012

Ray Vickson

Using the factored form allows you to avoid derivatives completely. Squares are always ≥ 0, and if you can get a value of 0 that must be a minimum---end of story.

RGV

13. Oct 30, 2012

Dick

No, knowing the function can expressed as (y^2-3x^2)^2 tells everything you need to know about how it behaves near the origin. For one thing, it's always nonnegative, yes?

14. Oct 30, 2012

Zondrina

Oh! I see. So (y2-3x2)2 ≥ 0 no matter what. So the point (0,0) gives us 0 so that (0,0) gives a min.

So taking a wild crack at this, if in theory I got a scenario where f ≤ 0 rather than f ≥ 0, it would be a maximum instead? If f = 0 it would be a saddle?

15. Oct 30, 2012

SammyS

Staff Emeritus
The function, $\displaystyle f(x,\,y)=9x^4 - 6x^2y^2 + y^4\,,\$ is an interesting function in that the minimum at (0, 0) is not an isolated minimum.

16. Oct 30, 2012

Dick

If f(x,y)<=0 and f(0,0)=0 then (0,0) would be a local max. Right? I'm not sure what you mean by f=0 and why that would be a saddle. Suggest you rethink the question.

17. Oct 30, 2012

Zondrina

Yes that's what I was trying to say, but failed ^

So

f(x,y)<=0 and f(0,0)=0 then (0,0) max
f(x,y)>=0 and f(0,0)=0 then (0,0) min
f(x,y) = 0 and f(0,0)=0 then (0,0) is a saddle?

EDIT : Or can I not determine a saddle in this manner.

18. Oct 30, 2012

Dick

19. Oct 30, 2012

Zondrina

Okay.. slowing down a few notches here. Looking back at my argument for part (2), a saddle occurs when no matter what neighborhood I take around my origin, there will be points Q and P where f(Q) > f(0,0) and f(P) < f(0,0). So f = 0 means nothing to me without considering neighborhoods!

20. Oct 30, 2012

Dick

That's pretty much it. You might want to look up the definition of saddle point to clarify it further.