# Homework Help: 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.

21. Oct 30, 2012

### Zondrina

Thanks a bundle for your help. Yeah I'm going to go review a few definitions now and write out a few study notes revolving around this.

Thanks again.