Solving for p at (0,0): Origin Singularity in f(x_1,x_2)

[Number2Pencil]

Homework Statement

Given the function
<br /> f(x_1,x_2) = (x_1 - x_2^2)(x_1 - px_2^2)<br />

where p is a constant parameter, for what value of p will the origin (0,0) be a singular point of this function?

Homework Equations

The Attempt at a Solution

I thought that singular meant either a discontinuity or shooting to infinity. I'm not really sure what to do here. I thought since singular point is related to curve smoothness and slope, I should take the gradient, and I got:

<br /> \frac{\delta f}{\delta x_1} = 2x_1 - px_2^2 - x_2^2<br />
<br /> \frac{\delta f}{\delta x_2} = -2 x_2 (x_1-px_2^2) - 2(x_1 - x_2^2)p x_2<br />

Where at the point (0,0), the value of p really makes no difference. Is my approach wrong?

 

[D_Tr]

If you get the answer, I would like to know it! Maybe it asks for something else and you misheard / incorrectly copied the problem statement? I am saying this because at (0,0) you have a stationary point and the problem makes total sense if you replace "singular" with "saddle", for example.

 

[Number2Pencil]

I talked to my professor, and he said "the singularity will hold irrespective of the value p." So it looks like I was correct about p not really making a difference.

But I would still like to know WHY the point (0,0) is a singular point. Any thoughts?

 

[pbuk]

I thought that singular meant either a discontinuity or shooting to infinity.

No, you are thinking of a singularity. A singular point is something entirely different.

 

[Mark44]

<br /> \frac{\delta f}{\delta x_1} = 2x_1 - px_2^2 - x_2^2<br />
<br /> \frac{\delta f}{\delta x_2} = -2 x_2 (x_1-px_2^2) - 2(x_1 - x_2^2)p x_2<br />

Tip: For partial derivatives, use \partial instead of \delta, like so:
$$\frac{\partial f}{\partial x_1}$$

 

[Number2Pencil]

No, you are thinking of a singularity. A singular point is something entirely different.

Well my professor must have been confused in his response as well, because he definitely wrote the word "singularity"

Could you provide any tips on how to solve this problem? From your link, it looks like the singular point is where the function intersects with another point of the function. Nothing I have tried can make that happen.

 

[vela]

How did your professor define singular point? That's where you should start.

 

[pbuk]

Well my professor must have been confused in his response as well, because he definitely wrote the word "singularity.

The two terms can sometimes be used in each other's place, the context can be used to tell which is meant: the function you quoted clearly does not have a discontinuity at (0, 0) for any value of p so that is not what is being looked for.

Could you provide any tips on how to solve this problem? From your link, it looks like the singular point is where the function intersects with another point of the function.

Not necessarily. You have calculated the partial derivatives, how can these be used to identify a singular point (Wikipedia may be more helpful than Mathworld for this)?

 

[Number2Pencil]

The prof didn't bother defining singular point, and yes I did pay attention in class! hah

Thanks for the tip all, I had no idea singular point COULD be different than singularity. Sure enough I found something that seems helpful on Wikipedia:

http://en.wikipedia.org/wiki/Singular_point_of_a_curve

The singular points are those points on the curve where both partial derivatives equal 0. So in my case, any value of p would result in both partial derivatives being zero, meaning the origin point (0,0) is always a singular point.