Determining whether functions are sign definite, sign semidefinite, or sign indefinite

[member 731016]

Homework Statement

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Relevant Equations

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For this problem,

I am confused how they get all their negative definite, positive definite, and negative semidefinite domains. I agree that $V_2$ is positive definite by definition. However, for $V_3$ I think they made a mistake since by definition, $V_3$ is negative simi-definite.

I'm confused by $V_1$ since Wolfram alpha gives,

So it must be sign indefinite since it is neither negative simi-definite nor positive simi-definite by definition.

Thanks for any help!

 

[andrewkirk]

The value of V1(x,y) will never be positive, being the the negative of the sum of two squares of real numbers. So it is either negative definite or negative semi-definite.
For it to be negative semi-definite, there must be some x,y pair that is not 0,0, for which it gives zero. That means that both terms must be zero. That means that we must have x=y for the second term to be zero. For the first term to be zero we must have $x=k\pi$ for some integer $k$. The answer makes the requirement that $x^2+y^2<\pi^2$. Let's discuss that requirement below. But given that requirement, the only integer $k$ that gives a point in that disc is $k=0$, which is the point $x=y=0$. So there are no nonzero points in the domain $D$ at which the function gives zero. So it is negative definite rather than semidefinite on that domain.

Now why does it restrict the function to that domain? I suspect there is some important context missing, either in the question or in the preceding material, where it defines these terms. In particular, I suspect your text is using the second definition from this page, which makes a function negative definite if there is some neighbourhood of the origin on which the function is always negative except at the origin. The answer identifies such a neighbourhood, being the open disc of radius $\pi$ so, under that definition, the function is negative definite.