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

• member 731016
member 731016
Homework Statement
Relevant Equations
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!

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.

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