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

  • #1
member 731016
Homework Statement
Please see below
Relevant Equations
Please see below
For this problem,
1717971710447.png

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,
1717971693519.png

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

Thanks for any help!
 
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  • #2
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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