Condition for a point to lie inside a closed curve

Suppose f(x,y) represents a polynomial function in x and y in which the coefficients of the highest powers of x and y are positive. Prove that a point P(a,b) lying inside a closed curve f(x,y) must satisfy f(a,b)<0.

#Calculus #Advice #MathProblem #Math

Easy Math Editor

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Comments

I don't get this one. What if $f(x,y) \equiv -x^2-y^2+1$ ? $f(x,y) = 0$ is the same curve as $x^2+y^2-1 = 0$ ; that is, the unit circle. But for the origin $P(0,0)$ which is inside the curve, we have $f(P) = f(0,0) = 1 > 0$ . Am I missing something?

Ivan Koswara - 7 years, 7 months ago

I think we have to keep the coefficients of highest power positive. If x and y have the same power and different signs then I'm not sure what is to be done. For, the time being lets try the first case only.

Sorry to have missed out on that detail.

Sambit Senapati - 7 years, 7 months ago

If the highest powers have different signs, the curve $f(x,y)=0$ is most unlikely to be closed.

Mark Hennings - 7 years, 7 months ago

And you are going to need to restrict your attention to polynomial functions (or at least ones for which $f(x,y) \to \infty$ as $(x,y) \to \infty$ , since otherwise curves like $e^{-x^2} + e^{-y^2} = 1.5$ will cause you problems as well.

Mark Hennings - 7 years, 7 months ago

I have modified the problem.

Sambit Senapati - 7 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$