Singularities of two variables rational functions

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Homework Help Overview

The discussion revolves around the behavior of rational functions of two variables, particularly focusing on singularities and limits at the origin. The original poster questions whether a specific quotient involving a positive polynomial can maintain a positive lower bound in a defined punctured rectangle.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the conditions under which a rational function can be positive and discuss the implications of limits approaching the origin. There are inquiries about the nature of singular points and the behavior of functions near these points.

Discussion Status

The conversation includes various interpretations of rational functions and their properties. Some participants have offered insights into the requirements for positivity in rational functions, while others are questioning the definitions and implications of singularities. There is no explicit consensus on the conclusions yet.

Contextual Notes

Participants are considering the implications of positive polynomials and the behavior of functions at singular points, with some noting the need to eliminate problematic functions that do not conform to the expected behavior. The discussion also touches on the limits of functions as they approach the origin.

hedipaldi
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Homework Statement



Let p(x,y) be a positive polynomial of degree n ,p(x,y)=0 only at the origin.Is it possible that
the quotient p(x,y)/[absolute value(x)+absval(y)]^n will have a positive lower bound in the punctured rectangle [-1,1]x[-1,1]-{(0,0)}?

Homework Equations


The Attempt at a Solution


When the limit at the origin is 0,there is no positive lower bound.If the limit is infinity (which is not seems to be possible) there is such lower bound.This quotient is a special case of rational function in each quadrant,and i try to analyse it.Can someone help?
 
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f:R^2->R, f(x,y)=1+sin(1/(x^2+y^2)) for (x,y)!=(0,0), f(0,0)=1?
Where is the point of the two variables and "positive"?
 
This is not a rational function,it is not a quotient of two polynomials.
 
hedipaldi said:

Homework Statement



Does a positive rational function of two variables can have singular points at which the limit is nor final nor infinity?
What is that supposed to mean?
 
for example (x^4+3X^2*Y+X*Y)/(X^2+6Y^2)
 
I think the point of the restriction to positive f is to eliminate functions like
f(x,y) = \frac{(x - a)(y - b)}{(bx - ay)(bx + ay - 2ab)}
where something clearly goes wrong at (a,b).

Clearly for a rational function to be positive, numerator and denominator must either both be negative or both be positive. That seems to require that the numerator and denominator have the same irreducible factors, and that for a given factor either both have it to an even power, or both have it to an odd power.
 
Oh rational function, sorry.

$$f(x,y)=\frac{x^2+y^4}{y^2+x^4}$$
Where x4 just reduces the singular line to a single point and y4 makes it positive everywhere where the function is defined.

Approaching (0,0) as (h,h) gives a limit of 1, approaching it as (2h,h) gives a limit of 4, so the limit does not exist.
That seems to require that the numerator and denominator have the same irreducible factors
If you consider a factor with power 0 as factor.
 
mfb said:
If you consider a factor with power 0 as factor.

True.
 

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