Singularities of two variables rational functions

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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)}?

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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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Answers and Replies

  • #2
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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"?
 
  • #3
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This is not a rational function,it is not a quotient of two polynomials.
 
  • #4
vela
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Homework Statement



Does a positive rational function of two variables can have singular points at wich the limit is nor final nor infinity?
What is that supposed to mean?
 
  • #5
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for example (x^4+3X^2*Y+X*Y)/(X^2+6Y^2)
 
  • #6
pasmith
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I think the point of the restriction to positive f is to eliminate functions like
[tex]f(x,y) = \frac{(x - a)(y - b)}{(bx - ay)(bx + ay - 2ab)}[/tex]
where something clearly goes wrong at [itex](a,b)[/itex].

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.
 
  • #7
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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.
 
  • #8
pasmith
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If you consider a factor with power 0 as factor.

True.
 

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