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Singularities of two variables rational functions

  1. Dec 23, 2012 #1
    1. The problem statement, all variables and given/known data

    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)}?
    2. Relevant equations



    3. 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?
     
    Last edited: Dec 23, 2012
  2. jcsd
  3. Dec 23, 2012 #2

    mfb

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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"?
     
  4. Dec 23, 2012 #3
    This is not a rational function,it is not a quotient of two polynomials.
     
  5. Dec 23, 2012 #4

    vela

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    What is that supposed to mean?
     
  6. Dec 23, 2012 #5
    for example (x^4+3X^2*Y+X*Y)/(X^2+6Y^2)
     
  7. Dec 23, 2012 #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.
     
  8. Dec 23, 2012 #7

    mfb

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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.


    If you consider a factor with power 0 as factor.
     
  9. Dec 23, 2012 #8

    pasmith

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    True.
     
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