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Two variables polynomial

  1. Dec 16, 2012 #1
    Hi
    Let p(x,y)≥0 be a polynomial of degree n such that p(x,y)=0 only for x=y=0.Does there exist a positive constant C such that the inequality p(x,y)≥C (IxI+IyI)^n (strong inequality!) holds for all -1≤x,y≤1?
    The simbol I I stands for absolute value.
     
  2. jcsd
  3. Dec 22, 2012 #2

    chiro

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    Hey hedipaldi.

    What does (IxI + IyI)^n refer to? (what are the I's)?
     
  4. Dec 22, 2012 #3
    Thise are absolute values.It means [abs.val(x)+abs.val(y)]^n
     
  5. Dec 22, 2012 #4

    chiro

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    If (0,0) is the only root then it means that everything is greater than 0.

    The only thing now is to consider the makeup of a 2D polynomial.

    If the double polynomial (or bivariate polynomial) has a structure p(x,y) = (a_n*x^n + a_(n-1)*x^(n-1) + ... + a0)*(b_n*y^n + b_(n-1)*y^(n-1) + ... + b0) and consider the behaviour in the region |x|, |y| <= 1.
     
  6. Dec 22, 2012 #5
    Does the limit p(x,y)/[absvalue(x)+absvalue(y)]^n nesecarily exist (finite or +infinite)?
    This will solve my problem.
     
  7. Dec 22, 2012 #6

    chiro

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    What limit are you thinking of? (In other words what does x and y tend to)?
     
  8. Dec 22, 2012 #7
    x and y tend to o. i.e (x,y) tends to (0,0).
     
  9. Dec 22, 2012 #8

    chiro

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    It will tend to zero because all polynomials (including bi-variate ones) are continuous.

    Continuity implies that lim x->a, y->b f(x,y) = f(a,b) = 0 for (a=0,b=0).
     
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