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Lower bound

  1. Dec 24, 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
    I observed that p(x,y) must have even degree.Also if the quotient tend to infinity at the origin the answer is yes.Otherwise p(x,y) must be hogeneous,and this may imly that the quotient has a positive lower bound.I need help for progressing
     
  2. jcsd
  3. Dec 25, 2012 #2

    haruspex

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    Have you tried a very simple example, like x^2+y^2?
     
  4. Dec 25, 2012 #3
    This is not a counter example.It has a positive lower bound near the origin.
     
  5. Dec 25, 2012 #4

    haruspex

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    ... and therefore it is possible. Are you sure the wording of the OP is as you intend?
     
  6. Dec 25, 2012 #5
    As i understood,the meaning is to show that for every such p(x,y) there exists such C.
    How do you understatd the wording?
     
  7. Dec 25, 2012 #6
    The original wording is attached:Q.5
     

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  8. Dec 25, 2012 #7

    haruspex

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    The original wording makes more sense. To express it you should have written "Is it guaranteed that..."
    If I have any helpful thoughts I'll post again.
     
  9. Dec 26, 2012 #8
    Thank's
     
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