Lower bound

  • Thread starter 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


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
 

Answers and Replies

  • #2
haruspex
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Have you tried a very simple example, like x^2+y^2?
 
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This is not a counter example.It has a positive lower bound near the origin.
 
  • #4
haruspex
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This is not a counter example.It has a positive lower bound near the origin.
... and therefore it is possible. Are you sure the wording of the OP is as you intend?
 
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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?
 
  • #6
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The original wording is attached:Q.5
 

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  • #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.
 
  • #8
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Thank's
 

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