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Lower boundby hedipaldi
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#1
Dec2412, 05:59 PM

P: 206

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
Dec2512, 05:27 AM

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Have you tried a very simple example, like x^2+y^2?



#3
Dec2512, 06:05 AM

P: 206

This is not a counter example.It has a positive lower bound near the origin.



#4
Dec2512, 06:26 PM

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P: 9,921

Lower bound



#5
Dec2512, 06:34 PM

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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
Dec2512, 06:58 PM

P: 206

The original wording is attached:Q.5



#7
Dec2512, 11:31 PM

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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
Dec2612, 07:41 AM

P: 206

Thank's



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