Lower bound


by hedipaldi
Tags: bound
hedipaldi
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#1
Dec24-12, 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
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haruspex
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#2
Dec25-12, 05:27 AM
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Have you tried a very simple example, like x^2+y^2?
hedipaldi
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#3
Dec25-12, 06:05 AM
P: 206
This is not a counter example.It has a positive lower bound near the origin.

haruspex
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#4
Dec25-12, 06:26 PM
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Lower bound


Quote Quote by hedipaldi View Post
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?
hedipaldi
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#5
Dec25-12, 06:34 PM
P: 206
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?
hedipaldi
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#6
Dec25-12, 06:58 PM
P: 206
The original wording is attached:Q.5
Attached Thumbnails
001.jpg  
haruspex
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#7
Dec25-12, 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.
hedipaldi
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#8
Dec26-12, 07:41 AM
P: 206
Thank's


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