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two variables polynomial |
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| Dec16-12, 11:46 AM | #1 |
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two variables polynomial
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. |
| Dec22-12, 01:35 AM | #2 |
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Hey hedipaldi.
What does (IxI + IyI)^n refer to? (what are the I's)? |
| Dec22-12, 06:09 AM | #3 |
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Thise are absolute values.It means [abs.val(x)+abs.val(y)]^n
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| Dec22-12, 05:58 PM | #4 |
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two variables polynomial
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. |
| Dec22-12, 07:06 PM | #5 |
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Does the limit p(x,y)/[absvalue(x)+absvalue(y)]^n nesecarily exist (finite or +infinite)?
This will solve my problem. |
| Dec22-12, 07:24 PM | #6 |
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What limit are you thinking of? (In other words what does x and y tend to)?
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| Dec22-12, 07:47 PM | #7 |
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x and y tend to o. i.e (x,y) tends to (0,0).
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| Dec22-12, 07:55 PM | #8 |
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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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