Prove that the polynomial is $g(x,y)(x^2 + y^2 -1)^2 + c$
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This is from a Brazilian math contest for college students (OBMU):
Let $f(x,y)$ be a polynomial in two real variables such that the polynomials
$$frac{partial f}{partial x}(x,y)$$
$$frac{partial f}{partial y}(x,y)$$
are divisible by $x^2+y^2-1$. Prove that there's a polynomial $g(x,y)$ and a constant $c$ such that
$$f(x,y) = g(x,y)(x^2+y^2 -1)^2 +c$$
real-analysis multivariable-calculus contest-math multivariate-polynomial
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up vote
0
down vote
favorite
This is from a Brazilian math contest for college students (OBMU):
Let $f(x,y)$ be a polynomial in two real variables such that the polynomials
$$frac{partial f}{partial x}(x,y)$$
$$frac{partial f}{partial y}(x,y)$$
are divisible by $x^2+y^2-1$. Prove that there's a polynomial $g(x,y)$ and a constant $c$ such that
$$f(x,y) = g(x,y)(x^2+y^2 -1)^2 +c$$
real-analysis multivariable-calculus contest-math multivariate-polynomial
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This is from a Brazilian math contest for college students (OBMU):
Let $f(x,y)$ be a polynomial in two real variables such that the polynomials
$$frac{partial f}{partial x}(x,y)$$
$$frac{partial f}{partial y}(x,y)$$
are divisible by $x^2+y^2-1$. Prove that there's a polynomial $g(x,y)$ and a constant $c$ such that
$$f(x,y) = g(x,y)(x^2+y^2 -1)^2 +c$$
real-analysis multivariable-calculus contest-math multivariate-polynomial
This is from a Brazilian math contest for college students (OBMU):
Let $f(x,y)$ be a polynomial in two real variables such that the polynomials
$$frac{partial f}{partial x}(x,y)$$
$$frac{partial f}{partial y}(x,y)$$
are divisible by $x^2+y^2-1$. Prove that there's a polynomial $g(x,y)$ and a constant $c$ such that
$$f(x,y) = g(x,y)(x^2+y^2 -1)^2 +c$$
real-analysis multivariable-calculus contest-math multivariate-polynomial
real-analysis multivariable-calculus contest-math multivariate-polynomial
edited 2 days ago
Jean Marie
28.1k41848
28.1k41848
asked 2 days ago
Rafael Deiga
657310
657310
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