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The following system of three nonlinear algebraic equations is to be solved for $~x, y, z~ $ as functions of the variables $~u, v,w~$:

$u = x + y + z$

$v = x^2 + y^2 + z^2$

$w = x^3 + y^3 + z^3$

$a)$ Prove or find a counter example: for each $(u,v,w)$ near $(0,2,0)$, there is a unique solution $(x,y,z)$ near $(-1,0, 1)$.

$b)$ Is the Implicit Function Theorem applicable for $(u,v,w)$ near $(2, 4, 8)$ and $(x,y, z)$ near $(0, 0, 2)$?

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1 Answer

Consider the function $f:\mathbb{R}^6 \rightarrow \mathbb{R}^3$ defined by your three equations. The Jacobian can be written

$$J = \left[ \frac{\partial f_i}{\partial u_j} \left| \frac{\partial f_i}{\partial x_j} \right.\right] = \left[ A | B \right]$$

with $u=(u,v,w)$ and $x=(x,y,z)$ (pardon the abuse of notation). The implicit function theorem guarantees the existence of a unique $(x,y,z)$ for any $(u,v,w)$ in the neighborhood of a region where $f=0$, whenever $B$ is invertible.

So you should calculate $B$, which is the matrix of partial derivatives of $f$ with respect to $x$, $y$ and $z$, and see if it is invertible at:

  1. $(u,v,w) = (0,2,0)$ and $(x,y,z) = (-1,0,1)$

  2. $(u,v,w) = (2,4,8)$ and $(x,y,z) = (0,0,2)$

For completeness, you should also check that $f(u,v,w,x,y,z)=0$ is satisfied at those points.

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thanks a lot you saved my life.. –  smoke Feb 16 '12 at 19:37
    
@smoke Feel free to click the 'accept' button. –  Chris Taylor Feb 16 '12 at 19:58
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