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Prove that the following system:

\begin{align}F(x,y,z)&=2x^2+y^2+1-z^2=0\\ G(x,y,z)&=2x^2+2y^2-z^2=0\end{align}

can be solved for $z$ and $y$ as functions of $x$.

Furthermore, provide the values of the points that allow the parameterization.

An attempt:

According to the implicit function theorem as long as the determinant of the jacobian given by $\frac{\partial(F,G)}{\partial(y,z)}$ ($\partial(F,G)$ represents the partial derivative) is not equal to $0$, the parametrization is possible. $$\frac{\partial(F,G)}{\partial(y,z)}=4yz$$

Meaning that all points where $z$ and $y$ are not $0$.

My solution:

Same as previous, conclude that according to the implicit function theorem it is all points where $z$ and $y$ are not $0$. However, test the possibilities where $y=0$ and $z=0$.

  • $y=0$: by subtracting $F-G$ you get $1=0$, which is impossible.

  • $z=0$: gives $G=2x^2+2y^2=0$ which gives $x=y=0$, which is impossible for $F$, as it gives $1=0$.

My question:

Is it necessary to consider the possibilities of $y=0$ and $z=0$. These possibilities should be impossible as either one makes the determinant of the jacobian $0$ which if you were to use Cramer's rule to find the partial derivative: $\frac{\partial(y)}{\partial(x)}$ would lead to dividing by zero.

Furthermore, it leads to the same conclusion. Is there a reason to why it is important and are there cases where it would not lead to the same conclusion?

Mina Farag
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  • Note that if we substract $G-F$, we obtain $y^2-1=0$, si $y=\pm1$ – Alain Remillard Mar 17 '20 at 10:40
  • The implicit function theorem says nothing about a point at which the Jacobian vanishes, so to be thorough you do have to check those cases separately. – amd Mar 17 '20 at 20:37
  • @amd The implicit function theorem is more or less a consequence of Cramer's rule which implies that at points where the determinant of the jacobian is 0, the denominator needed to solve for ∂(y)/∂(x) is will also be zero, hence those points should not be considered at all. For further clarification, the denominator of the quotient that produces the value of ∂(y)/∂(x) and ∂(z)/∂(x) is actually the determinant of the Jacobian, hence as long as the determinant is not 0, we are not dividing by 0. This is why the solution is bizarre to me, why consider y=0 or z=0 when they lead to division by 0? – Mina Farag Mar 17 '20 at 22:07
  • Nonzero Jacobian is only a sufficient condition. See this answer for a simple example of an implicit function $F:\mathbb R^2\to\mathbb R$ for which the $F_y$ vanishes at a point near which $y$ can be expressed as a function of $x$. – amd Mar 18 '20 at 19:47

1 Answers1

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Personally I'm not sure about those ideas but I have this solution.

Cnsider the difference of G and F: $$ G - F = y^2 -1 $$ $$ y^2 - 1 = 0 $$ This implies that $y^2 = 1$ giving: $$ F(x,y,z) = 2x^2 + 2 - z^2 = 0 $$ $$ G(x,y,z) = 2x^2 + 2 - z^2 = 0 $$

This simply means $$ z^2 = (2x^2 + 2) $$ $$ z = \pm(2x^2 + 2)^\frac{1}{2} $$

Let $t = x$ then we have one of four solutions: $$(x,y,z) = (t,1,\pm(2t^2+2)^\frac{1}{2})$$ $$(x,y,z) = (t,-1,\pm(2t^2+2)^\frac{1}{2})$$

Though for a proof of existence of a solution it would suffice to just show that $(t,1,(2t^2+2)^\frac{1}{2})$ when put in to F and G is a solution for both.

just_floating
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