4
$\begingroup$

I am stuck at this problem and I would be glad if somebody helped me out:

By the implicit function theorem, it should be shown that: $$f(x,y,z) := z^{3}+2xy-4xz+2y-1 .$$

The zero level set $f^{-1}(0)$ in a neighborhood $U$ of $(x_0, y_0) = (1,1)$ can be rewritten through a differentiable function $z=g(x,y)$ where $g(1,1)=1$. Then also the partial derivatives: $g_x(1,1), g_y(1,1)$ should be calculated.

The partial derivative in respect to $z$ is: $f_z = 3z^2- 4x$ and it holds that $3z^2 - 4x \ne 0$, because otherwise the invertibility would not be given anymore. So if $(x_0, y_0)=(1,1)$ then $z^3- 4z+4=0$ gives $3$ solutions, of which none are such that $3z^2-4 = 0$. So the zero level set in $U$ can be rewritten through $g(x,y)$.

Now the remaining task is to solve $z^3 +2xy-4xz+2y-1=0$ for $z$. I sit around at this end:

$$z^3- 4xz=-2xy-2y+1 = z(z^2- 4x)=-2xy-2y+1 \Rightarrow z = \frac{-2xy-2y+1}{z^2-4x} .$$

According to Wolfram Alpha, there are 3 exact solutions.

$\endgroup$
  • $\begingroup$ I think you need to check that $f_z$ is non-zero at $(1,1,1)$. Isn't that point given? $\endgroup$ – Dylan Moreland Dec 6 '11 at 0:37
  • $\begingroup$ $z^{3}+2xy-4xz+2y-1=0$ has three solutions for $z$ for fixed $x$ and $y$ because it is a cubic equation. For $x=y=1$ the solutions are $z=1$ and $\frac{-1 \pm \sqrt{13}}{2}$. But you seem to be missing part of the question. $\endgroup$ – Henry Dec 6 '11 at 0:39
  • $\begingroup$ @Dylan Moreland Thanks! $f_{z}$ is -1 at (1,1,1). But why (1,1,1)? I don't see where you want me to go with this, either... $\endgroup$ – Tashi Dec 6 '11 at 0:42
  • $\begingroup$ @Tashi $g$ is spitting out $z$-values, and you're told that it's spitting out $1$ at $(1,1)$. Maybe I'm misinterpreting the question? $\endgroup$ – Dylan Moreland Dec 6 '11 at 0:45
  • $\begingroup$ Thanks @Dylan Moreland: I understood what you mean with $f_{z}$ non zero at (1,1). So there is no need to find $g(x,y)$? How to calculate $g_{x}(1,1)$ and $g_{y}(1,1)$ without knowing $g(x,y)$ ? $\endgroup$ – Tashi Dec 6 '11 at 1:09
2
$\begingroup$

By implicit function theorem, if we can show that $f_z(1,1)\neq 0$, then "the zero level set $f^{−1}(0)$ in a neighborhood $U$ of $(x_0,y_0)=(1,1)$ can be rewritten through a differentiable function $z=g(x,y)$", as you would like to show. To see this, we have $$f_z(1,1)=3z^{2}-4x\Big|_{(x,y,z)=(1,1,1)}=3-4=-1\neq 0.$$ Now, sinc $f^{−1}(0)$ can be written as $z=g(x,y)$ in a neighborhood $U$ of $(x_0,y_0)=(1,1)$, we have $f(x,y,g(x,y))=0$, or equivalently, $$g(x,y)^{3}+2xy-4xg(x,y)+2y-1=0.$$ Now, take the partial derivative of the above equation with respect to $x$, we obtain $$3g^2(x,y)g_x(x,y)+2y-4g(x,y)-4xg_x(x,y)=0.$$ Evaluate it at $(x,y)=1$ and by $g(1,1)=1$, we have $$3g_x(1,1)+2-4-4g_x(1,1)=0,$$ which implies that $g_x(1,1)=-2$. Similarly, take the partial derivative of the above equation with respect to $y$, evaluate it at $(x,y)=1$, we obtain $$3g^2(1,1)g_y(1,1)+2-4g_y(1,1)+2=0.$$ Since $g(1,1)=1$, we get $g_y(1,1)=4$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.