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I have a question regarding the Implicit Function Theorem which I'll ask by way of an example...

Can the equation $\sqrt{x^2+y^2+2z^2}=\cos(z)$ be solved uniquely for $y$ in terms of $x$ and $z$ near $(0,1,0)$? For $z$ in terms of $x$ and $y$?

Applying the theorem in the case of $y$, with $F(x,y,z)=\sqrt{x^2+y^2+2z^2}-\cos(z)$, we have $F(0,1,0)=0$, and $\frac{\partial F}{\partial y}(0,1,0)\neq 0$, so this satisfies the Theorem and I know that I can solve for $y$ in terms of $x$ and $z$ on some neighborhood of $(0,1,0)$.

My question, however, involves solving for $z$ in terms of $x$ and $y$. In this case $\frac{\partial F}{\partial z}(0,1,0)= 0$, and so the hypothesis of the theorem is not satisfied. Is this sufficient information to tell me that $F$ can not be solved uniquely for z in terms of x and y in some neighborhood of $(0,1,0)$, or does it just mean that the theorem fails in this case and I need to look at other ways of determining whether I can or can not solve uniquely for $z$?

mookid
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mrmingus
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    The implicit function theorem only gives sufficient conditions, therefore you cannot say anything about the uniqueness of $y$ in the first example. Indeed, when you solve for $y$ you'll notice that there are two solutions for it, $y=\pm\sqrt{\cos^2(z)-x^2-2z^2}$. See link, the theorem is an "if" statement, hence sufficiency (not "if and only if"). – user72272 Oct 31 '14 at 01:18
  • As far as the first example is concerned, only the positive solution $y=\sqrt{\cos^2(z)-x^2-2z^2}$ works for the point in question, (0,1,0). It is my understanding that provided the hypothesis of the Theorem are satisfied, namely F is of class C1, F=0 at the point in question (in this case (0,1,0)), and that the partial derivative of F with respect to one of its variables (in this case y) evaluated at the specified point is not equal to 0, then F can be solved uniquely for y in some neighborhood (perhaps very small), of the point in question. – mrmingus Oct 31 '14 at 01:31

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As you say, the theorem does not apply and you have to look closer. This is a simple example. $$ x^2+y^2=1 $$ cannot be solved for $y$ around $(1,0)$. But $$ x^2+y\,|y|=1 $$ can.

Julián Aguirre
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  • My bad, I saw an absolute value and blindly assumed it wasn't $y$-differentiable. I'll remove my comment. – Git Gud Oct 31 '14 at 12:08
  • @GitGud It is a good exam question actually; the differentiability of $x|x|$. – JP McCarthy Oct 31 '14 at 12:15
  • Thanks. That helps. I was thinking that maybe a way to show that F can not be solved uniquely for z in terms of x and y near (0,1,0) would be to look at the Taylor expansion of $cos(z)$. If we're looking at a small neighborhood of (0,1,0), then $cos(z)$ could be approximated by 1. Then we can solve for z in terms of x and y, yielding: $z=\pm\sqrt{1-x^2-y^2}$. The point (0,1,0) then satisfies both the positive and negative solution for z, and so z is not uniquely determined on a neighborhood of (0,1,0). Does this make sense? Is the Taylor expansion justified here? Is there another way? – mrmingus Oct 31 '14 at 12:43