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Implicit Function Theorem

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Level sets in R2.jpg

Given $u = f(x)$ for $f: {\bf R}^n \rightarrow {\bf R}$, then for each $c \in {\bf R}$ the level set relative to $c$ of $f$ is

$\{ x: f(x) = c \} \subset {\bf R}^n$ (in the domain of $f$).

In other point these are points with the same altitude.

Is it always a curve?

Could be a point - if this is a local maximum or minimum.

Example. Let

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

paraboloid. The $0$-level curve for $c = 0$ is a point $( 0, 0 )$.

Level sets in R2 not curve.jpg

What would be a level set that is not a curve, but that isn't a maximum or a minimum?

Answer: A saddle point.

Ex. Let

$$z = x^2 - y^2,$$

the $0$-level set is

$( x, y ): x^2 - y^2 = 0$, or $x^2 = y^2$, or $x = \pm y$, i.e. two lines.

What do these two examples have in common? They both are critical points, i.e.

$$f'( a, b ) = 0,$$

so that the tangent plane is horizontal.

Theorem (Implicit Function Theorem). Let

$$f: {\bf R}^2 \rightarrow {\bf R}, f(a) = c, a \in {\bf R}^2, c \in {\bf R}.$$

Suppose $f$ is differentiable on an open disk $D$ around $a$, and that $f'$ is continuous on $D$. Suppose further $f′(a) \neq 0$. Then the intersection of the $c$-level set with $D$ is a curve (i.e., it can be parametrized).

The name, Implicit Function Theorem, makes sense as an implicitly ($f(a) = c$) represented function can be represented explicitly.

Circle implicit function.jpg

Ex.

  • Implicit: $x^2 + y^2 = 1,$
  • Explicit:

$$\begin{array}{} y = g(x) &= \pm ( 1 - x^2 )^{\frac{1}{2}} \\ &= + ( 1 - x^2 )^{\frac{1}{2}} {\rm \hspace{3pt} if \hspace{3pt}} y > 0 \\ &= - ( 1 - x^2 )^{\frac{1}{2}} {\rm \hspace{3pt} if \hspace{3pt}} y < 0. \end{array}$$

How does the theorem apply here?

The theorem is easy to modify to get an analogue for parametric surfaces: If

$$f: {\bf R}^3 \rightarrow {\bf R}, f(a) = c, a \in {\bf R}^3, c \in {\bf R}$$

and $f$ is differentiable on an open ball $B$ around $a$, and $f'$ is continuous on $B$, and $f'(a) \neq 0$, then the intersection of the $c$-level set with $B$ is a surface.

Now, we have the theorem for curves in dimension $1$ and for surfaces in dimension $2$. The $n$ dim analogue is obvious.