# Help me understand and remember this Theorem

• Jan 8th 2011, 08:59 AM
CSM
Help me understand and remember this Theorem
I am really struggling with a theorem in my book (Tao; Analysis II).
I don't think I can properly remember this theorem because it contains so much, plus I dont think I really get the Theorem. So my question to you guys is, can you make this theorem understandable for me. Maybe give some simple examples. Show it's usefulness. Tell me how to remember it, etc.

Any help is really appreciated.

The Implicit Function Theorem:
Let $E$ be an open subset of $\mathbb{R}^n$ and let $f:E\rightarrow \mathbb{R}$ be a continous differentiable function and $y=(y_1,y_2,..,y_n)$ a point in $E$ with $f(y)=0$ and $\frac{\partial f}{\partial x_n}(y)\neq 0$.
Then there exists an open subset $U$ of $\mathbb{R}^{n-1}$ which contains $(y_1,y_2,...,y_{n-1})$ and there exists an open subset $V$ of $E$ that contains $y$ and a function $g:U\rightarrow \mathbb{R}$ such that $g(y_1,...,y_{n-1})=y_n$, and:
$\{(x_1,...,x_n)\in V : f(x_1,...,x_n)=0\}$= $\{(x_1,...,x_{n-1},g(x_1,...,x_{n-1})):(x_1,...x_{n-1})\in U\}$
In other words $\{x\in V:f(x)=0\}$ is a graph of a function on $U$. Furthermore $g$ is differentiable in $(y_1,..,y_{n-1})$ and we have:
$\frac{\partial g}{\partial x_j}(y_1,...,y_{n-1})=-\frac{\partial f}{\partial x_j}(y)\slash \frac{\partial f}{\partial x_n}(y)$ for all $1\leq j \leq n-1$.
• Jan 8th 2011, 10:56 AM
Drexel28
Quote:

Originally Posted by CSM
I am really struggling with a theorem in my book (Tao; Analysis II).
I don't think I can properly remember this theorem because it contains so much, plus I dont think I really get the Theorem. So my question to you guys is, can you make this theorem understandable for me. Maybe give some simple examples. Show it's usefulness. Tell me how to remember it, etc.

Any help is really appreciated.

The Implicit Function Theorem:
Let $E$ be an open subset of $\mathbb{R}^n$ and let $f:E\rightarrow \mathbb{R}$ be a continous differentiable function and $y=(y_1,y_2,..,y_n)$ a point in $E$ with $f(y)=0$ and $\frac{\partial f}{\partial x_n}(y)\neq 0$.
Then there exists an open subset $U$ of $\mathbb{R}^{n-1}$ which contains $(y_1,y_2,...,y_{n-1})$ and there exists an open subset $V$ of $E$ that contains $y$ and a function $g:U\rightarrow \mathbb{R}$ such that $g(y_1,...,y_{n-1})=y_n$, and:
$\{(x_1,...,x_n)\in V : f(x_1,...,x_n)=0\}$= $\{(x_1,...,x_{n-1},g(x_1,...,x_{n-1})):(x_1,...x_{n-1})\in U\}$
In other words $\{x\in V:f(x)=0\}$ is a graph of a function on $U$. Furthermore $g$ is differentiable in $(y_1,..,y_{n-1})$ and we have:
$\frac{\partial g}{\partial x_j}(y_1,...,y_{n-1})=-\frac{\partial f}{\partial x_j}(y)\slash \frac{\partial f}{\partial x_n}(y)$ for all $1\leq j \leq n-1$.

I'm not precisely sure what you want! It has a lot of uses. Intuitively it says that if you have an implicit equation $f(x,y,z)=0$ in $\mathbb{R}^3$ and the function satisfies the above conditions then you can "solve" the equation for $z$ in the sense that $z=g(x,y)$. Does that help?
• Jan 8th 2011, 12:09 PM
CSM
Quote:

Originally Posted by Drexel28
I'm not precisely sure what you want! It has a lot of uses. Intuitively it says that if you have an implicit equation $f(x,y,z)=0$ in $\mathbb{R}^3$ and the function satisfies the above conditions then you can "solve" the equation for $z$ in the sense that $z=g(x,y)$. Does that help?

Hi Drexel28, first of all thanks for trying to help me. Any help is appreciated because I think that I don't really understand the theorem at all.

I get what you say. This is what I encounter when I google implicit functions. But doesn't the theorem say much more than that? And is the theorem really needed for expressing your z in a g(x,y) form?
[I am going to restate the theorem for n=3 and see if I can grasp it then].
• Jan 8th 2011, 12:15 PM
CSM
The Implicit Function Theorem for $n=3$:
Let $E$ be an open subset of $\mathbb{R}^3$ and let $f:E\rightarrow \mathbb{R}$ be a continous differentiable function and $y=(y_1,y_2,y_3)$ a point in $E$ with $f(y)=0$ and $\frac{\partial f}{\partial x_n}(y)\neq 0$.
Then there exists an open subset $U$ of $\mathbb{R}^{2}$ which contains $(y_1,y_2)$ and there exists an open subset $V$ of $E$ that contains $y$ and a function $g:U\rightarrow \mathbb{R}$ such that $g(y_1,y_2)=y_3$, and:
$\{(x_1,x_2,x_3)\in V : f(x_1,x_2,x_3)=0\}$= $\{(x_1,x_2,g(x_1,x_2)):(x_1,x_2)\in U\}$
In other words $\{x\in V:f(x)=0\}$ is a graph of a function on $U$. Furthermore $g$ is differentiable in $(y_1,y_2})$ and we have:
$\frac{\partial g}{\partial x_j}(y_1,y_2})=-\frac{\partial f}{\partial x_j}(y)\slash \frac{\partial f}{\partial x_3}(y)$ for all $1\leq j \leq 2$.
• Jan 8th 2011, 12:17 PM
CSM
Can you maybe give me an example exercise in which I HAVE to use the implicit function theorem to it's full potential (but for n=3 or 4)?
• Jan 8th 2011, 12:18 PM
Drexel28
Quote:

Originally Posted by CSM
Hi Drexel28, first of all thanks for trying to help me. Any help is appreciated because I think that I don't really understand the theorem at all.

I get what you say. This is what I encounter when I google implicit functions. But doesn't the theorem say much more than that? And is the theorem really needed for expressing your z in a g(x,y) form?
[I am going to restate the theorem for n=3 and see if I can grasp it then].

No, you're one-hundred percent correct. It does say more than that, but that's the first thing one usually thinks of. Can you think of a specific part of the statement that you take issue with?
• Jan 8th 2011, 01:55 PM
Opalg
Quote:

Originally Posted by CSM
I am really struggling with a theorem in my book (Tao; Analysis II).
I don't think I can properly remember this theorem because it contains so much, plus I dont think I really get the Theorem. So my question to you guys is, can you make this theorem understandable for me. Maybe give some simple examples. Show it's usefulness. Tell me how to remember it, etc.

Any help is really appreciated.

The Implicit Function Theorem:
Let $E$ be an open subset of $\mathbb{R}^n$ and let $f:E\rightarrow \mathbb{R}$ be a continous differentiable function and $y=(y_1,y_2,..,y_n)$ a point in $E$ with $f(y)=0$ and $\frac{\partial f}{\partial x_n}(y)\neq 0$.
Then there exists an open subset $U$ of $\mathbb{R}^{n-1}$ which contains $(y_1,y_2,...,y_{n-1})$ and there exists an open subset $V$ of $E$ that contains $y$ and a function $g:U\rightarrow \mathbb{R}$ such that $g(y_1,...,y_{n-1})=y_n$, and:
$\{(x_1,...,x_n)\in V : f(x_1,...,x_n)=0\}$= $\{(x_1,...,x_{n-1},g(x_1,...,x_{n-1})):(x_1,...x_{n-1})\in U\}$
In other words $\{x\in V:f(x)=0\}$ is a graph of a function on $U$. Furthermore $g$ is differentiable in $(y_1,..,y_{n-1})$ and we have:
$\frac{\partial g}{\partial x_j}(y_1,...,y_{n-1})=-\frac{\partial f}{\partial x_j}(y)\slash \frac{\partial f}{\partial x_n}(y)$ for all $1\leq j \leq n-1$.

Apologies in advance if this is too simple-minded to be useful. But the implicit function theorem is essentially a geometric result, and it's important to visualise the geometry before you worry about the analytic statement.

Start with the elementary, two-dimensional, case. The equation $f(x,y)=0$ represents a curve in the plane. If $\frac{\partial f}{\partial y} = 0$ at a point $(x_0,y_0)$ on the curve, then the tangent at that point is vertical and it may not be possible to express y uniquely as a function of x in a neighbourhood of that point. But if $\frac{\partial f}{\partial y} \ne 0$ then y can be expressed as a function of x in some neighbourhood of $(x_0,y_0)$, and $\frac{dy}{dx} = -\frac{\partial f}{\partial x}\big\slash\frac{\partial f}{\partial y}$ in that neighbourhood.

Simplest possible example: $f(x,y) = x^2+y^2-1$. The curve is the unit circle, and the tangent is vertical at the points $(\pm1,0)$, where $\frac{\partial f}{\partial y} = 0$. In a neighbourhood of any other point $(x_0,y_0)$ on the circle, you can express y (locally) as a function of x, $y=\sqrt{1-x^2}$ if $y_0>0$, or $y=-\sqrt{1-x^2}$ if $y_0<0$. What's more, the derivative is given by $\frac{dy}{dx} = -\frac{\partial f}{\partial x}\big\slash\frac{\partial f}{\partial y} = -\frac xy$.

Now move on to three dimensions. The equation $f(x,y,z)=0$ represents a surface in 3D-space. If $\frac{\partial f}{\partial z} = 0$ at a point $(x_0,y_0,z_0)$ on the surface, then the vertical line at that point is tangent to the surface and it may not be possible to express z uniquely as a function of x and y in a neighbourhood of that point. But if $\frac{\partial f}{\partial z} \ne 0$ then z can be expressed as a function of x and y in some neighbourhood of $(x_0,y_0,z_0)$, and $\frac{\partial z}{\partial x} = -\frac{\partial f}{\partial x}\big\slash\frac{\partial f}{\partial z}$ in that neighbourhood.

Exercise: work through the "simplest possible example", the unit sphere, given by the function $x^2+y^2+z^2-1$. The setup is exactly analogous to the 2D example of the unit circle. if z lies on the "equator" of the sphere then $\frac{\partial f}{\partial z} = 0$ and z cannot locally be expressed uniquely in terms of x and y. In a neighbourhood of any other point on the sphere, there is such an expression for z, and the partial derivatives $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ are given by $\frac{\partial z}{\partial x} = -\frac{\partial f}{\partial x}\big\slash\frac{\partial f}{\partial z} = -\frac xz$ and $\frac{\partial z}{\partial y} = -\frac{\partial f}{\partial y}\big\slash\frac{\partial f}{\partial z} = -\frac yz$.

I think that if you can retain the picture of that simple example, then the implicit function theorem takes on a much more friendly appearance.
• Jan 8th 2011, 03:55 PM
CSM
Opalg, thank you very much for your explanation. It really helps. The way you portrayed makes me understand the theorem better because I did not trip over open subsets in your posts. But the hardest part for me is to understand why this must be true: $\frac{\partial f}{\partial x_n}(y)\neq 0$. What does it mean if $\frac{\partial f}{\partial x_n}(y)\eq 0$ (geometrically)?

I already figured that if it is non-zero then the Jacobian will be invertible (for the Inverse function theorem).