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Please give me an intuitive explanation of 'implicit function theorem'. I read some bits and pieces of information from some textbook, but they look too confusing, especially I do not understand why they use Jacobian matrix to illustrate this theorem.

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You you tried looking in Wkipedia? They do have a whole page devoted to the subject... –  Mariano Suárez-Alvarez Mar 10 '11 at 16:49
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A quick summary: Linear algebra tells us exactly when we can uniquely solve for a subset of the variables from a system of linear equations (we need a subset of columns of the matrix to form a non-singular matrix). Implicit function theorem tells the same about a system of locally nearly linear (more often called differentiable) equations. That subset of columns of the matrix needs to be replaced with the Jacobian, because that's what's describing the "local linearity". –  Jyrki Lahtonen Jul 6 '12 at 5:18

6 Answers 6

up vote 21 down vote accepted

Let's use a simple example with only two variables. Assume there is some relation $f(x,y)=0$ between these variables (which is a general curve in 2D). An example would be $f(x,y) =x^2+y^2-1$ which is the unit circle in $\mathbb{R}^2$. Now you are interested to figure out the slope of the tangent to this curve at some point $x_0,y_0$ on the curve [with $f(x_0,y_0)=0$].

What you can do is to change $x$ a little bit $x = x_0 + \Delta x$. You are interested then how $y$ changes ($y= y_0 + \Delta y$); remember that we are interested in points on the curve with $f(x,y)=0$. Using Taylor expansion on $f(x,y)=0$ yields (up to lowest order in $\Delta x$ and $\Delta y$) $$f(x,y)= \partial_x f(x_0,y_0) \Delta x + \partial_y f(x_0,y_0) \Delta y =0.$$ The slope is thereby given by $$ \frac{\Delta y}{\Delta x} = - \frac{ \partial_x f(x_0,y_0)}{\partial_y f(x_0,y_0)}.$$ As $\Delta x \to 0$ higher order terms in the Taylor expansion (which we neglected) vanish and $\frac{\Delta y}{\Delta x}$ becomes the slope of the curve implicitly defined via $f(x,y)=0$ at $(x_0,y_0)$.

More variables and higher dimensional spaces can be treated similarly (using Taylor series in several variables). But the example above should provide you with enough intuition and insight to understand the 'implicit function theorem'.

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The implicit function theorem really just boils down to this: if I can write down $m$ (sufficiently nice!) equations in $n + m$ variables, then, near any sufficiently nice solution point, there is a function of $n$ variables which give me the remaining $m$ coordinates of nearby solution points. In other words, I can, in principle, solve those equations and get the last $m$ variables in terms of the first $n$ variables. But (!) in general this function is only valid on some small set and won't give you all the solutions either.

Here's a concrete example. Consider the equation $x^2 + y^2 = 1$. This is a single equation in two variables, and for a fixed $x_0 \ne 1, y_0 \ne$ satisfying the equation, there is a function $f$ of $x$ such that $x^2 + f(x)^2 = 1$ for $y$ near $x_0$, and $f(x_0) = y_0$. (Explicitly, for $y_0 > 0$, $f(x) = \sqrt{1 - x^2}$, and for $y_0 < 0$, $f(x) = -\sqrt{1 - x^2}$.) Notice that the function doesn't give you all the solution points — but this isn't surprising, since the solution locus of this equation is a circle, which isn't the graph of any function. Nonetheless, I have basically solved the equation and written $y$ in terms of $x$.

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Nice explanation Zhen. Could you please tell me if there's a formal statement of this version of the theorem that you've mentioned somewhere. Thanks. –  Keivan Aug 3 '12 at 23:59

There are basically two interpretations of (part of) the implicit function theorem(IMFT).

One is that it tells you under what conditions we have solutions to some equation of the form $f=0$, which has been mentioned in Zhen Lin's answer.

More precisely,

let $E$ be an open set of ${\Bbb R}^{n+m}={\Bbb R}^n\times{\Bbb R}^m$ and $f:E\to {\Bbb R}^n$ be a $C^1$ mapping such that $f(a,b)=0$, where $(a,b)\in {\Bbb R}^n\times {\Bbb R}^m$. Put $A=f'(a,b)$ and assume that $A_x$ is invertible, where $A_x$ denotes the restriction of the linear map $A$ on ${\Bbb R}^n\times\{0\}$. $\tag{*}$

Then we can "solve" the equation $f(x,y)=0$ for each of those $y$ near $b$, which means in an open neighborhood $W\subset{\Bbb R}^m$ of $b$, we have an implicitly defined function $g:W\to {\Bbb R}^n$ such that $f(g(y),y)=0$ for every $y\in W$. In particular, $f(g(b),b)=f(a,b)=0$. The word "solve" might be kind of confusing. After all, what does it mean by solving an equation $f=0$? Basically it means one can write some variables in terms of others. On the other hand, the IMFT only tells you the existence of $g$ but not what $g$ looks like.

Another way to look at the IMFT is that it tells you when a set defined by $$ S=\{z\in{\Bbb R}^d|f(z)=0\} $$ is locally a graph of some function. (Here "being locally a graph" means one can find an open set $U\subset {\Bbb R}^d$ such that $U\cap S$ is a graph.) Under the assumption $(*)$, we would have the following conclusion

there exsit open sets $U\subset{\Bbb R}^{n+m}$ and $W\subset {\Bbb R}^m$, with $(a,b)\in U$ and $b\in W$ such that $$ U\cap S=G:=\{(g(y),y)\mid y\in W\}, $$ where $g$ is a function from $W$ to ${\Bbb R}^n$.

Note that we call $G$ the graph of the function $g$.

Using the notations in the second interpretation, the first one basically tells you $$ G\subset U\cap S. $$

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Here is a nice explanation given by one of our Professor at an Undergraduate training camp.

Please see under the section geometry for the Implicit function theorem.

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The other answers have done a really good job explaining the implicit function theorem in the setting of multivariable calculus. There is a generalization of the implicit function theorem which is very useful in differential geometry called the rank theorem.

Rank Theorem: Assume $M$ and $N$ are manifolds of dimension $m$ and $n$ respectively. If $F : M \to N$ is a smooth map, $p \in M$ and $F_{*} : T_qM \to T_{F(q)}N$ has rank $k$ in a neighborhood of $p$, then there are coordinates $(x^1, \dots, x^k, \dots, x^m)$ centered around $p$ and $(v^1, \dots, v^n)$ centered around $F(p)$ such that in local coordinates, $F$ is given by the equation $$ F(x^1, \dots, x^k, \dots, x^m) = (x^1, \dots, x^k, 0, \dots, 0)$$

Essentially, the rank theorem tells us that if the total derivative of $F$ in a neighborhood of $p$ has rank $k$, then locally around $p$, we can think of $F$ as a linear map with rank $k$.

Here is an example of how you use the rank theorem in to prove a version of the implicit function theorem in differential geometry.

Assume $M$ is a manifold of dimension $n+k$ and $N$ is a manifold of dimension $k$. Assume $\Theta : M \to N$ is a smooth map and $\Theta_{*} : T_pM \to T_{\Theta(p)}N$ has rank $k$ (i.e $p$ is a regular point). Since maximal rank is an open condition, it follows that $F_*$ has rank $k$ in a neighborhood of $p$. By the rank theorem, there are coordinates $(x^1, \dots, x^{n+k})$ defined on the open set $U$ and centered around $p$ and coordinates centered around $\Theta(p)$ such that $$\Theta(x^1, \dots, x^{n+k}) = (x^1, \dots, x^k).$$ If we write $q = \Theta(p)$, then the above equation tells us that $$\Theta^{-1}(q) \cap U = \{ p \in U : x^{1}(p) = \dots = x^{k}(p) = 0 \}$$ which implies that $ (x^{k+1}, \dots, x^{k+n}) : \Theta^{-1}(q) \cap U \to \mathbb{R}^n$ is an open topological embedding. This defines an $n$-dimensional smooth structure on $\Theta^{-1}(q) \cap U$, and in coordinates, $\iota : \Theta^{-1}(q) \cap U \to U$ is given by $(x^{k+1}, \dots, x^{k+n}) \mapsto (0, \dots, 0, x^{k+1}, \dots, x^{k+n})$ which is smooth.

In summary, if $p$ is a regular point and $q = \Theta(p)$, then there are coordinates $(x^1, \dots, x^{n+k})$ for $M$ around $p$ such that $(x^{k+1}, \dots, x^{k+n})$ are coordinates for $\Theta^{-1}(q)$ around $p$ (once suitably restricted of course). This is kind of a differential geometers formulation of the normal implicit function theorem.

EDIT: In the comments, David pointed out that the statement of the rank theorem above was wrong. The reason is that rank $ \geq k $ is an open condition and rank $< k $ is a closed condition, which I didn't really understand when I wrote this answer 2 years ago. Things should be fixed now

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could it be possible for you to explain your above statement of the theorem by an example on $\mathbb{R}^2$ or $\mathbb{R}^3$? so that I can see what is really going on. though I know diff geom but I need to explain some one. –  Une Femme Douce Jan 23 '13 at 8:41
    
I am confused. Take $M = N = \mathbb{R}$. Take $f(x) = x^2$ and $p=0$. Then $f_{\ast}$ is $0$ at $p$, so of rank $0$, but $f$ is not locally constant. Did you mean to require that $\mathrm{rank} f_{\ast}$ be $k$ on an open neighborhood of $p$? –  David Speyer Sep 7 at 18:01

In short, it is the basis upon which mathematicians differentiate implicitly. In most Calculus I textbooks, this topic is cursorily addressed and students are asked to assume, by some version of chain rule logic, that differentiating equations will follow the logic of differentiating functions despite lacking a formal proof. This theorem, amongst other ramifications, purports to prove the basis upon which we differentiate equations implicitly.

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