Editing Implicit function theorem (section)

== Definitions ==
Let <math>f: \R^{n+m} \to \R^m</math> be a [[continuously differentiable]] function. We think of <math>\R^{n+m}</math> as the [[Cartesian product]] <math>\R^n\times\R^m,</math> and we write a point of this product as <math>(\mathbf{x}, \mathbf{y}) = (x_1,\ldots, x_n, y_1, \ldots y_m).</math> Starting from the given function <math>f</math>, our goal is to construct a function <math>g: \R^n \to \R^m</math> whose graph <math>(\textbf{x}, g(\textbf{x}))</math> is precisely the set of all <math>(\textbf{x}, \textbf{y})</math> such that <math>f(\textbf{x}, \textbf{y}) = \textbf{0}</math>.

As noted above, this may not always be possible. We will therefore fix a point <math>(\textbf{a}, \textbf{b}) = (a_1, \dots, a_n, b_1, \dots, b_m)</math> which satisfies <math>f(\textbf{a}, \textbf{b}) = \textbf{0}</math>, and we will ask for a <math>g</math> that works near the point <math>(\textbf{a}, \textbf{b})</math>. In other words, we want an [[open set]] <math>U \subset \R^n</math> containing <math>\textbf{a}</math>, an open set <math>V \subset \R^m</math> containing <math>\textbf{b}</math>, and a function <math>g : U \to V</math> such that the graph of <math>g</math> satisfies the relation <math>f = \textbf{0}</math> on <math>U\times V</math>, and that no other points within <math>U \times V</math> do so. In symbols,

<math display="block">\{ (\mathbf{x}, g(\mathbf{x})) \mid \mathbf x \in U \} = \{ (\mathbf{x}, \mathbf{y})\in U \times V \mid f(\mathbf{x}, \mathbf{y}) = \mathbf{0} \}.</math>

To state the implicit function theorem, we need the [[Jacobian matrix and determinant|Jacobian matrix]] of <math>f</math>, which is the matrix of the [[partial derivative]]s of <math>f</math>. Abbreviating <math>(a_1, \dots, a_n, b_1, \dots, b_m)</math> to <math>(\textbf{a}, \textbf{b})</math>, the Jacobian matrix is

<math display="block">(Df)(\mathbf{a},\mathbf{b})
= \left[\begin{array}{ccc|ccc}
 \frac{\partial f_1}{\partial x_1}(\mathbf{a},\mathbf{b}) & \cdots & \frac{\partial f_1}{\partial x_n}(\mathbf{a},\mathbf{b}) &
 \frac{\partial f_1}{\partial y_1}(\mathbf{a},\mathbf{b}) & \cdots & \frac{\partial f_1}{\partial y_m}(\mathbf{a},\mathbf{b}) \\
 \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
 \frac{\partial f_m}{\partial x_1}(\mathbf{a},\mathbf{b}) & \cdots & \frac{\partial f_m}{\partial x_n}(\mathbf{a},\mathbf{b}) &
 \frac{\partial f_m}{\partial y_1}(\mathbf{a},\mathbf{b}) & \cdots & \frac{\partial f_m}{\partial y_m}(\mathbf{a},\mathbf{b})
\end{array}\right]
= \left[\begin{array}{c|c} X & Y \end{array}\right]</math>

where <math>X</math> is the matrix of partial derivatives in the variables <math>x_i</math> and <math>Y</math> is the matrix of partial derivatives in the variables <math>y_j</math>. The implicit function theorem says that if <math>Y</math> is an invertible matrix, then there are <math>U</math>, <math>V</math>, and <math>g</math> as desired. Writing all the hypotheses together gives the following statement.