Editing Implicit function theorem (section)

== Application: change of coordinates ==
Suppose we have an {{mvar|m}}-dimensional space, parametrised by a set of coordinates <math> (x_1,\ldots,x_m) </math>. We can introduce a new coordinate system <math> (x'_1,\ldots,x'_m) </math> by supplying m functions <math> h_1\ldots h_m </math> each being continuously differentiable. These functions allow us to calculate the new coordinates <math> (x'_1,\ldots,x'_m) </math> of a point, given the point's old coordinates <math> (x_1,\ldots,x_m) </math> using <math> x'_1=h_1(x_1,\ldots,x_m), \ldots, x'_m=h_m(x_1,\ldots,x_m) </math>. One might want to verify if the opposite is possible: given coordinates <math> (x'_1,\ldots,x'_m) </math>, can we 'go back' and calculate the same point's original coordinates <math> (x_1,\ldots,x_m) </math>? The implicit function theorem will provide an answer to this question. The (new and old) coordinates <math>(x'_1,\ldots,x'_m, x_1,\ldots,x_m)</math> are related by ''f'' = 0, with
<math display="block">f(x'_1,\ldots,x'_m,x_1,\ldots, x_m)=(h_1(x_1,\ldots, x_m)-x'_1,\ldots , h_m(x_1,\ldots, x_m)-x'_m).</math>
Now the Jacobian matrix of ''f'' at a certain point (''a'', ''b'') [ where <math>a=(x'_1,\ldots,x'_m), b=(x_1,\ldots,x_m)</math> ] is given by
<math display="block">(Df)(a,b) = \left [\begin{matrix}
 -1 & \cdots & 0 \\
 \vdots & \ddots & \vdots \\
 0 & \cdots & -1
\end{matrix}\left|
\begin{matrix}
\frac{\partial h_1}{\partial x_1}(b) & \cdots & \frac{\partial h_1}{\partial x_m}(b)\\
\vdots & \ddots & \vdots\\
\frac{\partial h_m}{\partial x_1}(b) & \cdots & \frac{\partial h_m}{\partial x_m}(b)\\
\end{matrix} \right.\right] = [-I_m |J ].</math>
where I<sub>''m''</sub> denotes the ''m'' × ''m'' [[identity matrix]], and {{mvar|J}} is the {{math|''m'' × ''m''}} matrix of partial derivatives, evaluated at (''a'', ''b''). (In the above, these blocks were denoted by X and Y. As it happens, in this particular application of the theorem, neither matrix depends on ''a''.) The implicit function theorem now states that we can locally express <math> (x_1,\ldots,x_m) </math> as a function of <math> (x'_1,\ldots,x'_m) </math> if ''J'' is invertible. Demanding ''J'' is invertible is equivalent to det ''J'' ≠ 0, thus we see that we can go back from the primed to the unprimed coordinates if the determinant of the Jacobian ''J'' is non-zero. This statement is also known as the [[inverse function theorem]].

=== Example: polar coordinates ===
As a simple application of the above, consider the plane, parametrised by [[polar coordinates]] {{math|(''R'', ''θ'')}}. We can go to a new coordinate system ([[cartesian coordinates]]) by defining functions {{math|1=''x''(''R'', ''θ'') = ''R'' cos(''θ'')}} and {{math|1=''y''(''R'', ''θ'') = ''R'' sin(''θ'')}}. This makes it possible given any point {{math|(''R'', ''θ'')}} to find corresponding Cartesian coordinates {{math|(''x'', ''y'')}}. When can we go back and convert Cartesian into polar coordinates? By the previous example, it is sufficient to have {{math|1=det ''J'' ≠ 0}}, with
<math display="block">J =\begin{bmatrix}
 \frac{\partial x(R,\theta)}{\partial R} & \frac{\partial x(R,\theta)}{\partial \theta} \\
 \frac{\partial y(R,\theta)}{\partial R} & \frac{\partial y(R,\theta)}{\partial \theta} \\
\end{bmatrix}=
 \begin{bmatrix}
 \cos \theta & -R \sin \theta \\
 \sin \theta & R \cos \theta
\end{bmatrix}.</math>
Since {{math|1=det ''J'' = ''R''}}, conversion back to polar coordinates is possible if {{math|1=''R'' ≠ 0}}. So it remains to check the case {{math|1=''R'' = 0}}. It is easy to see that in case {{math|1=''R'' = 0}}, our coordinate transformation is not invertible: at the origin, the value of θ is not well-defined.