Implicit Functions#

The Implicit Function Theorem

Let \(F: U \subseteq \mathbb{R}^m \times \mathbb{R}^n \to \mathbb{R}^n\) be a real vector function which is \(k\)-times continuously partially differentiable (\(k \ge 1\)) on an open set \(U\).

If there exists some \(\boldsymbol{p} \in U\) such that

\[F(\boldsymbol{p}) = F(p_1, \dotsc, p_{m+n}) = \boldsymbol{0}\]

and there exist \(n\) columns \(\boldsymbol{j}_{i_1}(\boldsymbol{p}), \dotsc, \boldsymbol{j}_{i_n}(\boldsymbol{p})\) in the Jacobian matrix \(J_F(\boldsymbol{p})\) (where \(i_1, \dotsc, i_n\) are the corresponding indices) such that the \(n \times n\)-matrix

\[\begin{bmatrix}\vert & \vert & \vert \\ \boldsymbol{j}_{i_1}(\boldsymbol{p}) & \cdots & \boldsymbol{j}_{i_n}(\boldsymbol{p}) \\ \vert & \vert & \vert\end{bmatrix}\]

is invertible, then there exist an open neighborhood \(V \subseteq \mathbb{R}^m\) of \((p_{i_1'}, \dotsc, p_{i_m'})\) (where \(i_1', \dotsc, i_m'\) are the indices of the remaining \(m\) columns \(\boldsymbol{j}_{i_1'}(\boldsymbol{p}), \dotsc, \boldsymbol{j}_{i_m'}(\boldsymbol{p})\)), an open neighborhood \(W \subseteq \mathbb{R}^n\) of \((p_{i_1}, \dotsc, p_{i_n})\) and a unique real vector function \(f: V \to W\) such that for all \(\boldsymbol{z} \in U\) with \((z_{i_1'}, \dotsc, z_{i_m'}) \in V\) and \((z_{i_1}, \dotsc, z_{i_n}) \in W\) we have \(F(\boldsymbol{z}) = \boldsymbol{0}\) if and only if \(f(z_{i_1'}, \dotsc, z_{i_m'}) = (z_{i_1}, \dotsc, z_{i_n})\).

In this case, \(f\) is \(k\)-times continuously partially differentiable on \(V\) and its Jacobian matrix is given as follows:

\[J_f(z_{i_1'}, \dotsc, z_{i_m'}) = -\begin{bmatrix}\vert & \vert & \vert \\ \boldsymbol{j}_{i_1}(\boldsymbol{z}) & \cdots & \boldsymbol{j}_{i_n}(\boldsymbol{z}) \\ \vert & \vert & \vert\end{bmatrix}^{-1} \begin{bmatrix}\vert & \vert & \vert \\ \boldsymbol{j}_{i_1'}(\boldsymbol{z}) & \cdots & \boldsymbol{j}_{i_m'}(\boldsymbol{z}) \\ \vert & \vert & \vert\end{bmatrix}\]

Example: \(F(x,y) = 2x + 3y +3\)

Consider the real scalar field \(F: \mathbb{R}^2 \to \mathbb{R}\) defined as follows:

\[F(x,y) = 2x + 3y +3\]

It is infinitely continuously partially differentiable on \(\mathbb{R}^2\) with the following Jacobian matrix:

\[J_F(x, y) = \begin{bmatrix}2 & 3\end{bmatrix}\]

For \(\boldsymbol{p} = (0, -1)\), we have:

\[F(\boldsymbol{p}) = F(0, -1) = 0 \qquad J_F(\boldsymbol{p}) = \begin{bmatrix}2 & 3\end{bmatrix}\]

Example: Solving in Terms of \(x\)

The "column" \(\boldsymbol{j}_2\) of \(J_F(\boldsymbol{p})\) is just \(3 \ne 0\) and so the "matrix"

\[\begin{bmatrix} \vert \\ \boldsymbol{j}_2(\boldsymbol{p}) \\ \vert \end{bmatrix} = \begin{bmatrix} 3 \end{bmatrix}\]

is "invertible".

We thus know that there is an open interval \(V\) around \(0\), an open interval \(W\) around \(-1\) and a unique real function \(f: V \to W\) such that for all \(\boldsymbol{z} = (x, y) \in V \times W\) we have \(F(x, y) = 0\) if and only if \(y = f(x)\).

Furthermore, \(f\) is infinitely continuously differentiable on \(V\) with the following derivative:

\[\begin{aligned}f'(x) & = -\begin{bmatrix}\vert \\ \boldsymbol{j}_2(x,y) \\ \vert \end{bmatrix}^{-1} \begin{bmatrix}\vert \\ \boldsymbol{j}_1(x,y) \\ \vert \end{bmatrix} \\ & = -\begin{bmatrix}3\end{bmatrix}^{-1} \begin{bmatrix}2\end{bmatrix} \\ & = -\begin{bmatrix}\frac{1}{3}\end{bmatrix} \begin{bmatrix}2\end{bmatrix} \\ & = -\frac{2}{3}\end{aligned}\]

We can also verify this explicitly, since \(F(x, y) = 0\) if and only if \(y = -\frac{2}{3}x - 1\). If we define \(f: \mathbb{R} \to \mathbb{R}\) as \(f(x) = -\frac{2}{3}x - 1\), we have actually found a concrete example. Specifically, \(V = \mathbb{R} = W\) and we easily see that \(F(x, y) = 0\) if and only if \(y = f(x)\). Moreover, we have \(f'(x) = -\frac{2}{3}\).

Example: Solving in Terms of \(y\)

The "column" \(\boldsymbol{j}_1\) of \(J_F(\boldsymbol{p})\) is just \(2 \ne 0\) and so the "matrix"

\[\begin{bmatrix} \vert \\ \boldsymbol{j}_1(\boldsymbol{p}) \\ \vert \end{bmatrix} = \begin{bmatrix} 2 \end{bmatrix}\]

is "invertible".

We thus know that there is an open interval \(V\) around \(-1\), an open interval \(W\) around \(0\) and a unique real function \(f: V \to W\) such that for all \((y, x) \in V \times W\) we have \(F(x,y) = 0\) if and only if \(x = f(y)\).

Furthermore, \(f\) is infinitely continuously differentiable on \(V\) with the following derivative:

\[\begin{aligned}f'(y) & = -\begin{bmatrix}\vert \\ \boldsymbol{j}_1(x,y) \\ \vert \end{bmatrix}^{-1} \begin{bmatrix}\vert \\ \boldsymbol{j}_2(x,y) \\ \vert \end{bmatrix} \\ & = -\begin{bmatrix}2\end{bmatrix}^{-1} \begin{bmatrix}3\end{bmatrix} \\ & = -\begin{bmatrix} \frac{1}{2} \end{bmatrix} \begin{bmatrix} 3 \end{bmatrix} \\ & = -\frac{3}{2}\end{aligned}\]

We can also verify this explicitly, since \(F(x, y) = 0\) if and only if \(x = -\frac{3}{2}y - \frac{3}{2}\). If we define \(f: \mathbb{R} \to \mathbb{R}\) as \(f(y) = -\frac{3}{2}y - \frac{3}{2}\), we have actually found a concrete example. Specifically, \(V = \mathbb{R} = W\) and we easily see that \(F(x, y) = 0\) if and only if \(x = f(y)\). Moreover, we have \(f'(y) = -\frac{3}{2}\).

Example: \(F(x, y) = \mathrm{e}^y + y^3 - x^3 - x^2 -1\)

Consider the real scalar field \(F: \mathbb{R}^2 \to \mathbb{R}\) defined as follows:

\[F(x, y) = \mathrm{e}^y + y^3 - x^3 - x^2 -1\]

It is infinitely continuously partially differentiable on \(\mathbb{R}^2\) with the following Jacobian matrix:

\[J_F(x, y) = \begin{bmatrix}-3x^2 - 2x & \mathrm{e}^y + 3y^2\end{bmatrix}\]

For \(\boldsymbol{p} = (0, 0)\), we have:

\[F(\boldsymbol{p}) = F(0, 0) = 0 \qquad J_F(\boldsymbol{p}) = \begin{bmatrix}0 & 1\end{bmatrix}\]

Example: Solving in Terms of \(x\)

The "column" \(\boldsymbol{j}_2\) of \(J_F(\boldsymbol{p})\) is just \(1 \ne 0\) and so the "matrix"

\[\begin{bmatrix} \vert \\ \boldsymbol{j}_2(\boldsymbol{p}) \\ \vert \end{bmatrix} = \begin{bmatrix} 1 \end{bmatrix}\]

is "invertible".

We thus know that there is an open interval \(V\) around \(0\), an open interval \(W\) around \(0\) and a unique real function \(f: V \to W\) such that for all \(\boldsymbol{z} = (x, y) \in V \times W\) we have \(F(x, y) = 0\) if and only if \(y = f(x)\).

Furthermore, \(f\) is infinitely continuously differentiable on \(V\) with the following derivative:

\[\begin{aligned}f'(x) & = -\begin{bmatrix}\vert \\ \boldsymbol{j}_2(x,y) \\ \vert \end{bmatrix}^{-1} \begin{bmatrix}\vert \\ \boldsymbol{j}_1(x,y) \\ \vert \end{bmatrix} \\ & = -\begin{bmatrix}\mathrm{e}^{f(x)} + 3 f(x)^2\end{bmatrix}^{-1} \begin{bmatrix}-3x^2 - 2x\end{bmatrix} \\ & = \frac{3x^2 + 2x}{\mathrm{e}^{f(x)} + 3 f(x)^2}\end{aligned}\]

We can also simplify this a bit to get rid of \(\mathrm{e}^{f(x)}\). Since \(F(x, f(x)) = 0\) for all \(x \in V\), we have:

\[F(x, f(x)) = \mathrm{e}^{f(x)}+f(x)^3 - x^3 - x^2 - 1 = 0\]

From that, we obtain:

\[\mathrm{e}^{f(x)} = -f(x)^3 + x^3 + x^2 + 1\]

Plugging this into the expression for \(f'(x)\), we get:

\[f'(x) = \frac{3x^2 + 2x}{-f(x)^3 + 3f(x)^2 + x^3 + x^2 + 1}\]

Example: Solving in Terms of \(y\)

The "column" \(\boldsymbol{j}_1\) of \(J_F(\boldsymbol{p})\) is just \(0\) and so the "matrix"

\[\begin{bmatrix} \vert \\ \boldsymbol{j}_1(\boldsymbol{p}) \\ \vert \end{bmatrix} = \begin{bmatrix} 0 \end{bmatrix}\]

is not "invertible". Therefore, the theorem cannot tell us anything.

Example: \(F(x, y) = x^2 + y^2 - 1\)

Consider the real scalar field \(F: \mathbb{R}^2 \to \mathbb{R}\) defined as follows:

\[F(x, y) = x^2 + y^2 - 1\]

It is infinitely continuously partially differentiable on \(\mathbb{R}^2\) with the following Jacobian matrix:

\[J_F(x, y) = \begin{bmatrix}2x & 2y\end{bmatrix}\]

Example: Solving in Terms of \(x\)

For an arbitrary point \(\boldsymbol{p} = (x_0, y_0)\) such that \(x_0^2 + y_0^2 - 1 = 0\) with \(y_0 \ne 0\), we have:

\[F(\boldsymbol{p}) = F(x_0, y_0) = 0 \qquad J_F(\boldsymbol{p}) = \begin{bmatrix}2x_0 & 2y_0\end{bmatrix}\]

The "column" \(\boldsymbol{j}_2(\boldsymbol{p})\) of \(J_F(\boldsymbol{p})\) is just \(2y_0 \ne 0\) and so the "matrix"

\[\begin{bmatrix} \vert \\ \boldsymbol{j}_2(\boldsymbol{p}) \\ \vert \end{bmatrix} = \begin{bmatrix} 2y_0 \end{bmatrix}\]

is "invertible".

We thus know that there is an open interval \(V\) around \(x_0\), an open interval \(W\) around \(y_0\) and a unique real function \(f: V \to W\) such that for all \(\boldsymbol{z} = (x, y) \in V \times W\) we have \(F(x, y) = 0\) if and only if \(y = f(x)\).

Furthermore, \(f\) is infinitely continuously differentiable on \(V\) with the following derivative:

\[\begin{aligned}f'(x) & = -\begin{bmatrix}\vert \\ \boldsymbol{j}_2(x,y) \\ \vert \end{bmatrix}^{-1} \begin{bmatrix}\vert \\ \boldsymbol{j}_1(x,y) \\ \vert \end{bmatrix} \\ & = -\begin{bmatrix}2y\end{bmatrix}^{-1} \begin{bmatrix}2x\end{bmatrix} \\ & = -\begin{bmatrix}\frac{1}{2y}\end{bmatrix} \begin{bmatrix}2x\end{bmatrix} \\ & = -\frac{x}{y}\end{aligned}\]

We can also verify this explicitly. For \(y_0 \gt 0\), we have \(f(x) = \sqrt{1-x^2}\) and for \(y_0 \lt 0\), we have \(f(x) = -\sqrt{1 - x^2}\). In both cases, we easily see that \(F(x, y) = 0\) if and only if \(y = f(x)\). Moreover, we have \(f'(x) = -\frac{x}{\pm\sqrt{1-x^2}} = -\frac{x}{f(x)}\), matching our result.

Example: Solving in Terms of \(y\)

For an arbitrary point \(\boldsymbol{p} = (x_0, y_0)\) such that \(x_0^2 + y_0^2 - 1 = 0\) with \(x_0 \ne 0\), we have:

\[F(\boldsymbol{p}) = F(x_0, y_0) = 0 \qquad J_F(\boldsymbol{p}) = \begin{bmatrix}2x_0 & 2y_0\end{bmatrix}\]

The "column" \(\boldsymbol{j}_1(\boldsymbol{p})\) of \(J_F(\boldsymbol{p})\) is just \(2x_0 \ne 0\) and so the "matrix"

\[\begin{bmatrix} \vert \\ \boldsymbol{j}_1(\boldsymbol{p}) \\ \vert \end{bmatrix} = \begin{bmatrix} 2x_0 \end{bmatrix}\]

is "invertible".

We thus know that there is an open interval \(V\) around \(y_0\), an open interval \(W\) around \(x_0\) and a unique real function \(f: V \to W\) such that for all \((y, x) \in V \times W\) we have \(F(x,y) = 0\) if and only if \(x = f(y)\).

Furthermore, \(f\) is infinitely continuously differentiable on \(V\) with the following derivative:

\[\begin{aligned}f'(y) & = -\begin{bmatrix}\vert \\ \boldsymbol{j}_1(x,y) \\ \vert \end{bmatrix}^{-1} \begin{bmatrix}\vert \\ \boldsymbol{j}_2(x,y) \\ \vert \end{bmatrix} \\ & = -\begin{bmatrix}2x\end{bmatrix}^{-1} \begin{bmatrix}2y\end{bmatrix} \\ & = -\begin{bmatrix} \frac{1}{2x} \end{bmatrix} \begin{bmatrix} 2y \end{bmatrix} \\ & = -\frac{y}{x}\end{aligned}\]

We can also verify this explicitly. For \(x_0 \gt 0\), we have \(f(y) = \sqrt{1-y^2}\) and for \(x_0 \lt 0\), we have \(f(y) = -\sqrt{1 - y^2}\). In both cases, we easily see that \(F(x, y) = 0\) if and only if \(x = f(y)\). Moreover, we have \(f'(y) = -\frac{y}{\pm\sqrt{1-y^2}} = -\frac{y}{f(y)}\), matching our result.

Example: \(F(x, y, z) = \sin (x + y - z^2) - \frac{1}{\sqrt{2}}\)

Consider the real scalar field \(F: \mathbb{R}^3 \to \mathbb{R}\) defined as follows:

\[F(x, y, z) = \sin (x + y - z^2) - \frac{1}{\sqrt{2}}\]

It is infinitely continuously partially differentiable on \(\mathbb{R}^3\) with the following Jacobian matrix:

\[J_F(x, y, z) = \begin{bmatrix}\cos(x + y - z^2) & \cos(x + y - z^2) & -2z\cos(x + y - z^2)\end{bmatrix}\]

For \(\boldsymbol{p} = \left(\frac{\uppi}{4}, 0, 0\right)\), we have:

\[F(\boldsymbol{p}) = F\left(\frac{\uppi}{4}, 0, 0\right) = 0 \qquad J_F(\boldsymbol{p}) = \begin{bmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0\end{bmatrix}\]

Example: Solving in Terms of \(y\) and \(z\)

The "column" \(\boldsymbol{j}_1(\boldsymbol{p})\) of \(J_F(\boldsymbol{p})\) is just \(\frac{1}{\sqrt{2}} \ne 0\) and so the "matrix"

\[\begin{bmatrix} \vert \\ \boldsymbol{j}_1(\boldsymbol{p}) \\ \vert \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} \end{bmatrix}\]

is "invertible".

We thus know that there is an open set \(V\) around \((0, 0)\), an open interval \(W\) around \(\frac{\uppi}{4}\) and a unique real scalar field \(f: V \to W\) such that for all \(\boldsymbol{z} = (y, z, x) \in V \times W\) we have \(F(x, y, z) = 0\) if and only if \(x = f(y, z)\).

Furthermore, \(f\) is infinitely continuously partially differentiable on \(V\) with the following Jacobian matrix:

\[\begin{aligned}J_f(y, z) & = -\begin{bmatrix}\vert \\ \boldsymbol{j}_1(x,y,z) \\ \vert \end{bmatrix}^{-1} \begin{bmatrix}\vert & \vert \\ \boldsymbol{j}_2(x,y,z) & \boldsymbol{j}_3(x,y,z) \\ \vert & \vert \end{bmatrix} \\ & = -\begin{bmatrix}\frac{1}{\sqrt{2}}\end{bmatrix}^{-1} \begin{bmatrix}\frac{1}{\sqrt{2}} & 0\end{bmatrix} \\ & = -\begin{bmatrix}\sqrt{2}\end{bmatrix} \begin{bmatrix}\frac{1}{\sqrt{2}} & 0\end{bmatrix} \\ & = \begin{bmatrix}-1 & 0\end{bmatrix}\end{aligned}\]

We can also verify this explicitly, since \(F(x, y, z) = 0\) locally if and only if \(x = -y + z^2 + \frac{\uppi}{4}\). If we define \(f: \mathbb{R}^2 \to \mathbb{R}\) as \(f(y, z) = -y + z^2 + \frac{\uppi}{4}\), we have actually found a concrete example. Specifically, \(V = \mathbb{R}^2\) and \(W = \mathbb{R}\), and we easily see that \(F(x, y, z) = 0\) if and only if \(x = f(y, z)\). Moreover, we have \(J_f(y, z) = \begin{bmatrix}-1 & 2z\end{bmatrix}\), which at \((0, 0)\) gives \(\begin{bmatrix}-1 & 0\end{bmatrix}\).

Example: Solving in Terms of \(x\) and \(z\)

The "column" \(\boldsymbol{j}_2(\boldsymbol{p})\) of \(J_F(\boldsymbol{p})\) is just \(\frac{1}{\sqrt{2}} \ne 0\) and so the "matrix"

\[\begin{bmatrix} \vert \\ \boldsymbol{j}_2(\boldsymbol{p}) \\ \vert \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} \end{bmatrix}\]

is "invertible".

We thus know that there is an open set \(V\) around \(\left(\frac{\uppi}{4}, 0\right)\), an open interval \(W\) around \(0\) and a unique real scalar field \(f: V \to W\) such that for all \(\boldsymbol{z} = (x, z, y) \in V \times W\) we have \(F(x, y, z) = 0\) if and only if \(y = f(x, z)\).

Furthermore, \(f\) is infinitely continuously partially differentiable on \(V\) with the following Jacobian matrix:

\[\begin{aligned}J_f(x, z) & = -\begin{bmatrix}\vert \\ \boldsymbol{j}_2(x,y,z) \\ \vert \end{bmatrix}^{-1} \begin{bmatrix}\vert & \vert \\ \boldsymbol{j}_1(x,y,z) & \boldsymbol{j}_3(x,y,z) \\ \vert & \vert \end{bmatrix} \\ & = -\begin{bmatrix}\frac{1}{\sqrt{2}}\end{bmatrix}^{-1} \begin{bmatrix}\frac{1}{\sqrt{2}} & 0\end{bmatrix} \\ & = -\begin{bmatrix}\sqrt{2}\end{bmatrix} \begin{bmatrix}\frac{1}{\sqrt{2}} & 0\end{bmatrix} \\ & = \begin{bmatrix}-1 & 0\end{bmatrix}\end{aligned}\]

We can also verify this explicitly, since \(F(x, y, z) = 0\) locally if and only if \(y = -x + z^2 + \frac{\uppi}{4}\). If we define \(f: \mathbb{R}^2 \to \mathbb{R}\) as \(f(x, z) = -x + z^2 + \frac{\uppi}{4}\), we have actually found a concrete example. Specifically, \(V = \mathbb{R}^2\) and \(W = \mathbb{R}\), and we easily see that \(F(x, y, z) = 0\) if and only if \(y = f(x, z)\). Moreover, we have \(J_f(x, z) = \begin{bmatrix}-1 & 2z\end{bmatrix}\), which at \(\left(\frac{\uppi}{4}, 0\right)\) gives \(\begin{bmatrix}-1 & 0\end{bmatrix}\).

Example: Solving in Terms of \(x\) and \(y\)

The "column" \(\boldsymbol{j}_3(\boldsymbol{p})\) of \(J_F(\boldsymbol{p})\) is \(0\) and so the "matrix"

\[\begin{bmatrix} \vert \\ \boldsymbol{j}_3(\boldsymbol{p}) \\ \vert \end{bmatrix} = \begin{bmatrix} 0 \end{bmatrix}\]

is not "invertible". Therefore, the theorem cannot tell us anything.

Example: \(F: \mathbb{R}^3 \to \mathbb{R}^2\)

Consider the real vector function \(F: \mathbb{R}^3 \to \mathbb{R}^2\) defined as follows:

\[F(x, y, z) = \begin{bmatrix} x^2 + y^2 - z^2 - 8 \\ \sin(\uppi x) + \sin (\uppi y) + \sin (\uppi z)\end{bmatrix}\]

It is infinitely continuously partially differentiable on \(\mathbb{R}^3\) with the following Jacobian matrix:

\[J_F(x, y, z) = \begin{bmatrix}2x & 2y & -2z \\ \uppi \cos(\uppi x) & \uppi \cos(\uppi y) & \uppi \cos(\uppi z)\end{bmatrix}\]

For \(\boldsymbol{p} = (2, 2, 0)\), we have:

\[F(\boldsymbol{p}) = F(2, 2, 0) = \boldsymbol{0} \qquad J_F(\boldsymbol{p}) = \begin{bmatrix}4 & 4 & 0 \\ \uppi & \uppi & \uppi \end{bmatrix}\]

Example: Solving in Terms of \(x\)

The matrix formed by columns \(\boldsymbol{j}_2(\boldsymbol{p})\) and \(\boldsymbol{j}_3(\boldsymbol{p})\) of \(J_F(\boldsymbol{p})\) is

\[\begin{bmatrix} \vert & \vert \\ \boldsymbol{j}_2(\boldsymbol{p}) & \boldsymbol{j}_3(\boldsymbol{p}) \\ \vert & \vert \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ \uppi & \uppi \end{bmatrix}\]

and is invertible.

We thus know that there is an open interval \(V \subseteq \mathbb{R}\) around \(x = 2\), an open neighborhood \(W \subseteq \mathbb{R}^2\) around \((y, z) = (2, 0)\) and a unique function \(f: V \to W\) such that for all \((x, y, z) \in V \times W\) we have \(F(x, y, z) = \boldsymbol{0}\) if and only if \((y, z) = f(x)\).

Furthermore, \(f\) is infinitely continuously differentiable on \(V\) with the following derivative:

\[\begin{aligned}f'(x) &= -\begin{bmatrix} \vert & \vert \\ \boldsymbol{j}_2(x,y,z) & \boldsymbol{j}_3(x,y,z) \\ \vert & \vert \end{bmatrix}^{-1} \begin{bmatrix}\vert \\ \boldsymbol{j}_1(x,y,z) \\ \vert \end{bmatrix} \\ &= -\begin{bmatrix} \partial_2 F_1(x, f_1(x), f_2(x)) & \partial_3 F_1(x, f_1(x), f_2(x)) \\ \partial_2 F_2(x, f_1(x), f_2(x)) & \partial_3 F_2(x, f_1(x), f_2(x)) \end{bmatrix}^{-1} \begin{bmatrix}\partial_1 F_1(x, f_1(x), f_2(x)) \\ \partial_1 F_2 (x, f_1(x), f_2(x)) \end{bmatrix} \\ & = -\begin{bmatrix} 2f_1(x) & -2f_2(x) \\ \uppi \cos (\uppi f_1(x)) & \uppi \cos (\uppi f_2(x)) \end{bmatrix}^{-1} \begin{bmatrix} 2x \\ \uppi \cos(\uppi x) \end{bmatrix} \\ & = -\frac{1}{2 \uppi f_1(x) \cos (\uppi f_2(x)) + 2\uppi f_2(x) \cos (\uppi f_1(x))} \begin{bmatrix}\uppi \cos (\uppi f_2(x)) & 2 f_2(x) \\ -\uppi \cos(\uppi f_1(x)) & 2 f_1(x)\end{bmatrix}\begin{bmatrix} 2x \\ \uppi \cos(\uppi x) \end{bmatrix} \\ & = \frac{1}{f_1(x) \cos(\uppi f_2(x)) + f_2(x) \cos(\uppi f_1(x))} \begin{bmatrix} -x \cos(\uppi f_2(x)) - f_2(x) \cos(\uppi x) \\ x \cos(\uppi f_1(x)) - f_1(x) \cos(\uppi x) \end{bmatrix}\end{aligned}\]

Example: Solving in Terms of \(y\)

The matrix formed by columns \(\boldsymbol{j}_1(\boldsymbol{p})\) and \(\boldsymbol{j}_3(\boldsymbol{p})\) of \(J_F(\boldsymbol{p})\) is

\[\begin{bmatrix} \vert & \vert \\ \boldsymbol{j}_1(\boldsymbol{p}) & \boldsymbol{j}_3(\boldsymbol{p}) \\ \vert & \vert \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ \uppi & \uppi \end{bmatrix}\]

which is invertible.

We thus know that there is an open interval \(V \subseteq \mathbb{R}\) around \(y = 2\), an open neighborhood \(W \subseteq \mathbb{R}^2\) around \((x, z) = (2, 0)\) and a unique function \(f: V \to W\) such that for all \((y, x, z) \in V \times W\) we have \(F(x, y, z) = \boldsymbol{0}\) if and only if \((x, z) = f(y)\).

Furthermore, \(f\) is infinitely continuously differentiable on \(V\) with the following derivative:

\[\begin{aligned}f'(y) &= -\begin{bmatrix} \vert & \vert \\ \boldsymbol{j}_1(x,y,z) & \boldsymbol{j}_3(x,y,z) \\ \vert & \vert \end{bmatrix}^{-1} \begin{bmatrix}\vert \\ \boldsymbol{j}_2(x,y,z) \\ \vert \end{bmatrix} \\ &= -\begin{bmatrix} \partial_1 F_1(f_1(y), y, f_2(y)) & \partial_3 F_1(f_1(y), y, f_2(y)) \\ \partial_1 F_2(f_1(y), y, f_2(y)) & \partial_3 F_2(f_1(y), y, f_2(y)) \end{bmatrix}^{-1} \begin{bmatrix}\partial_2 F_1(f_1(y), y, f_2(y)) \\ \partial_2 F_2 (f_1(y), y, f_2(y)) \end{bmatrix} \\ & = -\begin{bmatrix} 2f_1(y) & -2f_2(y) \\ \uppi \cos (\uppi f_1(y)) & \uppi \cos (\uppi f_2(y)) \end{bmatrix}^{-1} \begin{bmatrix} 2y \\ \uppi \cos(\uppi y) \end{bmatrix} \\ & = \frac{1}{f_1(y) \cos(\uppi f_2(y)) + f_2(y) \cos(\uppi f_1(y))} \begin{bmatrix} -y \cos(\uppi f_2(y)) - f_2(y) \cos(\uppi y) \\ y \cos(\uppi f_1(y)) - f_1(y) \cos(\uppi y) \end{bmatrix}\end{aligned}\]

Example: Solving in Terms of \(z\)

The matrix formed by columns \(\boldsymbol{j}_1(\boldsymbol{p})\) and \(\boldsymbol{j}_2(\boldsymbol{p})\) of \(J_F(\boldsymbol{p})\) is

\[\begin{bmatrix} \vert & \vert \\ \boldsymbol{j}_1(\boldsymbol{p}) & \boldsymbol{j}_2(\boldsymbol{p}) \\ \vert & \vert \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ \uppi & \uppi \end{bmatrix}\]

which is not invertible. Therefore, the theorem cannot tell us anything.

Proof

TODO