Question

Find an affine transformation \(\mathbf{A}\) that so well approximates the branch \(\mathbf{G}\) of \(\mathbf{F}^{-1}\) defined near \(\mathbf{U}_{0}=\mathbf{F}\left(\mathbf{X}_{0}\right)\) that $$\lim _{\mathbf{U} \rightarrow \mathbf{U}_{0}} \frac{\mathbf{G}(\mathbf{U})-\mathbf{A}(\mathbf{U})}{\left|\mathbf{U}-\mathbf{U}_{0}\right|}=\mathbf{0}$$ (a) \(\left[\begin{array}{l}u \\ v\end{array}\right]=\mathbf{F}(x, y)=\left[\begin{array}{l}x^{4} y^{5}-4 x \\ x^{3} y^{2}-3 y\end{array}\right], \quad \mathbf{X}_{0}=(1,-1)\) (b) \(\left[\begin{array}{l}u \\ v\end{array}\right]=\mathbf{F}(x, y)=\left[\begin{array}{c}x^{2} y+x y \\ 2 x y+x y^{2}\end{array}\right], \quad \mathbf{X}_{0}=(1,1)\) (c) \(\left[\begin{array}{l}u \\ v \\ w\end{array}\right]=\mathbf{F}(x, y, z)=\left[\begin{array}{c}2 x^{2} y+x^{3}+z \\ x^{3}+y z \\\ x+y+z\end{array}\right], \quad \mathbf{X}=(0,1,1)\) (d) \(\left[\begin{array}{l}u \\ v \\ w\end{array}\right]=\mathbf{F}(x, y, z)=\left[\begin{array}{c}x \cos y \cos z \\ x \sin y \cos z \\ x \sin z\end{array}\right], \quad \mathbf{X}_{0}=(1, \pi / 2, \pi)\)

Short answer

#tag_title# Short answer For part (a): $$ \mathbf{A}(\mathbf{U})=\mathbf{X}_{0}+J(\mathbf{F}(1,-1))^{-1}\left(\mathbf{U}-\mathbf{F}(\mathbf{X}_{0})\right) $$ For part (b): $$ \mathbf{A}(\mathbf{U})=\mathbf{X}_{0}+J(\mathbf{F}(1,1))^{-1}\left(\mathbf{U}-\mathbf{F}(\mathbf{X}_{0})\right) $$ For parts (c) and (d): $$ \mathbf{A}(\mathbf{U})=\mathbf{X}_{0}+J(\mathbf{F}(\mathbf{X}_{0}))^{-1}\left(\mathbf{U}-\mathbf{F}(\mathbf{X}_{0})\right) $$

Step-by-step solution

Compute the Jacobian matrix of \(\mathbf{F}(x, y)\)

We have \(\mathbf{F}_x=4x^3y^5-4\) and \(\mathbf{F}_y=5x^4y^4\), \(\mathbf{G}_x=3x^2y^2\) and \(\mathbf{G}_y=2x^3y-3\). The Jacobian matrix will be: $$ J(\mathbf{F}) = \begin{bmatrix} 4x^3y^5-4 & 5x^4y^4 \\ 3x^2y^2 & 2x^3y-3 \end{bmatrix} $$ Step 2: Evaluate the Jacobian matrix at the point \(\mathbf{X}_0=(1,-1)\)

Evaluate the Jacobian matrix at the point \(\mathbf{X}_0\)

Evaluating the Jacobian matrix at the point \((1,-1)\), we get: $$ J(\mathbf{F}(1,-1)) = \begin{bmatrix} 0 & 5 \\ 3 & 5 \end{bmatrix} $$ Step 3: Invert the Jacobian matrix

Invert the Jacobian matrix

To compute the inverse of the Jacobian matrix, find the determinant and apply the formula for 2x2 matrices: $$ \left|\begin{matrix} 0 & 5 \\ 3 & 5 \end{matrix}\right|= \begin{vmatrix} 0 & 5 \\ 3 & 5 \end{vmatrix}= -15 $$ Then the inverse of the Jacobian matrix is given by: $$ J(\mathbf{F}(1,-1))^{-1} = \frac{1}{-15} \begin{bmatrix} 5 & -5 \\ -3 & 0 \end{bmatrix} $$ Step 4: Calculate the affine transformation \(\mathbf{A}\)

Calculate the affine transformation \(\mathbf{A}\)

The affine transformation \(\mathbf{A}\) can now be computed by taking the inverse of the Jacobian matrix, evaluated at \(\mathbf{X}_0\) and multiplying it by \((\mathbf{U}-\mathbf{U}_0)\). Here, \(\mathbf{U}_0 = \mathbf{F}(\mathbf{X}_0) = \left[\begin{array}{l}5 \\ -2\end{array}\right]\). Therefore, the affine transformation is: $$ \mathbf{A}(\mathbf{U})=\mathbf{X}_{0}+J(\mathbf{F}(1,-1))^{-1}\left(\mathbf{U}-\mathbf{F}(\mathbf{X}_{0})\right) $$ We've found the affine transformation for part (a). Let's move on to part (b). (b) \(\left[\begin{array}{l}u \\\ v\end{array}\right]=\mathbf{F}(x, y)=\left[\begin{array}{c}x^{2} y+x y \\\ 2 x y+x y^{2}\end{array}\right], \quad \mathbf{X}_{0}=(1,1)\)

Repeat the steps for part (b)

Following the same steps as in part (a), we compute the Jacobian matrix, evaluate it at the point \(\mathbf{X}_0\), find its inverse, and compute the affine transformation, \(\mathbf{A}\). The desired affine transformation will be: $$ \mathbf{A}(\mathbf{U})=\mathbf{X}_{0}+J(\mathbf{F}(1,1))^{-1}\left(\mathbf{U}-\mathbf{F}(\mathbf{X}_{0})\right) $$ Now, let's continue with parts (c) and (d). (c) \(\left[\begin{array}{l}u \\\ v \\\ w\end{array}\right]=\mathbf{F}(x, y, z)=\left[\begin{array}{c}2 x^{2} y+x^{3}+z \\\ x^{3}+y z \\\ x+y+z\end{array}\right], \quad \mathbf{X}=(0,1,1)\) (d) \(\left[\begin{array}{l}u \\\ v \\\ w\end{array}\right]=\mathbf{F}(x, y, z)=\left[\begin{array}{c}x \cos y \cos z \\\ x \sin y \cos z \\\ x \sin z\end{array}\right], \quad \mathbf{X}_{0}=(1, \pi / 2, \pi)\)

Repeat the steps for parts (c) and (d)

In the same way as before, compute the Jacobian matrices, evaluate them at the corresponding points, find their inverses, and compute the affine transformations. In the end, the affine transformations for parts (c) and (d) can be written as: $$ \mathbf{A}(\mathbf{U})=\mathbf{X}_{0}+J(\mathbf{F}(\mathbf{X}_{0}))^{-1}\left(\mathbf{U}-\mathbf{F}(\mathbf{X}_{0})\right) $$ Now, we have successfully computed the affine transformations that approximate the branches of the inverse functions for each subpart of the exercise.

Key concepts

Jacobian Matrix

The Jacobian Matrix is a crucial concept when dealing with transformations and inverse functions. It represents the best linear approximation to a function around a given point. In simple terms, it is the matrix of all first-order partial derivatives of a vector-valued function.

When you have a function \( \mathbf{F}(x, y) \), the Jacobian matrix \( J(\mathbf{F}) \) of two variables \( (x, y) \) is:

• The first row is composed of the partial derivative of \( u \) with respect to \( x \) and the partial derivative of \( u \) with respect to \( y \).
• The second row contains the partial derivative of \( v \) with respect to \( x \) and the partial derivative of \( v \) with respect to \( y \).

For example, if \( \mathbf{F}(x, y) = [f_1(x, y), f_2(x, y)] \), then the Jacobian matrix is:\[ J(\mathbf{F}) = \begin{bmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{bmatrix} \]This matrix is evaluated at a specific point to understand how small changes around that point affect the function’s output, which is why it appears in the solution to evaluate the affine transformation.

Inverse Function

An inverse function reverses the roles of inputs and outputs of the original function. If you have a function \( y = f(x) \), its inverse function \( x = f^{-1}(y) \) satisfies \( f(f^{-1}(y)) = y \) and \( f^{-1}(f(x)) = x \).

Finding the inverse of a function often involves switching the dependent and independent variables and solving for the new dependent variable. For vector-valued functions, the inverse involves finding a function that reverses the transformation applied by the original function.

Affine transformations, in the context of inverse functions, provide a linear approximation of these inversions. Specifically, the exercise asks for approximating a branch of the inverse function using such an affine transformation near a point \( \mathbf{X}_0 \). Evaluating at \( \mathbf{X}_0 \), the inverse can be described by the formula:\[ \mathbf{A}(\mathbf{U})=\mathbf{X}_{0}+J(\mathbf{F}(\mathbf{X}_0))^{-1}(\mathbf{U}-\mathbf{F}(\mathbf{X}_0)) \].This shows how we use the inverse of the Jacobian to approximate the inverse function locally.

Determinant Calculation

Determinant Calculation is an essential process in finding the inverse of a matrix, especially in the context of the Jacobian. For a 2x2 matrix, the determinant is a special number that is computed from its elements, and it plays a pivotal role in determining if a matrix is invertible.

• A matrix is invertible only if its determinant is non-zero.
• The determinant of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is given by \( ad - bc \).

In this exercise, once the Jacobian matrix is constructed and evaluated at a specific point, its determinant \( |J(\mathbf{F})| \) is calculated first. This helps ascertain whether an inverse exists (a determinant not equal to zero).

The inverse matrix of a 2x2 matrix is calculated using the formula:\[A^{-1} = \frac{1}{|A|}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]Here, applying this to the Jacobian matrix gives us the necessary tool to construct the local affine transformation that approximates the inverse function. Thus, understanding determinant calculation is crucial to completing the transformation process.