Question

Let \(f=\left(f_{1}, f_{2}\right)\) be the mapping of \(R^{2}\) into \(R^{2}\) given by $$ f_{1}(x, y)=e^{x} \cos y, \quad f_{2}(x, y)=e^{x} \sin y $$ (a) What is the range of \(f ?\) (b) Show that the Jacobian of \(f\) is not zero at any point of \(R^{2}\). Thus every point of \(R^{2}\) has a neighborhood in which \(f\) is one-to-one. Nevertheless, \(f\) is not one-toone on \(R^{2}\). (c) Put \(a=(0, \pi / 3), b=f(a)\), let \(g\) be the continuous inverse of \(\mathbf{f}\), defined in a neighborhood of \(\mathrm{b}\), such that \(\mathrm{g}(\mathrm{b})=\mathrm{a} .\) Find an explicit formula for \(\mathrm{g}\), compute \(\mathbf{f}^{\prime}(\mathbf{a})\) and \(\mathbf{g}^{\prime}(\mathbf{b})\), and verify the formula (52). ( ) What are the images under \(\mathrm{f}\) of lines parallel to the coordinate axes?

Short answer

Based on the given exercise, we analyzed the mapping \(f:\mathbb{R}^2\to\mathbb{R}^2\), computed the range of \(f\), showed that the Jacobian of \(f\) is nonzero at every point in \(R^2\), found the continuous inverse of \(f\), computed \({f'}(a)\) and \({g'}(b)\) and verified formula (52), and finally found the images under \(f\) of lines parallel to the coordinate axes. The range of \(f\) is \(\mathbb{R}^2\), and its continuous inverse \(g(u,v)\) is given by \(g(u,v) = \left( \frac{1}{2}\ln(u^2 + v^2), \arctan\frac{v}{u} \right)\). We verified the formula (52) by showing that \((f')^{-1}(a)=g'(b)\). The images under \(f\) of lines parallel to the \(x\)-axis are curves given by the equations \(u = e^x\cos c\) and \(v = e^x\sin c\), while the images under \(f\) of lines parallel to the \(y\)-axis are circles with center at the origin and radius \(e^k\), where \(k\) is a constant.

Step-by-step solution

Compute the range of f

Given the mapping \(f:\mathbb{R}^2\to\mathbb{R}^2\) by: \[ f_1(x, y) = e^x\cos y \quad \text{and} \quad f_2(x, y) = e^x\sin y \] To find the range of \(f\), we can look at values of \(f_1(x,y)\) and \(f_2(x,y)\) for \(x, y \in\mathbb{R}\). Notice that \(e^x\) can take any positive value and \(\cos y, \sin y\) lie in the range \([-1, 1]\). Therefore, the range of \(f_1\) and \(f_2\) is \(\mathbb{R}\), and the range of \(f\) is \(\mathbb{R}^2\).

Show that the Jacobian of f is not zero at any point of \(R^{2}\)

Compute the partial derivatives of \(f_1\) and \(f_2\): \[ \frac{\partial f_1}{\partial x} = e^x \cos y, \quad \frac{\partial f_1}{\partial y} = -e^x \sin y, \quad \frac{\partial f_2}{\partial x} = e^x\sin y , \quad \frac{\partial f_2}{\partial y} = e^x \cos y \] The Jacobian matrix of \(f\) is given by: \[ J_f(x,y) = \begin{pmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{pmatrix} = \begin{pmatrix} e^x\cos y & -e^x\sin y \\ e^x\sin y & e^x\cos y \end{pmatrix} \] The determinant of the Jacobian matrix is: \[ \det(J_f(x,y)) = (e^x \cos y)(e^x \cos y) - (-e^x \sin y)(e^x \sin y) = (e^x)^2 (\cos^2 y + \sin^2 y) = e^{2x} \] Since \(e^{2x} > 0\) for all \(x \in \mathbb{R}\), the Jacobian is nonzero at every point in \(R^2\).

Find the continuous inverse of f

Since \(f\) has a nonzero Jacobian, it has a continuous inverse \(g\) in a neighborhood of \(b = f(a)\). We have \(a = (0, \pi/3)\) and \(b = f(a) = (f_1(a), f_2(a)) = (\frac{1}{2}, \frac{\sqrt{3}}{2})\). Using the fact that \(f_1^2(x,y) + f_2^2(x,y) = e^{2x}(\cos^2 y + \sin^2 y) = e^{2x}\), we find that \(x = \frac{1}{2}\ln(f_1^2(b)+f_2^2(b)),\) and \(y = \arctan\frac{f_2(b)}{f_1(b)}\). Thus, the continuous inverse \(g (u,v) =(x,y)\), defined in a neighborhood of \(b\), is given by \[ g(u,v) = \left( \frac{1}{2}\ln(u^2 + v^2), \arctan\frac{v}{u} \right). \]

Compute \({f'}(a)\) and \({g'}(b)\) and verify the formula (52)

Computing the partial derivatives of \(g(u,v)\), we get: \[ \frac{\partial g_1}{\partial u} = \frac{u}{u^2 + v^2}, \quad \frac{\partial g_1}{\partial v} = \frac{v}{u^2 + v^2}, \quad \frac{\partial g_2}{\partial u} = -\frac{v}{u^2 + v^2}, \quad \frac{\partial g_2}{\partial v} = \frac{u}{u^2 + v^2} \] Now evaluating \(f'(a)\) and \(g'(b)\): \[ f'(a) = J_f(0,\pi/3) = \begin{pmatrix} 1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & 1/2 \end{pmatrix}, \quad g'(b) = \begin{pmatrix} 1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & 1/2 \end{pmatrix} \] Finally, we verify the formula (52) which states \((f')^{-1}(a)=g'(b)\): \[ (g'(b))(f'(a)) = \begin{pmatrix} 1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & 1/2 \end{pmatrix} \begin{pmatrix} 1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & 1/2 \end{pmatrix}= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \] The above computations verify the formula (52).

Find the images under f of lines parallel to the coordinate axes

1. For lines parallel to the \(x\)-axis, we have \(y=c\), where \(c\) is a constant. So we get \(f_1(x,c) = e^x\cos c\) and \(f_2(x,c) = e^x\sin c\). The image of these lines under \(f\) are curves with the equations \(u = e^x\cos c\) and \(v = e^x\sin c\). 2. For lines parallel to the \(y\)-axis, we have \(x=k\), where \(k\) is a constant. So we get \(f_1(k,y) = e^k\cos y\) and \(f_2(k,y) = e^k\sin y\). The image of these lines under \(f\) are circles with center at the origin and radius \(e^k\).

Key concepts

Jacobian Matrix

In mathematical analysis, the Jacobian matrix is a critical tool used to understand how multivariable functions behave. It is a matrix of all first-order partial derivatives of a vector-valued function. To illustrate, let's consider the function given in the exercise, which maps \(R^2\) into \(R^2\) with components \(f_1\) and \(f_2\).

The Jacobian matrix of this function, denoted as \(J_f(x,y)\), is structured as:
\[J_f(x,y) = \begin{pmatrix}\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y}\end{pmatrix}\]
This matrix helps us analyze the local behavior of the function near any point in its domain. In the context of the exercise, we computed the determinant of this matrix to show that it is nonzero at any point in \(R^2\), which implies the function \(f\) is locally invertible around every point of \(R^2\).

It's useful to note that the Jacobian is an essential component in determining the properties of a function such as whether it's a diffeomorphism in a certain region, which would necessitate a non-zero determinant of the Jacobian matrix.

Inverse Functions

An inverse function reverses the effect of the original function. It maps the output back to the input, reinstating the original value prior to the function's application. In our exercise, \(f\) has an inverse function \(g\) since the Jacobian determinant of \(f\) is never zero, satisfying a necessary condition for invertibility.

Consider the point \(a=(0, \pi / 3)\) and its image \(b=f(a)\). The continuous inverse \(g\) is defined in a neighborhood of \(b\), indicating there exists a local inversion allowing us to recover \(a\) from \(b\) via \(g\). The explicit formula for \(g\) delineates the relationship between the coordinates of \(b\) with those of the original point \(a\) before \(f\) was applied. Specifically, \(g\) recovers \(x\) by taking the natural logarithm scaled by \(1/2\) of the sum of the squares of \(u\) and \(v\), the components of \(b\), and \(y\) is calculated using the \(\arctan\) function.

Understanding the concept of inverse functions is vital across numerous fields, including calculus and differential equations, where reversing operations often plays a key role in problem-solving.

Range of a Function

The range of a function represents all the possible values a function can output, based on its domain. In our exercise, since the functions \(f_1(x, y) = e^x\cos y\) and \(f_2(x, y) = e^x\sin y\) both operate on all real numbers in \(\mathbb{R}^2\), the output values could seem to be quite varied at first glance.

However, upon closer examination, we observe that the exponential function \(e^x\) only outputs positive values, while \(\cos y\) and \(\sin y\) oscillate between -1 and 1. This interplay ensures that the range for both \(f_1\) and \(f_2\) remains within the real plane \(\mathbb{R}^2\), covering every possible point in it. Consequently, the range of \(f\), which is the set of all possible \(\(f_1, f_2\)\), is the entirety of \(\mathbb{R}^2\).

This concept is critical in understanding the output limitations or possibilities for any given function, and it assists in visualizing the function's behavior over its entire domain.

Partial Derivatives

Partial derivatives represent the rate at which a multivariable function changes as you vary one of its input variables, keeping all the other variables constant. They are foundational in the study of multivariable calculus and are deeply connected to the concept of the gradient and the Jacobian matrix.

For the function \(f\) in the exercise with the component functions \(f_1\) and \(f_2\), the partial derivatives are computed with respect to each variable independently. For instance, \(\frac{{\partial f_1}}{{\partial x}}\) measures how \(f_1\) changes as \(x\) changes while \(y\) is held constant.

From the solution provided, we derived the partial derivatives and built the Jacobian matrix to assess the function's local invertibility. Both concepts, partial derivatives for the individual behavior of variables and the Jacobian matrix for the collective behavior, together provide a powerful framework for analyzing and understanding the dynamics of multivariable functions.