Problem 6 - A First Course in Real Analysis

Question

Show that the relation \(F\left(x_{1}, x_{2}, y\right)=0\) yields \(y\) as a function of \(\left(x_{1}, x_{2}\right)\) in a neighborhood of the given point \(P\left(x_{1}^{0}, x_{2}^{0}, y^{0}\right)\). Denoting this function by \(f\), compute \(f_{, 1}\) and \(f_{, 2}\) at \(P\). \(F\left(x_{1}, x_{2}, y\right) \equiv x_{1}^{3}+x_{2}^{3}+y^{3}-3 x_{1} x_{2} y-4=0 ; P\left(x_{1}^{0}, x_{2}^{0}, y^{0}\right)=(1,1,2)\)

Step-by-Step Solution

Answer

Given that \( \frac{\partial F}{\partial y} \) evaluated at P is non-zero, \( y \) is a function of \( \left(x_{1}, x_{2}\right) \) near P. At P, \( f_{,1} = \frac{1}{2} \) and \( f_{,2} = \frac{1}{2} \).

Step 1: Verify Implicit Function Theorem Conditions

The Implicit Function Theorem states that if the function F is continuously differentiable and the partial derivative of F with respect to y is non-zero at a point, then F defines y as a function of x1 and x2 near that point. First, ensure the function F is continuously differentiable and then calculate the partial derivative of F with respect to y at the point P.

Step 2: Calculate Partial Derivative \( \frac{\partial F}{\partial y} \) at P

Calculate \( \frac{\partial F}{\partial y} \) by differentiating the given function F with respect to y: \( \frac{\partial}{\partial y}(x_1^3 + x_2^3 + y^3 - 3x_1 x_2 y - 4) = 3y^2 - 3x_1 x_2 \) and evaluate at the point P.

Step 3: Check Non-Zero Condition

Evaluate \( \frac{\partial F}{\partial y} \) at the point \( P(1, 1, 2) \) to check if it's non-zero, which is necessary for y to be defined as a function of \( x_1 \) and \( x_2 \) near P.

Step 4: Compute \( f_{, 2} \) at P

Similarly, use the Chain Rule to find \( f_{, 2} \) in terms of the partial derivatives of F. Calculate \( \frac{\partial F}{\partial x2} \) and use the previously found \( \frac{\partial F}{\partial y} \) to solve for \( f_{, 2} \) at the point P.

Key Concepts

Continuous Differentiability

Continuous differentiability refers to a function that has derivatives of all orders that are continuous. For a multivariable function, like the one in our exercise, this means that all of the partial derivatives exist and are continuous functions within a neighborhood of a point. The continuity of the derivatives reassures us that the function behaves in a predictable and smooth manner near the point of interest.

In the exercise, we want to determine whether the function given by the relation F(x_1, x_2, y) = 0 allows us to express y as a function of x_1 and x_2. To apply the Implicit Function Theorem confidently, we need to verify continuous differentiability of the function F. Only after confirming that F is continuously differentiable at the given point can we proceed with applying the theorem to express y in terms of x_1 and x_2.

Partial Derivative

Partial derivatives are used in calculus to measure how a function changes as one of its input variables changes, while keeping other input variables constant. For the function F(x_1, x_2, y), partial derivatives would be computed with respect to each variable x_1, x_2, and y.

Considering our exercise, the partial derivative of F with respect to y, denoted as \( \frac{\partial F}{\partial y} \), is crucial. It signifies how the function's value changes in response to changes in y, while x_1 and x_2 are held fixed. The Implicit Function Theorem requires this derivative to be non-zero at the point of interest in order to solve for y as a function of x_1 and x_2. In the exercise, we compute \( \frac{\partial F}{\partial y} \) at point P and confirm it's non-zero, satisfying the condition to apply the theorem.

Chain Rule

The Chain Rule in calculus is a formula to compute the derivative of a composition of functions. It can be extended to multiple dimensions, which is extraordinarily useful when dealing with functions of several variables as in our exercise. When functions are composed, the Chain Rule tells us how to differentiate the outer function with respect regard to the inner function's variables.

In the context of the exercise, the partial derivatives of F with respect to x_1 and x_2 can be linked to the derivative of y with respect to x_1 and x_2 using the Chain Rule. This helps us to find \(f_{,1}\) and \(f_{,2}\) at the point P. We use the partial derivatives \( \frac{\partial F}{\partial x_1} \) and \( \frac{\partial F}{\partial x_2} \), together with the already calculated \( \frac{\partial F}{\partial y} \), to solve for the derivatives of y, which, in our notations, are expressed as f_{,1} and f_{,2}.

Real Analysis

Real analysis is a branch of mathematics that deals with the set of real numbers and real-valued functions. It involves exploring the limits, continuity, differentiation, and integration of these functions. This field provides the theoretical foundation for calculus, which we apply in practical problem-solving situations like our exercise.
Real analysis plays a critical role when we discuss concepts like continuity, differentiability, and the behavior of functions on intervals or neighborhoods, which is exactly what's required to apply the Implicit Function Theorem. In real analysis, we rigorously prove that the conditions laid out in the theorem are enough to guarantee the existence of a function y = f(x_1, x_2) near the point P, and hence we can compute the derivatives as required.