Problem 46 - Calculus Early Transcendentals

Question

(a) Maximize \(\sum_{t=1}^{n} X_{i} y_{i}\) subject to the constraints \(\sum_{t=1}^{n} x_{t}^{2}=1\) and \(\Sigma_{i=1}^{n} y_{i}^{2}=1.\) (b) Put $$x_{i}=\frac{a_{i}}{\sqrt{\Sigma a_{j}^{2}}} \quad \text { and } \quad y_{i}=\frac{b_{i}}{\sqrt{\Sigma b_{j}^{2}}}$$ to show that $$\sum a_{i} b_{i} \leqslant \sqrt{\Sigma a_{j}^{2}} \sqrt{\Sigma b_{j}^{2}}$$ for any numbers \(a_{1}, \ldots, a_{n} b_{1}, \ldots, b_{n}\) . This inequality is known as the Cauchy-Schwarz Inequality.

Step-by-Step Solution

Answer

The maximum sum is 1, proving the Cauchy-Schwarz Inequality.

Step 1: Set up the optimization problem

We are tasked with maximizing the expression \( \sum_{t=1}^{n} X_{i} y_{i} \) subject to the constraints \( \sum_{t=1}^{n} x_{t}^{2}=1 \) and \( \Sigma_{i=1}^{n} y_{i}^{2}=1 \). This is a standard optimization problem subject to constraints which suggests the use of Lagrange multipliers.

Step 2: Apply the Cauchy-Schwarz Inequality

The expression \( \sum_{i=1}^{n} x_i y_i \) by the Cauchy-Schwarz Inequality is at most \( \sqrt{\sum_{t=1}^{n} x_t^2} \times \sqrt{\sum_{t=1}^{n} y_t^2} \). Given the constraints, it simplifies to \( \sqrt{1} \times \sqrt{1} = 1 \). Thus, the maximum possible value of the sum is 1, achieved when \( x_i = y_i \) for all \( i \).

Step 3: Substitute and simplify

Substitute \( x_i = \frac{a_i}{\sqrt{\sum a_j^2}} \) and \( y_i = \frac{b_i}{\sqrt{\sum b_j^2}} \) into the expression. This gives:\[ \sum_{i=1}^{n} \frac{a_i}{\sqrt{\sum a_j^2}} \cdot \frac{b_i}{\sqrt{\sum b_j^2}} = \frac{\sum_{i=1}^{n} a_i b_i}{\sqrt{\sum a_j^2} \cdot \sqrt{\sum b_j^2}} \]

Step 4: Conclude the proof for Cauchy-Schwarz Inequality

By the result from Step 2, this fraction is at most \( 1 \), hence:\[ \sum a_i b_i \leq \sqrt{\sum a_j^2} \cdot \sqrt{\sum b_j^2} \]This confirms the Cauchy-Schwarz Inequality.

Key Concepts

Lagrange multipliers

Understanding Lagrange multipliers is crucial for tackling constrained optimization problems. Imagine you have a function you want to maximize or minimize, subject to certain constraints represented by equations. The Lagrange multiplier technique is a strategy used to find local maxima and minima of a function subject to equality constraints.

The idea is to transform the original problem into an unconstrained problem, by introducing additional variables, called Lagrange multipliers, one for each constraint. These multipliers help incorporate the constraints into the function we want to optimize. We construct a new function called the Lagrangian, which combines the original function and the constraints, weighted by the Lagrange multipliers.

For instance, if we want to maximize the function \( f(x, y) \) subject to the constraint \( g(x, y) = 0 \), we form the Lagrangian \( \mathcal{L}(x, y, \lambda) = f(x, y) + \lambda \cdot g(x, y) \). Solving the system of equations derived from setting the gradient of \( \mathcal{L} \) to zero gives us the solutions for the variables and the multipliers.

Optimization problem

Optimization problems are everywhere in mathematics and the real world, involving the selection of the best solution from a set of available alternatives. In the context of calculus and higher-level mathematics, these problems often require finding the maximum or minimum value of a function given certain constraints.

There are various methods to solve optimization problems. Basic problems without constraints can be solved using derivative tests. However, when constraints are present, as in the sum of squared variables equalling one, more advanced techniques like Lagrange multipliers are used. These methods leverage calculus to effectively find optimal solutions that satisfy all given conditions.

In the exercise given, maximizing the sum \( \sum_{t=1}^{n} x_i y_i \) under the constraints \( \sum_{t=1}^{n} x_t^2 = 1 \) and \( \sum_{t=1}^{n} y_t^2 = 1 \) demonstrates such a problem. By using Lagrange multipliers, we can determine the point where the function achieves its maximum value within the specified limits.

Mathematical proof

Mathematical proofs are a fundamental aspect of mathematics, allowing us to verify the truth of statements and theorems rigorously.

In this exercise, the Cauchy-Schwarz Inequality is proved using the setup and constraints of an optimization problem. The proof starts by using an inequality involving sums and proceeds by showing how substituting particular forms for variables — as shown in the step with the vectors \( x_i \) and \( y_i \) — satisfies both the inequality and constraints.

Proofs often involve a series of logical steps and calculations. In this scenario, by substituting specific values related to vectors in our exercise, we confirm that the sum \( \sum a_i b_i \) is less than or equal to \( \sqrt{\sum a_j^2} \cdot \sqrt{\sum b_j^2} \). This step-by-step process verifies the statement of the Cauchy-Schwarz Inequality, highlighting the power and precision of mathematical proofs.

Vector spaces

Vector spaces are a central concept in linear algebra, serving as a foundation for many areas in mathematics. A vector space is a set equipped with two operations: addition and scalar multiplication. These operations must follow specific rules, such as associativity, commutativity, and distributivity.

In the context of the exercise, we deal with vectors \( x = (x_1, x_2, \ldots, x_n) \) and \( y = (y_1, y_2, \ldots, y_n) \). These can be considered points in a vector space, constrained by the condition that their squared components sum to one. This constraint makes them unit vectors, lying on a sphere in the vector space.

In linear algebra, the Cauchy-Schwarz Inequality, as explored in the exercise, is an essential property associated with dot products in vector spaces. It states that the absolute value of the dot product of two vectors is less than or equal to the product of their magnitudes, which can be visualized as a statement about the maximum overlap two vectors can have in any direction. Understanding vector spaces and their properties is vital for grasping the underlying principles of the inequality.