Problem 16 - Advanced Engineering Mathematics

Question

Verify by direct integration that the functions are orthogonal with respect to the indicated weight function on the given interval. Let \(\left\\{\phi_{n}(x)\right\\}\) be an orthogonal set of functions on \([a, b]\) suchthat \(\phi_{0}(x)=1\) and \(\phi_{1}(x)=x\). Show that \(\int_{a}^{b}(\alpha x+\beta) \phi_{n}(x) d x=0\) for \(n=2,3, \ldots\) and any constants \(\alpha\) and \(\beta\).

Step-by-Step Solution

Answer

The functions are orthogonal for \(n=2,3,\ldots\) by the given orthogonality conditions.

Step 1: Understanding Orthogonality

The concept of orthogonality implies that two functions \(f(x)\) and \(g(x)\) are orthogonal with respect to the weight function \(w(x)\) on the interval \([a, b]\) if \(\int_{a}^{b} f(x)g(x) w(x)\, dx = 0\). We need to show this condition for the given functions and interval.

Step 2: Setup the Integral

We are given \(\phi_0(x) = 1\) and \(\phi_1(x) = x\). Our task is to show that \(\int_{a}^{b} (\alpha x + \beta)\phi_n(x) \, dx = 0\) for \(n = 2, 3, \ldots\). The given set of functions \(\{\phi_n(x)\}\) is orthogonal on \([a, b]\).

Step 3: Evaluate the Integral

Because of the orthogonality, for \(n \geq 2\), \(\phi_n(x)\) must be orthogonal to both \(\phi_0(x)\) and \(\phi_1(x)\). Thus, \(\int_{a}^{b} \phi_n(x) \, dx = 0\) and \(\int_{a}^{b} x\phi_n(x) \, dx = 0\). Substituting these results, we have \(\int_{a}^{b} (\alpha x + \beta) \phi_n(x) \, dx = \alpha \int_{a}^{b} x\phi_n(x) \, dx + \beta \int_{a}^{b} \phi_n(x) \, dx = 0 \).

Key Concepts

Weight Function

A weight function, denoted as \( w(x) \), is a significant component when investigating orthogonal functions. It acts as a multiplier in the integral, providing additional context or "weight" to the data. This concept is particularly valuable in physical applications where different parts of the domain might contribute differently to a process. For example, in the context of orthogonality, the weight function ensures that certain sections of the function's domain have more influence when determining orthogonality.In our exercise, the weight function is considered implicitly. When we compute the integral \( \int_{a}^{b} f(x)g(x)w(x)\, dx \), we often take \( w(x) = 1 \), implying an equal weight distribution across the domain. However, in more complex problems, \( w(x) \) might vary and drastically change the analysis. Recognizing the role of \( w(x) \) helps in understanding why some functions can interact to achieve an integral result of zero, indicating orthogonality.

Integration

Integration is a crucial mathematical process for solving problems involving orthogonal functions. In the context of this exercise, integration is used to evaluate whether the given functions are orthogonal over a specific interval. The integral formula \( \int_{a}^{b} f(x)g(x)\, dx \) helps sum up the product of the functions over the domain from \( a \) to \( b \).In our problem, integration is applied to confirm that across the interval \([a, b]\), the linear function \( (\alpha x + \beta) \) is orthogonal to the higher order functions \( \phi_n(x) \) for \( n \geq 2 \). This means calculating \( \int_{a}^{b} (\alpha x + \beta) \phi_n(x)\, dx \) and verifying it equals zero.Key points to remember:

• If the functions involve coefficients as in our case with \( \alpha \) and \( \beta \), extract them before carrying out integration.
• Ensure that the calculations respect the orthogonality condition established early in the problem.

Orthogonality

Orthogonality in the context of functions extends the notion of perpendicularity from vectors to the realm of function spaces. Two functions \( f(x) \) and \( g(x) \) are considered orthogonal over an interval \([a, b]\) if their weighted integral equals zero: \( \int_{a}^{b} f(x)g(x) w(x)\, dx = 0 \). This indicates that within the interval, there is no 'overlap' between the functions.In our exercise, orthogonal functions \( \phi_n(x) \) need to satisfy this condition with a linear function \( (\alpha x + \beta) \). By establishing \( \int_{a}^{b} \phi_n(x)\, dx = 0 \) and \( \int_{a}^{b} x\phi_n(x)\, dx = 0 \) for \( n \geq 2 \), the problem leverages orthogonality to show there is no contribution from these terms when calculating the integral:- The integral demonstrates that individual higher-order terms do not influence the final result.- Orthogonality assures independent function behavior in systems, crucial in places like signal processing or solving differential equations.This fundamental concept showcases why understanding orthogonality is vital for tackling complex mathematical problems.