Problem 10 - Advanced Engineering Mathematics

Question

For \(m \neq n\) $$\begin{aligned} \int_{0}^{p} \sin \frac{n \pi}{p} x \sin \frac{m \pi}{p} x d x &=\frac{1}{2} \int_{0}^{p}\left(\cos \frac{(n-m) \pi}{p} x-\cos \frac{(n+m) \pi}{p} x\right) d x \\ &=\left.\frac{p}{2(n-m) \pi} \sin \frac{(n-m) \pi}{p} x\right|_{0} ^{p}-\left.\frac{p}{2(n+m) \pi} \sin \frac{(n+m) \pi}{p} x\right|_{0} ^{p}=0. \end{aligned}$$ For \(m=n\) $$\int_{0}^{p} \sin ^{2} \frac{n \pi}{p} x d x=\int_{0}^{p}\left(\frac{1}{2}-\frac{1}{2} \cos \frac{2 n \pi}{p} x\right) d x=\left.\frac{1}{2} x\right|_{0} ^{p}-\left.\frac{p}{4 n \pi} \sin \frac{2 n \pi}{p} x\right|_{0} ^{p}=\frac{p}{2}$$ so that $$\left\|\sin \frac{n \pi}{p} x\right\|=\sqrt{\frac{p}{2}}.$$

Step-by-Step Solution

Answer

The integral is 0 for \(m \neq n\) and \(\frac{p}{2}\) for \(m=n\), with the norm being \(\sqrt{\frac{p}{2}}\).

Step 1: Break Down the Given Integral Expression

First, we need to interpret and solve the given integral \( \int_0^p \sin \frac{n \pi}{p} x \sin \frac{m \pi}{p} x \, dx \). This expression is evaluated using the trigonometric identity for the product of sines: \( \sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)] \).

Step 2: Apply Trigonometric Identity

Substitute the identity \( \sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)] \) into the given integral:\[ \int_0^p \sin \frac{n \pi}{p} x \sin \frac{m \pi}{p} x \, dx = \frac{1}{2} \int_0^p \left( \cos \frac{(n-m) \pi}{p} x - \cos \frac{(n+m) \pi}{p} x \right) \, dx \]

Step 3: Evaluate the Integral for Different Frequencies

When \( m eq n \), evaluate each integral separately:- \( \left. \frac{p}{2(n-m)\pi} \sin \frac{(n-m)\pi}{p} x \right|_0^p = 0 \) because the sine function at these arguments is zero.- \( \left. \frac{p}{2(n+m)\pi} \sin \frac{(n+m)\pi}{p} x \right|_0^p = 0 \) for the same reason.Thus, the entire expression evaluates to 0 when \( m eq n \).

Step 4: Consider the Case When \( m = n \)

For \( m = n \), the integral becomes \( \int_0^p \sin^2 \frac{n \pi}{p} x \, dx \). Use the identity \( \sin^2 A = \frac{1}{2} - \frac{1}{2} \cos 2A \) to simplify the integral:\[ \int_0^p \left( \frac{1}{2} - \frac{1}{2} \cos \frac{2n \pi}{p} x \right) \ dx. \]

Step 5: Calculate the Simplified Integral

Evaluate each part:- \( \left. \frac{1}{2} x \right|_0^p = \frac{p}{2} \).- \( \left. \frac{p}{4n\pi} \sin \frac{2n\pi}{p} x \right|_0^p = 0 \) because the sine function is zero at the boundaries.This results in \( \frac{p}{2} \).

Step 6: Find the Norm of the Sine Function

To find the norm, \( \left\| \sin \frac{n \pi}{p} x \right\| = \sqrt{\text{integral value}} = \sqrt{\frac{p}{2}} \).

Key Concepts

Trigonometric Integrals

Trigonometric integrals involve finding the integral of functions that include trigonometric terms such as sine or cosine. In our problem, we are particularly interested in integrals of the form \( \int_{0}^{p} \sin \frac{n \pi}{p} x \sin \frac{m \pi}{p} x \ dx \). This specific type is important because it determines whether two sine functions are "orthogonal", which means they do not overlap in frequency.

Understanding this integral starts by using a trigonometric identity for the product of sine functions. By converting our expression using \( \sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)] \), we can break down this complex integral into more manageable parts. Each part becomes easier to solve because cosine functions have simple properties.

• If \( m eq n \), the integral evaluates to zero. This is known as orthogonality; it occurs because the two sine waves are distinct frequencies.
• If \( m = n \), we instead find the integral of \( \sin^2 \), which does not vanish.

The integral thus evaluates to \( \frac{p}{2} \) for \( m = n \). These processes help us uncover the relationships of these functions within a given range.

Trigonometric Identities

Trigonometric identities play a crucial role in simplifying the integrals of trigonometric functions. In our example, we use the identity \( \sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)] \). This identity helps transform the product of two sines into a more straightforward expression involving cosines, which are easier to integrate.

One significant identity we also used is \( \sin^2 A = \frac{1}{2} - \frac{1}{2} \cos 2A \). This identity simplifies integrals of squared sine functions. Thereby, it turns the integral into one involving a constant and a cosine term, both straightforward to integrate over any interval.

• Cosine's Easy Integration: Integrating cosine leads to sine, which is zero at integral boundaries when multiplied by certain factors.
• Large Role of Amplitude Factors: Factors \( \frac{1}{2} \) or \( \frac{1}{2} \cos 2A \) modify the amplitude of these functions to ensure correct output after integration, precise for definite integrals.

Understanding these identities immensely aids us in reducing complex integrals to simpler ones, thus making them approachable for computation.

Square-integrability

Square-integrability is a concept that defines the behavior of a function over an interval when the square of the absolute value of the function has a finite integral over that interval.

In terms of the sine functions, when we say they are square-integrable over \([0, p]\), we mean the integral of the square over this interval is finite. For our sine function \( \sin \frac{n \pi}{p} x \), its square-integrability is expressed as \( \int_{0}^{p} \sin^2 \frac{n \pi}{p} x \, dx = \frac{p}{2} \). This outcome not only confirms integrability but also provides a critical value used in further mathematical applications, like finding the norm of the function.

Function Norm: The norm of the sine function is derived from this integral value, specifically \( \left\| \sin \frac{n \pi}{p} x \right\| = \sqrt{\frac{p}{2}} \). The norm provides a sense of the 'size' or 'magnitude' of the function over its interval.
Square-integrable functions remain fundamental in analysis as they ensure convergent integrals, allowing complex functions to be represented in useful forms. They ensure the resulting values probabilistically meaningful and useful in various areas of mathematics and physical applications.