Problem 11 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

For \( m \neq n \), the integral is 0. For \( m = n \), the integral is \( \frac{p}{2} \). Norms are \( \sqrt{p} \) and \( \sqrt{\frac{p}{2}} \) respectively.

Step 1: Examine the Integral with Different Indices

The problem starts by considering the integral \( \int_{0}^{p} \cos \frac{n \pi}{p} x \cos \frac{m \pi}{p} x \, dx \) for \( m eq n \). By using the trigonometric identity for the product of cosines, it becomes two separate integrals with frequencies \( n-m \) and \( n+m \), resulting in zero due to the integral of sine over one period.

Step 2: Simplify the Integral for Same Indices

For the case \( m = n \), the function \( \cos^2 \frac{n \pi}{p} x \) is expressed in terms of 1 and \( \cos \) using the identity \( \cos^2 \theta = \frac{1}{2} (1 + \cos 2\theta) \). Integrating gives \( \int_{0}^{p} \cos^2 \frac{n \pi}{p} x \, dx = \frac{p}{2} \) since the integral of the \( \cos \) term evaluates to zero.

Step 3: Evaluate Integrals with Constant Terms

The integral \( \int_{0}^{p} 1 \cdot \cos \frac{n \pi}{p} x \, dx \) evaluates to zero because any integral of a complete sine cycle over its period is zero. The integral \( \int_{0}^{p} 1^2 \, dx = p \) is straightforward, covering the whole interval.

Step 4: Determine the Norms

The norm \( \|1\| \) is the square root of the integral \( \int_{0}^{p} 1^2 \, dx = p \), leading to \( \sqrt{p} \). For \( \cos \frac{n \pi}{p} x \), the norm is derived from \( \sqrt{\int_{0}^{p} \cos^2 \frac{n \pi}{p} x \, dx} = \sqrt{\frac{p}{2}} \).

Key Concepts

Orthogonality of Functions

In the context of Fourier series, the orthogonality of functions refers to the property that certain functions, when integrated over a defined interval, result in zero. This essential feature enables Fourier series to represent functions as sums of sines and cosines. When functions are orthogonal, they are independent of each other over that interval. For example, consider the integral \( \int_{0}^{p} \cos \frac{n \pi}{p} x \cos \frac{m \pi}{p} x \, dx \) for \( m eq n \). Through the process of integration by applying appropriate trigonometric identities, it is found to equal zero. This zero result confirms that the cosine functions with different indices are orthogonal over the interval \([0, p]\). This property is central to efficiently breaking down more complex functions into simpler trigonometric components.
Orthogonality simplifies computations in Fourier analysis, as it ensures that each term of a Fourier series contributes independently to the representation of the function. This means that the coefficients in the series can be calculated individually, without interference from the others.

Trigonometric Identities

Trigonometric identities play an instrumental role in solving integrals involving trigonometric functions, especially in Fourier series analysis. One such identity is the product-to-sum formula: \( \cos A \cos B = \frac{1}{2} (\cos(A+B) + \cos(A-B)) \). This identity helps transform the product of cosine functions into a sum, facilitating simpler integration. In the given exercise, this identity is applied to \( \cos \frac{n \pi}{p} x \cos \frac{m \pi}{p} x \), allowing one to separate the original problem into two distinct cases which can be more easily integrated.
Another commonly used identity is \( \cos^2 \theta = \frac{1}{2} (1 + \cos 2\theta) \). This is crucial when dealing with \( \cos^2 \frac{n \pi}{p} x \) as it transforms the function into a constant and another cosine function, simplifying the integration process. These identities are fundamental for reducing the complexity of trigonometric integrals by converting harder problems into sums of easier ones.

Integration Techniques

Various integration techniques are used to solve integrals involving trigonometric functions. These techniques allow students to manage more complex problems through strategic simplification. Here, let's explore some practical approaches:

• **Substitution Technique:** Useful for intricate integrals, substitution involves changing variables to simplify the integral into a basic form more straightforward to compute. However, for our cosine integrals, substitution isn't the main tool.
• **Integration by Parts:** Though not directly applied here, it's a valuable method for functions expressed as products, reducing them to simpler integral forms.
• **Definite Integrals Over Periods:** For trigonometric functions, notably sines and cosines, integrating over their period often results in clear outcomes. When computing \( \int_{0}^{p} \cos \frac{n \pi}{p} x \, dx \), the integration yields zero due to the orthogonality of the function across its full cycle.

Mastering these techniques can reveal the underlying patterns and guide students through problem-solving, especially with trigonometric-based integrals. Integration strategies simplify the process, allowing for computations that are otherwise complex or lengthy.