Problem 9 - Advanced Engineering Mathematics

Question

Show that the given set of functions is orthogonal on the indicated interval. Find the norm of each function in the set. $$ \\{\sin n x\\}, n=1,2,3, \ldots ; \quad[0, \pi] $$

Step-by-Step Solution

Answer

The set \( \{\sin nx\} \) is orthogonal on \([0, \pi]\); each function's norm is \( \sqrt{\frac{\pi}{2}} \).

Step 1: Understanding Orthogonality

The given set of functions \( \{\sin nx\} \) is considered orthogonal if the integral of the product of two different functions over the interval \([0, \pi]\) is zero. Essentially, this means that for \( n eq m \), \( \int_0^{\pi} \sin nx \sin mx \,dx = 0 \).

Step 2: Orthogonality Verification

To verify orthogonality, we calculate \( \int_0^{\pi} \sin nx \sin mx \,dx \). Using the identity \( \sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)] \), this integral becomes \( \frac{1}{2}\int_0^{\pi} [\cos((n-m)x) - \cos((n+m)x)] \, dx \). For \( n eq m \), both terms integrate to zero over \([0, \pi]\), confirming that the functions are orthogonal.

Step 3: Norm Calculation of Functions

The norm of a function \( f(x) \) over an interval \([a, b]\) is defined as \( \|f\| = \sqrt{\int_a^b [f(x)]^2 \, dx} \). For \( \sin nx \), we evaluate \( \int_0^{\pi} \sin^2 nx \, dx \). Using the identity \( \sin^2 x = \frac{1}{2}(1 - \cos 2x) \), this simplifies to \( \frac{1}{2}\int_0^{\pi} (1 - \cos 2nx) \, dx \).

Step 4: Integration and Norm Finalization

The integration \( \int_0^{\pi} 1 \, dx = \pi \), and \( \int_0^{\pi} \cos 2nx \, dx = 0 \) since \( 2n \) is an integer, leading to full sine waves over the interval, yielding zero. Therefore, \( \int_0^{\pi} \sin^2 nx \, dx = \frac{\pi}{2} \), providing that the norm is \( \|\sin nx\| = \sqrt{\frac{\pi}{2}} \).

Key Concepts

Understanding Sine Functions

Sine functions are a fundamental block in trigonometry. These functions describe oscillating movements, like waves. When visualizing a sine function, imagine a smooth wave extending infinitely. The standard sine function is written as \( \sin x \), where \( x \) represents the angle in radians. This function has a few key characteristics:

• Amplitude: The maximum height of the wave crest. For \( \sin x \), this is 1.
• Period: The length of one complete wave cycle, for \( \sin x \), this period is \( 2\pi \).
• Reflection Symmetry: The sine function is symmetrical around the origin.

In the context of this exercise, the functions \( \{\sin nx\} \) are scaled and periodically adjusted versions of the standard sine function. Here, \( n \) multiplies the angle, increasing the frequency of oscillations. Consequently, the period of \( \sin nx \) changes to \( \frac{2\pi}{n} \). This property is crucial as it allows such functions to become orthogonal under certain conditions.

The Role of Integration in Orthogonality

Integration is a core concept in calculus and is crucial for verifying orthogonality in functions. At its simplest, integration can be viewed as the process of finding the area under a curve within a given interval. In mathematics, it's often written as \( \int \) followed by the function and the limits of integration.
To determine if the sine functions are orthogonal, we calculate integrals over the interval \([0, \pi]\). For two different sine functions, say \( \sin nx \) and \( \sin mx \), they are orthogonal if their product's integral is zero:

• \( \int_0^{\pi} \sin nx \sin mx \,dx = 0 \)

This step involves applying trigonometric identities to simplify the product, such as using the identity:

• \( \sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)] \)

Thus, through integration and employing these identities, one verifies that sine functions are orthogonal when their coefficients differ, i.e., \( n eq m \).

Exploring Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for every value of the included variables. These identities are vital tools in simplifying expressions and solving trigonometric equations, especially when dealing with orthogonal functions.
In this particular exercise, one useful identity is the product-to-sum formula:

• \( \sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)] \)

This identity allows us to transform a product of sine functions into a difference of cosine functions. It is instrumental when verifying orthogonality and calculating the norm of functions over an interval. By turning the product into a sum, it becomes easier to integrate each term separately. For example, for the integral \( \int_0^{\pi} \sin nx \sin mx \,dx \), applying this identity results in integrals of cosine functions, both of which simplify considerably under the limits of integration. Knowing these identities is essential for tackling complex trigonometric problems and shows the interconnectedness of these mathematical concepts.