Question

Determine the scalar product of the functions \(f(x)=\sin (2 n \pi x / L)\) and \(g(x)=\cos (2 p \pi x / L)\) on the domain \(0 \leq x \leq L\), where \(n \neq p\) are positive integers greater than zero.

Short answer

Answer: Yes, the functions are orthogonal over the given domain.

Step-by-step solution

Recall the definition of scalar product

The scalar product (also known as the dot product or inner product) of two functions f(x) and g(x) is defined as the integral of the product of the functions over a given domain. In our case, the domain is 0 ≤ x ≤ L.

Write down the given functions

We are given the functions: 1. \(f(x) = \sin(\frac{2n\pi x}{L})\) 2. \(g(x) = \cos(\frac{2p\pi x}{L})\) where n ≠ p are positive integers greater than zero.

Set up the integral for the scalar product

To find the scalar product, we will calculate the integral of the product of f(x) and g(x) over the domain 0 ≤ x ≤ L: \(\int_0^L f(x)g(x) dx = \int_0^L \sin(\frac{2n\pi x}{L})\cos(\frac{2p\pi x}{L}) dx\)

Solve the integral

We will first use the product-to-sum identity: \(\sin(A) \cos(B) = \frac{1}{2}[\sin(A-B) - \sin(A+B)]\) where \(A = \frac{2n\pi x}{L}\) and \(B = \frac{2p\pi x}{L}\) to simplify the integrand: \(\int_0^L \sin(\frac{2n\pi x}{L})\cos(\frac{2p\pi x}{L}) dx = \frac{1}{2}\int_0^L [\sin(\frac{2(n-p)\pi x}{L}) - \sin(\frac{2(n + p)\pi x}{L})] dx\) Now, we will integrate each term of the expression with respect to x: \(\frac{1}{2}\int_0^L [\sin(\frac{2(n-p)\pi x}{L}) - \sin(\frac{2(n + p)\pi x}{L})] dx = \frac{1}{2}[ \frac{L}{2(n - p)\pi}(-\cos(\frac{2(n-p)\pi x}{L}))\Big|_0^L - \frac{L}{2(n + p)\pi}(-\cos(\frac{2(n+p)\pi x}{L}))\Big|_0^L]\)

Evaluate the integral

Now we will evaluate the integral at x = L and x = 0: \(\frac{1}{2}[ \frac{L}{2(n - p)\pi}(-\cos(\frac{2(n-p)\pi x}{L}))\Big|_0^L - \frac{L}{2(n + p)\pi}(-\cos(\frac{2(n+p)\pi x}{L}))\Big|_0^L] = \frac{L}{4\pi}[\frac{-\cos(2(n - p)\pi) + \cos(0)}{n - p} - \frac{-\cos(2(n + p)\pi) + \cos(0)}{n + p}]\)

Simplify the expression

Since the cosine function is even, \(\cos(2(n - p)\pi) = \cos(2(n + p)\pi) = 1\), and as \(\cos(0) = 1\), we can simplify our expression as: \(\frac{L}{4\pi}[\frac{-1 + 1}{n - p} - \frac{-1 + 1}{n + p}] = 0\) Since the scalar product is 0, which means the two functions are orthogonal over the given domain, and our task here is done.

Key concepts

Scalar Product

The scalar product, also known as the inner product or dot product, is a crucial concept when dealing with functions, especially in the field of functional analysis. It is a measure that helps determine how two functions relate to each other over a specific interval or domain. The scalar product of two functions, say \( f(x) \) and \( g(x) \), on a given domain, is calculated by integrating the product of these functions over that domain:

• Set up the integral: \( \int_0^L f(x) g(x) \, dx \)
• This process essentially combines the functions and "sums up" their joint behavior across the interval.

For example, finding the scalar product of the functions \( \sin(\frac{2n\pi x}{L}) \) and \( \cos(\frac{2p\pi x}{L}) \) over \( 0 \leq x \leq L \) helps to determine whether they have perpendicular tendencies in the function space. These particular trigonometric functions can exhibit orthogonality, meaning their scalar product could equal zero on this interval.
This orthogonality implies that the two functions do not influence each other in terms of magnitude over the given domain, and it simplifies a lot of calculations in physics and engineering.

Trigonometric Integrals

Integrating trigonometric functions can sometimes be tricky because of their oscillating nature. Thankfully, there are identities and methods that can simplify these integrals.

• To integrate products of sine and cosine functions, such as \( \sin(ax) \cos(bx) \), direct integration might not be straightforward.
• In our example, the function to integrate is \( \sin(\frac{2n\pi x}{L}) \times \cos(\frac{2p\pi x}{L}) \).

Instead of directly integrating this product, we can use trigonometric identities to turn it into a form that is easier to manage. By properly applying these identities, we can separate complex products into sums or differences that are far simpler to integrate. This tactic can significantly reduce the effort required for solving these integrals and helps simplify the process into a series of basic integral calculations.

Product-to-Sum Identities

The product-to-sum identities are incredibly helpful when simplifying expressions involving products of sine and cosine functions. By converting these products into sums, they make difficult integration problems much more manageable.

• For the functions \( \sin A \) and \( \cos B \), the identity is: \( \sin A \cos B = \frac{1}{2}[\sin(A - B) - \sin(A + B)] \).
• This identity allows for the separation of a product of two trigonometric functions into a difference of two simpler sine functions.

In our exercise, the application of this identity helped convert \( \int_0^L \sin(\frac{2n\pi x}{L})\cos(\frac{2p\pi x}{L}) \, dx \) into a more straightforward form that could be integrated directly.
The resulting simpler integrals could then be handled with basic integration techniques. These transformations not only reduce calculation time but also provide clearer insights into the behavior of complex trigonometric functions within the given domain. By highlighting key algebraic manipulations, product-to-sum identities turn daunting trigonometric integrals into something far more approachable and understandable.