Chapter 9: Problem 2

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

Expert verified

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.

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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

Step by step solution

Recall the definition of scalar product

Write down the given functions

Set up the integral for the scalar product

Solve the integral

Evaluate the integral

Simplify the expression

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