Question

For the following exercises, use this information: The inner product of two functions \(f\) and \(g\) over \([a, b]\) is defined by \(f(x) \cdot g(x)=\langle f, g\rangle=\int_{a}^{b} f \cdot g d x\). Two distinct functions \(f\) and \(g\) are said to be orthogonal if \(\langle f, g\rangle=0\). Show that \(\\{\sin (2 x), \cos (3 x)\\}\) are orthogonal over the interval \([-\pi, \pi]\).

Short answer

Yes, \(\sin(2x)\) and \(\cos(3x)\) are orthogonal over \([-\pi, \pi]\).

Step-by-step solution

Define the Inner Product

The problem states that the inner product of two functions \(f(x)\) and \(g(x)\) is \(\langle f, g \rangle = \int_{a}^{b} f(x) g(x) \, dx\). Here, our functions are \(f(x) = \sin(2x)\) and \(g(x) = \cos(3x)\), with the interval \([-\pi, \pi]\).

Set Up the Integral

We need to find \(\langle \sin(2x), \cos(3x) \rangle = \int_{-\pi}^{\pi} \sin(2x) \cos(3x) \, dx\). This involves integrating the product of these two functions over the given interval.

Simplify the Integral Using Trigonometric Identities

To simplify the integral \(\int_{-\pi}^{\pi} \sin(2x) \cos(3x) \, dx\), use the identity \(\sin(A) \cos(B) = \frac{1}{2}[\sin(A+B) + \sin(A-B)]\). Thus, \(\sin(2x) \cos(3x) = \frac{1}{2}[\sin(5x) + \sin(-x)]\).

Integrate Each Term

Now integrate \(\int_{-\pi}^{\pi} \frac{1}{2}[\sin(5x) + \sin(-x)] \, dx\). This becomes 1. \(\frac{1}{2} \int_{-\pi}^{\pi} \sin(5x) \, dx\) 2. \(\frac{1}{2} \int_{-\pi}^{\pi} \sin(-x) \, dx\).

Evaluate the Integrals

Both integrals \(\int_{-\pi}^{\pi} \sin(5x) \, dx\) and \(\int_{-\pi}^{\pi} \sin(-x) \, dx\) evaluate to zero because the sine function is odd, and integrating an odd function over a symmetric interval around zero results in zero.

Determine Orthogonality

Since \(\langle \sin(2x), \cos(3x) \rangle = 0\), \(\sin(2x)\) and \(\cos(3x)\) are orthogonal over \([-\pi, \pi]\).

Key concepts

Inner Product of Functions

The inner product is a powerful tool in mathematical analysis and has specific applications in fields like functional analysis and signal processing. When we talk about the inner product of two functions \(f(x)\) and \(g(x)\), we essentially measure the degree to which these two functions are aligned over a certain interval. This is somewhat akin to finding the dot product of vectors.

Given the functions \(f(x) = \sin(2x)\) and \(g(x) = \cos(3x)\), their inner product over the interval \([-\pi, \pi]\) is calculated as \(\langle f, g \rangle = \int_{-\pi}^{\pi} f(x) g(x) \, dx\). If the inner product equals zero, it indicates that these functions are orthogonal over the specified interval.

Here’s the breakdown:

• Calculate the integral \(\int_{-\pi}^{\pi} \sin(2x) \cos(3x) \, dx\) to find the inner product.
• Use trigonometric identities to simplify the function into a more integrable form.
• Evaluate the integral to determine orthogonality.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables. These identities are essential for simplifying expressions within calculus, especially when dealing with products or sums of trigonometric functions.

When simplifying the integral \(\int_{-\pi}^{\pi} \sin(2x) \cos(3x) \, dx\), we use the identity:
\[ \sin(A) \cos(B) = \frac{1}{2} [\sin(A+B) + \sin(A-B)] \]
This transforms the product \(\sin(2x) \cos(3x)\) into
\[ \frac{1}{2} [\sin(5x) + \sin(-x)] \]
With these identities:

• The expression is divided into separate sine functions.
• This simplifies the calculation and makes the term integrable over the given interval.

Interval Integration

Integration over an interval is a method to compute the net area under a curve represented by a function \(f(x)\), between two points \(a\) and \(b\). In this problem, the interval is \([-\pi, \pi]\).

For the functions involved, we have two integrals to evaluate:

• \(\int_{-\pi}^{\pi} \sin(5x) \, dx\)
• \(\int_{-\pi}^{\pi} \sin(-x) \, dx\)

Both of these integrals evaluate to zero. This is because the sine function is an odd function, meaning it is symmetric about the origin. Therefore, an integral of an odd function over a symmetric interval around zero will always result in zero.

To evaluate such integrals:

• Recognize that the sine and cosine functions have symmetrical properties.
• Exploit these properties to simplify and solve integrals efficiently.
• Confirm that the orthogonality condition is satisfied when the integral evaluates to zero.

Orthogonality indeed tells us much about the independence and mutual exclusivity of the functions over the interval.