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 \(\langle f, g\rangle=0\). Evaluate \(\int_{-\pi}^{\pi} \sin (m x) \cos (n x) d x\)

Short answer

The integral evaluates to 0, proving the functions are orthogonal.

Step-by-step solution

Set Up the Inner Product Integral

Given the functions \(f(x) = \sin(mx)\) and \(g(x) = \cos(nx)\), we need to compute the integral of their product from \(-\pi\) to \(\pi\):\[\langle \sin(mx), \cos(nx) \rangle = \int_{-\pi}^{\pi} \sin(mx) \cos(nx) \, dx\]

Apply the Product-to-Sum Formula

To simplify the integrand \(\sin(mx)\cos(nx)\), use the identity \(\sin(A)\cos(B) = \frac{1}{2} [\sin(A+B) + \sin(A-B)]\). Substitute \(A = mx\) and \(B = nx\):\[\int_{-\pi}^{\pi} \frac{1}{2} [\sin((m+n)x) + \sin((m-n)x)] \, dx\]

Split and Simplify the Integral

Split the integral into two separate integrals:\[\frac{1}{2} \left( \int_{-\pi}^{\pi} \sin((m+n)x) \, dx + \int_{-\pi}^{\pi} \sin((m-n)x) \, dx \right)\]Both integrals involve sine functions integrated over full periods \([-\pi, \pi]\).

Evaluate Each Integral

The integral of \(\sin(kx)\) over one full period \([-\pi, \pi]\) is zero:\[\int_{-\pi}^{\pi} \sin(kx) \, dx = 0, \, \text{for any integer } k\]Therefore:\[\int_{-\pi}^{\pi} \sin((m+n)x) \, dx = 0\]\[\int_{-\pi}^{\pi} \sin((m-n)x) \, dx = 0\]

Conclude the Result

Substitute back the evaluated integrals into the expression from Step 3:\[\frac{1}{2} (0 + 0) = 0\]Thus, the inner product \(\langle \sin(mx), \cos(nx) \rangle = 0\), proving that the functions are orthogonal.

Key concepts

Inner Product

Imagine you have two functions and you want to measure how "compatible" they are in a mathematical sense. This is where the inner product comes in. It's a way to multiply two functions to see how much they overlap over a certain range.

The inner product of two functions, say \(f(x)\) and \(g(x)\), over an interval \([a, b]\) is written as \( \langle f, g \rangle \). This is defined using a definite integral:

• \( \langle f, g \rangle = \int_{a}^{b} f(x) \cdot g(x) \, dx \)

This formula gathers the "alignment" of the two functions throughout the interval from \(a\) to \(b\). If the result is 0, the functions are orthogonal, meaning they are completely independent of each other over that interval.

Orthogonal functions are like two different TV channels set to different frequencies—they don't interfere with each other.

Product-to-Sum Formula

Breaking down complex expressions is sometimes easier by using identities from trigonometry, like the product-to-sum formula. This helps for simplifying the product of sine and cosine functions, making integrals much easier to solve.

The product-to-sum formula states:

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

In practice, if you have a product like \( \sin(mx) \cos(nx) \), you can substitute \(A = mx\) and \(B = nx\). This converts the expression into a sum, making it simpler to integrate.

This is especially useful because the resulting sine terms often integrate easily over certain intervals, dramatically simplifying your calculations.

Definite Integration

Integration is a core concept in calculus, used to determine the accumulated area under a curve across a given interval. When we speak about definite integrals, we mean calculating this over specific bounds from \(a\) to \(b\).

For functions like sine, the integral can often result in zero over symmetric intervals, like \([-\pi, \pi]\), because the areas above and below the x-axis cancel each other out.

Definite integration is especially powerful because it calculates real, finite values that describe quantities such as areas and can certify conditions like orthogonality between functions. It turns abstract functions into tangible numbers, allowing us to conclude relationships like independence and orthogonality through careful evaluation.