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Magazine Chapter 12: Problem 10
Show that the given set of functions is orthogonal on the indicated interval. Find the norm of each function in the set. $$ \left\\{\sin \frac{n \pi}{p} x\right\\}, n=1,2,3, \ldots ; \quad[0, p] $$
Short Answer
Expert verified
The functions are orthogonal on \([0, p]\), and the norm of each function is \(\sqrt{\frac{p}{2}}\).
Step by step solution
01
Understanding Orthogonality
Two functions, \(f(x)\) and \(g(x)\), are orthogonal on an interval \([a, b]\) if their inner product is zero. The inner product of two functions is given by \(\int_a^b f(x)g(x) \, dx\). Our task is to show that \(\sin\left(\frac{n\pi}{p}x\right)\) and \(\sin\left(\frac{m\pi}{p}x\right)\) are orthogonal for \(n eq m\) on the interval \([0, p]\).
02
Set up the Inner Product
For two different functions \(\sin\left(\frac{n\pi}{p}x\right)\) and \(\sin\left(\frac{m\pi}{p}x\right)\), the inner product is \(\int_0^p \sin\left(\frac{n\pi}{p}x\right) \sin\left(\frac{m\pi}{p}x\right) \, dx\).
03
Utilize Trigonometric Identity
Use the identity \(\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]\) to rewrite the integral: \[ \int_0^p \sin\left(\frac{n\pi}{p}x\right) \sin\left(\frac{m\pi}{p}x\right) \, dx = \frac{1}{2} \int_0^p [\cos\left(\frac{(n-m)\pi}{p}x\right) - \cos\left(\frac{(n+m)\pi}{p}x\right)] \, dx \].
04
Solve the Integral
Solve the two integrals separately: \[ \int_0^p \cos\left(\frac{(n-m)\pi}{p}x\right) \, dx = 0 \] and \[ \int_0^p \cos\left(\frac{(n+m)\pi}{p}x\right) \, dx = 0 \] when \(n eq m\), as they are integrals of complete periods of the cosine function, whose net area over a period is zero.
05
Conclude Orthogonality
Since both integrals equal zero when \(n eq m\), the inner product is zero, thereby proving the functions are orthogonal on \([0, p]\).
06
Find the Norm of Each Function
The norm of a function \(f(x)\) is given by \( \|f\| = \sqrt{\int_0^p f(x)^2 \, dx} \). Here, \(f(x) = \sin\left(\frac{n\pi}{p}x\right)\).
07
Calculate the Norm
Using the identity \(\sin^2 A = \frac{1 - \cos 2A}{2}\), we find \[ \int_0^p \sin^2\left(\frac{n\pi}{p}x\right) \, dx = \frac{1}{2} \int_0^p (1 - \cos\left(\frac{2n\pi}{p}x\right)) \, dx = \frac{p}{2} \]. Thus, the norm is \(\|\sin\left(\frac{n\pi}{p}x\right)\| = \sqrt{\frac{p}{2}}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trigonometric Identities
Trigonometric identities are useful tools that help simplify the calculations and transformations of trigonometric functions. In this exercise, we make use of the identity \( \sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)] \) to aid in showing the orthogonality of functions.
This identity is particularly useful in integration, where the product of sine functions can be expressed in terms of cosine functions.
This identity is particularly useful in integration, where the product of sine functions can be expressed in terms of cosine functions.
- Instead of directly integrating a product of sines, which can be complicated, we convert it to a combination of cosines.
- The integral of a cosine function over its complete period equals zero, which significantly simplifies the problem.
Inner Product
The inner product of two functions is a key concept in understanding orthogonality. It is similar to the dot product in vector spaces but extends to functions.
The inner product between two functions, say \( f(x) \) and \( g(x) \), over an interval \([a, b]\) is computed as \( \int_a^b f(x)g(x) \, dx \).
The inner product between two functions, say \( f(x) \) and \( g(x) \), over an interval \([a, b]\) is computed as \( \int_a^b f(x)g(x) \, dx \).
- This gives a single scalar value representing the 'overlap' or 'interaction' between the two functions over the interval.
- When this inner product is zero, it indicates that the functions are orthogonal, meaning they do not influence each other over the stated interval.
Function Norms
Function norms are critical to understanding the 'size' or magnitude of functions. In functional spaces, just as in vector spaces, norms provide a measure of distance or length.
The norm of a function \( f(x) \) over the interval \([a, b]\) is given by the square root of the inner product of a function with itself: \( \|f\| = \sqrt{\int_a^b f(x)^2 \, dx} \).
The norm of a function \( f(x) \) over the interval \([a, b]\) is given by the square root of the inner product of a function with itself: \( \|f\| = \sqrt{\int_a^b f(x)^2 \, dx} \).
- This measure allows us to understand how large or small a function is in the interval \([a, b]\).
- In the exercise, we calculated the norm of \( \sin\left(\frac{n\pi}{p}x\right) \) to be \( \sqrt{\frac{p}{2}} \), indicating the uniform size of all these sine functions over the given interval.
Integration Techniques
Integration techniques play a pivotal role in solving problems involving function spaces, like proving orthogonality.
In this exercise, integration is used to determine both the inner product and the norm of the given functions.
In this exercise, integration is used to determine both the inner product and the norm of the given functions.
- The use of a trigonometric identity transforms a complex sine product into simpler cosine terms that are easier to integrate.
- Solving these integrals requires understanding that integration over full periods of sine and cosine functions results in zero net area, simplifying computations.
