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Magazine Chapter 12: Problem 15
Determine the half-range cosine series expansion of the function \(f(t)=2 t-1\),
valid for \(0
Short Answer
Expert verified
The half-range cosine series of \( f(t) = 2t - 1 \) is derived, consisting solely of cosine terms, with coefficients determined by integration.
Step by step solution
01
Identify the Problem
We need to find the half-range cosine series expansion for the function \( f(t) = 2t - 1 \) over the interval \( 0 < t < 1 \) and sketch both \( f(t) \) and the periodic extension over \( -2 < t < 2 \).
02
Understanding half-range expansions
Half-range cosine series is used to expand functions over a half-interval \((0, L)\) as a cosine series, assuming even periodic extension. For such a function \( f(t) \), we can express it as a series involving cosine terms only.
03
Calculate Fourier Coefficients
For the Fourier cosine series, calculate coefficients using: \[ a_0 = \frac{2}{L} \int_0^L f(t) \, dt, \quad a_n = \frac{2}{L} \int_0^L f(t) \cos\left(\frac{n\pi t}{L}\right) \, dt \] Here, \( L = 1 \), and \( f(t) = 2t - 1 \).
04
Calculate the Coefficient \( a_0 \)
For \( a_0 \): \[ a_0 = 2 \int_0^1 (2t - 1) \, dt = 2 \left[ t^2 - t \right]_0^1 = 2(1^2 - 1) = 0 \] Thus, \( a_0 = 0 \).
05
Calculate the Coefficient \( a_n \)
For \( a_n \): \[ a_n = 2 \int_0^1 (2t - 1) \cos(n\pi t) \, dt = 2 \left[ \frac{2}{n^2\pi^2} - \frac{1}{n\pi} \right](-1)^n \] After computation, \( a_n = \frac{4(-1)^n}{n^2\pi^2} - \frac{2(-1)^n}{n\pi} \).
06
Construct the Cosine Series
The half-range cosine series expansion becomes: \[ f(t) \approx \sum_{n=1}^{\infty} \left( \frac{4(-1)^n}{n^2\pi^2} - \frac{2(-1)^n}{n\pi} \right) \cos(n\pi t) \] This series represents the even extension of \( f(t) \).
07
Sketch the Graphs
To sketch the graphs, plot \( f(t) = 2t - 1 \) over \( 0 < t < 1 \) as a straight line, then mirror it across the y-axis for even extension. The periodic part should repeat this pattern for \( -2 < t < 2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Half-Range Expansion
When working with functions, sometimes we only focus on an interval, such as from 0 to a certain point like 1. This is where a half-range expansion comes into play. It helps us represent a function across this half-interval by extending it in a specific way.
For instance, with the function given as \( f(t) = 2t - 1 \) on the interval \( 0 < t < 1 \), we can perform a half-range expansion. The idea is to create a series that mirrors the function on this interval using only cosine terms, making it a beautifully even, periodic function across a wider range, such as \( -2 < t < 2 \).
To achieve the half-range expansion, ensure the function is even, meaning symmetrical about the y-axis, before proceeding to expand it as a series of cosine functions.
For instance, with the function given as \( f(t) = 2t - 1 \) on the interval \( 0 < t < 1 \), we can perform a half-range expansion. The idea is to create a series that mirrors the function on this interval using only cosine terms, making it a beautifully even, periodic function across a wider range, such as \( -2 < t < 2 \).
To achieve the half-range expansion, ensure the function is even, meaning symmetrical about the y-axis, before proceeding to expand it as a series of cosine functions.
Cosine Series
The cosine series is a special type of series used in Fourier analysis to describe a function using only cosine terms. It's particularly handy for working with functions that need an even extension.
For our function \( f(t) = 2t - 1 \), we're creating a cosine series over the interval \( 0 < t < 1 \). Only cosine functions are used, because cosine is an even function, which matches well with even extensions needed for half-range expansions.
The cosine series for our example takes the form:
For our function \( f(t) = 2t - 1 \), we're creating a cosine series over the interval \( 0 < t < 1 \). Only cosine functions are used, because cosine is an even function, which matches well with even extensions needed for half-range expansions.
The cosine series for our example takes the form:
- Start with the basic formula involving the cosine functions.
- Calculate the necessary coefficients that multiply each cosine term.
Fourier Coefficients
Fourier coefficients are the secret ingredients in building a Fourier series, whether it's a general Fourier series, a sine series, or a cosine series. They determine the weight of each term in the series.
For the cosine series of \( f(t) = 2t - 1 \), the coefficients are found from integrals involving the original function and cosine terms. Specifically:
For the cosine series of \( f(t) = 2t - 1 \), the coefficients are found from integrals involving the original function and cosine terms. Specifically:
- The coefficient \( a_0 \) represents the average term in the series and is calculated by \( a_0 = \frac{2}{L} \int_0^L f(t) \, dt \). For the given function, this turns out to be zero.
- The coefficients \( a_n \) provide the weights for the cosine terms at different frequencies, calculated by \( a_n = \frac{2}{L} \int_0^L f(t) \cos(\frac{n\pi t}{L}) \, dt \). For our function, these coefficients simplify into terms involving powers of \( \pi \).
