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Magazine Chapter 12: Problem 31
Find the half-range cosine and sine expansions of the given function.
$$
f(x)=\left\\{\begin{array}{ll}
x, & 0
Short Answer
Expert verified
Calculate the Fourier coefficients for both series using integrals over function sections.
Step by step solution
01
Identify the Interval for Half-Range Expansion
First, we observe that the function is defined on the interval \(0 < x < 2\). For half-range expansions, we use either the cosine or sine expansion, which generally involve even or odd extensions of the function, respectively, across a determined interval, such as \([0, L]\) or \([-L, L]\). Here, the maximum value is 2, so we consider \(L = 2\).
02
Setting Up the Half-Range Cosine Series
The half-range cosine series uses even extension, implying \(f(x) = f(-x)\). Each coefficient \(a_n\) is given by \(a_n = \frac{2}{L} \int_0^L f(x) \cos\left(\frac{n\pi x}{L}\right) \, dx\). For our function and \(L=2\), this leads to the integrals over two sections, from 0 to 1, where \(f(x)=x\), and from 1 to 2, where \(f(x)=1\).
03
Calculate \(a_0\) for Cosine Expansion
The zeroth coefficient \(a_0\) is given by \(a_0 = \frac{2}{L} \int_0^L f(x) \, dx\). This becomes \(a_0 = \frac{1}{2} \int_0^1 x \, dx + \frac{1}{2} \int_1^2 1 \, dx\). Calculate these integrals separately: \[\frac{1}{2} \left(\frac{x^2}{2} \bigg|_0^1\right) + \frac{1}{2} [x]_1^2\], resulting in \(a_0 = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}\).
04
Calculate \(a_n\) for Cosine Expansion
For \(n \geq 1\), \(a_n = \frac{2}{L} \left( \int_0^1 x \cos\left(\frac{n\pi x}{2}\right) \, dx + \int_1^2 \cos\left(\frac{n\pi x}{2}\right) \, dx \right)\). Solve each integral using integration by parts and computing specifically from these bounds.
05
Setting Up the Half-Range Sine Series
The half-range sine series uses odd extension, implying \(f(x) = -f(-x)\). Each coefficient \(b_n\) is calculated with \(b_n = \frac{2}{L} \int_0^L f(x) \sin\left(\frac{n\pi x}{L}\right) \, dx\). With \(L=2\), compute \(b_n = \int_0^1 x \sin\left(\frac{n\pi x}{2}\right) \, dx + \int_1^2 \sin\left(\frac{n\pi x}{2}\right) \, dx\).
06
Calculate \(b_n\) for Sine Expansion
Use integration by parts for these integrals. Compute \(b_n = \frac{2}{2} \left( \int_0^1 x \sin\left(\frac{n\pi x}{2}\right) \, dx + \int_1^2 \sin\left(\frac{n\pi x}{2}\right) \, dx \right)\). Similar to \(a_n\), calculate each part and compile into series form.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Half-Range Expansions
Half-range expansions are a technique used to simplify Fourier series by focusing on a specific half of the interval, either the positive or negative half. This is especially helpful for functions that display symmetry across an interval. When we deal with a given function defined on an interval from 0 to a particular value \(L\), rather than from \(-L\) to \(L\), we form either a cosine or sine series as a half-range expansion.
- Cosine Series: This involves an even extension where the function mirrors itself evenly across the interval.
- Sine Series: This is used for odd extensions and reflects the function in an odd manner across the interval.
Cosine Series
A cosine series is leveraged when extending a function in an even manner, leading to a symmetry about the vertical axis. This is particularly useful for periodic functions observed in physics and engineering.
- Even Extension: In the cosine series, the assumption \(f(x) = f(-x)\) helps in simplifying the Fourier expansion by focusing solely on cosine terms.
- Coefficient Calculations: For a function defined over \([0, L]\), the coefficients \(a_n\) are calculated as: \[a_n = \frac{2}{L} \int_0^L f(x) \cos\left(\frac{n\pi x}{L}\right) \, dx\]
Sine Series
The sine series involves extending a function using an odd pattern, resulting in symmetry about the origin. Such extensions are crucial for odd periodic functions.
- Odd Extension: This series applies when \(f(x) = -f(-x)\), capturing symmetry around zero and simplifying representations solely using sine terms.
- Coefficient Formulation: Similar to the cosine series, the sine series uses: \[b_n = \frac{2}{L} \int_0^L f(x) \sin\left(\frac{n\pi x}{L}\right) \, dx\]
Integration by Parts
Integration by parts is a fundamental technique used to integrate products of functions. It's especially relevant when solving Fourier series problems, as shown in the computation of coefficients for half-range expansions.
This method is based on the product rule for differentiation and is represented as:\[\int u \, dv = uv - \int v \, du\]
This method is based on the product rule for differentiation and is represented as:\[\int u \, dv = uv - \int v \, du\]
- Choosing Functions: Selecting \(u\) and \(dv\) from the given integral is crucial. Typically, you would let \(u\) be a polynomial and \(dv\) as trigonometric functions, like sine or cosine.
- Application in Series: Integration by parts allows simplification of more complex integrals in cosine or sine series, leading to manageable computations for determining series coefficients.
