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Magazine Chapter 6: Problem 4
A function \(f(x)\) is defined by \(f(x)=\pi-x \quad 0
Short Answer
Expert verified
(a) \(f(x) = \sum_{n=0}^{\infty} a_n \cos(nx)\); (b) \(f(x) = \sum_{n=1}^{\infty} b_n \sin(nx)\).\
Step by step solution
01
Understand the Problem
The function given is periodic with period \(2 \pi\) and we are asked to express it in terms of half-range cosine and sine series.
02
Define the half-range functions
For half-range series, define the function over the interval \([0, \pi]\).\(f(x) = \pi - x \) for \(0 < x < \pi\).
03
Form the Cosine Series
For the half-range cosine series (only even terms are considered), we need to find coefficients \(a_n\). The general form of the cosine series is \[ f(x) = \sum_{n=0}^{\infty} a_n \cos(nx).\]
04
Compute Cosine Fourier Coefficients
The cosine coefficients are given by the integral:\[ a_n = \frac{2}{\pi} \int_0^{\pi} (\pi - x) \cos(nx) \,dx. \]Calculate \(a_0\) separately:\[ a_0 = \frac{2}{\pi} \int_0^{\pi} (\pi - x) \,dx = \frac{\pi^2}{2}.\]For \(a_n\), \(n \eq 0\):\[ a_n = \frac{2}{\pi} \left[ \pi \int_0^{\pi} \cos(nx) \,dx - \int_0^{\pi} x \cos(nx) \,dx \right].\]Simplify using integration by parts.
05
Form the Sine Series
For the half-range sine series (only odd terms are considered), we need to find coefficients \(b_n\). The general form of the sine series is \[ f(x) = \sum_{n=1}^{\infty} b_n \sin(nx).\]
06
Compute Sine Fourier Coefficients
The sine coefficients are given by the integral:\[ b_n = \frac{2}{\pi} \int_0^{\pi} (\pi - x) \sin(nx) \,dx. \]Simplify as follows:\[ b_n = \frac{2}{\pi} \left[ \pi \int_0^{\pi} \sin(nx) \,dx - \int_0^{\pi} x \sin(nx) \,dx \right].\]Here, continue by solving the integral using integration by parts.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
cosine series
A cosine series represents a function using only cosine terms. This is particularly useful for even functions, as cosine is an even function. The half-range cosine series takes an original function defined on \[0, \pi\] and expands it periodically.
To form the cosine series for a given function \( f(x) = \pi - x \), you compute the coefficients \( a_n \). The general form of the cosine series is:
\[ f(x) = \sum_{n=0}^{\infty} a_n \cos(nx). \]
The coefficients \( a_n \) are determined by:
\[ a_n = \frac{2}{\pi} \int_0^{\pi} (\pi - x) \cos(nx) \,dx.\]
Key points to remember:
To form the cosine series for a given function \( f(x) = \pi - x \), you compute the coefficients \( a_n \). The general form of the cosine series is:
\[ f(x) = \sum_{n=0}^{\infty} a_n \cos(nx). \]
The coefficients \( a_n \) are determined by:
\[ a_n = \frac{2}{\pi} \int_0^{\pi} (\pi - x) \cos(nx) \,dx.\]
Key points to remember:
- For \( a_0 \), the formula simplifies to: \[ a_0 = \frac{2}{\pi} \int_0^{\pi} (\pi - x) \,dx = \frac{\pi^2}{2}. \]
- For \( a_n \) where \( n \eq 0 \), you need to use integration by parts for terms involving \( x \. \)
sine series
A sine series represents a function using only sine terms. This method is suitable for odd functions because sine is an odd function. For the half-range sine series, we need to compute the coefficients \( b_n \).
The general sine series form is:
\[ f(x) = \sum_{n=1}^{\infty} b_n \sin(nx). \]
The coefficients \( b_n \) are calculated by:
\[ b_n = \frac{2}{\pi} \int_0^{\pi} (\pi - x) \sin(nx) \,dx. \]
We can simplify this to:
\[ b_n = \frac{2}{\pi} \[ \pi \int_0^{\pi} \sin(nx) \,dx - \int_0^{\pi} x \sin(nx) \,dx \]. \]
Just like with the cosine series, the integral involving \( x \sin(nx) \) requires using integration by parts.
The general sine series form is:
\[ f(x) = \sum_{n=1}^{\infty} b_n \sin(nx). \]
The coefficients \( b_n \) are calculated by:
\[ b_n = \frac{2}{\pi} \int_0^{\pi} (\pi - x) \sin(nx) \,dx. \]
We can simplify this to:
\[ b_n = \frac{2}{\pi} \[ \pi \int_0^{\pi} \sin(nx) \,dx - \int_0^{\pi} x \sin(nx) \,dx \]. \]
Just like with the cosine series, the integral involving \( x \sin(nx) \) requires using integration by parts.
integration by parts
Integration by parts is a vital technique used for solving integrals involving the product of functions. Given two functions \( u(x) \) and \( v(x) \), the integration by parts formula is:
\[ \int u \, dv = uv - \int v \, du. \]
This method is especially useful in Fourier analysis. For example, to compute the coefficients \( a_n \) or \( b_n \):
Applying integration by parts for \( a_n \) in the example:
Let \( u = x \) and \( dv = \cos(nx) \,dx \). Then,
\[ du = dx \] and \[ v = \frac{\sin(nx)}{n}. \]
The integral part becomes:
\[ \int_0^{\pi} x \cos(nx) \,dx = x \frac{\sin(nx)}{n} \Bigg|_0^{\pi} - \int_0^{\pi} \frac{\sin(nx)}{n} \,dx. \]
This process allows you to find the needed integral values in the cosine and sine series.
\[ \int u \, dv = uv - \int v \, du. \]
This method is especially useful in Fourier analysis. For example, to compute the coefficients \( a_n \) or \( b_n \):
- Select \( u \) such that its derivative simplifies the integral.
- Choose \( dv \) as the remaining part of the integrand.
Applying integration by parts for \( a_n \) in the example:
Let \( u = x \) and \( dv = \cos(nx) \,dx \). Then,
\[ du = dx \] and \[ v = \frac{\sin(nx)}{n}. \]
The integral part becomes:
\[ \int_0^{\pi} x \cos(nx) \,dx = x \frac{\sin(nx)}{n} \Bigg|_0^{\pi} - \int_0^{\pi} \frac{\sin(nx)}{n} \,dx. \]
This process allows you to find the needed integral values in the cosine and sine series.
