Question

You are given \(f(x)\) on an interval, say \(0< x< b\). Sketch several periods of the even function \(f_{c}\) of period \(2 b,\) the odd function \(f_{s}\) of period \(2 b,\) and the function \(f_{p}\) of period \(b\), each of which equals \(f(x)\) on \(0< x< b\). Expand each of the three functions in an appropriate Fourier series. $$f(x)=x^{2}, \quad 0< x< 1$$

Short answer

Sketch even, odd, and periodic extensions, then use Fourier series formulas with calculated coefficients.

Step-by-step solution

- Define the given function

We are given the function: \( f(x) = x^2 \) on the interval \( 0 < x < 1 \).

- Sketch the even function \( f_c \)

The even extension of \( f(x) \) is done by mirroring \( f(x) = x^2 \) over the y-axis and then periodically repeating it with period \( 2b = 2 \). Therefore, on \( -1 < x < 0 \), \( f_c(x) = ( - x )^2 = x^2 \). The even extension over one period can be written as: \( f_c(x) = x^2, \text{ for } -1 < x < 1 \). Sketch the period from \( -1 \) to \( 1 \) and repeat.

- Sketch the odd function \( f_s \)

The odd extension of \( f(x) \) is done by negating \( f(x) \) over the y-axis and then periodically repeating it with period \( 2b = 2 \). Therefore, on \( -1 < x < 0 \), \( f_s(x) = -( - x )^2 = - x^2 \). The odd extension over one period can be written as: \( f_s(x) = \begin{cases} x^2, & 0 < x < 1 \-x^2, & -1 < x < 0 \end{cases} \). Sketch the period from \( -1 \) to \( 1 \) and repeat.

- Sketch the periodic function \( f_p \)

The periodic extension of \( f(x) = x^2 \) is by periodically repeating it with period \( b = 1 \). For \( f_p(x) \), the function is simply \( x^2 \) repeated every 1 unit. Thus, \( f_p(x) = x^2 \) for \( 0 < x < 1 \), then repeat.

- Find the Fourier series for \( f_c \)

The Fourier series for an even function of period \( 2 \) is as follows: \[ f_c(x) = a_0 + \sum_{n=1}^{\infty} a_n \cos\left( \frac{n\pi x}{b} \right) \]. Calculate the coefficients: \[ a_0 = \frac{1}{b} \int_{0}^{b} f(x)\, dx = \frac{1}{1} \int_{0}^{1} x^2\, dx = \frac{1}{3} \]. \[ a_n = \frac{2}{b} \int_{0}^{b} f(x) \cos\left( \frac{n\pi x}{b} \right) dx = 2 \int_{0}^{1} x^2 \cos(n \pi x) dx \]. Evaluate the integral using integration by parts.

- Find the Fourier series for \( f_s \)

The Fourier series for an odd function of period \( 2 \) is as follows: \[ f_s(x) = \sum_{n=1}^{\infty} b_n \sin\left( \frac{n\pi x}{b} \right) \]. Calculate the coefficients: \[ b_n = \frac{2}{b} \int_{0}^{b} f(x) \sin \left( \frac{n\pi x}{b} \right) dx = 2 \int_{0}^{1} x^2 \sin(n \pi x) dx \]. Evaluate the integral using integration by parts.

- Find the Fourier series for \( f_p \)

The Fourier series for \( f(x) \) with period \( b = 1 \) is: \[ f_p(x) = a_0 + \sum_{n=1}^{\infty} a_n \cos(2n\pi x) + b_n \sin(2n\pi x) \]. Calculate the coefficients: \[ a_0 = \int_{0}^{1} f(x) dx = \int_{0}^{1} x^2 dx = \frac{1}{3} \]. \[ a_n = \int_{0}^{1} f(x) \cos(2n\pi x) dx \]. \[ b_n = \int_{0}^{1} f(x) \sin(2n\pi x) dx \]. Evaluate the integrals using integration by parts.

Key concepts

Even Functions

A function is called even if it satisfies the condition: \( f(x) = f(-x) \) for all \( x \) in its domain. This means that the graph of an even function is symmetric with respect to the y-axis. In our example, we extend the given function \( f(x) = x^2 \). Since \( ( - x )^2 = x^2 \), the extension remains the same on both sides of the y-axis. The periodic extension therefore mirrors around \( x = 0 \) and repeats every 2.

Odd Functions

A function is considered odd if it satisfies the condition \( f(-x) = - f(x) \) for all \( x \) in its domain. This implies the graph of an odd function is symmetric with respect to the origin. In the case of \( f(x) = x^2 \), the odd extension is created by reflecting and then negating the function on the negative side, resulting in \( -x^2 \). This function then repeats with a period of 2.

Periodic Functions

A periodic function repeats its values in regular intervals or periods. For a function \( f(x) \), if there exists a period \( T > 0 \) such that \( f(x + T) = f(x) \) for all \( x \), then \( f(x) \) is periodic with period \( T \). In this exercise, the given function \( f(x) = x^2 \) is periodically extended to have a period of 1. This means the function \( f(x) \) repeats itself every unit interval.

Fourier Coefficients

Fourier series approximates a periodic function using sines and cosines. To do this accurately, coefficients \( a_n \) and \( b_n \) are calculated. These are integrals that represent the weights of the cosine and sine terms. For even functions, only cosine terms are non-zero, and for odd functions, only sine terms are non-zero. To find these coefficients, integrals like: \[ a_n = \frac{2}{b} \int_{0}^{b} f(x) \text{cos}\bigg(\frac{n \pi x}{b}\bigg) \, dx \]. are used.

Integration by Parts

Integration by parts is a method used to solve integrals involving products of functions. It's based on the formula: \[ \int u \, dv = uv - \int v \, du \]. In the Fourier series, we often come across integrals like \( \int x^2 \cos(n \pi x) dx \) which can be evaluated using integration by parts. Here, we set \( u = x^2 \) and \( dv = \cos(n \pi x) dx \) to simplify the integral step-by-step.