Question

You are given \(f(x)\) on an interval, say \(0

Short answer

The Fourier series expansions are: \(f_c(x) = \frac{1}{2} + \sum_{n=1}^{\infty} \frac{4(-1)^n}{n^2 \pi^2} \cos(n \pi x)\) \(f_s(x) = \sum_{n=1}^{\infty} \frac{2((-1)^{n+1} - 1)}{n^2 \pi^2} \sin(n\pi x)\) \(f_p(x) = \frac{1}{3} + \sum_{n=1}^{\infty} (a_n \cos(2n \pi x) + b_n \sin(2n \pi x))\)

Step-by-step solution

Define the function f(x)

The given function is defined as \[f(x) = x^2\] on the interval \(0 < x < 1\).

Sketch the functions

Sketch the following functions: 1. Even function \(f_c(x)\) of period \(2\) which is symmetric about the vertical axis. 2. Odd function \(f_s(x)\) of period \(2\) which is antisymmetric about the vertical axis. 3. Periodic function \(f_p(x)\) of period \(1\) that repeats every interval.

Express f(x) as the even function f_c(x)

The even extension of \(f(x)\) is defined by \[f_c(x) = \begin{cases} x^2 & \text{if } 0 < x < 1 \ (2-x)^2 & \text{if } 1 < x < 2 \ \text{repeat every 2} \ \end{cases}\]

Expand f_c(x) in Fourier series

The Fourier series of the even function \(f_c(x)\) is given by \[f_c(x) = a_0 + \sum_{n=1}^{\infty} a_n \cos\left(\frac{n\pi x}{b}\right) \] Calculate \(a_0\) and \(a_n\): \[a_0 = \frac{1}{2}\int_0^1 x^2 dx + \frac{1}{2}\int_1^2 (2-x)^2 dx = \frac{1}{2}\] \[a_n = \int_0^1 x^2 \cos\left(\frac{n\pi x}{1}\right) dx = \frac{4(-1)^n}{n^2 \pi^2}\]

Express f(x) as the odd function f_s(x)

The odd extension of \(f(x)\) is defined by \[f_s(x) = \begin{cases} x^2 & \text{if } 0 < x < 1 \ -(2-x)^2 & \text{if } 1 < x < 2 \ \text{repeat every 2} \ \end{cases}\]

Expand f_s(x) in Fourier series

The Fourier series of the odd function \(f_s(x)\) is given by \(f_s(x) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{b}\right)\) Calculate \(b_n\): \[b_n = \int_0^1 x^2 \sin\left(\frac{n\pi x}{1}\right) dx\ = \frac{2((-1)^{n+1} - 1)}{n^2 \pi^2}\]

Express f(x) as the periodic function f_p(x)

The periodic extension of \(f(x)\) is defined by repeating it every period \[f_p(x) = \begin{cases} x^2 & \text{if } 0 < x < 1 \ (x+1)^2 & \text{if } -1 < x < 0 \ \text{repeat every 1} \ \end{cases}\]

Expand f_p(x) in Fourier series

The Fourier series for the function \(f_p(x)\) is a combination of sine and cosine functions as: \[f_p(x) = a_0 + \sum_{n=1}^{\infty} a_n \cos\left(\frac{2n\pi x}{1}\right) + b_n \sin\left(\frac{2n\pi x}{1}\right)\] Calculate \(a_0\), \(a_n\) and \(b_n\): \[a_0 = \int_0^1 x^2 dx = \frac{1}{3}\] \[a_n = \int_0^1 x^2 \cos\left(2n\pi x\right)dx \] \[b_n = \int_0^1 x^2 \sin\left(2n\pi x\right)dx \]

Key concepts

Even Function

An even function, by definition, satisfies the property that for every value in its domain, the function's output is the same for both positive and negative inputs. Mathematically, this is expressed as \( f(x) = f(-x) \). This means that the graph of an even function is symmetric about the y-axis.
In our example, when extending the function \( f(x) = x^2 \) to form an even function \( f_c(x) \) of period 2, we need to ensure that this symmetry holds. We achieve this by defining:

• \( f_c(x) = x^2 \) for \( 0 < x < 1 \)
• \( f_c(x) = (2 - x)^2 \) for \( 1 < x < 2 \)
• Repeat this definition over subsequent intervals.

By utilizing the symmetry about the y-axis, we ensure that the function remains consistent when translated periodically.

Odd Function

An odd function is characterized by its antisymmetric nature around the origin. Specifically, for each input \( x \) in its domain, the function meets the condition \( f(-x) = -f(x) \). Graphically, this implies a rotational symmetry around the origin (180 degrees).
In our problem, to express \( f(x) = x^2 \) as an odd function with period 2 (let's call it \( f_s(x) \)), we follow these steps:

• \( f_s(x) = x^2 \) for \( 0 < x < 1 \)
• \( f_s(x) = -(2 - x)^2 \) for \( 1 < x < 2 \)
• Repeat this definition in subsequent intervals.

By extending the function to meet this odd condition, it ensures that every instance of the function respects the requirement \( f_s(-x) = -f_s(x) \).

Periodic Function

Periodic functions are those that repeat their values at regular intervals over an infinite domain. The smallest such interval is called the period. If \( f(x) \) is a periodic function with period \( T \), it means \( f(x + T) = f(x) \) for all \( x \).
In this case, we must construct a periodic extension of the function \( f(x) = x^2 \) with period 1 (denote this as \( f_p(x) \)).
The periodic extension is:

• \( f_p(x) = x^2 \) for \( 0 < x < 1 \)
• \( f_p(x) = (x + 1)^2 \) for \( -1 < x < 0 \)
• Repeat this definition every period of 1.

Such periodic behavior ensures that the function maintains its defined shape over all intervals of its domain.

Function Expansion

Function expansion in the context of Fourier Series allows decomposition of a periodic function into a sum of sines and cosines. This can simplify the analysis and manipulation of the function.
For even and odd functions, the steps are slightly different:

• **For Even Functions:** Expand using only cosine terms: \( f_c(x) = a_0 + \sum_{n=1}^{\infty} a_n \cos\left(\frac{n\pi x}{b}\right) \)
  The coefficients are calculated where:
• \( a_0 = \frac{1}{2} \int_0^1 x^2 \, dx + \frac{1}{2} \int_1^2 (2-x)^2 \, dx = \frac{1}{2} \)
• \( a_n = \int_0^1 x^2 \cos\left(\frac{n\pi x}{1}\right) \, dx = \frac{4(-1)^n}{n^2 \pi^2} \)
• **For Odd Functions:** Use sine terms only: \( f_s(x) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{b}\right) \)
  Where
• \( b_n = \int_0^1 x^2 \sin\left(\frac{n\pi x}{1}\right) \, dx = \frac{2((−1)^{n+1} - 1)}{n^2 \pi^2} \)
• **For General Periodic Functions:** Utilize both sine and cosine terms: \( f_p(x) = a_0 + \sum_{n=1}^{\infty} a_n \cos\left(\frac{2n\pi x}{1}\right) + b_n \sin\left(\frac{2n\pi x}{1}\right) \)
• Where \( a_0 = \int_0^1 x^2 \, dx = \frac{1}{3} \)
• \( a_n = \int_0^1 x^2 \cos\left(2n\pi x\right)dx \)
• \( b_n = \int_0^1 x^2 \sin\left(2n\pi x\right)dx \)

Function expansion helps transform complex periodic functions into simpler trigonometric series which can be easily analyzed and manipulated.