Question

Let $f(x)={\sin }^{2}x,0<x<\pi .$ Sketch (or computer plot) the even function $f_c$ of period $2\pi ,$ the odd function $f_s$ of period $2\pi ,$ and the function $f_p$ of period $\pi ,$ each of which is equal to $f(x)$ on $(0,\pi ).$ Expand each of these functions in an appropriate Fourier series.

Short answer

The Fourier series for $f_c$ is: $\frac{1}{2}-\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{2}{n^2-1}\cos (nx)$, for $f_s$ is: $\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n+1}\frac{2}{n(n^2-1)}\sin (nx)$, and for $f_p$ is: $\frac{1}{2}-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2-1}\cos (2nx)$

Step-by-step solution

Introduction and Problem Understanding

Understand that you need to find three different periodic functions based on the given function $f(x)={\sin }^2(x)$ on $(0,\pi )$. These functions are: the even function $f_c$ of period $2\pi$, the odd function $f_s$ of period $2\pi$, and the function $f_p$ of period $\pi$. Then, you need to compute the Fourier series for each.

Find the Even Extension $f_c(x)$

The even extension is obtained by reflecting $f(x)$ across the y-axis. Therefore, $f_c(x)$ for $-\pi <x<\pi$ can be written as: $$f_c(x)=\{\begin{array}{lll}{\sin }^2(x)& 0<x<\pi {\sin }^2(-x)={\sin }^2(x)& -\pi <x<0\end{array}$$ Since ${\sin }^2(x)$ is already an even function, no change is necessary.

Compute the Fourier Series for $f_c(x)$

The Fourier series for an even function with period $2\pi$ only contains cosine terms. We need to calculate the coefficients, given by: $$a_0=\frac{1}{\pi }{\int }_0^{\pi }{\sin }^2(x)dx$$ and $$a_n=\frac{1}{\pi }{\int }_0^{\pi }{\sin }^2(x)\cos (nx)dx$$ with $n=1,2,3,\dots$. After solving, the Fourier series is: $$f_c(x)=\frac{1}{2}-\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{2}{n^2-1}\cos (nx)$$

Find the Odd Extension $f_s(x)$

The odd extension is obtained by reflecting and then negating $f(x)$ across the y-axis. Therefore, $f_s(x)$ for $-\pi <x<\pi$ can be written as: $$f_s(x)=\{\begin{array}{lll}{\sin }^2(x)& 0<x<\pi -{\sin }^2(-x)=-{\sin }^2(x)& -\pi <0\end{array}$$.

Compute the Fourier Series for $f_s(x)$

The Fourier series for an odd function with period $2\pi$ only contains sine terms. We need to calculate the coefficients given by: $$b_n=\frac{1}{\pi }{\int }_0^{\pi }{\sin }^2(x)\sin (nx)dx$$ with $n=1,2,3,\dots$. After solving, the Fourier series is: $$f_s(x)=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n+1}\frac{2}{n(n^2-1)}\sin (nx)$$

Find the Periodic Extension $f_p(x)$

The periodic extension $f_p(x)$ is simply repeating $f(x)={\sin }^2(x)$ with period $\pi$. Thus, over a single period: $$f_p(x)={\sin }^2(x)$$

Compute the Fourier Series for $f_p(x)$

Since $f_p(x)$ has a period of $\pi$, its Fourier series contains both sine and cosine terms. We compute the coefficients: $$a_0=\frac{2}{\pi }{\int }_0^{\pi }{\sin }^2(x)dx$$ $$a_n=\frac{2}{\pi }{\int }_0^{\pi }{\sin }^2(x)\cos \text{(}2nx\text{)}dx$$ $$b_n=\frac{2}{\pi }{\int }_0^{\pi }{\sin }^2(x)\sin (2nx)dx$$ resulting in the Fourier series: $$f_p(x)=\frac{1}{2}-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2-1}\cos (2nx)$$

Key concepts

Even Function

An even function is symmetric about the y-axis. This means that if you reflect the graph of the function across the y-axis, it remains unchanged. Mathematically, a function f(x) is even if it satisfies the condition:

• f(x) = f(-x) for all x in the domain.

For the given function f(x) = sin²(x), it is already even because sin²(x) = sin²(-x). Therefore, the even extension of the function over the interval (-π, π) is straightforward. Understanding even functions makes it easier to handle specific terms in Fourier series, as only cosine terms will be present.

Odd Function

An odd function is antisymmetric about the origin, meaning its graph is reflected and inverted across the y-axis. Mathematically, a function f(x) is odd if it satisfies:

• f(x) = -f(-x) for all x in the domain.

For the function f(x) = sin²(x) on (0, π), the odd extension involves negating the reflected part. This concept is important while dealing with Fourier series, as odd functions only contain sine terms.

Periodic Function

A periodic function repeats its values at regular intervals. The smallest interval after which the function repeats is called the period. Mathematically, a function f(x) is periodic if there exists a positive number T such that:

• f(x + T) = f(x) for all x in the domain.

In this exercise, you deal with functions of periods 2π and π. For instance, the function f_p(x) = sin²(x) with period π means it repeats every π interval. Periodic functions are fundamental in Fourier series because these series decompose a periodic function into a sum of sines and cosines.

Integration

Integration is a fundamental concept in calculus used to find areas under curves or solve differential equations, among other applications. In the context of Fourier series, integration helps compute the coefficients of the series. Each coefficient represents a specific frequency component of the function. For example:

• The constant term a_0 is found using $a_0=\frac{1}{\text{period}}\times \text{integral of the function over one period}$.
• The cosine coefficients a_n are given by $a_n=\frac{2}{\text{period}}\times \text{integral of the function times cosine over one period}$.
• The sine coefficients b_n are found similarly.

Understanding integration is crucial for finding these coefficients and thus the complete Fourier series representation of the function.

Sinusoidal Functions

Sinusoidal functions include sine and cosine functions, and they form the basis of Fourier series. A sinusoidal function has the form A*sin(Bx + C) or A*cos(Bx + C), where A, B, and C are constants representing amplitude, frequency, and phase shift, respectively. In Fourier series, any periodic function can be expressed as a sum of sinusoidal functions. For example, the Fourier series of an even function with period 2π consists entirely of cosine terms because cosine is an even function. Similarly, the series for an odd function contains only sine terms, as sine is an odd function.