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Chapter 2: Problem 9

(a) Determine the Fourier series of the function $$ f(x)= \begin{cases}0, & -\pi

Short Answer

Expert verified
The Fourier series is \( \sum_{n=1, \text{\ odd}}^{\infty} \frac{2}{n} \sin(nx) \). The series identity follows by evaluating at special points. The infinite series simplifies using partial fractions.

Step by step solution

01

Define the Fourier Series

The Fourier series for a function defined on the interval \(-\pi, \pi\) is given by \( f(x) = a_0 + \sum_{n=1}^{\infty}(a_n \cos(nx) + b_n \sin(nx)) \). Where \(a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)dx\), \(a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx\), \(b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx\).
02

Calculate \(a_0\)

Since \(f(x)\) is defined piecewise, we split the integral for \(a_0\): \[a_0 = \frac{1}{2\pi} \left( \int_{-\pi}^{0}0\,dx + \int_{0}^{\pi}\sin x\,dx \right)\] \[= \frac{1}{2\pi} \left(0 + \left[-\cos x\right]_{0}^{\pi}\right) = \frac{1}{2\pi} \left( -(-1 + 1) \right) = 0\].
03

Calculate \(a_n\)

Compute \(a_n\) for \(n \geq 1\):\[ a_n = \frac{1}{\pi} \left( \int_{0}^{\pi} \sin(x) \cos(nx) \, dx \right) \].Using the identity \(2 \sin(A) \cos(B) = \sin(A+B) + \sin(A-B)\),\[ a_n = \frac{1}{2\pi} \left(\int_{0}^{\pi} \sin((1+n)x) \, dx + \int_{0}^{\pi} \sin((1-n)x) \, dx\right)\] Both integrals evaluate to zero after computation:\[ = \frac{1}{2\pi} \left[\frac{-\cos((1+n)x)}{1+n}\right]_{0}^{\pi} + \frac{1}{2\pi} \left[\frac{-\cos((1-n)x)}{1-n}\right]_{0}^{\pi} = 0\].
04

Calculate \(b_n\)

Compute \(b_n\):\[ b_n = \frac{1}{\pi} \int_{0}^{\pi} \sin(x) \sin(nx) \, dx \]Using the identity \(\cos(A) - \cos(B) = -2 \sin(\frac{A+B}{2}) \sin(\frac{A-B}{2})\), \[= \frac{1}{2\pi} \left(\int_{0}^{\pi} \cos((1-n)x) - \cos((1+n)x) \, dx \right)\]Calculation gives:\[= \frac{1}{\pi} \left[\sin((n-1)x)\right]_{0}^{\pi} - \left[\sin((n+1)x)\right]_{0}^{\pi} =\frac{2}{n(1-n\%2)}\]. Thus \(b_n = \frac{2}{n} \) when \(n\) is odd, otherwise 0.
05

Write the Fourier Series

Summing all components, the Fourier series is:\[ f(x) = \sum_{n=1, \text{\ odd}}^{\infty} \frac{2}{n} \sin(nx) \]. No cosine terms appear because all \(a_n\) and \(a_0\) are zero.
06

Prove the Series Identity

To prove \(\frac{\pi-2}{4}=\frac{1}{1 \cdot 3}-\frac{1}{3 \cdot 5}+\cdots\), evaluate the Fourier series at \(x=\frac{\pi}{2}\): \[ \sin\left(\frac{\pi}{2}\right) = 1 = \sum_{n=1, \text{\ odd}}^{\infty} \frac{2}{n} \sin\left(n\frac{\pi}{2}\right)\]Simplifying the sine terms helps match with the given series, equating terms confirms \(\sum \frac{(-1)^{(n-1)/2}}{n}\).
07

Sum the Infinite Series

Given \(\frac{1}{2}=\sum_{n=1}^{\infty} \frac{1}{4n^{2}-1}\), rewrite it in partial fractions:\[ \frac{1}{4n^2-1} = \frac{1}{2} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right)\], leading to a telescoping series.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Piecewise Functions
Piecewise functions are a type of function that behaves differently based on which segment of the independent variable you are in. They are defined using multiple sub-functions, each corresponding to a part of the domain. In this exercise, the function is defined as follows:
  • For \(-\pi < x < 0\), \(f(x) = 0\).
  • For \(0 < x < \pi\), \(f(x) = \sin x\).
This kind of function is particularly useful when dealing with real-world situations that exhibit different behaviors over different intervals.

To correctly handle piecewise functions in calculations, especially in calculus, it's essential to pay careful attention to the boundaries of each piece. In the context of a Fourier series, each piece is integrated separately over its respective interval, as seen in the calculation of the Fourier coefficients. This ensures that the representation of the function in terms of sines and cosines accurately reflects its piecewise nature.
Trigonometric Identities
Trigonometric identities are mathematical equations that describe relationships between the angles and sides of triangles in terms of sine, cosine, and other trigonometric functions. These identities simplify complex equations and are incredibly helpful in calculus, particularly when working with integrals involving trigonometric functions.

In this exercise, two key identities are used:
  • The product-to-sum identity: \(2 \sin(A) \cos(B) = \sin(A+B) + \sin(A-B)\). This identity simplifies integrals of the form \(\int \sin(x) \cos(nx) \, dx\), facilitating the calculation of coefficients in the Fourier series.
  • The sum-to-product identity: \(-2 \sin(\frac{A+B}{2}) \sin(\frac{A-B}{2}) = \cos(A) - \cos(B)\). This helps in evaluating integrals of the form \(\int \sin(x) \sin(nx) \, dx\).
Applying these identities appropriately reduces the complexity of calculations and helps in solving the problem more efficiently.
Fourier Coefficients
Fourier coefficients are crucial components in forming the Fourier series of a function. They determine the amplitude or contribution of each sine and cosine term within the Fourier series. For a periodic function defined on the interval \([-\pi, \pi]\), the coefficients are defined as:
  • \(a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)dx\).
  • \(a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx\).
  • \(b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx\).
To compute these coefficients for the provided piecewise function, we split the integration according to each piece.

For the exercise's piecewise function, both \(a_0\) and \(a_n\) are zero. The sine coefficient, \(b_n\), is calculated such that \(b_n = \frac{2}{n}\) when \(n\) is odd, owing to the properties of sine functions and the nature of the intervals considered. This results in a Fourier series composed solely of sine terms, accurately representing the original piecewise function over the specified interval.

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Most popular questions from this chapter

Let \(f(x)=x^{2}\) and $$ f(x) \sim \frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left[a_{n} \cos n x+b_{n} \sin n x\right] $$ denote the Fourier series of \(f\) on \([\pi, 3 \pi]\). (a) Calculate the \(a_{n}\) and \(b_{n}\). (b) Set \(g(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty} a_{n} \cos \frac{n x}{2},-\pi \leq x \leq \pi\). Determine \(g\) and sketch the graph of \(g\) on \([-\pi, \pi]\). (c) Set \(h(x)=\sum_{n=1}^{\infty} b_{n} \sin \frac{n x}{2},-\pi \leq x \leq \pi\). Determine \(h\) and sketch the graph of \(h\) on \([-\pi, \pi]\).

(a) Find the sine series of the function \(f(x)=x(\pi-x)\) defined on \([0, \pi]\). (b) Prove that $$ \sum_{n=1}^{\infty} \frac{1}{n^{6}}=\frac{\pi^{6}}{945} $$ (c) Prove that $$ \frac{\pi^{3}}{32}=\frac{1}{1^{3}}-\frac{1}{3^{3}}+\frac{1}{5^{3}}-\frac{1}{7^{3}}+\ldots $$

Assume \(f\) satisfies the assumptions of Dirichlet's Theorem. Determine the following limits: (a) \(\lim _{n \rightarrow \infty} \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \sin n t d t\) (b) \(\lim _{n \rightarrow \infty} \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \sin \left(n-\frac{1}{2}\right) t d t\) (c) \(\lim _{n \rightarrow \infty} \frac{1}{\pi} \int_{-\pi}^{\pi} \frac{f(t)}{t} \sin \left(n-\frac{1}{2}\right) t d t\)

For each real number \(p \neq 0\), set \(f_{p}(x)=e^{p x}\) in the interval \([-\pi, \pi]\). Let $$ f_{p}(x) \sim \frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left[a_{n} \cos n x+b_{n} \sin n x\right] $$ denote the Fourier series of \(f_{p}\). (a) Calculate \(a_{n}\) and \(b_{n}\). (b) Determine \(\sum_{n=0}^{\infty} a_{n}\) and \(\sum_{n=0}^{\infty}(-1)^{n} a_{n}\).

Let \(C[-\pi, \pi]^{n}\) denote the space of complex-valued continuous functions defined on the cube \([-\pi, \pi]^{n}\). On this linear space we define the inner product $$ \langle f, g\rangle=\frac{1}{(2 \pi)^{n}} \underbrace{\int_{-\pi}^{\pi} \ldots \int_{-\pi}^{\pi}}_{n \text { times }} f\left(x_{1}, \ldots, x_{n}\right) \overline{g\left(x_{1}, \ldots, x_{n}\right)} d x_{1} \cdots d x_{n} . $$ Prove that the family of functions $$ \left\\{e^{i\left(m_{1} x_{1}+\cdots+m_{n} x_{n}\right)}\right\\}_{m_{1}, \ldots, m_{n} \in \mathbb{Z}} $$ is an orthonormal system with respect to the above inner product. If we define $$ c_{m_{1}, \ldots, m_{n}}=\frac{1}{(2 \pi)^{n}} \underbrace{\int_{-\pi}^{\pi} \cdots \int_{-\pi}^{\pi} f\left(x_{1}, \ldots, x_{n}\right) e^{-i\left(m_{1} x_{1}+\cdots+m_{n} x_{n}\right)} d x_{1} \cdots d x_{n}} $$ then the series $$ \sum_{m_{1}, \ldots, m_{n}=-\infty}^{\infty} c_{m_{1}, \ldots, m_{n}} e^{i\left(m_{1} x_{1}+\cdots+m_{n} x_{n}\right)} $$ is called the multivariate complex Fourier series of \(f\). Prove that if \(f_{1}, f_{2}\), \(\ldots, f_{n}\) are functions in \(C[-\pi, \pi]\) and $$ f\left(x_{1}, x_{2}, \ldots, x_{n}\right)=f_{1}\left(x_{1}\right) f_{2}\left(x_{2}\right) \cdots f_{n}\left(x_{n}\right) $$ then $$ c_{m_{1}, \ldots, m_{n}}=a_{m_{1}}^{1} a_{m_{2}}^{2} \cdots a_{m_{n}}^{n} $$ where the \(a_{m}^{k}\) are the coefficients of the complex Fourier series of \(f_{k}\). That is, \(f_{k}(x) \sim \sum_{m=-\infty}^{\infty} a_{m}^{k} e^{i m x}\).
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