Problem 5 Determine the Fourier series for... [FREE SOLUTION]

Chapter 7: Problem 5

Determine the Fourier series for the function defined by $$ \begin{aligned} f(x) &=2 x \quad 0 \leq x \leq 2 \pi \\ f(x+2 \pi) &=f(x) & \end{aligned} $$

Short Answer

Expert verified

The Fourier series is $ f(x) = 2 + \sum_{n=1}^{\infty} \frac{4}{n} \sin(nx)$

Step by step solution

Write the Fourier series formula

The Fourier series for a function defined on [0, 2\pi] and periodic with period 2\pi is given by$f(x) = a_0 + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx))$

Compute the coefficient a_0

The coefficient a_0 is computed using:$a_0 = \frac{1}{2\pi} \int_{0}^{2\pi} f(x) \text{d}x$Substitute f(x) = 2x to get:$a_0 = \frac{1}{2\pi} \int_{0}^{2\pi} 2x \text{d}x = \frac{2}{2\pi} [\frac{x^2}{2}]_{0}^{2\pi} = \frac{1}{\pi} [\pi^2] = 2$

Compute the coefficients a_n

The coefficients a_n are computed using:$a_n = \frac{1}{\pi} \int_{0}^{2\pi} f(x) \cos(nx) \text{d}x$Substitute f(x) = 2x to get:$a_n = \frac{2}{\pi} \int_{0}^{2\pi} x \cos(nx) \text{d}x$Using integration by parts, let u = x\ and \text{d}u = \cos(nx)\text{d}x\ to get:$a_n = \frac{2}{\pi} [\frac{x\sin(nx)}{n}]_{0}^{2\pi} - \frac{2}{\pi} \int_{0}^{2\pi} \frac{\sin(nx)}{n} = 0\text{} \text{(since both terms integrate to zero over}[0, 2\pi] ) = 0$

Compute the coefficients b_n

The coefficients b_n are computed using:$b_n = \frac{1}{\pi} \int_{0}^{2\pi} f(x) \sin(nx) \text{d}x$Substitute f(x) = 2x to get:$b_n = \frac{2}{\pi} \int_{0}^{2\pi} x \sin(nx) \text{d}x$Using integration by parts, let u = x and dv = \text{sin} (nx )\text{d}x to get:$b_n = -\frac{2}{n} [x \text{cos} (nx)]_{0}^{2\pi} + \frac{2}{n} \int_{0}^{2\pi} \text{cos} (nx) \text{d}x) = \frac{4}{n}\text{ (since the second term integrates to zero over}[0, 2\pi] }$

Write the Fourier series

Substitute a_0, a_n, b_n into the Fourier series formula to get:$ f(x) = 2 + \sum_{n=1}^{\infty}\left( 0 \cos(nx) + \frac{4}{n} \sin(nx)\right) = 2 + \sum_{n=1}^{\infty} \frac{4}{n} \sin(nx)$

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Key Concepts

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

Periodic Functions

A periodic function repeats its values at regular intervals, called periods. In this problem, we consider the function defined as $f(x)=2x$ over the interval $[0, 2\pi]$, with the periodic condition $f(x+2\pi)=f(x)$. This means that after every interval of $2\pi$, the function returns to its original value. Periodic functions are essential in Fourier analysis because they can be represented by a sum of sine and cosine functions, which are also periodic.

Fourier Coefficients

Fourier coefficients are the constants $a_0$, $a_n$, and $b_n$ in the Fourier series. They are determined through integration and characterize the amplitude of the sine and cosine components of the function.
For our function:

The coefficient $a_0$ is found using the integral $a_0 = \frac{1}{2\pi} \int_{0}^{2\pi} f(x) \, \text{d}x$.
The coefficients $a_n$ and $b_n$ are found using $a_n = \frac{1}{\pi} \int_{0}^{2\pi} f(x) \cos(nx) \, \text{d}x$ and $b_n = \frac{1}{\pi} \int_{0}^{2\pi} f(x) \, \sin(nx)\text{d}x$ respectively.

Substituting $f(x)=2x$, the calculations are simplified using integration by parts.

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It is based on the product rule for differentiation and is given by the formula: \[ \int u \, \text{d}v = uv - \int v \, \text{d}u \] For our problem, integration by parts is used twice:

To find $a_n$, set $u = x$ and $\text{d}v = \cos(nx)\text{d}x$.
To find $b_n$, set $u = x$ and $\text{d}v = \sin(nx)\text{d}x$.

This simplifies the integrals, and we find that many terms equal zero over the interval $[0, 2\pi]$, simplifying our results.

Trigonometric Series

A trigonometric series is a sum of sine and cosine functions. The Fourier series is a type of trigonometric series used to represent periodic functions.
The general form of the Fourier series is: \[ f(x) = a_0 + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx)) \] For the given function, substituting the calculated Fourier coefficients, we obtain:

$a_0 = 2$
$a_n = 0$ for all $n$
$b_n = \frac{4}{n}$

Therefore, the Fourier series simplifies to: \[ f(x) = 2 + \sum_{n=1}^{\infty} \frac{4}{n} \sin(nx) \] This indicates the sine components of the series are weighted by $\frac{4}{n}$.

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91影视

Determine the Fourier series for the function defined by $$ \begin{aligned} f(x) &=2 x \quad 0 \leq x \leq 2 \pi \\ f(x+2 \pi) &=f(x) & \end{aligned} $$

Short Answer

Step by step solution

Write the Fourier series formula

Compute the coefficient a_0

Compute the coefficients a_n

Compute the coefficients b_n

Write the Fourier series

Key Concepts

Periodic Functions

Fourier Coefficients

Integration by Parts

Trigonometric Series

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