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Chapter 3: Problem 7

Gegeben ist die Funktion \(f(x)=\frac{4}{\pi}\left(\pi x-x^{2}\right), x \in[0, \pi]\). Setzen Sie die Funktion ungerade zu einer \(2 \pi\)-periodischen Funktion fort und berechnen Sie die FOURIER-Reihe der Funktion.

Short Answer

Expert verified
Extend \(f(x)\) to an odd, \(2\pi\)-periodic function and find its Fourier series by calculating the sine coefficients \(b_n\) using integration.

Step by step solution

01

Understanding the Task

We are given a function \( f(x) = \frac{4}{\pi} (\pi x - x^2) \) defined on the interval \([0, \pi]\). Our goal is to extend this function to be odd and \(2\pi\)-periodic, and then find its Fourier series.
02

Extending the Function to be Odd

An odd function \( f(x) \) must satisfy \( f(-x) = -f(x) \). Thus, for \( x \in [-\pi, 0] \), we define \( f(x) = -\frac{4}{\pi} (\pi (-x) - (-x)^2) = \frac{4}{\pi} (\pi x - x^2) \). This makes the function odd on the interval \([-\pi, \pi]\).
03

Creating a \(2\pi\)-Periodic Function

To make \( f(x) \) \(2\pi\)-periodic, we extend it such that \( f(x + 2\pi) = f(x) \) for all \( x \). We use the odd extension from Step 2 on every interval of length \(2\pi\).
04

Calculating the Fourier Series Coefficients

The Fourier series of an odd \(2\pi\)-periodic function consists of sine terms only. The sine coefficients are given by: \[ b_n = \frac{2}{T} \int_{0}^{T} f(x) \sin\left(\frac{n\pi x}{T}\right) \, dx \] where \( T = 2\pi \). Thus:\[ b_n = \frac{1}{\pi} \int_{0}^{\pi} \left(\frac{4}{\pi}(\pi x - x^2)\right) \sin(nx) \, dx \].
05

Evaluating the Integral for \( b_n \)

We need to evaluate: \[ b_n = \frac{4}{\pi^2} \int_{0}^{\pi} (\pi x - x^2) \sin(nx) \, dx \] This integral can be solved using integration by parts. Calculate the two parts separately:- For \( \int x \sin(nx) \, dx \), use the identity for integration by parts.- For \( \int x^2 \sin(nx) \, dx \), use the identity twice as it repeats the process.
06

Writing the Final Fourier Series

After calculating \( b_n \), write the Fourier series as: \[ f(x) \sim \sum_{n=1}^{\infty} b_n \sin(nx) \].Insert the calculated values of \( b_n \) into the series.

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Key Concepts

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

Odd Function Extension
When extending a function to be odd, the target is to make sure the function satisfies the condition: \( f(-x) = -f(x) \). An odd function has a specific symmetry: it looks the same upside down. This makes it mirror around the origin instead of across the y-axis, unlike even functions. For the given function \( f(x) = \frac{4}{\pi} (\pi x - x^2) \) on \([0, \pi]\), we have to define it for \([-\pi, 0]\) such that it meets the odd function criteria.
**Steps to Create Odd Extension**
  • Identify the existing function in the given interval.
  • For each positive \(x\), find its corresponding negative \(x\).
  • Define the odd extension using: \( f(-x) = -f(x) \).
  • Apply this across the negative interval \([-\pi, 0]\).
Once this is done, repeating the structure can help construct a periodic function.
Periodic Function
A periodic function is one which repeats at regular intervals, described by its period \(T\). This means that for a function \(f(x)\), \(f(x + T) = f(x)\) at all times. For this exercise, the requirement is for the function to be \(2\pi\)-periodic.
**Constructing a Periodic Function**
  • Understand the initial interval where the function is defined and made odd: \([-\pi, \pi]\).
  • Repeat this interval across the real line to meet the condition, \(f(x + 2\pi) = f(x)\).
  • Ensure the constructed function holds its properties (odd symmetry in this case) over each repeated segment.
This ensures the function repeats the same pattern every \(2\pi\) distance along the \(x\)-axis.
Integration by Parts
Integration by parts is a pivotal method used in calculus to solve integrals that are products of two functions. The formula for integration by parts is given by: \[ \int u \cdot dv = uv - \int v \cdot du \] where \(u\) and \(dv\) are parts of the original integral that can be strategically chosen.
**Applying Integration by Parts**
  • Choose \(u\) and \(dv\) such that \(du\) and \(v\) are simpler to compute.
  • Differentiate \(u\) to obtain \(du\) and integrate \(dv\) to obtain \(v\).
  • Substitute back into the integration by parts formula to solve the integral.
In this exercise, applying integration by parts simplifies the calculation of sine coefficients, especially in tackling integrals like \( \int x \sin(nx) \, dx\) and evaluating each component systematically.
Sine Coefficients
In the context of Fourier series, sine coefficients (denoted as \(b_n\)) are specific components that help express a periodic odd function as a sum of sine terms. For any \(2\pi\)-periodic odd function, the Fourier series only contains sine terms because sine is an odd function itself.
**Determining Sine Coefficients**
  • The formula for sine coefficients in the Fourier series is: \[ b_n = \frac{2}{T} \int_0^T f(x) \sin \left(\frac{n\pi x}{T} \right) \, dx \] where \(T\) is the period (here \(2\pi\)).
  • For this function, it becomes: \[ b_n = \frac{4}{\pi^2} \int_0^\pi (\pi x - x^2) \sin(nx) \, dx \]
  • Compute this integral for each \(n\) to find the Fourier series expansion in sine terms.
By determining these coefficients, one can construct the complete Fourier series, representing the original function.

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Most popular questions from this chapter

Berechnen Sie den Konvergenzradius der Reihe \(\sum_{k=0}^{\infty}\left(\frac{1-i}{2-i}\right)^{k}(z-i)^{k}\) und geben Sie den Konvergenzkreis an.

Berechnen Sie die FOURIER-Reihe der ungeraden \(2 \pi\)-periodischen Funktion $$ f(x)=\left\\{\begin{array}{llr} y=x & \text { f眉r } & -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \\ y=\pi-x & \text { f眉r } & \frac{\pi}{2} \leq x \leq \frac{3 \pi}{2} \end{array}\right.. $$ Nutzen Sie das Ergebnis und die PARSEVALsche Gleichung zur Berechnung des Wertes der Reihen $$ \sum_{k=1}^{\infty} \frac{1}{(2 k-1)^{2}} \quad \text { bzw. } \quad \sum_{k=1}^{\infty} \frac{1}{(2 k-1)^{4}}. $$

Berechnen Sie den Wert der Reihe \(\sum_{k=3}^{\infty}\left(\frac{3}{4}\right)^{k}\).

Bestimmen Sie den Konvergenzradius der Potenzreihe \(\sum_{k=1}^{\infty} \frac{2^{k}}{k !}(x-5)^{k}\) und geben Sie das Konvergenzintervall an. Untersuchen Sie die Konvergenzeigenschaften an den Randpunkten des Konvergenzintervalls.

Zeigen Sie unter Nutzung der arctan-Reihe die G眉ltigkeit der Beziehung $$ \frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+-\cdots=\sum_{k=0}^{\infty}(-1)^{k} \frac{1}{2 k+1} $$ und geben Sie eine Zahl \(n \in \mathbb{N}\) an, so dass der Fehler bei der Berechnung von \(\frac{\pi}{4}\) durch \(\sum_{k=0}^{n}(-1)^{k} \frac{1}{2 k+1}\) kleiner als \(10^{-5}\) wird.
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