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Chapter 2: Problem 1

Set \(f(x)=1-x^{2}\) in the interval \([-\pi, \pi]\) and let $$ f(x) \sim \frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left[a_{n} \cos n x+b_{n} \sin n x\right] $$ be the Fourier series of \(f\). (a) Calculate the \(a_{n}\) and \(b_{n}\). (b) To what values does the Fourier series of \(f\) converge at the points \(x=5 \pi\) and \(x=6 \pi\) ? Explain.

Short Answer

Expert verified
(a) \(a_0 = \frac{2\pi}{3}, a_n = 0, b_n = 0\) for \( n \geq 1 \). (b) At \(x=5\pi\), series converges to \(1-\pi^2\); at \(x=6\pi\), to \(1\).

Step by step solution

01

Identify the function

The function given is \( f(x) = 1 - x^2 \) on the interval \([ -\pi, \pi ]\). The task is to find the Fourier coefficients \( a_n \) and \( b_n \).
02

Calculate a_0 (constant term)

The coefficient \( a_0 \) is computed using the integral \[ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} (1 - x^2) \, dx \]. Calculate the integral: \( \int 1 \, dx = x \) and \( \int x^2 \, dx = \frac{x^3}{3} \). Evaluating these between \(-\pi\) and \(\pi\) gives \[ a_0 = \frac{1}{\pi} \left[ (x - \frac{x^3}{3}) \right]_{-\pi}^{\pi} = \frac{1}{\pi} \left[ \pi - \frac{\pi^3}{3} \right] - \left[ -\pi + \frac{\pi^3}{3} \right] = \frac{2\pi}{3}. \]
03

Calculate a_n (cosine coefficients)

To find \( a_n \) for \( n \geq 1 \), use \[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} (1 - x^2) \cos(nx) \, dx. \] Integration by parts may be necessary. Evaluate \( \int \cos(nx) \, dx = \frac{\sin(nx)}{n} \) and for \( x^2 \cos(nx) \), integration by parts is used. After solving, since \( 1 - x^2 \) is an even function and \( \cos(nx) \) is also even, the result from symmetry analysis and integration by parts leads to \( a_n = 0 \) for all \( n \geq 1 \).
04

Calculate b_n (sine coefficients)

Because \( f(x) = 1 - x^2 \) is an even function, while \( \sin(nx) \) is odd, their product \( (1 - x^2) \sin(nx) \) is odd. Thus the integral \[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} (1 - x^2) \sin(nx) \, dx \] is zero, as the integral of an odd function over a symmetric interval around zero is zero.
05

Converge at x=5\pi and x=6\pi

The Fourier series converges to the average of the left and right limits of the function at the endpoints outside the original interval, due to periodicity. Since \( f(x) = 1 - x^2 \) is periodic with period \( 2\pi \) and extension of \( f(x) = 1 - (x \mod 2\pi)^2 \), it behaves differently for different periods. At \( x=5\pi \), it maps to \( x=\pi \) and at \( x=6\pi \), it maps to \( x=0 \), thus \( f(\pi) = 1 - \pi^2 \) and \( f(0) = 1 \).

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Key Concepts

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

Fourier Coefficients
Fourier coefficients are the heart of a Fourier series. They allow us to represent a function as a series of sines and cosines. If you imagine the series as a sort of infinite sum, the coefficients dictate how much of each sine and cosine wave you need to add together to approximate your function.
  • Constant term (a_0): This is the average value of the function over one period. It is determined using the integral \[ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} (1 - x^2) \, dx \]. For our function, this results in \( \frac{2\pi}{3} \).
  • Cosine coefficients (a_n): These coefficients are obtained by integrating the product of the function and \( \cos(nx) \). For even functions like f(x) = 1 - x^2 with even \( \cos(nx) \), these coefficients turn out to be 0 because their integrals vanish.
  • Sine coefficients (b_n): Found by integrating the product of the function and \( \sin(nx) \). However, because \( \sin(nx) \) is odd and does not match the function's symmetry, these coefficients also become 0 when the integration is performed over symmetric intervals.
Periodic Function
A periodic function repeats its values in regular intervals or periods. For the Fourier series, understanding periodicity is crucial because it dictates how the series will reconstruct the function.
  • The base period for the function \(f(x) = 1 - x^2\) in the interval \([-\pi, \pi]\) is \(2\pi\). This means that outside of this interval, the function will repeat according to this period.
  • When using Fourier series, the function doesn't just resume its shape but continues to follow the periodic repetition from wherever it is mapped within any \(2\pi\) interval.
  • Understanding the period allows us to recompute values at points such as \(x=5\pi\) or \(x=6\pi\), by effectively "folding" the function back into its base period. This concept is imperative for determining series convergence at any given point.
Integration by Parts
Integration by parts is a technique useful for integrating products of functions, specifically from calculus, and it often simplifies evaluating integrals that appear in Fourier coefficient calculations.
  • When integrating expressions like \((1-x^2)\cos(nx)\), breaking down the integral into more manageable parts is required. Here, you would let one part of the function be \((1-x^2)\) and the other part \(\cos(nx)\).
  • The technique involves choosing which function to differentiate and which to integrate in order to make the problem simpler.
  • For the Fourier series of \(f(x)\), since \(1-x^2\) results in zero contribution to both cosine and sine coefficients over symmetric intervals, integration by parts is a powerful tool in confirming these results.
Convergence of Series
The convergence of a Fourier series refers to how well the infinite series of sine and cosine terms approximates the original function as more terms are added.
  • The Fourier series of a periodic function like \(f(x)\) converges if it approaches a specific value at each point as you sum more terms. Generally, at every point \(x\), the series converges to the average of the left-hand and right-hand limits of the function.
  • In the exercise, at \(x=5\pi\) and \(x=6\pi\), the expected convergence values are \(1-\pi^2\) and \(1\), respectively, because these points map to \(\pi\) and \(0\) within the base period.
  • For Fourier series, convergence properties depend heavily on the function's smoothness and symmetry, with the series reflecting discontinuities in the so-called Gibbs phenomenon when those occur.

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Most popular questions from this chapter

Let \(f \in E\) be an even function satisfying \(\int_{-\pi}^{\pi} f(t) d t=5\). Define the function \(F\) by $$ F(x)=\int_{-\pi}^{x} f(t) d t, \quad-\pi \leq x \leq \pi . $$ Let $$ F(x) \sim \frac{A_{0}}{2}+\sum_{n=1}^{\infty}\left[A_{n} \cos n x+B_{n} \sin n x\right] $$ denote the Fourier series of \(F\) and set $$ G(x)=\frac{A_{0}}{2}+\sum_{n=1}^{\infty}\left[A_{n} \cos n x+B_{n} \sin n x\right] $$ Calculate \(G(-\pi), G(\pi)\), and \(G(0)\).

Set \(f(x)=|\sin x|\) and let $$ f(x) \sim \frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left[a_{n} \cos n x+b_{n} \sin n x\right] $$ denote the Fourier series of \(f\) on \([-\pi, \pi]\). (a) Calculate the \(a_{n}\) and \(b_{n}\). (b) Set \(g(x)=\sum_{n=1}^{\infty}\left[-n a_{n} \sin n x+n b_{n} \cos n x\right]\). Determine \(g\) and sketch the graph of \(g\) on \([-\pi, \pi]\). (c) Calculate the sums $$ \sum_{n=1}^{\infty} \frac{1}{4 n^{2}-1}, \quad \sum_{n=1}^{\infty} \frac{(-1)^{n}}{4 n^{2}-1}, \quad \sum_{n=1}^{\infty} \frac{1}{\left(4 n^{2}-1\right)^{2}}, \quad \sum_{n=1}^{\infty} \frac{n^{2}}{\left(4 n^{2}-1\right)^{2}} . $$

Find the complex Fourier series of \(f(t)=\frac{1}{1-\frac{1}{2} e^{-i t}}\) on the interval \([-\pi, \pi]\).

Let $$ f(x)= \begin{cases}A x+B, & -\pi \leq x<0, \\ \cos x, & 0 \leq x \leq \pi .\end{cases} $$ For what values \(A\) and \(B\) does the Fourier series of \(f\) converge uniformly to \(f\) on all of \([-\pi, \pi]\) ?

For each natural integer \(n\) we define $$ f_{n}(x)=1+\sum_{k=1}^{n}[\cos k x-\sin k x] . $$ Calculate the value of the integral \(\int_{-\pi}^{\pi}\left|f_{n}(x)\right|^{2} d x\).
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