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Chapter 7: Problem 75

The quantity \(a_{n}=\frac{1}{\pi} \int_{-\pi} f(x) \sin n x d x\) plays an important role in applied mathematics. Show that if \(f^{\prime}(x)\) is continuous on \([-\pi, \pi],\) then \(\lim _{n \rightarrow \infty} a_{n}=0 .\)

Short Answer

Expert verified
The limit of \(a_n\) as \(n\) approaches infinity is zero.

Step by step solution

01

Introduction to the Problem

We are given the quantity \( a_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x \, dx \) and need to show that \( \lim_{n \rightarrow \infty} a_{n} = 0 \). It involves integrating a product of functions over an interval.
02

Integration by Parts

To evaluate \( a_n \), we use integration by parts. Choose \( u = f(x) \) and \( dv = \sin(nx) \, dx \). Then \( du = f'(x) \, dx \) and \( v = -\frac{1}{n} \cos(nx) \).
03

Apply Integration by Parts

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), compute:\[ a_n = \frac{1}{\pi} \left[ -\frac{f(x)}{n} \cos(nx) \right]_{-\pi}^{\pi} + \frac{1}{\pi} \int_{-\pi}^{\pi} \frac{f'(x)}{n} \cos(nx) \, dx \]
04

Evaluate the Boundary Terms

Evaluate \([-\frac{f(x)}{n}\cos(nx)]_{-\pi}^{\pi}\). Since \(\cos(nx)\) is periodic and \(f(x)\) is continuous over \([-\pi, \pi]\), this term approaches zero as \( n \to \infty \).
05

Estimate the Integral Term

For the integral term, apply the fact that \(\frac{1}{\pi} \int_{-\pi}^{\pi} \frac{f'(x)}{n} \cos(nx) \, dx\) can be bounded. Since \( \|f'(x)\| \) is bounded by assumption and \(|\cos(nx)| \leq 1\), the entire integral is bounded by \( \frac{\|f'(x)\|_{max}}{n} \cdot 2\pi\), which approaches zero as \(n\) grows.
06

Conclude the Limit

Since both the boundary term and integral term tend to zero as \(n\) grows, \(\lim_{n \to \infty} a_n = 0\).

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Key Concepts

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

Fourier Series
The Fourier series is a way to represent a function as a sum of simple sine and cosine terms. It is particularly useful in analyzing periodic functions. This makes it a highly significant tool in both mathematics and engineering. When dealing with a Fourier series, we break down a complex signal into simpler, easily understandable components.
  • The Fourier series allows us to express any periodic function as an infinite sum of sines and cosines.
  • In our problem, the term \(a_n\) is derived from this concept, and it's part of developing a full Fourier series representation.
  • Specifically, \(a_n\) is the coefficient for the sine terms in the Fourier series.
Understanding the behavior of these coefficients as \(n\) approaches infinity is crucial. It tells us about the smoothness and continuity of the original function. This is why it is important to show that \(\lim_{n \rightarrow \infty} a_n = 0\). It implies that these sine components cease to have any significant influence, simplifying the function's representation even further.
Limit of a Sequence
A sequence is an ordered list of numbers. The concept of limit is fundamental in calculus and analysis. It describes the behavior of sequences or functions as they approach a certain value or infinity.
  • The limit of a sequence \(\{a_n\}\) tells us what the sequence approaches as \(n\) becomes very large.
  • In this exercise, we prove that the sequence of \(a_n\) approaches zero as \(n \rightarrow \infty\).
  • This is done by applying integration by parts, which transforms our integral and allows us to evaluate its limit more easily.
Understanding how to find the limit of sequences can help us determine the long-term behavior of a function, as well as its convergence properties.
Continuity of Derivative
In calculus, a function is said to have a continuous derivative if its derivative does not have any abrupt changes. This continuous smoothness is required in many mathematical proofs and is a vital condition in exploring certain functions.
  • A continuous derivative indicates that the change of a function at any given point is uniform and predictable.
  • In our problem, the continuity of the derivative \(f'(x)\) over \([-\pi, \pi]\) plays a key role.
  • It ensures that when we estimate the integral term from integration by parts, we can apply bounding methods effectively.
The existence of a continuous derivative simplifies many calculations and limits potential errors when applying calculus techniques. It's one of the primary reasons why we can assert that certain integrals and limits will behave in a predictable manner, leading to correct conclusions.

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