Question

True or False? In Exercises \(63-66,\) determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) converges for \(|x|<2,\) then \(\int_{0}^{1} f(x) d x=\sum_{n=0}^{\infty} \frac{a_{n}}{n+1}\)

Short answer

The statement is True

Step-by-step solution

Identify the Given Statement

The given statement says that if the power series \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) converges for \(|x|<2,\), then the integral \(\int_{0}^{1} f(x) d x=\sum_{n=0}^{\infty} \frac{a_{n}}{n+1}\).

Consider the Variables

A variable 'x' is being referred to - the variable for the function itself and 'n' - the index for the power series term.

Identify Function Convergence Range

It's mentioned that the given power series converges for \(|x|<2,\). This is the interval of convergence.

Evaluate the Statement

The given statement is indeed true. Let's break it down: It is known that a power series can be integrated term by term over any interval lying within its interval of convergence. Here, the interval of integration [0,1] lies within the interval of convergence (-2,2). Thus the given statement is True. We can integrate the power series term by term from 0 to 1 which results in \(\sum_{n=0}^{\infty} \frac{a_{n}}{n+1}\).

Key concepts

Integral of Power Series

Understanding how to integrate a power series is crucial when dealing with functions expressed in this form. A power series is an infinite sum of the form \( f(x) = \sum_{n=0}^{\infty} a_n x^n \), where \( a_n \) are the coefficients and \( x \) is the variable. When integrating such a series, it is essential to ensure that the integration takes place within the interval of convergence, which, in our example, is \( |x|<2 \).

The integral of a power series \( \int f(x) \, dx \) within its interval of convergence is found by integrating each term individually. That is, we apply the integral to each term \( a_n x^n \) in the series, resulting in \( \int a_n x^n dx = \frac{a_n}{n+1} x^{n+1} + C \), where \( C \) is the constant of integration. When evaluating from \( 0 \) to \( 1 \) for instance, the constant \( C \) will cancel out, leaving us with \( \sum_{n=0}^{\infty} \frac{a_n}{n+1} \) as a series representation of the integral.

Interval of Convergence

The interval of convergence is a fundamental concept when working with power series. It is the set of all x-values for which the series converges to a finite number. Outside this interval, the series may diverge or simply be undefined.

In practice, we determine the interval of convergence by using various tests for convergence, like the ratio test or root test, upon the terms of the series. In the given exercise example, the power series \( f(x) = \sum_{n=0}^{\infty} a_n x^n \) converges for \( |x|<2 \) which means the interval of convergence is \( (-2,2) \). The boundaries of this interval, known as the radius of convergence, are critical as they determine where the power series transitions from convergent to divergent behavior.

It is important for students to grasp that this interval is the safe zone where operations like integration and differentiation can be performed on the series term by term without worrying about the convergence issues.

Term-by-Term Integration

Term-by-term integration allows us to take a complex power series and integrate it as if we're dealing with a simple polynomial. This process is legal within the function's interval of convergence. Provided the series \( \sum_{n=0}^{\infty} a_n x^n \) converges for \( x \) in some interval, we can integrate it term by term over any interval that lies entirely within the convergence interval.

Here's how we approach it: if we want to integrate from \( 0 \) to \( 1 \) for the power series given in the exercise, we integrate each term \( a_n x^n \) to get \( \frac{a_n}{n+1} x^{n+1} \), and since we're integrating within the radius of convergence, we won't face divergence issues. Summing these integrated terms gives us the series \( \sum_{n=0}^{\infty} \frac{a_n}{n+1} \) as the integral of the original power series between \( 0 \) and \( 1 \). This maneuver simplifies otherwise complex integrations and highlights the power and usefulness of power series in solving integrals.