Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the convergence set for the given power series. Hint: First find a formula for the nth term; then use the Absolute Ratio Test. $$ x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !}+\frac{x^{9}}{9 !}-\cdots $$

Step-by-Step Solution

Answer

The convergence set is \((-\infty, \infty)\).

Step 1: Identify the General Term

First, let's identify the general term for the given power series. Observe the pattern in the terms: the term has alternating signs, only odd powers of \(x\), and the denominator includes factorials of odd numbers. Thus, the general term \(a_n\) can be expressed as:\[ a_n = (-1)^n \frac{x^{2n+1}}{(2n+1)!} \]

Step 2: Apply the Absolute Ratio Test

Next, apply the Absolute Ratio Test to find the radius of convergence. The Absolute Ratio Test involves calculating:\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]Substituting our expression for \(a_n\):\[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(-1)^{n+1} \frac{x^{2(n+1)+1}}{(2(n+1)+1)!}}{(-1)^n \frac{x^{2n+1}}{(2n+1)!}} \right| = \frac{|x|^{2n+3}}{(2n+3)!} \times \frac{(2n+1)!}{|x|^{2n+1}} \]Simplifying, we get:\[ |x|^2 \frac{1}{(2n+3)(2n+2)} \]

Step 3: Evaluate the Limit

Evaluate the limit of the expression obtained in Step 2:\[ \lim_{n \to \infty} |x|^2 \frac{1}{(2n+3)(2n+2)} = 0 \]Since the limit is 0 for any finite value of \(|x|\), the series converges for all \(x\).

Step 4: State the Convergence Set

Based on the Absolute Ratio Test applied to the series, the series converges for all values of \(x\) because the limit derived from the Ratio Test is 0 for all \(x\). Thus, the power series converges everywhere on the real line \((-\infty, \, \infty)\).

Key Concepts

Convergence Set

A convergence set refers to the set of all values for which a given power series converges. In the context of this power series, our aim is to determine where the series converges by using a method such as the Absolute Ratio Test. For the series:

• Starting with the formula for the general term: \[ a_n = (-1)^n \frac{x^{2n+1}}{(2n+1)!} \]

We observe that by employing the Absolute Ratio Test, the series converges not just for some, but for all values of \(x\), due to a global convergence solution where the limit evaluates to zero. Thus, the series converges everywhere on the real number line, which means the convergence set is \((-\infty, \, \infty)\). This is a special property that not all power series possess.

Absolute Ratio Test

The Absolute Ratio Test is a powerful tool for analyzing the convergence of a series. It provides a way to find the radius of convergence, which is crucial for determining the convergence set. The test proceeds by considering the limit:\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]The general term derived here is:\[ a_n = (-1)^n \frac{x^{2n+1}}{(2n+1)!} \]By substituting into the ratio test expression, we obtain:\[\left| \frac{a_{n+1}}{a_n} \right| = |x|^2 \frac{1}{(2n+3)(2n+2)}\]As \(n\) tends towards infinity, the fraction tends towards zero. Thus, the power series converges for any value of \(x\). The result from this test showcases the use of simplification and factorial manipulation in determining convergence properties.

Factorials

Factorials are mathematical expressions that represent the product of all positive integers up to a certain number. They are denoted and calculated as \(n! = n \cdot (n-1) \cdot (n-2) \cdots 1\). In this series, factorials appear in the denominator of each term involving odd numbers:

• For the generalized term: \[ (2n+1)! \]
• This expression grows extremely fast as \(n\) increases, causing terms with higher \(n\) to diminish quickly, which strongly impacts convergence assessments.

Factorials are crucial in controlling the growth of the terms within a series, thereby influencing convergence greatly. They ensure that for large \(n\), the terms shrink rapidly enough for the series to be potentially convergent.

Alternating Series

Alternating series are characterized by terms that switch signs between positive and negative. This is evident in expressions such as:

• The term \((-1)^n\) controls the switch in sign from term to term.

In the power series:\[x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\]The alternating nature helps in convergence analysis. Alternating series often have the potential to converge where non-alternating series do not, because the alternating signs lead to cancelation effects of terms. Importantly, even if individual terms do not decay rapidly enough, the alternating nature can still ensure that the sum of terms approaches a finite value. This property complements the results found with tests like the Absolute Ratio Test.