Problem 52 - Thomas Calculus

Question

a. Find the interval of convergence of the power series. $$ \sum_{n=0}^{\infty} \frac{8}{4^{n+2}} x^{n} $$ b. Represent the power series in part (a) as a power series about \(x=3\) and identify the interval of convergence of the new series. (Later in the chapter you will understand why the new interval of convergence does not necessarily include all of the numbers in the original interval of convergence.)

Step-by-Step Solution

Answer

a: Interval of convergence is \((-4, 4)\). b: New series interval is \((2, 8)\).

Step 1: Simplify the General Term of the Series

The given power series is \( \sum_{n=0}^{\infty} \frac{8}{4^{n+2}} x^{n} \). To simplify the general term, note that \(4^{n+2} = 4^n \times 16\). Thus, the general term becomes \( \frac{8}{16} \times \frac{1}{4^n} \times x^n = \frac{1}{2} \times \left(\frac{x}{4}\right)^n \). Therefore, the power series becomes \( \sum_{n=0}^{\infty} \frac{1}{2} \left( \frac{x}{4} \right)^n \).

Step 2: Determine the Radius of Convergence

The series is a geometric series with common ratio \( r = \frac{x}{4} \). Remember the criteria for convergence of a geometric series: \(|r| < 1\). Thus, we need \( \left| \frac{x}{4} \right| < 1 \), leading to \(-1 < \frac{x}{4} < 1 \). Solving this inequality gives \(-4 < x < 4\). Thus, the radius of convergence is \( r = 4 \).

Step 3: Test the Endpoints for Convergence

We need to check if the series converges at the endpoints \(x = -4\) and \(x = 4\).- At \(x = -4\), the series becomes \( \sum_{n=0}^{\infty} (-1)^n \frac{1}{2} \), which diverges as it does not tend to zero.- At \(x = 4\), the series becomes \( \sum_{n=0}^{\infty} \frac{1}{2} \), which also diverges.Thus, the interval of convergence is \((-4, 4)\).

Step 4: Change the Power Series to Center at \(x = 3\)

To represent the series about \(x = 3\), use the substitution \( u = x - 3 \), such that \(x = u + 3\). Substituting into the original series gives the new series as \( \sum_{n=0}^{\infty} \frac{1}{2} \left( \frac{u+3}{4} \right)^n \). This shifts the center of the series to \(x = 3\).

Step 5: Determine the New Interval of Convergence

For the new series, the convergence criterion is \( \left| \frac{u+3}{4} \right| < 1 \). Solving \(-4 < u+3 < 4\), we find that \(-1 < u < 5\). Substituting back \( u = x - 3 \) gives \( -1 < x-3 < 5 \), which becomes \( 2 < x < 8 \) after solving.Therefore, the interval of convergence for the new series is \((2, 8)\).

Key Concepts

Power Series

A power series is a series of the form \( \sum_{n=0}^{\infty} a_n (x - c)^n \), where \( a_n \) are coefficients and \( c \) is the center of the series. It is similar to a polynomial, but it includes infinitely many terms. The variable \( x \) is raised to increasing integer powers and is multiplied by coefficients that can vary with each term.

One of the main interests in power series is determining the values of \( x \) for which the series converges, known as the interval of convergence. The expression \( (x-c)^n \) shows that the convergence depends on how "far" \( x \) is from \( c \). Depending on the nature of the series, \( x \) can be any number in a certain interval or just at specific points.

• They can represent functions in a localized manner, providing a useful tool in calculus and analysis.
• The convergence and divergence behavior of a power series is crucial for their application.

Geometric Series

A geometric series is a type of series where each term after the first is found by multiplying the previous one by a fixed, non-zero number known as the common ratio. Its general formula is \( a + ar + ar^2 + ar^3 + \dots \), which can be compactly rewritten as \( \sum_{n=0}^{\infty} ar^n \).

For a geometric series to converge, the absolute value of the common ratio \( r \) must be less than 1 (\(|r| < 1\)). If this condition is met, the series converges to the sum \( \frac{a}{1-r} \). However, if \(|r| \geq 1\), the series diverges.

• It is one of the simplest series and provides a foundational understanding of series behavior.
• The geometric series is particularly important in calculus, especially for understanding radii of convergence for power series.

Radius of Convergence

The radius of convergence of a power series is the distance from the center \( c \) within which the series converges. It is directly connected to the interval of convergence.

To find the radius of convergence, often the ratio or root tests are used, particularly the ratio test. For a power series \( \sum a_n (x - c)^n \), you apply the condition \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L \) to find the radius \( R = \frac{1}{L} \). In the context of geometric series, if \( r = \frac{x}{4} \), then ensuring \( |r| < 1 \) helps to find the radius by solving the inequality.

• The radius provides clarity on how far the series "reaches" in terms of convergence.
• Each endpoint must be tested separately to see if convergence occurs there, often using different methods like the alternating series test.

Endpoints Testing

Endpoints testing is a critical step to determine the exact interval of convergence. Once the radius of convergence is found, the endpoints of this interval may or may not include points where the series converges.

Testing these points involves substituting each endpoint into the series and analyzing the result:

• For each substitution, observe whether the series converges or diverges.
• This can involve using the test for divergence, ratio test, alternating series test, etc.

In the given problem, both endpoints at \( x = -4 \) and \( x = 4 \) led to divergent series:

• At \( x = -4 \), the terms alternated in sign, leading to no limit.
• At \( x = 4 \), the terms stayed constant, again showing divergence.

Understanding endpoint behavior refines the original conclusion we draw from the radius of convergence, ensuring the interval is correctly identified.