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: \((-4, 4)\); b. New interval: \((-1, 7)\)

Step 1: Identify the Power Series

The given power series is \( \sum_{n=0}^{\infty} \frac{8}{4^{n+2}} x^n \). Simplify the coefficient \( \frac{8}{4^{n+2}} \). We have \( \frac{8}{4^2} = \frac{1}{2} \), so the term becomes \( \frac{1}{2} \left(\frac{1}{4}\right)^n \). The power series can be rewritten as \( \sum_{n=0}^{\infty} \frac{1}{32} \left( \frac{x}{4} \right)^n \).

Step 2: Use the Ratio Test for Convergence

The ratio test will help find the interval of convergence:Calculate the ratio between successive terms:\[\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{1}{32} \left( \frac{x}{4} \right)^{n+1}}{\frac{1}{32} \left( \frac{x}{4} \right)^n} \right| = \left| \frac{x}{4} \right|.\] For convergence, \( \left| \frac{x}{4} \right| < 1 \).

Step 3: Solve the Inequality for Convergence

Solve the inequality from the ratio test:\[\left| \frac{x}{4} \right| < 1 \Rightarrow -1 < \frac{x}{4} < 1\Rightarrow -4 < x < 4.\]The series converges for \( x \) in the interval \((-4, 4)\).

Step 4: Check Endpoint Convergence

Testing the endpoints, \( x = -4 \) and \( x = 4 \):For \( x = -4 \), the series becomes \( \sum_{n=0}^{\infty} (-1)^n \frac{1}{32} \), which diverges.For \( x = 4 \), the series becomes \( \sum_{n=0}^{\infty} \frac{1}{32} \), which also diverges.Thus, the interval of convergence is \((-4, 4)\).

Step 5: Translate the Series About \( x=3 \)

We want the series centered about \( x=3 \). Set \( x = 3 + u \), so \( u = x - 3 \).The series is \( \sum_{n=0}^{\infty} \frac{1}{32} \left( \frac{u+3}{4} \right)^n \). Expand it using binomial expansion if necessary and adjust it into a standard power series form.

Step 6: Determine New Interval of Convergence

Applying the previous convergence criteria using \( u \): For \( u = x - 3 \), we have the condition \( \left| \frac{x-3}{4} \right| < 1 \).Solving this gives:\[-4 < x-3 < 4 \Rightarrow -1 < x < 7.\]Therefore, the interval of convergence for the series centered at 3 is \((-1, 7)\).

Key Concepts

Interval of Convergence

The interval of convergence of a power series tells us the range of values for which the series converges. Convergence means the series reaches a finite sum, even though there are infinitely many terms. It's crucial to know the interval because outside it, the series may not behave as expected.

To find the interval, we often rely on the ratio test, which gives a simple inequality. Solving this inequality helps us determine the interval. However, the interval can sometimes include just the open interval, or extend to include one or both endpoints depending on further testing. Checking the endpoints separately is key as convergence at these points cannot be determined solely by the ratio test.

• The given series converged for \( x \) values in the interval \((-4, 4)\).
• Testing endpoints ensures if the series converges or diverges there.
• In this case, both endpoints \( x = -4 \) and \( x = 4 \) result in divergence.

Ratio Test

The ratio test helps us find when a power series converges by comparing the size of successive terms. It's a convenient tool for power series because it simplifies the decision about convergence.

We calculate the absolute value of the ratio of successive terms \(a_{n+1}/a_n\), and examine if it approaches a value less than one. If it does, the series converges. If it’s greater than one, the series diverges, and if it equals one, the test is inconclusive.

• For our series, the ratio test condition \( \left| \frac{x}{4} \right| < 1 \) leads to the inequality which defines convergence.
• This resulted in a convergence interval of \((-4, 4)\) under the series centered at zero.

Series Expansion

Series expansion is the process of expressing functions as infinite sums of simpler terms. For power series, it often involves polynomial-like terms being summed up.

In the given exercise, the series is expanded using terms involving powers of \( (x/4) \). Such expansions help us approximate functions and understand their behavior. Sometimes, it's necessary to translate the expansion to be centered at another value, which involves changing the variable correctly.

• The original series is written as \( \sum_{n=0}^{\infty} \frac{1}{32} \left( \frac{x}{4} \right)^n \).
• Translating the series to center around \( x=3 \) involves expressing it in terms of \( u = x-3 \).

Center of Convergence

The center of convergence refers to the value around which a power series is centered. It is like the 'origin' or starting point from which the series expansion is tracked.

Initially, the given series is centered at \( x=0 \). However, when the expansion needs to be around a different point, we shift our perspective. Centering at \( x=3 \) involves a change in variable substitutions where \( x \) is replaced by \( 3 + u \) (with \( u = x-3 \)), transforming the whole series.

• This re-centering changes the interval of convergence, showing how the same function can behave differently when checked from another point.
• The new interval for the series centered at \( x=3 \) is \((-1, 7)\).