Problem 79 - Calculus Early Transcendentals

Question

Find a power series that has (2,6) as an interval of convergence.

Step-by-Step Solution

Answer

Question: Find a power series with an interval of convergence containing (2,6). Answer: One possible power series with an interval of convergence containing (2,6) is \[ \sum_{n=0}^{\infty} \frac{1}{2^n} (x-4)^n \]

Step 1: Choose the center of the power series

First, let's choose a center for the power series that is within the interval (2,6). A good choice is the midpoint of the interval, which is at x=4. Therefore, we can create the power series around this point.

Step 2: Define the formula of the power series

Now let's create a power series with its center at x=4, which looks like this: \[ \sum_{n=0}^{\infty} a_n (x-4)^n \] where \(a_n\) are the coefficients we need to determine.

Step 3: Apply the ratio test for convergence

To find the radius of convergence, we will use the ratio test. The ratio test states that if \(\lim_{n\to\infty} |\frac{a_{n+1}}{a_n}| < 1\), then the series converges. Applying the ratio test to our power series, we can get: \[ L =\lim_{n\to\infty} |\frac{a_{n+1}(x-4)^{n+1}}{a_n(x-4)^n}| = |(x-4)|\lim_{n\to\infty} |\frac{a_{n+1}}{a_n}| \]

Step 4: Define the radius of convergence

We want \(L < 1\) for the interval of convergence. This can be rewritten as: \[|x-4| < \frac{1}{\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|}\] The right side of this inequality represents the radius of convergence, \(R\). Thus, we have the inequality: \[|x-4| < R\]

Step 5: Calculate the endpoints and coefficients

The interval of convergence is between (2, 6), so the distance from the center to each endpoint is 2, meaning \(R = 2\). Now, choose any coefficients \(a_n\) for our power series that satisfy the ratio test, making them constant for simplicity. For example, let \(a_n = \frac{1}{2^n}\). Then we have: \[\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}| = \lim_{n\to\infty}|\frac{\frac{1}{2^{n+1}}}{\frac{1}{2^n}}| = \frac{1}{2} < 1\]

Step 6: Write down the final power series

With the given coefficients, the power series becomes: \[ \sum_{n=0}^{\infty} \frac{1}{2^n} (x-4)^n \] This power series has (2,6) as its interval of convergence, as required in the exercise.