Problem 29 - Calculus Early Transcendentals

Question

Use the geometric series $$f(x)=\frac{1}{1-x}=\sum_{k=0}^{\infty} x^{k}, \quad \text { for }|x|<1$$ to find the power series representation for the following functions (centered at 0 ). Give the interval of convergence of the new series. $$f(3 x)=\frac{1}{1-3 x}$$

Step-by-Step Solution

Answer

Answer: The power series representation of f(3x) is $$f(3x) = \sum_{k=0}^{\infty} 3^k x^k$$ and its interval of convergence is $$- \frac{1}{3} < x < \frac{1}{3}$$

Step 1: Substitute 3x into the series representation for f(x)

Replace x with 3x in the given geometric series representation for f(x): $$f(3x) = \frac{1}{1 - 3x} = \sum_{k=0}^{\infty} (3x)^k$$

Step 2: Simplify the series representation

Simplify the given series by applying the exponent to both the base 3 and x separately: $$f(3x) = \sum_{k=0}^{\infty} 3^k x^k$$

Step 3: Find the interval of convergence for the new series

To determine the interval of convergence, we need to find the values of x for which the series converges. In the original geometric series, |x| < 1 for it to converge. So, for our new series, we would have: $$|3x| < 1$$ Now, divide both sides by 3 to solve for x: $$|x| < \frac{1}{3}$$ This result states that the interval of convergence for the power series representation of f(3x) is: $$- \frac{1}{3} < x < \frac{1}{3}$$