Problem 32 - 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. $$f\left(x^{3}\right)=\frac{1}{1-x^{3}}$$

Step-by-Step Solution

Answer

Answer: The power series representation of the function $$f\left(x^{3}\right)=\frac{1}{1-x^{3}}$$ is $$\sum_{k=0}^{\infty} x^{3k}$$. Its interval of convergence is \(|x|<1\).

Step 1: Substitute \(x^3\) into the geometric series representation

Instead of using \(x\) in the given geometric series representation, we will use \(x^3\). So, the function will become:$$f\left(x^{3}\right)=\frac{1}{1-x^{3}}=\sum_{k=0}^{\infty} (x^{3})^{k}$$

Step 2: Simplify the power series representation

Now, we will simplify the exponent of \(x\) in the power series:$$\sum_{k=0}^{\infty} (x^{3})^{k} = \sum_{k=0}^{\infty} x^{3k}$$This is the power series representation of the function $$f\left(x^{3}\right)=\frac{1}{1-x^{3}}$$.

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

To find the interval of convergence for the new power series, we will use the fact that the interval of convergence for the original geometric series is \(|x|<1\). Since we replaced \(x\) with \(x^3\), we need to find the interval for \(x^3\). The inequality will be:$$|x^3|<1$$Taking the cube root of both sides, we get:$$|x|<1^{\frac{1}{3}}$$Since \(1^{\frac{1}{3}}=1\), the interval of convergence for the new power series representation is still \(|x|<1\). So, the power series representation for the function $$f\left(x^{3}\right)=\frac{1}{1-x^{3}}$$ is $$\sum_{k=0}^{\infty} x^{3k}$$ with an interval of convergence of \(|x|<1\).