Problem 31 - 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. $$h(x)=\frac{2 x^{3}}{1-x}$$

Step-by-Step Solution

Answer

## Short Answer The power series representation for the function \(h(x)=\frac{2 x^{3}}{1-x}\) is given by \(h(x) = \sum_{k=0}^{\infty} 2x^{k+3}\) and the interval of convergence is \(|x| < 1\).

Step 1: Manipulate the geometric series to match h(x)

First, let's multiply the given geometric series representation by \(2x^3\) to match the given function \(h(x)\): $$2x^3 \cdot f(x) = 2x^3 \cdot \sum_{k=0}^{\infty} x^{k}$$ Now, let's perform the multiplication: $$2x^3 \cdot f(x) = \sum_{k=0}^{\infty} 2x^{k+3}$$ So, we have the power series representation for \(h(x)\): $$h(x) = \sum_{k=0}^{\infty} 2x^{k+3}$$

Step 2: Find the interval of convergence

The original geometric series has an interval of convergence \(|x| < 1\). Since our function, \(h(x)\), is derived from this geometric series, its interval of convergence will remain the same: $$|x| < 1$$ However, to be more precise, we can use the Ratio Test for the power series if needed. For our case, we can see that the interval of convergence remains the same. Thus, the power series representation for \(h(x)\) is: $$h(x) = \sum_{k=0}^{\infty} 2x^{k+3}$$ with the interval of convergence: $$|x| < 1$$