Problem 30 - 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. $$g(x)=\frac{x^{3}}{1-x}$$

Step-by-Step Solution

Answer

Question: Determine the power series representation of the function \(g(x) = \frac{x^3}{1-x}\) centered at 0 and its interval of convergence. Answer: The power series representation of \(g(x)\) is \(g(x) = \sum_{k=0}^{\infty} x^{2+3k}\), and the interval of convergence is \(|x|<1\).

Step 1: Find the relationship between g(x) and f(x)

First, we need to express \(g(x)\) in terms of \(f(x)\). Looking at the given functions, we can see that by multiplying both the numerator and denominator of \(g(x)\) by \(x^2\), we can get the relationship as follows: $$g(x) = x^2 f(x^3)$$

Step 2: Substitute the power series representation of f(x) into g(x)

Now, we can substitute the power series representation of \(f(x)\) into the expression for \(g(x)\). $$g(x) = x^2 \sum_{k=0}^{\infty} (x^3)^{k}$$

Step 3: Simplify the power series representation#g(x)

Simplify the expression by combining the terms involving \(x\): $$g(x) = \sum_{k=0}^{\infty} x^{2+3k}$$

Step 4: Determine the interval of convergence for the new series

Since the given series for \(f(x)\) converges for \(|x|<1\), we want to find the interval of convergence for \(f(x^3)\), or when \(|x^3|<1\). Taking the cube root of both sides, we can see that the interval of convergence remains the same: \(|x|<1\). So, the power series representation for \(g(x)\) is $$g(x) = \sum_{k=0}^{\infty} x^{2+3k}$$ and the interval of convergence is \(|x|<1\).