Question

Use the power series $$\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n} x^{n}$$ to determine a power series, centered at 0 , for the function. Identify the interval of convergence. $$ f(x)=\ln \left(1-x^{2}\right)=\int \frac{1}{1+x} d x-\int \frac{1}{1-x} d x $$

Short answer

The power series representation of the function \(f(x)=\ln \left(1-x^{2}\right)\) is \(\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{n+1} x^{n+1}\) and the interval of convergence is \(-1<x<1\)

Step-by-step solution

Power Series Expansion of the Integral Terms

Use power series representation to expand the integral terms in the given equation: \(\int \frac{1}{1+x} d x = \int \sum_{n=0}^{\infty}(-1)^{n} x^{n} d x\) and \(\int \frac{1}{1-x} d x = \int \sum_{n=0}^{\infty} x^{n} d x\)

Integrate the Power Series

Integrate the expanded power series from step 1 term-by-term: \(\int \frac{1}{1+x} d x = \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+1} x^{n+1}\) and \(\int \frac{1}{1-x} d x = \sum_{n=0}^{\infty}\frac{1}{n+1} x^{n+1}\)

Subtract the Series

Subtract the second power series from the first one, based on the representation of the function \(f(x)\): \(f(x) = \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+1} x^{n+1}-\sum_{n=0}^{\infty}\frac{1}{n+1} x^{n+1} = \sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{n+1} x^{n+1}\)

Determine the Interval of Convergence

The interval of convergence is the same as that of the original series, since the original series converges for \(-1<x<1\), the series of \(f(x)\) will also converge for \(-1<x<1\). Therefore, the interval of convergence is \(-1<x<1\)