example video example express 1/(1 - x)^2 as a power series by differentiating the equation below. what is…

Explanation:

Step1: Differentiate the power - series

The power - series $\frac{1}{1 - x}=\sum_{n = 0}^{\infty}x^{n}=1 + x+x^{2}+x^{3}+\cdots$ for $|x|\lt1$. Differentiating term - by - term using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $\frac{d}{dx}(\frac{1}{1 - x})=\frac{1}{(1 - x)^{2}}$ and $\frac{d}{dx}(1 + x+x^{2}+x^{3}+\cdots)=1+2x + 3x^{2}+\cdots=\sum_{n = 1}^{\infty}nx^{n - 1}$.

Step2: Rewrite the series

If we replace $n$ by $n + 1$ in $\sum_{n = 1}^{\infty}nx^{n - 1}$, let $m=n - 1$, then $n=m + 1$. When $n = 1$, $m = 0$. So $\sum_{n = 1}^{\infty}nx^{n - 1}=\sum_{m=0}^{\infty}(m + 1)x^{m}$.

Step3: Determine the radius of convergence

According to the theorem, the radius of convergence of a differentiated power - series is the same as the radius of convergence of the original power - series. Since the radius of convergence of $\frac{1}{1 - x}=\sum_{n = 0}^{\infty}x^{n}$ is $R = 1$, the radius of convergence of $\frac{1}{(1 - x)^{2}}=\sum_{n = 0}^{\infty}(n + 1)x^{n}$ is also $R = 1$.

Answer:

The power - series for $\frac{1}{(1 - x)^{2}}$ is $\sum_{n = 0}^{\infty}(n + 1)x^{n}$, and the radius of convergence $R = 1$. In the blanks:

• The first blank: $2x$
• The second blank: $nx^{n - 1}$
• The third blank: $(n + 1)x^{n}$
• The fourth blank: $1$