Q. 7 - Calculus

Question

In Example 1 we used Theorem 8.11 to find the Maclaurin series for $\frac{1}{(1-x{)}^2}.$ Explain how Theorem 8.11 could be used to find the Maclaurin series for $\frac{1}{(1-x{)}^2}$, where $k$ is a positive integer greater than $2$.

Step-by-Step Solution

Answer

Since $\frac{d^{k-1}}{d{x}^{k-1}}\left(\frac{1}{1-x}\right)=\frac{(k-1)\text{ ! }}{(1-x{)}^k},$ so to find the Maclaurin series for$\frac{1}{(1-x{)}^k},$ we  take the$(k-1{)}^n$ derivative of the series $\frac{1}{1-x}$term by term and divide each term by $k !$

Step :1 Given Information

Given function :$\frac{1}{(1-x{)}^2}$

Step 2 : Explaining how Theorem 8.11 could be used to find the Maclaurin series

The Maclaurin series for $\frac{1}{1-x}$ is $\boxed{\sum _{k=0}^{\infty }{x}^k}$

So, the Maclaurin series for $\frac{1}{(1-x{)}^2}$ can be found by simply differentiating the Maclaurin series for $\frac{1}{1-x}$.

This is because $\frac{d}{dx}\left(\frac{1}{1-x}\right)=\frac{1}{(1-x{)}^2}$

Similarly, since $\frac{d^{k-1}}{d{x}^{k-1}}\left(\frac{1}{1-x}\right)=\frac{(k-1)\text{ ! }}{(1-x{)}^k}$, so to find the Maclaurin series for $\frac{1}{(1-x{)}^k}$, we could take the ${(k-1)}^n$ derivative of the series $\frac{1}{1-x}$ term by term and divide each term by $k!$