Q 60. - Calculus

Question

In exercises 59-62 concern the binomial series to find the maclaurin series for the given function .

$\frac{1}{(1+x{)}^2}$

Step-by-Step Solution

Answer

The maclaurin series for the given function is

$(1+x{)}^{-2}=1-2x+3x^2-4x^3+\cdots$

Step 1. Given information

We have been given 

$f\left(x\right)=\frac{1}{(1+x{)}^2}$

to find the maclaurin series by using binomial series 

Step 2.Defining the series

For any non- zero constant  p, the Maclaurin series for the function  $g\left(x\right)=(1+x{)}^p$ is called the binomial series which is given by  $\sum _{k=0}^{\mathrm{\infty }} \left(\begin{array}{l}p\\ k\end{array}\right)x^k$  where the binomial coefficient is

 

$\left(\begin{array}{l}p\\ k\end{array}\right)=\left\{\begin{array}{r}\frac{p(p-1)(p-2)\cdots (p-k+1)}{k!}\text{ if }k>0\\ 1,\text{ if }k=0\end{array}\right.$

Step 3. Binomial series for the given function is

So for the function $f\left(x\right)=\frac{1}{(1+x{)}^2}$, the binomial series is

$(1+x{)}^{-2}=\sum _{k=0}^{\mathrm{\infty }} \left(\begin{array}{l}-2\\ k\end{array}\right)x^k$

implies that ,

$\begin{array}{r}(1+x{)}^{-2}=\left(\begin{array}{c}-2\\ 0\end{array}\right)x^0+\left(\begin{array}{c}-2\\ 1\end{array}\right)x^1+\left(\begin{array}{c}-2\\ 2\end{array}\right)x^2+\left(\begin{array}{c}-2\\ 3\end{array}\right)x^3+\cdots \\ =1-2x+\frac{-2(-3)}{2!}{x}^2+\frac{-2(-3)(-4)}{3!}{x}^3+\cdots \\ =1-2x+3x^2-4x^3+\cdots \end{array}$

Step 4. The maclaurin series for given function is

The maclaurin series of given function $f\left(x\right)=\frac{1}{(1+x{)}^2}$ is