Question

How do you find the power series representation of $\dfrac{\left( 1+x \right)}{{{\left( 1-x \right)}^{2}}}$ ?

Answer 1

To solve the above question we will first write the expansion of the term ${{\left( 1-x \right)}^{-1}}$as ${{\left( 1-x \right)}^{-1}}=1+x+{{x}^{2}}+.......$ and since we can write ${{\left( 1-x \right)}^{-1}}$ as $\dfrac{1}{\left( 1-x \right)}$ so, we will differentiate the obtained expansion to obtain the square in the denominator and then we will multiply with x on both the side and then we will get the expansion of the terms $\dfrac{1}{{{\left( 1-x \right)}^{2}}}and\dfrac{x}{{{\left( 1-x \right)}^{2}}}$ both and hence, to obtain the expansion of $\dfrac{\left( 1+x \right)}{{{\left( 1-x \right)}^{2}}}$ we will add the expansion of $\dfrac{1}{{{\left( 1-x \right)}^{2}}}and\dfrac{x}{{{\left( 1-x \right)}^{2}}}$ both.

Complete step by step answer:
We can see from question that we have to write the power series representation of $\dfrac{\left( 1+x \right)}{{{\left( 1-x \right)}^{2}}}$, which means we have to write the expansion of $\dfrac{\left( 1+x \right)}{{{\left( 1-x \right)}^{2}}}$.
Since, we know that the expansion of ${{\left( 1-x \right)}^{-1}}$ is given as ${{\left( 1-x \right)}^{-1}}=1+x+{{x}^{2}}+.......\text{up to infinite terms}\text{.}$
And, we can write ${{\left( 1-x \right)}^{-1}}$ as $\dfrac{1}{\left( 1-x \right)}$.
So, we will get:
$\Rightarrow \dfrac{1}{\left( 1-x \right)}=1+x+{{x}^{2}}+.......\text{up to infinite terms}\text{.}$
Now, after differentiating both the side of the above expansion we will get:
$\Rightarrow \dfrac{d\left( \dfrac{1}{\left( 1-x \right)} \right)}{dx}=\dfrac{d\left( 1+x+{{x}^{2}}+....... \right)}{dx}$
Now, we know that $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$, and $\dfrac{d\left( A+B \right)}{dx}=\dfrac{dA}{dx}+\dfrac{dB}{dx}$
$\Rightarrow -1\times {{\left( 1-x \right)}^{-1-1}}\times \left( -1 \right)=\dfrac{d\left( 1 \right)}{dx}+\dfrac{dx}{dx}+\dfrac{d{{x}^{2}}}{dx}+\dfrac{d{{x}^{3}}}{dx}+......$
$\Rightarrow \dfrac{1}{{{\left( 1-x \right)}^{2}}}=1+2x+3{{x}^{2}}+......$
$\Rightarrow \dfrac{1}{{{\left( 1-x \right)}^{2}}}=1+2x+3{{x}^{2}}+.......\text{up to infinite terms}\text{. ----}\left( 1 \right)$
Now, after multiplying both side of the above equation (1) by x, we will get:
$\Rightarrow \dfrac{x}{{{\left( 1-x \right)}^{2}}}=x\left( 1+2x+3{{x}^{2}}+.......\text{up to infinite terms}\text{.} \right)$
$\Rightarrow \dfrac{x}{{{\left( 1-x \right)}^{2}}}=\left( x+2{{x}^{2}}+3{{x}^{3}}-.......\text{up to infinite terms}\text{.} \right)---\left( 2 \right)$
Now, after adding equation (1) and (2) we will get:
$\begin{aligned} & \Rightarrow \dfrac{1}{{{\left( 1-x \right)}^{2}}}+\dfrac{x}{{{\left( 1-x \right)}^{2}}}=\left( 1+2x+3{{x}^{2}}+.......\text{up to infinite terms}\text{.} \right) \\\ & +\left( x+2{{x}^{2}}+3{{x}^{3}}+.......\text{up to infinite terms}\text{.} \right) \\\ \end{aligned}$
Now, after simplifying the LHS , we will get:
$\Rightarrow \dfrac{1+x}{{{\left( 1-x \right)}^{2}}}=\left( 1+2x+3{{x}^{2}}+.......\text{up to infinite terms}\text{.} \right)+\left( x+2{{x}^{2}}+3{{x}^{3}}+.......\text{up to infinite terms}\text{.} \right)$
$\Rightarrow \dfrac{1+x}{{{\left( 1-x \right)}^{2}}}=\left( 1+2x+3{{x}^{2}}+.......\text{up to infinite terms}\text{.} \right)+\left( x+2{{x}^{2}}+3{{x}^{3}}+.......\text{up to infinite terms}\text{.} \right)$
Now, after adding the corresponding similar term of the equation, we will get:
$\Rightarrow \dfrac{1+x}{{{\left( 1-x \right)}^{2}}}=\left( 1+3x+5{{x}^{2}}+7{{x}^{3}}.......\text{up to infinite terms} \right)$
Hence, the power series representation of $\dfrac{\left( 1+x \right)}{{{\left( 1-x \right)}^{2}}}$ is given as: $\left( 1+3x+5{{x}^{2}}+7{{x}^{3}}.......\text{up to infinite terms} \right)$
This is the required solution.

Note: When we add two series which consists of the infinite terms then we add coefficient terms with the same power terms. Also, note that that expansion of ${{\left( 1+x \right)}^{n}}$, where n is negative or fraction is given as: ${{\left( 1+x \right)}^{n}}=1+nx+\dfrac{n\left( n-1 \right)}{2!}{{x}^{2}}+\dfrac{n\left( n-1 \right)\left( n-2 \right)}{3!}{{x}^{3}}+......$.