Question

How do you find the maclaurin series expansion of $f\left( x \right)={{\left( 1-x \right)}^{-2}}$ ?

Answer 1

Hint : Maclaurin series are that type of expansion which consists of non-negative integer power of the variable. In this question, we are going to write the geometric series for $f\left( x \right)={{\left( 1-x \right)}^{-1}}$ and then on further simplifying, it is very easily calculated for $f\left( x \right)={{\left( 1-x \right)}^{-2}}$ .
For a geometric progression series,

& f\left( x \right)=1+x+{{x}^{2}}+.... \\\ & \sum\limits_{n=0}^{n=\infty }{{{x}^{n}}=\dfrac{1}{1-x}} \\\ \end{aligned}$$ And hence the series can be found. However the expression for a maclaurin series is given by $$\sum\limits_{n=0}^{n=\infty }{\left( {{f}^{\left( n \right)}}\left( 0 \right)\dfrac{{{x}^{n}}}{\left| \\!{\underline {\, n \,}} \right. } \right)=}f\left( 0 \right)+{{f}^{'}}\left( 0 \right)x+{{f}^{''}}\left( 0 \right){{x}^{2}}+......+\dfrac{{{f}^{k}}\left( 0 \right)}{\left| \\!{\underline {\, k \,}} \right. }{{x}^{k}}+...$$ **Complete step-by-step answer** : In order to solve the question, we need to know what a maclaurin’s expansion is. Now maclaurin series is that type of expansion which consists of non-negative integer power of the variable. Moreover ,the expression for a maclaurin series is given by the following equation: $$\sum\limits_{n=0}^{n=\infty }{\left( {{f}^{\left( n \right)}}\left( 0 \right)\dfrac{{{x}^{n}}}{\left| \\!{\underline {\, n \,}} \right. } \right)=}f\left( 0 \right)+{{f}^{'}}\left( 0 \right)x+{{f}^{''}}\left( 0 \right){{x}^{2}}+......+\dfrac{{{f}^{k}}\left( 0 \right)}{\left| \\!{\underline {\, k \,}} \right. }{{x}^{k}}+...$$ So, let’s first of all write which function’s expansion we need to find, i.e. $$f\left( x \right)={{\left( 1-x \right)}^{-2}}$$ Now taking the geometric series for the function $${{x}^{n}}$$ $$\sum\limits_{n=0}^{n=\infty }{{{x}^{n}}=\dfrac{1}{1-x}}$$ Now as we know that the derivative of $$\dfrac{1}{1-x}$$ is $$\dfrac{1}{{{\left( 1-x \right)}^{2}}}$$ , $$\dfrac{1}{{{\left( 1-x \right)}^{2}}}=\dfrac{d}{dx}\left( \dfrac{1}{1-x} \right)$$ Replacing the term $$\left( \dfrac{1}{1-x} \right)$$ with its equivalent We get, $$\dfrac{1}{{{\left( 1-x \right)}^{2}}}=\dfrac{d}{dx}\left( \sum\limits_{n=0}^{n=\infty }{{{x}^{n}}} \right)$$ Now as we know that the interval for convergence is $$\left| x \right|<1$$ And inside the interval of convergence, we can easily differentiate the summation term by term $$\dfrac{1}{{{\left( 1-x \right)}^{2}}}=\sum\limits_{n=0}^{n=\infty }{\dfrac{d}{dx}}\left( {{x}^{n}} \right)=\sum\limits_{n=1}^{n=\infty }{n{{x}^{n-1}}}$$ Now changing the index of summation to $$n=0$$ $$\dfrac{1}{{{\left( 1-x \right)}^{2}}}=\sum\limits_{n=0}^{\infty }{\left( n+1 \right){{x}^{n}}}$$ Therefore, the maclaurin series expansion for $$f\left( x \right)={{\left( 1-x \right)}^{-2}}$$ is $$\sum\limits_{n=0}^{\infty }{\left( n+1 \right){{x}^{n}}}$$ **Note** : Alternatively, this can also be solved by taking the binomial expansion We know that the binomial series tells that $${{\left( 1+x \right)}^{n}}=1+nx+\dfrac{n\left( n-1 \right)}{\left| \\!{\underline {\, 2 \,}} \right. }{{x}^{2}}+\dfrac{n\left( n-1 \right)\left( n-2 \right)}{\left| \\!{\underline {\, 3 \,}} \right. }{{x}^{3}}+....$$ Now for the given function, we can replace $$x$$ by $$-x$$ and $$n$$ by $$-2$$ Then, we get the given function as $$\begin{aligned} & {{\left( 1+x \right)}^{-2}}=1+\left( -2 \right)\left( -x \right)+\dfrac{\left( -2 \right)\left( -3 \right)}{\left| \\!{\underline {\, 2 \,}} \right. }{{\left( -x \right)}^{2}}+\dfrac{\left( -2 \right)\left( -3 \right)\left( -4 \right)}{\left| \\!{\underline {\, 3 \,}} \right. }{{\left( -x \right)}^{3}}+\dfrac{\left( -2 \right)\left( -3 \right)\left( -4 \right)\left( -5 \right)}{\left| \\!{\underline {\, 4 \,}} \right. }{{\left( -x \right)}^{4}}+... \\\ & \Rightarrow {{\left( 1+x \right)}^{-2}}=1+2x+3{{x}^{2}}+4{{x}^{3}}+5{{x}^{4}}+.........+n{{x}^{n-1}}+.... \\\ \end{aligned}$$ Therefore, we get the same form for function $$f\left( x \right)={{\left( 1-x \right)}^{-2}}$$ as $$\sum\limits_{n=0}^{\infty }{\left( n+1 \right){{x}^{n}}}$$