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,
 \[\begin{align}
  & f\left( x \right)=1+x+{{x}^{2}}+.... \\
 & \sum\limits_{n=0}^{n=\infty }{{{x}^{n}}=\dfrac{1}{1-x}} \\
\end{align}\]
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
 \[\begin{align}
  & {{\left( 1+x \right)}^{-2}}=1+\left( -2 \right)\left( -x \right)+\dfrac{\left( -2 \right)\left( -3 \right)}{\left| \!{\underline {\,
  2 \,}} \right. }{{\left( -x \right)}^{2}}+\dfrac{\left( -2 \right)\left( -3 \right)\left( -4 \right)}{\left| \!{\underline {\,
  3 \,}} \right. }{{\left( -x \right)}^{3}}+\dfrac{\left( -2 \right)\left( -3 \right)\left( -4 \right)\left( -5 \right)}{\left| \!{\underline {\,
  4 \,}} \right. }{{\left( -x \right)}^{4}}+... \\
 & \Rightarrow {{\left( 1+x \right)}^{-2}}=1+2x+3{{x}^{2}}+4{{x}^{3}}+5{{x}^{4}}+.........+n{{x}^{n-1}}+.... \\
\end{align}\]
Therefore, we get the same form for function \[f\left( x \right)={{\left( 1-x \right)}^{-2}}\] as
 \[\sum\limits_{n=0}^{\infty }{\left( n+1 \right){{x}^{n}}}\]