Question

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

Answer 1

Hint: Maclaurin series is just a special case of Taylor series centered at $x=0$. It is given as $f\left( 0 \right)+\dfrac{{f}'\left( 0 \right)}{1!}x+\dfrac{{f}''\left( 0 \right)}{2!}{{x}^{2}}+\dfrac{{f}'''\left( 0 \right)}{3!}{{x}^{3}}+........\infty $ where, ${f}'\left( 0 \right),{f}''\left( 0 \right),{f}'''\left( 0 \right)$ are the first, second and third derivatives of the function$f\left( x \right)$ at $x=0$. Here, $1!,2!,3!$ are the factorials of 1,2 and 3 respectively. Therefore, we shall calculate a few terms and then observe the pattern occurring to find our final solution.

Complete step-by-step solution:
This series follows a pattern consisting of the derivative of the function, x raised to a certain power and the factorial of the number.
Let us differentiate the function $f\left( x \right)$ now. We shall use the property of differentiation given as $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ several times to differentiate the function.
$\begin{align}
  & \Rightarrow {f}'\left( x \right)=\dfrac{d}{dx}\left( \dfrac{1}{{{\left( 1+x \right)}^{2}}} \right) \\
 & \Rightarrow {f}'\left( x \right)=\dfrac{-2}{{{\left( 1+x \right)}^{3}}}\dfrac{d}{dx}\left( 1+x \right) \\
\end{align}$
$\Rightarrow {f}'\left( x \right)=\dfrac{-2}{{{\left( 1+x \right)}^{3}}}$ …………….. (1)
Again differentiating the function with respect to x, we get:
 $\Rightarrow {f}''\left( x \right)=\dfrac{d}{dx}{f}'\left( x \right)$
$\begin{align}
  & \Rightarrow {f}''\left( x \right)=\dfrac{d}{dx}\left( \dfrac{-2}{{{\left( 1+x \right)}^{3}}} \right) \\
 & \Rightarrow {f}''\left( x \right)=-2.\left( \dfrac{-3}{{{\left( 1+x \right)}^{4}}} \right)\dfrac{d}{dx}\left( 1+x \right) \\
\end{align}$
$\Rightarrow {f}''\left( x \right)=-2.\left( \dfrac{-3}{{{\left( 1+x \right)}^{4}}} \right)1$
$\Rightarrow {f}''\left( x \right)=\dfrac{6}{{{\left( 1+x \right)}^{4}}}$ …………………… (2)
Third time differentiating the function with respect to x, we get
 $\begin{align}
  & \Rightarrow \dfrac{d}{dx}{f}''\left( x \right)=\dfrac{d}{dx}\dfrac{6}{{{\left( 1+x \right)}^{4}}} \\
 & \Rightarrow {f}'''\left( x \right)=6\dfrac{d}{dx}\dfrac{1}{{{\left( 1+x \right)}^{4}}} \\
\end{align}$
$\Rightarrow {f}'''\left( x \right)=6\left( \dfrac{-4}{{{\left( 1+x \right)}^{5}}} \right)\dfrac{d}{dx}\left( 1+x \right)$
$\Rightarrow {f}'''\left( x \right)=\left( \dfrac{-24}{{{\left( 1+x \right)}^{5}}} \right)1$
$\Rightarrow {f}'''\left( x \right)=\dfrac{-24}{{{\left( 1+x \right)}^{5}}}$ ………………. (3)
From (1), (2) and (3), we substitute the value of x equals to zero to calculate the value of the first, second and third derivative of function at point $x=0$.
$\Rightarrow {f}'\left( 0 \right)=\dfrac{-2}{{{\left( 1+0 \right)}^{3}}}$
 $\Rightarrow {f}'\left( 0 \right)=\dfrac{-2}{1}=-2$
Now, ${f}''\left( 0 \right)=\dfrac{6}{{{\left( 1+0 \right)}^{4}}}$
$\Rightarrow {f}''\left( x \right)=6$
And, ${f}'''\left( 0 \right)=\dfrac{-24}{{{\left( 1+0 \right)}^{5}}}$
$\Rightarrow {f}'''\left( x \right)=-24$
Also, $f\left( 0 \right)=\dfrac{1}{{{\left( 1+0 \right)}^{2}}}$
$\Rightarrow f\left( 0 \right)=1$
Hence, the value of the first, second and third derivative of function at point $x=0$ is -2, 6 and -18 respectively. We analyze these results and find that they follow a particular pattern.
The McLaurin series is given as $f\left( 0 \right)+\dfrac{{f}'\left( 0 \right)}{1!}x+\dfrac{{f}''\left( 0 \right)}{2!}{{x}^{2}}+\dfrac{{f}'''\left( 0 \right)}{3!}{{x}^{3}}+........\infty $.
Substituting the values in this equation, we get
 $1+\dfrac{-2}{1}x+\dfrac{6}{2\times 1}{{x}^{2}}+\dfrac{-24}{3\times 2\times 1}{{x}^{3}}+........\infty $
Therefore, the MacLaurin series is given as $1-\dfrac{2}{1!}x+\dfrac{6}{2!}{{x}^{2}}-\dfrac{24}{3!}{{x}^{3}}+........\infty $

Note: We must have prior knowledge of basic differentiation in order to apply the basic properties like chain rule of differentiation while dealing with mathematical problems. The calculations in this problem must be done carefully and all the negative signs must be taken care while performing rigorous multiplication.