Question

How to find the Maclaurin series and the radius of convergence for $f\left( x \right)=\dfrac{1}{{{\left( 1+x \right)}^{2}}}$?

Answer 1

In order to find the solution to the given question that is to find the maclaurin series and the radius of convergence for $f\left( x \right)=\dfrac{1}{{{\left( 1+x \right)}^{2}}}$, first use the concept that the Maclaurin series is given by $f\left( 0 \right)+\dfrac{{f}'\left( 0 \right)x}{1!}+\dfrac{{f}''\left( 0 \right){{x}^{2}}}{2!}+...+\dfrac{{{f}^{n}}\left( 0 \right){{x}^{n}}}{n!}$ and find the value of $f\left( 0 \right)$, ${f}'\left( 0 \right)$ and ${f}''\left( 0 \right)$. After that substitute these values in the formula of Maclaurin’s series and identify the pattern if it’s forming any power series. At last determine the radius of the convergence with the help of the ratio test for series that is Suppose $L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|$ then the ratio test states that: if L < 1 then the series converges absolutely; if L > 1 then the series is divergent; and if L = 1 or the limit fails to exist, then the test is inconclusive, because there exist both convergent and divergent series that satisfy this case.

Complete step by step solution:
According to the question, given function in the question is as follows:
$f\left( x \right)=\dfrac{1}{{{\left( 1+x \right)}^{2}}}$
In order to find the it’s Maclaurin’s series, use the concept that the Maclaurin series is given by $\Rightarrow f\left( 0 \right)+\dfrac{{f}'\left( 0 \right)x}{1!}+\dfrac{{f}''\left( 0 \right){{x}^{2}}}{2!}+...+\dfrac{{{f}^{n}}\left( 0 \right){{x}^{n}}}{n!}$
First start by finding the value of $f\left( 0 \right)$ on the first few derivatives, shown below:
$\Rightarrow f\left( 0 \right)=\dfrac{1}{{{\left( 1+0 \right)}^{2}}}=1$
After this find the derivative of ${f}'\left( x \right)$ to find the value of ${f}'\left( 0 \right)$
$\Rightarrow {f}'\left( x \right)=-\dfrac{2\left( x+1 \right)}{{{\left( x+1 \right)}^{4}}}$
$\Rightarrow {f}'\left( x \right)=\dfrac{-2}{{{\left( x+1 \right)}^{3}}}$
Now we find the value of ${f}'\left( 0 \right)$, by substituting $x=0$ in the above equation, we get:
$\Rightarrow {f}'\left( 0 \right)=\dfrac{-2}{{{\left( 0+1 \right)}^{3}}}=-2$
After this find the derivative of ${f}''\left( x \right)$ to find the value of ${f}''\left( 0 \right)$
$\Rightarrow {f}''\left( x \right)=\dfrac{6{{\left( x+1 \right)}^{2}}}{{{\left( x+1 \right)}^{6}}}$
$\Rightarrow {f}''\left( x \right)=\dfrac{6}{{{\left( x+1 \right)}^{4}}}$
Now we find the value of ${f}''\left( 0 \right)$, by substituting $x=0$ in the above equation, we get:
$\Rightarrow {f}''\left( 0 \right)=\dfrac{6}{{{\left( 0+1 \right)}^{4}}}=6$
Substitute the value of $f\left( 0 \right)$, ${f}'\left( 0 \right)$ and ${f}''\left( 0 \right)$ in the first few terms of the formula of Maclaurin’s series, we get:
$\Rightarrow 1+\dfrac{\left( -2 \right)x}{1!}+\dfrac{\left( 6 \right){{x}^{2}}}{2!}+...$
Simplify it further by opening the brackets, we get:
$\Rightarrow 1-\dfrac{2x}{1!}+\dfrac{6{{x}^{2}}}{2!}+...$
We know the value of first few factorials that is $1!=1$ and $2!=2$, applying these values in the above equation, we get:
$\Rightarrow 1-\dfrac{2x}{1}+\dfrac{6{{x}^{2}}}{2}+...$
$\Rightarrow 1-2x+3{{x}^{2}}...$
We can see the similarity here. The above expression can be written in the form of the power series that is:
$\Rightarrow \sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n+1}}n}{{x}^{n-1}}$
To determine the radius of the convergence, apply the ratio test for series in the above equation where the ratio test states that: suppose $L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|$ then if L < 1 then the series converges absolutely; if L > 1 then the series is divergent; and if L = 1 or the limit fails to exist, then the test is inconclusive, because there exist both convergent and divergent series that satisfy this case.
$\Rightarrow \displaystyle \lim_{n \to \infty }\dfrac{{{\left( -1 \right)}^{n+1+1}}\left( n+1 \right){{x}^{n-1+1}}}{{{\left( -1 \right)}^{n+1}}n{{x}^{n-1}}}$
$\Rightarrow \displaystyle \lim_{n \to \infty }\dfrac{{{\left( -1 \right)}^{n+2}}\left( n+1 \right){{x}^{n}}}{{{\left( -1 \right)}^{n+1}}\left( n \right){{x}^{n-1}}}$
$\Rightarrow \displaystyle \lim_{n \to \infty }\left( -1 \right)\dfrac{\left( n+1 \right)}{\left( n \right)}x$
$\Rightarrow \displaystyle \lim_{n \to \infty }\left( -1 \right)\left( 1+\dfrac{1}{n} \right)x$
We know that the $\displaystyle \lim_{n \to \infty }\left( \dfrac{1}{n} \right)=0$, applying this in the above equation, we get:
$\Rightarrow \left( -1 \right)\left( 1 \right)x$
After solving the brackets, we get:
$\Rightarrow -x$
By the ratio test, $|x|<1$ for the series to be convergent. Therefore, the radius of convergence is $1$.
Therefore, The Maclaurin series of the function $f\left( x \right)=\dfrac{1}{{{\left( 1+x \right)}^{2}}}$ is given by $\Rightarrow \sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n+1}}n}{{x}^{n-1}}$ and the radius of convergence is $1$.

Note: Students can go wrong while determining the values of ${f}'\left( 0 \right)$ and${f}''\left( 0 \right)$. In order to find the derivative of ${f}'\left( 0 \right)$, students first find the value of$f\left( 0 \right)$and then take its derivative which is completely wrong and leads to the wrong answer. So, it’s important to remember that first find the derivative of ${f}'\left( x \right)$ to find the value of ${f}'\left( 0 \right)$ by substituting $x=0$ in ${f}'\left( x \right)$.