Question

Write the Maclaurin series for $\tan ^{-1} x$.

Answer 1

Hint : In this question, we need to find the Maclaurin Series for ${\tan ^{ - 1}}x$ .For this, we will find some derivatives of ${\tan ^{ - 1}}x$ as if $f\left( x \right) = {\tan ^{ - 1}}x$ then we will find the values of ${f'}\left( x \right),{f^{''}}\left( x \right),{f^{'''}}\left( x \right),{f^{iv}}\left( x \right),....$ After that we will find out the values of $f\left( 0 \right),{\text{ }}{f'}\left( 0 \right),{\text{ }}{f^{''}}\left( 0 \right),{\text{ }}{f^{'''}}\left( 0 \right),{\text{ }}{f^{iv}}\left( x \right){\text{, }}.....$ And finally we will use the formula of Maclaurin series to evaluate our answer.
Formula of Maclaurin series is:
$f\left( x \right) = f\left( 0 \right) + {f'}\left( 0 \right)x + \dfrac{{{f^{''}}\left( 0 \right)}}{{2!}}{x^2} + \dfrac{{{f^{'''}}\left( 0 \right)}}{{3!}}{x^3} + \dfrac{{{f^{iv}}\left( 0 \right)}}{{4!}}{x^4} + ......$

Complete step-by-step answer :
Here we are given the function as ${\tan ^{ - 1}}x$
i.e., $f\left( x \right) = {\tan ^{ - 1}}x$
and we have to find the Maclaurin’s series for ${\tan ^{ - 1}}x$
Now, we know that Maclaurin series for a function $f\left( x \right)$ is given by:
$f\left( x \right) = f\left( 0 \right) + {f'}\left( 0 \right)x + \dfrac{{{f^{''}}\left( 0 \right)}}{{2!}}{x^2} + \dfrac{{{f^{'''}}\left( 0 \right)}}{{3!}}{x^3} + \dfrac{{{f^{iv}}\left( 0 \right)}}{{4!}}{x^4} + ......{\text{ }}$
So, let us first evaluate the values of some of its derivatives as
${f'}\left( x \right),{\text{ }}{f^{''}}\left( x \right),{\text{ }}{f^{'''}}\left( x \right),{\text{ }}{f^{iv}}\left( x \right),....$
So, we have $f\left( x \right) = {\tan ^{ - 1}}x{\text{ }} - - - \left( 1 \right)$
Differentiating $f\left( x \right)$ with respect to $x$ we get
${f'}\left( x \right) = \dfrac{{d\left( {{{\tan }^{ - 1}}x} \right)}}{{dx}}$
As we know that, $\dfrac{{d\left( {{{\tan }^{ - 1}}x} \right)}}{{dx}} = \dfrac{1}{{1 + {x^2}}}$
So, we have
${f'}\left( x \right) = \dfrac{1}{{\left( {1 + {x^2}} \right)}} = {\left( {1 + {x^2}} \right)^{ - 1}} - - - \left( 2 \right)$
Again differentiating ${f'}\left( x \right)$ with respect to $x$ we get
${f^{''}}\left( x \right) = \dfrac{d}{{dx}}\left( {{{\left( {1 + {x^2}} \right)}^{ - 1}}} \right)$
$\Rightarrow {f^{''}}\left( x \right) = - {\left( {1 + {x^2}} \right)^{ - 2}}2x - - - \left( 3 \right)$
Now, differentiating ${f^{''}}\left( x \right)$ with respect to $x$ we get,
${f^{'''}}\left( x \right) = \dfrac{d}{{dx}}\left( { - {{\left( {1 + {x^2}} \right)}^{ - 2}}2x} \right)$
Using product rule i.e., $\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{d\left( v \right)}}{{dx}} + v\dfrac{{d\left( u \right)}}{{dx}}$ we get
${f^{'''}}\left( x \right) = - {\left( {1 + {x^2}} \right)^{ - 2}} \cdot 2{\text{ }} + {\text{ }}2x\left( {2{{\left( {1 + {x^2}} \right)}^{ - 3}} \cdot 2x} \right)$
Taking common ${\left( {1 + {x^2}} \right)^{ - 2}}$ , we get
${f^{'''}}\left( x \right) = {\left( {1 + {x^2}} \right)^{ - 2}}\left( { - 2 + 8{x^2}{{\left( {1 + {x^2}} \right)}^{ - 1}}} \right)$
$\Rightarrow {f^{'''}}\left( x \right) = {\left( {1 + {x^2}} \right)^{ - 2}}\left( { - 2 + \dfrac{{8{x^2}}}{{\left( {1 + {x^2}} \right)}}} \right)$
On simplifying it, we get
$\Rightarrow {f^{'''}}\left( x \right) = {\left( {1 + {x^2}} \right)^{ - 2}}\left( {\dfrac{{ - 2 - 2{x^2} + 8{x^2}}}{{\left( {1 + {x^2}} \right)}}} \right)$
$\Rightarrow {f^{'''}}\left( x \right) = \left( {6{x^2} - 2} \right){\left( {1 + {x^2}} \right)^{ - 3}}{\text{ }} - - - \left( 4 \right)$
Now, differentiating ${f^{'''}}\left( x \right)$ with respect to $x$ we get,
${f^{iv}}\left( x \right) = \dfrac{d}{{dx}}\left( {\left( {6{x^2} - 2} \right){{\left( {1 + {x^2}} \right)}^{ - 3}}} \right){\text{ }}$
Using product rule, we get
${f^{iv}}\left( x \right) = \left( {6{x^2} - 2} \right)\left( { - 3{{\left( {1 + {x^2}} \right)}^{ - 4}}2x} \right) + {\left( {1 + {x^2}} \right)^{ - 3}}\left( {12x} \right)$
Taking common ${\left( {1 + {x^2}} \right)^{ - 3}}$ , we get
${f^{iv}}\left( x \right) = {\left( {1 + {x^2}} \right)^{ - 3}}\left[ {\dfrac{{\left( {6{x^2} - 2} \right)\left( { - 6x} \right)}}{{\left( {1 + {x^2}} \right)}} + 12x} \right]$
$\Rightarrow {f^{iv}}\left( x \right) = {\left( {1 + {x^2}} \right)^{ - 3}}\dfrac{{\left[ { - 36{x^3} + 12x + 12x + 12{x^3}} \right]}}{{\left( {1 + {x^2}} \right)}}$
On simplifying it, we get
${f^{iv}}\left( x \right) = {\left( {1 + {x^2}} \right)^{ - 4}}\left( { - 24{x^3} + 24x} \right)$
$\Rightarrow {f^{iv}}\left( x \right) = - 24x\left( {{x^2} - 1} \right){\left( {1 + {x^2}} \right)^{ - 4}}{\text{ }} - - - \left( 5 \right)$
and so on….
Now, we will find out the values of $f\left( 0 \right),{\text{ }}{f'}\left( 0 \right),{\text{ }}{f^{''}}\left( 0 \right),{\text{ }}{f^{'''}}\left( 0 \right),{\text{ }}{f^{iv}}\left( 0 \right){\text{,}}.....$
So, from equation $\left( 1 \right)$ on putting $x = 0$ we get
$f\left( 0 \right) = {\tan ^{ - 1}}\left( 0 \right){\text{ = 0 }}$
From equation $\left( 2 \right)$ on putting $x = 0$ we get
${f'}\left( 0 \right) = \dfrac{1}{{\left( {1 + {0^2}} \right)}} = {\left( {1 + {0^2}} \right)^{ - 1}} = 1$
Similarly, from equation $\left( 3 \right)$ on putting $x = 0$ we get
${f^{''}}\left( 0 \right) = - {\left( {1 + {0^2}} \right)^{ - 2}}2\left( 0 \right) = 0$
Now, from equation $\left( 4 \right)$ on putting $x = 0$ we get
${f^{'''}}\left( 0 \right) = \left( {6{{\left( 0 \right)}^2} - 2} \right){\left( {1 + {0^2}} \right)^{ - 3}} = - 2$
and from equation $\left( 5 \right)$ on putting $x = 0$ we get
$\Rightarrow {f^{iv}}\left( 0 \right) = - 24\left( 0 \right)\left( {{0^2} - 1} \right){\left( {1 + {0^2}} \right)^{ - 4}} = 0$
Putting all these values in Maclaurin series, we get
${\tan ^{ - 1}}x = 0 + x\left( 1 \right) + \dfrac{{{x^2}}}{{2!}}\left( 0 \right) + \dfrac{{{x^3}}}{{3!}}\left( { - 2} \right) + \dfrac{{{x^4}}}{{4!}}\left( 0 \right) + ......$
$\Rightarrow {\tan ^{ - 1}}x = x - 2\dfrac{{{x^3}}}{{3!}} + ......$
Now we know that,
$n! = n \cdot \left( {n - 1} \right) \cdot \left( {n - 2} \right) \cdot ......3 \cdot 2 \cdot 1$
So, expanding $3! = 3 \cdot 2 \cdot 1 = 6$ we get
$\Rightarrow {\tan ^{ - 1}}x = x - 2\dfrac{{{x^3}}}{6} + ......$
$\Rightarrow {\tan ^{ - 1}}x = x - \dfrac{{{x^3}}}{3} + ......$
Therefore, the Maclaurin series for ${\tan ^{ - 1}}x$ is given as,
${\tan ^{ - 1}}x = x - \dfrac{{{x^3}}}{3} + ......$
So, the correct answer is “${\tan ^{ - 1}}x = x - \dfrac{{{x^3}}}{3} + ......$”.

Note : Students should take care while finding all the derivatives. Also, they should note that all even values will be equal to $0$ ,so we have the Maclaurin series in odd order only as well as there is an alternative sign between the terms. Also, they can find more functions to increase the expansion.