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.