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 $$f^{\prime }\left(x\right),f^{{}^{″}}\left(x\right),f^{{}^{‴}}\left(x\right),f^{iv}\left(x\right),....$$ After that we will find out the values of $$f\left(0\right),f^{\prime }\left(0\right),f^{{}^{″}}\left(0\right),f^{{}^{‴}}\left(0\right),f^{iv}\left(x\right)\text{, }.....$$ And finally we will use the formula of Maclaurin series to evaluate our answer.
Formula of Maclaurin series is:
$$f\left(x\right)=f\left(0\right)+f^{\prime }\left(0\right)x+{\displaystyle \frac{f^{{}^{″}}\left(0\right)}{2!}}{x}^2+{\displaystyle \frac{f^{{}^{‴}}\left(0\right)}{3!}}{x}^3+{\displaystyle \frac{f^{iv}\left(0\right)}{4!}}{x}^4+......$$

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:
$$f\left(x\right)=f\left(0\right)+f^{\prime }\left(0\right)x+{\displaystyle \frac{f^{{}^{″}}\left(0\right)}{2!}}{x}^2+{\displaystyle \frac{f^{{}^{‴}}\left(0\right)}{3!}}{x}^3+{\displaystyle \frac{f^{iv}\left(0\right)}{4!}}{x}^4+......$$
So, let us first evaluate the values of some of its derivatives as
$$f^{\prime }\left(x\right),f^{{}^{″}}\left(x\right),f^{{}^{‴}}\left(x\right),f^{iv}\left(x\right),....$$
So, we have $$f\left(x\right)={\tan }^{-1}x---\left(1\right)$$
Differentiating $$f\left(x\right)$$ with respect to $$x$$ we get
$$f^{\prime }\left(x\right)={\displaystyle \frac{d\left({\tan }^{-1}x\right)}{dx}}$$
As we know that, $${\displaystyle \frac{d\left({\tan }^{-1}x\right)}{dx}}={\displaystyle \frac{1}{1+x^2}}$$
So, we have
$$f^{\prime }\left(x\right)={\displaystyle \frac{1}{\left(1+x^2\right)}}={\left(1+x^2\right)}^{-1}---\left(2\right)$$
Again differentiating $$f^{\prime }\left(x\right)$$ with respect to $$x$$ we get
$$f^{{}^{″}}\left(x\right)={\displaystyle \frac{d}{dx}}\left({\left(1+x^2\right)}^{-1}\right)$$
$$\Rightarrow f^{{}^{″}}\left(x\right)=-{\left(1+x^2\right)}^{-2}2x---\left(3\right)$$
Now, differentiating $$f^{{}^{″}}\left(x\right)$$ with respect to $$x$$ we get,
$$f^{{}^{‴}}\left(x\right)={\displaystyle \frac{d}{dx}}\left(-{\left(1+x^2\right)}^{-2}2x\right)$$
Using product rule i.e., $${\displaystyle \frac{d\left(uv\right)}{dx}}=u{\displaystyle \frac{d\left(v\right)}{dx}}+v{\displaystyle \frac{d\left(u\right)}{dx}}$$ we get
$$f^{{}^{‴}}\left(x\right)=-{\left(1+x^2\right)}^{-2}\cdot 2+2x\left(2{\left(1+x^2\right)}^{-3}\cdot 2x\right)$$
Taking common $${\left(1+x^2\right)}^{-2}$$ , we get
$$f^{{}^{‴}}\left(x\right)={\left(1+x^2\right)}^{-2}\left(-2+8x^2{\left(1+x^2\right)}^{-1}\right)$$
$$\Rightarrow f^{{}^{‴}}\left(x\right)={\left(1+x^2\right)}^{-2}\left(-2+{\displaystyle \frac{8x^2}{\left(1+x^2\right)}}\right)$$
On simplifying it, we get
$$\Rightarrow f^{{}^{‴}}\left(x\right)={\left(1+x^2\right)}^{-2}\left({\displaystyle \frac{-2-2x^2+8x^2}{\left(1+x^2\right)}}\right)$$
$$\Rightarrow f^{{}^{‴}}\left(x\right)=\left(6x^2-2\right){\left(1+x^2\right)}^{-3}---\left(4\right)$$
Now, differentiating $$f^{{}^{‴}}\left(x\right)$$ with respect to $$x$$ we get,
$$f^{iv}\left(x\right)={\displaystyle \frac{d}{dx}}\left(\left(6x^2-2\right){\left(1+x^2\right)}^{-3}\right)$$
Using product rule, we get
$$f^{iv}\left(x\right)=\left(6x^2-2\right)\left(-3{\left(1+x^2\right)}^{-4}2x\right)+{\left(1+x^2\right)}^{-3}\left(12x\right)$$
Taking common $${\left(1+x^2\right)}^{-3}$$ , we get
$$f^{iv}\left(x\right)={\left(1+x^2\right)}^{-3}\left[{\displaystyle \frac{\left(6x^2-2\right)\left(-6x\right)}{\left(1+x^2\right)}}+12x\right]$$
$$\Rightarrow f^{iv}\left(x\right)={\left(1+x^2\right)}^{-3}{\displaystyle \frac{\left[-36x^3+12x+12x+12x^3\right]}{\left(1+x^2\right)}}$$
On simplifying it, we get
$$f^{iv}\left(x\right)={\left(1+x^2\right)}^{-4}\left(-24x^3+24x\right)$$
$$\Rightarrow f^{iv}\left(x\right)=-24x\left(x^2-1\right){\left(1+x^2\right)}^{-4}---\left(5\right)$$
and so on….
Now, we will find out the values of $$f\left(0\right),f^{\prime }\left(0\right),f^{{}^{″}}\left(0\right),f^{{}^{‴}}\left(0\right),f^{iv}\left(0\right)\text{,}.....$$
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
$$f^{\prime }\left(0\right)={\displaystyle \frac{1}{\left(1+0^2\right)}}={\left(1+0^2\right)}^{-1}=1$$
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
$${\tan }^{-1}x=0+x\left(1\right)+{\displaystyle \frac{x^2}{2!}}\left(0\right)+{\displaystyle \frac{x^3}{3!}}\left(-2\right)+{\displaystyle \frac{x^4}{4!}}\left(0\right)+......$$
$$\Rightarrow {\tan }^{-1}x=x-2{\displaystyle \frac{x^3}{3!}}+......$$
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
$$\Rightarrow {\tan }^{-1}x=x-2{\displaystyle \frac{x^3}{6}}+......$$
$$\Rightarrow {\tan }^{-1}x=x-{\displaystyle \frac{x^3}{3}}+......$$
Therefore, the Maclaurin series for $${\tan }^{-1}x$$ is given as,
$${\tan }^{-1}x=x-{\displaystyle \frac{x^3}{3}}+......$$
So, the correct answer is “$${\tan }^{-1}x=x-{\displaystyle \frac{x^3}{3}}+......$$”.

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.