Question

How do I find the Maclaurin series for $\dfrac{\arcsin \left( x \right)}{x}$?

Answer 1

In this question we have to find the Maclaurin series of the given function $\dfrac{\arcsin \left( x \right)}{x}$. The Maclaurin series is given by the expression $f\left( x \right)=f\left( 0 \right)+\dfrac{{{f}^{1}}\left( 0 \right)}{1!}x+\dfrac{{{f}^{2}}\left( 0 \right)}{2!}{{x}^{2}}+\dfrac{{{f}^{3}}\left( 0 \right)}{3!}{{x}^{3}}...$.
We will first find the Maclaurin series for the term $\arcsin \left( x \right)$ and then divide it with $x$ to get the Maclaurin series for $\dfrac{\arcsin \left( x \right)}{x}$.

Complete step by step answer:
We have the given expression as:
$\Rightarrow \dfrac{\arcsin \left( x \right)}{x}$
We will start by finding the Maclaurin series for $\arcsin \left( x \right)$ and then divide by $x$.
The general form of the Maclaurin series is:
$f\left( x \right)=f\left( 0 \right)+\dfrac{{{f}^{1}}\left( 0 \right)}{1!}x+\dfrac{{{f}^{2}}\left( 0 \right)}{2!}{{x}^{2}}+\dfrac{{{f}^{3}}\left( 0 \right)}{3!}{{x}^{3}}+...$
Therefore, we will consider $f\left( x \right)=\arcsin x$.
The first derivative of the term can be found out as:
$\Rightarrow f'\left( x \right)=\dfrac{d}{dx}\left( \arcsin \left( x \right) \right)$
On using the formula and taking the derivative, we get:
$\Rightarrow f'\left( x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}......(1)$
Along with the higher derivatives. But since the calculations will get tedious, we will use another method for calculating the powers which is by using the Binomial expansion. The binomial expansion is:
$\Rightarrow {{\left( 1+x \right)}^{n}}=1+nx+\dfrac{n\left( n-1 \right)}{2!}{{x}^{2}}+\dfrac{n\left( n-1 \right)\left( n-2 \right)}{3!}{{x}^{3}}+....$
Therefore, now equation $\left( 1 \right)$ becomes:
$\Rightarrow f'\left( x \right)=\dfrac{d}{dx}\left( \arcsin \left( x \right) \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$
On rearing the root term into exponential form, we get:
$\Rightarrow {{\left( 1-{{x}^{2}} \right)}^{-\dfrac{1}{2}}}$
Now the above expression is in the form of the left-hand side of the binomial expansion. Therefore, on expanding, we get:
$\Rightarrow 1+\left( -\dfrac{1}{2} \right)\left( -{{x}^{2}} \right)+\dfrac{\left( -\dfrac{1}{2} \right)\left( -\dfrac{3}{2} \right)}{2!}{{\left( -{{x}^{2}} \right)}^{2}}+\dfrac{\left( -\dfrac{1}{2} \right)\left( -\dfrac{3}{2} \right)\left( -\dfrac{5}{2} \right)}{3!}{{\left( -{{x}^{2}} \right)}^{3}}+...$
On simplifying the terms, we get:
$\Rightarrow 1+\dfrac{1}{2}{{x}^{2}}+\dfrac{\dfrac{3}{4}}{2}{{x}^{4}}+\dfrac{\dfrac{15}{8}}{6}{{x}^{6}}+...$
On rearranging the fractions and simplifying them, we get:
$\Rightarrow 1+\dfrac{1}{2}{{x}^{2}}+\dfrac{3}{8}{{x}^{4}}+\dfrac{5}{16}{{x}^{6}}+..$
Now, to get the power series, we will integrate the terms from the limit of $0$ to $x$ with respect to $t$.
\Rightarrow \arcsin x=\int\limits_{0}^{x}{\left\\{ 1+\dfrac{1}{2}{{t}^{2}}+\dfrac{3}{8}{{t}^{4}}+\dfrac{5}{16}{{t}^{6}}+... \right\\}}dt
Now we know that $\int{1}dt=t$ and ${{\int{t}}^{n}}dt=\dfrac{{{t}^{n+1}}}{n+1}$ therefore, on using the formula, we get:
$\Rightarrow \arcsin x=\left[ t+\dfrac{1}{6}{{t}^{3}}+\dfrac{3}{40}{{t}^{5}}+\dfrac{5}{112}{{t}^{7}}+... \right]_{0}^{x}$
On putting the limits, we get:
$\Rightarrow \arcsin x=x+\dfrac{1}{6}{{x}^{3}}+\dfrac{3}{40}{{x}^{5}}+\dfrac{5}{112}{{x}^{7}}+...$

Which is the required Maclaurin series for $\arcsin \left( x \right)$.
Now the Maclaurin series for $\dfrac{\arcsin \left( x \right)}{x}$ can be found as:
$\Rightarrow \dfrac{\arcsin x}{x}=\dfrac{1}{x}\left( x+\dfrac{1}{6}{{x}^{3}}+\dfrac{3}{40}{{x}^{5}}+\dfrac{5}{112}{{x}^{7}}+... \right)$
On simplifying, we get:
$\Rightarrow \dfrac{\arcsin x}{x}=1+\dfrac{1}{6}{{x}^{2}}+\dfrac{3}{40}{{x}^{4}}+\dfrac{5}{112}{{x}^{6}}+...$, which is the required solution.

Note: It is to be remembered that Maclaurin series is a power series which gives us the approximation of a function for its values really close to zero. It is to be remembered that the Taylor series and the Maclaurin series are similar series. Maclaurin is a type of Taylor series which uses zero as a single point in its derivative.