Question

How‌ ‌do‌ ‌you‌ ‌find‌ ‌the‌ ‌Maclaurin‌ ‌series‌ ‌of‌ ‌$f\left(‌ ‌x‌ ‌\right)=\ln‌ ‌\left(‌ ‌1+{{x}^{2}}‌ ‌\right)$?‌ ‌

Answer 1

In this problem we need to calculate the Maclaurin series for the given equation. We know that the Maclaurin series is nothing but the Tylor series expansion of a function about $0$. Mathematically we can write it as $f\left( x \right)=f\left( 0 \right)+{{f}^{'}}\left( 0 \right)x+\dfrac{{{f}^{''}}\left( 0 \right)}{2!}{{x}^{2}}+\dfrac{{{f}^{\left( 3 \right)}}\left( 0 \right)}{3!}{{x}^{3}}+...+\dfrac{{{f}^{\left( n \right)}}\left( 0 \right)}{n!}{{x}^{n}}+...$. To calculate the Maclaurin series, we need to calculate the first, second, third, ${{n}^{th}}$ derivative of the given equation. After finding the derivatives we need to calculate the value of derivatives at zero. The above process is somewhat lengthy and not an easy process to do. But we need to follow this process to get the Maclaurin series. For some type of function, the Maclaurin series is already defined and we are free to use them. We have the Maclaurin series for the function $\ln \left( 1+x \right)$. So, we will use the Maclaurin series of $\ln \left( 1+x \right)$ to calculate the Maclaurin series of a given function and we will simplify the obtained equation to get the required result.

Complete step-by-step solution:
Given that, $f\left( x \right)=\ln \left( 1+{{x}^{2}} \right)$.
We have Maclaurin series for the function $\ln \left( 1+x \right)$ as
$\begin{aligned} & \ln \left( 1+x \right)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-... \\\ & \Rightarrow \ln \left( 1+x \right)=\sum\limits_{k=1}^{n}{\left( \dfrac{{{\left( -1 \right)}^{k+1}}{{x}^{k}}}{k} \right)} \\\ \end{aligned}$
From this equation the Maclaurin series of the given function is calculated by substituting the variable $x$ with ${{x}^{2}}$, then we will get
$\begin{aligned} & f\left( x \right)=\ln \left( 1+{{x}^{2}} \right) \\\ & \Rightarrow f\left( x \right)={{\left( {{x}^{2}} \right)}^{1}}-\dfrac{{{\left( {{x}^{2}} \right)}^{2}}}{2}+\dfrac{{{\left( {{x}^{2}} \right)}^{3}}}{3}-.... \\\ \end{aligned}$
Simplifying the above equation, then we will get
$\Rightarrow f\left( x \right)={{x}^{2}}-\dfrac{{{x}^{4}}}{2}+\dfrac{{{x}^{6}}}{3}-.....$
We can also write the above expression as
$\Rightarrow f\left( x \right)=\sum\limits_{k=1}^{n}{\left( \dfrac{{{\left( -1 \right)}^{k+1}}{{\left( {{x}^{2}} \right)}^{k}}}{k} \right)}$

Note: In this problem we have used the Maclaurin series of function $\ln \left( 1+x \right)$ which is predefined. We have Maclaurin functions for some of the remaining functions also. For example
${{e}^{x}}=\sum\limits_{n=0}^{\infty }{\dfrac{{{x}^{n}}}{n!}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+....$.
$\cos x=\sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}{{x}^{2n}}}{\left( 2n \right)!}}=1-\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}-\dfrac{{{x}^{6}}}{6!}+.....$.
$\sin x=\sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}{{x}^{2n+1}}}{\left( 2n+1 \right)!}}=x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}-\dfrac{{{x}^{7}}}{7!}+.....$.
So, we can use the above formulas directly when they are in need.