Question

How do you do the taylor series expansion of $\arctan \left( x \right)\And x\sin x$?

Answer 1

To find the taylor series expansion of $\arctan \left( x \right)$, we know that differentiation of $\arctan \left( x \right)$ with respect to x is equal to $\dfrac{1}{1+{{x}^{2}}}$ and we also know the relation that $\dfrac{1}{1-c}=1+c+{{c}^{2}}+{{c}^{3}}+......$ so in this expansion series if we put c as $-{{x}^{2}}$ then we can find the taylor series of $\arctan \left( x \right)$. And the Taylor series expansion of $x\sin x$ is found by writing Taylor series expansion of $\sin x$ and then multiplying x with that series.

Complete step by step answer:
First of all, we are going to find the taylor series expansion of $\arctan \left( x \right)$. We know that the derivative of $\arctan \left( x \right)$ with respect to x is equal to:
$\dfrac{d\left( \arctan x \right)}{dx}=\dfrac{1}{1+{{x}^{2}}}$ ………… Eq. (1)
We also know the relation which is equal to:
$\dfrac{1}{1-c}=1+c+{{c}^{2}}+{{c}^{3}}+......$
Now, if we substitute c as $-{{x}^{2}}$ in the above then the above equation will look like:
$\begin{aligned} & \Rightarrow \dfrac{1}{1-\left( -{{x}^{2}} \right)}=1+{{\left( -{{x}^{2}} \right)}^{1}}+{{\left( -{{x}^{2}} \right)}^{2}}+{{\left( -{{x}^{2}} \right)}^{3}}+...... \\\ & \Rightarrow \dfrac{1}{1-\left( -{{x}^{2}} \right)}=1-{{x}^{2}}+{{x}^{4}}-{{x}^{6}}+..... \\\ \end{aligned}$
We know that, negative sign multiplying with negative sign gives positive sign so L.H.S of the above expression will look like:
$\Rightarrow \dfrac{1}{1+{{x}^{2}}}=1-{{x}^{2}}+{{x}^{4}}-{{x}^{6}}+.....$
Now, using the above relation in eq. (1) we get,
$\dfrac{d\left( \arctan x \right)}{dx}=1-{{x}^{2}}+{{x}^{4}}-{{x}^{6}}+.....$
Integrating with respect to x on both the sides we get,
$\begin{aligned} & \Rightarrow \int{\dfrac{d\left( \arctan x \right)}{dx}}dx=\int{\left( 1-{{x}^{2}}+{{x}^{4}}-{{x}^{6}}+...... \right)dx} \\\ & \Rightarrow \arctan x=\int{\left( 1-{{x}^{2}}+{{x}^{4}}-{{x}^{6}}+...... \right)dx} \\\ \end{aligned}$
We know the integration of ${{x}^{n}}$ with respect to x where n can be any whole number is equal to:
$\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$
Using the above integration in the R.H.S of the above equation which is written just above this integration we get,
$\begin{aligned} & \Rightarrow \arctan x=x-\dfrac{{{x}^{2+1}}}{2+1}+\dfrac{{{x}^{4+1}}}{4+1}-\dfrac{{{x}^{6+1}}}{6+1}+....... \\\ & \Rightarrow \arctan x=x-\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{5}}}{5}-\dfrac{{{x}^{7}}}{7}+...... \\\ \end{aligned}$
Hence, we have written the taylor series expansion of $\arctan x$ as follows:
$\arctan x=x-\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{5}}}{5}-\dfrac{{{x}^{7}}}{7}+......$
Now, we are going to find the taylor series expansion of $x\sin x$. For that, we know the taylor series expansion of $\sin x$ is given as:
$\sin x=x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}-\dfrac{{{x}^{7}}}{7!}+......$
Now, multiplying x on both the sides of the above equation we get,
$x\sin x=x\left( x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}-\dfrac{{{x}^{7}}}{7!}+...... \right)$
Multiplying x with each term of the bracket on the R.H.S of the above equation we get,
$x\sin x=x.x-\dfrac{{{x}^{3}}}{3!}.x+\dfrac{{{x}^{5}}}{5!}.x-\dfrac{{{x}^{7}}}{7!}.x+......$
We know that when the base is same and the two bases with different exponents are multiplied then the result of such multiplication is the same base with the addition of the exponents so in the above the base “x” is same and two x with different powers are multiplied then we get,
$\begin{aligned} & \Rightarrow x\sin x={{x}^{1+1}}-\dfrac{{{x}^{3+1}}}{3!}+\dfrac{{{x}^{5+1}}}{5!}-\dfrac{{{x}^{7+1}}}{7!}+...... \\\ & \Rightarrow x\sin x={{x}^{2}}-\dfrac{{{x}^{4}}}{3!}+\dfrac{{{x}^{6}}}{5!}-\dfrac{{{x}^{8}}}{7!}+...... \\\ \end{aligned}$
Hence, we have found the taylor series expansion for $x\sin x$ as:
$x\sin x={{x}^{2}}-\dfrac{{{x}^{4}}}{3!}+\dfrac{{{x}^{6}}}{5!}-\dfrac{{{x}^{8}}}{7!}+......$

Note:
The mistake that could be possible in the above solution is that while writing the taylor series expansion of $\sin x$ which is given as:
$\sin x=x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}-\dfrac{{{x}^{7}}}{7!}+......$
You might forget to write the factorial sign after 3, 5 and 7 so make sure you have written the factorial signs after the numbers 3, 5 and 7.