How Do We Derive and Verify the Taylor Series for \(\arctan(x)\)?

[evinda]

Hello! (Cool)

I want to find the Taylor series of $f(x)=\arctan(x), x \in [-1,1], \xi=0$
$$f'(x)=\frac{1}{1+x^2}$$

According to my notes:

$$\frac{1-(-t^2)^n}{1+t^2}=1-t^2+ \dots + (-1)^{n-1} t^{2n-2}$$

So, $\frac{1}{1+t^2}=1-t^2+ \dots +(-1)^{n-1}t^{2n-2}+\frac{(-t^2)^n}{1+t^2} \Rightarrow \int_0^x \frac{1}{1+t^2}=\int_0^x [1-t^2+ \dots +(-1)^{n-1}t^{2n-2}]dt+\int_0^x \frac{(-t^2)^n}{1+t^2} dt $

Then,they find that $\arctan(x)=x-\frac{x^3}{3}+ \dots + (-1)^{n-1} \frac{1}{2n-1} x^{2n-2}+ \int_0^x \frac{(-t^2)^n}{1+t^2}dt$

Then,to check if $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{2n-1} x^{2n-1}$ is the right Taylor series, they check if $\int_0^x \frac{(-t^2)^n}{1+t^2}dt \to 0$.
Firstly, how do we get this relation: $$\frac{1-(-t^2)^n}{1+t^2}=1-t^2+ \dots + (-1)^{n-1} t^{2n-2}$$ ?

Also, why do we check if $\int_0^x \frac{(-t^2)^n}{1+t^2}dt \to 0$,to see if we found the right power series or not??

 

[Prove It]

You recognised that $\displaystyle \begin{align*} \frac{d}{dx} \left[ \arctan{(x)} \right] = \frac{1}{1 + x^2} \end{align*}$. Good.

Now notice that $\displaystyle \begin{align*} \frac{1}{1 + x^2} = \frac{1}{1 - \left( -x^2 \right) } \end{align*}$, which is easily recognised as the closed form of the geometric series $\displaystyle \begin{align*} \sum_{n = 0}^{\infty} \left( -x^2 \right) ^n = \sum_{n = 0}^{\infty} \left( -1\right) ^n \, x^{2n} \end{align*}$, which is convergent where $\displaystyle \begin{align*} \left| -x^2 \right| < 1 \implies \left| x \right| < 1 \end{align*}$.

Now integrate this series to get the series for $\displaystyle \begin{align*} \arctan{(x)} \end{align*}$. This will have the same radius of convergence, but you will need to test the convergence of the endpoints.