# How do you do the taylor series expansion of arctan(x) and xsinx?

##### 1 Answer
Oct 14, 2015

I will do the one for $\arctan x$. Maybe someone else can do $x \sin x$. (With that one, if you know the taylor series for $\sin x$, simply multiply all the terms by $x$.)

So, notice how $\frac{d}{\mathrm{dx}} \left[\arctan x\right] = \frac{1}{1 + {x}^{2}}$, and recall how you should know the following relation by this point in your class:

$\textcolor{g r e e n}{\frac{1}{1 - b} = {\sum}_{n = 0}^{N} {b}^{n} = 1 + b + {b}^{2} + {b}^{3} + \ldots}$
(this is an important relation; know this!)

Knowing that performing operations on a Taylor series parallels performing operations on the function which the series represents, we can start from here and transform the series through a sequence of operations.

Let us consider a Taylor series centered around $a = 0$.

1. Substitute $- {x}^{2}$ for $b$.

We should consider the properties of limits with sums:

${\lim}_{b \to - {x}^{2}} \left[1 + b + {b}^{2} + {b}^{3} + \ldots\right] = {\lim}_{b \to - {x}^{2}} {\sum}_{n = 0}^{N} {b}^{n}$.

Now, let's rewrite $\frac{1}{1 - b}$ as:

$\frac{1}{1 - b} \implies \frac{1}{1 - \left(- {x}^{2}\right)} = {\sum}_{n = 0}^{N} {\left(- {x}^{2}\right)}^{n}$

$= {\left(- {x}^{2}\right)}^{0} + {\left(- {x}^{2}\right)}^{1} + {\left(- {x}^{2}\right)}^{2} + {\left(- {x}^{2}\right)}^{3} + \ldots$

$= 1 - {x}^{2} + {x}^{4} - {x}^{6} + \ldots$

where we've done the substitution $b = - {x}^{2}$.

2. Integrate to achieve $\arctan x$.

Meaning that we can integrate sums term by term:

$\int \left[1 + b + {b}^{2} + {b}^{3} + \ldots\right] \mathrm{db} = \int \left[{\sum}_{n = 0}^{N} {b}^{n}\right]$.

We're almost there. Since $\int \frac{1}{1 - \left(- {x}^{2}\right)} \mathrm{dx} = \arctan x$, we need to perform $\int \left[{\sum}_{n = 0}^{N} {\left(- {x}^{2}\right)}^{n}\right]$. Thus, the Taylor series centered around $a = 0$ is:

$\implies \int \left[1 - {x}^{2} + {x}^{4} - {x}^{6} + \ldots\right] \mathrm{dx}$

$= \textcolor{b l u e}{x - {x}^{3} / 3 + {x}^{5} / 5 - {x}^{7} / 7 + \ldots}$

That's all!