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

##### 1 Answer

I will do the one for

So, notice how

#color(green)(1/(1-b) = sum_(n=0)^N b^n = 1 + b + b^2 + b^3 + ...)#

(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

**1.** Substitute

We should consider the properties of limits with sums:

#lim_(b->-x^2) [1 + b + b^2 + b^3 + ...] = lim_(b->-x^2) sum_(n=0)^N b^n# .

Now, let's rewrite

#1/(1-b) => 1/(1 - (-x^2)) = sum_(n=0)^N (-x^2)^n#

#= (-x^2)^0 + (-x^2)^1 + (-x^2)^2 + (-x^2)^3 + ...#

#= 1 - x^2 + x^4 - x^6 + ...#

where we've done the substitution

**2.** Integrate to achieve

Meaning that we can integrate sums term by term:

#int [1 + b + b^2 + b^3 + ...]db = int [sum_(n=0)^N b^n]# .

We're almost there. Since

#=> int [1 - x^2 + x^4 - x^6 + ...]dx#

#= color(blue)(x - x^3/3 + x^5/5 - x^7/7 + ...)#

That's all!