Question

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

Answer 1

I will do the one for `arctanx`. Maybe someone else can do `xsinx`. (With that one, if you know the taylor series for `sinx`, simply multiply all the terms by `x`.)

So, notice how `d/(dx)[arctanx] = 1/(1+x^2)`, and recall how you should know the following relation by this point in your class:

`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 `a = 0`.

1. Substitute `-x^2` for `b`.

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)` as:

`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 `b = -x^2`.

2. Integrate to achieve `arctanx`.

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/(1 - (-x^2))dx = arctanx`, we need to perform `int [sum_(n=0)^N (-x^2)^n]`. Thus, the Taylor series centered around `a = 0` is:

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

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

That's all!