Question

How do you find the maclaurin series expansion of `f(x) = x^2 arctan (x^3)`?

Answer 1

`sum_{i=0}^infty (-1)^{n} x^{6n+5}/{2n+1}`

Explanation:

Substitute `y=x^3` (note that when `x\to 0`, also `x^3` and thus `y` tend to 0, so it's a good substitution).

Write down the McLaurin series for `arctan(y)`, which is known: it's the sum, with alternate sign, of the odd powers of `y` divided by the power itself:

`y-y^3/3+y^5/5-y^7/7+y^9/9+...`,

or in a compact way,

`sum_{i=0}^infty (-1)^{n} y^{2n+1}/{2n+1}`

Now substitute back `y=x^3`:

#sum_{i=0}^infty (-1)^{n} (x^3)^{2n+1}/{2n+1}= sum_{i=0}^infty (-1)^{n} x^{6n+3}/{2n+1}#

Multiply the whole expression by `x^2`:

#sum_{i=0}^infty (-1)^{n} x^2*x^{6n+3}/{2n+1}= sum_{i=0}^infty (-1)^{n} x^{6n+5}/{2n+1}#