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

1 Answer
Oct 15, 2015

#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}#