Question

What is the Maclaurin Series for `f(x) = x^2ln(1+x^3)`?

Answer 1

`f(x) = x^5-x^8/2+x^11/3 - x^14/4...`

`" " = sum_(n=1)^(oo) \ (-1)^(n+1) \ x^(3n+2)/n`

Explanation:

We start with the well known Maclaurin Series for `ln(1+x)`:

`ln (1+x) = x-x^2/2+x^3/3 - x^4/4 +...`

If we replace `x` in the series by `x^3` then we get:

`ln (1+x^3) = (x^3)-(x^3)^2/2+(x^3)^3/3 - (x^3)^4/4 + ...`
`" " = x^3-x^6/2+x^9/3 - x^12/4...`

So then we have:

`f(x) = x^2ln(1+x^3)`
`" " = x^2{x^3-x^6/2+x^9/3 - x^12/4... }`
`" " = x^5-x^8/2+x^11/3 - x^14/4...`
`" " = sum_(n=1)^(oo) \ (-1)^(n+1) \ x^(3n+2)/n`