Question

How do you find the maclaurin series expansion of `(x/(1+x^3))`?

Answer 1

Use the Maclaurin series for `1/(1-t)` and substitution to find:

`x/(1+x^3) = sum_(n=0)^oo (-1)^n x^(3n+1)`

Explanation:

The Maclaurin series for `1/(1-t)` is `sum_(n=0)^oo t^n`

since `(1-t) sum_(n=0)^oo t^n = sum_(n=0)^oo t^n - t sum_(n=0)^oo t^n = sum_(n=0)^oo t^n - sum_(n=1)^oo t^n = t^0 = 1`

Substitute `t = -x^3` to find:

`1/(1+x^3) = sum_(n=0)^oo (-x^3)^n = sum_(n=0)^oo (-1)^n x^(3n)`

Multiply by `x` to find:

`x/(1+x^3) = sum_(n=0)^oo (-1)^n x^(3n+1)`

This is a geometric series with common ratio `-x^3` so it converges if `abs(x) < 1` and has radius of convergence `1`.