Question

How do you find the maclaurin series expansion of `x^3/(1+x^2)`?

Answer 1

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

`x^3/(1+x^2) = sum_(n=0)^oo (-1)^n x^(2n+3)`

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^2` to find:

`1/(1+x^2) = sum_(n=0)^oo (-x^2)^n = sum_(n=0)^oo (-1)^n x^(2n)`

Then multiply by `x^3` to get:

`x^3/(1+x^2) = x^3 sum_(n=0)^oo (-1)^n x^(2n) = sum_(n=0)^oo (-1)^n x^(2n+3)`

This is a geometric series with common ratio `-x^2`, hence the radius of convergence is `1`.