Question

How do you find a power series representation for `x^3/(2-x^3)` and what is the radius of convergence?

Answer 1

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

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

with radius of convergence `root(3)(2)`

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^3/2`

Then we find:

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

Multiply by `x^3/2` to find:

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

This is a geometric series with common ratio `x^3/2` so converges when `abs(x^3/2) < 1` which is when `abs(x/root(3)(2)) < 1`, which is when `abs(x) < root(3)(2)`. So the radius of convergence is `root(3)(2)`.