Question

Radius of Convergence for Power Expansion?

Expanded as a power series in terms of `x - 2/3`, what is the radius of convergence of `f(x) = 1/(x(1-x))`?

Answer 1

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

with radius of convergence `R=1/3`

Explanation:

We have:

`f(x) = 1/(x(1-x)) = 1/x + 1/(1-x)`

Substitute: `t = (x-2/3)`

`f(t) = 1/(t+2/3)+ 1/(1-(t+2/3)) = 1/(2/3+t) + 1/(1/3-t) = 3/2 1/(1+3/2t)- 3/(1-3t)`

Both terms can be expressed as sum of a geometric series with ratio: `(-(3t)/2)` and `(3t)` respectively:

`1/(1+(3t)/2) = sum_(n=0)^oo ((-3t)/2)^n` for `abs ((3t)/2) < 1 => t in (-2/3,2/3)`

`1/(1-3t) = sum_(n=0)^oo (3t)^n` for `abs (3t) < 1 => t in (-1/3,1/3)`

So we have that in the intersection of the two intervals, that is for `t in (-1/3,1/3)`

`f(t) = 3/2sum_(n=0)^oo ((-3t)/2)^n-3 sum_(n=0)^oo (3t)^n`

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

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

and substituting back `x`:

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

with radius of convergence `R=1/3`