Question

How do you find a power series representation for `f(x)= 1/(1+4x^2)` and what is the radius of convergence?

Answer 1

`f(x) = sum_(n=0)^oo (-4x^2)^n` with radius of convergence `1/2`

Explanation:

Consider the power series:

`sum_(n=0)^oo (-4x^2)^n = 1 - 4x^2 + 16x^4 - 64x^6 +...`

Then:

`(1+4x^2)(sum_(n=0)^oo (-4x^2)^n)`

`=sum_(n=0)^oo (-4x^2)^n + 4x^2 sum_(n=0)^oo (-4x^2)^n`

`=sum_(n=0)^oo (-4x^2)^n - sum_(n=1)^oo (-4x^2)^n`

`=(-4x^2)^0 = 1`

provided the sum `sum_(n=0)^oo (-4x^2)^n` converges.

So

`sum_(n=0)^oo (-4x^2)^n = 1/(1+4x^2) = f(x)`

This is a geometric sequence, so will converge if the common ratio has absolute value `< 1`.

That is:

`abs(-4x^2) < 1`, so `x^2 < 1/4`, so `abs(x) < 1/2`

In general

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

which converges if `abs(a) < 1`.