Question

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

Answer 1

`1/(1-x)^2=1+2x+3x^2+...`

Explanation:

We are given

`f(x)=1/(1-x)^2`

This is fairly similar to `1/(1-x)`, for which we know a power series:

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

The radius of convergence for this power series is `x in (-1,1)`.

While it would be easy to say that

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

This is not a valid representation of a power series.

Usually, some power series arise from derivatives. It'd be worth a shot to try this one, too.

`"d"/("d"x) [1/(1-x)] = "d"/("d"x) [1+x+x^2+...]`

By the quotient rule,

`"d"/("d"x) [1/(1-x)] = - ("d"/("d"x) [1-x])/(1-x)^2=color(red)(1/(1-x)^2`

As `"d"/("d"x) x^k = kx^(k-1)`:

`"d"/("d"x) [1+x+x^2+...] = 0 + 1 + 2x + 3x^2 + ... = sum_(k=0)^oo kx^(k-1)`

Hence the power series representation of `f(x)` is

`1/(1-x)^2 = sum_(k=0)^oo kx^(k-1)`

with radius of convergence `x in (-1,1)`.