Question

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

Answer 1

Recall that:

The power series for `1/(1-x)` is `sum_(n=0)^N x^n`.

What you can do is take the derivative of both sides:

`d/(dx)[sum_(n=0)^N x^n] = d/(dx)[1/(1-x)] = color(green)(d/(dx)[1 + x + x^2 + ...])`

When we take the derivative, note that the first term will disappear (`d/(dx)[1] = 0`), so the first `n` shifts from `n=0` to `n=1`.

`sum_(n=1)^N nx^(n-1) = 1/(1-x)^2 = color(green)(?)`

We can shift it back to `n = 0` for conventional purposes.

How you do that is consider that you are currently starting at `n=1`. That means `n-1 = 0`. To get the exponent to become `0` using `n = 0` below `sum`, `n-1` becomes `n`.

When `n = 1` below `sum`, `nx^(n-1)` starts at `1x^0`, so when `n = 0` below `sum`, `nx^"stuff"` becomes `(n+1)x^"stuff"`.

`color(highlight)(sum_(n=0)^N (n+1)x^(n) = 1/(1-x)^2) = color(green)(?)`

So, all we need to do is do it for the explicit series to get:

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

`= color(blue)(1 + 2x + 3x^2 + 4x^3 + ...)`

The radius of convergence is based on the idea that the sum of magnitudes `|a| >= 1` will not converge if they are all positive.

`1/(1-x)^2 > 1` when `|x| >= 1`.

Therefore, the radius of converge is:

`color(blue)(x in (-1, 1))`

or you can write it as:

`color(blue)(|x| < 1)`