Question

How do you write `(x+1)/(1 - x)^2` in a power series representation?

Answer 1

First, notice how we can rewrite this as:

`x/(1-x)^2 + 1/(1-x)^2`

If you recall, you should have been taught that:

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

Notice how then you can take the derivative (remember the chain rule with `-x`) to get:

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

`= color(highlight)(1 + 2x + 3x^2 + 4x^3 + 5x^4 + ...)`

That gives you the second half of the answer. Now, here is something rather interesting we can do. We can multiply by the whole series by `x`.

`color(darkblue)(x/(1-x)^2) = x * 1/(1-x)^2 = color(darkblue)(x*d/(dx)[sum_(n=0)^N x^n])`

Therefore, we can modify `1 + x + x^2 + x^3 + ...` to get the power series for the first half, `(x)/(1-x)^2`, by taking the derivative of every term and multiplying every term by `x`.

NOTE: `\mathbfx` and `\mathbf(d/(dx))` are NOT commutative, so `\mathbf(d/(dx)[x*f(x)])` is NOT the same as `\mathbf(x*d/(dx)[f(x)])`.

`color(darkblue)(x/(1-x)^2) = x*[0 + 1 + 2x + 3x^2 + 4x^3 + ...]`

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

Finally, we can add on the series for `1/(1-x)^2` to get:

`= color(darkblue)([" "" "x + 2x^2 + 3x^3 + 4x^4 + ...]) + color(highlight)([1 + 2x + 3x^2 + 4x^3 + 5x^4 + ...])`

`= color(blue)(1 + 3x + 5x^2 + 7x^3 + 9x^4 + ...)`

`= color(blue)(sum_(n=0)^N (2n+1)x^n)`