How do you find a power series representation for (1+x)/((1-x)^2)?

1 Answer
Sep 16, 2015

Fair warning---I expect this to be a long answer!

I got

sum_(n=0)^(N) (1+x)/(1-x)^2 = 1 + 3x + 5x^2 + 7x^3 + ...

for some large finite N.


So, something that I believe you have already been taught is that:

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

for some large finite N. Notice how:

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

and that:

color(darkred)(-1/(1-x)) = -sum_(n = 0)^N x^n (2)

Analogizing from (1):

x/(1-x)^2 = x*["Power Series for "1/(1-x)^2] (3)

Remember these four relationships, because we will be referring back to them.

First, notice how you can rewrite this as:

(1+x)/(1-x)^2 = color(green)(1/(1-x)^2 + color(red)(x/(1-x)^2)) (4)

Already you may see how things could unfold. Making use of (0) in conjunction with (2), we get:

color(darkred)(-1/(1-x) = -1 - x - x^2 - x^3 - ...)

Now, making use of (1), we get:

color(green)(1/(1-x)^2) = d/(dx)[-1 - x - x^2 - x^3 - ...]

= color(green)(-1 - 2x - 3x^2 - ...)

This becomes the left half that we are looking for. Now for the right half. Using (3), we get:

color(red)(x/(1-x)^2) = x[-1 - 2x - 3x^2 - 4x^3 - ...]

= color(red)(-x - 2x^2 - 3x^3 - ...)

Next, we can add them together (like-terms with like-terms) according to (4):

"Power Series of " (1+x)/(1-x)^2 = "Power Series of " color(green)(1/(1-x)^2 + color(red)(x/(1-x)^2))

prop [-1 - 2x - 3x^2 - 4x^3 - ...] + [-x - 2x^2 - 3x^3 - ...]

prop color(green)(-1 - 3x - 5x^2 - 7x^3 - ..).

And finally, notice how we started with 1/(1-x), but we manipulated it before taking the derivative to get 1/(1-x)^2 by multiplying it by -1 to get the power series for -1/(1-x).

The negative sign carried through all our operations.

Therefore, to get the true power series, we have to undo the multiplication we did on 1/(1-x) by dividing by -1 at the end, thus dividing the previous result by -1. So, finally, we get:

color(blue)(sum_(n=0)^(N) (1+x)/(1-x)^2 = 1 + 3x + 5x^2 + 7x^3 + ...)

(that is why I purposefully put the prop symbol to emphasize that we were not at the final answer yet)

Wolfram Alpha agrees with this answer.