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

Question

Christopher Agresta · Accepted Answer

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.

Aurora Alvarado · Answer

undefined To find a power series representation for $ \frac{{1+x}}{{(1-x)^2}} $, we can start with the geometric series formula $ \frac{1}{{1-x}} = \sum_{n=0}^{\infty} x^n $. Taking the derivative of both sides, we get $ \frac{1}{{(1-x)^2}} = \sum_{n=1}^{\infty} nx^{n-1} $. Then, multiplying by $ 1+x $, we have $ \frac{{1+x}}{{(1-x)^2}} = \sum_{n=1}^{\infty} nx^{n-1} (1+x) $. Finally, we simplify this expression to get $ \frac{{1+x}}{{(1-x)^2}} = \sum_{n=1}^{\infty} (n+1)x^n + \sum_{n=1}^{\infty} nx^{n-1} $.