 In this video, we're going to discuss the solution of question eight for the final exam for math 1220, in which case we're given the function f of x equals 3 over 2 plus x minus x squared, and we're asked to find a power series representation for f. So this will involve some techniques we learned from section 11.9 in the textbook and the corresponding lecture notes there. So given that we have a rational function, we can find its power series representation by trying to compare the rational function to a geometric series of some kind. And so when we look upon this, because we have this quadratic denominator, I feel inclined that we're going to have to do a partial fraction decomposition, first of all, because to compare it to the geometric series, we have a linear factor, and so we have the factor first, this quadratic factor. So we're looking for factors of 2 plus, we have to factorize 2 plus x minus x squared right here. So we need factors of that. So let's see, this would factor probably as 2 plus x times 1 minus x. That wrote really just a quick guess there. Let's see if it works there. If you foil it out, 2 times 1 is 2, x times negative x is a negative x squared, and then you'll get 2 times negative x plus x. That doesn't quite match up, so it looks like our signs are opposite. So just kind of guessing and checking. We could have used a reverse foil technique right now to do this a little bit better. But if we try that one, 2 minus x times 1 plus x, I think this one works this time. So if you multiply it out foil, you'll get 2 plus 2x minus x minus x squared. That combined to give us 2 plus x minus x squared. So we have the factorization of the denominator, that's good. And so that then informs us that the partial fractions would look like a over 2 minus x plus b over 1 plus x, like we see right there. And so then clearing the denominators, this would look like 3 is equal to a times 1 plus x and then b times 2 minus x. And so attempting to annihilate values of x here, we could first start off with x equals 2. That would annihilate the b right here, and that would leave us with 3 equals 3a, that is a equals 1, that's nice. If we choose x to equal negative 1, that will annihilate the a, leaving just b behind. So we get 3 equals, in that case, we're going to also get, notice you're going to get 2 minus a negative 1, that's again a 3 there. So 3 equals 3b divided by 3, you could be likewise equals 1, that's nice. And so coming up here, we see that our fraction can be broken up as 1 over 2 minus x and 1 over 1 plus x. And so now that with this partial fraction decomposition in hand here, we can then start rewriting this thing for the geometric series. We're going to be using the observation, right, that 1 over 1 minus u is equal to the sum of u to the n as n ranges from 0 to infinity. This is the basic formula we're going to use here, and this is something you'd want probably for a note card or a note cheater, whatever resource you have available here. And so for the first one, we're going to factor out the 2, so we already have an equal sign there. So we get 1 half times 1 over 1 minus x over 2, and then for the next one, we just write it as a double negative, so we get 1 over 1 minus a negative x, like so. So the first one will be written as 1 half the sum where n goes from 0 to infinity, and then here we're going to get an x over 2 to the n. And then for the second one, we get the sum where n equals 0 to infinity of negative x to the n, like so. And so this right here would be an acceptable answer. You could put this on the line right here. This is a linear combination of power series, which itself is a power series. Although the more proper answer would be to actually combine these as a single power series for which if we combine like terms here, we're going to get the sum where n ranges from 0 to infinity. And as the coefficient of x to the n, we're going to get 1 half to the n plus 1. Didn't quite get myself enough space there. I'll back this up. And then we're going to get a negative 1 to the n, and this all sits in front of x to the n. So notice what happened here is that this negative x to the n, we broke it up as negative 1 to the x to the n. And this x to the n over 2 to the n, we broke that up as 1 half to the n times x to the n. So factoring out the common x to the n, but also the first one had this factor of 1 half, so when you distribute it through the sum, the powers of 1 half are going to increase by 1. And so this what you see in white would be the correct answer, but had you instead inserted our previous answer written in yellow, this part right here, this uncombined form, this would also be considered correct for this one.