 Welcome back to our lecture series Math 1220, Calculus II for students at Southern Utah University. As usual, I'm your professor today, Dr. Angel Misseldine. This will be the first video in our series as part of lecture 44, which will talk about the topics that can be found in 11.9 of James Stewart's Calculus textbook. Admittedly in the previous video, we had started talking about section 11.9, but today our central focus is going to be on what we call a power series representation of a function. The ideas in this lecture are going to be based upon the following observation if we have a geometric series. We're going to take the sum, we're in ranges from zero to infinity of the terms x in here. This is a geometric series where x is acting as the constant ratio for this geometric series. In expanded form, we'll get something like this, 1 plus x plus x squared plus x cubed plus x of the fourth plus x of the fifth, et cetera, et cetera, et cetera. What we know is that this geometric series, we convergent so long as the absolute value of x is less than 1, this just means the ratio is small for a geometric series. On that domain, for this interval negative 1 to 1, we see that our geometric series is equal to the rational function 1 over 1 minus x. In particular, this is equality here. This rational function equals this infinite polynomial whenever x is adaptive values less than 1. This is equality. We could say that the function is represented by the power series on the interval of convergence. Modifying this geometric series example, we get something like the following. Consider the following geometric series, n equals 1 to infinity of the sum, negative 3x to the n minus 1 over 2 to the 3n. We want to find a rational function which agrees with this geometric series on its interval of convergence. We're going to begin by trying to rewrite our sum right here. n equals 1 to infinity. The starting value doesn't really have much significance when it comes to the convergence of any series, let alone geometric series. We notice you have a slight mismatch right here. We have an n minus 1 on top. We have a 3n on the bottom. We would love it if all those exponents were the same. Let's deal with the bottom first, right? Because we have a 2 to the 3n, we can factor the exponent and we actually would get 2 cubed to the n on the bottom. Then we have a negative 3x to the n minus 1 right here. Now, of course, 2 cubed is the same thing as 8, so we can just make that substitution right here. Notice we have n 8 on the bottom. We have n minus 1, negative 3x is on the top, so I'm going to just take one of the 8s out on the bottom. If we just remove one of the 8s, that will be a 1 8 there. We get the sum as n equals 1 to infinity. We're going to get a negative 3x to the n minus 1 on top. We are going to get an 8 to the n minus 1 on the bottom, and thus combining them together, we get the sum of a negative 3x over 8 raised to the n minus 1 power. Now we see, in fact, we do have this geometric sequence. As such, it'll then, by the formula, well, we're going to have a 1 8 out in front just because of the coefficient of 1 8 that's already there. Then with our geometric series formula, where we get a over 1 minus r, the r is just the constant ratio. What is the thing that we're taking powers of? That's this guy right here. This is our r value. We're going to see we have a negative 3x over 8 as our r. Then the a here is just the first term of the sequence. When we plug n equals 1 into this expression right here, what do we get? We're going to end up with a negative 3x over 8 to the 1 minus 1 power. Now good news, that power is just going to be 0, and something raised to the 0 power is just a 1. And so we end up with 1 8 times 1 over 1 plus 3x over 8, like so. I'm not a big fan of fractions instead of fractions. So let's times the big fraction here by 8 over 8. This then gives us, we're going to have 8 on top from this green aid right here. We're going to get 8 on the bottom from this 8. But then when you distribute this 8 onto the two pieces, we end up with 8 plus 3x like so. And then the 8 on top cancels with the factor of 8 on the bottom, in which case then we see that our geometric series can be written as the function 1 over 3x plus 8. So this rational function is equal to this geometric series, but there's a stipulation of course, it's only going to be true on the interval of convergence when x is between negative 8 thirds and positive 8 thirds. Where do we get that? Well, we're just using the observation that the ratio needs to be small. That is to say negative 3x over 8 needs to be small. And as you start solving for this inequality, you're going to get the absolute value of x is less than 8 thirds, in which case it didn't follows immediately from there. So on this interval, when x is between negative 8 thirds and positive 8 thirds, the geometric series we see right here is in fact identical to this function right here. So we could say that this geometric series represents this rational function so long as x is between negative 8 thirds and 8 thirds.