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Main content
AP.CALC:
LIM‑8 (EU)
,
LIM‑8.E (LO)
,
LIM‑8.E.1 (EK)
,
LIM‑8.F (LO)
,
LIM‑8.F.1 (EK)

Video transcript

- [Instructor] So we have this function that's equal to 2 minus 8x squared plus 32x to the fourth minus 128x to the sixth. And it just keeps going and going. So it's defined as an infinite series. And what I want to explore in this video, is there another way to write this function so it's not expressed as an infinite series? Well, some of you might be thinking, well, this looks like a geometric series on the right hand side, an infinite geometric series, and we know what the sun of an infinite geometric series is if it converges. So maybe that's a way that we can express this. So let's try to do that. So first let's just confirm that this is an infinite geometric series. And in order for it to be a geometric series, each successive term has to be some common ratio times the previous term. So to go from 2 to negative 8x squared, what do you have to multiply by? Well you have to multiply by negative 4x squared. Now let's see if you multiply negative 8x squared times negative 4x squared, what do you get? Well, negative 4 times negative 8 is positive 32. x squared times x squared is x to the fourth. So that works. And then you multiply that times negative 4x squared, and you indeed would get negative 128x to the sixth. So this indeed looks like an infinite geometric series on the right-hand side. In fact, we can rewrite f of x as being equal to the sum from n equals 0 to infinity of, you have your first term, and then you have your common ratio, negative 4x squared to the nth power. Let's confirm that works when n equals 0 this is going to be 1. So 2 times 1 is 2, and that indeed is our first term there. And then to that, you're gonna add it to when n is equal to 1. So that's just going to be two times negative 4X squared, which is indeed this second term right over here. And so this looks like it works. Now what is the sum of an infinite geometric series like this? Well it's going to be a finite value, assuming the absolute value of your common ratio is less than 1. So first of all, let's just think about under what conditions is the absolute value of our common ratio less than 1? And then we could say, okay, that helps us to find a radius of convergence. And then if x is in that zone, or if it's in that interval, then we can figure out a non-infinite geometric series way of expressing this function. So if we just think about under what circumstances will this converge, will it come out to a finite value? That's a situation in which the absolute value of your common ratio is less than 1. And so let's see if we can simplify this a little bit. No matter what x is, it's always going to be not, x squared is always going to be non-negative. And so the only, so this entire expression is always going to be negative. And so if you take the absolute value of it, this is going to be evaluate as 4x squared, which is always going to be positive. So this is equivalent to 4x squared, which needs to be less than 1. Or we could say that x squared needs to be less than 1/4. Or we could say that x needs to be less than 1/2 and greater than negative 1/2. One way to think about it is anywhere in this interval, if you square it, you're going to be less than 1/4. At 1/2, if you square it, it's equal to 1/4. And at negative 1/2, if you square it, it's equal to 1/4. But for lower absolute values, it's going to be less than 1/4. And so that's what this interval right here says. Another way to think about it is the absolute value of x needs to be less than 1/2. And so we've just defined an interval over which this infinite geometric series will converge. You could say this has a radius of convergence of, let me write it this way, radius of convergence, convergence of 1/2, you can go 1/2 above 0 and 1/2 below 0. But now that we've set the conditions under which this would converge, let's rewrite it. So this function is going to be equal to, we know what the sum of an infinite geometric series is. It's going to be equal to the first term over 1 minus your common ratio, 1 minus negative 4x squared. And so we can rewrite our function as f of x is equal to 2 over 1, subtract a negative 1 plus 4x squared for the absolute value of x is less than 1/2. We have the interval over which we converge and there you have it. We are done.
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