Geometric series as a function | Infinite sequences and series | AP Calculus BC | Khan Academy

So we have this function that's equal to two minus eight x squared plus 32 x to the fourth minus 128 x to the sixth, and just keeps going and going. So it's defined as an infinite series, and what I want to explore in this video is: 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 sum 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.

First, let's just confirm that this is an infinite geometric series. 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 8 x squared, what do you have to multiply by? Well, you have to multiply by negative 4 x squared.

Now let's see if you multiply negative 8 x squared times negative 4 x 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 4 x squared, and you indeed would get negative 128 x to the sixth. So this indeed looks like an infamous 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 4 x 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.

Then to that, you're going to add it to when n is equal to 1, so that's just going to be 2 times negative 4 x squared, which is indeed this second term right over here. 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 one.

So first of all, let's just think about under what conditions the absolute value of our common ratio is less than one, and then we could say, "Okay, that helps us to find a radius of convergence." 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.

If we just think about under what circumstances this will 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 one. So let's see if we can simplify this a little bit. No matter what x is, it's always going to be—negative x squared is always going to be non-negative. So this entire expression is always going to be negative, and if you take the absolute value of it, this is going to evaluate as 4 x squared, which is always going to be positive.

So this is equivalent to 4 x 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 one-half and greater than negative one-half. One way to think about it is, anywhere in this interval, if you square it, you're going to be less than one-fourth. At one-half, if you square it, it's equal to one-fourth, and at negative one-half, if you square it, it's equal to one-fourth. But for lower absolute values, it's going to be less than one-fourth, 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 one-half. 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 of one-half. You can go one-half above zero and one-half below zero.

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 4 x squared.

So we can rewrite our function as f of x is equal to 2 over 1 subtract a negative 1 plus 4 x squared, for the absolute value of x is less than one-half. We have the interval over which we converge, and there you have it, we are done.