 Welcome to the GVSU Calculus screencasts. In this episode, we'll calculate the values of a few functions defined by power series. Remember that a power series is a series of the form sum, say, k goes from 0 to infinity of c sub k x to the k, where these c sub k's are just real numbers for each k. And in this screencast, we're going to evaluate a few special functions defined by power series at a few points. As an example, let's define f of x by the power series sum k goes from 0 to infinity, 3 to the k x to the k. This power series defines f as a function whose inputs are those values of x for which the series converges. And if we input a specific value of x into our function f into this power series, then we end up with a series of real numbers. For example, f of 1 quarter would be the series that we obtain by substituting 1 quarter in for x. So f of 1 quarter is the sum k goes from 0 to infinity of 3 to the k times 1 quarter to the k. And if we look at a few terms of this series of real numbers, when k is 0, we get 1. When k is 1, we get 3 quarters. When k is 2, we get 9 sixteenths and so on. So f of 1 quarter is the sum of the series 1 plus 3 fourths plus 9 sixteenths plus 2764ths and so on, if that series converges. Now in fact, this series for f of 1 quarter should be familiar to us. Pause the video for a moment, look at the series, figure out why it's familiar, and then evaluate f of 1 quarter by finding the sum of this series. And resume the video when you're ready. f of 1 quarter is the sum as k goes from 0 to infinity of 3 to the k times 1 quarter to the k, and we can combine these terms together into 3 fourths to the k. We should be able to recognize this as a geometric series with a ratio of 3 quarters. Remember the sum of the geometric series with ratio r is 1 over 1 minus r, so f of 1 quarter is 1 over 1 minus 3 fourths for 4. So in this case f of 1 quarter is just 4. This calculation of f of 1 quarter provides us some insight into finding a general rule for f of x. Why don't you pause the video for a moment and use that idea that we used to find f of 1 quarter to find a rule for f of x by summing this series. And for which values of x should we expect the series for f to converge, or in other words, what's the domain of f? Now the thing to note here is that f of x is the sum k from 0 to infinity of 3 to the k x to the k, and again we can put those two terms together into just 3x quantity to the kth power. And we can recognize then that f is a geometric series with a ratio 3x. We know that the geometric series converges as long as the ratio is less than 1 in absolute value. So our power series for f will converge as long as the ratio, in this case 3x, in absolute value is less than 1, or for x's that are between negative 1 third and 1 third. And in these situations we can sum this geometric series with ratio r, the sum of the geometric series with ratio r is 1 over 1 minus r, and f of x is just equal to 1 over 1 minus 3x. Let's look at another example. Let's let f of x be defined by the power series sum k goes from 0 to infinity x to the k over k factorial. If we evaluate this function of x equals 1, we get f of 1 is the sum as k goes from 0 to infinity of 1 to the k over k factorial. That's just 1 plus 1 plus 1 half plus 1 sixth plus 1 twelfth and so on. Again, this series for f of 1 should be familiar. Pause the video for a moment, reflect on what this series looks like and how it's familiar to you, and then sum the series to find the value of f of 1. Resume the video when you're ready. If you remember our Taylor series, the Taylor series expansion we found for e to the x is the sum k goes from 0 to infinity of x to the k over k factorial, and this expansion was valid for every value of x. That means that our function f of x is really just e to the x. As a result, f of 1 is just e to the first power or e. Let's look at one more example. Let f of x be the power series sum k goes from 0 to infinity minus 1 to the k, x to the k over k squared plus 1. If we evaluated a half substituting 1 half in for x, we get f of 1 half is the sum k goes from 0 to infinity minus 1 to the k, 1 half to the k times 1 over k squared plus 1, and that's equal to the sum of the series 1 minus a quarter plus 1 20th minus 1 80th, if the series converges. This series for f of 1 half is not one that we know, but we do have some techniques that we've developed that allow us to approximate the sum of the series as close as we like. Pause the video for a moment and find a value of n, so the nth partial sum of this series approximates f of 1 half to 3 decimal places. And while you're at it, why does f of 1 half even exist? Resume the video when you're ready. Note that the series for f of 1 half is an alternating series, so we can apply the alternating series test. The sequence of positive terms here 1 over 2 to the k times 1 over k squared plus 1 decreases to 0, so the alternating series test tells us that the series for f of 1 half converges, and so f of 1 half does exist. We can use the alternating series estimation theorem to find a value of n so that the nth partial sum s sub n of our series for f of 1 half is within 3 decimal places of f of 1 half. Remember that the absolute value of the difference between s sub n, the nth partial sum, and the sum of the series, which is f of 1 half in this case, is less than the n plus first positive term, c sub n plus 1. And in our case, c sub n plus 1 is 1 over 2 to the n plus 1 times 1 over n plus 1 squared plus 1, because remember we're using the series for f of 1 half. All we need to do now is find a value of n so that that c sub n plus 1 is less than 0.001. This we can do by trial and error, just substituting different values in for n until we find one that works. And we can see that by just trial and error, a value of n equal 5 makes c sub n plus 1 about 0.004 something, which is certainly less than 0.001. That means that the fifth partial sum of our series for f of 1 half approximates f of 1 half to 3 decimal places. And just using a little bit of technology, we can find the fifth partial sum of the series for f of 1 half. It turns out to be about 0.79. So f of 1 half is about 0.79 to 3 decimal places. To summarize, a power series is just a series of the form sum k goes from 0 to infinity, c sub k, x to the k, where c sub k, these are just real numbers. And a power series defines a function whose domain is the set of all the values of x for which the power series converges. We use power series then to define new functions. One of the problems with power series is that it's difficult, if not outright impossible, to find exact values for functions defined by power series at given inputs. But we can use tools like the alternating series estimation theorem to approximate values of power series at given inputs, assuming that the series that we get is an alternating series. That concludes our screencast on evaluating power series at points. We hope you'll come back again soon.