 Welcome to the Grand Valley State University Calculus screencasts. In this video, we'll talk about deriving new power series from old ones. Remember that a power series is a series of the form some c sub k x to the k, where c sub k is a real number. In this screencast, we'll see how we can take power series that we already know and use those to derive power series for other functions. So let's start first with a power series that we know. Let f of x be the sum x to the k as k goes from 0 to infinity. We know that for any value of x, f of x is a geometric series with ratio x. So f of x is 1 over 1 minus x, and the series converges to 1 over 1 minus x on the interval from negative 1 to 1. We're going to use this series representation for f to quickly determine power series representations for some other functions. We'll use this power series representation for f of x to quickly determine power series representations for two other functions. First, let's consider the problem of finding the power series representation for the function 1 over 1 plus x to the 4. To do this, we're going to exploit a connection between f of x and 1 over 1 plus x to the 4. Note that f of the opposite of x to the 4 is 1 over 1 minus the opposite of x to the 4, which is 1 over 1 plus x to the 4. So if we can find a power series for f of the opposite of x to the 4, we found a power series representation for 1 over 1 plus x to the 4. Let's just evaluate our power series representation for f of x at the opposite of x to the 4. And we see that f of the opposite of x to the 4 is the sum, as k goes from 0 to infinity, of the opposite of x to the 4 to the k, just replacing x with the opposite of x to the 4. And so we see that f of the opposite of x to the 4 is the sum as k goes from 0 to infinity, negative 1 to the k, x to the 4k. Now, the power series representation for f of x converges when x is between negative 1 and 1, and we're replacing x with the opposite of x to the 4, so the power series for f of the opposite of x to the 4 will converge when the opposite of x to the 4 is between negative 1 and 1, or when x to the 4 is between negative 1 and 1, which is the same as when x is between negative 1 and 1. In the end, then, we see that the power series for f of the opposite of x to the 4, which is 1 over 1 plus x to the 4, is the sum, k goes from 0 to infinity, minus 1 to the k, x to the 4k, which is 1 minus x to the 4th, plus x to the 8th, minus x to the 16th, and so on. And this power series representation is valid for all x's between negative 1 and 1. So we found a power series representation for a new function using a power series representation that we already knew. Let's look at a second example. Consider the problem of finding a power series representation of 4x cubed over 1 plus x to the 4th squared. Now, it may not be clear where to start here, but we're going to use the power series representation for 1 over 1 plus x to the 4th that we just derived. And the key idea is to recognize that if we differentiate 1 over 1 plus x to the 4th, we get the opposite of g of x. So to find the power series for g of x, we would just have to know how to differentiate the power series for 1 over 1 plus x to the 4th. We use an important property of power series. If we have a power series representation for function f of x that converges absolutely to f of x on some interval, then if we differentiate that series term by term, that new series will converge to the derivative of f on that same interval of convergence. In other words, if we differentiate the series for the opposite of 1 over 1 plus x to the 4th term by term, we'll obtain the series for the derivative of the opposite of 1 over 1 plus x to the 4th. And differentiating term by term will give us where we want to go. We start with g of x, which is the opposite of the derivative of 1 over 1 plus x to the 4th. We substitute in that power series we found for 1 over 1 plus x to the 4th. Then we differentiate the power series term by term. We then wind up with g of x being the sum k goes from 1 to infinity, negative 1 to the k plus 1 times the derivative of x to the 4k, which is 4k x to the 4k minus 1. And this power series representation for g of x is valid on the same interval that the power series representation for 1 over 1 plus x to the 4th was. It is on the interval from negative 1 to 1. Now you may be wondering about a couple of things here. This negative 1 to the k plus 1 comes from taking negative 1 to the k and multiplying by the negative sign that's outside of the sum. And then the sum starts at k equals 1 because the k equals 0 term in the sum before we differentiate is a constant term. And when we differentiate a constant, we get 0, so we just don't include that. What we see here is that we started with an old power series representation this time for 1 over 1 plus x to the 4th. And we obtained a power series for a new function just by differentiating term by term. Let's look at one more example. Suppose we want to integrate e to the x squared dx. One of the problems is we don't know an elementary anti-derivative of e to the x squared. So instead, what we can do is look for a power series representation of the integral of e to the x squared dx. At the start, we know that e to the x is the sum k goes from 0 to infinity x to the k over k factorial, and this is valid for all real numbers x. This was the Taylor series expansion for e to the x centered at 0. We can then find the power series for e to the x squared just by substituting x squared in for x in this power series for e to the x. So we have e to the x squared is the sum as k goes from 0 to infinity x squared to the k over k factorial. That's the sum as k goes from 0 to infinity x to the 2k over k factorial. And this is valid for all values of x. Now we use another important property of power series. If we have a power series representation for a function f that converges absolutely to f on some interval negative r to r, then we can integrate this series for f term by term, and that series converges to the power series representation for the integral of f of x dx. In other words, we can integrate a series term by term, and that series, the resulting series, converges to the integral of our function. And we're going to use this idea to find a power series representation of the integral of e to the x squared dx. Just as we did with the differentiation problem, we're going to integrate term by term. To integrate e to the x squared dx, we integrate the power series representation for e to the x squared dx. And then we integrate term by term, and when we integrate x to the 2k over k factorial, we can take out the 1 over k factorial, and an antiderivative for x to the 2k is x to the 2k plus 1 over 2k plus 1. So we get the integral of e to the x squared dx has the power series expansion, some k goes from 0 to infinity, x to the 2k plus 1 over k factorial times 2k plus 1 plus a constant. And this is valid for all values of x. So once again, we started with a power series that we knew, and by integrating this time, we came up with a power series representation of a new function. So to summarize what we've just done, we took power series representations for known functions, and we were able to use those series representations to quickly find power series representations for other functions. And in this screencast, the methods we used were composition, for example, finding the series representation for e to the x squared based on the series representation of e to the x. We also differentiated series term by term to get a series for the derivative of a function, and we integrated series term by term to get a series representation for an antiderivative of a function. And that concludes our screencast on using power series to get new series from old ones. We hope you come back again soon.