 While we need calculus to write down meaningful series, part of the point of developing the theory of series is to extend our calculus. This is based around the following idea. If we can express a function as a composition f of g of x, where we already know the series expansion for f of x, we can avoid the endless differentiation by using simple algebra. However, this does require that we know the series for some basic functions. If all you know about infinite series are the McLaren series for a couple of functions, you'll probably fail the section on series. But among the things you should know about infinite series are a couple of common McLaren series. In particular, it's very useful to know the McLaren series for 1 over 1 minus x, e to the x, sine x, and cosine x. So let's use this to find a McLaren series for sine of x squared. We'll write down the first four terms of the series, and find the interval of convergence. Now we could do this by taking repeated derivatives, but we already know the McLaren series for sine of x, so let's see if we can make use of it. So our McLaren series for sine of something is the sum from 0 to infinity of minus 1 to the n over 2n plus 1 factorial times something to the power 2n plus 1. We'd like to find this series for sine of x squared, so we'll substitute, and obtain our series for sine of x squared. Now the McLaren series converges for all x, which is to say it converges for everything between minus infinity and infinity. In this case, since our everything is x squared, we can substitute that in, which tells us that the series converges for everything between minus infinity and infinity. The first four terms of the series will be for n equals 0, 1, 2, or 3. If n equals 0, the term is going to be... If n equals 1, the term is going to be... If n equals 2, the term is... And if n equals 3, the term is... How about some calculus? Suppose I have a convergent power series over some interval between a and b. Then if I want to find the derivative of the power series, it's going to be the series consisting of the derivative of each of the individual terms, at least in that interval between a and b. And a very important idea to keep in mind, we always need to check the end points for convergence. Similarly, suppose I have a convergent power series over some interval, then the antiderivative of that power series is going to be... The series whose terms are the antiderivative of the individual terms of the original series. And again, we should always check the end points for convergence. Because we find the integral of a series by integrating the individual terms, and the derivative of a series by differentiating the individual terms, we sometimes refer to this process as term-wise integration or term-wise differentiation. And again, it's important to remember that we always check the end points for convergence. For example, let's find the first five terms of the McLaren series for 1 over 1 minus x squared, and then determine the interval of convergence for the series. So rather than going through this endless differentiation, let's try to find 1 over 1 minus x squared by integrating or differentiating something we already know the McLaren series for. Fortunately, as good calculus students, you know the McLaren series for several important functions. So you know the McLaren series for e to the x. And maybe we'll be lucky, and the derivative or the integral of e to the x will be 1 over 1 minus x squared. And since the derivative of e to the x is 1 over 1 minus x squared, no, actually it isn't. Oh, well, the anti-derivative is 1 over, no, that doesn't work either. Well, you know what they say, if your first guess isn't right, give up and join the circus. Well, actually the only people who say that are people who manage circuses. Let's try some of the other functions we know the McLaren series for. We know the McLaren series for sine of x. So let's see if it's derivative or anti-derivative of what we want. And again, we find that neither the derivative nor the anti-derivative give us 1 over 1 minus x squared. We know the McLaren series for cosine of x. And again, that doesn't do us any good. But before we do anything drastic, let's try that last McLaren series for 1 over 1 minus x. And finally, we do have success in that the derivative is in fact 1 over 1 minus x squared. So 1 over 1 minus x squared is the derivative of 1 over 1 minus x. We can replace the 1 over 1 minus x with its power series, then differentiate the power series term-wise. And for good measure, we'll write down the first couple of terms of the series. Remember, we also want to check on that interval of convergence. The original series converges on the interval minus 1 to 1. So the new series will also converge on this interval, but we do need to check the end points. At x equals negative 1, our series becomes, which is going to be a divergent series. At x equals 1, our series becomes, which is also divergent. So the interval of convergence does not include either end point. And we can say that the series converges over the interval minus 1 to 1. So let's try to find the indefinite integral of e to the x squared. So let's pull in our McLaren series for e to the x. We know the first five terms of the McLaren series for e to the x. So instead of writing down x, we'll write down x squared. And that'll give us the first five terms of the McLaren series for e to the x squared. To find the integral, we'll integrate term-wise and get the first five terms of our series. Now if we want to write this down as an infinite series, we'll start with our McLaren series, replace x with x squared and do a little algebraic simplification. We'll integrate term-wise, and that gives us our indefinite integral. And don't forget that constant of integration.