Function as a geometric series

- [Instructor] We're asked to find a power series for f, and they've given us f of x is equal to six over one plus x to the third power. Now, since they're letting us pick which power series, you might say, well, let me just find the Maclaurin series because the Maclaurin series tends to be the simplest to find 'cause we're centered at zero. And so you might immediately go out and say all right, well, let me evaluate this function at zero, evaluate its derivative at zero, its second derivative at zero, so on and so forth. And then I can use the formula for the Maclaurin series to just expand it out. But very quickly, you will run into roadblocks because the first, evaluating this at f, at x equals zero is pretty straightforward. Evaluating the first derivative is pretty straightforward. But then once you start taking the second and third derivatives, it gets very hairy, very fast. You could do a simplification, where you could say, well, let me find the Maclaurin series for f of u is equal to six over one plus u, where u is equal to x to the third. So you find this Maclaurin expansion, in terms of u, and then you substitute for x to the third. And actually, that makes it a good bit simpler, so that is another way to approach it. But the simplest way to approach it is to say, hey, you know what, this form right over here, this rational expression, it looks similar. It looks like the sum of a geometric series. Let's just remind ourselves what the sum of a geometric series looks like. If I have a plus a times r, so a is my first term, r is my common ratio, plus, I'm gonna multiply it times r again, plus a times r squared, plus a times r to the third power, and I keep going on, and on, and on forever. We know that this is going to be equal to a, our first term, over one minus our common ratio, and this just comes from a, the sum of a geometric series. And notice that what we have here, our f of x, our definition of f of x, and the sum of a geometric series look very, very similar. If we say that this, right over here, is a, so a is equal to six. And if negative r is equal to x to the third, or we could say, let me rewrite this. I could write this denominator as one minus negative x to the third. And so now, you could say, okay, well, r could be equal to negative x to the third. And just like that, we can expand it out. Well, if a is equal to six, and r is equal to negative x to the third, well, then we could just write this out as a geometric series, which is very straightforward. So let's do that. And I will do this in, I'll do this in this nice pink color. So the first term would be six, plus six times our common ratio, six times negative x to the third. And so actually, let me just write that as negative six x to the third, and then we're gonna multiply by negative x to the third again. So that's going to be, if I multiply this times negative x to the third, that's gonna be positive six times x to the sixth power. And then I'm gonna multiply it by times negative x to the third again, so it's gonna be minus six times x to the ninth power, and I'm gonna go on, and on, and on. So it's gonna be, and then I could keep going, I multiply it times negative x to the third. I will get six x to the twelfth power. Now, we can go on, and on, and on, and on forever. And so the key here was, and this is the Maclaurin series expansion for our f of x, but the key is to not have to go through all of this business and just to recognize that hey, the way this function was defined is it looks a lot like the sum of a geometric series. And it can be considered the sum of a geometric series, and we can use that to find the power series expansion for our function. This is a very, very, very useful trick.