 In the last video, we saw that connecting a function to its derivative or anti-derivative for one of, where one of those functions has a known power series representation can then be used to create a power series representation for the other function, either the derivative or anti-derivative of the function we don't have a power series representation for yet. And so the idea is we connected one over one minus x with one over one minus x squared. And so I wanna mention that if you wanted to do a higher power like one minus x cubed, we could just keep on taking derivative, derivative, derivative, derivatives over and over and over again. And therefore, for any repeated linear factor in the dominator, we can get a power series representation by taking subsequent derivatives. And because we can do repeated factors, we can actually do any partial, or we can do any rational function. We have all the tools necessary. It just requires maybe doing polynomial division and partial fraction decomposition. We can do, we can find the power series representation for any rational function. Also, if you have an irreducible quadratic, my recommendation is to use complex numbers. But if you don't wanna do that, you can use, well, we'll talk about this video when you do if you have an irreducible quadratic. Before we see that though, let's actually take a look at the one you see on the screen here. Let's find a power series representation for the natural log of one plus x. And we're gonna use this to find, and we're gonna find this radius convergence as well as the power series representation. So like we saw previously, what I'm talking about right now is that we want to connect this function, y equals the natural log of one plus x. We wanna connect this with a function which we could find a power series representation for. So for example, if we take the derivative, the derivative would be one over one plus x for which we have a power series representation of this one because we can compare this rational function to the geometric series where you think it has one over one minus negative r, negative x there, and therefore r is equal to negative x. And so the power series representation would then be take the sum where it goes from zero to infinity. We're gonna get negative one to the n times x to the n in expanded form. This looks like one minus x plus x squared minus x cubed plus x to the fourth minus x to the fifth, continue on. So we get the alternate geometric series right here. And so recognizing that this function is the derivative of the function we started with, we can get back up to the natural log of one plus x by taking the antiderivative. So we just need to integrate this function right here. And so doing that, taking the integral of y prime dx here, this would give us, you're gonna have some constant. I always like to put that in the front because as this is an infinite series, I put at the end, there is no end. So put it in the front, c plus x minus x squared over two plus, whoops, plus x cubed over three minus x to the fourth over four plus x to the fifth over five and then repeats, right? You wanna do enough terms that you can establish a pattern going on here. And so we get c plus a sum, whoops, right at sigma again. And we're gonna go from n equals, how do we wanna describe this one? Well, since it starts off at x here, I'm actually gonna take n equals one to infinity. And then what's the coefficient? We're gonna have an x right here. The coefficient is just, well, it's alternating, right? So we need to have some alternating factor, negative one to the n plus one. We want n plus one on the top so that it starts positive and not negative because this should be positive, negative, positive, negative, like so. So we're gonna get negative one to the n plus one times x to the n, like so. And if you don't wanna do plus one, you could do a minus one right here. It doesn't really make much of a difference. You just have to make sure you start with positive. But what happens in the denominator? The denominator seems just to match up with whatever the exponent was, right? Like you see here, you get x cubed over three. And so that's the pattern we've established. The denominator needs to be just the exponent n right there. And so we get this power series representation, but it has one defect. We don't know what the plus c is, but we can figure it out, right? Because we know this is supposed to equal one plus, or the natural log of one plus x. So plug in some strategic number, like say x equals zero is a really great choice because it's the center of the power series. On the left hand side, we're gonna get the natural log of one plus zero, which is the natural log of one, which is itself zero. On the right hand side, because we chose the center of the power series, we should just get the constant term c. And so we see that c equals zero. And therefore, this is an instance which we would have been justified if we had forgotten the plus c because the plus c does turn out to be zero. But we don't wanna do it because, we don't wanna get it right because we forgot something or we made a mistake. We wanna do it because we did it correctly. That way we're liable to get truth the next time around. And therefore we see we have the following power series representation. Because our function, the natural log of one plus x is the antiderivative of a function with the known power series representation, we are able to find a power series representation for that function. We wanna do the same thing basically here. If you wanna find a power series representation for f of x equals arc tangent of x, well the idea is we know it's derivative. Why f prime of x is gonna equal one over one plus x squared. We wanna find a power series representation for this function and we can handle that actually. We haven't done this one yet, but we can handle it, it's no big deal because rewriting this as one over one minus a negative x squared, we see that our ratio is negative x squared. And therefore the geometric series will be an alternating geometric series. We get n equals zero to infinity. We're going to get negative one to the n times x to the two n. So I actually fudged a little bit, sorry about that. We did do this one earlier. This was actually done in a previous example. You can actually see the link to the video right now. If you did wanna see the details of that. In expanded form this looks like one minus x squared plus x to the fourth minus x to the sixth, continue on. So I did wanna sort of make a comment here to stop. If you're trying to find a rational, some type of power series representation for any rational function, let's say you had something like three over x plus one times one plus x squared or something like that. The idea is you're gonna have to do some type of partial fraction decomposition, a over x plus one plus bx plus c over one plus x squared, something like that, right? You're gonna have some partial fraction decomposition. You need to figure out what the numerator is, but in terms of the power series representation, the numerator doesn't matter too much because it was multiplied by the numerator when done. If you have linear factors, that's great. You can use a geometric series. If you have a repeated linear factor, like we were talking about earlier, then you just use derivatives, a geometric series to go from there. If you have an irreducible quadratic like one plus x squared, you get a geometric series like we do here. If the quadratic's a little bit more complicated than that, you'll have to complete the square. And so if you throw in polynomial division, we actually have all the skills necessary to handle every single rational function. Every rational function can find a power series representation using these techniques of polynomial division, partial fraction decomposition, geometric series, and derivatives. All of those are taken care of. So what we're trying to do in this video is extend it from rational functions to other functions that are somehow related to rational functions via derivatives. We saw that with the natural log because its derivative is a rational function. And arc tangent, its derivative is also a rational function, this one over one plus x squared. So we're turning back to our problem. If we wanna find a power series representation for arc tangent, what we need to do is we need to find the anti-derivative of this power series right here. So that's gonna give us c plus x minus x cubed over three plus x to the fifth over five minus x to the seventh over seven and the pattern will continue on and on and on and on. So this is gonna equal c plus the sum where we're gonna get n equals one. So we do start at one right here. So n equals one going off towards infinity. It is alternating, right? So we're gonna get negative one. We're gonna get negative one to a power. It needs to start off as an odd. So it needs to, we want it to go positive, negative, positive, negative, positive, negative. So we can actually start off at n if we so chose. That would be perfectly acceptable here. But then what do we do for it? We have x's, we have to get some type of power. What's the exponent gonna be here? It's only odd powers, right? It goes one, three, five, seven. How can you capture odd powers? Well, here's a nice little trick that helps you out here. I'm actually gonna go back and start this thing at zero. And as a consequence of starting at zero, actually that actually works out a little bit greater for us because I guess if we start at one, we actually need like an n minus one right there. If we start off at zero, then we're gonna get negative one to the n. That's a positive at zero, so that's great. And then here's the trick I'm trying to tell you. Take the number two n plus one. Notice that whatever number you choose for n, two n will always be an even number. Therefore, if you add one to it, you're gonna get an odd number. So two n plus one's always odd. If you start at n equals zero, you're gonna get two times zero, which is zero plus one is one. That gives you the first one, that's great. And then the denominator always matches up with the exponent again. And so we're gonna get two n plus one in the denominator like that. I guess parenthesis isn't necessary there. So that one might be a little bit harder to grasp for us, but if your exponents are always even, just have a two n as your exponent. If your exponents are always odd, take a two n plus one as your exponent. Your denominator has to match up with the exponent. That's quite common when you take the integral of a power series. And as it is alternating, we get a negative one to the n. And remember here that since we wanna start positive, negative, positive, negative, if our initial value is n, that's an even number, we're gonna take negative one to the n. So now we have a power series representation, but what about the plus c again? We'll choose the center of the power series, x equals zero. You can get arc tangent of zero, which is itself equal to zero. And then the right hand side is just gonna equal c. The whole advantage of choosing the center is that all of these powers of x will disappear when you plug in the center. So we see that the c is again zero and we now have our power series representation. Arc tangent of x will equal the sum as n ranges from zero to infinity of the sequence negative one to the n, x to the two n plus one over two n plus one. So in these last through videos, we've seen exactly all the things we need to do to find a power series representation for any rational function and then also things that are related to rational functions via anti-derivative such as arc tangent and the natural log.