# Maclaurin series of cos(x)

Approximating cos(x) with a Maclaurin series (which is like a Taylor polynomial centered at x=0 with infinitely many terms). It turns out that this series is exactly the same as the function itself! Created by Sal Khan.
Video transcript
In the last video, we hopefully set up some of the intuition for why - or I should say what - the Maclaurin series is all about, and I said at the end of the videos that a Maclaurin series is just a special case of a Taylor series. In the case of a Maclaurin series, we're approximating this function around x is equal to 0, and a Taylor series, and we'll talk about that in a future video, you can pick an arbitrary x value - or f(x) value, we should say, around which to approximate the function. But with that said, let's just focus on Maclaurin, becuase to some degree it's a little bit simpler, and that by itself can lead us to some pretty profound conclusions about mathematics, and that's actually where I'm trying to get to. So let's take the Maclurin series of some interesting functions and I'm gonna do functions where it's pretty easy to take the derivatives, and you can /keep/ taking their derivatives over and over and over and over and over again. So let's take the Maclaurin series of cosine of x, so if f(x)=cos(x), then - before I even apply this formula, that we somewhat derived in the last video, or at least got the intuitive for in the last video - let's take a bunch of derivatives of f(x), just so we have a good sense of it. So, if we take the first derivative, if we take the first derivative, derivative of cos(x) = -sin(x) if we take the derivative of that, if we take the derivative of that, derivative of sin(x) is cos(x), and we have the negative there, so it's -cos(x) so if we take the derivative of that, so this is the third derivative of cos(x), now it's just going to be positive sine of x, and if we take the derivative of that, we get cos(x) again. We get cosine of x again. So if we take the derivative of that, this is the fourth derivative, I should, I should use this notation but you get the idea, we'll get cos(x) again. And if you look at what we talked about in the last video, we want the difference - we want the function, and we want it's various derivatives evaluated at 0, so let's evaluate it at 0. So f(0), cos(0) is 1, cosine of zero is one. Whether you're talking about zero radians or zero degrees, doesn't matter, sine of zero is zero, so this is f prime of - f prime of zero, is zero. And then cos(0) is, once again, one, but we have the negative out there, so it becomes negative one. So f - the second derivative evaluated at zero is negative one. Let's take the third derivative, the third derivative evaluated at zero well, sine of zero is just zero, and then the fourth derivative evaluated at zero, cosine of zero is one. So f prime prime prime at zero is now equal to one. So you see an interesting pattern here - one, zero, negative one, zero, one, then you go to zero, then you go to negative one, zero. So if we were to apply this to find it's Maclaurin representation, what would we get? Let me do my best attempt at this. So we would get, our polynomial would be - so our polynomial approximation of cosine of x is going to be f(0), f(0) is one, and then we have one plus f'(0) times x. But f'(0) is just zero, so we're not going to have this term over there, it's going to be zero times x, I won't even take the trouble of writing it down, it would be this zero time x, then plus f prime prime or second derivative, which is negative one, so I'll write negative - negative, this is a negative one right here, this is a negative one, times x squared, times x squared, over 2 factorial - over two factorial, which in this case is just going to be two. But I'll just write it down here as two factorial, to make the pattern a little bit more obvious, and then we go to the next term, the third derivative evaluated at zero but the third derivative evaluated at zero is just zero, so this term won't be there as well, then you go to the fourth derivative, the fourth derivative evaluated at zero is positive one, so this coefficient right here is going to be a one, and so you're going to have one times x to the fourth over four factorial, so plus x to the fourth over four factorial, and I think you start seeing a pattern now. You have sign switches - and you would see this if we kept going, so you can verify it for yourself if you don't believe me - so you have a positive sign, a negative sign, a positive sign, and then a negative sign, so on and so forth, and this is, uh, one times x to the zeroth power, then you jump two to x to the squared, jump two to x to the fourth, and so if we kept that up, we'd have a positive sign, now we have a negative sign, it would be x to the sixth over six factorial, then you have a positive sign, x to the eighth over eight factorial, and then you'd have a negative sign, x to the tenth over ten factorial, and you can just keep going that way. And if you kept going with this series, this would be the polynomial representation of cosine of x. And it's frankly just kind of cool that if can be represented this way. It's a pretty simple pattern here for a trigonometric function. Once again, it kind of tells you that all of this math is connected. And we'll see, two or three videos from now, it's connected in far more profound ways then you can possibly imagine.