 In this video, we're gonna find some Taylor series for the function sine of X. And we're gonna start off with the Maclaurin series. So sine of X centered at X equals zero, the Maclaurin series. Now, we saw in the previous video in this series, the Maclaurin series and for, we saw a couple Taylor series for the function e to the X, the natural exponential. And the natural exponential has the great fact that if you take the nth derivative of e to the X, you always end up with e to the X as well. It's higher derivatives are predictable. Now, sine is actually a great alternative here. Let's take f of X this time to be sine of X. This is a great function to go to next because sine of X also has some predictable derivatives, right? Notice the following. If we take the first derivative of sine, you end up with cosine of X. If you take the second derivative, you're gonna end up with negative sine. If you take the third derivative, you're gonna end up with negative cosine. And then when you take the fourth derivative of sine, the derivative of negative cosine is actually gonna be a sine. And now you can see that this thing looped back to where it started with. And so the derivatives of the sine function are cyclic. That is, it goes in a cycle. Every four derivatives, you'll go back through the same loop. Sine, cosine, negative sine, negative cosine. Sine, cosine, negative sine, negative cosine. Sine, cosine, negative sine, negative cosine. And because of that, because of that cyclic nature, we can predict what's gonna happen as we go through higher powers here. And so notice if we start plugging in X, we plug in X equals zero here for our center. Well, f of zero is sine of zero and sine of zero is zero. Then we plug into the first derivative, f prime of zero would be cosine of zero. Cosine of zero is one. Then the next one, the second derivative, we're gonna get the second derivative of zero is gonna be negative sine of zero, which is still zero. And then the next one, the third derivative, evaluate zero, you're gonna get negative kind of, negative cosine of zero, cosine of zero is one. And so you get negative one. And so then when you go to the fourth derivative and evaluate this at zero, you'll get back zero. And again, this cyclic nature just repeats itself. You'll get, the coefficients here are always gonna look like zero, one, zero, negative one, zero, one, zero, negative one, zero, one, zero, one, zero, one, zero, negative one. It's just gonna repeat itself over and over and over and over again. And therefore the Maclaurin series for sine will look like the sum as it ranges from zero to infinity. The general formula, right, is the nth derivative evaluate at your center. We're gonna plug in the number zero right here. Divide this by factorial. Don't forget the factorial. Anytime it's by x to the n. Now what's going on here is gonna be the following idea. If we write this in expanded form, if we write this in expanded form, well, like we see right here, f of zero is equal to zero. And therefore the constant term is gonna look like zero over zero factorial. Then the next one we're gonna get plus one over one factorial times x. The next one we're gonna get zero over two factorial times x squared. The next one we're actually gonna get minus one over three factorial times x cubed. That's one complete cycle. I'm gonna put a second cycle just to be very clear what's going on here. Next we'll get zero over four factorial times x to the fourth plus one over five factorial times x to the fifth plus zero over six factorial times x to the sixth. And then finally we're gonna get minus one over seven factorial times x to the seventh. And then the pattern again repeats itself over and over and over again. We see one cycle of four right here. We see another cycle of four right here. And then it'll just repeat itself over and over and over and over again. Now, whenever there was a coefficient of zero, you actually don't need that term because this is gonna vanish. And so you see that the zeros occur will add zero at x equal to x squared at x to the fourth, x to the sixth and x will be at x to the eighth, et cetera. These things zero out at even powers of x. All the even powers of x, they're gone. When we look at odd powers of x, those things have survived. You'll notice that it goes positive, negative, positive. And, oh, that was a zero there, sorry. Here's the positive. And then the next one's a negative. And it'll continue to do that. Positive, negative, positive, negative, positive, negative. Only the even powers of x, sorry, only the odd powers of x seem to survive. And we get this alternating sum. So it's gonna look like x over one factorial. Then we're gonna get minus x cubed over three factorial. Then we're gonna get x to the fifth over five factorial. Then we're gonna get minus x to the seventh over seven factorial. The next term that doesn't vanish would be x to the ninth over nine factorial. Then minus x to the 11th over 11 factorial. And this continues on and on and on. So one way of trying to find a closed formula for this, because this right here does give us the Maclaurin series. But we can simplify this in the following way. This becomes the same thing as the sum as n equals zero to infinity. It's an alternating sum. So we're gonna have to have a factor of negative one. We're gonna have a power of x and then there's gonna be a factorial in the denominator. Now the factorial in the denominator will just be the same degree as the power of x right there. And as we have only odd powers, we're gonna make the powers be two n plus one. So notice we choose n to be zero. You're gonna get zero plus one, which is one. When n equals one, you get two plus one, which is three. When n equals two, you get two times two plus one, which is five. So the sequence two n plus one will grab all of the odd numbers. So we get x to the two n plus one, and then we have two n plus one factorial in the denominator. This is an alternating sequence. It starts off positive, negative, positive, negative. So we're gonna get negative one to the n. And therefore we've now discovered the Maclaurin series for sine of x. And that's pretty cool. It's kind of like e to the x's because you have powers of x over n factorials. But you only have the odd powers. You don't have any of the even powers and it's an alternating sum. So those are some important differences there. And I wanna point out to you that for sine here, it's Maclaurin series only involves the odd powers, which is kind of curious because sine is an odd function, isn't it, right? Sine has the property that sine of negative x is equal to negative sine of x. This is what we call an odd function. We get this because when you take like an odd power, negative x cubed is the same thing as negative x cubed. And so sine kind of acts in this regard as an odd monomial. And so what a quinketing that the powers of x are odd for an odd function, that's not a quinketing, right? That's actually something we'll talk about a little bit more later. What I wanna do for the rest of this video though is actually kind of do this exercise again for sine. So we wanna find a Taylor series. But this time let's center it at pi thirds. Because of some of the details we did on the previous slide, we can kind of speed this process up a little bit. We know what the derivatives of sine are gonna look like. We need to look at the zero derivative and evaluate that at pi thirds. That's gonna be sine of pi thirds, which remembering the values from your unit circle, that's gonna be three over two. If you do the first derivative of pi thirds, remember that's gonna be cosine of pi thirds. And that's when you get one half. If we do the second derivative at pi thirds, remember the second derivative here is gonna be negative sine. We evaluate this at pi thirds, and therefore we get negative root three over two. And then finally, if you do the fourth derivative, no, sorry, we're on the third derivative right now. The third derivative of sine was negative cosine. We evaluate that at pi thirds, and that gives us then a negative one half. And so this one's a little bit more difficult to express because we don't have as much simplification as we did before. But from what we saw before, you take the sum as from it, where n ranges from zero to infinity, you're going to get the nth derivative evaluated at pi thirds. This sits above the in factorial. Don't forget the in factorial. You're also gonna get x minus pi thirds to the n. That's gonna show up. And so if we write this out in expanded form, this will look like the root three over two plus one half times x minus pi thirds. Then you're gonna get, I guess the next one's minus, root three over two times x minus pi thirds squared minus one half times x minus pi thirds cubed. And then this pattern will just repeat itself over and over and over again at that point. Oh, I'm sorry, I forgot the in factorials didn't die. I told you not to do it, and then I did it myself. So actually they don't start picking up until the second term because zero factor and one factor are both one. So you're gonna get a two times two right there. You're gonna get a six times two right there. So if you prefer, you can write that as a four and you can write this as a 12. Again, if you want to, or you can leave it kind of factored. And so this gives us the Taylor series. If you want a little bit more compact formula, what you can do is you can kind of break up the even terms together and you can then, you can combine the odd terms together because the evens and odds kind of behave differently for sign here. And so if you wanna gather all the even stuff together, you would get this sum, we're in ranges from zero to infinity. We're going to get x minus pi thirds to the two end. So we're only grabbing the even powers there. And in that situation, all of the evens always have a root three over two in front of it. So you can describe three times two times two in factorial. But it's also an all-terrain sum. It's gonna go positive, negative, positive, negative, positive, negative. So you get this negative one to the end. This only will describe the even terms or only grabbing even powers of x two to the end. If you did that with the odd terms, and equals zero to infinity, that one also is gonna be alternating negative one to the end. You don't get a square root of three this time, but you are gonna get a two in the bottom times, since these are now odd, you're gonna get two in plus one factorial. And that's gonna be multiplied by this x minus pi thirds. And this will go to the two in plus one power. And so in this situation, we're just grabbing the odd terms in this sum. And so this one's a little bit more messy than sine was, but well, the Maclaurin series for sine I should mention. That's generally what's gonna happen. The Maclaurin series is generally much cleaner as a power series than these ones centered somewhere else. And that's one of the reasons why we really like the Maclaurin series so much. Now I should mention that in both of these examples, we didn't actually say it. What would be the radius of convergence in this situation? What's the radius of convergence for this one right here? And so this one gets a little bit trickier because we have to look at the consecutive terms right here. But what's gonna happen? What happens if you try to do the ratio test? You take a n plus one over a n, right? In this situation, again, things get a little bit more funky because you're going to get f of n plus one. Well, I guess we only have to worry about the odd. We only have to worry about the odd positions here. And so because of that, because all the even ones just disappear. And so you're gonna end up with something like negative one to the n plus one times x to the two n plus three. I replaced the n with n plus one distributed. And then you have this two n plus three factorial down here. And then if you take one over a to the n, you end up with this two n plus one factorial over the denominator, you get negative one to the nx to the two n plus one. That didn't turn out very well, two n plus one. Now you have to be a little bit more careful as you simplify this. The negative ones you can just ignore since they're inside absolute values. When you cancel out the powers of x, you are gonna take away a two n plus one and that'll leave actually an x squared behind. And the same similar things happen right here. When you cancel out this, you're gonna end up with a two n plus one, two n plus three and a two n plus two, like so. And so you end up with the absolute value of x squared over two n plus three and two n plus one. But much like we did with the natural exponential as n goes to infinity, there's no choice of x for which the square on top can compensate for this because we're gonna go off towards the absolute value of x squared over infinity. And there's no real value that can compete with dividing by infinity there. This will go off to zero. And therefore the limit of the ratios is always one which tells us that then the radius of convergence will be infinity as well. So we see that this in the chlorine series and also for the Taylor series, we see that this will also have a radius of convergence equal to infinity, just like the natural exponential.