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Sine Taylor series at 0 (Maclaurin)

Sine Taylor Series at 0 (Maclaurin). Created by Sal Khan.
Video transcript
In the last video we took the Maclaurin Series of Cosine of x we approximated it using this polynomial and we saw this pretty interesting pattern. Let's see if we can find a similar pattern if we try to approximate sine of x using a Maclaurin series. And once again a Maclaurin series is really the same thing as Taylor Series where we are centering our approximation around x is equal to 0 .so this is a special case of a Taylor Series so let's take f of x in this sitiuation to be equal to sine of x f of x is now equal to sine of x and lets's do the same thing that we did with cosine of x . Let's just take the different derivatives of sine of x really fast. so if you have the first derivative of sine of x is just cosine of x . The second derivative of sine of x is derivative of cos x which is negative sine of x the third derivative is going to be the derivative of this so I will just write 3 in parentheses there in stead of doing all the prime prime prime. so the third derivative is derivative of this which is negative cosine of x . the fourth derivative the fourth derivative is the derivative of this which is positive sine of x again. so you see just like cosine of x it kind of cycles after you take the derivative enough time. and we care in order to the Maclaurin series we care about evaluating the function and each of these derivatives at x is equal to 0 . so let's do that. so for this let me do this in a different color not that same blue. so i'll do it in this purple color. so f that's hard to see I think. so let's do this in the other blue color. so f of 0 in this situation is 0 and f the first derivative evaluated at 0 is 1. cosine of 0 is 1 negative sine of 0 is going to be 0. so f prime prime the second derivative evaluated at 0 is 0. the third derivative evaluated at 0 is negative 1. cosine of 0 is 1 you have a negative out there it is negative 1 and the fourth derivative evaluated at 0 is going to be 0 again. we could keep going once again seems like there is a pattern 0 1 -1 0 then you are going to go back to positive 1 so on and so forth . so let's find it's polynomial representation using Maclaurin Series. just a reminder this one up here this was approximately cosine of x and you will get closer and closer to cosine of x I am not rigorously showing you how close and that is definitely the exactly the same thing as cosine of x but you get closer and closer and closer to cosine of x as you keep adding terms here and if you to infinity you are going to be pretty much at cosine of x now let's do the same thing for sine of x . so i'll pick a new color. this green should be nice. so this is our new p of x. so this is approximately going to be sine of x as we add more and more terms and so the first term here f of 0 that's just going to be 0 so we are not even going to need to include that. the next term is going to f prime 0 which is 1 times x . so it's going to be x then the next term is f prime the second derivative at 0 which we see here is 0. let me scroll down a little bit it is 0 so we won't have the second term this third term right here the third derivative of sine of x evaluated at 0 is negative 1 so we are now going to have a negative 1 . let me scroll down so you can see this negative 1 this is negative 1 in this case times x the third over 3 factorial. so negative x the third over 3 factorial and then the next term is going to be 0 because that's the fourth derivative . that's the fourth derivative evaluated at 0 is the next coefficient. we see that that's going to be 0 so it's going to drop off and what you are going to see here and actually maybe I haven't done enough pep terms for you. for you to feel good about this let me do one more term right over here just so it becomes clear f of fifth derivative of x is going to be cosine of x again. the fifth derivative let me do that in the same color just so that it's consistent. the fifth derivative the fifth derivative evaluated at 0 is going to be 1 so the fourth derivatives evaluated at 0 is 0. then you go to the fifth derivative evaluated at 0 is going to be positive 1 and if I kept doing this it would be positive 1 times I would have to write 1 as a coefficient times x to the fifth over 5 factorial so there is something interesting going on here and for cosine of x I had 1 essentially 1 times x to the zero then I don't have x to the first power I don't have x to the odd powers actually then I just have x to all the even powers and whatever power it is I am dividing it by that factorial and then the sign keeps switching and this is ,I shouldn't say this is an even power because 0 really isn't , well I guess you can view it as an even number cuz it.. I won't go into all of that but it's essentially 0 2 4 6 so on and so forth so this is interesting specially when you compare to this . this is all of the odd powers this is x to the first over 1 factorial I didn't write it here there's x to the third over 3 factorial plus x to the fifth over 5 factorial. ya zero would be an even number . anyway I don't.almost. my brain is in a different place right now and you could keep going if we kept this process up you would then keep switching signs x to the seventh over 7 factorial plus x to the ninth over 9 factorial so there is something interesting here you once again see this kind of complimentary nature between sine and cosine here. you see almost this..they kind of they are filling each other's gaps over here cosine of x is all of the even powers of x divided by that power's factorial sine of x when you take it's polynomial representation is all of the odd powers of x divided by it's factorial and you switch signs. In the next video I'll do e to the x and what's really fascinating is that e to the x starts to look like a little bit of a combination here. but not quite and you really do get the combination when you involve imaginary numbers and that's when it starts to get really really mind-blowing.