Finding Taylor or Maclaurin series for a function
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Maclaurin series of sin(x)
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 a Taylor series, where we are centering our approximation around x is equal to 0. So it's just a special case of a Taylor series. So let's take f of x in this situation to be equal to sine of x. And let'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 the sine of x is the derivative of cosine of x, which is negative sine of x. The third derivative is going to be the derivative of this. So I'll just write a 3 in parentheses there, instead of doing prime, prime, prime. So the third derivative is the derivative of this, which is negative cosine of x. 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 times. And we care-- in order to do 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. I'll do it in this purple color. So f-- that's hard to see, I think So let's do this 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 then the fourth derivative evaluated at 0 is going to be 0 again. And we could keep going, but once again, it seems like there's a pattern. 0, 1, 0, negative 1, 0, then you're going to go back to positive 1. So on and so forth. So let's find its polynomial representation using the Maclaurin series. And just a reminder, this one up here, this was approximately cosine of x. And you'll get closer and closer to cosine of x. I'm not rigorously showing you how close, in that it's definitely the exact 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 go to infinity, you're 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're not even going to need to include that. The next term is going to be f prime of 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're 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 to the third over 3 factorial. And then the next term is going to be 0, because that's the fourth derivative. The fourth derivative evaluated at 0 is the next coefficient. We see that that is going to be 0, so it's going to drop off. And what you're going to see here-- and actually maybe I haven't done enough 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 the fifth derivative of x is going to be cosine of x again. The fifth derivative-- we'll do it in that same color, just so it's consistent-- the fifth derivative evaluated at 0 is going to be 1. So the fourth derivative evaluated at 0 is 0, then you go to the fifth derivative evaluated at 0, it's going to be positive 1. And if I kept doing this, it would be positive 1-- I have to write the 1 as the coefficient-- times x to the fifth over 5 factorial. So there's something interesting going on here. For cosine of x, I had 1, essentially 1 times x to the 0. Then I don't have x to the first power. I don't have x to the odd powers, actually. And then I just essentially have x to all of the even powers. And whatever power it is, I'm dividing it by that factorial. And then the signs keep switching. 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, because-- well 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, especially 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. This is x to the third over 3 factorial plus x to the fifth over 5 factorial. Yeah, 0 would be an even number. Anyway, 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 sines. X to the seventh over 7 factorial plus x to the ninth over 9 factorial. So there's something interesting here. You once again see this kind of complimentary nature between sine and cosine here. You see almost this-- they're kind of 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 its polynomial representation, is all of the odd powers of x divided by its factorial, and you switch sines. 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.
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