 Welcome back everyone. In this video, we are going to prove that sine of x is equal to its Maclaurin series, which you see right here, the sum where n ranges from zero to infinity of negative one to the n times x to the two n plus one over two n plus one factorial. And so in a previous video, we have already proven that this is the Maclaurin series for sine. You should hopefully be able to see a link to that video right now on the screen. What we plan to do right now is actually prove that sine is equal to its Maclaurin series. Now the basic strategy we're going to do here is to use Taylor's inequality where we can show that the remainder, rn of x, which has a reminder here, this is the difference between the function f of x with its Taylor polynomial approximations. This is bounded above by m times the absolute value of x minus a to the n plus one over n plus one factorial, where in this situation x minus a is less than a fixed number d and m is greater than or equal to the absolute value of the n plus first derivative of the function evaluated at x. Again, where x is no more than d units away from the center of the Taylor series. Now we're working with a Maclaurin series, so we do get a little bit of simplification here. a value is going to equal zero, so we're going to have that the absolute value of x is less than d. And so we have to look for some upper bound m that's greater than or equal to the n plus first derivative of x when x is absolute value is less than d. Our inequality will also look like rn of x will be less than or equal to m times the absolute value of x to the n plus one over n plus one factorial, like so. So that that's going to be our inequality we're going to work with here. So now we have to investigate this upper bound m derivatives here. Now we did a previous video about e to the x. e to the x has the beautiful property that's equal to all its all entire derivatives, so that helps us a lot in this calculation. Sine is pretty friendly in that regard as well, because sine, if we take the n plus first derivative of sine of x, you're going to get one of four options. You're going to either get plus or minus sine, or you're going to get plus or minus cosine, because the derivative of sine is cosine, its derivative is negative sine, its derivative is negative cosine, its derivative is positive sine. I hope I said those correctly. And so you're just going to get those numbers right there. The derivatives of sine are always plus or minus sine or plus or minus cosine. And so if we take absolute values of these things, absolute value of the derivatives of sine, we're going to take absolute value of these things, which makes the plus minus completely irrelevant. We need a number that simultaneously bounds sine and cosine, and we can use the number one right there. That's our m value, because the number one will bound above all of the derivatives of sine. It doesn't matter which one you use, we could always use one. And so that's a great tool right there. So our remainder, rn of x, this will be bounded above by one times x, the absolute value of x to the n plus one over n plus one factorial. So that is, this is going to look like the absolute value of x to the n plus one over n plus one factorial. And like we commented in the previous video, if you take the limit as n goes to infinity of the absolute value of x to the n plus one over n plus one factorial, this limit will equal zero. You should reference the link you can see right now if you want some more details there. So this limit is going to be zero as n goes to infinity. So as n gets bigger and bigger and bigger, this tells us that rn of x will approach zero as n goes to infinity right here. So what this tells us is that for a fixed d, right, so for a fixed d right here, this thing will go up towards zero. So this tells us that on the interval negative d to d, we get equality. We get equality right here. That sign of x equals its Maclaurin series so long as x is between d and negative d. So how big can d be? One thing you're going to notice is that this, this, this, this statement right here, right, these didn't depend on d whatsoever. Even though d was fixed, we could allow d to be bigger and bigger and bigger and bigger. In fact, allow d to go towards infinity, always be in a finite number along the process though. But as d goes off towards infinity, this inequality will still happen, right? And I should mention that since it's bounded below by zero rn of x, the squeeze theorem applies here, right, rn of x, it's bigger than zero, but it's bigger, it's greater than equal to zero, but it's also less than equal to zero, right, as you take the limit. And therefore, by the squeeze theorem, rn of x is going to approach zero. And when rn of x approaches zero, that tells you that your series is equal to its, sorry, your function is equal to its Maclaurin series. So in fact, we see that sign of x is equal to its Maclaurin series right here. So I would invite the viewer right here to try to mimic what we did here to show that cosine is equal to its Maclaurin series. The argument's getting very much the same, there's really not much that changes here whatsoever. So if you want a slightly more challenging example, try cauch or cinch, try the hyperbolic sine function or hyperbolic cosine. There's just a little bit different, right? You'll have to compute what the Maclaurin series that are. It's not much different than computing the Maclaurin series or sine or cosine. So you can see a link on the screen to a video for finding sine and cosine of the Maclaurin series if you want to. Try it for cinch and cauch and you'll see that you can show that these functions are likewise equal to their Maclaurin series. Now one has to be cautious, not every function's equal to its Maclaurin series. And in the next video, I'll show you a counter example, a function which is, which is, which is nowhere equal to its Maclaurin series except at the center. So one has to be very, very careful that we actually do show that a function's equal to its Maclaurin series. We don't just want to assume it and Taylor's inequality is the tool we use in that case.