6.3 Taylor and Maclaurin Series

[STARTS AT 37:35]

Professor: For each of the following functions, find the Maclaurin series and its interval of convergence. Now, I don't need to find the interval-- or the Maclaurin series-- because we know the Maclaurin series for this is n equals 0 to infinity of x to the n over n factorial, based on our previous work. So first, let's see where this converges. 

Let's go with our ratio test. On the ratio test the absolute value of x to the n plus 1 over n plus 1 factorial divided by x to the n over n factorial, that would be an x, and that would be an n plus 1, x over n plus 1, which converges to 0. It converges to 0. So our-- and this is true always, because it doesn't depend on the value of x-- so our R is infinity. Our interval of convergence is negative infinity to infinity. So it converges everywhere. 

However, do we know that it converges to f of x? We don't necessarily know that. So let's say if we can show that the remainder term goes to 0. So the maximum value of any derivative of e to the x, on the absolute value of x being less than or equal to some value, b-- we're just going to pick a number, call it b-- the maximum value is e to the b. We are going to make this value equal to our M. That's what we're going to use. That's our maximum value to try out. 

So the absolute value of Rn of x is less than or equal to e to the b over n plus 1 factorial absolute value of x to the n plus 1. Well, by the divergence test-- and also, you might consider the squeeze theorem-- this is going to have to go to 0, because we already showed that that series converges. Because that really is represent-- this portion, or under this sequence, is represented by this series. Because this series converges-- because it converges right here everywhere, any value of x-- then we know that this sequence must go to 0, from the divergence test. Therefore, e to the x is equal to that series. And that is true for all x. 

Next, let's look at sin x. This will be the last thing we consider. Now, we've said this is represented at x equals zero. Actually, let me go back up here. Close to zero, that is a good approximation. But we just showed that they are equal. That's important to note. So here, this is approximately-- it's represented very well by negative 1 to the n x to the 2k-- 2n plus 1-- over 2n plus 1 factorial. 

Now, if we take the ratio test again-- if we take our ratio test, that is negative 1 to the n plus 1, this will end up being 2 to the-- x to the 2n plus 3 2n plus 3 factorial negative 1 to the n x to the 2n plus 1 over 2n plus 1 factorial. This is equal to-- I'll go ahead and write this as negative one of the n plus 1 x to the 2n plus 3 over 2n plus 3 factorial times-- we'll flip over the second part, take the reciprocal there-- negative 1 to the n x to the 2n plus 1 over 2n plus 1 factorial, which is then equal to-- I'm going to go ahead and drop off the negative 1 to the n-- so the negative 1 term, the absolute value of that is just 1. 

So this is going to be the absolute value of x squared. And with the factorials, we get left with 2n plus 3 2n plus 2. And this converges to 0. So our radius of convergence is infinity. Our interval of convergence is negative infinity to infinity. But now we have to see if they are equal. This does converge. But does it converge to sin x. 

Well, our n of x, we want to consider that, this is equal to the n plus 1-th, here evaluated at some c in that-- in our interval plus 1 factorial x to the n plus 1. Now, because any derivative of sin is bounded by 1, we can go ahead and say, the absolute value of Rn of x is less than or equal to x to the n plus 1 over n plus 1 factorial. 

Now, by the same reasoning earlier, this converges to 0. Therefore, that power series converges to sin for all values of x. x to the 2n plus 1 over 2n plus 1 factorial. 

[ENDS AT 44:30]