Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Some standard formulas that get used for Taylor series is cosine of X. So if we thought about figuring out what cosine of X equaled as far as a Taylor series goes, we get F prime of X equaling negative sine XF double prime of X negative cosine XF triple prime of X, sine XF to the 4th X cosine X. If we then evaluate what happens when our a IS0F is 0 is 1F prime is 0 is 0F double prime is 0 is -1 F triple prime IS00F to the 4th 0 is one. What we can see is that every other term is going to be 0, and so when we plug it into the formula, we'd have 1 / 0 factorial X to the 0 + 0 / 1 factorial X to the first plus. -1 / 2 factorial X ^2 + 0 / 3 factorial X ^3 + 1 / 4 factorial X to the 4th plus... So when we write it in a summation, we're going to realize that every other term is going to go to zero. That's gone, and this one's gone. So what we're really focusing are things with even exponents. 024. So my X is going to be to the two N power. Every other term is going from negative to positive. So we're actually going to need a -1 in there. We want our first term to be positive. So when it's zero, we'll have it to the north, power -1 to the north. And then the first, the next term after that to the first would make that one a negative. The next one would be positive. So X to the two N over 2 N factorial. We're going to look at the same concept with sine X. So if we start with sine X, we get our first four derivatives. We find those locations for F of 0 and F prime of 0, etcetera. Now we can see that we have quite a few terms that are going to go to 0 every other term. So now when we look at this, we see X is to the 1st and then X is to the third. So now we're talking about the odd powers. So we're going to have X to the two n + 1. Over one and then three. So 2N plus one factorial in the denominator, I need the first term to be positive, so -1 to the N So when N is zero -1 to the zero is a positive. But for the next term out -1 to the first, we give us that negative. So this is the summation that's equivalent to sine X. Now if we had sine Pi X / 2. Find our F prime or F double prime or F triple prime or F to the 4th. Evaluate these all at zero, and we can see that we're going to get 0 / 0 factorial X to the 0 + π halves over one factorial X + 0 / 2 factorial X ^2 + -π ^2 ^3 / 3 factorial X ^3 plus... So when I wrote this, I got n = 0 to Infinity of -1 to the N Pi halves to the 2N plus 1X to the two n + 1 / 2 N plus one factorial. The solutions manual wrote it a little more complex. They wrote -1 to the N π to the two N plus 1X to the two n + 1, the two to the 2N plus one of the denominator, and two n + 1 factorial. Those are really equivalents. When I kept taking the derivative the inside, I left that whole term together so I could take it to an exponent power. Now the reality is if we came back here to the sine X formula and we figured out our summation, we could actually stick in to this control copy. If I look at this formula and now instead of the XI put in Pi X / 2, so instead of this X I'm going to put in Pi X / 2 to the two N plus one. We could see that that's Pi halves to the 2N plus one and X to the two n + 1. Which would be exactly the same thing we had just by realizing we're replacing this with the equivalent of what our X was in the generic formula, we could save ourselves a lot of this work if we see the pattern this next one. If F of X is X ^2 / 2 - 1 plus cosine X, we showed that cosine X is really n = 0 to Infinity of -1 to the NX to the two n / 2 N factorial. So this is going to turn into X ^2 / 2 -, 1 + 1 -, X ^2 / 2. So these first four terms are all going to cancel, and we're going to end up with just X to the 4 / 4 factorial minus X to the 6 / 6 factorial. So we could think of this function as the summation when n = 2 to Infinity of -1 to the two X to the 2 * n. Over 2 * n factorial. So if we started with two, we'd get a positive 1X to the 4 / 4 factorial. If we went N equal 3 next, we get a negative X to the 6th over six factorial. So that's an equivalent equation. If we started with F of X equaling X ^2 sine X, we know that our sine X is the summation formula. So when I multiply it by X ^2, all I've really done is take this X value here and increase the exponent by two. So X ^2 sine X is just n = 0 to Infinity of -1 NX to the two n + 3 all over 2N plus one factorial. Thank you and have a wonderful day.