 So we are rolling with lecture 49, how'd I do, okay? Actually, I think today is day 59 of a 71-day semester, so we're getting there. So basically, this chapter is the end of this course, because when you get in chapter 9, that is Math 242. So there's a lot to do and some, I think, really interesting stuff ahead. Let's finish up, power series, at least this particular version of power series, we'll continue with that kind of another look at power series, more specific to certain functions and using higher order derivatives of those functions. Actually in a sense, figure out how it is that your calculator probably does sign problems and cosine problems and inverse tangent problems and all those other things that we think are little machines are so smart, but they are stamped with an algorithm that we can figure out, in fact we can use, we probably don't use the 9th derivative or the 11th or the 13th derivative, but it can be done and we'll see how it is that calculators probably come up with some of those values. We are differentiating and integrating power series, so let's say that we have a function that in kind of closed form as opposed to expanded form, c sub n x minus a to the n, so we've got a power series in x minus a. If we start at zero, then our first term is just a constant and does not have an x minus a in it. If we let this thing run to infinity, not that it converges for all values of x, but we can certainly find the values for which they do converge. If we want to take the derivative, which we got to this yesterday, but we're going to use this today, so I thought it would probably be better if we began class today with this, so the derivative of the left side, f prime of x, we should be able to take the derivative of the right side. How's that going to look? n c n x minus a, or minus? n minus 1, and where do we want to start n? I think we decided we could in a sense start it at zero, but we lose that first term, that constant term, so we might as well start it at one to generate the actual term. We're going to use both of these today, so let's also get the integral. So we should be able to integrate the left side and integrate the right side. What do we get out of that? Bring c sub n along for the ride, x minus a to the n plus 1, all over n plus 1, n starting at zero. And when we do the integration, there's the possibility of a constant. Try to get at least one example today of how it is that we evaluate that constant and how that changes from one problem to another. So we can integrate kind of the argument of the description of the kind of infinitely termed polynomial if you want to refer to it as that, and we can differentiate and integrate that. So I think we had an example that we had started. We have a function that looks like this. I posed the question, what does this look like? How is that related to our original function one over one minus x? Somebody said it's that function squared, which it is, but we're going to have a hard time squaring the sigma notation description. It just so happens that this is the derivative of that. I think we validated that, didn't we? Derivative of that is, so the function is 1 minus x to the negative first, negative 1, 1 minus x to the negative second times the derivative of what's inside, which is this thing, right? One over 1 minus x, that quantity squared. So we already have a power series for this. In fact, we started class with that yesterday. What is the power series for 1 over 1 minus x? I'm going to leave the 1 out, because the ratio is x, right? And if we're going to start it at n, then we better start in at 0, because we don't want an x in our first term. We want the first term to be x to the 0, which is 1, because that is our first term, and the ratio is x. So if that's 1 over 1 minus x as a power series, and we want the derivative of that, which is equal to this function that we're handed, so we can take the derivative of this side, we also better be able to take the derivative of this side. And what is that? And we're going to lose that first term, right? Derivative of the 1 is going to be 0. So the first thing we want to generate is the derivative of this. Derivative of x is 1, so we can start in at 1 and do that. Does that work? So we were handed this problem. We actually said a couple things that we think it is. We thought it was the original function, the 1 over 1 minus x, the quantity squared. Probably not going to get us anywhere, so it's got to be the derivative or the integral of something. This happens to be the derivative. So we just differentiate it in its closed form, and we should be in business. I don't think it's a waste of time to check a couple of values. When n is 1, what do we get? When n is 1, we get 1. Is that right? When n is 2, we get 2x to the 2 minus 1, so 2x. How are we doing so far? When n is 3, we ought to get a 3 and an x to 1 degree less than that. So the derivative of that is 0, which is certainly present here. Derivative of that is 1. Derivative of that is 2x. Derivative of that, so it looks like it's going to generate the right power series. Questions about that one? We started that one yesterday. We kind of left it, but we knew what we were going to do. We just ran out of time. So I do have light blue on today. Carolina won last night. I'm not a Carolina fan, but I'm an ACC fan. And do you know that NC State was not in the NCAA tournament, but we made as much money off of the NCAA tournament as Carolina did. In fact, we probably made more because they had expenses of traveling, too, feeding the team. We didn't travel because we didn't go. So we didn't have any expenses. What they do in the ACC is we had seven teams in the tournament. So every time you have a team play a game from your league, at least in our league, the ACC, because we share all the revenue equally, all 12 teams. So seven teams made it. Four teams didn't get past the first round. Duke played three games. Is that right? I think they played three times in the NCAA tournament. Carolina played six. Maryland played two, I think. So each one of those times we have a team playing the tournament, it becomes like a share of the total revenue. So I think it's either 16 or 17 shares that are equally divided amongst the ACC. So we actually made as much money off the tournament as anybody else. And more than teams that actually played. Not that I like it that way. I'd much rather have it that we go somewhere and play. That'd be a lot more fun. But it does help the league when the ACC team wins. And I don't know if I mentioned it in here, but I, when I was a high school basketball coach a long time ago, it feels like a previous life ago. Roy Williams was also a high school coach and we used to work camps together. And a lot of times we would play poker in the evenings when the camp sessions were over at Carolina. He was, I guess he had just graduated from Carolina. He was a walk on player there. We used to work camp and play poker in the evenings until wee hours of the evening. And he is a really good guy. It's kind of hard for me to cheer against Roy Williams, except when we play him. I really like to beat him. But when we're not in the game, I'm a Roy Williams fan. He's a good guy. Now they'll probably take this show off the air that I say that Roy Williams is a good guy. Next example, natural log of 1 minus X. Little disappointed by his language this year in some of the press conferences, but that's not typical. Roy Williams not the one that I know. That was kind of disappointing. We just did a derivative problem. Again, we're going to try to relate it to this one. We decided that was X to the N. So there's what we're trying to use to find a power series representation for natural log of 1 minus X. We just took the derivative of this side and took the derivative of this side. We're moving on to a new example. Any guesses? Okay, both of those I think we can use U substitution just to make sure we have what we think we have and integral. Let's see if we can integrate this thing, this left side description, 1 over 1 minus X. Is that, I'll put a couple question marks here, is that the function that we want? How many of you think it is? It's close. Okay, some of you are thinking through it and thinking we need a negative. So if we let U, which was the other suggestion, equal 1 minus X, dU would be negative 1 dx. So there's our U value. It looks like we need a negative 1 and a dx. So we'll multiply by negative 1 outside and negative 1 inside. So there's what we have, which is dU over U, which is natural log of U, which is natural log of 1 minus X. Now we're only going to choose values for which we can actually evaluate a natural log and the fact that we were handed this function to begin with, we're not going to kind of institute our own absolute value notation. So is it exactly what we want? It's not exactly what we want, but it is the negative of what we want. If it's the negative of what we want, can't we just find the integral and when we're done integrating just negate every term? That ought to work. So let's instead of finding this natural log of 1 minus X, let's find the negative of that, which we just decided that was the integral of this. So we should be able to integrate the power series for that. This is kind of, we've already done this work so we know that, but now that gave us the right to use this particular power series, how do we integrate it? What's the integral of X to the n over n plus 1 from n equals 0? So the first term of this is what? X to the 0, which is 1. We're going to integrate 1 so we want our first term of this to be X. n equals 0 generates X to the 1, right? X to the 1 over 1 and that's what we want so n equals 0. Keep in mind that's really not what we were asked to find, so at the end we're going to have to multiply both sides by negative 1. Let's write some of this out. Actually let's write it out up here. What would this first term would be? X, next term over 2, X cubed over 3 and so on. So we don't really want that, we want the negative of that, so we're going to multiply this side by negative 1. Therefore we need to multiply this side by negative 1. So every term was positive up here and now every term is negative. I'm missing something on it. Yeah, we did the integration right here so we should have a k value in there somehow. I'm kind of adding it in now so let's, that's actually a minus, but we're going to have a k value. I don't think we found a k value yet. Any ideas on how we can see what k is for this particular series? Somebody said yesterday when we wanted to find a c or a k value we could, okay, some kind of initial conditions, boundary conditions. I think somebody said yesterday just find a point, right, that we know is on that curve. How about when X is zero here? That's going to be awfully easy on the right side to evaluate when X is zero. In this piece right here on the left when X is zero what should we get? We should get zero, right? So when X is zero on the right side we better get zero on the left side. I don't know if I need to write that out. If zero is going to be on the left side and every term that has X in it is also zero because X itself is zero, then leave a whole lot of room for k, does it? k is zero, is that correct in this problem? So these two things give us the fact that k is zero so we don't need to include it on this particular problem anymore. We should check it out to make sure what it is. So I'm going to leave it out and there's our power series. Let's do a little bit more with this. Let's put in, I guess we didn't know this one. What's the interval of convergence for this? What was the ratio that we were dealing with in generating this particular power series? That's what we had. We started with this guy right here which the ratio was X and we want the absolute value of the ratio to be less than one so in the interval from negative one to one this is going to converge. We want to stay in that interval. Let's choose a value X equals a half and let's see what happens. The left side becomes what? One minus a half would be a half, natural log of a half and I guess I just have to put an X value of one half every time I see an X. On the right side, if X is a half take one of the tougher decisions last night was like at nine o'clock do you watch the game, the NCAA championship game or do you watch 24? That was tough. I did. Unfortunately they had commercials like at the same time. Why do they do that? Don't they know people are trying to watch both? Natural log of one half, that's the same thing as the natural log of two to the negative first and can't I take that negative one and bring it out in front. That's negative natural log two which is exactly the same thing as natural log of a half. Worried about Jack Bauer, he's been exposed to that. Oh sorry, you haven't seen it yet. Wait but you knew that from last time. Got a bunch of negative signs. Could we multiply both sides by negative one and say natural log of two is one half plus one half squared over two one half cubed over three. Check that out. Somebody's got your calculator with you and it's operating. So we've got a point five. We've got a half squared which is a fourth divided by two which is an eighth which is point one two five. A half cubed which is an eighth and eighth divided by three is one over twenty four. What's one over twenty four? One six repeating and then one half to the fourth would be one sixteenth divided by four one over sixty four two five. So we could get exact decimal values for all those. If you looked up natural log of two in your calculator what does it say that is? Okay, however we're doing with those. We add those together are we getting closer and closer to the value that our calculator is giving us for the natural log of two. I think we ought to be getting pretty close. If we continue this process should we expect it to get should we expect this process to get us closer and closer and closer to the actual value for the natural log of two? Why? Why should we expect that? It's all positive values. Okay and why is there? Let's get back to that point. Equal, okay? Because it's equal to that we said it was going to converge, right? So this thing if we let it go all the way to infinity it does converge to the actual value for natural log of two. So it is basically that. It is a convergent power series. If these are the values that we got by plugging in x equals one half which is in the interval of convergence it eventually does have to converge. How close are we with those four? Six, eight, two, two, nine, all right? So we've only got four terms. How many are we missing? Just an infinite number. Now they're not going to be huge but they're not minuscule either. I mean the next one is 1 over 30 to 1 over 160 I think is the next one. Does that sound right? I think that sounds right. So thing is they are convergent, they are useful. I'm not going to say this is how your calculator figures out natural logs but it's not too far away from how your calculator figures out natural logs. We'll see a little bit more of that when we get to Taylor and Maclaurin series. Okay, next example. This ties together some things that we have encountered along the way. Inverse tangent of x. Isn't it true that the inverse tangent of x is the integral of one over one plus x squared, right? Remember that? And we could validate that because of the derivative of inverse tangent is one over one plus x squared. Therefore the integral of one over one plus x squared must be back to the inverse tangent. So if we had this function, could we write a power series for that function? Is it something in that form? So we did some of this yesterday. Do we have to do any major conversions to get it in the form we want? It's pretty close, isn't it? Is that it? Right? Convert the plus to minus a negative. So this ought to be the first term. There's kind of the key to getting it in the right form. We've got to get a one in that position which we now have. So it looks like the ratio is negative x squared. So if we're going to write the sigma notation version of that, first term is one. I'm not going to write that down. Ratio is to the n, right? We want zero because we want the first term to be what? One. So negative x squared to the zero would be one. So that seems to be, we can do a little better than that but let's write this out. So the first term would be one. The next term would be negative x squared. Next term? Negative. What's the ratio? It's negative x squared. So we need to take this times negative x squared which would be positive, x to the fourth. That times negative x squared would be negative x to the sixth and so on, right? Because the ratio is negative and if your ratio is negative it's going to force the signs to alternate. So the first term is positive if you multiply by the negative ratio it makes this negative. If you multiply this by negative x squared it makes it positive, so on. We did a little bit of this yesterday. Let's separate the negative part, the alternating part from the actual kind of magnitude of the term itself and let's clean that up just a little. X squared to the n is really what? X to the two n. That's right, isn't it? I mean the negative one to the n causes the signs to alternate. If n starts at zero the first one ought to be positive which is correct and then it's just the exponents are going up by two. From n equals zero doesn't have any x's, the next one ought to have an x squared and x to the fourth, x to the sixth and so on. So that seems to be correct. So there's one over one plus x squared. What do we want out of that? To do the problem that was handed to us originally. We want to integrate it. We want a power series for inverse tangent which is the integral of one over one plus x squared. So let's integrate this side and integrate this side and let's see what we get with respect to x. We know what that is. That's inverse tangent. Now let's do the integration on the power series. Do we just bring the negative one to the n along for the right? When we integrate something are we going to change its sign? Its sign is going to stay the same, right? If it was negative and we integrate it it's still negative. So we don't want that to change but we do want x to the two n to become what? Let's see, right? Add one to the power. Divide by the new power. Okay, that's the next thing I was going to ask is that's what was happening to n. n was being doubled. Let's see if it gives us what we want. When n is zero, so we're going to integrate, there was our series so we're going to integrate the first term which is one. What should we get? Well at n equals zero we're going to get negative one to the zero which is one so it's still positive. When n is zero, two zero plus one is x to the one over one. Isn't that what we want? Now let's see if it gives us what we want here. When n is one, negative one to the one is negative so we do want that. When n is one that'd be two ones plus one that'd be three, is that correct? x to the three over two ones plus one. Isn't that going to give us what we want? So we really do just want the power, whatever the final power was to go up by one. So I think this is what we want. Right, right. So we could do what Noah said except then you're going to have to start it at one, right? Yes. But this will work. We just want to take the power which was two in, add one to it, divide by the new power, that should work. So it seems to be generating, I think that's probably a good thing to do is to see if it's generating the first few terms. Doesn't take but a few seconds to do that. Seems to be the correct power series. Don't think we need a arbitrary constant over here. So it should be x minus x cubed over three with n equals two, that ought to be x to the five over five. Is that right? So how do we evaluate what k is? We can't always assume that it's zero but it might be zero. Okay let's just pick a value. Let's say x is zero. So if x is zero, that's by the way a nice convenient value because it causes all these terms to disappear. What is the inverse tangent? What angle has zero for its tangent? Isn't that what inverse tangent is? You're searching for the angle. What angle has zero for its tangent? So here's a tangent graph. Right, I mean I realize we're looking at the inverse tangent but I'm asking it in terms of a tangent. If we know the tangent is zero, what angle has that for its tangent? Zero. Right? So the inverse tangent of zero is zero. So what's that say that k is? k is zero. So we can leave that out. So we have a power series for inverse tangent is, let's check it out and see if it works. We know it does with zero. That's not necessarily going to tell us anything. Inverse tangent of, I don't know what's a common value. What angle has one for its tangent? What's the inverse tangent of one? So you're trying to think of the angle such that when you take the tangent of that angle you get one. Five or four. Forty-five degrees is the degree equivalent but we're going to work in radians so it's five over four. So the inverse tangent of one is pi over four. Well, we put four x, we put in one on the left side so we better put in one on the right side. Let's see how we're doing with just these terms. Somebody take the number pi and divide it by four, what is that? Point seven, eight, that's an approximation for that. So I guess we're approximating now. So we've got one minus one-third. Remember that whole mentality of how these alternating series progress? One was too much, right? So then we subtract some away. Oh, we subtract too much, right? So then we better add something back in. But lo and behold, we add too much back in and let's subtract some away. What's one-seventh? Point one, four. So if we stop this thing here, which we actually should let it go on forever, how close are we to finding the inverse tangent of one? And we're not very far into this infinitely termed series. So if you would let it run all the way to infinity, you would get exactly the number pi divided by four. Ben? Yes, thank you. Is that what you had your hand up for? Right, because we've got only odd powers. Yes? How did you know to do pi over four? I just wanted to pick a value that I could put in on the right side. That would be pretty easy to compute. So zero was our first one, so we could find the k, because zero on the left gives you zero on the right, therefore k was zero. So I think an equally easy value is one. So if you put one everywhere there's an x, well then that means over here there's an x, you better put in one. What angle has one for its tangent? That'd be the angle pi over four. So the inverse tangent of one is pi over four. We could put in another one that's not quite so convenient, and maybe we ought to do that just to see what happens. But it's just easy to put in one. Jacob? If you use that number to try to find k, wouldn't k be one? Because it would be, you know, 0.7854 is equal to 0.7854k. It's plus k. Plus k. Let's look at it as zero. Oh, okay. That's good. You're okay? Yeah. Okay, because I wasn't really following where that question was headed, but if you're okay with it, then okay. We wanted to figure out a good way, I'm not saying this is a great way, but if this is pi over four, I think we ought to do one more example, but I don't want to leave this one yet. If this is the number pi divided by four, which we have the luxury of calculators and we can plug in, you know, pi, and it gives us pi to whatever ten or twelve decimal places. But if we didn't really know a value for pi, couldn't we take this side, multiply it by four, so now the left side is exactly this number pi, shouldn't we be able to do that with the right side as well? And this is still an equation, right? So four times, shouldn't that work? Shouldn't this thing give us a value for pi if we let it go long enough? So they have the value for pi figured out to what, hundreds of thousands of decimal places? I think this would be a way we could figure pi to hundreds of thousands of decimal places. Maybe. I wouldn't want to do it by hand, but we could set up a supercomputer to do this all the way out to several hundred thousand decimal places, and it ought to give us the value for pi. Let's do a not so common value, actually this is kind of common. I could, I'm going to opt to not do that, okay? So we could put a value square root of three right here. What angle has the square root of three for its tangent? You give me that trig look every time we go into trig. That vacant, distant trigonometric look. So if I put in square root of three, what angle, common angle, has the square root of three for its tangent? Sixty degrees would be the degree equivalent. What's that in radiance? Pi over three. So I could use square root of three, that's a little bit of a hassle on the right side, isn't it, to use the square root of three? Because everywhere there's an x, we've got to put in a square root of three, because I put it in over here. So if we did all this stuff on the right, everywhere there's an x, we plug in a square root of three, and we group as many of them as humanly possible, we should get a number that's very, very close to pi over three, okay? That's kind of a convenient value. Let's take a value that's a little easier to use on the right side, but maybe not quite as convenient. So let's say we want the inverse tangent of two, not a convenient value. What angle has two for its tangent? I don't know. But if we want to know what angle that is, we should be able to put two in for x on the right side. So without knowing what the inverse tangent of two is, we want to check this when we're done, but it should be two minus what? Eight-thirds plus, ah, we got a problem. What's the matter with this? It's not working. Don't we have an interval of convergence for this thing? And that would be what? Negative one to one? We can't necessarily trust this series to give us what we want because we're outside of that interval of convergence, right? So we'll do better than this with Taylor series. We'll get a series for which the inverse tangent converges all the time. One for sine, one for cosine, one for e to the x, one for natural log of x, but we are outside of the interval of convergence, so it's not going to work. Let's get about a minute of this. There is one more, I think, a very good example problem in the book. Let's get a first look at it. We don't have to do everything that they do in this example. But let's say we're handed this problem. How could we use what we've been doing today thus far to integrate one over one plus x squared? Okay? We could say it's one of these, which it doesn't take much. Seventh, x to the seventh. Yeah, we've already done the x squared because that's an inverse tangent. Now it's seventh. So the ratio is x to the seventh. So, and again, we could separate the negative one to the n. x to the seventh to the n would be x to the seven n. All right, we want to integrate that. So we want to integrate this side. Let's go ahead and integrate this side. What do we get? Everything stays the same sign. x to the seven n plus one over seven n plus one over seven n plus Now, the problem in the book asks you to do something else and that is evaluate this from zero to one half, which we would evaluate this thing at one half, which means you put a half in where there's an x. If you subtract from that what you get at zero, well at zero you're going to get zero so you don't have to worry about that. And then it talks about accuracy. What we end up with is an alternating series. Pretty clearly it's alternating. And if you want a certain level of accuracy, how do you get the level of accuracy that you want from an alternating series? It's the alternating series estimation theorem. What's it say? You're never any further away from the actual sum than the value of the next term. So if you want, I think in the book it says, ten to the negative seventh level of accuracy, then you search for the first term in your alternating series. You don't have to search very far in this one. I think it's the fourth or the fifth term, such that the magnitude of that term is smaller than ten to the negative seventh, which is point six zeros and one. I think it's the fourth or the fifth term. So you can go back and use things that we've already encountered, for example, the alternating series estimation theorem to help us approximate. Because it is an infinite number of terms. We're not going to be able to find them all by hand. All right, we will start Taylor and McLaurin series tomorrow.