 Welcome back. We're at lecture 54. I really thought we could start from the very first moment with chapter 8, section 8, but there's an interesting, not that they aren't all interesting, but a pretty unique kind of problem in the web assign that I think would be worth a few minutes of class time. And since it's due tonight, you probably share that thought that it might be good to look at it in class today. Somebody read it again. The function is x cubed. Is that right? The value is a equals negative 1. Center that a equals negative 1? Okay. So with Taylor series, and this is series now, not a polynomial, so we're going to let this run indefinitely. It doesn't run indefinitely, but we haven't really encountered that yet, so probably ought to see what happens. We want the nth order derivatives at this value, right? At a over n factorial x minus a, which in this case is minus negative 1, which is plus 1 to the n. We know we're going to need higher order derivatives, so let's get those. You'll start to get a clue that it's not going to go on forever as we go down this list. Excuse me. So the first derivative is 3x squared. Second derivative is 6x. Third derivative is 6, and here's why we kind of lose all the terms beyond this one because the fourth derivative and beyond would be what? Zero. So all the rest of them. So if the fourth derivative, fifth derivative, and sixth derivative, they're all zero, then those terms have a zero in them. Zero times whatever else is there, this and this, doesn't matter, they're all zero. So even though it's supposed to go on forever, and technically I guess it does, it repeats terms of zero all the way out to infinity, it does kind of stop. So for n equals zero, what do we have? That means the original function at negative 1, which is negative 1 cubed, right? Times x plus 1 to the zero over zero factorial, that's the first term, is that correct? When n is 1, we want the first derivative, so that'd be 3 times negative 1 squared, which would be 3, x plus 1 to the 1 over 1 factorial. You'll see what happens when we're done and we've got kind of an internal check that can we write this function as if it were a polynomial? Well, by all means we can, because it is a polynomial. Can you stand and walk upright and breathe like a human being? Yes, because you are a human being, okay? So it's kind of one of those that we can check it at the end to see if it really is what it's supposed to be. When n is 2, we want the second derivative at negative 1, which is negative 6, x plus 1 squared over 2 factorial for n equals 3, third derivative at negative 1, while the third derivative is 6, it's a constant, x plus 1 cubed over 3 factorial, which is 6. So this particular function, f of x equals x cubed, can be written, and I didn't go any further because if we went on to the fourth derivative at negative 1 at 0, so whatever else is there is 0. Every one of the terms beyond this, they're all 0. So negative 1 plus 3 times x plus 1 minus 3 times x plus 1 squared plus x plus 1 cubed. Does that look right? Now we know what f of x is. f of x is x cubed. So let's see if we do all the stuff that we're supposed to do. This isn't part of the problem. Technically you're done with the problem when you get here, but this is why I think it's probably worth looking at to see that it really is just x cubed on the right side. So we've got a negative 1, and we've got 3 times that, so that'd be 3x plus 3. We've got negative 3 times x plus 1, the quantity squared. So let's square it and then distribute the negative 3, which would be what? Negative 3x squared. The middle term is 2x, so minus 6x. The last term would be 1 times negative 3. So there's that binomial squared with the negative 3 distributed, and then we've got x plus 1 cubed, which is what? x cubed. How's that go? x plus 1, the quantity cubed. 3x squared. Right, the coefficients are going to be 1, 3, 3, 1 when you cube something, and the power of x descends, and then 1 us ends as you go from left to right. So 3x, and then that one is 1. The coefficients are 1, 3, 3, 1. x cubed, x squared, x to the first, x to the zero. This actually, if you checked it, this would have 1 to the zero, 1 to the first, because 1 is our second term of the binomial. So the ones really don't affect anything. Well, minus 1 plus 1, they knock each other out. 3x and 3x is 6x, and then we subtract 6x, so they knock each other out. Plus 3, minus 3, minus 3x squared, plus 3x squared. How about that? Isn't that what we started with? There isn't a point. That's why, in fact, we haven't done any of those. The point of Taylor, the main point, not that this is point less. I mean, it's kind of neat that you can write it in another form and generate it this way, but the main point of Taylor and McLaurin series are taking things that clearly are not polynomials. They're not even like polynomials. Sine of x is very unlike polynomials. But you can write the sine of x as if it were some infinitely termed polynomial. You can write cosine of x, which is not a polynomial, as if it were a polynomial. Tangent of x, inverse tangent of x, e to the x. So these things that are transcendental, that's kind of cool and interesting that we can write them as if they were simple things like polynomials. Can we write a polynomial as if it were a polynomial? Well, I would hope so. So in fact, we just did. Even though the form of it looked different, because we started with a polynomial, the polynomial we end up with is 100% equivalent to what we started with. And I guess of mild interest, since it's not like great interest, but it seems like these things ought to go on forever and supposed to go to infinity. But in fact, these don't because some of your higher order derivatives down the line are zero. So technically they go on forever, but after a certain point in time terms are zero, so it doesn't look like they go on forever. So it has some merit, but not a whole lot actually. Not very interesting when they're not something that is clearly not at all like a polynomial, but yet we are able to write it as if it were. All right, 8.8. It's called the binomial series. We'll do a little bit of background on this. Trying to get everybody in agreement that this is in fact how you could expand a binomial. So let's start with a kind of a generic binomial and then we're going to adapt it to one that's a little bit simpler, but also allows for exponents that aren't necessarily nice, clean, positive integers. So let's start with some things we know about how to expand binomials and some patterns that have been made clear to you over the years in your math and math related classes. So we know if we've got A plus B or any binomial that the first term of the expanded form is whatever is the first term of the binomial raised to that power. In this case, the next coefficient, or we probably didn't even remember it this way, or learn it this way, the middle term of something squared is twice their product is probably how you learned it, but we'll also be able to pick up another pattern there and then it is squared. So if it's squared, it's got one more term than the power to which we're raising the binomial. So it's got three terms. So we're at the place where we need the last term, which is the second term of the binomial to its higher power. So some things that are here that we don't necessarily think about on this. Here's an A squared then an A and then here's an A to the zero. So the A's are descending. Here's a B squared. We're going the other way. Here's a B and in here there's a B to the zero. So the A's are descending. The B's are ascending. And then the coefficient we could get in one to one from Pascal's Triangle or some other way. We just remember the coefficient. But it's also true that the coefficient is the power to which we're raising that binomial. Let's see if that's true as we work down the page. And we're not going to do a lot of these. But let's do enough where we can pick up patterns that we've used over the years and all agree that there's also another pattern that's there even though we maybe didn't use that pattern. So the first term would be, and we just did a cubing function, would be the first term of the binomial to the highest power, which in this case is cubed. This is cubed so we should expect four terms, one more than the power to which the binomial's being raised. And we know that our last term is going to be B cubed, which is the second term of the binomial to its highest power. What's going to happen to the A's as we go from left to right? They're going to be descending. So we're going to have an A squared here and then we're going to have an A and then this one technically has an A, but it's A to the zero. So cubed squared to the first and A to the zero. The B's should be ascending or ascending. So we've got a B to the zero here. We're going to have a B to the first, B to the second and B to the third. Now let's try to pick up on this simple pattern that if we're cubing the binomial, the coefficient here ought to be three, right? We saw that this was squared and the second coefficient was two. We're going to see that as true all the way down this page or at least as far as we go. Let's pick up the fact that we just did one of these and the coefficients were one, three, three, one, right? So here's our one, here's three. This is going to be three and then we've got another one here. So these can be determined probably the first way you did this was this Pascal's triangle. This would be the coefficients of a binomial to the zero. Here's the coefficients of a binomial to the first. One A plus one B. Here's a binomial squared. One A squared plus two AB plus one B squared. Here's a binomial cubed, one, three, three, one, and you can work your way down. Everybody's used that? Am I okay with that? Haven't used it? So how would I generate the next row for the coefficients of some binomial to the fourth? You add the... Right, you add the two above it. So there's one, these two would be four, these two would be six, and then four, and then one. Now, are you seeing though that the lead coefficient, excuse me, the coefficient of the second term, here was two, here is going to be three, and then the next one it's going to be four, right? So it is the power to which we're raising that binomial. I'm going to leave some space there because we're going to come back to this one and revisit it to see if there's another pattern. Does anybody remember another pattern? Or anybody use another pattern all the time in trying to develop something like A plus B, or any binomial cubed or to the fourth, or anybody have another pattern? Don't you feel like A plus B? To do, to come up with this? I mean, there are other ways of coming up with this as far as multiplying it out. You could do this one times A plus B and come up with that. You might revisit that thought that's working in your mind. Let's do this one and we'll revisit the cubed and we'll revisit the fourth and see if we see a common pattern other than the one that we're using. First term is what? It's going to have five terms and the last term is going to be B to the fourth. Here's our coefficients when we need them, one, four, six, four, one. But this is to the fourth, so our first coefficient is going to be four, A cubed B to the first, right? We'll get this second pattern in a second, but let's go ahead and get the fact that the next coefficient is going to be six, A squared, B squared. Another pattern is that the power of this term is to the fourth because you would add the powers of A and B. A is cubed, B is to the first, so the power is to the fourth, some of the power is to the fourth, some of the power is to the fourth, and so on. Another pattern that's evident. And we want A and B cubed, again the sum of the powers is to the fourth, so the power of that term is four, the power of every term is four. Let's see if we can use each term to generate its successor. So, and I'm going to do so with exponents. Let's start with this one, and then we'll backtrack to the previous one and we'll go ahead to the term to the fourth. If I want to generate this value, this three, I could take this exponent, this power, take it times the lead coefficient that's already there, which in all cases is going to be one, so in essence just this power, divide it by the number of this term, I'll clarify that in other numbers before we're done with this, but the number of this term is one, that should generate the coefficient of the next term. Let's see if it works as we work our way to the right. I'm going to take the power of A times the coefficient that's already there, divide it by the number of this term, which is two, the two's reduced, and that should generate the coefficient of the next term. Has anybody used that pattern before? Is it like the derivative divided by the exponent? It kind of will eventually be related, and you should expect that in this chapter, that it's going to have something to do with derivatives. But at this point in time, in fact it has something to do with derivatives and factorials. But at this point in time, it's just the power of the first term of the binomial times the coefficient that's there divided by the number of the term, and that generates this. We should be able to do it here also, should be one times the three that's already there divided by three, because this is a third term, and that should generate the next coefficient, which it does. Let's see if it works down here. So I'm trying to generate the next coefficient. Four divided by the first term, there's the next coefficient. Three times the four that's already there divided by two, is that the next coefficient? Seems to work. Two times the six that's already there divided by three, because this is the third term. There's our 12 divided by three, which is four, which is the next coefficient. So this is not needed. This pattern kind of renders that unnecessary, because that gets a little cumbersome after a while anyway. If you're raising something to the 17th power, we can start that, and it's not any worse, really. It's just higher numbers than this right here. Take the one times the four that's already there divided by four, because this is the fourth term, and that should generate the next coefficient. That pattern, okay, seems to be there, and if you want to check it, it does work here also. Two divided by one is that one. One times two divided by two generates the next coefficient, which is one. So if we had a plus b, I don't want to get ridiculous with this, but let's say we have it to the 12th. We don't need Pascal's Triangle. We can just go ahead and start writing these terms out. The first term we know is a to the 12th. What's the next term? 12 a to the 11. b to the 1. All right, and let's use this pattern to generate the next coefficient without having to visit this guy again. So it should be 11 times the 12 that's here divided by two, because this is the second term, and I'm trying to generate the next coefficient. So what is the next? 66 a to the 10th. b to the 2nd. And let's do one more. The next coefficient should be 10 times 66 divided by three, 220. And if you doubt that, then go ahead and do your Pascal's Triangle down to the row where it's got a 12 in it. It's got a one and then a 12. And check this out. So that's a to the 9th, b to the 3rd. So that should work. Now let's relate it to a little bit more as far as how we got there rather than just the power and the other coefficient and so on. Let's take, let me write this one on another page because that's going to get cluttered. Already is cluttered. So we want something that's not necessarily tied to the preceding coefficient or the preceding term. We want to come up with something to work whether we find the 3rd term and the 4th term in order to get the 5th term or not. Maybe we want to go right immediately to the 5th term. So right now we're kind of stuck in that recursive mode where we have to find the predecessor to find the successor. So we decided that to get this coefficient we were going to take basically, I said four divided by one, but that gave us this. Now the four divided by one, here's the way I wrote it the first time through. We wrote three times the four that's already there divided by two. I'm going to take that and write that slightly different that's not so dependent on the predecessor and I think we'll get a pattern that we don't need the one before it. So I wrote three times four. Now it's really three times four over one. Four is the same thing as four over one. I know that seems a little, you know, that's what's the big deal. But I want to get the one in the denominator and I also want this new one to come into the denominator. So I've got three times four. Might even be better if it were four times three. I know that sounds stupid. But for the purposes, the four times the three is how we're going to see how the pattern works. But in the denominator it looks like we have what working? Factorials are starting. So what did we do? This generates the sixth. Now in the previous problem, I took the sixth times the two and divided it by three. I'm just going to take all that but I'm going to write it slightly different. So isn't this whole thing six? Right? So that's where we got the six. So it's three times four times two times one. There's our six and what did we do with that? Didn't we multiply it by two? Didn't we have six times two divided by the number of this term? Well, there's our two. So here's our six. We multiplied that by two and we divided this by what? Three. Three because this was the third term. Well, if I divide it by three, I'll just throw a three into the denominator. Now the denominator's kind of having that factorial look, right? Do you see a pattern developing in the numerator? Okay, it's kind of a strange thing but the first one, it was four. Then it's four times three. Then it's four times three times two. Let's see if we can generate the next one without having to tie it to the one before. So this is four over one. Then the next coefficient was four times one number less than four over two factorial. The next coefficient was four times three times two all over three factorial. So if that pattern is going to persist, what's the next coefficient? Four times three times two times one all over four factorial. Well, four, three, two, one over one, two, three, four, that's one. That's our last coefficient, which means that we're done with this process. Does that work? So let's jump down a little bit to this and then we can see how this ties in with the stuff in this chapter. And then we'll try to put it into something that we think is going to work all the time. The next coefficient, if we do this right, should be six over one, right? Now, without tying it to this term, based on the pattern that we saw happen right here, what should the next one be? Six times five over two. Which would be 15. You can check that out. It's not like going all the way down the page of Pascal's Triangle. I think you'll find that coefficient to be true. So if we use this pattern again, six, five, four over three factorial. What's that one? Well, the six and the three factorial knockout. So 20. Did you get that, Sarah? Yeah, I did. It made me really happy. Good job. I'm happy that you're happy. What's the next one? Six, five, four, three. Is that a 20 again? Is that right? 15. So it's either got to be 15 or 20. Either the 20 is going to be repeated or we've kind of reached this point of symmetry and now we kind of backtrack. So 15a squared b to the fourth. The next coefficient should be six, five, four, three, two. Which better be six if the symmetry thing is working. And the last coefficient, the numerator and denominator should be the same, but if you just take this one step further and instead of five factorial, we're going to have six factorial. The numerator and denominator are both six factorial, so the last coefficient should be one. So this is a pattern that appears to be working. If we were going to try to put this into words about where to stop. Like here we stop at six and one factorial. Six and one number less than six over two factorial. The denominator seems to be pretty clear. One factorial, two factorial, three. So that's going to be easy to put into symbols. The numerator not quite so easy. So what was six originally? It was the power to which the binomial was being raised. So where do we stop, I guess, is the way to kind of clarify. So, and let's compare it to, I guess we could compare it to the power of A in that term. So in this case we would want six, but we would want to stop there. Here we'd want six times five, and we'd want to stop at five. So I guess you could just say what is the power of the first term of the binomial? That's the number that we want to stop with, right? When this is four, we want to stop at four. When this is three, we want to stop at three, and so on. So how do we describe what that is as you work your way down the expansion of the binomial? Well, whatever this is, if this is A plus B to the N, and then we're going to kind of, in a way, make it simpler, in a way, make it slightly more complicated, so the net gain is about zero. But if we have something, let's say N here, then we would want our first coefficient to be N, our first exponent to be N, and then for our next coefficient, which is the first non-one coefficient, we would want this to be what? Times A to the N minus one, B to the first. I'm going to switch pages here. That's getting a little cluttered. Isn't this proceeding like it's supposed to proceed? Is that right? So tell me what the next one is. Is that right? This is probably a little better way to compare it because we don't get tangled up with the numbers. We actually see the letters. So here, A is to the N minus two, but we stopped at N minus one. The next term should be N times N minus one times N minus two, and we should stop there. Sorry, there should be a B squared here. So the value that we're stopping at, in this case is N minus two, even though the power to which the first term of the binomial that it's being raised is to the N minus three. So how would you classify what that is? Is that N minus two factorial? How would you compare that? I heard you rephrase my question because I don't think it was a very good question. How would I write what N minus two is compared to what the power that A is being raised to in this expansion? Isn't the two one number greater than this number? And isn't the one one number greater than this? So would this be a fair description of what that coefficient should be in each case? So if the power of the term is N minus three, we want this to be one number greater. So N minus K plus one, where this is A to the K power. So let's see if this is the pattern. We'll get a cleaner look because this one has B's in it and eventually the one we're going to work with is instead of A plus B to a power, it's going to be one plus X to a power, so that makes things a little bit cleaner. But we just want that number to be one larger than that. So if this, for example, is three, then we would have N minus three plus one. That's the way it was, right? So it actually becomes N minus two is where we ended this. When this is two, this is one number greater. It's going to be easier than this when we get to the one plus X to the K, because we're actually going to throw it into the Taylor expansion. But we're going to see that pattern, that N minus K plus one, because K, the power of A in this case, and the last number that we wrote to generate the coefficient are not the same, but they're going to differ by one. So here's where we, I guess, kind of simplify things a little bit. We want to take a binomial. Let's take a simpler binomial this time. Instead of A plus B, we want one plus X. The one really helps things because one to higher powers is just one. So then we're going to have, the only thing that's present is going to be powers of X. What's that sound like? If all we have are numbers and powers of X, that's power series, right? So it is related to what we've done thus far in this chapter, and that takes all the binomial expansion stuff and does related to the stuff we've done thus far. Kind of classifying each term as we progress. So the eventual goal is to get this in this kind of a form where we're establishing out here what this thing is as it progresses out to the right. So to get there, we're going to use a Taylor series, I guess technically a Maclaurin series because we're going to center it at zero for this function. So we're going to need higher derivatives of this function. So we want to put it in that form. So over here, let's get derivatives, and then we want to evaluate eventually those derivatives at zero. So what's the derivative of 1 plus X to the k times 1? Derivative of what's inside? Derivative of 1 plus X would just be 1. Now you'll start to see some of the patterns that we saw with just kind of generic binomials. You'll start to see them develop here. What's the derivative of this derivative? Okay, don't we take this exponent times the coefficient that's already there? Wouldn't that be k times k minus 1? 1 plus X to the k minus 2. What's the next derivative? Don't we take this exponent times the stuff that's already there? Out in front in terms of the coefficient, would that be k times k minus 1 times this exponent which is k minus 2 times 1 plus X to the k minus 3? Do you see that accumulation type pattern coming? Let's get one more and then we'll generate what this Taylor series looks like and we'll see a very similar result to what we saw with a plus b to the n or a plus b to the k. So the next derivative would be the stuff that's already there times this new exponent 1 plus X is to the k minus 4. That's getting kind of crowded over there so let's bring this over here. We want the original function at 0. We want all these at 0. So the original function at 0 is what? 1. The first derivative at 0 is, okay, this is just 1 to, right now, as far as we know, integer powers, which is going to be 1. The second derivative is k times k minus 1. Third derivative, and this is as far as we need to go. I think the pattern is pretty clear. Third derivative is k times k minus 1 times k minus 2 1 plus X to the k minus 3 at 0, 1 plus 0 to the k minus 3. That's just going to be 1. So we're finding terms to replace in this position. The nth order derivative is at 0. There's our first coefficient. There's our second coefficient. There's the third coefficient. There's the next one. I think the pattern is pretty clear what happens from that point. So what's the rest? Well, we've got an n factorial in the denominator. We've already dealt with stuff that had that and we've got powers of X. We've dealt with stuff that had that in it. So the only new stuff are these coefficients in terms of these nth order derivatives at 0, which in turn become our coefficients. So let's write it out in expanded form and let's see if we can come up with kind of a closed version of that before we finish today. So the first term is going to be, well, our original function at 0 was 1 X to the 0 over 0 factorial. Not a big mystery there. The n equals 1 term, well, what's the first derivative at 0? That was K over 1 factorial X to the 1. Second derivative term. So we've done n equals 0, n equals 1. Now we're at n equals 2. K times K minus 1. 2 factorial X to the 2. Can't we kind of pick up the pace a little bit now? Is the pattern that's present pretty clear? What's the next term? K times K minus 1, K minus 2 over 3 factorial. So now let me go back to a pattern that we addressed earlier. Now it's a little more critical that we see the pattern. When the power is 2, how do we know when to stop that? That's kind of what we have to put into symbols here in just a few seconds. Because we like for stuff to agree because that makes it easier. nth derivative, n factorial, X to the n, that makes this whole process a lot easier to me. But we've got some agreement with the factorial and the power. We don't have that same agreement as far as where we stop with this K, K minus 1, K minus 2. Where do we stop that? Let's try to describe that. So there's 1 plus X to the K and this thing can go on forever. We've seen what happens when it is just a simple polynomial 1 plus X to the third. Those aren't all that interesting. It becomes interesting when it's something that's not a polynomial, but yet we're going to try to write it as a polynomial type expression. So the things that are clear, when n is 3, we want a 3 factorial. When n is 3, we want X to the 3. X to the n. All we have to do is describe generically how do we, where do we stop with the K's and the K minus 1's and the K minus 2's. We've got it started. Now we just have to say where do we stop? We stop at K minus 3 when it's 4. We stop at K minus 2 when n is 3. We stop at K minus 1 when n is 2. We stop at K minus 4's in the symbols. K minus n plus 1. Does that work? Now it's technically, I guess, if maybe some of you were thinking this, you were thinking about stopping it at one number less than n. That's also correct. Right? So if n is 4, you're going to stop it at n minus 1. K minus n minus the negative 1 is really plus 1. So that's the way you're going to see it in the text. But if this is what you were thinking, you're thinking correctly. It's that same pattern. So this is the binomial series expansion and we don't have very much time. In fact, we have no time at all to look at how we're going to use this, but at least we were able today to get it developed. So this is something of the form so we might do a little bit with this tomorrow, but we'll spend most of the time tomorrow reviewing for Friday's test.