 So let me, I guess, pick up what we left off. We left there questions or issues to be solved. So no point in asking because you never answered. Okay. So if you remember what we were talking about last time, this is like a power series which are of the form of some constant coefficient, let me call it a spoon, something like that. Or maybe it's like an infinitely long polynomial. The polynomial has infinite degree. We have a constant term which could be zero. We have a linear term, could be zero. We have quadratic term, given term, and so on. Maybe it's been shifted by some translation. So instead of being x minus the center could be something like that. Okay? So we have, so these give us some kind of, so in some sense this is a series where the series depends on a variable but in another sense it's a kind of a function if it converges. Right? So it's two different ways of looking at the same thing. So let me do I guess over an example. So these might, so there's three possibilities. Either it converges to zero, there's sort of no point to such a series but I could give you something that slopes up really fast and will only converge it actually to zero. Or it converges on some interval between about x minus c less than some radius and maybe this is less than or equal to, maybe it's not a little fancy, whatever. Or it converges whatever, so this R is called the radius of convergence, the interval might include one or both end points, it might not so you have to check that separately, okay? And it's possible that c is zero. So I guess this case is just R and c, radius. It's called a radius rather than a width or I have to width because in fact if x is a complex number then this is an actual circle. So these make perfect sense for complex numbers which we'll deal with later in the course but when these are complex numbers this is an actual radius because it's a circle or a disk on which this converges. But in our case we're only talking about intervals because you know the circle in line, you guys are waiting very, very long for that. So maybe I should do an example or two and maybe I should show them in front of my, okay? So say we have, say we have something like and it doesn't really matter what end starts at in terms of where does it converge because we don't really care about what it does at the end but for definite miss let's just start it at zero, say I have something like minus x, well actually let's absorb that. Let's say I have 2x minus 1 to the end, 5 to the end. So there's a series for example and we want to know for what x, so what is, again I want to emphasize that this is defining a function and we want to know where does this function make sense. This is also defining, if you want to look at it one way, it's defining a function of x which only makes sense possibly for certain values of x, it doesn't make sense for others. Or you can look at it as a whole bunch of series, either way, it's the same question. So here we go, yes the ratio test is a good idea. So I start with the ratio test which means that I want to know what happens as n goes to infinity of the n plus first term divided by the nth term and I'm going to write it this way. So the n plus first term, unfortunately I've used a here, oh well, sorry, this a is not valid. So we want to take the limit, I need to do a little more room, right limit, when we change n to n plus one and divide by what happens when we divide by that, so I can put n to the end there, 2x minus 1 to the end there. And then I use some algebra to simplify this thing. So this 2x minus 1 to the n plus 1 divided by that guy leaving me just the 2x minus 1 on the top, still need to take the limit. So I take care of that, the 5 to the n, 5 to the n plus 1 on the bottom divides the 5 to the n on the top, giving me a 5 here, the n plus 1 here divides that and nothing cancels and then I take the limit. So in the limit, 2x minus 1 over 5 has no n's in it so that doesn't change and this limit is 1. So that's easy, so that tells me that 2x minus 1 over 5 is my ratio, the limit of my ratio. So for very large n, this looks like 2x minus 1 over 5. No, sorry, this over the next one looks like 2x minus 1 over 5, we go down by this factor every time. So this will converge if this ratio is less than 1. So this will be less than 1 exactly when 2x minus 1 absolute value is less than 5, which is the same thing as saying 2x minus 1 is between 5 and minus 5 and I'm going to put this up just a little bit. No, I guess I know, I'm going to the next board. So 2x minus 1 is between plus or minus 5. You can sort of read off the answer from this. Well, let me do it again. It's not absolute, it's going to be absolute out. So I add 1 to both sides and now I divide it by 2. So for sure, this series will converge absolutely when x is between minus 2 and 3. The question is, so and it diverges when x is bigger than 3 and when x is less than minus 2. But we want to know what happens if x equals 3 or x equals minus 2 because in that case, this ratio will be 1. So we have to handle those separately. Is it clear to everyone why I have to do this? So we handle those two things separately. When x is say 3, the series becomes, let's do x equals 3 here, when x equals 3, the series becomes 6 minus 1 to the end and x equals 5 over n times 5 to the end. We have to write this as 6 minus 1 and so this handles that and this diverges because it's far from on first the series with b equals 1. x equals minus 2 and the series becomes minus 4, minus 1, which is minus 5 to the end. Maybe I should write it that way, minus 4 to the end. Lastly, there's a little problem here anyway when n equals 0, this thing blows up. So maybe it should start at 1. So again, this is minus 5 to the end over 5 to the end because we have minus 1 to the end, which is an alternating series. It's decreasing, the limit, and n goes from infinity to 1 over n, it's 0, so it's 2. That means that the interval of convergence looks wrong. The interval of the convergence does not include 3, but it does include minus 2, almost always, but not always. One end will be an alternating series and the other one won't, the reason it's not always is maybe if the power of x is even, so instead of having an n here and a 2n, something like that, then it won't be an alternating series. You square or not just 1, you get 1. Right? Is that clear? So this is fairly straightforward, fairly mechanical. Just go through the ratio test and you check both ends and there you go. Not a lot of deep thought involved here, just a lot of manipulation. Any questions on how to do this kind of thing? Of course, they can be a little more complicated, maybe the series needs to do the interval test to check one of the ends or something a little more complicated, but the idea is always the same. Due to the ratio test, you solve the resulting inequality then you check the two series that correspond to the end and you get an n. You guys are very blunt. So if there's no questions on these, let me move along. Is it clear? Okay. So, there are a couple of power series, well, we can just start with 1. There's a couple of power series that we know what they add up to. So if I take the most simple power series where all the coefficients are 1, this is the geometric series with the ratio x. And so, in fact, we know that this adds up to minus x, when x is less than 1. If x equals 1, you can check for the diverges in both cases. If x equals plus 1, then we get 1 plus 1 plus 1 plus 1, which is plus infinity. If x equals minus 1, then we get 1 minus 1 plus 1 minus 1 plus 1 and then the partial sums alternate between 1 and 0 so that doesn't converge. But for x less than 1, this series equals that function. So, we know, we know exactly, and it's important now because we're adding it up that we know where it starts. It still converges if we start here at x to the 10th, but it doesn't converge to this. So, in fact, let's ask that. What is the sum of x to the n that starts at x to the 10th? Is some function we know, what function is it? Hint, I'm not going to tell you. Is the geometric series, what's the ratio? The ratio is x, but a, the a in the geometric series is not 1. This guy is a geometric series with ratio x. So, remember, a geometric series in general, I need a place to write. So, a general geometric series looks like this, right? This is the sum, well, in this case, this guy, this guy is a geometric series with ratio 1 and a 1. This guy is also with ratio x and a equals 1. This guy is also a geometric series, but it's a ratio 1. Somebody said x, and what's the first term? x to the 10th. So, this sum is x to the 10th over 1 minus x. You can see that in two ways. One way, you can see this is a geometric series with ratio x and a factor x to the 10. You can also see that this is the same, we write it here. This is the same as I can factor the x to the 10th out if the other series I already have. Yeah. If it's a geometric series starting at zero, yeah. Because a, so a geometric series is always, let's go back to this one, geometric series looks like if we're starting at n equals zero, something like that. This is r to the zero and r to the 1. So, if we start index at zero, then of course a is the first term. Sometimes it looks a little funny, but yeah. So, if I change this index to a 3, then it's not a geometric series starting at zero. We have to fiddle it. So, either you can factor out the a, r, q or r. I do a lot. I'm going to erase this. So, people okay with this? Are people a little bit, anything on geometric series still? He is. She's not. You're in between anything and not. I can't. I mean, if you have a question, I'm happy to answer it. But if you are so lost that you can't formulate the question, it makes life a lot harder for people to understand. Oh, okay. So, we have this. So, today is really geometric series day, but hiding in the context of power series. So, suppose I had, instead of that, something that looked like, well, let me not write it as a sum. Suppose I have a polynomial that looked like x square plus x to the 4 plus x to the 6. Let's start it. Let's not start it. Plus x to the 8, like that. Can we figure out what this sounds up to? Hint? Yes. Again, it's a geometric series. May not quite look like a geometric series, but it is. What's the ratio? X square. X square. What's A? What's the leading term? X square. X square. So, what's the sum? Not very hard. So, again, it's just a geometric series. We can write this as a geometric series. We can write it either n equals 2, n equals 1 to infinity, to the x to the 2n. This doesn't look quite like a geometric series the way it's written, but if you think of this x to the 2n as x square to the n, and then I want to factor out the first x square because this begins it. So, again, it's the same manipulation that's all along. So, we can plug in things here and get new series, but we can shift it. Use one very similar to the one I just did a while ago, but not exactly to say, okay, so suppose I want this gup, I'm not going to put the factor of n in. So, it's almost that one, but not quite because I'm not putting in the factor of n. How can I turn this into something that looks like that? So, I rewrite this, and now it's a geometric series in the same old, same old. So, this is 1 plus 2x minus 1 over 5 plus 2x minus 1 over 5 squared plus blah, blah, blah, which sums to, we can actually add this up now because it's a geometric series, the leading term here is 1, so it's 1 over 1 minus that. Now, this is, you know, you could do a little algebra to make it purdier, so I'm going to multiply by 5 top and bottom. This is 5 and 6 minus 2x. Yeah, I'm sorry, why did I put a limit on it? Oh, I should, this doesn't make any sense if this is not true. Well, if this does not converge, yeah, that's a 1. It's a special way to write 1. I need this. I mean, it still is a formula. It's just that the formula is not defined. Yeah, if that's greater than 1, it doesn't add up. If x is 3, then this does not sum to what I come up with. So, you know, often we don't write this until we're ready to use it. Here, I'm just, so just like when you write tangent of x, you don't usually write x not equal to 5 over 2, 3 pi over 2, blah, blah, blah, because the domain is implicit or when I write a square root of x minus 1, again, I don't say where x is less than 1 because it doesn't, you know, doesn't make sense of it, I'm sorry, where x is greater than 1 because I know from the formula that it doesn't make sense when x is less than 1. So, the same thing here. This is here, this is implicit in the writing of this if you know about series. So, sometimes you write this, sometimes you don't. Is this clear? Is it, number one was his objection clear? And then my rebuttal, you're stupid, don't say that. So, it was a good point, it's just something we usually leave off because we know we can figure it out when we need to know what it is. But if I just blindly start using this for x equals 7, I'm going to get very strange answers. Okay, so, this function is just another way to write this function. But on the domain, so what is this? This is, let me just write it this way, on this domain. This equality only holds on this domain. This one holds on a bigger set of values. This one blows up when x is 3. This one also blows up when x is 3, but it also blows up a lot of other times. So, in some sense, in some sense, these power series are more restricted because we have functions which are defined on much larger domains when we write them this way, then they are in terms of series. And we will see in a little while that they're still useful on the domains where they're defined. Right now, it just seems like we're writing stuff we already know in another way. And we won't get past that until after these. So, one thing that we've done here in order to manipulate these series into new series, I guess let me do one other one. Obviously, if I have an alternate like that, which is the sum of 0 and infinity minus 1, what's this series have? Yeah? Yes? Yes? Let me write it this way just to emphasize. So, if the thing alternates in sign, it's still a heightened geometric series. In fact, a lot of these power series are really just geometric series in a hidden format. There are other ones that we will manipulate, but for today, almost everything we're doing is some kind of geometric series in a hidden form. So, what we've done so far is multiply these series by some terms like this, sorry, like this. Or make a substitution x to the 2n, I mean x, replace the x with an x squared or the minus x or something like that. Already that gives us a bunch of different power series that look slightly distinct. We can also manipulate them further. Suppose I have something like, suppose I have something like, so instead of giving you a series and having to guess what it is, let's start with a series we know and see and derive something new. So, suppose I start with this one, yeah, not writing what I'm thinking of, but I start with this guy and I decide to take the derivative of each term and I'll get something slightly different. So, derivative of 1 is 0, derivative of x, x is 1, derivative of x squared is 2x, derivative of x cubed is 3x squared, derivative of x to the 4th is 4x cubed, et cetera. In fact, let me write the other representation in the middle here. So, if I take the derivatives, in general, I have nx to the n minus 1 for n equal to 1 to infinity, the general term is nx to the n minus 1, but that should equal the derivative of this. So, the derivative of this is 1 over 1 minus x squared times, minus, at the minus 1, right, the derivative of 1 over 1 minus x is 1 minus x to the power minus 1, so I get minus 1 minus x to the power minus 2 times the derivative of minus x. So, now, we actually have another power series sitting here that we know how to add up. We could shift this one a little bit. This is the same, if we want to start at 0, then this n is this n, did I get that right? Something's wrong here, yeah. The limit of the term, not if it's less than 1, again, this only makes sense when the absolute value of x is less than 1. If x is bigger than 1, this is all garbage. So, when x makes, when this sum makes sense, this stuff is okay. When the sum doesn't make sense, it's nonsense. Because you're adding infinities and taking derivatives of infinite things, and you just come up with craziness. You can make some amount of sense out of it, but you have to work very hard. And so, we just ignore that stuff. So, you can do stuff with it, but you have to be very careful, and that's well beyond what we do in this class. So, this manipulation only makes sense when x is less than 1. So, now we have another series here that we can obtain from an old one by taking derivatives. We can also, I mean, and again, I can get another series by taking a derivative again, but I can get a series for 1 over 1 minus x cubed, and so on, right? We can also go another way. Actually, we use that word. Suppose, again, only for certain values of x, suppose I wanted to represent the arc tangent as a series. Well, I remember, I hope you remember, that this is the integral of 1 over 1 plus x squared, the x plus a constant. Let's forget about the constant right now, or if you like, I can put a t here and integrate it into here and that. So, we can use this same kind of idea to derive a series for the arc tangent. What would we do? So, we have to do it in stages. What do we need first? First term. Well, no, I don't want to get the first term and then the second term. But, let's use the fact that we already know another series. So, can I somehow, by standard manipulations that I already know, transform this into this? So, can I replace this x with something to get it to be a 1 over 1 plus x squared? Overplace what do I make? Negative x squared, starting at 0 equals 0, the infinity of 1 plus x squared. Negative x squared, the n, goes by straight substitution. Again, this is only good for x less than 1 in absolute value, which I can do a little algebra to make it per year. So, this is minus 1 to the n x to the 2 n, okay? Now, we're close. Now, we integrate. So, of course, this is relying on a big theorem that we have improvement by playing this through, which is that we can integrate term by term. So, and so I'm claiming that this is the same thing as if I integrate each term. So, again, let me write out, well, just to emphasize what we're doing, we're integrating each term of this series here. So, when n is n is 0, I have a 1. When n is 1, I have a minus x squared. When n is 2, I have a plus x to the 4th minus x cubed. That's what I'm doing, but I'm actually going to do it in this form. It's easier. So, when I integrate, what happens? Well, I pick up a constant because I get a constant in variation. When I integrate 1, I get an x. When I integrate x squared, I get a minus x cubed over 3. When I integrate x to the 4th, I get an x to the 5th. When I integrate x to the 6th, I get an x to the 7th over 7th. And so on. If we want to do it sort of all in one shot, minus 1 to the n is a constant because we're integrating with respect to x. So, this is some constant that I'm not telling you yet. And then when I integrate, I get minus 1 to the n. That's just some number, plus 1 or minus 1. And then x to the 2n, it's x to the 2n plus 1, but I have to divide by 2. We did some magic trick here. Well, it's written right there. I can't write it again. So, we don't know at the moment what this constant of integration is, but we do. What is this constant of integration in this case? How can we figure it out? Okay, again, I'm claiming that the arc tan is some number that we don't know what we do plus this series. Yes, that's it, yeah. Good guess, though. So, how can we figure it out? Yeah. Exactly. So, we know that the arc tan, so this is 4x small. None of this stuff makes any sense if x is bigger than 1. But if x is less than 1, then this is true. We know that arc tan of 0 has to be that same constant plus some n equals 0 to infinity minus 1 to the n, 0 to the 2n plus 1 over 2n plus 1. That's all 0. So, c is whatever the arc tangent of 0 is. What angle has a tangent of 0? 0. So, c is 0. So, in fact, this constant doesn't matter. It's 0. So, what we've just done is figured out that the arc tangent is, so, 2n plus 1 is just a clever way to write odd. So, this is like this, x minus x cubed over 3 plus x 5 over 4 or 5 minus x 7 over 7 plus blah, blah. So, what good is this? Well, one thing that this is good for is if we don't, if we aren't able to actually calculate a value of the arc tangent directly. I mean, why do you think, well, your calculator is now smarter, but what do you think your calculator is doing when you ask for an arc tangent? It doesn't just magically know it. It needs to calculate it. Current calculators use a smarter version of this, but you can just calculate this series. If I want to know the arc tangent of 0.03, I plug in 0.03 and I compute a series like this with enough terms to whatever calculator, whatever accuracy I want. So, this is what's going on, well, not exactly this because there are better ways, but something like this, this is what's going on inside the calculator when you want to calculate the arc tangent. You transform something transcendental into something arithmetic. Newton actually tried to figure out the series for, I don't know, maybe the tangent by trying to solve, by trying to invert this series, but he was a smart guy and he made good progress on that. That's a little harder than what we want to do. Other things we can do is we can multiply and divide them and so on, so we'll do more of this manipulation of function of series after the plan. Oh, one other announcement, I'm going to hold next to office hours tomorrow. I still have office hours today, but I'm going to hold office hours tomorrow as well.