 So, we're not going to do that yet, but I'm leaving it on now, whether I want it to give me a half step like last time. But I'll do that very soon. So, one thing I want to point out, last time I wrote down the series for E to the X wrong. It was sort of morally right, but I divided this by two. So, I'm doing something that should be obvious from the McGoran series. I shouldn't have to do any work. We're almost over. Of course, I can start taking derivatives at zero, logging them in, figure out the series from first principles. But why would I do that? So, what should I do if I do this? Huh? Separate them into two. Think of this as X cubed times e to the X squared over two. Good, and then, I'm sorry? So, nobody knows? Okay, so let's make it be an easier problem. Suppose the problem was the series for U, E to the U. Yeah. All right, this is X cubed times a whole bunch of jump. And a whole bunch of jump. I don't need that much room. I'll just do this. This would be sum n to the zero to the infinity X squared over two. But this is actually right. We just let's clean it up a little bit. This is X to the two n, and that's the two to the n. And this is sum sum out X to the two n. This two to the n is on the bottom. So I'll write it down here. So this would be X squared over two all to the n to the term. And then n times the value. And then this is X cubed times every single quantity. So that'll just raise this power by three. So, I mean, while we can, in principle, figure this out, let's just check it and confirm it by starting the hard way. So this is, so just emphasize this. This is when n is zero, I get X cubed over plus. And then when n is one, I get X to the fifth over two if I'm doing that right. And then what X is, when n is two, I get X to the seventh over two squared, which is four times two that you have right. This is two squared times two, I think so. So let's just, and now let's write it the hard way. So I have X cubed e to the X squared over two, that's my n. That's from zero to zero. And now if I take the derivative, you see it's already awful. So here I'll use the computer to take the derivative. So since I don't want to do all the work, I will just ask if I can do all the work. This stuff is for later, which is my thing for now. Okay. So I want to take the derivative of X cubed over two squared over two with respect to X. So it's that. Now I want the derivative of that with respect to X. Okay. So if we notice that the X, so here's F. You can't see the same thing. Okay. So F is there. This is F prime of zero, but see that's zero because I have an X squared at the end of it. F double prime of zero. This is all zero because of the X's. After the prime of zero is 6 e to the 1 half X squared plus a bunch of crap is zero, double prime of zero is, this coefficient is one because this coefficient, so I have zero plus zero plus zero plus the third derivative divided by three factorial. So the first non-zero term in the series is non-zero. And then the next term, zero, because there's an X in everything, but when I take the derivative of that and I put out an X here and so that's 60 e to the 1 half X squared. So that's another 60. Not another 60. A 60. So that would be zero, a factorial over 60. I mean 60 over five times four is a half in this term. So you could do this, but it's awful. So it's much easier to take something that you know that's simple and combine it. Okay, well, this is not why I brought the computer here. So, side trips. So what I thought was trying to do last time when the computer fought with me, actually the computer was finding the series for say e to the X. So there's the first part. Can you see this? There's a graph. So forgive my editing here. This is the series for e to the X, just the first five terms. If I want to look at, compare, well, let's see. So here's the series for e to the X, e to the zero. If I change this and I add in the first two terms, well, let's just take for three terms, one plus X. So this is the first quadratic term. This is the constant of one. This is one plus X, which is the tangent line in zero. This parabola here, which is starting to look a lot like e to the X, is one plus X, well, one plus X plus X squared over two. If I go to three terms, that's the purple one. It starts to hug the graph a little more. If I go to, say, 13 terms, get some horrible looking polynomial, let me actually just do the 13. And you can barely tell the difference out to negative five. The green curve, you can't even see the red curve because the green curve is sitting on top of it. Can you see them in the back? We don't care. So we can see, actually, let's just see the convergence here. So each of these colored lines is more and more terms of the Taylor series, and it's starting to hug the graph of e to the X more and more and more. If we play the same game with the sign instead of the exponential, maybe I'll only do five. So this is the graph of the sign. And you can see more and more terms are starting to hook on and get closer and closer to the graph of the exponential if instead of doing five, I'll do 15. It starts to really match up. Maybe there's too many going on. So let's do the last one. So here's the sign curve here. These are, this is the degree 12 Taylor polynomial, the degree 14, the degree... So they start to match more and more on a wider and wider range. So let me do one other case. If you remember, this guy converges, the infinite series converges for all X, for both of these ones that I did. But if we do say the log of 1 plus X, so let me do the same thing here, the log of 1 plus X. The red curve is the log of 1 plus X. The green curve is its tangent line. If we take the polynomial, the second degree Taylor polynomial, it starts to get closer. Again, the green curve is our two terms of the table polynomial, the red curve is the things we want. So let's say five terms of the series. It starts to get closer in here. But notice that it just sort of feels away in here. Let's go a little further away. If I start taking more and more terms, so instead of taking the degree 5, let's take 50 terms of the Taylor series. It matches really well, and then it just falls off. And that's because the Taylor series only converges between here and here. So you can see maybe it matches really well, but then it just falls off. So this bump is the best that the Taylor series can do. And if you try to go beyond 1, even though the graph is defined, the Taylor series is no good because it doesn't work. Even if I take, this is the degree 50 series, let me just show you what I'm doing. This polynomial that I'm trying to approximate the log with, it converges really well when it converges on the outside of the graph. What I wasted, I don't know, quarter of the class, trying to make it work last time. Any questions? Notice that also we don't have to do this stuff as zero. So for example, suppose I wanted to approximate the log of 3 by a series. I could use this same series, but expand it not at zero, expand it at 3. I mean, not to take 50 terms and emphasize that it's based at 3. And notice that, so here I'm taking a Taylor series centered at x equals 3, not at x equals 0. This is a Taylor series, not a Florence series. So the series that I'm doing here is centered at x equals 3 and it looks more like that. So I'm expanding my series here around x equals 3 and the same kind of stuff happens with the log. It still falls off if I take lots of terms, so instead of 5 terms, let's take 50 terms. It still falls off but it falls off here because it converges around x equals 3 with the radius of convergence that reaches all the way out to about 7. So the radius of convergence is important. It's not just some garbage that we may can do if you want to use this. No. So what the radius is really doing is saying something weird happens at the edge where I can convert. Maybe that weird thing in a real number. So the log, for example, this function is going to be log of 1 plus x blows up to x to minus 1. So this series can't converge this way beyond x equals minus 1 because this function is infinite. And just because of the way series works it can't go any further this way even. So it will blow up. Which is 4 away from 3. So the series will also not work well beyond x equals 7. If I move over and generate it at 10 I should get something that's good between minus 1 because something happens at minus 1 that hurts it. Now there are other series where the thing that goes wrong actually happens in a complex plane so the function is nice for all real numbers but the series can't go very far because it's the complex number. Okay, any other questions about this stuff? I'm going to come through the way of this stuff now. Cross track, sorry. For the sine, bar in series for e to the x for the sine of x we know the series for the cosine of x by taking the derivative of the sine when we're writing it again we know the series for the log which is really the scalar series for the log basic. Okay, we also know geometric series but to be able to extend this repertoire a little bit k is some number. So 1 plus x to the fourth power you know this. Maybe xq plus x to the c. And in general 1 plus x to the f because 1 plus n minus 1 plus x to the f where n choose n choose j is n factorial. You've all seen this before I think. Has anyone not seen this before? Did you guys actually like go to high school and stuff? It's amazing. Okay, so this is something that you're supposed to have seen but you haven't. This is called the binomial theorem. How we can expand power of binomials a plus b to the f power. So this is also related to something you may have seen called Pascal's triangle. This is just Pascal's triangle. These numbers are the numbers you get out of Pascal's triangle. 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, or just written in a slightly different way. So if I go to the 10th line Pascal's triangle and I want the third entry then that value will be 10 factorial divided by 3 factorial divided by 7 capital. Okay? So this is just a true fact that you should know but I guess you don't. So notice that this looks a lot so let me just write that again in a slightly more compact notation. 1 plus x instead of using n here I'm going to use k. This is n equals 0 to k of n choose k x to the n. I'm just rewriting by a switch from n to k so let's just put it back in case. You don't have to change your notes. This is the same. If I expand this out so this is the same as that. If k is an integer left between 0 and n. Sorry, that's silly. If k is a whole number. Yes, we have that. But what if k is a fraction? And we make sense of this. We can certainly, so this looks a lot like a series. It's not a series because it stops at the k terms. But we can make it still work even if k isn't a whole number. But instead of doing that let's just put it aside. But I want to think about this. It will not be like a polynomial that stops at the k term. It will be an infinite series. But maybe it still works. So let's just make a series for 1 plus x to the k x. And let's see if it works. So f of x is 1 plus x to some power. And as long as the power isn't 1 I'm okay. And then we can take its derivative of 0 1. If we take the derivative of k comes down f prime of 0. If I take the next derivative of k comes down and I evaluate it to 0 times k minus 1. Let's keep going this way. k minus 1 times k minus 2 1 minus x k minus 3. And if I evaluate it to 0 I get k times k minus 1. And this pattern will continue. If k isn't an integer if k is not a positive number this will never stop. If k is a positive number eventually I'll start getting 0s here. Because k minus something is zero. But if k is just like negative 3 or 1 half this process will continue. And in general I'll have what will be nth derivative. But it will start with k and the k minus 1 will still be there. When will I stop? k minus n. Here I stop from k minus n plus 1. I won't use k minus n. k minus n minus 1 is where I stop. And then I have an x. Then I have a 1 plus x to the k minus n power. I take the derivative of 12 times I will have 12 times 11 times 10 9, 8, 7, 6, 5, 4 until I get down to k minus 11. And then I will be into the k minus 12th power. Oh and I plug in zero here. I just get this jump. So what does this mean? This means we know the series. So my series for 1 plus x to the k power is going to be f of zero f prime of zero times x plus f of the prime of zero divided by 2 means that squared f triple prime of zero divided by three times x squared times x cubed pa-ba-ba-ba But we know the formula for all of these yet. Well it's k minus n minus 1 which is k minus n but then get one back That means that the series in this case is 1 plus k times x plus k times k minus 1 over 2 times the x square plus k times k minus 1 k minus 2 over 3 factorial x cubed plus 0.111 So like this in summation notation 1 plus x to the k is we start at zero we go as long as we need to 1 k k minus 1 k minus 2 stop just before k minus n divided by n factorial n multiplied by x to the n So that's sort of a mouthful so we can use the same notation to write it a little shorter where here we have to change the meaning of k choose n a little bit to allow for fractions So here we're going to take k choose n to be this even if k is a fraction or even if k is negative and if n is bigger than k it's 0. So this agrees with the previous meaning there because what does this n minus k do it kills off all the terms sorry my n's and k's the creative rule makes them match So this k minus n factorial kills off everything here in the case back to earlier the n is still here so this is the same as that if k is a whole number bigger than n if k is a whole number smaller than n it's going to be 0 so we're done and if k is a fraction well this notion of k choose n because choose n things out of a pile of k things doesn't make sense to say choose n things out of a pile of log 5 things I don't even know what log 5 things means 3.5 things means well that makes some amount of sense but we just define it to be this and it agrees completely and as we've seen we get a series and this series allows us to do things so this is called the binomial series this series allows us to do things like write a series for 1 plus x the square root of 1 plus x for the cube root so this captures the series for fractional powers maybe I should do an example so sort of in the same vein as this if we wanted to write a series for 1 over the square root of 9 plus x close what I want a series for this I can do this without having to take all of the derivatives and go through all of the work I can just turn this into something that looks like this by a little bit of algebra so what would I do you're not telling me all of the algebra what k would I use negative 1 half so this is 9 plus x for negative 1 half well this is still not the same as that I need a 1 there but to get a 1 there no problem I can just divide by 9 the same kind of trick we do when we were doing trig substitutions 1 plus x over 9 minus 1 half but then I have a 9 get inside for a minute this is getting closer and now I can take well I know what 9 to the minus 1 half is that's a third square root of 1 over the square root of 9 so this is 1 third to the minus 1 half and now this is in the same form that I can use the binomial series form I think of this as my u and I do 1 plus u there so my series is well should I write it in some notation so it's 1 third of what I get where I use an x over 9 instead of an x there minus 1 half choose n which really means that list of terms which I'll write out in a second x over 9 which would be just to write out a few terms to emphasize so this will be minus 1 half first term then my next term is well here we have minus 1 half minus 1 so this is minus 1 half times minus 3 half divided by 2x okay you can be divided by a piece of motion yes I did sorry but then again 1 minus 1 half x square minus 3 half minus 5 half over 3 factorial x cubed plus ba-ba-ba-ba so we can get a series for this without having to start crashing through this through this through this through you could also just take derivatives through this through this and you should get the same answer I don't want to do that because okay in order to get other series from series that's a little more complicated to get to general terms but suppose I just want for a few terms suppose I want the series for e to the x on the sign of x if I take them they're going to be a horrible mess because I use the product tool but as long as we are in the region where both series converge oh I didn't point it out if you do the ratio test on this this converges for x between plus and minus 1 the endpoints are hard they depend on the value of x and it may or may not include x equals 1 or minus 1 depending on k so suppose I have this these both converge everywhere so it's okay even now or so then if I could multiply these series to this it's a little complicated to multiply these things together because this is like multiplying infinitely long polynomials so let's just write out some terms and see what we get so the first thing to e to the x is 1 plus or 2 x cubed over 3 factorial the sign of x is 3 factorial plus x to the fifth or 5 factorial minus x to the seventh or seventh factorial so what is the first non-zero term of this series is there a constant term? no it's not 1 it's 0 because okay the first non-zero term is not 1 either way if I try to multiply these things together then you FOIL it out instead of it's not FOIL it's sort of like FOIL by the x a power of 2 a 1 times none of these gives me a power of 2 because the next power when I multiply by 1 is a cube but I can kick up a power of 2 by multiplying this x times this x and it gives me an x squared this x times this x cubed gives me an x to the fourth I'm not ready for that yet but I will get an x cubed term 1 times minus x cubed just get this x cubed times this x and so on so let me just go do you want me to do a few more terms or are you okay? do you want me to do a few more terms or are you okay? right so this series is a little hard to write out the general terms so this is uh oh so when I put this this gives me a 6 minus 1 is a 5 so this gives us at least the first few terms by multiplying things down it doesn't give us a general formula because it's a little less yes there's a minus and then the fifth power will be coming from 1 times this x squared times this yep sorry there are two x to the fourth terms I only see x times x oh and the other one you have to write c times this one so I have x times this one and then I also have this one this one times this one you're right so this is gone you have to be a little more careful than I'm being right now okay so just go power by power and notice that eventually it stops right we have to look at all the ways to add up to 3 all the ways to add up to 4 all the ways to add up to 5 and since I missed one I got one wrong but we can multiply you can also divide series in this way but you could I'm not going to go through it but we could figure out the series for the tangent by dividing the series for the sine by the series for the cosine I'll do that in a bit but again it's kind of messy we have to do long division of the polynomial remember they taught me that in 6th grade if you didn't understand it well if you do long division of these infinite polynomials the series happens so we can do a particular series as long as they can work so one of the things that I'm going to mention if you remember like the alternate series we had something that told us how far off would you be in the ten terms or five terms or whatever so our series like this and we want to know what is the error if I take a different center still the same so this is my function and I want to know how far how much off will I be somewhere in ten terms at least two numbers the answer is kind of like the answer in the alternating series test but not quite so this is often called this is called the remainder the answer is if we know a bound we have a number that tells us how big the n plus first derivative is so this is kind of like the other thing that everybody created the error when we did the integral the improper integrals where you needed to know no sorry the Taylor series or the trapezoid rule this is kind of like the k and that that is so what this is saying n plus first derivative of x less than some number m for x between 0 and a the remainder is number a is less than the term you can use but with the derivative of value for this n so this is less than m factorial in fact it's not just a it's at all of the answers less than does that make sense so if we could bound the derivatives so another way that you might see this is that it's the term you can use you can use to evaluate that some number of t is the guy that makes it big