 The exams are mostly graded, but we don't have the numbers added up yet. I'll get the results when we post it on the web over the weekend. We have to get them all collated and everything. We don't need to talk about it. So there's also a very short web assignment. There's only six problems, so it's short. Everybody's tired. The exams are awful. I didn't see a crime during. So it couldn't have been happening. Did you want me to add those two? So we're going to have 11 problems? Maybe. It's hard to ask the riveted questions. So we have new stuff to do, so we can have more harder stuff. So actually some good news is, although it's a little bit easier from here on out. It's sort of like the difficulty in this class goes sort of like this. And we're at about right here right now. It's a little bit more, well maybe we're here. It depends on how you see it. We're somewhere right around here. What? But notice this is five. That's one. And then here we find it. So the reason that we did chapter eight before we did chapter seven. Chapter seven is easier, essentially. I mean it's different. It's a little easier than the stuff in chapter eight. Most students find the stuff in chapter eight difficult. And that's the reason that we did it earlier because people also find it difficult to concentrate after Thanksgiving. So I try to move the easier stuff so that after Thanksgiving it's not like my head is exploding. The semester's ending. Because we still have a little bit more, I don't think it's more difficult than what we've been doing. It's about the same level. So if you remember what we've been doing is we were talking about power series. And this is when we have something that looks like some constant x to the n, and we might as well start at zero with anywhere else. We have a series that just looks like plus a constant and x squared. It's possible that we could make a slight substitution and let this be something like, it's the same. It's just a matter of where we put the order. So we'll focus on this form for now. Although those will come back. And what we did on Monday, if you can remember back that far before the traumatic event that wiped your memory, is, who is? My mom. Your mom. She did pretty well. I'm great at her job. She got great. Okay. So we have this thing and we already know what some of these are like because we know the geometric series to derive a bunch more. So we did things like making substitutions, like taking derivatives and integrating and so on. So for example, one thing that we came up with this at the end of the class, should I go through this again? Okay. So I'll just do this one quickly. This is field review. So one thing we can do is we can make a substitution. Here replace this x with a minus x. So just by doing substitution, this series is just where I, everywhere I see an x here I put a minus x. So there's no x there. Okay. This x becomes a minus x. This x squared becomes a minus x quantity squared, which is still positive. This x becomes a minus x cubed and so on. Which would be this series. That's an n. N equals zero to infinity minus one x to the n. And then we can do little tricks with it like integrate it. So we know that the r tan of one over one px so this is something that you know, right? This is the stuff you do know. Yeah, okay. And we can integrate this series term by term. So I'll just do it in this form rather than doing this, but we're going to go with this as an x. This is a minus x over x squared over two and so on. We already checked that the constant in this case is zero so I won't bother with it. And so that gives us minus one no, two n. He told me he could do it but he was six. What happens to the minus one? When we integrate x to the two n it becomes a two n plus one. But we have to divide by two n plus one. So if we do this integration we well it still starts with zero. So all I did is integrate this thinking of n and just some number I don't know. It's a constant. So the integral of minus one to the n is just minus one to the n. The integral of x to the two n is x to the two n plus one but I divide by it. So there's the integral which is the series n over what did I do wrong? One, minus x cubed over three plus x to the fifth over five minus x to the seven plus seven plus nine. So this is what we did at the end of the last class. Now there is a constant of integration that comes out of this but we checked already that it's zero because the tangent of zero equals that constant and in this case the zero because the arc tangent is zero. So we know that in fact when this series converges the arc tangent can be represented as a series. This only converges well, when does this converge? So this is something that was on the test. Maybe you got it right maybe you didn't. So this is a power series. What is the radius of convergence of this? What do we do to check? Well we should know already because what is the radius of convergence of that? One, yes so this converges for absolute x less than one. This one also only converges for absolute x less than one because that's what we started with but sometimes the less thans become less than or equal to so the integral of convergence on this one does not include well actually does not include minus one it's strict inequality here but on this one actually we do pick up minus one plus one sorry but let's just check. So by the ratio test we need to look at the limit as n goes to infinity of the absolute value now here I want to replace n by n plus one yeah so now what we're doing I already know the answer but let's just confirm we have a series in our hands it's a power series it's reasonable to ask for what values of x does this make sense but I want to know what is the interval of convergence and so I'm going to confirm first off that for x less than one absolute value I still have the same convergence so I'm going to look at by the ratio test what is a n plus one divided by a n take the limit I want that less than one so in this case my n plus first term is what I get when I plug in an n plus one for n here so that would be there's this minus one in the end but I don't care about it let me write it anyway x to the well when n is n plus one that's two times n plus one plus one more so that would be two n plus three so if you want to see that let me write that and then here same thing two n plus one plus one and now I have to divide by a n so that would give me minus one to the end I don't have enough room minus one to the end x to the two n plus one and then on the top I have the two n right that's the ratio that I want to calculate so this is two n plus three this is two n plus one so let me write equals here so the minus one to the n plus one over minus one to the end gives me a minus one but I'm taking an absolute value so forget it two n plus one over two n plus one plus one is two n plus one over two n plus three x to the two n plus three over x to the two n plus one is x square so I take the absolute value and I take the limit as n goes to infinity so what is this limit? it's the absolute of x square but x is always possible so this is just x square so this will convert what x square is less than one in absolute value that's the same thing that x is less than one right? but now I still need to check what happens when x is one and what happens when x is minus one so if x equals one because the ratio test gives me no information when the ratio is one I know it diverges if x is bigger than one it converges when x is less than one but I don't know about when x equals one or minus one so when x is one the series becomes the sum minus one to the n I'm one over two to the n plus one so does this converge or better? what? no, it's not a p-series it's an alternating series and the limit is zero so this is an alternating series it's decreasing and the limit as n goes to infinity of one over two n plus one equals zero so it converges and we also have to check what if x is minus one then I get minus one minus one minus one to the n minus one to the two n plus one over two n plus one this is x x is minus one so n is an even number minus one to the n is plus one and if n is an even number two n plus one is an odd number right? so about two n plus one is always an odd number this is always an odd number let me just write it this way minus one to the n minus one to the odd over two n plus one so this alternates minus one to an odd power is always minus one so this actually equals minus one to the n to the minus one well this is still an alternating series it just maybe starts with a minus if you want this to be minus one to the n plus one and so this still converges so that means that the interval of convergence picked up the two n's so this is through r tan of x and then a little more yes then this series minus one to the n x to the two n plus one over two n plus one starting at zero and this happens for x plus or minus one so what can happen when you calculate these series and pick up other ones is you can slightly change the interval of convergence not a lot, but the n's can either add on or fall off because it's very delicate what happens at the end of the series which is why we have to do this special testing the ratio test is like a very crude implement the ratio test it gives you a definite answer and you deal with the delicate work it's like a sledgehammer it's not really good for eating little fine bones out of sand and you have to do something more delicate like this to see what happens when you have little fine things to work with okay not actually what I'm talking about today except I used half the class on that so that's good so we have this business where we can take an old series that we know such as the geometric series and produce functions of other ones but what if we want to go the other way what if we have a function that we know and we can see what they add up to they're functions that we know sometimes what if we have a function that we know and we want to produce a series for it this is useful because the series allow us to do operations just with multiplication and addition for things which are expressed in more complicated ways so it allows us to treat things that are transcendental or functions that are more complicated in terms of simpler things but only when it converges so what if so we have some function that we want let me take e to the x but what I'm going to tell you works for like any nice function so we have our function so that's my example and I want a series do you want a series? they raise the question I want to produce a series for a known function do I have any hope of doing that otherwise I wouldn't be asking so let's just start going what do we know well let's just look at zero so if x is zero we know e to the x 1 so already we know the first term of the series near zero the series has to start with a 1 if it started with a 3 it would be wrong we know our series is going to start with a 1 if there is a series it has to start with a 1 is everybody clear on that? no it has to look like sums of powers of x's it has to look like a polynomial so it won't be e to the n because that would do no good it's like saying if I want a series for e first calculate e something not useful so so what I said is if e to the x is some constant plus another constant times x plus another constant times x squared plus another constant times x cubed well then e to the this one has to be 1 yeah I don't I'm saying if there is a power series for it this is what it has to look like so it didn't say that there will be a power series for it but in fact there has to be because at least with a radius of convergence zero because I know that e to the zero is one so if I let all of these terms be zero there will be one plus zero plus zero plus zero plus zero so that means for sure the power series has to look like one plus something it just might not converge for any value except x equals zero because I know the first term is one and that works so now let's see if we can make it a little bit so see if we can make it a little better so what would the second term be so how can we figure out what the second term would be why did this work why did I figure out that if e to the x let me just say it this way instead let me say it this way instead so I claim that if this is true then c zero is one since one is e to the zero and if I plug in zero here I get c zero plus zero plus zero plus zero so the first term has to be one if this is true then this has to be true there's no other choice so how might I be able to get at what this terms well if I put one equal to x then I will get so it's a good try it's a good try so then it will tell me that e is one from the c zero plus c one plus c two plus c three that doesn't tell me a lot the sum of all these numbers has to be one point seven one eight that's not so useful so I mean this is not an obvious thing right this was figured out by Taylor and Lauren in the 18th century they were smart guys so so try taking the derivative maybe because I said we were to Taylor because there is yeah because the first term is a constant it doesn't change the constant means no matter what x is just the same the first term doesn't have an x in it it's a number okay so so he suggests taking the derivative because I gave a clue so let's take the derivative we know that e to the x well we want e to the x to be c zero well it's already one and so if this is true and the derivative well the derivative of one is zero the derivative of c one x is c one the derivative of c two x square is two c two x the derivative of c three x two is three c three three x square and so on and in general this will be the sum of n times what this guy is the sum of c n x to the n then this guy would be the same as that one but I put it in in front of everyone and I take the power down by one and I take the derivative I bring the power down well that's good because we know what the derivative of e to the x is c to the x and so now we can play the same trick that we just used we know that e to the zero is also c one so e to the zero which is one is also c one so now we know two terms this is not us we have another surprise exam next one okay so we know that we know that because when we took the derivative we managed to do the same trick as before we now know the value of the derivative so what might work to get this guy yeah take the second derivative so we make the derivative of this so if we take the derivative e to the x this is the derivative of what's up here so that will be well this is two c two and then here I get two times three c three x the next term that is written here will be four times what is it already this one is four c four and I'll get four times three c four c four it's not even supposed to be x squared plus stuff and in general this will be the sum of n n minus one whatever those c's were before in the next term sum n n minus one whatever those c's were before this is two and again I can plug in I took the derivative of this one I don't know yet how'd I get this is the one from this one c one is one so the trick was we started here saying I don't know anything except the first term but I know the function at zero so now I know the first term now I want the second term well if I take the derivative the first term goes away and the second term becomes the first term so that was what I did here I see your question in a second and then I take it again so okay c one is s do you agree with me or no I mean c one is one okay so forget everything and just look at this line e to the x is c one plus two times some crap times x plus three times something times x squared plus four times something times x cubed let x be zero and this is sum number plus zero plus zero etc e to the zero for x in this line I get the number that I want plus a bunch of zeros so the number that I want okay you have a question you're good now and then well it works so good once maybe it will work great another time so we take the derivative again and now we can plug in e to the zero is twice c two it's not c two but it's twice c two again plus zero plus zero plus zero so that means that c two twice c two is one so c two is a half and then I can do it again yeah so I took the derivative of this line this line here I had a three c through x squared so when I took the derivative of that the two came down gave me a two times three and then here I have a four times three c four x squared and the next term that I did write is a five times four c five x cubed so in general because I took the derivative twice this is what I got and we just keep going so tell me what c three will be one six one six well we'll c four b then so we just keep doing this c n will be one over in fact okay because every time I take a derivative I bring the power down and so when I take the derivative of this this is a six when I take the derivative of this this is four times three times two and so on five times four times three but they take the derivative again it's five times four times three times two take the derivative again it's five times four and so on and in general each derivative brings another power down and lets it play in the game and since we're only looking at the first term of the derivative we pick up what we want okay so what does that mean we've just shown that it works so we've just shown that if even the x is some power series then it has to be we didn't show that it is this one we just said that if it is one it has to be that one but now we can check when that one converges so if this if there's any answer to this question the answer has to be that it can't be anything else so how do we check we use the ratio test so we have to say when does this converge we use the ratio test and that's an easy one take the limit then it goes to infinity of x to the n plus one over n plus one factorial to y to the i x to the n over n factorial the x's cancel leave me one x on top and this is n plus one factorial over n factorial n factorial this is n plus one times n factorial so everything cancels except the first one and this one is zero no matter what x is so since zero is always less than one no matter what x is it converges everywhere so if we wanted to know what e to the seventh was we could just add up the powers of seven dividing them by the dividing them by the dividing them by the factorial so this gives us another representation of e to the x now notice that this trick didn't really rely on much about e to the x I just used e to the x to take it through this was easy we went through we could do this with any function provided that we know how to evaluate the function at zero exactly what I did works for any function maybe it gives us a series that doesn't convert everywhere but we can do this process for any function whatsoever so let me just do it for another function I'll do it a little faster okay we'll just leave that I still have that work I'll do it with just an ordinary function I'll do it with a specific function rather than just writing the f and do it with the f later so suppose what function should I use sine so cosine just because well I can do the sine so for the sine I'm going to do the same thing but I'm going to do it a little more efficiently this process was long it doesn't have to be so long so we know that so I'm going to say that the sine is some series but this would be cn x to the n so since the sine from zero to zero this means that c1 or c0 is zero was I off by c0 the concern that I get is zero because when I plug this in when I plug in x equals zero here I get c0 plus a bunch of zeros when I plug in x equals zero here I get c0 is zero now we look at what's the derivative of the sine the cosine of zero is one since the cosine of zero is one when I take the derivative of this so already we know that the sine near zero is x we've already seen that if you've taken physics they use this all the time if you remember back to calculus one the sine is about x because it's about the same as its derivative so now we do it again I take the second derivative of the sine this is the derivative of the cosine which is minus the sine minus the sine of zero zero that tells me that two c2 is zero now I do it again minus the cosine minus the cosine of zero minus one that tells me that three factorial which is six c3 is minus one so c3 is minus one so that means that what we have so far is that the sine of x is about the same minus x cubed over six and there's more stuff this is for x small okay so now we want to keep going take the derivative again so the derivative of minus the cosine is the sine zero so that next term is zero this is factorial three times two it's important that it's three times two and in fact this tells us that four factorial c4 is zero but that's the same as saying c4 is zero because every time I take a derivative I multiply the coefficient by the power of the original term that I started with so I just keep building up this is really the fourth power in the series and so when I take the derivatives to peel it off I have four times three times two times one so if I do it again on the fifth power we already have it but okay the fifth power is one let me write it which means that the next term is a plus one over five factorial wait where am I yeah that's right did I do that too fast but it's one over five factorial so if we just keep going notice what happens to the derivatives I have a zero a one a zero a minus one a zero because this story repeats every four times this is the sine this is a cosine this is a minus sine this is a minus cosine this is a cosine minus cosine so that tells me what the series is the series is the odd powers divided by the odd numbers factorial with an alternating sine so let me just write it in words first so this is not a very mathematical statement it's just so it's x minus x cubed over three factorial x five five factorial x seven it's a little bit it takes a little bit of a leak to realize that rather than putting odds here I can express every odd number as two n plus one let's see I want I can write this as starting at zero x to the x to the two n plus one over two n plus one factorial that's an alternative so I make that mistake all the time so you shouldn't feel that since I'm starting at zero because I want x to the one to be the first term if we do the same thing for the cosine two right now because we don't have many left then we'll get x to the even powers alternating so we still need to know the interval of convergence maybe this is only true for x equals zero it's actually true for bigger than x equals zero so let's I'll stop there and do that on one way