 Okay yesterday, I said that we would probably finish section 8.7. That did not happen I don't know that in fact that will happen today, but we're gonna make a good stab at it And if we do then we'll get a start in section 8.8 According to the syllabus we're not necessarily supposed to be done with this Binomial series stuff in 8.8 prior to the test on Friday That's probably gonna be the case we will probably have started 8.8, and We'll just have to clarify. What is your responsibility for the test? I can't tell you at this moment We left a problem Set up but pending so let's finish that There are a couple of web assigns web assign is due tonight is that right tomorrow, okay? So we're still prior to the deadline. There are a couple things that Nicole and I talked about as far as web assign problems that are a little Kind of strange, but once you see what it is that I think they want you to compare it to It's not necessarily that bad of a problem, but we'll look at some of those web assigned problems, too But let's finish we wanted to find the interval of convergence for the sine function Using the ratio test So we ended class yesterday with this So let's pick up today So there is the without the alternating signs part, which is not going to matter because it's absolute value There's the n plus first term We're going to take it and compare it using division to the nth term And we want to go way out there to the right and see what happens Let's Kind of put stuff where it belongs to n plus three is up here in plus three factorial We're going to have an x to the two n plus one And a two n plus one Factorial so how can I rewrite some of this stuff so we can evaluate this limit way out to the right? Oh Okay, let's let's seems to be the first thing is to compare these two so it's going to leave a and x Squared in the numerator right this has x The power of x here is two more than the power of x here Okay, and how are we going to? deal with Two n plus one factorial and two n plus three factorial this seems to be larger So let's see what that looks like Two n plus three what would be the next one as we work our way back to one? next one and Then what two n back to one Kind of lost my one there and then two n plus one factorial is actually part of that right? Two n plus one times two in all the way back to one, so it looks like we're going to leave a what two n plus three And a two n plus two down here is that correct? Because the two n plus one factorial is going to cancel with the rest of those and We'll leave those two terms down here, so the ends are positive. I think we can And bring that part out and really we could absolutely value of x squared I don't think we have to worry about that being negative, but what's going to happen to this as In approaches infinity what happens to that? that approach is zero which kind of overtakes this doesn't it this x squared so we've got the whole thing Approaching zero and we would expect the limit if it converges To be where? Less than one. I think we're less than one all the time, right? So there's no certain value of x or certain values of x that cause this thing to converge it converges for all x So our interval of convergence I don't think that's a surprise that this particular Taylor series or Maclaurin series for sine of x converges for all values of x So if we let it go all the way to infinity doesn't matter what you put in for x The fact that we were stuck with two Terms that each had in them in the denominator basically kind of forced this thing thing to converge For all values of x so we want to put in pi over three for x. That's fine. If you want to put in five pi over six Any value you want to for x and it should converge eventually If this converges if the sine converges for all values of x and we have a way of getting from the sine Taylor series or Maclaurin series to the cosine which would be what taking its derivative Shouldn't it have the same interval of convergence? It should or we could validate that the same way we did this but the We would find the same thing to be true I Can find that sheet and a nice summary sheet on intervals of convergence now that we're Kind of leaving that this is in your book, but it does spell out all the Series that we've looked at thus far The first one was an infinite geometric series The first term was one and the ratio is x. So that's an a over one minus r format And that's what the series looks like and there's its interval of convergence Because we wanted what the absolute value of the ratio to be less than one. That's where that came from Let's go to the last one. We did an inverse tangent in pretty much the same way we developed this one by integrating that and that we treated as a A over one minus r the ratio was negative x squared and then we integrated everything So that's how we arrived at that The ratio being negative x squared must be less than one in absolute value And then we have to check the end points. It just so happened that it converged at the end points I don't know that we ever did that but it does converge at both end points so those two are the kind of the outliers here in the List all the rest of them e to the x converges for all values of x Sine of x and cosine of x converge for all values of x so inverse tangent and this Kind of that first infinite geometric series. We looked at in terms of a power series. They have a Fairly limited interval of convergence. All the others are wide open I don't want to Skip over those Web assigned questions, so let's go to those and then we'll pick up with the error Estimate or the remainder when we stop and have a Taylor polynomial as opposed to a Taylor series We can get that what what are the problems that we said we were going to look at That the three to the n over five to the n that one That would be the one we're talking. That's the same one. We're talking about So there is a web assigned problem if somebody has that handy that I think there's a three to the end There's a five to the end and then there's an n factorial if I remember right anybody have that do you have it Cali? Is that the right problem? Yeah Three to the end five to the end and an n factorial Number eight number eight I've got that I just read that sign right there on top All right This is kind of a Testy little problem, and I think the next one is that the natural log of two Problem these both kind of fall in the same category. I think they want us to compare This one to this one What is that one? What's the one on the right? Go ahead. That's e to the x x to the n over n factorial is e To the x and you could have an expanded form or the sigma notation closed form. I think We just looked at this paper right here there's our Definition or at least in terms of sigma notation this Taylor series McLauren series approach That's what e to the x is Three to the n over five to the n since they're both to the end You could say that's really three-fifths to the n Over n factorial and then it's kind of a comparison in contrast with This thing over here that we know So when we have an x in this position X to the n over n factorial That's e to the x. We don't have an x occupying that position. We have a three-fifths So that should be if everything else is the same what should this be equal to e to the three-fifths So when you see it in that fashion, it's not That difficult of a problem, but you've got to say well, what am I trying to do with this? What am I comparing it to? I'm comparing it with this Here it's x to the n here. It's three-fifths to the n. That's e to the x. This must be e to the three-fifths Now if that's not believable Take your calculator and raise e to the three-fifths power add the first seven or eight terms of This series together and see if they're beginning to mesh In fact, that might be a good one for us when we talk about error To stop it at let's say n equals five And then we'll try to evaluate what the error might be what we're missing by not going all the way out to infinity All right, so that's problem eight The very next problem is the one that has natural log Is that correct? Has the natural log of two in it, so what's it look like? Plus natural log of two squared over two factorial minus Natural log of two cubed over three factorial. Okay. Thank you So that's over one factorial Let's try to take this same Approach on this problem where we've got something to the end Pretty clearly it looks like we have in factorial right in the denominator. There's a zero factorial One factorial two factorial so the denominator appears to be in factorial And then we're as we go off to the right. We're gaining powers of what? What is it that's being raised to higher powers as we work our way to the right? Okay, I heard a couple things I heard negative natural log of two and I heard natural log of two So aren't we as we work our way to the right the thing that is raised To higher powers is negative natural log of two here. It's to the zero here It's to the one here negative natural log of two is squared. That's why it's positive Now we could separate out the negative part, but I don't think I really want to because I want to compare it to this So once again x to the end over in factorial is e to the x We don't have x to the end, but we do have something To the end over in factorial as in goes from zero to infinity So that should be if this is e to the x this guy ought to be e to the Negative natural log of two How's that work the same thing? Okay. Now we've still got a little bit of work to do When we have e as the base and natural log as an exponent We really like to have e to the natural log of something That becomes the something right at this moment. We don't have e to the natural log of something We have e to the negative Natural log we could throw that down to the denominator. That's the easy way out Let what can we do with this negative one? That's out in front Right we can replace it in this position Right if it's in the exponent position we can bring it out in front So if it's out in front, we can put it back so to speak in the exponent position So that negative one that's out in front is now the exponent of two and what is two to the negative first? That's one over two and what's e to the natural log of something? It is that something so this problem has an answer one hand so both of these problems kind of force us to compare the given problem with the sigma notation power series for e to the x We don't have x to the n over n factorial, but we have something to the n over n factorial Now was there another problem we looked at or is that those are the two that? Anybody else have another web assign question or issue before we leave that? All right. Well if you come across it bring it up, but let's go on to the remainder When we had alternating series we decided that the error For the sum even if we stop it let's say at n equals 7 we're never any further away from the Actual infinite sum then the value the magnitude of the next term right so if we stopped at n equals 7 We would want to examine the 8th term As far as the error is concerned This is really not that much different. Well, I thought I printed this up, but maybe I didn't All right, this will get us set up and we'll just add in what the remainder actually is So we've got a function. We're going to represent it with a Taylor polynomial so we're going to stop it at n equals something n equals 5 n equals 13 whatever and We know that that's not exactly the sum unless we go to infinity So there's some remainder or error Associated with stopping the Taylor polynomial at n Being a specific number as opposed to letting it go to infinity. So T and T sub n is the nth degree Taylor polynomial So we're stopping that at I Don't want that. Let me see if I can correct it without messing it up too much. I Want to stop it at n so I can't use n over here Let's change that to I So where we had n's before we're going to have an I we want to start I at 0 and Let it run to n, but we're going to stop it at n. We're not going to take a limit now We're not letting it go to infinity So we know what that is. That's the Taylor if a is 0 It's the Maclaurin if we started at 0 and stop it at n. We've got a Taylor polynomial Not a Taylor series If we're going to have some type of convergence Which we don't always have but for e to the x and sine of x and cosine of x it's going to converge for all values of x so if we know it converges then eventually this remainder that we're Deleting not including as part of the Taylor polynomial It's going to be minuscule and in fact as we work our way further out to the right The value of this error or remainder gets smaller and smaller and approaches zero if we've got convergence What is? F the entire f of x is equal to the sum of its Taylor series So we've got to let it go all the way out to infinity in order to actually have it equal to the f of x so let's focus on the remainder Associated with having stopped the Taylor series at a specific value of n So error or remainder for t sub n the remainder associated with the nth Taylor polynomial So I thought I had this printed out, but let me so we're going to be off by a certain amount Kind of depends on the nature of the series, but the magnitude of our error so absolute value I'm going to go ahead and put in a value that is not the way it's written in the book But let's talk about what that z actually means In your book. You're going to see a capital M right here. So we want the maximum Value of the n plus first derivative So we want to know what the error is at its worst now. I use this z because If we're going to maximize the value of the n plus first derivative We want to pick a value somewhere between x and a or possibly x or a So as to maximize the value of the n plus first derivative Now the reason I like to write it like this because it to me it reminds me of what we had earlier in the Alternating series To me it looks a whole lot like the next term If we're stopping it in then the next term is really in plus one So we will have taken the derivative in plus one times We're raising this binomial x minus a to the n plus one and Instead of an n factorial. We have an n plus one factorial. So it has that look of the kind of the next term of the Taylor series we've stopped it at n equals seven. We want to examine the eighth term. It's slightly different because we do want to get a maximized value for the n plus first derivatives somewhere in that interval from x to a It's easy to do with signs and cosines if you start with a sign or cosine when you Take derivatives of signs and cosines. What do you get you get cosines and signs? What's the maximum value that you're ever going to get for a sign or a cosine? Is one so it's it's easy with those you just plug a one in here, and then you see what these are like and That'll give you an indication of the maximum value. I mean the error is probably smaller than that But this would be the error at its kind of its upper bound So we do want to think of the next term Not a hundred percent the next term because of the way this is handed to us right here We want to maximize the value of the n plus first derivative in the interval that we're working Let's do that problem that was ended up being e to the three-fifths and Let's see how close we can get. I don't know might be a bad problem, but it kind of seems fairly tame So let's write out the third Taylor polynomial, and then we'll see what kind of error is associated with that by examining For the most part the next term because the function itself is T three plus the remainder Associated with it, and we want to have a maximum value for the error So what is t three? What's the n equals zero term? one What's the n equals one term and the n equals two term in 50ths that sound right? Got some agreement. We've got some disagreement nine over 50 right nine over 25 over two Nine over 50. I don't know why they wouldn't trust you on that. I didn't trust you yesterday on that That's for n equals zero and one and two so our last term is n equals three. What do we get? We get a 27 five cubed 125 and that's divided by 750 Somebody take your calculator if you have it out and ready and working and let's see what we have So far We know if we did the whole Taylor series the sum all the way out to infinity was supposed to be what e to the Three-fifths somebody take that and figure that out on a calculator as well We're not always going to have the luxury of knowing what the sum is all the way out to infinity, but we do on this problem What is this and if we let this thing go all the way out to infinity? So we're not doing too badly But again most of the time you don't know how you're faring. We just happen to have the luxury of knowing what this is So we want to find the remainder associated With this third Taylor polynomial so we want the Fourth derivative of our function right the n plus first derivative Evaluated at Well, we want to maximize the value of the fourth derivative. Let's just leave it at that. We don't really have a Used this to get our answer so let's see if we can use that in comparison to help us find the error over n plus 1 factorial which would be 4 factorial Normally it's x minus a the absolute value of x minus a to the n plus first so for us Let me just put these So our a value is 0 right because we have a technically an x minus 0 to the n in this problem So we have a We plugged in a three-fifths where there was an x, but that was an x which was an x minus 0 So we're going to do the fourth derivative at 0 and this will become an x minus 0 to the fourth right Now let's fill in go ahead this is really What we used to kind of pattern this after is really 1 over n factorial x minus 0 to the n so our a value is really 0 I'm going to make a change in that because that's not exactly that's not kind of our final Resting place, but I'm putting that in there because That was our a value. So that's a kind of a placeholder right now But I think it's what we want here, and I think it's what we want here We do want a fourth derivative fourth derivative of what by the way e to the e to the x Right, and then we're going to let x be three-fifths when we're done So our fourth derivative of e to the x is e to the x The question is do we really want to evaluate this at 0? Well, we want to evaluate it this z value that I wrote earlier is The value in our interval what's our interval our interval is between x and a in this case It's between 0 and three-fifths That's going to maximize the value of this term So do we know what the fourth derivative is we do it's e to the x Do we know what x is x in this problem is? What did we do for x where we saw an x we put in a three-fifths? Three-fifths minus zero so that's three-fifths now the question is we want to maximize that n plus first derivative By putting in a value either zero or three-fifths or some value between zero and three-fifths Where's that that's an element of yes Right So z is going to be chosen somewhere in that interval or either of the endpoints So do I choose zero or three-fifths or some value between zero and three-fifths? So as to maximize the value of the fourth derivative Well x to the zero x to the three-fifths, which is larger X to the three-fifths. I mean not x to the three-fifths e to the three-fifths. Sorry So we have a choice zero or three-fifths or any value in between because we want to maximize the value of the fourth derivative Then it's arithmetic from here three to the fourth Eighty-one five to the fourth. That's twenty-five six twenty-five all that over four factorial Which is twenty-four so this is an upper bound for our error. We know we don't have much error In fact the error looks like it's What point zero zero six possibly? We're probably going to get a value. That's a little bit larger than that here You're never going to get the exact Error because if you did then you'd get you'd be able to add it to the Taylor polynomial and you'd have the exact value It's an upper bound for the error It's got a kind of a Embedded issue that we're here. We are trying to find e to the three-fifths And we've got to use e to the three-fifths to find it and there are other kind of embedded Problems in the book. I've actually kind of skipped over them because There's another one that forced us to find e squared well if we could find a e squared for a Taylor polynomial for e Almost kind of defeats the purpose. So yes there that that to me is a Problem we kind of glaze over that now because we have this nice little machine that we pop in e to the three-fifths And we don't think much about it but This is supposedly Even though it's got a kind of an embedded problem in there supposed to be an upper bound for the error I don't think we're going to go wrong if e to the three-fifths is really a major issue here We could probably not be in trouble by going e to the zero that's much easier and Then this term would become one as opposed to 1.822. I don't let's just see what happens when we do that With this the rate the way it's written. What do we get? Did you run that through Katie? Okay, which we know that's larger than the error because we can compare this thing to this thing and we know it's point zero zero six Let's see what happens if instead of using e to the three-fifths We use e to the zero which is one. So it's just 81 divided by 625 Times 24 it's probably a more realistic approach to this actually What is that? Okay, not enough probably So according to what we're supposed to do we're supposed to maximize the value of the fourth derivative Unfortunately putting in e to the zero doesn't maximize the value of the fourth derivative So as even though that's easier. It's probably not going to suffice in this problem Yeah, I also see that as a Potential flaw in this situation But in magnitude our error is Slightly less than that, but that's an upper bound for the error Okay, we can I think start this next section Did you come up with another problem? Are we okay with the ones we looked at anybody else? Gosh, we've been on 8.7 for what three weeks now Maybe we can actually go on to the next section. I kind of had an idea. This would happen Which is why I've tried to keep us a slightly ahead on the syllabus because I knew when we got to 8.7 especially We're not done yet with 8.7. Sorry. I just thought of another problem Probably the last thing in 8.7, let's go ahead and just decide that it is the last thing is the product of Taylor series or quotient so let's say we had e to the x and We wanted to multiply it by sign of x and we wanted to see what that looks like in terms of a of a power series e to the x somebody tell me what that looks like the first few terms of e to the x Well, it's x to the n over n factorial So one Plus here's the problem with products of power series is that you've got these dots out here Sign of x. What is that? Well sign is even or odd Sign is an odd function. So what's it look like as it progresses? One now one would be inherently even right because it's x to the zero over zero factorial. So that's even So it's x to the one over one factorial now. We're rolling Again, the Confounding part is the dot dot dot So if we're doing a multiplication Why don't we do kind of a distribution one term at a time? And when we stop seeing those terms in that column then that column is pretty safe We're on to the to the higher orders higher powers of x So if we take let's take this x and distribute it to each one of these What do we get when we take x and distribute it to each one of these terms? Everybody all right with that? Now we could do more of that but We're trying to stabilize the first few terms of this series. Let's actually let's get one more. So it'd be x to the Fifth or factorial and that's going to go on so we're done with that one Now let's distribute this and make sure we keep things in the right column So we've got a constant column, which we don't have an occupant of that right now and we're not going to have one right? Our x to the first column x squared x cubed x to the fourth and so on So what happens when we multiply this guy by this guy? So I want to line up my x cubes to keep track of them What's the next one? minus X to the fourth so I'm lining my extra the fourths up over Three factorial this guy times this guy minus x to the fifth over three factorial Minus x to the sixth Over three factorial and so on so we're done with that one Let's go on to this one and then we'll kind of see which ones have actually stabilized so this one times this one over Five factorial Plus x to the seventh Something happened here. I'm not liking this column. Are we okay? We lost our factorials didn't we let's revisit this one. I'm not that's not sitting real well with me So we took this one times this one there. We're good this times this same denominator. What should this denominator be? three factorial two factorial and This one should be three factorial Three factorial right lost some of the terms in the denominator All right now. We're back to this one this guy times this one We did this one times this one now. We have x to the seventh over what? x to the eighth over Five factorial three factorial Now is Five factorial times two factorial is that ten factorial? No No, we could call it that but kind of like to get things right So when I say things are starting to stabilize Our first term there's going to be no other occupants right of that first column x is the only one So we're not going to write it all but let's start up to write what this series is it's x We've got an x squared We aren't going to have any other members of that one How many x cubed do we have and what will we have any other terms that will occupy this column? That's it, isn't it? Everything else that we come across will be of higher power So how many x cubed do we have well? Here's a half x cubed and we're losing a sixth Is that right of x cubed? So what's our final count for x cubed? What's a half minus a sixth? a third X to the fourth are we stable there? We're not going to have any more Okay, so what is our we've got a one-sixth x to the fourth and then we're gonna do so we lose them Right we lose that term And then you kind of keep going with this process I know it seems kind of antiquated, but you can figure out a power series for a product of two power series But it's little tedious to do that and you could in a similar fashion I'm trying to avoid this one actually Set up a long division problem That would actually be worse because then you'd have to of the term that you're Dividing through into the other one. You're going to have to see how many times you think it will go through There's actually a Long division problem set up top of page 615 Where they have They wanted a tailor series for the tangent well, we have one for the sign we have one for the cosine Let's see what it looks like when we divide We can I mean it's not as it's not ridiculous, but it's I wouldn't classify it in the fun column again, the dot dots dots really enter in here So that would be the Divisor which in this case is the Cosine we want sine over cosine and in under here. We want the Sign Sign divided by cosine so like you would do any long division problem You would say how many times does one go into x and then you say well That's x if we're trying to keep things lined up and now you do the process where you take x times the one Bring it down x times minus one half x squared Which is what and you continue that process for as many of these as you have and what do you do with this stuff? That you bring down you subtract x minus x they drop out You've got this minus and negative so you're going to have some x cubes then what do you do? Then you again take one into the x cubed write it up here, so it's it's kind of the same It's a little more tedious. I guess to do the long division Which gets confounded by how many terms you actually include and how many you're deleting? But power series can be multiplied and can be divided all right now. We're done We will not do anything else with 8.7. We will start 8.8 tomorrow You