 Welcome back lecture 56 math 241. We're about to finish this thing up We have a little bit to do with binomial series today Just to kind of put this class into perspective not the students, but where we are we just took our fourth test on Friday Four of you got those back. I'll try to get the rest of them back tomorrow We have a couple days with applications of Taylor McLauren binomial series and a day to talk about the exam Let me say something real briefly because this is so-called dead week And I think you're starting to kind of at least think about gathering up stuff to get prepared for exams Probably your most important study guide for the final exam Would be what? Your tests okay. I think I've mentioned in here a couple times my philosophy of final exams which was Came from when I was a student that if it's not important enough to be on a Individual test then it probably doesn't have any business on the final exam So those things that didn't quite make it because they weren't important enough Have I ever taken a final exam as a student where those things were on the exam? Yes Did I think that was stupid for them to be on the final exam? Absolutely. Yes. That was dumb So The only thing or things that have a chance of being on the final exam that weren't addressed in some way shape or form on a test Would be what we're doing this week So binomial series is kind of a low-level problem made it to the most recent test But we need to wrap that up in case it's not just ideally 1 plus x to the k And then what we do in 8.9 Has a chance of being on the exam, but other than that it didn't make it to a test It's not on the exam now. I would study more than that Because sometimes problems There are three or four types of problems in a certain category and You know only one of those made it to the test you might want to look back at your notes and hand or the book and See what other kinds of problems fit into that category But yet didn't quite make it to the test so that but still your best study guide will be the four tests that we've had Uh Somebody had a hand up. Well, do were you thinking about a question? Did I see a question brewing? No, okay? All right. Well, let's wrap up 8.8 binomial series Where are we thus far? We are at this point that if we have 1 plus x to the k and if you wanted to do this battle kind of Any problem that we have that's in this form if you wanted to also do that problem in the Taylor or McLauren Format you could also do it that way and come up with a answer that is equivalent But we have developed thus far using higher-order derivatives and evaluating those at 0 1 plus x to the k and this was developed with integer values of k But the convenient thing is that work it works with non integers as well So we've seen that pattern. We saw some other patterns while we were developing this That kind of helped us get to this point in this one. I guess we could also It's a little awkward to put it in the closed form If you know it in this form you can at least get to this point in the problem and And We know that this is in factorial that this is the same thing. That's the power of x is also in This number seems to be what one Larger right in a sense that we're subtracting in and adding back one right to end this You may or may not have all these terms present for example We've already looked at n equals zero, but let's say In equals One if n is one you've got k minus one plus one So it really doesn't go k and then k minus one and k minus two and then come out here to this point Because what is k minus one plus one? It's just k So that's the only value. So this is here only to illustrate how we get from one term to the next It doesn't mean that this will always be present. This will always be present and all the other Successors will also be present. This just tells us where to start This tells us how to proceed if in fact there is an additional term and this tells us where to stop So for n equals one we have a k and a k only for n equals two K minus two plus one is k minus one So we have the first term and the second term this one and this one It's a little awkward at n equals zero if that bothers you you can say one plus This same thing and start in at one and that works as well If we have One plus x To let's say an integer, but not a like a real Boring problem like a positive integer, but let's say a negative integer. This is clearly not a polynomial expression How would that expansion go so that it on the right side? It has the appearance of a polynomial The first term is one Next term is Okay, negative three would be the k value next term which is Thank you negative four Does that work? next term negative three negative four negative five and So on right so we have this nice looking polynomial type expression and Clearly what we started with is not a polynomial in order to be a polynomial. What has to be true about the exponents We've been talking about polynomials for a couple of weeks positive Positive or at least non-negative, right? You can have a zero here's an x to the zero. That's legal Here's an x to the one x to the two So if they're non-negative integers, it's a pollen. It's a polynomial clearly. This is not a polynomial expression And we could simplify that stuff Suppose we had very similar to a test question That would be one plus x to what power Negative one-fifth so our k value Is negative one-fifth so we know it works with positive integers We're not even going to waste our time doing one of those we know it works with negative integers It should also work with rational numbers What should the expansion look like first term is one? Negative one-fifth. What do we do with that? That's the k value. What's the next term? Okay k minus one so one less than that is negative six fifths So there's the x squared term, so this ought to look somewhat familiar not that we use the same problem, but similar to that and what so the Enter or the radius of convergence we do one in this case We want the absolute value of x to be less than one if we have Anything that is going to be added to this interval of convergence Then we would basically check out our endpoints and see if it happens to converge there now There is kind of a general rule it mentions this on page 619 We have enough kind of rules. We know what to do if we need to Add to this if it's we check this out we get the interval we want to check out the endpoints We know how we can check those out to see if in fact that it converges at the endpoints Without memorizing a couple of additional cases, but if you choose to do so, there's where it is in the text All right, what if it's not in the nice convenient form of one plus x? So all these so far have been one plus x to the k So let's take a look at a problem and then when we're done. Let's take a look at how the picture looks and see if the picture makes the Interval of convergence or the radius of convergence Pretty evident Okay, there's the first indicator that it's not going to be kind of straight letter perfect one plus x to the k and Here's another indicator that it's not one plus x to the k But this isn't an awful example. It's just we've got to put it in in that form Now granted it's not going to be X it's going to be something else that's going to go in that position In terms of x and then we need to identify what the k value is and we might have some extra baggage as well So under this radical and there are a variety of ways of accomplishing what we're about to do So if you wanted to go about algebraically this process slightly differently Long as you feel comfortable that everything you're doing is legal There are a lot of ways to manipulate this But under the radical I'm going to factor out of four Which makes that one? What's the other term if I factor out of four? Okay, the square root of that four can come out of the two it's still in the denominator And that's close, but we want it in the form of one plus something something variable raised to the k let's bring the One-half out in front and then we have One over the square root of one what plus Negative x over four does that look right so the half is certainly part of our answer, but let's do the one plus Variable something in this case x or negative x over four raised to the k one plus negative x over four To what power? Negative one so for k it's going to be negative a half we've already done Substitutions like that, but I don't want to say x equals negative x over four because that doesn't look right But x is going to be replaced with negative x over four So the first term is one We want k Which is negative a hacks norm normally? It would be k times x, right? So there's our k and instead of x we have what so there's k times x What's the next term k? times Negative three-halves which is k minus one and so on right let's get one more term and then we'll kind of gather some Terms together, so we're still in the brackets here. Everything gets an extra one-half So we've got k k minus one k minus two All over three factorial our x value, which is now negative x over four Cute so the first term is one-half times one The next term is what? Go ahead and distribute the one-half So there's a two and there's a four so there's an eight and then it also gets this one-half Which is another two in the denominator and it's positive. Is that true? So one-sixteenth x Let's get one more positive Is that correct? There's two negatives There's a negative squared still going to be positive. So we're going to have a three. There's a Two a four an eight and a Sixteen that four gets squared. So there's a sixteen in the denominator We've got an eight in the denominator times this two gives us what 16 times 16 Is that right 256? Check me now my arithmetic is certainly substandard three over 256 We've got the negative x squared, which is just x squared and so on So we've got this nice clean polynomial for a very ugly non-polynomial function if we Took some different truncated versions of this Graph them we took this version and called it T1 because you could develop this by a tailor and get the same thing You would expect this particular truncated version of the polynomial to be linear It has a y-intercept of one-half and a slope of one-sixteenth. So fairly flat and Shifted up a half of a unit now it gets a little more difficult to call as far as what it is But we definitely know it's at least a parabola right and that parabola opens up and we could continue with the Different truncated versions of the complete series Here's an interesting picture That is in your book. So here's what we just Did a binomial series expansion on? There's T1 which we decided was a line right y-intercept of one-half and a slope of one-sixteenth We decided T2 Was a parabola that opened up? So we've got that pictured now as we do better we go from T1 to T2 to T3 What appears to be happening? Graphically more accurate looks like it's kind of matching The curve itself the curve itself is this guy right here So as we do go from T1 to T2 to T3 it seems like it matches the actual curve a little better There's the interval of convergence You can see from negative 4 to 4 that all of these Kind of match the original curve or the original solution curve But again T3 is doing a better job of matching it for a longer period of time But you can clearly see that things begin to diverge As we get close to 4 and as we get close to negative 4 It's only going to get worse as we go beyond 4 to the right and to the left of negative 4 So the interval of convergence. I don't know maybe we should validate that as well How would we get the interval of convergence? When we had 1 plus x to the k How did we come up with the interval of convergence there? I already did this once today early It was that right when that was x We took that x value its absolute value and made it less than 1 Well, we don't really have an x occupying that position anymore not on this one So what should it be for this problem? Negative x there right should be the absolute value of The thing that was serving as the x in the 1 plus x to the k Which is x over 4 Less than 1 and we use that little algebraic Manipulation multiply through by 4 we get that interval of convergence That is hopefully pretty clearly illustrated on this picture that it's The graphs are matching fairly well For that interval and you can see that it doesn't seem to be the case Beyond that that they're going to do a very good job Okay, that's how to kind of do the problems in the 1 plus x to the k format How to adjust a problem that isn't handed to us in exactly that format Now, how could we use this? And let's let's go ahead and use this one because we have One over and let's backtrack to what we had what did we have? 16th x 3 over 256 x squared let's get one more term which would be Negative there and we're cubing a negative so those two negatives give us a positive we're going to have Three and five fifteen in the numerator Two four eight an eight Let's see what we're going to have in the denominator We're going to have an eight We're going to have a six from the three factorial We're going to have a four cubed 64 From the negative x over four the quantity cubed and then we've got a two out in front right from the one half So it should be 15 over What would that be? What is it? So there's the cube term and there are others that's Maybe one too many but at least we're not So this is supposed to converge for values that we choose from x from negative four to four Let's pick an x value in there and see if we can equate Or find the value on the left side Of course calculators make this so easy that there's virtually not a problem with putting any x value in there between negative four and four Too easy to do on a calculator, but let's see if we can More or less get close to that value with this particular polynomial type series How about negative three? All right, let's do that first actually we don't need to write anything down. That's a good suggestion If x is zero it certainly is in the interval of convergence and if x is zero on this side four minus zero is four the square root of four is two And if we've got x equals zero every term that has x will be gone, right? So all these terms Are gone and we're we have one half so it did a pretty good job when x is zero Wouldn't any of them have done a good job when x is zero here's our picture at x equals zero There's not a whole lot of separation from t1 to t2 to t3. In fact, they're all going through that point So we should expect it to be exact Now we're going to come out here to negative three and we do have some separation between the Approximations and the curve itself. Let's see how well we do it. So at x equals negative three We are in the interval of convergence, so the left side should be and this will be our way We can check our answer to see if we're even close Four minus x or in this case four minus negative three, which is four plus three. So there's What we're kind of headed for I hope we get reasonably close Now without a calculator, how many of you are going to be able to do this? We haven't we have an energy conservation Class and we just set aside our calculators trying to conserve energy We want that carbon footprint to be small as we can possibly make it So we don't have this guy. We can't use a calculator I'm going to have a hard time doing that, but this side's pretty easy Because it's just a polynomial polynomials are always pretty easy. I Think I can handle that. Could you handle that? I don't know it might get kind of complicated We might have to square that think we can handle that think we could square negative three I'm still not liking that without a calculator No, we've got multiplication in Addition to cubing do you think we could cube negative three? I think we could cube that but then we got a multiply And oh, and then we've got a divide I Think we can all do everything that we need to do on the right side right without a calculator We may not like it. We don't like necessarily doing arithmetic without punching the buttons. I like it a lot better in this one Does anybody in here know how to extract a square root? That's a dying art With young people. It's an old-fashioned process Called extracting a square root sounds like extracting a wisdom tooth or something sounds painful You do you know how to do that? Not really, but you could there you can get it down to like this square root two times the square root of 30 I mean we could approximate probably we've got some approximating techniques Newton's method could potentially work But well, I guess what I'm trying to say is you're never going to have anything worse than multiplication division and Squaring or cubing or raising to the fourth All right, let's see get a couple calculators around of course. We have a calculator. Why can't we do this one? I agree with that that thought process But let's to avoid doing the arithmetic that we could do Let's see what we get on the right side Somebody also do this on a calculator take one and divide it by the square root of seven and let's see how how we've done is this 0.3 seven 26 it's 27 times 15 Kind of doesn't to me seem at all out of character that this thing would be alternating Because one half clearly is too large right? If you stop here, that's Too large and then we subtract three sixteenths, which means we probably subtract away Too much and then we have to add something back in but we add too much back in so we have to subtract something away That's that's not shouldn't be a surprise. What did you do the right side? Not bad and especially the way That we saw the curves being beginning to separate from the actual solution curve itself the further we got away from the origin because it was centered at zero and If we went to negative four and four begin to separate and we're Kind of on the outer fringes of that but we can take something that is Kind of without a calculator nearly impossible for us to do and turns it into an arithmetic Problem that we would rather use a calculator for But if you were given one over the square root of seven and you approximated it to be that Decent approximation Okay, let me see check my notes and see if we've done justice to this. I think we have We have and on a rare occasion We will end early because we will we've got two days to do 8.9, which is plenty of time So I will see you tomorrow