 Welcome back lecture 47 Returned the other part of the tests today, so I think we can kind of Address some things collectively about test three Average score for this classroom was 84, which is very good lots of scores in the 90s several scores a handful of scores above the 90 so More than a hundred think 102 was the highest And we had I think two or three of those Just some things that I noticed in common from grading the tests first two problems I Don't really remember anything in common a few minor errors most of you got the first two problems completely, right? You do need to I Think I just wrote it in I don't think you took any points off if you said if you just kind of gave the equation but it is either y of x equals that if you have some Y function in terms of x or for the Spring problem. It's actually an x of t Again, if you didn't have it x of t. I think I wrote that in Changed your x's to t's or whatever But do kind of pay attention to if you're dealing with y double prime y prime and then functions of x It is your final answer is a y of x I Don't really remember anything specifically a couple of minor errors on the problem three Problem for you did have to find the spring constant which Doesn't work out wonderfully, but it's not ridiculous because it is a Can't factor it, but when you solve using the quadratic formula you get Alpha equals negative five and beta equals four and The c1 and c2 values if you use those letters c1 and c2 worked out to be pretty nice, too I couldn't get integers as solutions for every part of every problem. I think problem three solving for b which is negative sixteen-fifths and c which is negative sixty-nine twenty-fifths Those aren't delightful, but most of the problems had integer values In five so each of these which I said on the day of the test that each of these was worth ten points I really kind of intended for a to be kind of quick and easy Look at it Four and sixteen and sixty-four it looks like there's a common ratio The ratio is negative point four several people put point four if they alternate in sign the ratio has to be negative So it is an infinite geometric series the ratio is negative point four You should also in there at some point in time say that So that I know that you know that the absolute value of the ratio is less than one which Combined with the fact that it's an infinite geometric series those two facts make it converge And it converges to one which is the first term Over one minus the ratio, so it's one over one point four Which is equal to five sevenths, and then it's approximately equal to one point seven one Four two eight five seven something like that A variety of reasons for number two the simplest is that it's harmonic for a sorry part B Some of you chose to validate convergence on 5c separately from the decomposition, which is fine So you basically compared it to a p it is not itself a p-series, but it can be compared to a p-series and Then you decomposed it looked at the telescoping pattern and figured that the sum all the way to infinity was five-thirds D was a gimmie The limit is one-third as long as the limit is not zero you're finished it diverges E you were supposed to use the integral test So that would probably be the highly suggested method would be to go ahead and use the integral test since that was Recommended somebody use the integral test and another test, which is fine because you did use the integral test But you don't really need it in combination with any other test and F You can compare it to either one over three in or one over in Which is harmonic? Unfortunately the comparison test Doesn't quite get us what we want because it would need to be larger than the existing divergent series This one unfortunately is smaller But because it's very closely related to an existing divergent series. We can use the Limit comparison test, which is f and the bonus although we've never done third order The pattern is the same the tricky part about the bonus is There's not any real specific pattern that we have accumulated to this point in time for factoring a cubic polynomial I could tell by Some of your work on this problem that you used a graphing calculator, which is fine to see what the roots were actually With the coefficients the way they are you can almost guess one of the roots one of the roots is one You can just look at the coefficients and see that that would give you zero once you get one of the roots You can use long division to reduce the cubic to a quadratic and either factor the remaining quadratic or Quadratic formula, but there are three roots one's a double root and the other is a single root So you use that same pattern and go from there So the solutions are on the solution sheet you want to kind of compare and contrast what you have Not that there's just one way to do the problem But this is at least one correct way to do the problem And if you think it's graded too harshly then you can appeal the grading But realize that that also opens up the possibility that more points could be taken off I could say oh, you know what? That I forgot I thought that was two points, and I took one off and it's ten points. I should have taken eight off I've never done that, but it is a possibility but that was collectively I thought that was Excellent mastery of this material across the board 84 for a class average is excellent All right, let's finish up 8.5 today See where we left off We did look at some sequence of partial sums of the Bessel function Why don't we start with that and a diagram that's from your book That I think has significant carryover As we finish this chapter we'll be doing some things that are similar to this plus I want to Notationally I have something from last class that's slightly different from this that's in the book I Wrote the sequence of partial sums So I wrote basically s sub one equals one because that was the sum of the first one terms In your book. That's denoted as s sub zero because n starts at zero Okay, so it's the same thing I was talking about last class I don't know which one is better to be honest with you, but from this diagram Let's go ahead and use the notation that the authors have So this is the sum of the first one terms even though they call it s sub zero on the Bessel function Here it is. That's not really very close And you wouldn't expect it to be very close to the final curve because it's only one term They call s sub one yesterday. We called s or I mean at Wednesday. We called s sub two pick off the first two terms One minus x squared over four should be a parabola that opens down Which is closer to the actual curve for a while, but then begins to deviate from the actual vessel curve We call this the sum of the first three terms They call it s sub two because we're going up to n equals two n equals zero n equals one equals two Those are three terms so you can see that it is Where is s sub two? There it is Comes up here and that's kind of what a fourth degree polynomial should look like it potentially could have Four roots. I mean if you slid that down you could see four possible points of intersection with the x-axis But it's not unlike any other fourth degree polynomial. You can see the coefficients the Coefficient really, but it's the denominator term So it's a one over 64, but you can see the denominator beginning to grow which kind of connotes the fact that it's probably Convergent in fact this is convergent for all values of x S sub three Sixth degree you really can't see enough of it here to show you what it's doing, but s sub four You can see it come down here and start to turn Come down here and start to turn Now what is the whole thing look like if you had as many terms as you could possibly generate that would be this graph here at the bottom so the entire Bessel function which is given to us by this strange power series would look something like this now that one thing that I think is valuable about showing this it's kind of a Oscillatory function in a sense. It's more like a sine or a cosine with a variable Yet decreasing amplitude. So here's the amplitude of the first branch Here's the amplitude of the next one. Here's the amplitude of the next one. So it looks like this Oscillatory function like a sine or cosine or sine plus cosine type function That has a decaying amplitude, but it clearly is not one of those So it's it's a function and we don't necessarily have to use traditional Oscillatory functions to end up with a function that does in fact oscillate. So that's where we left off Yesterday Wednesday, sorry we determined that these were the three possibilities When we dealt with a power series That it converges only at a certain value namely the value that causes the binomial to disappear each time at x equals a Converges for all x which is what the Bessel series did and there's some interval of convergence Maybe including an endpoint or two or possibly Excluding both endpoints. So we'll get a couple more examples of this category today But we shouldn't expect anything else to happen. In fact, I think we've done all these examples. So I think this is appropriate Geometric series they are technically power series in acts because the power of this one grows by Consecutive integers as we allow end to scroll from zero to infinity The interval of convergence This is this series So we can see that the ratio is x and it converges If the absolute value of the ratio is less than one. Well, that means in this case that the absolute value of x Is less than one. There's the interval of convergence We did this example. It was Convergent for only a single value and that single value that caused this one to converge was zero That was our a value in this problem. We're going to write it in there. It looks kind of ridiculous To write it as x minus zero, but it is x minus a and a is zero and it converges at that a value only So if that's the interval of convergence Yes, it's not an interval because it's just the single value zero then that means the radius of convergence is zero Let's go to the last one. The last one is the Bessel function We decided it converges for all values of x Therefore the interval of convergence is negative infinity to positive infinity and the radius of convergence is infinite Think this is exactly the same example. We did x minus three to the n over n We ended up with kind of a first answer of the interval from two to four We checked the end points. We decided it converged at two and diverged at four So we included two in the interval of convergence We also decided this was centered at x equal three Which that interval is in fact centered at x equals three three being the a value in that problem And from that centering value of three Basically, we're going to swing an arc one unit left and one unit right so the radius of convergence is one Which kind of captures all the other? Values that cause this particular power series to be convergent So they don't necessarily converge for all values. This in fact is kind of rare when they do converge for all values This is a little more common For us to get an interval of convergence This is in your book any questions about this Nicole Why is the third one down and is equal to one? What's the difference? If we tried to start that one at zero we'd be dividing by zero Right So there really is no term for n equals zero. We can get the same power series. We just don't have a constant Most of them will have a constant first When the end is equal to zero Back to all the others have a constant when it is equal to zero This one we can't start it at zero. So it does not have a constant. So the first term of that of This would be X minus three to the one all over one. So that's the first term no lead Constant as a part of this power series Anything else? Let's finish this up with these two problems So these are examples both these examples we're going to do are in the book But there's some interesting things that happen in them that I think it's good to see them happen in an example before we Find them happening in another problem and we haven't seen that in an example And I'd like for you to kind of direct me and how we go about Determining if there is an interval of convergence if the endpoints are in fact included in the interval of convergence What's the first thing we would do to Determine convergence and divergence for this particular power series That might be helpful I Don't think that will be all that advantageous when we get to the Kind of beyond the first step. What's the first step of these problems? The ratio test So if this were a test question, I'm not going to say Use the ratio test to determine the interval of convergence So that's that's the first step is to use that and I didn't mean that to be facetious by the way I'm just you have to know that that's your first step because I'm not going to say to use that and How does that go? Okay, the n plus first term and we do want absolute value and it looks like this thing might alternate, right I Don't know if that would be all that you maybe we can do that on a second step We could split that up right negative three to the end Equals what negative one to the end times three to the end I don't know how helpful that's going to be but that's if you wanted to separate the alternating part from a non alternating part Because we are using absolute value Time up to you so that n plus first term is going to be negative three to the Plus one what else Okay, get everything on the same line instead of having a fraction within a fraction In plus two All right, so what can we do away with or simplify or somehow combine as we continue In plus one on the bottom In the top and get a square root of one I Wouldn't Can't do that. I mean we could do that but we'd kind of like to get the right answer so you better not You can't do that. Oh, no. No, you can't reduce parts of sums and differences that are under square roots and cube roots or Even for that matter any kind of binomial. You can't just take a part of it and knock it down Even though that was in plus one plus one So you couldn't separate in plus one plus one to in plus one square root of one You can but you can't reduce them because you got to have like terms You've got to be able to multiply and or divide by something to reduce something So if it's part of a summer difference, you're kind of stuck with that now we can deal with that We can put them both under the same radical and Get a simplified version of that, but that's not the way to simplify it What about negative three to the n plus one over negative three to the n? Negative three to the one and what about negative three if we've got absolute value just three How about the X's okay X to the one in the numerator and These ends we don't have to worry about them Being negative, so I'm going to move them outside of the absolute value and put them under the same radical We really don't in fact, we'll lose that as we go to the next step Because we've already in a sense taken the absolute value of it, right? And getting three Let's deal with This I'm probably you know the answer. I mean if n approaches infinity What's 600 trillion plus one divided by 600 trillion plus two? I mean aren't they practically the same number numerator and denominator So if they're the same number that's one in the square root of one would be one so we're probably going to get one Let's see off to the side if we can Divide everything by n In the top Divide everything by n So n over n is one one over n We know what happens to that as n approaches infinity don't both of these Approach zero right as n gets infinitely large, so we've got one over one the square root of which is one So let's just go ahead and go to an answer. This circled part is one And we've got a 3x. We've already taken the absolute value So we could leave it in there or we could say three Absolute value of x not going to make any difference. We're hoping that once we do the ratio test That what we get for an answer. So this L this limit is what? In order for this to converge We want the limit to be less than one. So here's L so we want Sorry, I can drop the absolute value now right should write that instead Dealing with the absolute value inequality 3x can be less than one as long as it's greater than negative one Divide all the way through by three So at this point in time on this problem, this is our interval of convergence Not a whole lot of values for which this particular power series Converges at least the interval isn't all that large What would be next? Test the end points So at x equals negative one-third I don't know if your mind operates this way, but my strange mind operates this way as I'm writing it down I'm trying to think What is their product going to amount to and if it is what I think it is on my first go around What if I have that number up there divided by the square root of n plus one Do we have we dealt with anything that looks like that prior to this? So that way you haven't kind of an idea where to go. What do you think about the numerator? We have something to the end and something else to the end can we take their product to the end? Yeah, I think we can so the numerator is one What do you think that if the numerator is one? Not a harmonic because of the square root Okay, so you want to use a comparison. Okay, I think we've kind of done that generically So we that would work that we get you there Can we use a p-series here? Isn't this very close to one over the square root of n? Not a whole lot of difference between n plus one and N when you're putting in infinitely large numbers. What do we know about this? This is really One over into the one half right, so it's p-series p is one half and It diverge or they diverge when p is what? Less than or equal to one So you could say this is a p-series What about this one? Well, although it's not really a Straight by the letter p-series. It's pretty darn close. So we could do what we could compare is this one Larger than this one which already converges Well, we're adding a little bit to the denominator, which makes it smaller. So that doesn't work, but then we could Use the limit comparison test Which you would have done anyway, so it's about the same amount of work But as I was writing that down, I'm thinking square root of n plus one That's the square root of n which is into the one half, which is a p-series So the limit comparison test would be Doesn't really matter the order The one in question Divided by the one that we know something about You can work this out. It's basically what we've already done and that's going to be one Because the limit is some finite value In this case since this diverged Then they both diverge so in other words, we don't want that endpoint So that's not included Let's check the other end point any questions on that before we move to the other endpoint Any guesses what's going to happen at the other end point? It's going to be alternating right this one ended up being one to the end which doesn't alternate at all one to the end For differing end values is just one all the time the other version has a high probability of being an alternating series So the other endpoint is x equals one-third So what does the series look like? These are both to the end in the numerator. So their product is negative one So it is alternating we could use the alternating series test How does that go? Two pieces to it two parts to it Got to be decreasing or ultimately decreasing so the n plus first term and we're not talking about the alternating part We're talking about the numerical part So the n plus first term would be that is that smaller than its predecessor it is Larger denominator smaller fraction and is the limit of that nth term description Way out to the right Is it zero it is So this is convergent by the alternating series test So our final Interval of convergence We did not want negative one-third. We do want one-third Centered at what? What's the center of that interval? Zero and if you go back to the original problem We didn't have an x minus a we just had an x which is technically x minus zero So it is centered at zero and what's the radius of convergence? One-third of a unit either way from zero the centering value. Okay, let's see if we can get through this next one A little bit quicker we do have a value other than Zero that this should be centered around What are you going to write down first? Directions are find the interval of convergence if in fact there is one and if there is one what's the radius of convergence? Ratio test everywhere. There's an n. We now want an n plus one. All right, see if we can get everything All in the same fraction rather than have a fraction within a fraction n plus one X plus two to the n plus one be an n down here x plus two to the n We're going to have a three to the n plus two in the denominator Multiplying by the reciprocal we get a three to the n plus one. All right. What leaves this what stays around? And where is it if it's staying around? Okay, x plus two in the numerator and that's going to stay in the absolute value symbols, right? Because we don't know if that's positive or negative Three goes where? Denominator can we bring that out front? So a three in the denominator, right? We have one more three in the denominator than we do the numerator and n plus one over n as n approaches infinity This is just one, isn't it? Does that work? We've got an x plus two. We don't know if it's positive or negative. We better keep the absolute value notation We've got a positive three in the denominator. We'll bring that out the other n plus one over n approaches one That'll be our answer. We want that to be what? Less than one we'll convert that To an algebraic inequality that kind of gets rid of or dispenses with the absolute value now what? Multiply through by three Subtract two so negative five less than x less than one now This is supposed to be centered at what value from the original problem Negative two is that centered at negative two it is centered at negative two Well, if negative if we looked at this interval negative five to one On a number line, which is what we have thus far Then our value which we say it's centered at actually should be in the center of this interval Okay, somebody want to answer that question how can you look at the original problem? The original problem this was had an x plus two and Generically that's supposed to be x minus a In this case to the end So what's the a value if x plus two is really x minus something? That'd be x minus negative two so there's our Value about which it is centered and what appears to be the radius of convergence Three right because we're three units in either direction negative two The last thing to check which we have just about enough time to do this and we'll be finished are the endpoints So our first endpoint is negative five So for x we replace yet with negative five See what we get So it looks like we have a negative three to the end Okay, what can we do with that to kind of get rid of some why don't we go ahead and call that? negative one to the end Times three to the end. That's the same thing as negative three to the end That way we can knock out three to the end with three to the end plus one which leaves a three where in the bottom I'm not liking this one. It is alternating. I don't think it's going to pass the alternating series test as it Are the terms getting smaller as we progress now, so the alternating series test It's going to fail in the first part So is in plus one over three is that less than in over three? No so diverges So we don't want that endpoint the other endpoint was What was it? One I think it's in more trouble, isn't it? I mean the other one had a chance because it was alternating this one's not alternating So there's three to the end and we'll reduce it with three to the end plus one which leaves a Any chance of that converging? How could we validate that? Right there's the description of the nth term in order for that to have a chance to converge that's got to be disappearing to zero Well, that's one third which is clearly not zero Therefore it diverges if the limit of the nth term does not go to zero it diverges So we don't want that point. We don't want that point. So the original Interval of convergence was in fact our final answer Which was what negative five? Two one That all right, so that should finish us up And we will forge ahead, but we will wait till a new week to do that Have a great week