 Welcome back math 241 lecture 46 out of Probably somewhere around 60 before we're all said and done. So we are in chapter 8 We will not go further than chapter 8. In fact chapter 9 starts math 242, which most of you will be Heading toward I would presume So We have quite a bit to do in chapter 8 still pretty loaded with things that we have not Encountered yet So let's go ahead today and also Friday in here. We'll do power series Still puts us just a little bit ahead. I guess according to the syllabus, but We're going to need it when we get to some of the later Parts of chapter 8 All right, so let's go ahead with power series Test grades by the way the those that I've graded I think I've get graded 10 Are pretty good couple of scores over a hundred Of course, you're only happy with that if it's on your paper Anybody That necessarily thrill you that you know person two seats down got a 102 Kind of thrill you I would think if you got one yourself All right power series So I intended to have all the tests graded, but I think I did a little more moving than grading yesterday. So I Got kind of closed out in available hours We will have a lot more series starting at zero now instead of starting at one one of the reasons is that we'll want a Some kind of a constant as the first term and instead of adjusting the exponent in here will Just start it at zero and we'll establish that first term as a constant. So notice the subscripts On the constants are different. So we're allowing for different coefficients on different terms as we progress and Then we're going to increase the power of x as we progress to the right. So it's kind of like an infinitely termed Polynomial is what we end up with So at n equals zero We get c sub zero x to the zero which is one So there's one of the reasons why we want to start at zero we get a constant n equals one c sub one x to the one n equals two c sub two x squared and so on so we get this polynomial looking thing We'll be able to deal with some some of the partial sums and we'll be able to get different Polynomials when we truncate, but if we let it run forever It is a power series, but it behaves in a lot of ways like a polynomial So we will determine When these things converge they'll converge for certain values of x I mean you can probably pick one off right away What is one value of x that would cause even this generic power series to converge? If x were what value? Zero if x were zero then all these terms disappear and it converges to c sub zero So you're going to get some values of x for which it converges and you'll see some Categories develop as we look at these problems some other values of x for which the series diverges Okay, there's another type let's get an example of this type in fact We've already seen this one can't get a whole lot simpler than this one So it's a good one to start with So we don't have any varying Coefficients as we go all we have is powers of x. We've already seen it So when x excuse me when n is zero x to the zero is one when n is one x to the first x squared x cubed Well, we say they converge for some values and diverge for other values What is this particular power series? What have we already classified this under that is if you were actually talking to me today Geometric and not all infinite geometric series converge when do they converge? Okay, so if the absolute value of the ratio is less than one We know that it converges What is the ratio on this particular infinite geometric series? X Because we're multiplying by x as we go So it will converge if the absolute value of x is less than one so we have a so-called interval of convergence from Negative one not including negative one All the way up to one but not including one So for any other value of x that we choose in this interval This is a convergent series for anything outside of this It diverges okay another type of power series and Really, this is a the first one that we looked at is a more Specialized version of the one we're about to look at so instead of just having Increasing powers of x as we progress to the right We have increasing powers of the binomial term x minus a So the first one ought to be c sub zero x minus a to the zero well, that's just c sub zero n is one When n is two n is three and we get a similar Type of expression except powers. It's a power series, but not a power series in x It's a power series in the binomial x minus a So why would this be a sub category of this? This is kind of a more generalized version this should work for any a value this only works for the a value of Zero right if a were zero in this one we'd really be back up here to this one So they're really all Power series of this type and if a happens to be zero it's a simpler looking power series now when we have varying Coefficients we're not going to have that nice luxury that we have here when the coefficients were all one where you can say okay What do I how do I progress from term one to term two term two to term three? We're multiplying by x we don't have that luxury here because of the differing Coefficients as we progress so this is not going to be unless it's a really specialized Problem it's not going to be an infinite geometric series. It is a power series. It's a power series in x minus a sometimes it is said to be centered x equals a and I think That will become more clear as we progress to this section why we would say that it's centered at x equals a All right, so there's the most generic version We'll see some like this, but it's the same thing except a is zebra. All right. We're going to look at Probably four examples maybe more. I don't know how many of those will get to today and then we'll kind of classify What happens in those? different problems into some Categories of answers and we'll see that all Types of answers will basically fall into one of three categories when we're done So the series n factorial x to the n now we think n factorial. That's getting larger as n gets larger x to the n May or may not be getting larger as x gets large as n gets larger because we really don't know what x is So the tendency is to see this and say well clearly that's divergent, but we don't know what x is So the question is When does this converge and we have no idea what after some examples? You'll see Solutions to problems as varied as they are Falling into categories, but we don't really know what's going to happen here So this is a test that has a lot of carry over throughout the rest of this chapter of the ratio test We're going to take the n plus first term divided by the predecessor the nth term Way out there to the right. We're going to see what we get We know that if L is less than one. What's our conclusion? Convergent L is greater than one Divergent and what happens if L is equal to one Okay, no conclusion. So the test fails All right for this problem the n plus first term would be what? okay x to the n plus one So everywhere. There's an end. We should now have an n plus one So there's the value of the n plus first term. I've got absolute value brackets started here. Do I really need those? Okay, well right unless x is negative so x is unknown so we better keep them around And you'll see in maybe not in this problem so much But is in in other problems the absolute value will certainly come into play So there's the n plus first term we want to divide it by its predecessor Which is the nth term which is normally the way it's Anded to us. What is in plus one factorial? How could we rewrite that? Good in plus one times in factorial I had a question after class yesterday About that if if that troubles you in any way just put in some numbers Eight factorial is eight times seven times six times five all the way back to one Which is really eight times what? seven factorial so Just enough to convince yourself that that is in fact true for n plus one factorial Just for the purposes of reduction x to the n plus one is x to the n times x to the one question Jacob Would in plus two factorial be in plus one times in plus two times in factorial There's a variety of things that could be So if you add in plus two Factorial if you wrote it out it would be that so you could say that is in plus two times what in plus one factorial or Depending on what you needed it for you could say it's in plus two times in plus one times In factorial that might be a little more advantageous Especially if you had one of these terms around numerator and denominator and you wanted to knock it out With a like term in the opposite position so it kind of just depends on what you need So back to the kind of planning ahead. We knew we had an in factorial here So it's advantageous to have one in the numerator as well We've got an x to the n down here So maybe it's advantageous not that we couldn't do the problem without it But it's nice to just kind of mark out like terms numerator and denominator And that's where we are in factorial over in factorial X to the n over x to the n in plus one is inside the absolute value brackets But it's unaffected by the absolute value so we can bring it out front So the x that remains Might be positive might be negative. So let's keep the absolute value notation around Does it really matter? What x is here? When n is allowed to increase without bound this term gets larger and larger and larger What is the only value for x? That would keep this thing from getting larger and larger and larger Zero so this only converges When x is zero for all other values of x this diverges So that's not actually a very interesting problem if it only converges for the single value of x equals zero But the reason for the example is to we're going to kind of see what Categories our answer falls into there's one of the categories It converges for a single value only in this case that single value is x equals zero Questions on that before it gets moved. So these kind of increase in difficulty as we go to I don't think any of them are Ridiculous there's one that's Kind of testy notice this starts at one so we're not going to have that Constant term we probably would want to start this one at one for what reason? Kind of problematic if we started it at zero because we'd have division by zero which would not be delightful So it n equals one we've got that to the first over one squared over two cubed over three and so on so it's kind of a Polynomial, but it's got an infinite amount of terms, so we want to do the ratio test to see when this converges If in fact it ever does I guess we could determine that now it would certainly converge at what x value? Three so if x were three this thing would clearly converge are there more values for which it converges? We don't know that and We'll see a problem that converges for all values That I think is our example that follows this so we do have some differing things that happen All right ratio test. What is the n plus first term? That work take the n plus first term divide it by the nth term you'll be experts at the ratio test Very soon if you're not already Should write this x minus three Okay, and then when we multiply by the reciprocal of the denominator And even if you did not do what Nicole said hopefully you would be able to say to yourself What's x minus three to the n plus one? Divided by x minus three to the end there's going to be one of them left in the numerator right because it's raised to one power Larger, but that's a cleaner reduction when you do separate it in the numerator In an n plus one we could in fact bring those out in front There's no question about them being positive when the end starts at one But the other which is x minus three is still variable We don't know if that's positive or negative or when it's positive and negative So let's leave it inside the absolute value notation what happens to n over n plus one as n gets infinitely large That's goes to one So one times the absolute value of x minus three is The absolute value of x minus three so there's our limit there's our L and We want to know where this converges So we would want L to be what? Less than one so there's our before this point. We've had in a lot of cases a numerical answer Yesterday we ended class. I think we ended class with an example that had e For our answer e being larger than one meant that the series Diverged here. We have absolute value of x minus three. This is only going to converge when that is less than one So that means that what is inside the absolute value notation? Is kind of trapped between negative one and one so that's the algebraic Solution to that absolute value inequality we can add three So in the middle we get x which is what we want Here we get four and here we get two So for all values of x between two and four This particular power series is a convergent power series So at this point in time we might make some adjustments to this but at this point in time the interval of convergence is That interval so for all values of x Anywhere from two to four but excluding two to four. We already knew one of those values, right? Didn't we already know three? From the original power series three was going to cause it to converge But apparently it's not just three. It's any value between two and four excluding two and four now If x were two What would this be? It'd be one and our answer To the ratio test the L value would be one now when the L value is one What does the ratio test tell us? It doesn't tell us anything. It doesn't give a conclusion So we probably ought to check out two Since it the ratio test fails to tell us what happens it to we ought to check that out ourselves Is the same thing true at four if x is four? The limit is one the ratio test fails to tell us what happens at x equals four We better check this out ourselves. So we're going to check out two and four separately So our end points x equals two and x equals four let's go all the way back to our original series So if x is two, I'll just go ahead and write it in two minus three to the end All over in well two minus three to the end is really what? negative one To the end Hey, that looks familiar Converge it because Okay, we've already done that right this is an alternating harmonic therefore it's convergent So even though the ratio test the limit was one fails to give us a conclusion We kind of do that by hand do that ourselves and We determined that it in fact converges at x equals two. Let's do the same thing at x equals four Go back to the original power series What is that one? Four minus three is one one to the end well one to the end is just one We know about that one, too. What do we know about one over n from n equals one to infinity? That diverges that is not alternating harmonic, but just the harmonic series so Initially we said our interval of convergence Based on the ratio test was from two to four excluding the end points Now we know we also include this In the area for which this particular power series converges. So our final interval of convergence is two To four we don't want to include four So there's the interval of Convergence I used the term centered at x equals a a while ago This should be centered at x equals three If that's the case right is our interval centered at x equals three That's right in the middle of two and four. So it is centered at x equals three and one more piece of terminology and that is the radius of convergence In this case is one so starting at that centering value which in this case is three We're going to go basically one unit to the left and one unit to the right So if we were swinging an arc of radius one, we would basically capture our interval of convergence So interval of convergence it is centered at the a value from the original series and the radius of Convergence how many units away from that centering value are we getting we're getting one unit away That okay All right. This will be our last example. I know there's quite a bit of time left But this one's a little bit of a battle This is called the Bessel function. There are some pictures of The Bessel function On page 595 kind of what the what this is being used to model the Bessel function Says that functions first arose when Bessel solved Kepler's equation for describing planetary motion Since that time these functions have been applied in many different physical situations including the temperature distribution in a circular plate and The shape of a vibrating drum head So I'm sure that piques your interest because that piqued mine But it's a neat function and we'll see something happen with this one that we haven't seen on another example So when does it converge? When does it diverge? So we're going to use the ratio test Can you see what a mess this might be? What is the n plus first term? negative one Which this really doesn't matter because we are taking absolute value So the alternating sine portion of the original and also this become irrelevant, but I'm going to go ahead and write it down X to the 2n should become x to the 2 in plus one Quantity n plus one that matters in this problem 2 to the 2n becomes 2 to the 2 times the quantity n plus 1 and This is kind of crazy looking in factorial squared becomes In plus 1 Factorial squared, so there's the n plus first term Everywhere there was an n we should have replaced it with an n plus 1 All right, let's go ahead in the denominator and write out the nth term the way it was given to us And let's see what happens Way out to the right as n gets infinitely large Before we start knocking out like terms Can I get rid of negative one to the n plus one and negative one to the end? I mean at best there they would be I mean it's going to be negative one right which is going to be done away with So there's no reason to carry those any further Let's leave things where they are right now x to the 2 times the quantity n plus 1 that would be x to the 2n plus 2 same thing down here 2 to the 2 times the quantity in plus 1 and let's write this as n plus 1 Factorial twice So that's in the numerator might be helpful I mean you might skip this step, but it might be helpful just because it'll get you thinking about Reductions or separations in the numerator that might be helpful to get rid of some things in the denominator Another absolute value bar over here All right, so in the numerator Let's go ahead and simplify some things as we go How might we want to rewrite this term? Okay x to the 2n times x squared and we want an x to the 2n to reduce with the other x to the 2n that's in the denominator similar fashion 2 to the 2n plus 2 ought to be 2 to the 2n Times 2 to the second while we're rewriting things. We've got an n plus 1 factorial and Another n plus 1 factorial. Let's rewrite both of those as Okay, so that's the numerator we want to multiply by the reciprocal of the denominator So here's our denominator. Let's just take things that are in the opposite position and Reverse those positions, so we're going to have an x to the 2n here 2 to the 2n In factorial in factorial good ol. Mr. Bessel now what Okay, now how many of those things can we bring out front that are not affected by absolute value We can bring out the one fourth and that n plus one is in the denominator, right? Two of them and we do have the square root of x squared. Well, we're really not worried about that either. Are we? I'll leave it, but it's it's really kind of does it's unaffected by the absolute value In is going to approach infinity any before we go to the next step any questions about the terms that Knocked each other out numerator and denominator or what remains Are you all right with that? What happens to this term right here as n gets infinitely large That goes to zero How about our old friend x here? We're really x squared if this term is going to zero For a single value of x that we choose this part out in front eventually overtakes this, right? So it'd be zero times whatever value we have here. It's not like this is Changing as this is changing and one could be getting smaller the other one getting larger. This is a fixed value for a Certain value of x so it looks like we're going to get zero That's a different kind of answer because we would expect this as the L value We would still have some work to do when is the limit the L value less than one When is this limit less than one? It's always less than one L which is zero Is always less than one so what does that say? About convergence for this particular power series Converge it for all values of x so the interval of convergence This one is still centered. Where is this centered and you could even go back to the original Version and see where it is centered. We didn't have an x minus a we just had an x, right? So it's centered at zero and what's the? radius of convergence Also infinite right is infinitely large so we have seen what types of solutions we have seen a single value of x as a solution it only converged at that one value and We have seen an interval two to four including two That was an interval of convergence and then we've seen one that can be centered for I mean convergent for all values of x that we choose So you'll see this Theorem and I think we've seen an example of each type in the book to this point For a given power series x minus a remember a could be zero and it'd be the simpler type There are only three possibilities for Convergence it converges at a single value namely the a value Converges for all values of x we just finished one of those and There is a positive number such that the series converges and this would be some interval Whether it includes or excludes the endpoints we have to kind of do those separately, but it ultimately ends up being some interval I think this is Probably let's not start another example, but let's deal a little bit further with this one I don't know how far back we need to go with that Let's go back all the way to the original just to make sure Let's see what and this is a good lead-in. I'm not trying to disuse up a few minutes of class That this has some nice carry over into what we do later in chapter 8 Let's look at a few of the partial Summs, so let's look at what the first term would look like Which in this case this would be in Equals zero right So when we enter in to this series try to generate the first term We'll enter in n equals zero and what do we get that to the zero? That x to the zero two to the zero zero factorial. I think we did we cover that did that come up yet? That zero factorial by definition is one Think when you put in n equals zero you're going to see that the first term is One so now we want to generate the first two terms So we're going to put in n equals zero Which we already have and also n equals one so it's going to be one What's the next term when n is one negative one to the first so it's going to be negative? x to the What power squared Right when n is one and when n is one, what's the denominator for? That's not a real complicated looking polynomial In fact this one isn't complicated either. I like this one the first one One if we were to graph that That'd be pretty simple. We could then graph this When the generating the n equals zero and the n equals one term What would that be if you graph that? Just the first two terms kind of a truncated vessel series Brabler opening down right shifted up one unit would be what that is What'd you say Jacob? Said it's fat kind of fat. Yeah, you weren't talking about me. Were you? Okay, I mean it's factual, but let's not put it out there on television Okay, so we went to n equals two Negative one to the two now. We're back to positive. So we all gonna are gonna alternate signs When in is to what do we get up here? X to the Four that's not a surprise is it because we have x to the two in so it's going to go up by multiples of two as we progress And when in is to what do we get in the bottom? That'd be four two to the fourth which is 16 We'd have two factorial two factorial is two squared, which is four sixteen times four sixty four that right Another polynomial If you would graph this one, I mean the original Bessel function looks kind of strange But I guess the reason for going through something like this. That's not really strange There's nothing strange about that one. There's really nothing strange about that one That's a fourth-degree polynomial Tell me something about fourth-degree polynomials kind of a any generic fourth-degree polynomial How many times might they hit the x-axis? Four right because they'd have four possible solutions How many changes in direction does a fourth-degree polynomial have Three right because you take the derivative which would be third degree and set the derivative equal to zero Which would have three possible solutions, so we all know what this is. That's we don't need to you know Debate this one, but this one although we don't graph a whole lot of fourth-degree polynomials a fourth-degree polynomial There's one two three four Probably is going to look something like that right and I started this kind of up one unit so there's the s sub three the seek the Sequence of partial sums going all the way up to n equals two Without generating this I think I have this one written down There's when n equals zero n equals one n equals two We would expect the signs to alternate. We would expect the power of x to be what x to the sixth You probably see why convergence happens because these values the denominator begins to overtake the numerator So that eventually the denominator gets so big that it really doesn't matter what the numerator is That's why this converges for all values of x But that's a sixth-degree polynomial Six possible roots five changes in direction Not anything we want to graph on a regular basis, but it's not ridiculous Some of these are pictured together on page 596 You'll see a lot of these sequence of partial sums graphed as separate truncating polynomials and you'll see some similarities you'll see that they overlap in a very significant way And as you go further out you'll see that they overlap Even more as you work your way out further because this term is really kind of in a sense almost insignificant But that's the interesting diagram We'll see the same phenomenon when we get to Maclaurin and Taylor polynomials that as you Get certain polynomials You'll see some overlap and then if you get a Taylor series and let it run You'll see some interesting phenomenon there. So we're truncating here But we are gathering an additional term as we go and we'll see that illustrated later in this chapter All right, we won't meet in this class Physically tomorrow, and I'll see you in here on Friday