 We are back Math 241 lecture 55 if I remember right from what I saw out there good. I remember that We have started section 8 of chapter 8. We probably ought to do a little bit more. We're not going to finish it We'll kind of clarify by the end of class today what from this section you'll be responsible for on tomorrow's test We need to talk about content of the test today as well but There was a Nicole. What was that problem that you sent me an email about there's two of them But no just one sorry No, go ahead You should know me by now. This is the second semester. You've had me sorry the first one's x to the n over Okay, go ahead x to the n over n times 8 to the end Yeah And they want you to find the radius and then you go So the summation from One all the way to infinity. Okay, so we want the Interval of Convergence Give me a hint to where is this from so we when we hit that section in review. We'll just hit it a little bit Okay All right, so in 8.5 that was when they first started talking about power series And you were I think kind of first using the ratio test right to determine the interval of Convergence now if this is When we're done we should have an interval of convergence that is centered where? At zero because we don't have an x minus one or x plus two We just have x so it should be centered at zero And then we're going to go out two or three or eight or one-eighth of a unit. I don't know from here So one of the things we were going to review today is the ratio test so this will suffice We want to take a limit and this will be a test item tomorrow is using the ratio test Determining the interval on which it converges Maybe it possibly converges for all values of x maybe it converges only for a single value of x But those are the three choices So in the numerator we ought to put the n plus first term Denominator the nth term Get everything on the same line get rid of a complex fraction All right, what kind of falls out? And what's left about eight to the n over eight to the n plus one what is where? eight there's an extra eight on the bottom so and That is we can bring that out in front if we need to About x to the n plus one in the top and x to the end in the bottom X in the top and what happens to n over n plus one as in a purchase infinity That approaches one All right, so I think we've already done everything we need to as far as taking a limit as in a purchase infinity So we're done with that part actually and what do we have? absolute value of x Over eight is that all that's left And when we've evaluated everything we could take that eight out in front It's not that big a deal, but whatever we get for this in order for this to converge. So under what conditions? When we use the ratio test every time will this result converge? When this is less than one so algebraically we can take care of that in That fashion does that work? Multiply through by eight So there is our interval of convergence so far It is centered at x equals zero the Radius of convergence will be eight. Do we need to do anything else? Okay, we do need to check the end points It's not going to change the fact that it's centered at zero It's not going to change the fact that the radius of convergence is eight But it might change the interval in a sense that we want to include one or both of these end points So let's see what happens at x equals negative eight Now that we didn't address the fact that that started at one. Why did it need to start at one? Right, we would add a zero in the denominator. So that would have been a problem. So starting it at one was necessary What can be done with this because this is going to be you tomorrow about this time Okay Okay, split that up right And then also an eight to the end and the eight to the ends knock out. Is that what I was hearing also What are we left with? What do you think that is? It's the alternating harmonic and we've already dealt with that the harmonic itself is divergent, but the alternating harmonic is Convergent you need to justify that on the test I think if it's in this category, that's a good question and I was going to address that but it's good that you asked I think probably other people were ready to ask the same question Because we've dealt with this particular series probably Four or five times as we've gone through this chapter. I think you can just say it's an alternating harmonic But any other alternating series we should probably go through the two-part test, right? And we'll review that in a few minutes But I think it's okay to say this is alternating harmonic Therefore it is convergent. So we do want to include that particular end point at x equals eight So we get a to the end over itself Which to me looks like just the harmonic Therefore it's diverged. So our final interval of convergence. We do want to include Negative eight We do not want to include eight All the other stuff we gave is the same centered at zero Radius of convergence is eight Is that all right with that one and you said there was another one? number Number four on the web design and it's negative two to the end over the square root of n Negative two the whole thing to the end. Yeah Over the square root of n times x plus three to the end and Then it starts at one Goes to infinity So if we get and this is from the same section. Okay, so this should be another ratio test Any guesses? Okay centered at negative three right the a value is negative three Anything else? Okay, well, let's see. It's not that we okay ratio test again. I mean Let's just set it up see what we think happens negative two Do we want to do anything with that before? Well, let's just negative two to the n plus one and we're going to divide that by the nth term Let's go ahead and multiply by its reciprocal So we've got another x plus three in the numerator Let's go ahead and take we'll do everything all it in one step here. We've got another x plus three So that takes care of those two. How about these two? As n approaches infinity one and We've got an extra negative two only it's not going to be negative because we're in absolute value So an extra two in the numerator and a one in the denominator So we decided centered at negative three. What do you think's going to happen? Divide by two right like last problem We multiplied by eight to take care of the eight in the denominator and then what subtract three And then you could check the end points Does that work? Everybody okay with that? We're jumping off place all right, let's spend a few minutes on the binomial series and Then we'll kind of stop it short of finishing that section. We had a lot Leading up to this yesterday, but this is the place where we were able to end We decided that one plus x to the k using the Taylor or in fact McLaren series because it was centered at zero We decided this was an expanded form one Plus k times x K and k minus one X squared x cubed and so on so it doesn't seem like a big leap When k is a nice positive integer, but when this becomes helpful is when k is not a nice positive integer So the closed version of that And we probably ought to spend a minute or two is that So we start this coefficient out in front of the x to the n and which eventually gets divided by n factorial we started at k and then we kind of take Integers that are not necessarily integers numbers that are different from it Backing up one at a time. So k and then k minus one and k minus two We decided to stop this at two Even though the power and the factorial were both three So that was k minus n Plus one right is where we stopped that This is kind of awkward and we didn't address this yesterday because we got this right at the end of class and it's time to go But when n is zero X to the n that's x to the zero. That's not strange. We see that a lot zero factorial kind of a strange But by definition that's one But this thing kind of becomes strange we know what the first term is it's one And I've seen this strange Situation kind of alleviated a little bit so far. It's x to the zero over zero factorial, which is one But when we say k Times k minus one and we're going to stop this at k minus n plus one Well, what is in in a zero so k minus zero plus one Well, we're starting it at k and we're trying to end it at a number. That's one larger than k That doesn't look right The first term is one Okay, in other words, none of these terms are actually present not that it's zero, but They're not here yet. They don't really start to become part of the problem till we get to the second term and Third term and so on so when n equals one. It's pretty clear. Well, let's just try that one when n is one You got k Let me let me back up. Here's where we start we start at k And where do we end k minus one plus one? What is that? That's k K minus one plus one would be k So we start at k and we end at k. In fact, that's the only one on n equals one so as strange as that is at least it actually there's something there from this I've seen this written before because that's a little odd actually kind of a little awkward I've seen it kind of the foot the first term farmed out. It is one It is one here, but it's just odd to do that One is the first term and then actually start this process Where you start at k when n is one it's k minus one plus one So it ends at k starts at k and ends at k k is the only thing you write down and then the rest of them Kind of fall in line pretty nicely When n is two k minus two plus one is k minus one starts at k ends at k minus one Which is exactly what happens? So either way it works, but this might be a little Kinder to the eyes. I guess it just it's just odd when n is zero this k stuff is odd Strange not odd. All right. Well, let's say we had let's start very simply here Let's say we wanted to write And I'm not going to mess with integer powers Positive integer powers because we had an example of that yesterday and the question was well that seemed kind of stupid to do that And we all agreed that it was Not really a whole lot gained when it's a nice positive integer But when it's not in this case k is equal to one-half is this Original or this rewritten form Is this a polynomial? It is clearly not a polynomial the square root is not a polynomial in order to be a polynomial it has to be powers of x or x minus a to a non-negative integer This is non-negative But it's not an integer. So clearly this is not a polynomial Can we write it in some series that kind of resembles a polynomial? That's what this is all about So one plus x to the k and k is equal to one-half when we expand that. What do we get? What's first term one in general? What's the next term? K times x So let's keep track down here So our first term is one and for this particular series k is a half. So what a half x There's the next term K which is one-half times one-half minus one which is negative a half and Let's just continue down here instead of continuing up here and then rewriting it now that we've got the pattern going What's the next term? Maybe we don't have the pattern going Or it's so easy you don't want to respond Does that work? So the fact that we've got one negative here Means that this whole term is going to be negative now. We're two negatives right this term is back to positive So it looks like it's going to alternate right after we get past this one anyway So what's the end result? How many x squared do we have for this thing that clearly resembles a polynomial to? Describe this thing which is clearly not a polynomial What do we have here? negative 1 8 Is that right and here we've got two negatives, which is a positive it looks like we've got three eighths Divided by six three over 48 which would be what? 116 the next term should be negative right because there'll be three negatives here so we have this power series from this One plus x to the one half or one plus s x to the k So very polynomial looking those are always easier to deal with and things that are not polynomials Questions about that before we go to another example. We'll just do one more now These two are pretty simple. I'm keeping it one plus x exactly like what we developed If it's not one plus x we might have to do a little bit of work to get it in the form of one plus Something sorry, it's not what I want. So one plus x to the negative two So our k value here is negative to clearly again. This is not a polynomial Something to a negative power. It's got to be a non negative integer to be a polynomial It is negative so it falls apart. They're not a polynomial But let's write it as a polynomial type expression What would the next term be? In general, what is it? Kx so in this case, it's next term Next coefficient for x squared negative two One less than that negative three that many x squared Negative two negative three negative four x cubed And so on So what does it look like? negative 2x for the second term Should be back to positive Six over two three x squared. Is that right? And what do we have here? Negative four. Is that right? I don't know kind of looks like we have a little pattern going there If you would predict the next term, what would you predict it to be? Five x to the fourth. I think we could probably verify that too This I mean we are reviewing today This fits in with something we've had earlier that is potentially a test item tomorrow We use the binomial approach to this the One plus x to the k expansion. What's another way of getting one plus x To the negative second and you might want to think about it in terms of that How is that? How is that related? That's the translation of that by the way to that It's the whole thing squared. Okay, let's go to another answer because that's not going to help us It's the derivative, isn't it? If you took the derivative of this Is it equal to this? Yes So do we know something about? That and then we'll take its derivative and see if it matches up with what we just did according to the binomial expansion Isn't this really one over one minus? negative x Which is the a over one minus r format? So the first term is that starting to ring a bell now a little bit if it didn't earlier So if we were to expand this thing, what would it look like the first term would be? one That's the first term This is the ratio. So the next term must be negative x Think we have a little correction that we have to make right Not exactly what we want, but I think we can justify that in terms of an extra term that's there in the derivative We would multiply this by negative x and get x squared Multiply that by negative x and get That that by negative x and so on That's one over one plus x Well What's the derivative of one over one plus x? It's the derivative of one plus x to the negative first Which is there's that extra baggage right negative one times? One plus x to the negative second Which is negative one over one plus x squared now? That's not what we really wanted was it? We wanted positive one up there so what we're going to generate by taking the derivative of this is Actually the negative of what we generated over here. So the derivative of This which is negative one Sorry, that's not what I want the derivative of one over One plus x which is negative one is the negative of What we wanted So let's take the derivative and we know we need to negate everything to make it equal to what we just Found by the binomial expansion So what's the derivative of one zero? What's the derivative of negative x? Negative one what's the derivative of x squared? What's the derivative of negative x cubed and the derivative of x to the fourth and If we negate every one of those terms Aren't we right back to here? I think we got this a quicker and easier way Personally, I mean we just we didn't have to find something and then take its derivative and then negate all the terms We went right to our final answer, but if you'll compare those two Negative one times the one there it is negative one times two x Negative one times that one there it is so it is the same Function therefore the power series Should be the same however we derive it okay that'll suffice with 8.8 We're close to being done, but not quite done. So let's talk about the test which I Don't know probably Anything that is just one plus x So as long as we don't have to mess with that and convert it so that it's in that form an Example would be Let's suppose that we didn't have x there, let's suppose we had a Minus x over three and We had that to the one fourth This is off limits for the test because we really haven't done that yet now could we put it in the form of One plus something to the k. Yes, you can It's one plus Negative x over three To the one fourth so in that expansion Everywhere you see a k you would put in a one fourth in Everywhere this sounds odd, but everywhere you see an x you would replace the x with what is occupying that Position that x normally occupies So x gets replaced with negative x over three K gets replaced with one fourth that we're probably not there yet Okay, we weren't almost there, but not quite there, but anything one plus x To the half to the two thirds to the four fifths to the negative third Those I think would be fair game for this test because we have covered a couple like that So that could be a variety of things, but this will be exactly one plus x That's kind of where we are in section 8.8. All right. Well, let's back up to 8.4 Which is the first topic if you look at our core syllabus it shows that I guess technically 8.3 would be the first item for this test, but We were ahead at that point in time and 8.3 was actually on the last test so start this test starts at 8.4. That's alternating series We have an alternating series test How's that go? two parts You have this you're handed this alternating series. First of all, how do you recognize that it's alternating? Okay negative one to the n or negative one to the n plus one or something like that You see that in the argument to the right of the sigma notation All right, so we know it's an alternating series By the way, if you see in negative one To the two in that's not alternating what's negative one to the two in It's always one right so that's not that I'm going to trick you like that But don't allow yourself to be tricked by that it's not alternating because that's always even But if we know it's alternating What's the first thing we check? Okay, we want the terms to be ultimately decreasing right so the Value in magnitude so we don't care about the alternating part the n plus first term is smaller than It's predecessor so if it helps to do that Throw out the alternating part just the A sub n or b sub n whatever the rest of the argument is we want them to be ultimately decreasing That's going to help the cause if it's going to Converge and what's the other piece? Right the nth term description Way out there to the right Goes to zero so if they don't in magnitude get smaller and smaller and smaller and eventually approach zero Then it doesn't have a chance, but that's a little less restrictive when you think about it then a series That's not alternating if it's not alternating. This is not good enough to determine convergence Nicole Yes, actually a sub it doesn't matter, but I think you're probably correct I got a little either one would work because you're working your way out to the right, but it's the nth term description Thank you. Don't need to make it any more difficult than it is If we stop the alternating series at n equals five How could we determine an upper bound for the error associated with that truncated? alternating series We'd go to the sixth term, so you're never any further away from the actual sum Then the value of the so-called next term So if you stop it in you want to take a look at the n plus first term, so the error You want to look at the magnitude? We don't care if it's positive or negative the magnitude of the First term that was not included in this series and again, that's not the error. That's the upper bound So it's more than the error. So it's kind of the error at its worst Also in section four Absolute convergence was dealt with. I don't know. Let's just since we're reviewing Let's just kind of hit the quick and easy ones and we're considering whether or not this particular Series and it is an alternating series. Is it absolutely convergent? So we're really not only considering this We're also considering what? It's non alternating counterpart So if the question is is this series that we're handed the alternating series is this absolutely convergent It is not absolutely convergent because the alternating version Converges it's an alternating harmonic But the non alternating version diverges So they would both need to converge for it to be absolutely convergent So if the alternating and the non alternating Both converge, then it's absolutely convergent. What can I do to make this? Change the problem. How about if I did that Now it's absolutely convergent, right? This would be a I don't know p series q series lm series one of those p series what we called it and p is what? greater than one Therefore it converges Any test that I kind of specifically held you responsible for integral test Comparison limit comparison. I'm not specifically going to ask any of those now if they come in handy on a problem And you think it's useful In a problem you're certainly welcome to use them because we've covered them But i'm not going to specifically throw in an integral test on this one, right? So it would have to be Something we've had that's a pretty easy Case to battle through like this one p series p large p equals two p larger than one So this converges We could actually do the alternating series test on this so this Series is absolutely convergent But i'm not specifically going to say use the integral test to Validate convergence might be kind of in the background of another problem Also in 8.4 was our first look at the ratio test We've done a lot of those through the course of this chapter. We've done a couple of them today I'm not going to do a specific example problem right now of the ratio test and in fact As we go into 8.5 on power series And we're still doing power series Taylor and mclauren and the binomial. It's all kind of power series, but we're getting there in different ways Power series. We also use the ratio test In fact, those two problems that nicole Brought in today at the start of class are Problems from this section We use that to determine the interval of convergence and the radius Of convergence And remember that when you do that you have to check the end points kind of separately So we do get a generic interval Sometimes and we do need to check the end points Because we don't know what happens when the limit of the ratio test is one the test fails there So we have to check it separately If the l is Less than one we know that means convergence If the limit is greater than one, we know that means divergence When l is one The so-called ratio test fails So we have to do that by hand. So that's why we check the end points separately 8.6 was still more with power series, but its functions as power series I would recommend Kind of everybody coming into the testing situation with a With a toolbox not a literal toolbox, okay But a toolbox of functions that you know and you know what they look like Here's one of the tools in your toolbox Because we might need to use this to help us solve another problem We don't want to have to kind of redefine the wheel here All right, so what is one over one minus x that's the a over one minus r format so the first term would be One plus x Plus x squared plus x cubed right that's helpful to know if we see something else in that form We can kind of equate What's the first term? What's the ratio kind of fill it in here Uh another function that we developed through the course of this chapter That I think will be helpful to know is e to the x And then if we have e to the 2x or e to the negative 5x We don't do what daniel did last night and kind of Reestablish all these ugly first and second third and fourth derivatives. We just plug it into this one Or the sine or the cosine What is e to the x? What's it look like might just Be helpful to get it in this form Rather than the expanded form Now haven't we used that didn't we use that in a couple of web assign Questions also and that was helpful that we didn't really have x to the n We had something else to the n over n factorial instead of being e to the x It was e to the whatever that number was One of them was three fifths The other one was what negative natural log of two I think look back at those homework problems, but that This your knowledge of this becomes handy in those as well Now we didn't get this until um The sine and the cosine we didn't get these till we got to taylor series and mclauren series But they are power series. We're talking about the ones that you should know coming into the test Sine of x is odd How's that go odd powers odd factorials alternating? I don't know if it's all that helpful necessarily to know the closed Form i mean you could get there if you can come up with this you can get x to the Two n plus one that's a guaranteed odd number right if you double something and add one it's guaranteed to be odd And what alternating? That's not that big of a deal. I think you could come up with that In a few seconds, but knowing this is how the sine of x progresses in terms of powers of x. That's helpful Now let's suppose you remember the sign, but you draw a blank on the cosine Okay, how can you use the sign to come up with the power series for cosine? Take the derivative derivative of x is one derivative of this is one over three factorial Times three x squared Once you get it going you'll probably resurrect it anyway So the three is reduced. So that's what x squared Over two factorial now we're rolling. What's the next term? And this is an even function So it kind of makes sense that you're going to have twos and two factorials and fours and four factorials and so on But those are going to be handy for you to know Possibly The inverse tangent I think the inverse tangent is fairly easy to come up with If you don't remember what it is because the inverse tangent is the integral Of one over one plus x squared, right? And I feel confident that in a matter of a few seconds you could all write a power series for that First term is what? One the ratio is what? Negative x squared so you could come up with a power series for that and integrate it But it might also be to your advantage to Have that one committed to memory as well So what is the thing that we're going to integrate? First term is one Is that working for everybody? So when we integrate one we get x when we integrate negative x squared we get Integrate other way x to the three over three not three factorial just three Plus and now we've got the pattern going right So if you want to remember that one and kind of use that in our little library of functions that we have going Our little toolbox then that may be helpful in doing another problem. I think you can come up with that one pretty quickly so Daniel The problem that you were confronted with and he did the long way Let's make sure that we don't do that battle on the test because time is a little more precious on a test Than it is doing a problem outside of class. What was the sign of x to the fourth? Not the whole thing to the fourth just that right? Yeah, okay So what is that written out as a power series? Well, he did the first and second and third fourth derivatives, which I Wouldn't advise that because we have already done this battle For the sign of something The sign of something is that something minus what that's something cubed over Three factorial so Instead of having sign of x we have sign of x to the fourth so everywhere there's an x in here There should now be a what x to the fourth up here So what's it look like? x to the fourth minus x to the fourth cubed over three factorial x to the fourth to the fifth over five factorial And so on So don't blaze that trail again if the trail has already been blazed Yes inverse tangent the same thing it's fine except without the factorial on the bottom Like the same numbers. It's not it's not a factorial Uh, they alternate right? Yeah, so it is the same Thing except without the factorials. That's right. I'm right. Yeah And hopefully that's not going to be a point of confusion If it is then just derive it it doesn't take but 35 seconds to do it anyway Um Power series can be differentiated and integrated We've done that either in expanded form or the closed form Same derivative and integral rules Uh, maclaurin and taylor series we've hit those pretty hard all along T4 if you see a capital T sub four of x we want the fourth taylor Polynomial and how do you find the fourth taylor polynomial? Does it have four terms? Is that where we stop? You stop at n equals four So n equals zero n equals one n equals two n equals three and n equals four it could potentially have five terms Some of them from time to time drop out So every other one is dropping out t four could possibly only have Three terms or whichever terms happen to drop out Uh, um, that ought to do it. I think everything else we've dealt with Pardon that was 8.7 8.7 that was what you just talked about. Well, we've hit all the kind of the common taylor polynomials maclaurin series The things that I would hold you responsible for we've already dealt with Yesterday a lot today some so we don't really need to review them again All right, test will be tomorrow