Power Series and Convergence Tests

Alright, let's get on board. So, any questions? Like, can we review a few more examples of like convergence divergence? Absolutely, absolutely. But is there something that you want to point to? For example, which type you think you are the weakest the telescoping type, or the alternating, or comparison with the power function, geometric sequences, or what kind of sequences, or you just want them to be.Like the example you gave last time. Yeah. Oh, here. Okay, okay. Now, do you want to okay understand it now? Well, if that's the case, I am going to give you this now. Why don't we do natural log of the one plus x, and I want to use it to to estimate the natural log of a.One point zero four, no one point one, okay, and I want to see when you do your power expansion, you know how to do that, and eventually we're evaluating at one point one. I want you to decide how many terms to include so that we could have three significant figures, and you have to prove it. This is this has to do with the error estimate. We went over what would be the a.A particular upper bound for the error. Once you decide to truncate this infinite power expansion at a certain point, of course, I'm not saying it's alternating series, but if you do it, you're gonna find out it actually is. So, for how many terms? Yeah, to keep so that we could have ah to achieve three significant digits.We know how to count significant digits. This number is more enough. So let's suppose it's going to come out to zero and zero eight or four something, and then to have three significant digit here, you've got to make sure we round up to that digit here and throw away this one. Meaning your error cannot be bigger than zero point zero zero zero zero five. So this basically identifying the significant here must be combined with identification of the bulk figure.In order to decide exactly what's the error bound, right? Are you with me? Okay, so go ahead. You can you can tell me what you're doing to keep me posted. But I'll give you a couple of minutes. And I would also want to use it. Oh, okay, I'm stop. I'll stop talking.中文嗯。``` 中文嗯。```中文嗯。中文嗯。`````Ànd give me a signal once you want me to start talking.``` ```嗯，I don't know where to stop. Okay, well first we actually want to. Oh really? Huh, we want to get a power expansion. We're dealing with a power expansion.And the chapter to talk about the convergence. Oh my bad, I really should have a jump in earlier. If you don't know what the story is about yet, okay, we've been talking about power expansions. Now this one, at least we know in theory how we can do the power expansion. We can take the derivative and then take the second derivative. We know we're going to do the expansion not at zero because natural log of the and the x here is undefined at zero, but rather at one, which is fine.Except for, it's pretty tedious if we keep on taking the derivative. It's a lot easier if we just take one derivative. And then immediately we're getting this, and we're in our comfortable territory. We know how to do the power expansion of this, and that's a very typical geometric sequence. So can we do the power expansion for this one? Well, first of all, you need to understand the conceptual, eh? And this is really we think of it as as evaluating what is natural log of the one plus a.Delta X, and in particular, our Delta X is decently small. It's a point one. It is not infinitesimal, but if we can control the error range here, we can do it. Considering this as a power expansion. But just so that I don't need to rewrite this Delta X every time, I'll just call that Delta X the X now. So this is a function we're looking at. And can we get the power expansion? It's one over one plus X, please.```呃， one minus one over x。Plus one over x squared. Wait, wait, wait, wait, wait, wait, one over x. That's not the power expansion. In the power expansion, it's a polynomial, all the powers of x are positive. Hey, guys, you know you need to write it down in a conspicuous place by the end of your text or your notebook, and every now and then go go back and review it. There are two ways to do this expansion. When it comes down to words, you could at least look at this. One plus x to the ninety-first power, right? And let's do the power expansion accordingly. I want to announce again: whenever you're writing a power expansion, the powers must be positive integers, including the zero. Of course, we start from the constant term. So if you do it that way, what are we getting? I thought we've been over this enough. I'm not blaming anybody, but you really do need because again we're cramming, so there are too many formulas.Find your way to put them in a in a good place so that you could quickly thumb through the most important formulas. They're not in the book. So if you do this now, what you're getting is one plus the negative one choose one x, and plus the negative one choose two x squared, that, that, that, right? And I hope you still remember how do we do the combinatorics with the negative number.```Would it be, well, one minus x plus x squared over two.Minus x cubed every three factorial. 哎，等一下，等一下。Second， when you actually look at the ninety one choose one, yes, it is ninety one, right. But when you do the ninety one choose two, don't you have ninety one times ninety two divided by the two factorial? Right. Right. And after cancellation, what do we have? One. Yes, it's positive one now.Hey, does it come back to you? And then you do the ninety-one, two, three, and that's excused. Well, after you recover this way of thinking, we're gonna do it another way. I really need you to own this the sum of the geometric series.中文```- x cubed plus x The fourth. Uh huh. Now do we have an alternating series? Yeah, because the power. Yeah, and it doesn't converge very fast because there's no coefficient dividing by factorial that's helping it. However, though at that point here, I really want to give give you another perspective. I don't want to go ever forget this again. So actually, this is one of the geometric series, and what is a common ratio? The common ratio is negative x. In general, if we're doing it with one plus r plus r squared plus.The R cube, and we want to find out: at infinity, what do they add up to? That's a power expansion. And I think we've gone through this many times now. You notice a self-repeating pattern. So if you set it as a sum s now, and then as infinity because it goes on all the way to infinity, and you can see when you do rs, then that's almost itself. And so between these two, you'll be able to find out: at infinity, this geometric sequence really add up to one over the one.A reciprocal function is a sum of the geometric series. And of course, here we just need to identify the r here as a negative x. That's why when you plug it in, you start you end up with an alternating series. And dear kids, can you make sure that you write it down in a conspicuous place, not get mired under your notebook, but rather just write it at the back of your notebook where you have all your.I use the geometric thing to find that series. You did, huh? Then why did you not tell me? So you got this place. Oh, while you guys were doing the binomial expansion, I found the rest of the terms using the geometric series. Oh yeah, very good. Okay, good. So once we know this now, and we know this is really the expansion of this derivative of natural log one plus x, which is closed now. Remember this is.The natural log of the one plus x derivative's power expansion. What we need is the power expansion for the natural log of the one plus x. So let's get that. I hope you find this familiar. Natural log of one plus x minus x.Squared over two factorial plus. X cubed over three, da da da. And if you want to write a general formula, of course, it is equal to a. Yeah, that's a constant. Very good, Elaine is saying. Let's not hurry when we do the anti-derivative. Indeed, there's a constant there. However, what is natural log of one?It's zero. Yes, exactly. So I'm glad the conceptual that you remember to put it in there, which is indeed a very important conceptual step. But it turned out to be zero. And then add on to the negative one and the k minus one power and the x of the to the k power, and this is over the k. See if you agree with me.``` I got the same thing. Good, and Johnson. You. Yes. Okay, now do we know where to go next? We want to apply this power expansion to work out the individual function value.To go for an estimate, but in fact, it's a it's infinite expansion, and but we don't have an infinite lifetime to calculate. We have to choose to stop somewhere, and where do we choose to stop to guarantee that what is the error left out there? It's actually confined under three significant figures. So for that purpose here, you probably have to plug in a few terms and see to get a feeling what would be the final number because without which we don't even.But know what is the error, how big is the error. We just know three significant figure, but where that third significant figure lands here, which that's more point. That actually depends on the value of the function.``Ì believe up to the fifth term would be okay. How wide that? Tell me, sir. Since zero point one to the fifth power goes to five significant digits, and then divided by five is another. Ah ha! So you have decided whatever it is, and it lands on zero point da da da da da on this.Meaning, you have decided the three significant figure, and the the real significant figure starts to that small. But how did you know that? Well, you can't believe what I said earlier. I just said that very irresponsibly. As an example, I said, "If it were something, that's how you count the significant figures." So, did you work out the real figure yourself? You have to plug it in next year to actually calculate the future.Right.``` ````````Ìt is up to the fourth term because the fifth term has five zeros and is off the off the four significant digits needed. Hmm. I agree. In fact, earlier I.I didn't, the estimate not try to fool you, but I actually thought that's the case. Let's plug in a few numbers. The x here is zero point one, correct. So what we did is like point one, and then minus the point zero, zero five, and then you add on to that. They're just getting smaller and smaller. So approximately, we're looking something like a zero point zero eight or nine or something. At zero point nine eight three nine.Three, huh? Okay, okay, beautiful. But no matter what, we are sure though, and meaning the significant figure that we actually are going to gauge, we need the term to be falling under zero point zero zero zero five, or you can say five times ten to the negative fifth. And Johnson gave me exactly the correct answer. He identified for this alternating series, the error range is controlled by the very next term we're keeping out. It is the very term that we're keeping out.So he's saying, we only need to have an x to the fifth power because x itself is ten to the like like first. And however, though, it's the last term. Okay, can I actually just stop here? You you think we need to keep one more term, and let me actually write all the terms carefully. So, and the next term would be minus x to the fourth over the four. Ah, sorry, not four factorial, just four.And now we're adding on to the x to the fifth over the five, etc. etc. And Jonathan remember the correctly by these very special features of alternating series, and it easily checks out. The term is definitely monotonically decreasing and approaching zero. Remember it must be monotonic, and then basically if we cut it off, then the error is simply confined under the next term now. And you can plug it in, definitely checks out. It's a lot smaller than. Yes, my question would be: Can I stop here? If I stop just at the third term here, and then what is the error range? It would be ten to the negative fourth divided by a four, right? I think you can. Yeah, yeah, indeed we can. Because when you divide it by the four, exactly that helped a little. So in fact, this is already equal to two point five times.So that's ten to the ninth fifths. So that's already okay. I think Jonathan, you're overly diligent, and in fact, it's not so bad. Only three terms will do. But could somebody repeat for me? That's important. In order for the theorem about alternating series to to apply now, and what conditions are necessary? Did you guys clearly check the check those conditions? Wait, can you repeat what you just said? Well, in order for.For the theorem or the error range for the alternating series to hold, what conditions must we check about this current power expansion whether it qualifies for alternating series?```中文嗯。中文嗯。Are you checking your notes? I recommend that you think about what we did to prove the convergence and to find the error range. We paired the terms, we paired them in this this way, and then we paired them the other way, right? And we have to actually use a condition. Oh, if I do pair them, I have to pull out the negative, and inside will be negative. We have to use a condition that.Inside the circle within each parenthesis, the signs stay the same, which actually requires whatever you include in the each parenthesis here in terms of the magnitude. The first term must be bigger than the second term, otherwise, then there is no tell which sign negative or positive each parenthesis would adopt, right? So we need the function whatever this power to be monotonically decreasing approaching zero, because if eventually. This term that we're using, ICR range outside doesn't even approach zero, then there's no hope for convergence either. So there are two conditions. Number one, now basically we're doing the negative one to the k power ak. This ak could depend on x or may not depend on x, but no matter what, and without dependence on x, you just have an alternating number series. With a dependence on x, it's alternating functional series. But either way, for that to apply to for our rule for.A determination of the convergence for the alternating series to apply, and also for the error range to apply. We need to know that the magnitude of ak right now each ak is positive greater than equal to zero because we are putting out to the alternating part as explicitly the negative one raised to the k power. So we need that to be monotonically decreasing to zero. I didn't write down the words, but you have to write down monotonically decreasing to zero. This is just to secure when you group them in.To parentheses now, we have to be able to tell what is the sign of each parenthesis.````````````So each of these groups should also decrease down to zero. Yes. Alright. We're good now. Okay, so that's the alternating series. We also finish this now and cover. Do you want more examples, trickier ones? Yes. Okay, well, let's look at more. And what if I wanted to prove the convergence or divergence of this cosine cosine x, now, and we know how it looks like, right? It is actually one minus the theta squared over the two factorial plus the theta fourth over the four factorial.No, let's actually do this one. I actually want to have a cosine of the x plus half of the pi, and I want to find what's the power expansion of this guy. Because in fact, I want to do an estimate of what is going to be. We do know at the half of the pi now that's at the at the cosine half of the pi equal to zero, so we're going to do a power expansion in the.sinetian versus zero, so the job is to find out what is the cosine of the. Although you could actually do it using translating it into sine, but that that defeats the purpose. Um, let me see how I can. Oh, here, pi over four. Okay, that's good. Yes, yes, yes. This is good. That's.Exactly what we want, and I want you to estimate what is a cosine, for forty-seven degrees. I know this year I'm not trying to trick you into it because currently the major point is not, I tricking you to understand that whenever you're doing calculus here, we have to stick to radian measurement. We said that before, if you're using the degree now, even the derivative formula sine theta derivative gives you the cosine, that's not going to work because everything hinges on one simple fact that sine theta needs to be.Or over the theta needs to be approaching one, and this is proven graphically through the unit circle by comparing the arc length with the angle itself or the chord length. But for the arc length to equal to theta on the unit circle, the theta must be in radians. So keep that in mind. Not to mention, because we're defining the cosine sine as really the real and imaginary parts of the power expansion. And if you want to put the theta on the exponent, it must be a number. It cannot have any units.So remember, whenever you're doing calculus now, you might as well always stick to radians. At least whenever you're doing some plug-in or do any calculations, and I want to have an estimate, an proven meaning. You have to prove whatever the truncation will satisfy the error range now for cosine forty-seven degrees up to three significant. So you think about how to do that expansion. Of course, usually you just find a good.Point out the major components of this sentence. You can take the derivatives. In this case, it's not so hard, but I recommend we go for the sum of angle formula. Because if we do so, then immediately you see this really becomes the cosine of the x, and then the cosine of the pi over four, the minus sine x, sine pi over four. Correct. Okay, and then we do know the power expansion. The pi cosine sine power four both equal to a two. Of a root two, ah, sorry, ah, half of the root two, and the inside what you're getting is really cosine minus sine, and we know the power expansion of both. However, when you really bring them together, I'll let you write it and see what comes out. Hopefully, you're gonna see we don't end up the alternating series.中文嗯。``````嗯，嗯。Keep me posted. What you guys are getting for the cosine.It's one minus pi squared over two factorial ninety squared. Oh, don't plug in the number too early. You want to keep the x first. Okay. Otherwise, it just looks not marriage. And once we decide, according whether it's alternating series, is it monotonic, which term do we need to stop by, and then you could actually plug in the number and then actually calculate it. By the way.You really want to keep the procedure on AP because if you if you keep this generic term now, they're gonna award you one point for the correct formula. But the minute you start plugging in numbers, if you miscalculate the tiny little detail, you don't get any points for all the subsequent steps. So do it in letter form, and only when you get to the very last step, and then you can plug in the numbers. ```中文嗯。Yeah, yeah, I got the equation on the screen. Yes, you agree, huh? Yeah. Me too. Alright, now it is an alternating series, and could we apply the error term.Well, what in essence, I think you have the feeling it really should work, but nevertheless, if you really rigidly what we define as alternating series and you want to apply grouping them, it doesn't work because if you group them like this now and they don't keep the same sign. And sometimes the two terms inside each one, oh, oh, right.But if you actually group them in this manner, the two terms that you keep in each one, they are indeed always opposite of each other, but outside they're alternating. So you can't tell whether you're basically there's no inequality very easily available. So it doesn't quite fit, except for we do know that whatever the term is monotonically decreasing and surely approaching zero. So use your head. How do we finish this? Don't abandon the idea of alternating series.You want to tweak it. And you group four. Ah, well, I think you're grouping four terms. Yeah, I think you're. Yes, yes, very good. Meaning, you're gonna actually look at two nega and the two paws. Basically, you're always bundling them. You're always choosing a pair of a paws as if it's a one term. You never separate them. That's one term. That's one term. That's one term. And you would end up ah, it is alternating that. And when you're proving your theorem, you're really grouping the two terms on the outside.Now that's a legit idea. It works, but how do we apply that to actually get the error range and to decide we got the three significant figures?``` What I'm thinking is approximating x to be one third.And to begin with, to begin with, what would be what would be your error term that you're using to control the three significant figures. What I mentioned earlier, you correctly identified the error term here. It's the very term on the outside, right? Before after you truncated it, the next term is the error, and that's secured by the theorem about the.ErrorHuman:Well, to begin with, don't we also need to find the box figure so that we can speak with accuracy, which is the three significant figure, the next term we're throwing away. So, what is the box figure?Uh uh, could you repeat that, sir? I'm actually saying, and don't we need to know the bulk figure to judge where does the three sig figure? I mean the error term length. We need to set a standard to compare the term with. How small it's small enough, right? Right. And what is the desired error term?Ten to the negative fourth power. And you mean the five times the ten to the negative fourth power? Because it's gonna be whatever does it. What do you what do you mean? It's that, and how do we know that? You're right.But I would rather they do tell me the logic instead of just the answer. So fundamentally, you're saying whatever the number must, non-zero number must start from here, right? The answer must be zero point something. You are right. I just said the the number is close enough to cosine a quarter of the pi. Cosine a quarter of pi just equal to half a root two, so that's a one point four one four. What you're getting is close to zero point seven.七点零七，a little bigger, no, a little smaller because we're doing the cosine. So we know for sure the bulk figure approximately will be zero point seven something something, and we need to control the error to that point. So this is the error range we need to control by, and we're going to pick a term so that when it when you add them up together, that will be satisfied. So what do we need to pick? Don't overwork meaning. You want to pick the lowest one.I think we can cut it off here.Yes, yeah, correct. So let's plug in the number calculator. What is error term and amounts to?``````So what do we have? Zero point.zero zero zero zero seven. Yes, that's pretty much correct. So, I got that too. Yeah, I got that too. Very good. So now you see the denominator. The factorial's help. The moral lesson in this problem is that sometimes you're going to encounter something. Those are not on the surface very regularly or alternating series, but it's very close, and you can tweak it a little to make it work. And you did, you guys.Did that wonderfully. Do we often encounter this kind of alternating series? Yes, we often. And I still have a few minutes. I actually want to introduce a very important topic, and we're going to continue with the topic next time. But I want to say this: alternating series probably is the gentlest requirement for convergence among all the series. You know, it doesn't really take much for a series to converge. At the very least, each term must be approaching zero. And that's one of the two conditions. On top of that, there's only one meaning it must be monotonically decreasing. But if we're doing power expansion, very often that's already satisfied because, after all, we plug in a number that's decently small, and as you're raising it to higher powers, then necessarily, we're not necessarily. But very often, we're going to directly encounter a monotonically decreasing series. But I want you to pay attention to one thing: in the future, you're going to see a power expansion.Power expansion. We don't know what it's gonna converge for this one. The convergence factor is pretty powerful because of the factorial on the denominator. But according to the one plus the x, this power expansion we did before, and I want to isolate it so that you can pay attention to that. This is a power expansion we just worked out. Whether the series converges at all, albeit a power series, and there's still a requirement. For example, if I plug in the x equal to a four, would.It converges. It is still alternating series, but it is monotonically decreasing and does that approach zero? No, obviously not. In fact, it's highly divergent. Each term will be blowing up to infinity. So can we find out what is the maximum value of the x now, in order for this natural log of the one plus x to be convergent? I'm going to get rid of, whatever this on the side, just so that you can concentrate on this one single power.Series, and we want to find out what are the conditions on x for convergence.```中文```中文嗯。中文嗯。 I believe it should be less than or equal to one. Wonderful. What did you do? The groups must be decreasing. And how did you formally prove that that requires magnitude x to be smaller equal to one.Particularly, yeah, smaller one. I'm rather pleased. I think you're inventing a method I haven't even taught you. Did you just make your observation? Ah yes, since. Yeah. By inspection, if x is greater than one, then it will diverge. And I mean the function. In fact, the terms would increase when the x is big enough. That's correct. Let me show you.Formalize what to do that. But before that, Elaine, do you agree? Yeah, yeah, I agree too. Now, in order to prove the convergence or divergence, what we need to do is to take a ratio. By the way, this is called the ratio test. This is the first test, and we're going to learn altogether four tests. In fact, we learned earlier, we called that alternating series test. We cast it as two conditions: monotonic decreasing to zero. That's actually called alternating test, alternating series.And this one, it's not limited to the alternating series. In fact, it's actually for any single power series here. You take the ratio defined as the an x to be an plus one x divided by the an x, which is just a ratio between the next term and the previous term. It gives you the sense whether the term is increasing or decreasing. And we need that ratio to be strictly smaller than one.So that we know this is going to be monotonically decreasing, otherwise, if it's not even decreasing, it can't go to zero. It has got to blow up to infinity. However, though, in that it's proving more than strictly greater than one, and if it's strictly greater than one, now I wanted to come up with a counter example. This is homework. Come up with a counter example so that you find me a power series. Arbitrary, you don't have to.And make it equal to a function. It can be arbitrary power series, and I want this to be satisfied, meaning this function looks like it is truly decreasing. But nevertheless, you have to prove to me somehow it's not convergent. It's divergent. Actually, there will be. I mean, these counterexamples do exist. They're also very revealing. That's why I want you to find them. That actually means this is not good enough. We have to add in one more condition, and we will.But I want you to have a chance, but just so that in the future you don't misremember that tiny little nuance of the theorem. I find it it's most helpful if I force the kids to really do their own counter examples. Are you clear what I want you to do? I want you to construct a power series not necessarily alternating that satisfies this condition and yet it's not convergent. It's known to be divergent.We're good. Oh, you could even be sure that this an is approaching zero. We can make that add that in, and still you'll be able to find a counterexample where it is still divergent. So let's bring that in. I want to make it the strongest. We're good. Alright, sorry I didn't notice time. Keep up your good work, and look forward to seeing you. Bye. 对，哎，客气啊。

Power Series and Convergence Tests

Power Series and Convergence Tests

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