泰勒近似与误差估计

We are with a closure, and in a month also, we're going to start systematic problem, yeah, systematic problem solving training, and there are four very tiny topics, and I plan on cover each in one session. Hmm, no, in one week. I mean, one session, ah, theoretical background, one session problem solving, and these are very isolated little techniques. So we are done with the entire conceptual picture. I'm.Sort of, just chattering a little to wait for the other kids. Where is everybody? Oh, it is also about time. Then you should find your own AP prep book. I recommend that we go ahead because it's free online, and the older version doesn't really matter at all. I would prefer the older versions. I think that I should.Less dumb the down, but in any case, and you know pretty much the major conceptual, the overall picture now, and you can start practicing doing the free response problems. On college board, they give you a whole archive of the past year, and each year there are six problems. And I recommend doing them in problem types. For example, go to whatever you you probably remember the best. I'm just bash it out.And then later, we're gonna as you go deeper into review. Now you might find the material less and less readily available in your head. You have forgotten it, and then maybe you should actually do some review first and then continue with exercises. You know, I found like a few resources online, but like I didn't, I haven't found some of the ones you recommended. So I'm gonna check them out after the lesson's over. Yeah, that would be great. Hold on, so today, oh, I've.We will mention. But what about ID? Let's go ahead. So today, one of the four it's actually of the four topics I'm I'm talking about. It's called Euler's method doing the linear approximation, and this is particularly the AB material. Here is my pen. One second.Now, let's look at the question at hand. Now, in, for example, we have done plenty of this before. I want to get given the f of x that equals to um tangent x, and and then you take the natural log. No.I'll do tangent x one third power. Now I do one third power. You can write it in this manner now: take the cube root of the tangent and put it on the denominator. Come on, what? Hmm. And we want to well, apparently, at five pi over four.No doubt about it. Actually, can you find out what's the value at pi over four? Yes, yes, panel le. Eddie, good to see you. Now, I was actually talk about the structure of the following month material. We covered all the major conceptual.Topics now, and the remaining part is to do comprehensive problem solving. However, there are four specific application techniques that I want to cover in great detail, and each one takes a week. So fundamentally, these are small topics, but I do need your concentration so that I don't have to repeat myself. Well, today it's the first. It's called linear approximation Euler method. And usually the problem is posed in such a manner: we actually know the function value.At a specific value, it's a nice number, and we want to evaluate a number nearby. So we're actually looking at the function I'm giving you. It's nasty enough. It's one over the cube root of the tangent x, and but there is a good value. And then I'm waiting. What is the f of pi over four? Whenever you're dealing with trig, now I would say, run to the unit circle, put the angle on that.And be sure to tell me what you're what you're thinking. I I don't want to.It's a lot of time. I don't mean your mental processes are waste of time, but I want you to speed that up by talking to me. Because my goal is to finish one topic per week. I just put it on the unit circle, like a pi over four.``Ìs it one? Yes, it's very good. Ah, because the tangent is defined as the sine of a cosine, and we do that on the unit circle. Here's ah, sorry. Here's pi over four. That's at forty-five degrees. It's an angle by sector, and therefore the slope—that's the rise over run—that's also the meaning of the tangent. Now would be one, and in this case, when you plug it into our equation of.The final answer is one. So this is your. Come on. Hey, come on. Hey. Let me get our exes. I think it's a little too high. It's not perfect, but it will do. And we're dealing without line. What shoot? What are we talking about? My pen today. It's a little bad. What? Okay. Right, and that's point two forty-one. This is angle pi over four.And we do know, ten times ten times four equals one, therefore, the entire function value equals one. And we want to evaluate the number nearby. Let's actually find f one. How would we do it? Well, the way that we opened entire and the course of calculus now was from ah how to deal with the different orders of infinitesimal. We started by learning how to find the function value nearby. Now we know how to differentiate it. So if.If I want to do it in one step and only keep the lowest order infinitesimal, that's called linear approximation. Obviously, so how do we do it? You can find two points and make a line. Ah, you mean graph the function and find two points on the function and make a line. Yeah, like two points on the function. Like a tangent line, I guess. Ah, in that way.Use in the second line, right? What's like, which one is second? Oh, no, no, no, we're not talking about the second. I define as one of the cosine. The second line simply means you're connecting the two points. Here, here is a graph of the function. When you're connecting the two points, let's go the second line as opposed to the tangent line. You mean the tangent line, and that doesn't involve connecting.Two points. No matter what, we don't even know what is the shape of the graph. I do think you're thinking conceptually correctly, but you know, I since we covered all the conceptual point now, maybe today I'll be a bit of a, paging over. I'll just, well, talk you through and what is algorithm. What do we mean by the Euler method for linear approximation? What we do is that we find out well, you're given the function value and you want to find a nearby value. Step number one to define your DVA.The ocean, the delta x, yeah, that is equal to the one minus the pi over four. And if you plug in the three point one four or something, you're gonna see this number—it's about it's approximately zero point two something-ish—but it's not a huge number. But later we're gonna actually improve it by stepwise order method now. And well, but this is rudimentary one—we take only one step. And then we want to find the f one by looking at it. We go from the base value f of a.A quarter of the pie, and then plus the delta f. You can call that a df, but it's not infinitesimal. It's not infinitely small. It's just some small number, but basically, it's the change in the y value. And by definition, the change of the y value would be approximately because we do know that the df really just equal to the derivative at the value of the pi over four times the dx, right? But this is only true infinitesimally. If we actually go from.To a decent small but not even a small number, it's only an approximation. So the delta f, it's approximately. So I can't write equal sign anymore, but rather it's a linear approximation equal to the derivative at pi over four, and then multiplied by the delta x. By the way, and if you don't recognize it is really the power expansion, then then you haven't learned everything very well. We recognize it's actually the power expansion. We do know and the. The x term is actually the second derivative pi over the four over the two factorial, and the delta x squared, and plus the f triple derivative pi over four over the three factorial of the delta x cubed, and da da da, i'd infinitum. That's the full power expansion, correct? Yeah. Okay, beautiful. And now the so-called linear approximation. This is a must, meaning it's a perennial favorite on the talk.Here's test, yeah. So we do need to master. That's why, for this very reason, I'm saving the four frequently appearing topics here. They're not conceptually important. Well, everything is important. They're not conceptually that important, but they're actually just special techniques. They're called certain names and they're particular problem solving mechanisms that I want to cover at the very last after we do everything conceptually. But I want you to remember this very well. Keep a very clear head.So that's the whole thing. This is without any approximation. As a linear approximation, we only keep those two terms. That's it. Go ahead and calculate it, and we'll see next time how we improve it, make it more accurate. Because right now that delta x is really not very, very smart. Alright, go ahead, pencil write, and really tell me what numbers you are plugging in and what number you are getting, and how do we materialize? How do we actually?Find out that derivative.Eddie, are you doing your pencil work?Yes. Good, and by the way, if you need to, if you need the name, it's in your book. It's called linear approximation. Euler method.Would you mind sharing the numbers you're getting with me? I'm still calculating it. Okay.``` `````ànyoneI got one minus two, uh, minus two over three. Um, pi over, wait. Ah, first of all, let's elaborate on the derivative. I did you plug in a number for the derivative. What did what is the derivative of f of x? It's unavoidable. We have the derivative. Ah, I use.So I got negative one third ten x to the power of negative four thirds, times secant squared x. Brilliant. Yeah, I got the same thing. Excellent. And now we're plugging in a very nice value, which is a quarter of the pi, which means whatever that fancy power it just one, and what is the secant x and secant squared.I need chatting from the numbers. You got it right. That's just a two, right? Because the second of a forty-five degrees itself, it's root two. It's reciprocal of the cosine, which is one over root two. So this is a two now. So very good. The derivative at x equal to a quarter of the pi would simply equal to negative two thirds. But did I? Oh, that's why you are minimizing. So when we plug in, we assemble everything into this linear approximation. Simply going to be one and the minus.That is two thirds. And now we multiply by the delta x, which is one minus pi over four. If you do not want to resort to decimal right away, that's it. Remaining part is to plug in the pi with maybe three significant digits and then just resolve it. That is what it is. Wait, what are we using for delta x again? The delta x, it's actually simply the new function value minus the old one because we're trying to find f one. So the delta x is the.Defined as actually the one minus the power four. Correct. Doesn't make sense. Generically, what we have it's actually this f of if the x zero is a good point. Okay, we actually know that equal to the y. Zero, and we want to find a number nearby, which is x zero plus a delta x. What you're getting is simply going to be the y zero, and then plus the derivative at x zero times a delta x. And this is the formula capturing the Euler's method. But we do know this is only just the first two terms in our Maclaurin expansion. We know how to get it more accurate by including the higher derivatives. And that's it.Everything crystal, yeah. Okay, look at its number. Is it the underestimate or overestimate? You wouldn't know. Don't try to graph that function. You can, but honestly, we can cavitate. It's not so obvious. Well, you have to know from the overall structure. This is only approximation. When I push you to find it's over or underestimate, we have to actually consider the terms we throw out, right.Yeah. That is what we're doing.``````Please do keep me posted what argumental process is. Otherwise, this one is not a closed questions, a little open, and it's definitely not limited by the AP rubrics.To do with your understanding of the power expansion, and what are the terms beyond? Trying to find a pattern of derivatives to see the, you know, what the rest of the terms would do. Yeah, very good. And what do you mean by that specifically? The rest of the terms. Ah, whether what we're.Getting would be negative or not, like when we plug in pi over four to the rest of the terms. Right, but the rest of the terms are referring to. You mean the second order infinitesimal. Yeah, the expansion. That's right. Very good. But do you try it for all of the rest of the terms, or you try for the next lowest power alone. I try for the next lowest power. Alone, beautiful. So you're just telling me in a nutshell. You're trying to find out whether the second derivative is positive or negative, aren't you? Yeah, that's right on target. That's very good. By the way, intuitively, the second derivative, which is indeed the next term we're throwing out, right, would be plus the second derivative at pi over four divided by the two factorial delta x squared. It's exactly because of the delta x squared. Now we don't even care whether the delta x is positive or negative. Sometimes we do if it's odd power, but.It's a square anyway. It's already guaranteed to be positive. So the only term that concerns us would be the second derivative. But obviously, because second derivative greater than zero, meaning the function's concave up. And this is where Elaine, your earlier statement about the tangent line is very useful. That means if we're doing a linear approximation at the arbitrary point on the concave up curve now. And but notice here that concave up curve could be even decreasing. Okay, and in order to move from a.Specific point to a nearby point. You did not actually go through the segment, which is what I call the second line. That's a definition of a second line, meaning you're just drawing a line between the two points. But no, we don't have that kind of information. We only have the tangent line because we're using the first derivative. So you can see, the first derivative is if the curve is can't give up, then the curve always stays above the tangent line. Therefore, your Euler method linear approximation.It's always underestimated. But mapped onto what you learn inside algebraically. I mean, on the paper, what you have thrown out is a very nice term, which is a positive term. So obviously, it's underestimated. They're consistent, and even if the function decreasing, regardless the number you're evaluating, you're approximating it's on the left or on the right. You know, I could be asking for what is function value here on the right, on the left, as opposed to here on the right.Or I can be a binary function, which is actually on the descent. But if you draw a tangent line everywhere, the point on the tangent line is always going to be under the curve, so surely underestimate. You might as well put it down. For a concave up curve, correspondingly, the second derivative is positive, and the linear approximation is always under up underestimate. And vice versa, if it is a concave down curve, meaning if the second derivative.So, depending on the second derivative, and that's the concavity. And this is actually under or over estimate, and the correspondence would be positive, and that's concave up. So this concave up, and it's under estimate, because the curve stays above the tangent line, but it is negative. So that's concave down, and I'm drawing this a little symbol here in order for you to immediately visualize.It's bowl shape, and then it is obvious where the how the tangent line would compare with the curve. So this is an overestimate. Okay, makes sense. Ah, okay. Well, then, what if.Next time I'm a little dissatisfied, but instead of including higher power terms here, I can't prove it to you here because you don't have enough of the background knowledge yet. Even if you include higher powers in the microphone expansion, it could be volatile. In fact, meaning we don't know what are the higher powers, especially if you're dealing with a tangent, you can see these powers are getting bigger and bigger. They're not necessarily convergent, unlike polynomials. However, though there is. There's a trick we could play in order to improve our accuracy, and I want you to keep that number over here. And eventually, we're gonna actually compare with what we can get from the denominator. What we can do is actually the following: that power four is indeed very close, and it's not a nice number. Okay, so we're gonna actually approximate f two, which actually equals to. Uh, debate. We're still dealing. Oh, you know, this is horrible because all the numbers along the way are not.Meaning you can manage it without the calculator. Okay, get your calculator ready. Well, that's a contradiction though because we can do these calculate. Okay, let me actually make it more plausible. Now, why don't we combine it as a slope field with a slope field? Okay, I'm giving you a slope field now, which is why the derivative equal to, let's say, five of the x squared over the y.Cube. And multiply by one minus y. We don't want to solve solve this differential equation. Pretty complicated. It's solvable, but we don't want to bother. We want to do numerical. And remember, if you're just given a differential equation, you won't be able to solve the curve. What we're given is a slope.This field, this little field, it's even regarding x, but it's not very symmetrical in any way, neither even nor odd regarding y, and it's a pretty weird curve. So we still need a initial condition. So let's see, I would say the curve go through. I don't want to encounter any singularity here. So let's say it go through the point of, you know, for convenience, I'm going to change that into a two minus y.And they go through the point one, one. Right. What's what's this? Is that a five? That's five. That is a five. That is a five. Clear now. Yeah. Alright, and we actually want to find the what is going to be the y two. You know that.The difference between a given point, which is meaning our x is equal to two here. Earlier, we know one point, it just means one x equal one, the y equal one, right? We want to find that x equal two, and yet any condition that is not small. I mean, one is or so, a neutral number is not big, it's not small. And I'm not saying that randomly, okay, listening to the the math opinion. No, simply because how small or big depends on the power expansion. When you do the delta.Exit squared and the cube, and da da da da da. If you plug in one, that term is not even shrinking. So obviously, there's no reason for us to believe that it's small at all, right? So because the delta x in this case is one, and then it doesn't justify as a proposition to actually throw away any term at all because it's not infinitesimal. Do you guys follow me?Remember when you do your power expansion, you're looking at the derivative at a certain a now divided by the n factorial, and then this is n factorial n's derivative multiplied by the delta x to the n power, correct? And we go from zero all the way to infinity. It is true one of the conversion factor is n factorial, but you do know though when you take the derivative, sometimes you're going to generate because the power would be lowered, right. So there might be actually something on the numerator to cancel out the factorial on the denominator. Now, how do we depend on convergence? Meaning, we're justified to only keep the lowest order infinitesimal. We need to depend on the fact that the delta x to the n power is getting smaller as n increases, right? Yeah. But that's certainly not satisfied here. I'm actually.Giving you a very nasty number now because the delta x is not small. No matter it's one. So how do we actually use that definition and to go from it? You know that's very convenient if you want to do linear approximation. So we're gonna actually do a order, although not justifiable, but we're gonna do it. We're gonna get just a linear approximation, and then we can compare with another stepwise order method here to drastically improve the accuracy.Alright, so this is for comparison. I'm writing it in green now. What is a single-step linear approximation? Add it. What do we do?``````Go ahead. I'm listening. Remember how we do it: separate the differential equation. We're not solving it. We're using because we're interested in only one single value.The differential equation becomes a different way to give us what is going to be this term we need. You know, right now we're not required to solve that differential equation. We're only our only goal is to find the f of two. So you see, to move from the one to two, we're given the y zero and delta x equal one. All the unique component we need would be the derivative. We don't even know what the function is, but we're given the derivative right away, dependent on both x and y.So that's doable. ``Èlen, tell me what trajectory you are on. We just try finding the derivative term by just plugging in two. Oh no, one. Yes, we're plugging one and one. In fact, because this slow field is written in terms of both variables, yeah, we have to honor that.So it would be five. Ah, yes, you're we're calculating. So in fact, we're finding the derivative at one would equal to five times all of a ones. Yeah, you're right. Now what is f two nEleven. Ah. Don't we go for. Turn by turn, Ily, would you give it.中文嗯。Anyway, where are you? Do you see we're just applying this algorithm? Go ahead. Yeah. Um. We're just be like six. I I want you to spell.To the numbers for me, please. You're right. It is six. One plus five, right? One plus five, yeah, yeah, it is. Plus five times one, right? So F two is in comparison F one. The original function value is one. We add down to the slope times the run, give you the rise. So it is five times one now. Yeah, indeed, that equals six. We're gonna actually it's not very good for the reason I just stated. It's very crude now because.We throw away the higher powers while they're not even getting small. So, on the other hand, let's see how we could improve the accuracy, and we're going to compare just exactly how much of the deviation. A final answer would be compared with six. Now, the whole point is that this linear approximation only works well if your delta x is truly small. The smaller it is, the better the approximation works. And how do we make this one, which is not small, into small? Well, we break it down. We break it down into parts, and we're going to repeatedly do approximation of a nearby point. So, what if we carry out a five-step scheme now? No, alternatively, I'm going to actually use a different color. So, this is actually improvement improved. It is still linear, though it doesn't mean we're using the higher derivatives. It's improved linear approximation. App using five steps. What does that actually mean? We're going to.They go from the one to one to the article to one point two, and then you could use B approximation to create a y at one point two. But notice here, I'm calling that the y two because it's approximated. Okay, it's not the real value; it's only approximately true. But it's true than directly approximating f two. Oh, I should write the what f now since I call that f. Oh no, I put that y. So this is a rather.I should change this, not change here. So that's why. And then once we have this new point now, once I approximate the y value, I move from the original point to a new point, and then I can actually use that differential equation. Do it again to find out what about the x equal to one point four, and then I just keep moving. I'll show you the procedure. This is what I do on the AP, and you're going to actually write this three column table whether the x y.And the y derivative, because only after you find out the x and y now, you would have the y derivative, and they're in the same row because they are also concerning the same x. Now, once you have these three numbers ready, and you're able to actually find out what is going to be the x plus the delta x now. With the first row information, you can get this number, which is the y, and I'll call that the y one. This is the x zero y zero, okay, and then.That's going to be the y zero, and plus the derivative at zero, on a motor by a delta x. But once you have this pair of number, you can keep on creating the y derivative at one now. And then you can use the previous row to create the next row. Let's do it. I'm going to do the first two moves, and you guys take over. So we do what we did before. You go from one to one at the moment here. This table. I'm going to restart here. The table reads.One and one, and the first we plug into the derivative, and we figure out derivative equal to five, correct. And now we can move on to the one point two. Now this value would be the one plus the five derivative, and then multiplied by the zero point two, which actually gives me a two now. So at one point two, the y is already approximated at two now, and you plug in that pair of coordinates, the one point two and two, into this to find out what is the.At one point too. Would somebody please? Oh, this is awful, because we happen to fall into a trap. We hit a singularity of that two minus x. Now, ah, whatThen there's nothing wrong. It is absolutely right, except for I just give you a poor approximation because we happen to we won't be able to grow anymore. So for that purpose here, let me actually do four steps instead. Oh, maybe I can do more, but once I hit one point two, it's always going to happen. No, not not necessarily. Um, to make it doable, I was going to do ten steps, but it's way too tedious. Okay, so let's actually do four steps. So that means your delta. It's actually. Actually, a quarter. So the first number is on one point two, but one point two five, and the next number it would be one point five, etc. Etc. So, and like I said, this would be one point two five. Or you can call that five fourth. Then if I do so, we're getting the one, and then plus the five times the five fourth, which is the twenty five over four. So actually, that just give you twenty.nine over four, which is a little bigger. It's seven point two five. Wait a second. Yeah. Hmm. Oh shoot, yeah. I'm not multiplying five by the five fourth. I'm only.Multiplying that by one fourth, so that's all twenty five. That's fine. I thought that number is way too big, right? It's nine fourth, so it's a little bigger than two now. Because we step the further, this function is increasing. Do your own pencil work. I can see whether you agree with me. Yeah. Alright, let's continue. And then when you plug that in into that.Our equation. You're creating your next derivative at exactly one point two five. That number is a little nasty, but we can live with it. And do your own pen a pencil work and see whether your answer matches with mine.中文嗯。```Give me either a confirmation or some correction. See if I'm getting this right.Yeah, I got about the same thing. About the same thing, or exactly the same thing. Exactly the same thing. Excellent. I did you. I got the same thing as well. Okay, we noticed we're noticing something interesting now. The function went from increasing to decreasing, and in fact, this factor here would make the two minus y a so called attractor. Why? Because it actually means if you.You analyze the derivative now. It means whenever the y is greater than two, this actually means if the y is greater than two, then your derivative it's actually decreasing, right? The y derivative would be less than zero. And if the y is less than two, and then your derivative would be greater than zero. That means, if that number right now the y value, it's under two, then it will increase, would move toward two, but the minute it's over two, it would decrease back. No wonder.Earlier, had we done five steps, the following four would be trivial. In fact, that would be a wonderful solution and decently accurate. I didn't do it that way just because then you don't get a chance to practice. And so the remaining part is always because the y value would remain zero now, and then your function value is two. In fact, had we done five steps here, let me record that was was it obvious to you now? If we had done five steps here, then our answer, if we just look at the one move.To one point two, and right there, the y would equal to two. The derivative would equal to zero, and from there on, the y stays at two. So every single time one point four, one point six, and one point eight, and all the way to the two now, and the y would stay at a sorry derivative equal to zero, y would stay at two, and this is actually the approximation. And you can see this is a pretty far away from the first approximation we were getting. And when you really analyze the behavior, as I said, the.The derivative function gives you the assurance, meaning your number will not veer away too far from the two, because the minute you are greater than two, the derivative would pull you back, but if you are under two, the derivative would push you up. So this is already a lot more accurate, right? Now we're just seeing what if we actually do it with rigorously four steps. I'm going to actually really tedious you out for once. Now we do it only for once.We have done two one step now, and it's homework. Could you finish it with actually the well basically every single time you add one fourth to it, you're getting six fourth, and the next one be seven fourth, and next one be two. Work out these numbers here. I know they're nasty fractions. I will permit you to use a calculator, only for the calculation that these numbers when you submit homework. I want to see a full table including all the correct numbers, and eventually you see.Its final number is indeed. Oh, I mean the final one here. It's really very close to two. And apparently, we got enough of the reasoning to believe that two is the correct answer instead of the six. And you can see what is what was wrong in the earlier one. The method was right, except for the step. What was too big? The delta x was too big, and the solution is to break down and conquer, make it into smaller steps.Is it crystal clear? Yeah. Good. Next session. Still more exercises comprehensively. Meaning, I'm gonna connect it to other approximations, but then we'll be done with a linear approximation. You do need to master this well because it shows up to where there's one hundred percent certainty on your AP test. Okay, okay. Take care. Bye. Bye.

泰勒近似与误差估计

泰勒近似与误差估计

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