Logistic 方程、部分分式与三角恒等式

And I'm understanding correctly, you are the only one today. Can I have your video, please? If you are, and I'm only happier, we could actually move faster, and we're going to continue with what we did previously on logistic functions. We didn't quite finish solving it yet. We went into discussing what happens to separable differential.Equation and what happens to the magnitude, but any question concerning the previous material. Before today, we're going to really focus on the solution to the logistic function and the graph, and interpretation on the three starting positions. In this case, discussion on the initial condition will be vital. So it's all about talking about the slope field and graphing the solution. And but I repeat, any question from the previous session? I'm not exactly.Like confident on how to construct differential equations for a model. So that is actually a very wide ranging and deep topic. You know, after we handle this AP test now, you'll be invited to my modeling class, and which I don't always give. I actually have a fixed session for so-called math wizarding that I'm giving free sessions to different topics here, but every now and then, about every two years.There will be a free class on mathematical modeling, which involves how to set up the differential equation, how to set up sometimes not the continuous model but the discrete ones, and how to impose numerical method onto them. Whenever you can't really solve them analytically, etc. , etc. Applied to real world scenario. And it was in service of the modeling contest in high school after this summer. And I'm gonna give all the information that I know of concerning a wide range of mathematical.Medical computations having to do with advanced tools. So remind me that within the scope for what we're doing now, we're not going to do math modeling. It's a very deep topic, but are we decently clear how we set up for this specific scenario, that particular differential equation? Let me quickly recap: we're dealing with a logistic model, and Eddie, I'm turning on your recording. In fact, Elaine just posed a profound and extremely good question. She was asking for.How do we actually translate a particular ecology, a world, into a set of differential equations? My answer is, it's a very long story, and on AP calculus here, we are not required to do any more than the logistic model. That's regretful. Nevertheless, though, I want to be sure that you perform well on your. We did the scope of the AP requirement first, before we move on to something more. So last session.We set up the time derivative. I call that the p dot here. This one equals to the dp over the dt. Is simplistically a quadratic equation reaching a peak and coming down to zero at the two meaningful points. One is zero population, is the other is a carrying population. And it turns out we set up the number as actually a native alpha or actually absorbing the native as actually the population times a pop.Carrying population minus the instantaneous population, or the leading coefficient of the alpha. The real leading coefficient is a negative alpha. So, if you really want to write it out in this form as a polynomial, in order to pave the way—sorry, that's carrying capacity—for taking the derivative, you can do that. Um Elaine, at least are you crystal clear on how much where this comes from? Yeah, it's a minimal. Simple is the model. It doesn't have to be like that. For example, maybe it makes more sense to have the curve behave in maybe this manner. All of these are possible, and maybe it's not entirely symmetrical. Maybe the peak has reached a little to the left, and maybe something like this. Well, I don't really have a very strong refutation for any of these conjectures. I just said before we feel before there's enough of the evidence to show it's necessary to go.For a more complex model, we usually just default to the simplest, which is polynomial, as we did with quadratic equation. And we spent the previous session talking around solving differential equations, and we call that first order. And that's not linear because we have the appearance of the p squared, right? And nevertheless, though it is separable, so we're gonna actually separate out into the p and then divided by the p.And then P C minus the P. This is D P. Sorry, equals to the alpha of the DT. And at this moment, you're ready to take the antiderivative. The right-hand side is, you'll probably, easy to integrate an integral, take on that object constant C here. On the left-hand side, though, we won't be able to eyeball the answer unless we just break it down into partial fraction, which we did by writing it as actually, one of the P, an M plus the P C minus the P. Now, but if.You're adding the two terms. What you're getting is actually PC. So I would rather just do this and find the integral and then multiply, or rather, yes, multiply the PC onto the other side. So this would equal to the other PC t and then plus some constant. Be sure to go through the pencil work to recover this much and confirm you're getting the same. Every single step is important in this class because you will really be required to do this on your.There has never been any year that the test doesn't include logistic function. So, be sure to go through the procedure.```中文嗯。I don't really get where the PC came from. This one. Well, when you really add these two, what you're getting is not one over, right? When you recombining these two fractions, what actually do you have? Find the common denominator. P C minus P plus P over P times P C minus P. Meaning on the numerator, when you combine the two, what you get is just P C, and the denominator will be P minus P C minus P. Oh, I get it. Uh huh. So in order to not to太con.On this fraction or over the PC on the left hand side, we would rather just at this moment here multiply both side by the PC here. Notice I didn't change the constant because it's a it's unknown constant anyways. We could absorb the PC into it. So this is a broken down into partial fraction. At this moment, we are able to go forward and find the solution. Now I give you guys homework to really tease tease out the partial.For more complicated cases, and when you are able to do so, I haven't checked WeChat this afternoon. So if you send me homework within the past hours, oh beautiful Elaine, I see the one from you. Let me see, I did some work. I, Eddie, yes, you did it correctly. However, this is a standard answer that I can get from WeChat or from anywhere.In the textbooks, yeah, what I gave you guys, what I taught you, is a different way to do it. I was the homework. The purpose of the homework is to make sure that the two match, and you can see Elaine did it the other way. This way, the Elines, please check Elaine's homework. If you have access to each other at the moment, that's how I taught you. It's more flexible, it's more robust. In the future, when you're learning some very difficult integral here, using the complex integral.Would actually work. Meanwhile, your textbook formula will not. But I also acknowledge my method here will not be needed on AP. You're not going to see very difficult integration problem at all. It doesn't matter whether you really want to hit the test or you want to really learn good math here. Eddie, did you realize what was the purpose? My homework. What I actually assigned that homework. Sorry, yes or no. Yes.But why did you not do it that way? I can't make you out. I really can't make you out. Okay, I still can't. Can you type it into the WeChat if you can? But I know you're in the car. If it's too difficult, then that's fine. Forget about.Alright, now before we go on to solve the logistic function, and I do want to talk about homework in order to bridge the two solutions. So on one side, Elaine, you did really well. On one side here, we actually have a x squared minus x plus one. We go through the procedure of breaking it down into partial fraction. The minute you really see a integral problem on AP, I didn't realize we're supposed to do it the way.But I would. Okay, okay, beautiful. However, we're talking about it, but still do it down. Okay, so when you see such a problem on the AP test or even in your college test, here immediately your first thought goes to categorizing it into the factorizable, meaning you could break it down. The discriminant is greater than zero, you can break it down into two linear factors, or unfactorizable has got to be it is quadratic. Then, if it is. Unfactorizable. We do partial fraction, and which is actually the technique that we covered here. So this is a partial fraction technique. If it is unfactorizable, you do the other fundamental skill we learned about any quadratic equation, completing the square. In fact, ID is giving that standard answer, and eventually the function hinges on the form, which is a prototype of this one. It's a shifted this form here, and we do recognize the antiderivative.It happens to be our tangent. It is something you have to get familiar with and memorize. And however, though it doesn't mean we can't solve it using fourth factorization by using complex number. You then you did very well. In fact, these two complex numbers are the e of the i sixty. And forgive me, I'm going to use the notation of this polar form negative i sixty for a purpose. Okay, I'm not writing because you did it. You did it perfectly using the i square parenthesis. So.Instead, I'm going to do this way. What is equal to the x minus? I'll put that omega, and the other is omega conjugate. So that's minus omega, and that's plus the x minus the omega conjugate. However, in order to just practice our trick we learned now to make them add up to the one over here on the numerator, what we have to plug in on that side is actually one half of the omega negative. Because when.You multiply the omega with the omega conjugate. Apparently, as a pair of conjugate, they always give you one. So we could actually do this one, and the other one would be, uh, that's. Oh, actually, not just that. Here, well, what happens on the top here is that you're plugging in, and oh, sorry, I misread. That's why that's x. Sorry, to find the number of here, we plug.Plug in x equals to the omega conjugate into the two terms here. Into sorry, we're plugging x equal to the omega to find this number on the top. And however, if you do so on the other side, omega minus omega conjugate here obviously just give you the root three i. So absolutely, and you got that right. The coefficient here is a root three i, and that is minus root three i. Similarly, because now we're doing the omega conjugate minus omega, it would be the backward error. Okay.Once you get here, and then we're going to find the antiderivative of both components here. And again, Elaine did it very well. We actually have this. You three i a natural log of inside you're dealing with the x minus. The omega, and notice I wrote the absolute value. Elaine did it too. Without the absolute value, you can't even take that natural log. And then we're we're minusing through three i now, and natural log of.The magnitude of x minus omega. However, though here is something interesting, and in real numbers, the magnitude simply means getting rid of the negative sign. We know that it's actually the antiderivative because you can take the derivative natural log now, and it is going to be. This is in real defined on real numbers. Here it's an even function regarding natural log of the absolute value of x, so that you can see the derivative necessarily.The function, if it's even when x is negative, the derivative is still equal to one over i because when x is negative, one over i is also negative. However, though when you're subtracting a complex number, to begin with, what do we even mean by taking natural log of a complex number? What does that even mean? Well, let's think about a complex number that can be written as the some kind of a radius and then. Ei theta, when you take the natural log of such a number, there will be two parts. The result is also a complex number, right? So can we break it down into the real and imaginary component, please.伊人，What do you suggest？Contain i theta. Yes, absolutely. In fact, we do know natural log of the e i theta itself simply equal to the i theta. I totally agree. That is indeed imaginary term.``````Tell me what you are thinking.In polar form, a complex number has a polar magnitude, right? I thought about that. Yeah, it's already done. If you got here, then you're perfect, and there's no way we could further simplify it. That's as simplified as it is. Okay, in any case, so we're actually getting this much here. We broke it down into two numbers. Meaning natural log of a complex.It's meaningful. My question would be: When you really find the antiderivative of this complex number, do we add in the magnitude? We don't. We do not here. The reason that we had to add in the magnitude is only because for the real number we actually define the natural log of that. That's only meaningful if that is greater than zero. I didn't teach you that, so you learn you're not wrong by putting a magnitude there. You were following the rules. However, I was sort of anticipating that. If you do think further, you're going to realize, actually, it doesn't make sense. The reason is that the minute you put in the magnitude now around a complex number, it reduces to only a real number. It doesn't include any complex part anymore. Therefore, your natural log is also only real. When you take the derivative of a real valued function, we're not going to get a complex number. You see, it actually means something.Absolute value something else. This absolute value here is self-defeating. When you really take the derivative, this if you understand that as taking the magnitude of that complex number, in fact, we're not going to get this. We would only get the dependence on a real component instead of a complex number. So again, I didn't teach you that is why we need to talk about it. And then coming back to why we don't have the magnitude and not having the magnitude simply handles it will be fine.Actually, back to the real natural log of x, we wouldn't want the magnitude if we're permitting our answer itself to be complex. What does that mean? In case my natural log of x here, it's equal to seven negative seven, and what is natural log of the negative seven? Well, earlier it was undefined, with our high school perspective, but now we realize this is really a complex number itself. I can write it as natural log of the seven.Times e i pi，乘开开，开。So which really just equal to the natural log of the seven, which is actually equal to the I mean this is equal to the natural log of the absolute value of the negative seven, and then plus a i pi. There's that imaginary component. Which makes it totally correct, and then we don't need the absolute value anymore. It would be wrong if we do impose that absolute value.So earlier we put in the absolute value in order to expand out the domain of the x. Now back into all real numbers, otherwise that is confined to only positive. But knowing beyond the real numbers, yeah, in fact, the natural log of the negative number is also well defined, which includes this necessary imaginary part. That's why we do not want to put in the absolute value anymore. But however, though you need to know, in case you want to use this to show off on.Your AP, they will not call you wrong, but you cannot include the magnitude. Okay, as long as you're getting the correct final answer. However, if you do not use that, you just come back to use the real function. Then they will call you wrong if you do not add the absolute value here. So you have to realize what's the background the default domain we're talking about. So not to confuse yourself on the AP, be sure to always add the absolute value. But you cannot do that because this is.In the absolute value here, it's going to make that wrong. Meaning, you just cannot do this on the AP. This is for us; it's our private toolbox. Okay, which would prove very useful when you learn higher mathematics. At the time being here, know what we're supposed to do on the AP. That was much ado. I talked so much about the distinction between the two, but hopefully you would understand why. Now knowing that's not going to be the absolute value now. We'll just put a parenthesis here.哎， come on. And Eddie, do you follow what we're doing, and do you agree with what we're doing? Yeah. Okay, well, let's find out. And how do we combine this antiderivative, and how is that consistent with our tangent? This session is all about finding the equivalence between the two approaches here. So you really needed to. And watch out. What we're getting is, and first of all, I want to combine the two by pulling out the three i. And if I do so, what's left inside is natural log of the x minus omega divided by x minus omega conjugate. Makes sense. No, this here.This is not a real number. Because the numerator and the denominator are actually conjugate of each other. X itself is real, so numerator and denominator are conjugates. If the numerator and denominator are conjugates now, when you divide them, what are you getting? When you divide a pair of conjugates, let's suppose this numerator is unknown r e i some kind of theta. I don't know what they are.The numerator would be the R e negative i theta. Be sure you understand to begin with why these two are genuinely conjugates. Once you could convince yourself that it's so, and you could actually switch to this mindset and tell me what would be the quotient between the two.ReproductionR e two i theta over r. But why do we have the? Well, yeah.But can we cancel the R then? We can indeed. So if we do so, then what do we have? Like you said, we just end up actually e to i theta, right? Yeah. Alright, then knowing this is a pure imaginary number. Sorry, not a pure imaginary number, but real a unit complex number. Well, then when you take you plug this in, when you take the natural log, what do we have?Two i theta. Indeed, we just have two i theta, and remember this i up front. So we're going to actually multiply the two i together to get this. That's our final answer, isn't it? Yeah, I would say at this moment here it's looking good because back to real number one, number two, it's angle. Would that actually remind you, our tangent that you would get?By completing the square and go for the conventional way. Well then, I think behooves us to really investigate carefully what is that angle. Well, I recommend we draw a picture. We already know exactly what the omega is. This is our omega. Back to that picture we have drawn. It's a unit vector pointing to sixty degrees now, but on the top we're doing x minus it, which actually means you're concatenating. Here is the x now that's on the x axis. Okay.I don't know what X is the variable, and now we're minusing Omega. That means we're adding a negative Omega. This is a unit complex number with sixty degrees, with a unit one, right? And our job is to find out what is this angle theta. Go ahead and find it for me. In fact, you can find the tangent thereof, or you can find the cosine. Or the sign, or something thereof. However, I encourage you to find a tangent through the sign law. Maybe that's the easiest way to do it. Or you can directly find a tangent. Why? Because eventually, our goal is to actually write it in terms of an argument.``````You could keep me posted.```当然。嗯，嗯。Tell me what you're doing, please. I'm finding the side length. Good. That sounds the right thing to do.中文嗯。Give me posted. Tell me what you're doing. I'm still finding the side link. I.Okay. If you want to talk, you can talk, just so that you don't get hang on small details. Now I'm just doing the the work. Okay.```````````` 中文嗯。``````Can we talk about it? I'm afraid you got stuck somewhere. I'm trying to use sign lock. Yeah, use sign lock is one way to do it. Now we're interested in this side. However, this side doesn't—we do not know the angle. What we do know, angle.Would be actually the side of the one and the opposite side of the theta, and I'm going to actually show you how to do it using sine law and how to do using how to how to do it using the tangent. Okay, if we want to use sine law, what we have is actually the sine theta over the one would equal to the x over the sine. Oh, sorry, oh, I want to have flip it. So the one over the sine theta, and then would equal to the x over the sine.Of this angle, which is sixty degrees plus theta, although it's really the one twenty minus theta, it's the supplementary angle to that. But for supplementary angle, these two are the same. So this is how you could write your theta in terms of x. However, what we do know is that the sine of the sixty degrees plus the theta equal to the x sine theta, and you have to use the complex numbers or the sum.Of the angle formula to really spell this out, this actually equal to the cosine theta sine sixty degrees, which is the root of two, and then plus cosine sixty degrees, which is one half of the sine theta, equal to the x sine theta, and then it actually becomes a ratio of the tangent theta. When you gather all the sine theta to one side, this is minus one half sine theta now, and you could pull out your sine theta to end up with only x minus one half in the.And then you divide over by the cosine data. What we're getting is a tangent data, actually equal to the root three over two x minus one. See if you agree. I agree. Alright, now another way. It's a direct approach now. If I we actually want.To go for the coordinates because after all, we do know what are the two coordinates of this point, right? We do know this is gonna be minus one and that's sixty degrees. So the coordinates are the x minus one half because the this projection here is one half, and the altitude is root three over two, then minus the root three over two. Those are the pair of coordinates, and this will give you a direct expression of the tangent three.In fact, what I did here is to pretend that that's a geometric angle of the theta, which is defaulted to be positive. But if you go for the coordinate here, this is even better because the tangent theta even includes the fact that the theta is negative. Now here you're getting actually negative of the root three over two, and that's two x minus one. You see, this one we had to add that in manually, knowing that the geometry of the triangle only gives me that geometry.Metric angle, which is by default positive. So finally, we got our answer. Our answer is actually going to be the two root three and the arctangent of the root three over a two x minus one. Double check it, Elaine. With actually, and also Eddie, match the two answers here and see how they correspond to each other. Correspond to each other. Both are right, nevertheless. Though it takes some realization why they're both.Right. We're good. Okay, and look forward to seeing you. Keep up your good work, especially this today takes a lot of reviewing. We still haven't analyzed the solution to the logistic function. We shall do in our next session. There's no hurry. This is not your textbook, so you need to study it with some concentration and organize your notes so that you don't lose this part. We're good.Alright, I owe you guys ten minutes, eight minutes, and if we can, I'm going to tack it down to our next meeting. Thank you very much. Merry Christmas. Bye. Bye. Merry Christmas. Merry Christmas.

Logistic 方程、部分分式与三角恒等式

Logistic 方程、部分分式与三角恒等式

课程概览

本课重点

课程视频

自动转写文本