Logistic 增长、部分分式与渐近线

Well, we're gonna, the most important and your first and most nuanced mathematical model, a, in the form of a differential equation. It's called logistic model. There's an entire chapter on it. It's a perennial focus of the AP test, and I'm gonna spend that least two hours, even possibly three, on this very topic. And it's a general continuation from what we just did before, which a come on, which is a knowing the slow.Feel the solve the differential equation. It is a separable kind. Interpret the graph and answer all the questions. They're off, but before we start here, any question concerning homework or the previous sessions? We need all the material we learned so far in order to tackle this practical model. Hey Eddie, I thought you asked for leave today. No, so we are going to go skiing, but it's raining.So we can't go. Oh, I see. Sorry, I wish you could tomorrow. Good because today it's a very important session. We're talking about logistic model now. Logistic, like I said, it is a comprehensive mathematical model. You need to master in the form of differential equation. It's the only one that you're going to learn in AP calculus, and it is on the test every every year. So we learn how to actually solve a separable differential kind differential equation.But from real world, where do these where do these differential equations come from? So forget about the math for a minute. We're gonna look at a fish pond, and intuitively, we're dealing with actually the population fish population of course of the fish pond at this moment simplified into just a single kind of fish, or maybe we're just talking about the total population. We're not dealing with actually the mutual and food chain and ecology within the fish pond. Basically, we just.There is overall population, which is a function of time, which we call the t. And intuitively, the rate of change of the population, be it increasing or decreasing, depends on the dynamics between resources, food competition, and prey and predator, and the population reproduction itself, really just giving birth to small fish. So intuitively, we would know when the population is very, very low.And there's no parent fish to give a birth to small fish. Therefore, the rate of the population is a growth. It's also very, very small. We would imagine probably it would be sort of proportional to the population itself, but not linearly though. So at least we do know population also depends on the the rate of population growth, which is actually defined as a dp over dt. You you see, we begin to do differential equations now. Itself probably, it's a function.Of the population, because it depends. Because the population, more marriages and births, and definitely there will be more small fish. But at the same time, though, if the population is too big, exceeding the carrying capacity of the pond, basically the surrounding environment in terms of the oxygen concentration, food resources, it simply doesn't permit the population to grow anymore. So.Apparently, intuitively, there is also a so-called carrying capacity. This is the maximum population that can be sustained. Notice here, population. I said that can be sustained. You could really just in one afternoon adopt two tens of the little fish into the pond now, and immediately population would surge, but it's not going to last because immediately there is the exhaustion of oxygen and there is a mutual predator. Prey in relationship, and just overcrowding would lead to death of a mass death of the fish, and eventually the population would asymptote a single carrying maximum capacity, and that's how we're defining this carrying capacity. And this is defined by the environment itself. We need to know the ecology of the fish pond, and just how much of the.What the other kinds of, and living creatures are in that, and how much of the foliage it's underneath the water, et cetera, et cetera, algae, or something that could be eaten by fish. Now, so with a particular environment here, this PC comes, and then we're gonna go for the simplest kind of model. You know, mathematicians glory in drastic simplifications, and you know, there's that saying that we make everything as simple.As possible, but no simpler, meaning we can't make it simpler than reasonable. But we did reasonableness. We want to make it simple as possible. And I would say probably the simplicity depends by the population rate of growth on the population itself would probably be a quadratic. Do you agree? I pick somewhere in the middle, which is just the right amount of fish or number of fish here, so that there's a very vibrant population growth.And marriages and births, but at the same time, though, they're not exceeding the environmental supply. So this is actually when population zero, the rate of growth operates obviously zero, and when the population's at carrying capacity, it's also zero. And knowing these two, now we would also know what is the maximum rate of growth, and this is empirically determined, and maybe we would simply call that actually P down the max.Knowing these three pieces of information, we'll be able to write our quadratic equation. You see, we always go back to the algebra two skills that we learned before. Then could somebody write me this quadratic dependence of the p dot on p, please? Fitting into zero zero p c zero, and this is one half of the p c and p dot max. Maximum girls read.These are not variables; they are numbers. They are parameters.嗯，可以。We like write the roots out first. Very good. That's really the best start. So we do x times x minus p. You mean p? Because here there's yeah, that's a population, n times the p minus a PC. What a wonderful start. Up to this point, there's only one thing up for grabs. That's a leading coefficient, isn't it? Yeah. Depending on, after leading coefficient won't really change the two roots. Is the leading coefficient positive or negative? Negative. Negative. Very good. And then maybe we'll be we should write like we can.Write a script of this dialogue. Negative growth rate. So this this quadratic equation does have the domain going beyond the carrying capacity. But then in the front, what would be the coefficient? Could we like plug in the vertex into the equation? Absolutely. So.P dot max equals p times pc minus pc over two. And we're plugging in pc over two into the two position of the p, right? Oh yeah. That's right. So the p dot max, you know, that's a clumsy mouthful of a notation. Why don't I just call that r or r max? I'll just do that r max, meaning it's a rate, its maximum rate.Actually equals to it has got to one half of the PC, and then times the PC minus one half of the PC, which is another one half of the PC, and then times whatever that leading coefficient k here. If we do have a k in the front. So finally, what is our k?What's this P? This one. That just makes terrible. Oh. So do we divide the PC over two squared? Yeah, absolutely. On to the other side, so what we're having is actually four maximum rate divided by a PC squared, and that whole thing is serving simply as a.Coefficient, followed by the quadratic equation, written in the factored form. Perfect answer. I think that's really good. Do we all follow? Unusually, we just call that leading coefficient the alpha. It's a it's coefficient, so just bundle it up and know that we actually defined the alpha in this manner: four times the maximum growth rate divided by the carrying.In capacity squared. This is how we define alpha, and it just turned out to be the alpha p p c minus p na. This is a compact form of our dependence of the rate of change of the population, population growth rate on population itself, and it's quadratic dependence. Our clear. And notice here, and on the both two sides of the equation, the p and p dot are function of. So at this moment here, I'm going to make it look like a differential equation, or it is a differential equation. I notice my switch of the perspective from a function of a population. Yes, it is, but it's also a function of time. So I'm going to write it down as actually p as a function of time now. And p zero minus p, and this is a differential equation. The p dot is actually the rate of change of the heat. And could somebody show me the?The entire procedure of solving it. By the way, this is called the logistic model. The logistic model can be defined in so many ways. You can go from the motivation and say it's a simply quadratic population growth model. That's entirely correct. You could also say it's characterized by this very differential equation. That's correct. And later you could even just quote the solution which I'll reach by the end of today and say, "Oh, that's called that solution that function p is a function."P. H. Square logistic model. That's also legit. So logistic model. And during the solution process, you see why it has to do with logistic algorithms. Okay, how do we solve this? Oh, where did the alpha come from? The alpha is defined just to bundle up a bunch of complicated coefficients. To make the notation a little lighter.I don't quite follow where the coefficients came from. Ah, what do you mean? We we just landed like that. Oh, you mean why that coefficient is actually four m over the PC squared? Yeah. Oh, in fact, I need to read it here for us. She's saying now this curve must satisfy when you plug in the half of the current capacity, we are reaching the known maximum rate of growth, which we call RM.So I plug it into the hypothesized equation. At this moment, we don't actually know what this coefficient is yet. Okay, we're just going to actually call that to the k. But keeping the k, we're plugging in when p equal to one half of the PC. We plug it into this equation now, and what we're supposed to get is simply the maximum rate RM. So I'll be resolved the k for us.Okay. Really makes sense. Yeah, good. I'm glad you're asking. Ah, do take a minute to write down your notes and wrap your mind around it, and then we're gonna solve that differential equation.```Yeah, I think I'm good. Alright, Yula and Eddie. I'm good. Let's give it a shot. Well, we should realize the so-called P dot is really just dp over dt, so it is indeed a differential equation, and it is separable. Yeah. 嗯，In fact, except for that t, there's no appearance of t anywhere. So that makes it very easy to push the t separate the t on one end of the equation and push everything having to do the p to the other.By DT. And then we divide by terms of P. Very good. We divide over the P, and the P zero minus the P would equal to alpha DT. Why do I keep the alpha on the T side? Because the left side is complicated enough. There's no need to further complicated. So let's actually save the alpha on the simple side. It's a.Mathematical choice, okay. For convenience, you don't really have to. You can put the offer anywhere you want. Just a number. All right, at this moment, both sides are integrable. We can just find the anti-derivative of both sides of the equation. On the left, you you're going to get a function of p. On the on the right, a function of t. And at least that got got rid of the differential part, so that we end up with the outbreak equations.Eli, do you want to give it a try? We're finding the integral of alpha dt, right.You tell me, our goal is to actually solve the population as a functional time. Isn't this a standard procedure of solving a differential equation?I hope you can't see through. Usually, students can't. Okay, you're not supposed to have memorized this formula. In fact, we have never learned such a formula before. But remember, though, we did spend time decomposing these fractions. Yeah, you actually have a quadratic function on the denominator. Let me actually help you review. There's a technique called partial fraction.You write it down. What does that mean? If we start from this rational function regarding p now, I'll simply call that rational function. That's a function p which we're finding the anti derivative. You have a one over the p, and then p zero minus a p, right? And the denominator is already factored. This can be broken down into a p, basically the two single factors, each one.It's its own single fraction, and then plus. We're looking for some number to put on the top. It's a pure number now because each part is guaranteed to be a proper fraction, meaning with the power of the denominator necessarily strictly smaller than the power of the numerator. Ah, sorry, of the denominator. Sorry, say it again: the power of the numerator for each term here. We're separating necessarily strictly smaller than the power of the denominator, which makes it a proper rational function. Another gets to a proper fraction, so we have another question mark, and it's something else p zero minus the p. The question is, how would you find these two numbers? We could try like making one term like not important enough to like consider. Very good. Elaine is saying, since we're working on two numbers here, why don't we?Isolate one number at a time, trying to make the other term not count, for the time being, for the purpose being, and let's write, try to make one term vanish. How do we do so in order to find out just what number is on top of the P now? Isolating the first term.Multiply by p zero minus p. Very good. And what that actually means, you're isolating the other term. You're isolating this term. Fine. We're going to see what's going to happen. Elaine saying, "To isolate this question mark here, why don't we get rid of the denominator? Because we want to isolate the question mark, right? So we shall multiply." In your mind, you don't have to really.We do it, but in your mind, you're canceling that denominator. I'm put it here on the top. Meaning, on the left, you're also canceling that denominator. And why do we do that? What kind of a purchase would that give us? Oh, remember, there's an unresolved coefficient here. A, I haven't found out. We're trying to find what the what the B is.And I don't want to have to find A and B at the same time.呃。Because Elaine wants to isolate that question mark. That's one purpose. That's one reason. Another reason is if we actually multiply everything by that number on the top here, and you can see p zero minus p only.Here's on a. Our goal is to find the question mark, and this is identity, meaning the left-hand side always equal to the right-hand side. That gives us the the convenience of choosing maybe some value of the p to plug in, so that the term we don't want to be dealing with, which is the a term now, would vanish. You then you really should continue and tell us why we're doing this and how does that help us resolve the question mark. I don't really get why there's an N of V. A, like here. Yeah, because again, there actually a coefficient on the top of the numerator which we haven't found yet. Remember earlier, we said whatever that original fraction actually equals to some number over the p plus some other number. On to distinguish them, I can actually call them.These are unknown parameters at the moment. So they are just A and B. Parameters.And did you say you want the A to disappear?Make it disappear. You can plug in p equals a. Huh? Plug in p equals a. Why does that make a disappear? I would say it makes a show up even more. Eddie, we learned this in the previous group lesson. Do you remember?Remember, how to do it, what we did.Elena, did you actually wrap your mind around what we did in the previous lesson? We want to make that two not matter. There are two ways to think about it. I'm counting on you. Once we have rewritten everything like this now, can we pick a smart number to plug in to the P now, just so that immediately I have an explicit value for the question mark?The A, although unknown, would vanish, would cease to matter. P zero. Yeah, what a good idea. At this moment, we can just plug in the P E and the P zero, and you can see what you plug in. It's very denominator, whose numerator we're trying to find. So one way to think about it is to multiply the denominator so that this factor would be zero, and then you could say, "Oh, the question mark really just equal to the one over the P zero, isn't it?" So this.This quantity is one over p zero. It is simply what remains on the left. Or alternatively, I'm going to actually cancel this multiplication now. Oops, sorry, and just return everything to its pristine form. We want to look at what this question mark is. Then we really want this term to matter a lot more than the other term. How do we make that term matter more? Remember, we're in the domain of calculus now, which is populated by. Infinites and infinitesimals. To make the term really just the become the predominant one, don't we also do make the p into p zero, approach p zero, so that this becomes infinity? That's just peanuts. The left hand side though, it's also approach infinity. Now we're just comparing the two infinites. We're really just saying the bulk figure, which is captured in these two infinites here, will exactly match. And they already match the denominator infinity.So that just means the remaining coefficient should also match. Which actually means, well, remember it's only infinity at the location p is approaching p zero, so that's what we'll what you plug into the other p until you equate this one over p zero. It's a whole logic clear. Could you repeat that? The whole thing.Can you narrow down to which part you want me to repeat?We plug, we remember to plug the p zero into this p, and that gives us the numerator. Okay, makes sense. Yeah. And then, Eddie, does that fully make sense to you? Yeah.Well then you try, apply it, find out, find out. So what is the coefficient on top of the p now? Same logic.```嗯，Can you, like, can you talk about like the infinity part again? I, I still don't get it. Well, earlier we chose to remember motivating from the very start. Okay, we want to care about this question mark, and not the other one, so we need to make a to find a way to make.Make this term shine, and the other term vanish, or rather, at least shine in importance. And the way to do that is to actually make the p approach the location of that vertical asymptote, because in the vicinity of that very vertical asymptote, now, and the second term is really very, very big. Meanwhile, in the face of infinity, all the finite numbers are peanuts, and that legitimizes losing the the other term approximately. That's nothing. That's negative. So just imagine we compare the infinitives here with the infinitives on the left. You just need to match the two infinitives, which is only true at the location of the p equal to p zero. This gesture for plugging in the p equal to p zero effectively isolates one term and make the other term not matter for the time being. We can.Obviously, a by multiplying by p. I okay, we we can skip the multiplying part now. I think it's easier for me. I have shown the mechanism, but why don't we just simply plug it in and compare the coefficients? It amounts to the same thing, doesn't it? Yeah. Alright, so Elaine is saying, "Now we could apply it here and show us exactly how to do that," but.Let me ask, Ivy, are you comfortable now? Do you know how we did that? Um. Yeah. Alright, then you learn. You could try the first one.``````We could make p approach zero. What a wonderful idea! Now this time we want to isolate the importance of the first term now, and the convenient way to do that is to make the p approach zero. How does that? Well, then this infinity here would match with infinity on the left, and how does the coefficient there out?And means we keep the terms; those are already matched, right? And these are already matched here. We just need to make sure that whatever the coefficient we're looking for matches with the remaining part of the coefficient.Coefficient is one over p zero. Indeed, what did you do to get that? Ah, since p is approaching zero, the term on the denominator of the R P it approaches a p zero. Yes, the coefficient is one over p zero. Correct. So basically, then is saying we're plugging in the p approaches zero into this part now, which gives us the coefficient on the top. Every. Look at the results we get, and find the common denominator, and double check very carefully. Just by carefully finding the common denominator and compare what you get here with what's on the left-hand side of the equation, the process will convince you how every single term plays out. This technique is called partial fraction. It's a big deal, and we're going to practice before we move on.Yeah, it works. Give me some confirmation, Eli and Eddie. Once you're done.```Yeah, it works. Eddie.Yeah, have you confirmed? Very good. Well, let's look at a few more. And to save your trouble, no, a no, I'm not going to actually factor it for you. Just go for SQ that minus, um, I for x squared plus, ah, three x, on the numerator. Two x squared plus one. Yeah, this is a.Proper fraction, a rational function, proper in the way that the power of the numerator is already small, learn, and the denominator. So we're going to actually break it down into three partial fractions. Each one is completely factorized into a linear form. Once you know what they are, we're going to find what would be the coefficients on the top. Eventually, my goal is to find the antiderivative, and you tell me how you're going to do it. If you do not actually do what I'm suggesting.There is no easy way. The easy way to do finding the antiderivative of a rational function—the very first step is to break it down into partial fractions. So please go ahead and apply the principles we have discovered. Feel free to write on my board, or you can talk.Let me factor the denominator first. Yes, wonderful start. Go ahead and do that for us. So x times x minus three times x minus one. Yes, excellent. So if you want to eventually draw this, there are three vertical asymptotes. Each one is a single asymptote, meaning the infinity changes the sign when you cross that asymptote. They are located at zero, one, three. That's very good. So I can write them on the denominator now. Now the job is to isolate one term at a time, and I'm going to actually let Elaine, Ivy, and Eddie in that order do each of the three terms.``````Well, forget about the order. Whoever has the answer out with it. You can go ahead.So first, when we put the x minus one, x minus three on the numerator. What do you mean? Put it on the numerator. You mean multiply. But if we multiply both, that's not going to. If you.Multiply by a certain denominator. It's not going to make that term go away. What's wrong with the method I showed you? You really want to multiply it onto the numerator. It doesn't really matter. Isolate one infinite at a time. Just be happy with dealing with the infinites. It's easier. So do we make the do we make the x approach the root one at a time? Yeah, exactly.Okay.Oh, um, the second term. Indeed, oh, maybe you didn't write down the terms in my order. Maybe it was your first term. Yeah, in fact, we're we're trying to find out what that question mark is. And the Make X approach one highlights the importance of that term because that's the only term that's going to infinity. Very good. And what do we do now? We try to match it with the numerator in the other roots. Yeah, we're also plugging in x equal to one to the left-hand side, which includes here's a numerator that's a denominator, except for we don't need to plug into that very term that's already matched here.We save that we leave that out, which gives us. Negative three. Ah, you are playing the one, and the numerator gives you a three, right? And the denominator gives you the one.I am negative two, so I totally agree. We're getting a negative three humps. Well done, Ivy. Ivy and Elaine, do you agree? Did you follow the logic? Yeah, I follow. So for the second one, we can make x approach zero to isolate the first one. Beautiful, good idea. x is approaching zero. What what's happening on the left?Is it one fourth? You're plugging in zero. Why did I give you one fourth? Check the denominator. Oh, multiplied the denominator wrong. It's one over three. Yes, very good. Eddie, do you agree?Yeah, beautiful. Eddie, go ahead and finish the last one for us.You should expel your mental processes.You can make it a like substitute one, I guess. Why substituting one? Our goal is to make. But you have to actually make that term more, and substituting one now.The second term will be approaching infinity. How does that help you reveal the question mark? I give you two opportunities to listen to the logic, when Ivy and Elaine did it. Eddie, why did Ivy?We choose to plug in ISIS protein one to isolate that term on the top. I repeated this logic here several times now. In this class, in my teaching. If you start by something just.Let's ask. The whole class is waiting for you to talk.Would you like Ivy to explain to you why she chose to plug in x equals one? Yeah, Ivy, be kind enough to go ahead. Uh, I put x I chose x equals one so I could make the denominator like um approach infinity, and like. Ah, to make the make the denominator approach one.嗯，零。And why do we want to do that? Because we want to like, um, because then when we approach, oh, when we like, so we can like match it to the, um, the left side of the equation.And if the denominator of the term we're interested in is approaching zero, why does that actually enable us to match that term alone with the left-hand side? You like go ahead and say it. Since the denominators are a factor of the original denominator, that the other one, the the two denominators will both approach zero, and then this um, will be will be infinities.Scaled by some number, so we can compare the scaling factors for each of those. Yes, let me say that very clearly. ID, because if we're interested in this is about the first time I'm saying it. ID, you really need to pay attention. Otherwise, you might as well just ask for leave so that we could actually go on without you and not wasting time on it. I'm not saying this meanly. Alright, so we can plug in. Actually, if we're interested in finding the the number on the top here.I want to make this term matter more than the other terms, so much more that I don't worry about the other terms at all. How could I make that happen? You can only make that term approach infinity so that the other finite numbers would be peanuts, and then they won't matter anymore. Why it is so hard to wiggle that definitive term from you guys? I'm talking about Ivy and Elaine now. You have to make that term approach infinity so that it outshines the other terms. I'm Ivy and clear enough. Yes, so that you don't. We don't have to worry about the other coefficient on the top. That's why the point is not to make the denominator zero. The point is to actually magnify one term at the expense of the other terms to give us the convenience of worrying about one coefficient at a time. Then Eddie, go ahead and apply. How do we find the question mark now?嗯。```Ask if you're still blocked by something.呃，你 know the other kids are actually paying to listen to silence. At least you could be decent enough to ask questions if you don't have the answers.We repeated what's the logic behind it. We showed two examples. Then each one did a term. If not, that's not enough for you to learn. Your mind is not here.Do you want to leave and keep the recording? Ah. Yeah, please. Alright, we're gonna continue. And Ivy, would you quickly finish it for us? Um, I got nineteen over six.Show us the process. If I do so, these terms are not our concern anymore. They're negligible. The Latin side is also approaching infinity, but Latin side is the numerator we highlighted over the denominator. They already match this component. I don't need to touch it. I just need to pair the question mark with the rest of it, which means I'm plugging in a three into this and that. The residue, what we're getting would be actually on the top. That's nineteen on the bottom. It's three times a two. Therefore, we.19 over six. Did you follow? Yes. Earlier, I asked you twice. Every single time, when the kid finished, I asked, "Eddie, did you follow?" You could have said, "No, I didn't." I would respect it. I would be thrilled and explain it to you, but you said yes. Actually, you didn't learn it. You know how much I despise lies. I know it's a very hard habit for you to really just win yourself from. I'm not mocking at you.I like too, but I think you should have a really just put effort on just disowning that habit and learn something else. Try to be straightforward and honest. Would you? I'm waiting for your answer. This time, did you really learn? Yes. Let's actually forget about that term here.Calculate for me again. What is the coefficient on top of the second term. So we can make x approach one, and then, ah, we can plug in one to the top as well, which would give three. And then, um.嗯。嗯。嗯。嗯。The point is one minus three. You plug in what um one minus three. Yes, we plug into this term now to get actually negative two. Yeah. And then it would be three over negative negative three over two. No, there's something else.Why did you choose to plug into only this term here on the denominator? Oh, and then you had to plug into the x. Yes, you plug in everything else except for x minus one. Correct. So you do get the right answer, which is ninety-three halves. Alright, so this is a the technique of partial fraction. What do we achieve? You break it down into this form. Now we could eyeball the antiderivative. We do know.What would be the antiderivative of each of these now, right? That is natural log. Remember, it's natural log with the absolute value in it. So problem solved. Please write a note to yourself to your future self: partial fraction is a necessary step to integrate rational functions. Only after we break it down into partial fraction, can we integrate one term at a time and easily recognize all the antiderivatives in the form of natural log. That's crystal clear. Yeah. Before we go back, now, whatever you're seeing something that's a little different. Whatever my denominator is x squared minus x plus one, one over that. How do we find the antiderivative of this guy? Or in particular, before we even even worry about antiderivative here.Can we break it down into two partial fractions, each one is linear. Yeah, we can. We're not afraid of complex. Factor the denominator. Yeah, but the point is, you can factor it into real numbers because the discriminant b squared minus fourac it's actually negative. But remember, we're not afraid of complex numbers, guys. We factor them anyway. Just use complex numbers.Break the me to partial fractions anyway, find the antiderivative anyway, and then let's worry about how to interpret a antiderivative with complex numbers in that. Because that's not within your AP curriculum, but you're permitted to do it this way though on AP.嗯，可以用这个求解公式。Yes. Hmm. I got negative one plus minus. Negative one. Oh one. Yes. One plus minus the square of.Naive one squared minus four times one times naive one. Huh? Four times one times naive one. Oh, I wrote it wrong. One. Yes. Over two. Yes, and after simplifying. One.Plus minus the square root of three i over two, brilliant. Yeah, these are the two roots. And if you want to, you can write them there now. It's x minus x one, which is really just this one plus root three i, and x minus one over the one minus root three i, right? The worst we could do it to just write that. But before you do so, I want you to pay attention to: Do we recognize that complex number? Why don't you graph.Have that complex number, and see what would be the polar coordinates. Meaning, what's the magnitude? What's the angle? I want you to visualize the two roots. 中文嗯。```嗯，是，kind of like a thirty, sixty, ninety. Beautiful, and strictly speaking, and one is sixty degrees. This is actually the one plus the root three over two.One half, right? Sorry, I. That's actually if you write it the x one here as a that's e i sixty degrees. Well, the other one is a conjugate, so that's polar negative i sixty degrees here. We're talking about these two vectors. Answer it. Yeah. Would the graphical ah intuition help? Yes, trust me, it will. Alright, now let's see how do we find the two question marks on the top. How do we.Do this breakdown.These are the two routes, okay. So for example, I want to find the question second question mark. Then my idea is to make the x approach my omega two. Then whatever the top would actually equal to the other side when you actually make this x here approach omega two. Then actually you're just looking at omega two minus omega one, and what do we have?Go ahead and calculate what's that question mark.嗯，Naive, Square Root of Three, I. Beautiful. anybody agreeing ```I'd be very good. You can go ahead and find the other co-occurrence.Even are you getting the same? Yeah. Very good. What about the coefficient? I'll be other. Um, squared three I. That's it. So theoretically.We know what would be the antiderivative of both sides of the equation. Ah, in case we actually want to do something like this, you see, this is not far away from the logistic model. There, it's factorizable on the denominator. Here, the only difference is, a we have to factor with the complex numbers. But interestingly, the result when you go back to the real would be actually drastically different. And next session, we're gonna actually be very careful bashing this out instead of.Well, rather, Homer gets to. Oh, sorry, I don't need you raise it. Actually, find the antiderivative of this accordingly by doing each one individually, and then I want you to compare with what you would have if you compute the square. I would say keep the one here and make the rest here into. I'll tell you what to do. Validate for yourself, and this is equivalent to the original, and try to find the antiderivative.Correctly for this one, it folds into a different type. Then think about how do we bridge between the two seemingly different answers. Clear. Yeah. Alright. Take care. Bye. Bye. Happy holidays.

Logistic 增长、部分分式与渐近线

Logistic 增长、部分分式与渐近线

课程概览

本课重点

课程视频

自动转写文本