 So we're at lecture 43. Today's goal is to kind of gather things up, concepts, problems, expectations, all those things for a test that we have in this classroom tomorrow. It covers the supplement, at least the part of it that is appropriate for this class, 777, 7879. Hang on to that supplement because there are other things in there that you'll need in a future course, which I printed one out for you. I did not print out the entire supplement. I just printed out what was part of this course. Back to the book, the first three sections of chapter 8, even though on the syllabus it looks like we have still some things in 8.3 yet to do. We have finished 8.3, so it will be on this test. That'll actually, I think, you'll be glad that we have it on this test because it's one fewer or actually two fewer convergence tests that are in the rest of chapter 8. So we've got a lot of things going on in the rest of chapter 8, which that'll help the next test. So any question pending, any web assign question, or any concept before we kind of review? Yes, Jacob? I have just a general test layout question. Is it going to be enough to know if it converges or diverges? And to what value it converges? Or do you have to write all the, which test you used? You do have to write the test. And I'll mention this again. But some of them I hope will be pretty obvious that I won't need to tell you what test to use. Like if it looks like one that, let's say, could be integrated and you use the integral test, pretty obviously, you're going to show your work. You're not going to use the integral test on a piece of scrap paper and then just put divergent. So if you have your work for the integral and you determine that it doesn't have a solution, does not exist, therefore, the series diverges, then you could just put in parentheses by the integral test. I mean, it's pretty obvious that's how you got there. But that just lets me know that you know if the integral is divergent, so is the series associated with it. Some of them I will actually tell you to use the comparison test or the limit comparison test. So that, I hope, helps because then you don't have to go through this search process in your mind. Is it a geometric? Is it a p-series? Integral test, comparison test, limit comparison test. You know what's expected of you on that problem. But I'm not going to do that on all of them. So some of them you will decide, hopefully, those will be the ones that are real clear. Others I'll ask you to do the problem a certain way. Any other kind of questions like that or clarifications of second-order differential equations or sequence and series convergence and divergence? So we're not going to have one all the way through of the non-homogeneous code or whatever. Well, let's discuss that. If it is, it'll be a kind of a special, shorter version than the so-called typical one. So from the supplement, these were second-order linear non-homogeneous differential equations. We had three cases. I'll just go ahead and tell you because you don't like mysteries on tests. I don't like to give mystery-type questions on tests. All three cases will be used in some way, shape, or form in at least one of the problems. So it's not like you can say, well, I know the first two and I don't think he'll ask the third one. He will ask the third one. So these will all be covered. Now, not necessarily in the non-homogeneous, one of them might be into the homogeneous case, but all three of these cases. So if you have two distinct real roots to the characteristic equation, now, how much of that you show in going from the original second-order differential equation to the characteristic equation? If you want to just make that leap, that's fine. Or if you want to say, what are these things like that solve this, y equals e to the rx? And then go through y and y prime and y double prime and sub n and get the characteristic equation. That's up to you. But you don't have to show every step in getting to the characteristic equation. So if we have y in terms of x, then what would our solution look like that had two distinct real roots to the characteristic equation? Does that look familiar? I hope. So that's to the homogeneous part. If the left side is all the second-order stuff and the right side is 0, then you're done, unless there are some initial conditions. And if there are initial conditions, the y of 0 equals 7 and the y prime of 0 equals 4, then those will help you find c1 and c2. If you have a double root, it's real. So r1 is equal to r2. And we'll just say that it's r, or just r. So in fact, if you want to generate new solutions, and this is one of them, you can continue to modify. I think they called it a modification factor in the supplement. You can continue to multiply by x until you get something that actually does cause the left side when you put in second derivative, first derivative, original function to end up being 0. Again, you could have some initial conditions that would allow you then to solve for c1 and c2. Remember, if you do have to take the first derivative, that you do have a product here. So you have to use the product rule. If you're given y prime of 0 equals 5, then this term is a product rule, so don't forget that. Two complex roots. I guess the others are complex also, but I mean complex in a sense that they have an imaginary term. So the roots look something like that. So we end up with a negative under the radical and not real numbers. So we've got a completely different category. They start out looking like this, but they don't stay that way very long because of the fact that we have e to the i theta in there, and that then converts these exponential-looking things to trigonometric-looking things. So that's something that we used in converting this to this form. You're not going to be tested over that, but that's why it ends up looking very different. What do they look like when we have two complex roots that have the imaginary part as part of the answer? So I would, if I were you, know all three of these, and you will have a question on the test that kind of puts you in a position where you have to use all three of these. Yes, Nicole? I don't know if it was on homogeneous or the homogeneous equations, but actually I think it was the other one. When you have the first, I don't know what I'm trying to say. Two parts to the solution, the homogeneous part and the particular part? Right. When the particular part, the first part, when you take the derivative and it's the same as, I don't know, the homogeneous solution, don't you have to change the particular? Are we going to have any of those? That's possible. Can we do one of those? Yeah. Any questions on this? So the only thing that could kind of change the basics of these problems would be if you had some initial conditions. y of 0 equals something y prime of 0. And you can solve for c1 and c2. And you should plug those back in and then move on to the next problem. All right. These are the second order linear. What did I say? Yeah. The other one non-genius. Yeah, here we go. Just caught an error. Bring that up. These are homogeneous, right? Scrap that. So homogeneous means we've got all this stuff over here equal to 0. So if you were thinking that was incorrect, you were correct in thinking that was incorrect. So we want here non-homogeneous. So we've got all this stuff on the left side equals. And we have a subset of things. It's not wide open on the right side. But it's non-zero. And it comes from three different categories. Polynomial. So we get a generic version of that polynomial. If it's linear, our generic linear is what? What's a generic linear equation? Ax plus b. If it's a constant, then our generic constant is a or b or some letter. If it's quadratic, then our generic quadratic is x squared plus bx plus c. So we have to solve for those values potentially. So polynomial, exponential. So we need to, and this is what Nicole brought up a second ago, we need to make sure that the exponential on the right side is not also one of the homogeneous solutions. If it is, then we don't want to start with those because we know what that'll give us. That'll give us 0. And we just keep modifying it until we get one that is not. I'll make up some examples here or look one up of that type. And either sine of something, cosine of something, or sine plus cosine. So it's not wide open in terms of trigonometry. You will see some different problems. Excuse me. And as you're looking through the supplement, there is another technique in 7.8 that we did not do, variation of parameters. And you'll see some problems in here that have some things on the right side of the equal sign that we did not do, deal with, encounter. Don't be concerned if you're looking through the problem sets, if you see those, you're not responsible for those. So our technique is to put in the generic version of what we see on the right side, take first derivative, second derivative, plug them in, and equate coefficients. The coefficient of e to the 2x on the left side better be equal to the coefficient of e to the 2x on the right side, and so on. Term by term, equating coefficients of like terms on both sides of the equal sign. All right, so the first part of these that are non-homogeneous is to get the homogeneous solution. We just covered that. There are three categories, so the same three categories apply. That becomes part of your solution. Here's where the, what do we have on the right side question comes in. Do we have a polynomial, an exponential, or sine, or cosine, or both? Because we have to establish a generic version of what we see on the right side and get these three quantities, the original y sub p, plug it in where y is, first derivative of y sub p, plug it in where y prime is, second derivative of y sub p, plug it in where y double prime is, and then we start distributing, gathering like terms together, and equating like terms left side and right side of the equation. And then our final solution is y sub h plus y sub p. Then there is the possibility, again, it'd have to be a special problem that's shorter than most, where we could then put in some initial conditions and solve for some letters that still remain, the C1s and the C2s that still remain. But once we get to this stage, we'll have, we shouldn't have a's and b's and c's, right? We'll have those values established. So there won't be any doubt as to the values of the particular solution. There will be some C1s and C2s in the homogeneous solution that we could potentially solve for with some initial conditions or boundary conditions. All right, Nicole asked for a certain type of problem. Let me kind of build the problem as we go, OK? So I'll make sure it's one of that type. We're going to have something on the right side, but I'm going to wait and see what we get for the characteristic equation and then put this in so we get that situation that Nicole asked about. So the characteristic equation is what? And it looks like what? Minus 1, minus 1. So it looks like we have a double root. So the homogeneous part, and this is going to be non-homogeneous. I'm just waiting to fill this in here. The homogeneous part is what? C2x, e to the x. Does that work? And let's say over here we have 4 e to the x. So we've got an exponential. That's one of our three things on the right side. So here's our first piece, our second piece. We would go to the particular solution. So if we want to generate some e to the x's, don't we need to start with some e to the x's? We just don't know how many to start with. But what about that? What about a, e to the x? Can that be part of the particular solution? No, because that is part of the homogeneous solution. So if we put in some a, e to the x's, we're not going to be able to kick out e to the x's. We're going to kick out 0 because it's part of the homogeneous solution. So we scrap that. So we use our little modification factor. We continue to keep multiplying by x. How about that? That's not going to work either because that is also part of the homogeneous solution. So there is the first one that really has a chance of generating some e to the x terms. The other two won't generate anything because they're part of the homogeneous solution. They're going to kick out the 0. They're going to disappear. And there's a chance that this one too. We don't know that, but there's a chance that this one could not work. If, in fact, it doesn't work, then we continue to modify it by multiplying by x. But we would go ahead anyway with this, use the product rule to get the first derivative. And we're going to have the product rule again involved in finding the second derivative. But that's when you look at the term on the right side, if it is present in any way, shape, or form in the homogeneous solution, you better change it because it's not going to work. That's just going to be a waste of your time to say y sub p is 4 e to the x. Take it, it's first derivative. It's second derivative. You plug it in, you're going to get 0 on the left side. Yes? Your right side is like x sine of x. How do you set up a particular equation? So we have something over here. And on the right side, we have x sine of x. So we would find the homogeneous solution. We don't have that. So you go to the particular solution. What do we really have on the right side? A linear times a sine or cosine, or both. So we need to kind of generalize x. Well, we know our generic linear term is going to be ax plus b. We also need a, well, when we have sine of x, what do we have to allow for in the particular solution? Sines and cosines, right? Because we're using the original function, which has sines and cosines. It's derivative, which has cosines and sines. And the second derivative, which is back to sines and cosines. So we would want the cosine terms to all disappear. That can happen. But we would want the sine terms to remain. So in generic form, we would say sine of x would be what? A. Now, I'm sorry. I probably should use a c. I was thinking a step ahead where we're not going to need those anyway. So let's scrap that. Now, if we were to distribute this thing, we'd have ax times c times sine of x. Well, we don't know a, and we don't know c. But a times c is some unknown constant. So can we get rid of c? We can kind of get rid of c, because we're going to have unknown arbitrary coefficients anyway. We don't need to have product of a and c or a and b or b and c. It's a little redundant. Same thing with b times d times cosine. There's too many letters there. Or ax times d times cosine. So we don't need that one either. So that'll take care of it. So when you distribute stuff and you have two letters, you don't need two letters. There's just one coefficient. So that's the coefficient we would be searching for. Somebody might say, well, let's suppose that we do all this work. And there is a, I don't know if I like what I just said, to be honest with you. I'm always trying to think a couple steps ahead. Sometimes that, when I'm thinking ahead and writing behind, let's see if we can accommodate this. And this is what kind of troubles me right now. So let's say that we want there to be a 2 there. And let's say some other number here. And we want there to be a 4 here and a minus 3 here. I'm saying that the final answer looks something like this. That we're kind of allowing for a generic linear. And we know how to generate signs. We can do that with signs plus cosines. So I'm saying, get rid of the c and the d. But I may come back on that. Could we develop an answer that looks like this by kind of eliminating the c and the d? So we'd have 8x sine of x. We'd have 20x cosine of x. We're not going to be able to come up with something like this from this, are we? Isn't there only going to be one cosine or sine in the final? Not necessarily. We saw that happen when we had one of the terms missing. Like when we had the original y thing and we skipped the first derivative and we went then to second derivative. We saw that when that happened, we had either the sine or the cosine disappear. If we have both, I'm sorry, all present here, if we have y double prime, y prime, and y present, we're probably going to have something that has a sine and a cosine in it. I guess we are getting rid of the a's and b's because the example we did that happened. I don't think we did an example where there were both sines and cosines in them. Yeah, I may recant on getting rid of the c and the d because although there's one letter here, and if you track that back, that would be ax times c times sine of x. That's really a times c. I don't know that c and d are the same. That's where I think this falls apart, that I could factor them out and put the number that I factored out from here and here into here and have it absorbed in this unknown amount of x and this unknown constant. That makes me think that I want to go back to this, that we better allow for a different coefficient for sine of x and cosine of x. If we knew they were both the same, we could get rid of both of them and absorb them in the linear term, but we don't know that. I don't think we could generate something like this without having an unknown value here and here. So I'm going to come back to that. Might be a little redundant, but I'm just afraid that I'm not going to be able to generate something like this, which came from this, if I get rid of the c and the d. That's probably a little too healthy of a chunk for a test because if you start with this, what would be your next step? Find a derivative of that, and probably a little too lengthy for a test. And then find the second derivative, which is going to have probably more terms than this one, so probably a little too healthy. But I do think we need the a and the b and the c and the d. OK, in 7.9, we had applications. These could be applications of homogeneous or applications of non-homogeneous. So once we translate the problem, circuit problem, spring problem, whatever it is, into the form that it becomes a second order differential equation, it's the same stuff we've already looked at today. And these have similar things in their equation. So once we get everything translated, then we just use our solving techniques that we've accumulated to this point. So spring problem, what would be the coefficient of the second derivative term? Comes from force, one way of looking at force, not the Hooke's law, but just kind of force in general. Mass times acceleration is force. So there's mass times acceleration. We can kind of have that on one side of the equal sign. On the other side of the equal sign, we have other things, the restoring force from Hooke's law, which is, if it were on the right side, it'd be negative kx. If we bring it over the left side, it's kx, right? What other things could be on the right side of the equation for a spring problem that once we bring them over to the left side, they're going to appear over here? The damping constant, or damping coefficient, that term. And it says, in theory, we're supposed to assume that the damping term, the c value, is directly proportional to the velocity, right? So on the right side, it's what? Negative c dv over dt. And we bring that to this side, and it's plus, not dv over dt dx over dt, which is velocity. So there's our velocity, our damping coefficient, which is directly proportional to velocity. Here's our restoring force, negative kx. On the right side of the equation, we bring it to the left side. It becomes positive kx. And if we have something on the right side, what is it? Is it some coefficient or something? Some external force, right? So I would say for most of the cases that we really examined, we're going to have a homogeneous equation here. All right, similar terms on the circuit equation. And more than likely, we'll have something non-zero on the right side, because what goes on the right side? The electromotive force, so voltage or it'll tell you what belongs on the right side. And how do these terms go on the circuit? OK, it's 1 over c, right? Q. What is Q in this charge? Right? And what is Q prime? Sometimes they ask us to find, once we're done solving this second order differential equation for Q, they ask us to find something else, current. And that is first derivative, right? So what is L? Where are we going to pick up L? What is the unit? Or what is L representing? Inductance, and that is in entries, right? What is R? Resistance, and it is in ohms. And 1 over c, c is it is capacitance, and it is in ferrets. Right, so we've got Henry's ohms and ferrets here, and electromotive force over here. So they're the same. Once we get them set up, they're really the same thing. Their second order, either homogeneous differential equations or non-homogeneous. In fact, there is a kind of a nice little parallel on page 1178 in the supplement. It's got kind of a nice little parallel going from the spring system to the electric circuit system. But once you get it set up, it's stuff we've already reviewed. All right, back to the book. Really the purpose of sequences at the start of a unit of study on sequences in series, once you're beyond that, it has kind of outlived its purpose. When does a sequence converge? Remember, sequences are lists. Nothing is added. So if we want to see what happens to the sequence as we work our way out to the right to remember their commas, so we're just concerned with that nth term. Is the value of that nth term way out to the right? Is it getting closer and closer and closer to any certain number? If so, the sequence is convergent. Why is that important? Because in order for a series to be convergent, the sequence of partial sums is convergent. So we define convergence of series in terms of convergence of a sequence. So if we generate our own sequence, the sequence being the sum of the first one terms, the sum of the first two terms, the sum of the first three terms, and so on, we're not adding anything together in terms of this list. But as we work our way out here to the right, if we can develop a pattern in this particular sequence of partial sums converges, then the series associated with it converges. Yes? If it's n squared over 3n squared minus 1, we just know it's 1 third. It converges to 1 third because of the coefficient of one third. But if it's a series, then it's not convergent because it doesn't get a zero. Yeah, that's actually pretty good. Let's take a look at that. So let's say we have a sequence, and the nth term of the sequence is, give me what you just gave me. n squared over 3n squared minus 1. So we've got this sequence, a sub 1, when n is 1, what do we get? 1 1⁄2, a sub 2, 4 1⁄11. Let's get one more. A sub 3, OK? I don't know if it's clear if we have anything going on. Are we getting closer and closer to any one number? Well, we can look at the description of the nth term. We're not adding anything together here. We're not adding these terms together, but we're looking at the value of the nth term. It looks like it converges to 1 third, right? 1n squared over 3n squared. You could divide through by n squared, and you could validate this, but we've looked at that shortcut numerous times. So this is a convergent sequence. The value of the nth term gets closer and closer and closer to 1 third. Now we're switching horses to the series, so we're not talking about just values of individual terms, which is the sequence. Now we're adding terms together. So we want the first term added to the second term, added to the third term, and so on. So although the sequence is convergent, what about the series? This could be, I think, an easy test question. What's the first test that you have on this sheet, which I think this will be a handy study, is the nth term. If the limit of the nth term is not 0, haven't we already proved that? The limit of the nth term is clearly not 0. Therefore, this particular series diverges. Because when we add those terms way out to the right, aren't we adding in like a third, practically a third, to another third, to another third? We keep adding a third. Well, there's no way it has a chance of converging if that term that we're adding doesn't eventually disappear. In fact, even all those don't converge. So because, in this case, the limit of n squared over 3n squared minus 1, way out to the right, is equal to 1 third, which is not 0. Therefore, the series is divergent. So that's that first test. The limit of the nth term, as n approaches infinity, is non-zero. It does not have a chance to converge. Therefore, it diverges. Chandler? I was looking at a homework problem last night, and I mean, I've finally got it. But I don't know how. It was the sum of 3 over n. And that is divergent, and I just wanted to know why. OK. 1, 2, infinity? I guess. That'd be an easy test question. OK, it's p-series. If you want to call it that, you could factor the 3 out in front and then just look at the part that has to do with the summation. The 3 is not going to change the decision. 3 times a convergent thing is going to converge 3 times a divergent thing is going to diverge. So what about 1 over n? There could be 3, what I think are 3, really, really good, safe, correct answers. This is harmonic. We went through that whole process of harmonic series, kind of the harmonic, the basic harmonic, diverging. Somebody said p-series. It is a p-series, 1 over n to the p, and what? p is equal to 1. And in fact, when p is what? Less than or equal to 1, it's divergent. That'll work. I like the first one myself. Write down a word and you're done. There is another one. You could take the integral. It's 1 over that thing that's being scrolled, 1 over n, in this case, n's being scrolled from 1 to infinity. And it's not to any power other than 1. So in this case, 1 over n to the first is harmonic. And we could have variations of that. In this case, just a 3. And let me add to that. I don't think we've had this in an example. But if it happened to be an n over an n squared, that eventually is also harmonic in nature. So you had 2n plus 1 over 5n squared minus n minus 11. Basically, you have an n over an n squared. That behaves as a harmonic as well. n over n squared. Can't you simplify it to 1 over n? Yes. Well, you can as long as there's no baggage. So what I'm talking about is let's say you had 3n minus 1. You have baggage. And over here, you have 5n squared plus 2n minus 3. I mean, the dominant terms in the numerators 3n, the dominant term in the denominators 5n squared, that's harmonic right there based on the dominant terms. Clearly, out to the right, minus 1 doesn't have much meaning. And this gets overtaken, the plus 2n minus 3, those become secondary to this term, which becomes dominant. So that's also going to be harmonic in nature. But this is the type we're talking about. And any of those will work. This is a divert. This is an integral, which the limit does not exist. Therefore, the conclusion is the same. Let's finish with this one. Now, lots of things can go through your mind. I'm just trying to kind of hit a couple of these all at the same time. If it were just 7 over 3 to the n, you could bring the 7 out in front. And if we're left with a 1 over 3 to the n, that's a pretty good, that's infinite geometric, right? That's the same thing as 1 third to the n. So that'd be geometric. We don't have that, but I'm just trying those. There's our thought process that might get us there on another problem. What's the minus 2 due to the denominator, which makes the value of the fraction itself do what? If the fraction were getting smaller, wouldn't we have to make the denominator larger, right? We're making the denominator smaller, which makes the value of the fraction bigger. So yes, that's important to know that this is an infinite geometric series. The ratio is 1 third. The absolute value of the ratio is less than 1. Converges. If this one were even smaller than this one, then we would just use the what? Comparison test. It's not smaller. It's a little bit larger. So if you kind of think the comparison test is going to work, but it fails right at the end, what's the next bigger net that's going to capture convergence or divergence, even though comparison test? So this one fails. It'd be different if that were plus 2. Then we could use the comparison test. It's minus 2. We think it's a lot like this. We think it's going to converge. But we would use the limit comparison test. And how does that go? We want to see way out to the right. It doesn't matter which one goes in the numerator. Let's say the one in question. Didn't we determine that that one, which is a whole lot like this one, converges? You could use 1 over 3 to the end. It's fine. I'm just trying to make it as close to this one as I can. So I know this is convergent. If I get an answer, the one in question also converges, right? If the one that we know something about does something, converges or diverges, we get an answer to the limit problem. The one in question also does the same thing. So how's that limit go on this problem? Actually, we don't need these, right? So we get rid of the 7s. Divide everything by 3 to the end. That'll work. So that divided by 3 to the end. Now what happens as n approaches infinity? What happens to this term? That's 1. This is also 1. And what happens to this guy right here? 2 over 3 to the end as n approaches infinity. Zero? So we get a finite answer. If this one converges, which it does, this one must also converge. So we would just write a concluding statement. Well, now there's one that we haven't come to yet. So if you know some other tests like the ratio test. I remember you saying one of them because equal to 1 it fails. Yeah, and I think I may have mentioned the fact that we're going to encounter one later in this chapter that when the limit is 1, we don't get a conclusion. This is not it. It will happen later in the ratio test. So if we get an answer, a finite limit, the one in question does the same thing that the other one does that we know something about. Limit comparison test. And that's really the only one on the back of this sheet that we have done. So draw the line there. We have not done alternating series for this test. All right, see you tomorrow. So you don't have a sample test for this? I think there's one posted online. On your website? Yeah, for the distance add classes.