 We are at lecture 42, Math 241. We should finish back in the textbook, chapter 8, section 3 today on error estimates. When we add up a finite number of terms in the series, how can we tell are we reasonably close to the infinite sum or how far away might we be? So we'll do that today. And that shouldn't take the entire time. So I understand there are some web assigned questions. So let's at least get them to the point where you feel comfortable you can take them from there. What was that first one, Katie? n squared plus 4n plus 3 from 1 to n 30. n squared plus 4n plus 3 from, and this was from section 8.2. So I don't think they want you to use the kind of the remainder approach, the approximation approach that we're going to add today. So something from section 8.2, which we had infinite geometric series and telescoping series. What else was in 8.2? I don't know. We'll go back when we review. I know there are two of the main things in 8.2. This is not geometric. You're not multiplying by some r value as you progress from one term to the next. So my guess is that it's telescoping. It would probably be a little more obvious that it was of this type. If the denominator were factored, I think it's factorable. Let's stop that line of reasoning. Let me go back to the original. Can you tell from the original can we make a decision about convergence or divergence? OK, it's convergent. OK. So we could compare it to or use a limit comparison test to 1 over n squared or 2 over n squared, right? Each term of this compared to this, which is really this, about that guy right there that's framed. Convergent because, because I said it is, that's not good enough. It's a p-series, 1 over n to the p. Let's keep going. Not all p-series converge, so we better tell it somehow to justify that it's in the area where those ones that converge do, in fact, converge. p is equal to 2, therefore p is greater than 1. So it converges. So if that converges by the comparison test, this one, which is even smaller term by term, also converges. So the decision is not the hard part. The question asks, to what value does it converge? And we can get that same convergence decision without doing this off to the side. If we decompose it into partial fractions, and it is telescoping, and we see what terms remain and which ones knock each other out, then we can say it does converge. In fact, it converges to something, 7 over 19. What are the factors? OK, n plus 3, n plus 1. So off to the side, we decompose this thing. The n plus 1 term first, just as we write the telescoping, I think we'll see why that doesn't need to be first, but it might be a little better looking when it's first. Common denominator. So the denominators are equal. Excuse me. The denominators are equal. We need to equate the numerators. So let's group the n things together and the non-n things together. And then equate coefficients, left side and right side. How many n do we have? And what is left that doesn't have an n? So what are the equations that result to help us solve for a and b? Good. A plus b is 0 because we don't have any n on the left side. And we have a plus b of them on the right side. And it has to be 2 because that's the constant on this side must be equal to the constant on this side. So a is equal to negative b. And we can sub that n over here. Everywhere we see an a, we can put in a negative b. So negative 3b plus b is negative 2b is 2. So b is negative 1. And if a is the negative of b, then a is positive 1. So back to our original decomposition. So our original problem comes a, which is 1 over n plus 1. b is negative 1. So I'll just subtract the whole thing. At least now it has that telescoping kind of look. I don't know. Do we need to continue? Yes. So let's write out some terms. See what happens. When n is 1, what's the first term? When n is 2, what do we get? 1 third minus 1 fifth. Not looking too good so far, is it? So we're hoping to generate a term that's going to knock out the negative 1 fourth and then eventually knock out the negative 1 fifth. I think that's about to happen, isn't it? When n is equal to 3, what do we get? 4 minus 6. Minus? 1 sixth. 1 sixth minus? 1 seventh. 1 seventh, and so on. So let's see what terms will never drop out in which terms we think will eventually be gone based on the fact that we're going to generate that term. We're ever going to be able to knock out the 1 half. Now we're stuck with that. How about minus a fourth? OK, that one's gone. We're ever going to be able to knock out 1 third. We're stuck with that. Minus a fifth. Minus a sixth. We'll knock out with the next term that's here, right? Minus a seventh. Will we ever generate a positive 1 seventh? We will. So all the rest of the terms are going to eventually knock one another out. It's going to go on indefinitely. Even if it stopped after 100,000 terms, the value of these terms is so small that they're negligible. So it looks like we have a 1 half plus 1 third, which is, I don't want to finish the problem for you, but it is a certain number, right? So I would jump start the problem. So it is a certain number. Therefore, we could make a decision that it converges, right? Whatever that number is, it's kind of a mystery at this point. But it is a certain number. And because they all add up to a certain number, then we can say this converges. So you can make your decision after the fact. If, in fact, you know it's telescoping, and certain terms are going to drop out, and certain terms are going to remain. Is that all right for that one? To get it to a point where everybody can take it from there? OK, next one, Katie. 4 to the n plus 3 to the n over 12. So how do you read that if you're telling somebody what the problem is? How do you read that? I have no idea. Clearly. Yes, what do you say that is? The summation. The summation, or the sum, from n equals 1 to infinity of 4 to the n plus 3 to the n all over 12 to the n. So summation or sum tells us that we're going to use that sigma. And obviously, we've got an infinite amount of terms. OK, that was my question. So anybody that had success with this problem, that looks like a pretty good prospect to separate them. We've got a sum. So you can separate them and keep them all in the same sigma symbol, in the same argument here, or you can write two of them separately. Helpful? I think it's helpful. How is that helpful? Can you just change it to 1 third to the n? OK. They're both to the n. So it's the same thing as 4 twelfths to the n, which is the same thing as 1 third to the n. So that's helpful. I mean, I guess, what is that? That's geometric, so we can make a decision on that. Yeah, so they're both to the n, 3 to the n over 12 to the n. It's 3 over 12 to the n, which is 1 fourth. Infinite geometric, infinite geometric. So you can make a decision about convergence, and you can also find the sum if they, in fact, converge. Is that all right? Everybody feel confident from there? Yes, Nicole? When you're finding the sum, do you find the sum of each term individually? Yes, here. So they're both going to converge. So you can find the sum here, first term over 1 minus the ratio, right here. Yeah, and then find the sum separately here. So we've got those two added together. We'll just add those together. That's not an integral sign. That's an s, or something. Is that all right? And what was the other one? Are we hitting the ones that you had too, Nicole? Yeah, that was for the 2, 9. OK. The summation from n equals. Hey, all right, let's see. That's good. n equals 0 to infinity of. n equals 0. Yes. OK. 2 to the n times x plus 1 to the n. Nothing in the denominator? No. What section of the book is this from? 8, 2. All these are 8, 2. And then the question is for values of x. I don't know. I've got something that I want to try. I think it's going to get us somewhere. Alex? I'd plug in like 0 and 1 and divide those two to get r. OK. So your goal is seeing if this is geometric. Yeah, because when you plug in 0, they both go to 1. So it's just 1. And then when you plug in 1. For n. Yeah, for n, for n. And then you can do that. It's like what you get when you plug in 1. You divide that by 1. So it basically gets your r. All right. That sound good? My goal was eventually headed the same way. You actually plugged in numbers and saw terms and it looked like they were being multiplied by some common factor as you go. This is similar to the previous problem in the sense that we've got 2 to the n and we've also got x plus 1 to the n. So we could put those together, which I think is going to accomplish the same thing that Alex said by writing them out. So any series that has something in brackets raised to higher powers as you progress doesn't that thing in brackets become the r, right? So the first term, this is to the 0. The next time we encounter a term that we're going to add to it, it's to the 1. The next time it's to the 2. So each time we're going to multiply by 2 times the quantity x plus 1. Does that work? And that's what you have, right? So it looks like the ratio is 2 times the quantity x plus 1. And in order for this to converge, then the absolute value of the ratio, it is an infinite geometric series. If it's convergent, then the absolute value of the ratio is less than 1. So the absolute value of the ratio, there's the ratio, is less than 1. So it's going to be conditional, right? Certain values of x will allow this series to converge. We will revisit things like this later. In fact, we'll come up with these on our own. And when we develop some power series, this ends up being a power series that has not only 2 to higher powers, but also x plus 1 to higher powers. How do you solve that? Well, I didn't work like, I just did like the sum equals a over 1 minus r. And that got the right answer on it. OK. So you went ahead to this. You went ahead and said, OK, I'm going to assume that this ratio is in this area. So you went ahead to the sum. So how's the question phrased again? Good to put it. Put the sum in terms. Sum on a series for values of x. OK. So I guess they're assuming that it converges. These would be the conditions under which it would converge. So the sum would be the first term. So if you generated the first term, it is 1 over 1 minus the ratio. So there's the sum if you multiply. It's not, you don't multiply the r times 0. So it's 2x plus 1 to the 0, right? So to the 0 is 1. So the first term is 1. We don't use that very often. But occasionally, that comes into play. So first term is 1 over 1 minus the ratio. There's the ratio. You can, I don't know, simplify that. But if we had to determine the values for x that caused this to converge, which I guess they're assuming, wouldn't you, when you have an absolute value inequality, don't you do that? Does that look familiar? To solving an absolute value inequality? I don't think that's part of this problem. But had they asked the question a little differently, they might say, for what values of x does this particular series converge, we would need to solve that. Not part of this problem. This is the solution, apparently, to that problem. You set that equal to 0, then, to get your x value? Yes. I think they want this, don't they? What is the sum in terms of x? So you could simplify the denominator a little bit. But that's going to get you the answer. All right, let's finish up section 8-3 today. I used this diagram yesterday, but I wrote all over it. So let's start with this. We're trying to get an upper bound and a lower bound for the remainder for the error associated with adding a finite number of terms together in the infinite series. So with an uncluttered diagram, I think I can get us to the point where we left off yesterday. So here's the area under the curve from n all the way out to infinity would be this entire area. That's this guy right here. That's more than the terms we're missing. We're missing the n plus first term, which is right there. The n plus second term, which is right there. So we've got more area under the curve than we're actually missing from the terms that we're leaving out by stopping it in. So the remainder or the error, what are we lacking from by stopping this thing at n and not letting it go all the way to infinity? It's not this value, but it's bounded above by that value. This is more than the error. The error is the sum of the blocks. We've got the entire area under the curve right here. So that's our upper bound. We're really concerned with our error at the worst. What's our error at the worst? The other side of that is, suppose instead of starting it in like we did in the previous diagram, suppose we started in plus 1, well, we're still trying to find out a lower bound this time for the error associated or the remainder are associated with adding n terms together and not adding them all the way out to infinity. Notice this starts at n plus 1. So starting at n plus 1 and taking the area under the same curve, the same f of x curve, obviously it's got to be decreasing if it's going to converge. We don't quite capture all the terms that we're missing. We only have this area from n plus 1 to infinity. And the terms that we're actually missing, the blocks, there's the n plus first term. There's the n plus second term. They're actually larger than that. So we didn't quite capture all them with the area under the curve. So this becomes our lower bound for the error estimate. And up here, this becomes our upper bound. So up here, we got too much. We know that the actual error is less than that. Down here, we didn't get quite enough. So the actual error is a little bit more than that. So the page that we finished with yesterday, if I can find it, let's just use this one, which hasn't been written on. Here's our error, our remainder, associated with stopping our sum at n. We want to know the upper bound. We integrate, find the area under the curve from n, which is where we stopped all the way out to infinity. Our error is actually less than that. But we can't tell how much less. The lower bound is starting at n plus 1. So if we stop with the first 10 terms, which we will in our first example, n plus 1 is going to be 11. This is going to be a lesser area going from 11 to infinity than this is by going from 10 to infinity. But this is our lower bound for this remainder or this error associated with that. Now, this remainder, if we will take this remainder in here and add the sum of n terms to it, I can't do that to the thing in the middle of the inequality without also adding it to this term and also adding it to this term. So in all three terms of this inequality, add in s sub n, and the result is this. Where is the actual sum? Isn't this the actual sum? The remainder or error added into what we have come up with by adding the first n terms together, those two are actually equal to the exact sum. So where is the exact sum? What we got in the first 10 terms or first 42 terms, add in the lower bound for the error estimate. And it's somewhere between there and this sum of the first 10 or whatever terms, add in the upper bound for the error estimate. So it's somewhere in there. That'll give us a little better idea. And if you want to, you could take this value right here and this value right here and average them. That's probably even better than what we got by adding the first 10 terms or the first 20 terms together. You don't have to do that, but you could. So this gives us an upper bound and a lower bound for the actual exact sum all the way out to infinity. Our approximation plus the upper bound, our approximation plus the lower bound, and we're in business. What's that look like on a problem? So let's approximate. Is that going to converge? Why? The denominator is the same as the denominator. What kind of series is it? P series, one over n to the p. p is 3, which is greater than 1, therefore it's convergent. So it is a convergent series, so we should expect to be able to find the sum all the way out to infinity. So here's what we're really going to do. We're going to say the sum of this series, we're not going to be able to find it all the way out to infinity. We don't have any shortcut for this. It's not telescoping. It's not infinite geometric. So let's approximate it and then see how bad our approximation is. So we're not going to go all the way to infinity. We're going to stop at n equals 10. So I want to plus sign. I want to equal sign. Excuse me. So what's the first term? And we're going to write them all down and add them all up. And we have, in fact, a last term here because we're not letting it go to infinity. We're going to stop it at 10 cubed, right? 10 cubed, 1,000, right? 3 zeros. Now we could add these 10 terms up. It's not like a ridiculous amount. We're not adding 628 terms. We're adding 10 terms together. We can add them up on our calculator. If you add those together, you get approximately this value. How close is that value potentially? How far away, at the very worst, is that from the actual sum? So here's our, we're looking for r sub n now. Well, according to this, if I want an upper bound for the remainder or error, I just have to integrate the function from n, 10 in this problem, to infinity. And because we're going to use this value in a second problem, let's not use n equal 10 right now. Let's use just n. And we'll get an answer in terms of n, and then we can use it again. So the function associated with the description of the nth term, 1 over n cubed, is going to be 1 over x cubed. That's what we want to integrate. But in general, because we're going to use this again, let's leave this open and say we don't know what n is at this time. We know n in this problem is 10. So in improper integral, the bad piece, infinity, we delay that till the end of the problem. So this is technically not this problem, although it's going to help us on this problem and another problem. What's the integral of x to the negative third? Negative. Negative 1 half? OK, x to the negative 2, which we'll just go ahead and put the x squared down here. So we'll put in a. What's the nice thing about that term? Goes to 0. It's going to disappear as a approaches infinity. So let's go ahead and do that as a gets larger and larger and larger. This thing gets closer and closer to 0. So we've got minus the negative 1, which is 1 over 2 n squared. So that makes this problem a little bit easier that we've already integrated it. All we have to do is plug in n for this problem. So the remainder associated with the sum of the first 10 terms is less than or equal to. I'm going to go ahead and write this down. We've already written this down. We integrated that already and we got 1 over 2 n squared. The n is 10. So we added the first 10 terms together of this convergent series. We got some decimal number 1.197532. How far off might that be from the actual sum all the way out to infinity? This would be an upper bound for the error, which is what? 1 over 200? What is that as a decimal? 2 0's and a 5? That's not too bad. So we didn't miss much by stopping at 10. We're only off by this much. Suppose we wanted to do better than that. That's the second problem. Let's suppose. So we're basically, we've got an upper bound for the error there. The error is actually smaller than that, but that's the error at its worst, so to speak. Nicole? And for the lower bound, do you just subtract that much from what we got? Yeah, if we wanted the other end of this, we could go from 11 to infinity, which is that. Somebody tell me what that is. 11 squared is 121. 1 over 242. Is that right? So the actual error is somewhere in between there. But most of the time, we kind of want to know what is the error at its worst. At the very worst, the error is 0.005. The error is larger than that. We don't really know how much larger. So that doesn't really help us that much. This is the one that helps us. We kind of want to know the error at its worst. All right, new problem, same situation. How many terms are required to make sure that the sum is accurate to within some smaller decimal? That's exactly right. So what is n for the error? Put another 0 in there. This is where this comes in handy, that we already integrated this and not just evaluated it from 10 to infinity, but we got it in terms of n. So don't we want the, let's say this is the error at its worst. Is that what we want? That's not what we want. The other way around, right? We want the error. Here's the error. Didn't seem right when I was writing it to be less than that value. There we go. Thank you for the nice facial expressions there, because it didn't work for me either. All right, we need to solve for n. We want to know how many terms we have to add together. We know the sum of the first 10 is this close, potentially, to the actual value. Now we want to do a little bit better. How many terms do we have to add together? Multiply both sides by 2. Is that right? Now we want to flip both sides. What happens to the inequality when you flip both sides? It reverses, right? And what is the reciprocal of that? Reciprocal of 1 1,000 is 1,000, right? So n probably not going to be a nice whole number. Square root of 1,031 something. So how do we answer our question? How many terms do we have to add together to get within the desired level of accuracy? 32. Again, not awful. That's very, very accurate to only have to add 32 terms together. But I'd personally, I'd rather just add 10 together and then have a little less accuracy. So let's see if we can get a little closure to this problem, and then we'll be done with section 8.3. So we have the remainder associated with the sum of 10 terms. We got a what, 0.005 here and a 0.0041. Let's add in the sum of the first 10 terms. The sum of the first 10 plus the remainder associated with the sum of the first 10 is in fact the entire sum all the way out to infinity. What was the sum of the first 10? It was 1.197532. So if we add that value to it, we now have an upper bound for the infinite sum. Somebody checked my arithmetic here. 202532, is that correct? That's what I have written down. So I don't know how many decimal places of accuracy I used, but if we take what we accumulated for the sum of the first 10 and add it to the upper and lower bound, the actual sum all the way to infinity should be somewhere in here. That's not a whole lot more new information, to be honest with you. We already knew what kind of accuracy we had on the first problem. Somewhere less than 0.005, pretty good level of accuracy. I don't think that adds that much to what we knew originally. But it does give us kind of a lower bound and an upper bound for the actual series all the way to infinity. So that's why when Katie asked questions before class about some series problems, I wanted to know what section they came from. Because if they came from section 83, then they might be wanting us to do something like this, approximate the sum all the way out to infinity. But in fact, they were telescoping geometric and the other one was geometric. So question on that. In the middle, when you were solving for n, did we go to 0.005? That was a question. That was part two of the question. I think this is actually in your book. Part two says, how many terms are required to ensure that the sum is accurate to within 0.005? So it's kind of a new problem, new question, or a new part anyway to the original question. So if you have a desired level of accuracy, we might want to know how many terms we have to add together to get that. All right, well, let me just briefly, and we'll talk more about these topics tomorrow. The test, which we'll review for kind of officially tomorrow, begins with the supplement. So bring together, bring tomorrow, any questions that you have web assigned questions that we didn't get to today, any clarifications of any topics. I believe we start with the supplement. So this is the homogeneous, the three cases. We'll talk about those tomorrow. Two distinct real roots, double root, and complex roots. This was non-homogeneous. We'll have three categories on the right side of the equation, polynomials, exponentials, and sine, or cosine, or sine plus cosine. And these are applications of those, homogeneous and non-homogeneous. Back to the book, this was on sequences. So we're not talking about adding any terms together. We're talking about what happens to the value of the nth term as we go way out to the right. We made the leap from sequences to sums or series. The connection is, in order for a series to converge, the sequence of partial sums has to converge. So that's why it's important to look at what causes sequences to converge, because that's critical for series converging. And I think we had, I know we had infinite geometric series here, and also telescoping. And we'll have to do a more thorough job tomorrow if there's something else in there. We'll clean that up, too. And then what we finished up today. So that doesn't look like a lot, but it is a lot. There's three cases here, three possibilities for the particular solution here, and then some applications here, springs, circuits, and then what we've done thus far in sequences and series. So I think we really saw the second order differential equation was given. It would have to be, to take it from start to finish, it would have to be one that has some terms that kind of disappear, like the A or the B is 0, and they drop out. So it would have to be like a seven or eight minute problem tops, and we didn't really encounter very many of those. So it's possible. I'll try to give you a better answer tomorrow. But I'm not going to throw a 20 minute problem at you. All right, so we'll review for this test tomorrow.