Calculus, all content (2017 edition)
Suppose we want to find the sum of a convergent series, and can't do it directly. We can take a partial sum, but how do we know how far we are from the actual sum? We can use improper integrals for that!
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- at4:18,he says that the underestimate consists of the rectangles that are contained within the graph.So,shouldn't Rk be GREATER than the under estimate?Because the area under the graph is greater than the area of the rectangles contained in the graph.(11 votes)
- Sal means that Rk is an underestimate of an integral of the function. In other words Rk is less than the area under some part of the curve. But Rk is always the sum of the rectangles. It's greater than the area under one part of the curve, but less than the area under a different part of the curve.(4 votes)
- Since on the over-estimate we're starting the area/integral from k+1, shouldn't the Sk be S(k+1)?(6 votes)
- At3:46, I am confused. If he chose k+1 as the height, why does the rectangle fill in back to 'k'? Isn't it suppose to go to 'k+2', thus making it an over estimation?(6 votes)
- You're right, if he made the rectangle width start at k+1, and end at k+2 instead, he would have an overestimation of the integral from (k+2) to infinity. However at3:46, the time you mentioned, he was show trying to show an underestimation, so he ended the rectangle width at k. At5:26he does exactly what you mention and creates an overestimation instead.(1 vote)
- 4:45Sal can say this is an upper bound because, based on the sketch he drew, the particular riemann trapezoidal estimation was an underestimate. But this might not always be true, right? The estimate could sometimes be an overestimate. So I don't understand what to do from there:p(1 vote)
- That is exactly why he uses the rectangles (not trapezoids) relative to both k and k+1: in one direction the smooth curve is an overestimate relative to the rectangles (which are the true value), and in the other direction the smooth curve of the integral is an underestimate relative to the rectangles (which are the true value). Once he has the underestimate AND the overestimate, he knows he has the true value of the infinite sum trapped in the middle like a firefly between your hands. :)(7 votes)
- at5:30, why does he have to move over one interval to the right to make the rectangles? Also, does the rectangle from the interval from 0 to k represent S_k?(4 votes)
- Can someone explain why at1:52Sal shows the series as the sum of the partial sum of the function, with the first one being n=1 to K, but the second one is n=k+1 to infinity? I'm more confused on the second part of that, but also how he re-wrote the series as a sum of a series and a partial sum(2 votes)
- At approx5:25, Sal begins talking about the upper bound. Instead of using k+1 as the beginning of the bounding series, why not just take the left hand sum of the series? I'm having a hard time seeing how the series is an over estimate if there is a giant hole in this sum from k to k+1. Is there some difficulty in just taking the left hand sum?(3 votes)
- It seems our "overestimation" based on R_k+1 @6:58is smaller than our "underestimation" based on R_k @4:58? How can this be?
Well, @6:58we are over estimating the smaller integral starting at k+1. And @4:58we are underestimating the larger integral starting at k. But the way it is explained here it seems we are taking over and under estimates of the same integral by shifting the Riemann approximation.(1 vote)
- Is this topic (Series Convergence and Estimation) supposed to come 3rd in the Integral Calculus mission after Sequences and Series Intro & Mission Foundations?
I feel like it's assumed I know a lot of the subject matter that's a basis for being discussed in this topic but it's more or less all completely new to me :((2 votes)
- The mission is set up that way, you can leave feedback or press "I haven't learnt this yet" and you will proceed with the next question. It happens to me in many of the cases aswell, so don't feel discouraged because a topic seems unfamiliar. If you want to have fewer situations when unknown topics appear in the missions, you can do more practice in familiar topics. This works great for me when learning a new subject.(2 votes)
- When the series in divided into two parts, would the upper bound of the first part be K?(1 vote)
- Yes, the partial sum runs from n=1 to k. He is using "upper bound" in a different sense, though, so be careful of the distinction.(2 votes)
- I do not understand why Rk is less than the intergral of f(x) from k to intinity.4:30
thank you for your help(1 vote)
- Rk is the sum of f(n) from n = k +1 to ∞. He draws the function f(x) on the graph, then at each integer, x= k+1, k+2, k+3, ..., he draws a rectangle with height f(k+1), f(k+2), ..., and width 1, so that their area is equal to their height (since their base is 1). But you can see that every rectangle is under the curve because the curve is decreasing, so that the sum of the areas of the rectangles (which is exactly Rk) must be less than the total area under the curve (which is the integral of f(x) from k to ∞.).(2 votes)
- [Voiceover] So let's say S is the value that this infinite series converges to. We're going to assume that this series actually converges. And the definition of the series, each term is going to be a function of N. We're going to assume that this is the same type of series that we looked at when we looked at the integral test, or namely that this function is a continuous positive decreasing function over the interval that we care about. So it is continuous, it is positive, and it is decreasing. This isn't saying continuous plus decreasing. Let me just write it that way. Continuous positive and it is decreasing. So it could be a function that looks something like this. Now, my goal of this video is to see if we can estimate a range around S. So this is going to be very useful, because we've seen some infinite series where we are able to figure out what exactly does it converge to. But you could imagine there are many, many more where we're not going to be able to figure out exactly what it converges to. And instead we're going to have to do it using a computer or by hand. And in those cases, it's good to know how good our estimate is. And we also want as good of an estimate as possible with as little computation as possible. So let's think about how we can do that. Well, the way to tackle it, you could imagine, is let's split this up, this infinite sum, let's split it up into the sum of a finite sum. So let's say the first k terms. So n equals one to k of f of n. So this is very computable. If k is low enough and if f a simple enough function, you could probably do this by hand. But you could definitely do this with a computer. And then it's going to be that plus an infinite. Plus another infinite series. But now you're going to pick up at k plus one, the k plus one term. And you're going to go to infinity of f of n. So if we could put some bounds on this, then that'll allow us to put some bounds on this right over here. Because this is just the sum of the partial sum of the partial sum of the first k terms and the remainder that we get after to get us to the actual value. So you could see kind of what's left over after we take that partial sum. And this is easier to write for me than this right over here. So they key is, can we come up with some bounds for this? And to do that I'm going to go to this graph and use some of the same arguments we used, or the same conceptual ideas we used, for the integration test. There's two ways to conceptualize what the sum represents relative to this graph. As we'll see, it can represent an over estimate of the area between sum, x value, and infinity. And it could represent the underestimate of a different region. So let's look at that. So let's first think about the underestimate. Let's think about the underestimate. So if this right over here, let's say that this right over here is k. Actually, let me do it in a color that's... Let me do it in that yellow color. So let's say that this right over here is k. This is k plus one. Let's do this is k plus two. K plus 2, k plus 3, on and on and on. So one way to conceptualize this sum right over here is it could be the sum of the following rectangles. So the first term is this area, the area of this first rectangle. Because this area's height, or this rectangle's height, is f of k plus one. And its width is one, so f of k plus one times one. Its area is just going to be f of k plus one, which is exactly this first term right over here, where n is k plus one. And then the second term, by that same argument, could represent the area of this rectangle. The third term could represent the area of the rectangle. And we could just keep going on and on and on. So what are the sums of these areas of these rectangles? What are the sums of these terms representing? Well, you could view it as an estimate. You could view it as an estimate of the area under the curve between x equals k and x equals infinity. But it's going to be an underestimate. Notice these are all completely contained in that area. So one way to think about it is that our R sub k is less than or equal to its underestimate for the area between x equals k and infinity of f of x dx. So that essentially puts an upper bound on us. And this is already interesting, because now we can already say if S -- so we know that S is equal to this. Now if this is less than this thing right over here, we can say that S is going to be less than or equal to our partial sum plus this thing. Plus the improper integral from k to infinity of f of x dx. Notice if this is equal to this and now since this is less than this, this must be less than what we have on the right hand side. So just like that, if we're able to compute these two things and we are often able to compute these two things, we're able to put an upper bound on our actual sum. Now what about placing a lower bound on it? Well, we could conceptualize the same sum, the same R sub k. Instead we can conceptualize it this way, where the first term here represents not this rectangle, but it represents this rectangle. Notice it has the same height, but it's just shifted over one to the right. The second term represents this rectangle. This third term represents this rectangle. Why does that make sense? Well, the area of this first rectangle, this is going to be its height, which is f of k plus one times its width, which is just one, so it's going to be f of k plus one. So the area here is the first term, the area here is the second term, area here is the third term, the area here is the fourth term. One way to think about it, we just shifted all of those yellow rectangles one to the right. But now this is approximating a different region. This is approximating the area under the curve not from k to infinity, but from k plus one to infinity. And instead of being an underestimate, it's an overestimate. Now the curve is contained within the rectangles. So we could say that R sub k in this context, when we conceptualize it this way, is going to be greater than or equal to the improper integral not from k, but from k plus one to infinity f of x dx. And what does that allow us to do? Well, this places a lower bound on this, which will also place a lower bound on this. If this is greater than that, then this is going to be greater than this replaced with the improper integral. So we can write that. So S is going to be greater than or equal to S sub k plus the improper integral from k plus one to infinity f of x dx. Now, you might be saying, "Hey, Sal, this looks all "crazy, you have all this abstract notation here, you've "introduced an integral sign, this seems really daunting." But as we'll see in the next few videos, these are actually sometimes fairly straightforward to compute. This can be very straightforward to compute if our k is not too large. If it's even large, a computer can do it. And then these we can actually often compute sometimes numerically, but even more frequently, and that kind of defeats the purpose, but we can also compute them using our analytic tools, using, I guess you could say, the power of calculus. And so what this allows us to do is put a pretty neat band around what our actual value that we converge to is. And as we'll see, the higher our k, the better an estimate we get and the tighter arrange of our confidence for that estimate. And another way to write these two inequalities is to write a compound inequality that S is going to be less than or equal to this business. So copy and paste. And S is going to be greater than or equal to this business. So we could say this business is going to be less than or equal to S. So let me just copy and paste that. So copy and paste. Oops, that's not what I wanted to do. So copy and paste. Then we could write it just like that. And so the next series of videos will actually apply this and we'll see that it's pretty straightforward. It looks a little daunting right now.