AP®︎ Calculus BC (2017 edition)
Worked example: Series estimation with integrals
See how we can use improper integrals to approximate the infinite sum of 1/n².
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- 3:10, why did Sal have to introduce a new variable
b → ∞to take the limit?(16 votes)
- Sal made a mistake right there. He was correct to introduce a new variable, but what he did next was incorrect, and made it confusing as to why he introduced the variable.
The problem we're dealing with at that point in the video is evaluating an integral from 6 to infinity. That's an improper integral: the fundamental theorem of calculus tells us how to evaluate the integral from 6 to some other finite number (assuming there are no "blow-up" problems in between), but it doesn't tell us how to evaluate an integral that goes to infinity. We solve the problem by dividing it into two steps. First, we find the integral from 6 to some unspecified finite upper bound, which we'll call "b" in this case. And second, once we see what the integral would be with an upper bound of b, we find what the limit would be as b goes to infinity.
Sal's mistake here was that immediately after introducing the variable b, he again writes the integral with an upper bound of infinity, when he intended to write it with an upper bound of b for the reason explained in the previous paragraph. If this still doesn't make sense, you may want to go back and review the videos on improper integrals, which begin here:
- I know this is not too relevant, but didn't it bother anyone else that he forgot to change "n" to "x" when doing the integrals?
the integral of 1/n^2 dx is equal to x/n^2.
I know it's a very minor mistake, but it could be misleading.(20 votes)
- This is not just a mistake, this is a BIG ( and in some costly ) mistake. I'm reporting this mistake.(8 votes)
- Is there a series of videos about integrals with bounds at infinity as used in this video? I'm having trouble finding any(3 votes)
- The way you do such integrals is:
∫ f(x) over n to ∞ = lim c→∞ ∫ f(x) over n to c.
Then you do the integral in the usual way. Then you take the limit (which may or may not exist).
These are called improper integrals and Khan Academy does have videos on them.
- As k increases your estimate becomes more accurate. Could one find the exact sum by taking the limit as k→∞?(3 votes)
- Well, Euler proved that the sum from n=1 -> infinity of 1/n^2 is actually equal to (pi^2)/6
I'm not sure why that is, but you can definitely check out his proof of it on the internet :)(4 votes)
- Why do we have to split the sum into Sk and Sr in order to take the estimate? Why cant we start from n=1?(2 votes)
- We want to break the unknowable sum into a piece that is clear and known and a piece that can be estimated. Then we can use the estimated piece to set a bounds on all the values that the sum can possibly be. You usually cannot just start from n= 1 and keep adding terms to infinity.
known sum of first 4 terms+
estimate of the terms from 5 to ∞fair estimate
known sum of first 20 terms+
estimate of the terms from 21 to ∞better estimate
known sum of first 100 terms+
estimate of the terms from 101 to ∞even better(3 votes)
- At2:52Sal sets up an integral 1/n^2 dx. Shouldn't he be using f(x) not f(n)? His integral is with respect to x yet he integrates with respect to n.(2 votes)
- Yes, I actually reported this error many moons ago, but still have seen no correction made to the video. If you click "Report a Mistake" on the right side of the comment section you can view and submit errors you find in videos.(2 votes)
- At2:08, near the bottom, Sal writes that the summation is approximately __, but isn't it exactly equal to __ instead?(2 votes)
- This is because Sal wrote an approximate value from the calculator (he used 4 significant figures), and not an exact value.(1 vote)
- At about3:26Sal writes that the integrand is -n^-1. How'd he get the negative out in front of the n? I'm sure this is a simple question, I'm just blanking on this subject. I understand why there's a negative on the exponent and why it went from -2 to -1.(2 votes)
- So when integrating the format will be x^(n + 1)/(n + 1) which in this leads to n^-2 becoming n^(-2 + 1)/(-2 + 1) or n^-1/-1 which is simply -n^-1. So basically make sure when you're integrating this problem you don't forget about the denominator part of the problem!(1 vote)
- I'm confused because if you chose a different value of k, say 125 for example, when you apply the same integral and inequality don't you get a different number for the convergence?(2 votes)
- Not quite, although the series will continue to get larger with higher values of k, the integral you take on both the upper and lower bounds continues to get smaller, because of this you will still get the same bounds he gave. I tried it with k=10 and k=50 just to test it, the results are: for k=10 for the series you get 1.54 + (1/11) < S < 1.54 + (1/10) which is 1.630 < S < 1.664 and for k=50 you get 1.625 + 1/51 < s < 1.625 + 1/50 which is equal to 1.645 < S < 1.645.(1 vote)
- Sal says @1:23, "lets say we don't know the exact value" as if we should know how to take the exact value. Is the exact evaluation fo this expression shown on Khan academy(1 vote)
- Maybe he thought you should just know it - because it's reasonably famous - but I doubt he'd expect you to know the derivation. It's quite a difficult problem and was first solved by Euler. The exact value is π²/6
The problem (of determining the series limit) is known as the Basel problem. As far as I can tell there are no KA videos on it, but here's a link to the Wikipedia entry:
- [Voiceover] In the last video we saw that if we have an infinite series where each term is a function of N, and the function itself is a continuous, positive, decreasing function over the intervals we care about, and we assume that this series converges, then we can estimate that series, we can estimate the value that it converges to, using a finite number of terms and some integrals, and the way that we establish these integrals is because when we split up the sum into finite sum and then another infinite series, we're about to conceptualize that infinite series in two different ways. It could be an underestimate of this integral right over here, or it could be an overestimate of this integral right over here, and because of that it allows us to not only estimate S but to put bounds on it. If we can evaluate the left hand side here and the right hand side here for a given infinite series, then we can establish our bounds on S, so let's see how good these are by applying it to a particular infinite series. So let's say that the infinite series we want to apply it to, let me do it in the yellow color, so it's going to be, let's say, we go from N equals one to infinity of one over N squared, and let's say we didn't know how to find the exact value here, but we want to estimate it. Let's say we want to estimate it using the first five terms, so we'll do S sub five is equal to the sum of N equals one to infinity, sorry, to five of one over n squared, and let's see, this is going to be equal to, let's get a calculator out, this is going to be one over one squared, which is just one, plus one over two squared, which is one fourth, plus one over three squared, which is one ninth, plus one over four squared, which is one 16th, plus one over five squared, which is going to be one 25th, and we get this value right over here, so I'll just say this is, I'll say it's approximately, one point four six four. Approximately one point four six four, and then if we evaluate each of these integrals, so our K in this case is going to be five. I just picked that arbitrarily. We could probably get a better, or we would get a better estimate if we had a higher K, if we did the first 10 terms or 20 terms, and we would have a worse estimate if we did K equals three, but I just picked K equals five because it seems reasonable, something that I could compute. I actually didn't need a calculator to do that. I could have done it by hand. It would have just taken more time. Let's evaluate these integrals here. So the integral from five plus one, so it's six to infinity of one over N squared DX, and, let's see, that's going to be equal to, let's write it this way, this is going to be the limit, I'll do all of it in blue actually. This is going to be equal to the limit as, let's introduce a new variable, B approaches infinity of the integral from six to infinity of one over N squared DX, which is equal to the limit as B approaches infinity of, let's see, the antiderivative of this is negative N to the negative one, and we're going to evaluate that at B and at six, so this is equal to the limit as B approaches infinity of negative one over B, negative one over B minus negative one over six, so plus one over six. Well as B approaches infinity this term right over here goes to zero, so this is just going to be equal to one sixth, and we can use that exact same logic to evaluate this integral, so let's do that. Actually, I'll do it right over here. So the improper integral, I should say, from, and here we're going to evaluate not from six but from five, that's our K. Five to infinity of one over N squared DX. Well, the only difference here between these two integrals is this bottom bound. This was one over six, and by the same arugemnts, this one is going to end up being one over five. And so now we can put everything into this compound inequality. We're going to have one point four six four. One point one six four plus, and I'm being a little bit approximate here, it's really this S sub K, which we have the exact value for in our calculator, so it's one point four six four plus one sixth, plus one sixth, is less than or equal to, less than or equal to, is less than or equal to our sum, which we care about, which is less than or equal to one point four six four plus one fifth, plus one fifth, and so this one is fairly straight forward to calculate. This is going to be plus point two, so this is going to be less than or equal to one point two six four, so sum is going to be less than or equal to that, and on this side if we took our original expression, and then I added one sixth, one sixth, I get one point...I get one...let's see, what did I? Oh, this is supposed to be one... I wrote this one wrong. O.K., so this is going to be one point sixth three oh. Let me just write one point... So it's going to be one point six three oh, and on this side, my brain was malfunctioning. This should be one point one, I took away two instead of adding two, one point six six four, one point six six four, and so just like that you see, O.K., this thing must be like one point six three something. It could be one point four something. It could be one point five something. It could be one point six six, but this is already giving us a lot of precision, if we only cared about things to the 10ths place, this kind of, we're done. We know it's going to be one point six something, and if we go a little bit further this is already giving us a lot of precision, and you can imagine if you added a few more terms in the partial sum, you would get even more precision. So hopefully you found this to be kind of neat exercise. Not only did we get a good bit of precision, a good bit of precision with just taking a partial sum and the help of these integrals, but we got a good bound on it. We know that the sum that we're converging to is between these two values.