WEBVTT 00:00:07.000 --> 00:00:07.410 Hi. 00:00:07.410 --> 00:00:09.140 Welcome back to recitation. 00:00:09.140 --> 00:00:12.820 You've been learning about infinite series of constants. 00:00:12.820 --> 00:00:16.050 Adding up, you know, series of numbers. 00:00:16.050 --> 00:00:20.470 And one thing that we've talked about are divergent 00:00:20.470 --> 00:00:23.410 series and convergent series, and rates of divergence a 00:00:23.410 --> 00:00:24.470 little bit. 00:00:24.470 --> 00:00:26.785 So suppose you wanted to run an experiment. 00:00:26.785 --> 00:00:30.960 So suppose you started with the series, log n over n, and 00:00:30.960 --> 00:00:31.770 you started summing-- 00:00:31.770 --> 00:00:33.250 I picked n equals 3. 00:00:33.250 --> 00:00:35.480 It doesn't actually matter that much-- 00:00:35.480 --> 00:00:39.280 started summing at n equals 3, and going to infinity. 00:00:39.280 --> 00:00:42.690 And you set your computer to sort of keep track of what the 00:00:42.690 --> 00:00:45.790 partial sums are as it does this. 00:00:45.790 --> 00:00:49.170 So you wanted to produce a list of partial sums. And you 00:00:49.170 --> 00:00:54.270 know, maybe to get a good feel for this rate of divergence, 00:00:54.270 --> 00:00:57.860 what you want to do is let it run for a long time, collect a 00:00:57.860 --> 00:00:58.970 whole bunch of values. 00:00:58.970 --> 00:01:00.700 Say, you know, stop when it gets to some 00:01:00.700 --> 00:01:02.150 fairly large number. 00:01:02.150 --> 00:01:07.920 So suppose we run this, and we ask it to stop when it gets to 00:01:07.920 --> 00:01:11.750 a partial sum that's greater than or equal to 5,000. 00:01:11.750 --> 00:01:12.740 You know, whatever. 00:01:12.740 --> 00:01:15.780 A moderately large integer or number. 00:01:15.780 --> 00:01:17.020 Doesn't have to be an integer. 00:01:17.020 --> 00:01:19.500 So it's going to stop as soon as it reaches a partial sum 00:01:19.500 --> 00:01:21.060 that's bigger than 5,000. 00:01:21.060 --> 00:01:22.975 So what I want to know, is how long is that going to take? 00:01:22.975 --> 00:01:24.970 Now of course, that depends on your computer. 00:01:24.970 --> 00:01:27.740 So let's just say, you know, for sake of argument, that 00:01:27.740 --> 00:01:28.880 your computer-- 00:01:28.880 --> 00:01:31.460 this is generous to your computer. 00:01:31.460 --> 00:01:34.990 At least my computer doesn't go this fast-- that it takes 00:01:34.990 --> 00:01:40.150 10 to the minus 20 seconds to add an additional term to the 00:01:40.150 --> 00:01:41.670 partial sums. 00:01:41.670 --> 00:01:44.040 So suppose that your computer takes 10 to the minus 20 00:01:44.040 --> 00:01:46.960 seconds for each additional summand. 00:01:46.960 --> 00:01:49.390 How long, how many terms of the series, how long are you 00:01:49.390 --> 00:01:53.390 going to wait around, waiting for this sum to get to 5,000? 00:01:53.390 --> 00:01:55.770 So that's the question. 00:01:55.770 --> 00:01:59.145 Why don't you pause the video, take some time, try 00:01:59.145 --> 00:02:00.130 and work this out. 00:02:00.130 --> 00:02:02.750 This is probably not a quite good question for 00:02:02.750 --> 00:02:03.750 experimentation. 00:02:03.750 --> 00:02:07.800 I would advise against actually running it while you 00:02:07.800 --> 00:02:09.750 wait to continue with the video. 00:02:09.750 --> 00:02:12.520 But of course, you're welcome to try. 00:02:12.520 --> 00:02:15.700 But you can, you know, work this out, come back, and we'll 00:02:15.700 --> 00:02:16.950 talk about it together. 00:02:25.500 --> 00:02:26.200 Welcome back. 00:02:26.200 --> 00:02:27.990 Hopefully you had some fun with this problem. 00:02:27.990 --> 00:02:29.320 Let's let's talk about it. 00:02:29.320 --> 00:02:33.530 So what we want to do, is we want to look at this series 00:02:33.530 --> 00:02:35.670 and say, you know, about how long it's going to 00:02:35.670 --> 00:02:38.160 take to get to 5,000. 00:02:38.160 --> 00:02:41.240 What we want-- which partial sum of this series is about 00:02:41.240 --> 00:02:43.200 equal to 5,000 for the first time? 00:02:43.200 --> 00:02:45.540 So let's give a notation there. 00:02:45.540 --> 00:02:50.580 So let's call sN, big N now, s big N is equal to nth partial 00:02:50.580 --> 00:02:52.290 sum-- well, OK. 00:02:52.290 --> 00:02:53.740 Let me not make it the nth partial sum. 00:02:53.740 --> 00:02:58.200 Let me just make it the sum when the top is big N. So 00:02:58.200 --> 00:03:05.470 going from 3 to big N of log little n divided by n. 00:03:05.470 --> 00:03:09.730 So first of all, the answer could be a priori. 00:03:09.730 --> 00:03:11.550 The answer could be forever. 00:03:11.550 --> 00:03:15.830 It might be that series converges to some number less 00:03:15.830 --> 00:03:17.170 than 5,000. 00:03:17.170 --> 00:03:20.090 So if that were to happen, then, you know, no matter how 00:03:20.090 --> 00:03:22.460 long you waited around, you wouldn't get there. 00:03:22.460 --> 00:03:26.630 But we can check that that's not the case. 00:03:26.630 --> 00:03:29.910 So we can check that this series diverges. 00:03:29.910 --> 00:03:33.790 So how can we check that? 00:03:33.790 --> 00:03:44.390 So the series diverges by the integral test, because the 00:03:44.390 --> 00:03:53.650 integral of log x over x dx from, say, 3 to infinity, 00:03:53.650 --> 00:03:57.190 though of course the bounds, the lower bound, 3, is not 00:03:57.190 --> 00:03:58.130 crucially important. 00:03:58.130 --> 00:03:59.850 This is something we know how to integrate. 00:03:59.850 --> 00:04:06.690 This is equal to 1/2 log of x quantity squared, taken 00:04:06.690 --> 00:04:10.440 between, well, 3 and infinity. 00:04:10.440 --> 00:04:14.260 And anyway, you see that as x goes to infinity, log of x 00:04:14.260 --> 00:04:17.280 goes to infinity, and so log of x squared goes to infinity. 00:04:17.280 --> 00:04:18.530 So this diverges. 00:04:21.850 --> 00:04:24.440 So the series does diverge to positive infinity. 00:04:24.440 --> 00:04:26.520 And as a result of its divergence to positive 00:04:26.520 --> 00:04:28.240 infinity, we know that eventually, 00:04:28.240 --> 00:04:30.380 we do get to 5,000. 00:04:30.380 --> 00:04:32.150 OK, so this is, the answer is not, an 00:04:32.150 --> 00:04:33.300 infinite amount of time. 00:04:33.300 --> 00:04:35.410 The answer is some finite amount of time. 00:04:35.410 --> 00:04:37.730 So now the question is, how much. 00:04:37.730 --> 00:04:40.330 Now, the thing to remember about the integral test is not 00:04:40.330 --> 00:04:42.470 only can it tell you whether something converges or 00:04:42.470 --> 00:04:46.240 diverges, but it gives you a really good estimate of how 00:04:46.240 --> 00:04:47.980 quickly it converges or diverges. 00:04:47.980 --> 00:04:50.200 So I have over here a little picture. 00:04:50.200 --> 00:04:53.650 This is the function y equals log of x over x. 00:04:53.650 --> 00:05:00.010 So at x equals 1, this function is 0, and then it 00:05:00.010 --> 00:05:02.940 reaches a maximum, and then tapers down. 00:05:02.940 --> 00:05:04.140 And as x gets larger-- 00:05:04.140 --> 00:05:05.950 so it's a little hard to tell in this picture-- 00:05:05.950 --> 00:05:09.310 at x equals 6, this is still-- 00:05:09.310 --> 00:05:11.890 well, so the peak here is a little bit less than 0.4. 00:05:11.890 --> 00:05:14.590 At x equals 6, this, the function value is still bigger 00:05:14.590 --> 00:05:15.860 than a quarter. 00:05:15.860 --> 00:05:18.510 But it does eventually go down to 0, because x grows faster 00:05:18.510 --> 00:05:19.250 than log x. 00:05:19.250 --> 00:05:21.410 And of course if you were interested, an interesting 00:05:21.410 --> 00:05:24.220 exercise might be really precise, you know, where, to 00:05:24.220 --> 00:05:26.920 figure out precisely, what is this peak, and what's the 00:05:26.920 --> 00:05:29.780 x-value, and what's the y-value of that peak point 00:05:29.780 --> 00:05:31.160 before it starts turning down. 00:05:31.160 --> 00:05:34.340 If you wanted something fun to work out. 00:05:34.340 --> 00:05:34.630 And OK. 00:05:34.630 --> 00:05:36.940 But that doesn't actually have anything to do with our 00:05:36.940 --> 00:05:40.610 problem, which has to do with this integral test. So 00:05:40.610 --> 00:05:44.570 remember that the integral in question is the area under 00:05:44.570 --> 00:05:49.840 this curve from 3 onwards, and a partial sum is what you get 00:05:49.840 --> 00:05:53.790 when you take a bunch of these left endpoint rectangles. 00:05:53.790 --> 00:05:56.290 And because this function is decreasing, the rectangles 00:05:56.290 --> 00:05:58.200 completely cover the region. 00:05:58.200 --> 00:06:09.780 And so the integral test tells us that the integral from 3 to 00:06:09.780 --> 00:06:17.760 big N of our function, log x over x dx, is strictly less 00:06:17.760 --> 00:06:23.600 than, well, we want the partial sum that goes up to, 00:06:23.600 --> 00:06:29.160 so whose right interval here ends at big N. So that's 00:06:29.160 --> 00:06:31.830 actually the n minus first partial sum. 00:06:31.830 --> 00:06:35.470 So maybe what I'll do, is I'll add a plus 1 here, and I'll 00:06:35.470 --> 00:06:39.120 make this the big Nth partial sum is bigger than the sum 00:06:39.120 --> 00:06:42.630 from 3 to n plus 1 of the function. 00:06:42.630 --> 00:06:46.890 And also, if you turn this around, and you took right 00:06:46.890 --> 00:06:51.550 endpoint rectangles instead, you see that if you chop off 00:06:51.550 --> 00:06:58.580 the first rectangle, then what you get is the, you get an 00:06:58.580 --> 00:07:00.350 area less than the integral. 00:07:00.350 --> 00:07:03.080 So we also have an upper bound here. 00:07:03.080 --> 00:07:06.250 This is less than or equal to the very first term, which in 00:07:06.250 --> 00:07:14.830 our case is ln 3 over 3 plus that same integral. 00:07:21.930 --> 00:07:25.050 So this is this formula that Professor Jerison showed you 00:07:25.050 --> 00:07:27.960 in class that bounds the partial sums 00:07:27.960 --> 00:07:30.440 in between the integral. 00:07:30.440 --> 00:07:33.500 So here, where the integral, instead of being an improper 00:07:33.500 --> 00:07:35.690 integral, just goes up to n plus 1. 00:07:35.690 --> 00:07:39.530 So the integral test gives us, it tells us that this diverges 00:07:39.530 --> 00:07:43.640 if and only if this does, but it also gives us explicit 00:07:43.640 --> 00:07:48.780 bounds for those two expressions, how they relate 00:07:48.780 --> 00:07:49.360 to each other. 00:07:49.360 --> 00:07:52.210 And so it says that the rate of divergence of sN is very, 00:07:52.210 --> 00:07:54.060 very closely tied to the rate of 00:07:54.060 --> 00:07:56.280 divergence of this integral. 00:07:56.280 --> 00:07:57.020 Well, so what? 00:07:57.020 --> 00:07:57.350 So OK. 00:07:57.350 --> 00:07:59.840 So we said, we know, you know, about how big 00:07:59.840 --> 00:08:00.490 this integral is. 00:08:00.490 --> 00:08:01.250 How big is it? 00:08:01.250 --> 00:08:05.060 Well, we saw that it's about log x 00:08:05.060 --> 00:08:06.770 squared between the bounds. 00:08:06.770 --> 00:08:14.180 So this is equal to, well, it's about, it's 1/2 log of n 00:08:14.180 --> 00:08:23.080 plus 1 squared minus 1/2 log 3 squared is less than sN, which 00:08:23.080 --> 00:08:28.320 is less than log 3 over 3 plus the same thing. 00:08:34.160 --> 00:08:37.550 So this is log n plus 1 quantity squared 00:08:37.550 --> 00:08:42.050 minus 1/2 log 3 squared. 00:08:42.050 --> 00:08:46.530 And here this is log n plus 1 quantity squared. 00:08:46.530 --> 00:08:47.120 OK. 00:08:47.120 --> 00:08:50.120 So this right, this right hand side is equal to 00:08:50.120 --> 00:08:52.390 this right hand side. 00:08:52.390 --> 00:08:52.940 OK. 00:08:52.940 --> 00:08:54.440 So we have that s-- 00:08:54.440 --> 00:09:03.050 and this is a big N. So we have that sN is about bounded 00:09:03.050 --> 00:09:03.360 by these-- sorry-- is definitely bounded by these 00:09:03.360 --> 00:09:04.030 two expressions. 00:09:04.030 --> 00:09:05.450 Another thing to note is that the constants 00:09:05.450 --> 00:09:06.610 here are pretty small. 00:09:06.610 --> 00:09:06.810 Right? 00:09:06.810 --> 00:09:11.455 This 1/2 log 3 squared and log 3 over 3 times 1/2 log 3 00:09:11.455 --> 00:09:12.820 squared, those are pretty small. 00:09:12.820 --> 00:09:14.810 From this point on in this problem, I'm going to stop 00:09:14.810 --> 00:09:18.360 worrying about precise constants like that, and I'm 00:09:18.360 --> 00:09:21.350 going to start using an approximate sign. 00:09:21.350 --> 00:09:25.270 So another way to write this is to say that sN is 00:09:25.270 --> 00:09:26.910 approximately-- 00:09:26.910 --> 00:09:29.210 so this is like an equals sign, but 00:09:29.210 --> 00:09:30.520 both of them are curvy. 00:09:30.520 --> 00:09:31.980 So what I mean by this, it doesn't 00:09:31.980 --> 00:09:33.590 mean anything formally. 00:09:33.590 --> 00:09:35.490 So it's not a precise meaning. 00:09:35.490 --> 00:09:38.230 In this case, it's going to mean always 00:09:38.230 --> 00:09:39.290 about the same size. 00:09:39.290 --> 00:09:41.190 For purposes of this argument, it's going to 00:09:41.190 --> 00:09:42.210 be about the same. 00:09:42.210 --> 00:09:47.200 So this is about equal to 1/2 ln of n 00:09:47.200 --> 00:09:49.406 plus 1 quantity squared. 00:09:52.830 --> 00:09:53.690 OK. and what we want, So OK. 00:09:53.690 --> 00:09:53.860 So good. 00:09:53.860 --> 00:09:55.940 So we have a good approximation for about how 00:09:55.940 --> 00:09:57.860 big the nth partial sum is. 00:09:57.860 --> 00:09:58.580 Well, all right. 00:09:58.580 --> 00:09:59.790 So why? 00:09:59.790 --> 00:10:02.730 Well, because I know that what I want to do is I want to keep 00:10:02.730 --> 00:10:07.090 going until I reach a partial sum that has size about 5,000. 00:10:07.090 --> 00:10:09.250 So I want to know how, many terms do I need. 00:10:09.250 --> 00:10:09.790 Right? 00:10:09.790 --> 00:10:12.100 That's going to tell me how much time it requires, and I 00:10:12.100 --> 00:10:13.860 need to know how many terms there are. 00:10:13.860 --> 00:10:19.350 So I need to know when sN is about 5,000, how big is N? 00:10:19.350 --> 00:10:22.460 So coming back over here, so I need to take this equation, 00:10:22.460 --> 00:10:24.760 and I need to solve it for N in terms of sN. 00:10:24.760 --> 00:10:27.550 Because I know that sN is going to be about 5,000, so I 00:10:27.550 --> 00:10:29.510 want to know, how big is N going to be? 00:10:29.510 --> 00:10:29.790 So OK. 00:10:29.790 --> 00:10:30.340 So let's do that. 00:10:30.340 --> 00:10:32.350 So we can multiply by 2, take a square root, and 00:10:32.350 --> 00:10:33.210 exponentiate. 00:10:33.210 --> 00:10:34.220 And what we get-- 00:10:34.220 --> 00:10:37.370 so I'm going to move up here. 00:10:40.250 --> 00:10:47.440 So from this expression, what I get is that n plus 1 is 00:10:47.440 --> 00:10:50.280 approximately equal to-- so I'm going to multiply by 2, 00:10:50.280 --> 00:10:55.160 take a square root, and exponentiate, so that e to the 00:10:55.160 --> 00:10:59.150 square root of 2 sN. 00:11:03.060 --> 00:11:05.180 And you know what, I don't care about this plus 1. 00:11:05.180 --> 00:11:06.310 I'm going to forget about it. 00:11:06.310 --> 00:11:10.290 So N is approximately equal to e to the 2 square root of sN. 00:11:10.290 --> 00:11:14.130 Now in our case, we want to know particularly when sN is 00:11:14.130 --> 00:11:15.760 equal to about 5,000. 00:11:15.760 --> 00:11:27.150 So when sN is approximately equal to 5,000, this means 00:11:27.150 --> 00:11:30.020 that N is approximately-- 00:11:30.020 --> 00:11:34.090 well, 2 times 5,000 is 10,000, square root of 10,000 is 100, 00:11:34.090 --> 00:11:38.150 so N is approximately equal to e to the 100. 00:11:38.150 --> 00:11:41.040 Now e is between 2 and 3-- 00:11:41.040 --> 00:11:45.000 so OK, so you know, I don't, I can't really give a very 00:11:45.000 --> 00:11:47.100 precise estimate of this off the top of my head. 00:11:47.100 --> 00:11:50.080 But luckily I used a computer ahead of time 00:11:50.080 --> 00:11:51.400 to approximate it. 00:11:51.400 --> 00:12:02.260 So e to the 100 is about equal to 2 times 10 to the 00:12:02.260 --> 00:12:04.880 forty-third power. 00:12:04.880 --> 00:12:08.220 So that's the number, so that's about how big N is, 00:12:08.220 --> 00:12:10.380 order of magnitude. 00:12:10.380 --> 00:12:15.970 So we need to do about 2 times 10 to the forty-third terms. 00:12:15.970 --> 00:12:19.550 So now that's the number of terms. Now, if you remember, 00:12:19.550 --> 00:12:23.190 back in our question, we were asked, on the assumption that 00:12:23.190 --> 00:12:27.420 each term takes 10 to the minus 20 seconds, how long is 00:12:27.420 --> 00:12:28.070 this going to take. 00:12:28.070 --> 00:12:47.690 So if each term requires 10 to the minus 20 seconds, this 00:12:47.690 --> 00:12:52.850 means we're dealing with 2 times 10 to the 23 seconds. 00:12:52.850 --> 00:12:54.710 That's how long you'll have to wait. 00:12:54.710 --> 00:12:57.190 Now, this is a big number. 00:12:57.190 --> 00:12:57.710 Right? 00:12:57.710 --> 00:13:01.150 So let's think about how big a number this is. 00:13:01.150 --> 00:13:03.390 Because there are big numbers, and then there are 00:13:03.390 --> 00:13:05.220 ridiculously big numbers. 00:13:05.220 --> 00:13:08.810 So, a big number would be a number where like, maybe this 00:13:08.810 --> 00:13:11.680 10 to the minus 20 is a bad estimate, right? 00:13:11.680 --> 00:13:15.890 I mean, maybe you are emperor of the world, and you can 00:13:15.890 --> 00:13:19.410 harness all the computing power in the entire world to 00:13:19.410 --> 00:13:20.470 work on this problem. 00:13:20.470 --> 00:13:24.040 And so then, in that case, maybe you can do, say, 10 to 00:13:24.040 --> 00:13:29.940 the 25 computations per second, something. 00:13:29.940 --> 00:13:31.550 Or whatever. 00:13:31.550 --> 00:13:35.480 So how much of a dent would that make in this number? 00:13:35.480 --> 00:13:37.580 Like 2 times 10 to 23. 00:13:37.580 --> 00:13:37.760 OK. 00:13:37.760 --> 00:13:40.110 It's not going to be done before lunchtime, but how long 00:13:40.110 --> 00:13:40.700 will it take? 00:13:40.700 --> 00:13:42.950 So just to-- 00:13:42.950 --> 00:13:43.300 let's see. 00:13:43.300 --> 00:13:45.860 I have this written down here. 00:13:45.860 --> 00:13:46.330 OK. 00:13:46.330 --> 00:13:55.940 So for comparison, this is about equal to 50,000 times 00:13:55.940 --> 00:14:04.010 the age of the universe. 00:14:04.010 --> 00:14:07.680 So if you had, if you harnessed all the computing 00:14:07.680 --> 00:14:12.680 power on earth, you could probably get this done in a 00:14:12.680 --> 00:14:15.120 few billion years. 00:14:15.120 --> 00:14:18.530 So OK, so just a sense of just how slowly 00:14:18.530 --> 00:14:19.410 this thing is getting. 00:14:19.410 --> 00:14:22.740 If you wait, and you have all the computer power on earth 00:14:22.740 --> 00:14:27.400 adding terms of the series, in, say, a few billion years 00:14:27.400 --> 00:14:31.890 you will be able to figure out-- 00:14:31.890 --> 00:14:35.470 you will sum up terms that pass 5,000 in value. 00:14:35.470 --> 00:14:37.390 So OK. 00:14:37.390 --> 00:14:40.360 So I don't know. 00:14:40.360 --> 00:14:42.790 I find that amusing. 00:14:42.790 --> 00:14:46.530 And remember that the key here to this whole problem was just 00:14:46.530 --> 00:14:49.860 that we were using this integral test here. 00:14:49.860 --> 00:14:52.660 And specifically, we were using in the form that I have 00:14:52.660 --> 00:14:56.130 over here, this form that lets you bound the partial sums 00:14:56.130 --> 00:14:58.530 between two values of the integral. 00:14:58.530 --> 00:15:00.800 And because we have those very tight bounds, these 00:15:00.800 --> 00:15:05.160 approximations are all valid, and so we can really say, yes, 00:15:05.160 --> 00:15:07.270 this problem that I've described is really something 00:15:07.270 --> 00:15:11.990 you can't do in an afternoon or whatever. 00:15:11.990 --> 00:15:14.040 So I'll stop there.