WEBVTT 00:00:00.000 --> 00:00:01.988 align:middle line:90% [SQUEAKING] 00:00:01.988 --> 00:00:02.982 align:middle line:90% [RUSTLING] 00:00:02.982 --> 00:00:04.970 align:middle line:90% [CLICKING] 00:00:04.970 --> 00:00:14.260 align:middle line:90% 00:00:14.260 --> 00:00:17.320 align:middle line:84% CASEY RODRIGUEZ: So at the end of last time, 00:00:17.320 --> 00:00:24.790 align:middle line:84% let me just restate the theorem that we proved, 00:00:24.790 --> 00:00:32.840 align:middle line:84% which was the Weierstrass M-test, which 00:00:32.840 --> 00:00:35.450 align:middle line:84% I'll state in briefer form now as the following. 00:00:35.450 --> 00:00:39.090 align:middle line:90% 00:00:39.090 --> 00:00:45.110 align:middle line:84% Suppose fj are a sequence of functions from some subset 00:00:45.110 --> 00:00:56.870 align:middle line:84% S of R, so that for all j, there exists non-negative number M 00:00:56.870 --> 00:01:01.060 align:middle line:84% sub j, such that two things hold. 00:01:01.060 --> 00:01:05.575 align:middle line:84% First off, these M sub j's dominate the f sub j's. 00:01:05.575 --> 00:01:09.850 align:middle line:90% 00:01:09.850 --> 00:01:13.900 align:middle line:84% f sub j of x is less than or equal to M sub j. 00:01:13.900 --> 00:01:19.720 align:middle line:84% And two, these M sub j's are summable. 00:01:19.720 --> 00:01:31.870 align:middle line:90% 00:01:31.870 --> 00:01:39.700 align:middle line:84% Then the conclusion is, the sequence of partial sums. 00:01:39.700 --> 00:01:51.110 align:middle line:90% So converges uniformly. 00:01:51.110 --> 00:01:53.870 align:middle line:84% So there exists some function f from S to R 00:01:53.870 --> 00:01:56.270 align:middle line:84% so that the partial sums converge uniformly. 00:01:56.270 --> 00:02:02.110 align:middle line:90% 00:02:02.110 --> 00:02:04.090 align:middle line:84% So this is a sequence of functions built up 00:02:04.090 --> 00:02:10.300 align:middle line:84% by the fj's by just taking the sum of the first j 00:02:10.300 --> 00:02:11.560 align:middle line:90% equals 1 to n of them. 00:02:11.560 --> 00:02:14.120 align:middle line:90% 00:02:14.120 --> 00:02:18.930 align:middle line:84% And so, for example, and I kind of went through this last time. 00:02:18.930 --> 00:02:33.920 align:middle line:84% So example, if fj of x equals our old friend cosine of 160jx 00:02:33.920 --> 00:02:41.440 align:middle line:90% over 4 to the j, then-- 00:02:41.440 --> 00:02:49.940 align:middle line:90% and let's say on R. So S is R. 00:02:49.940 --> 00:02:52.145 align:middle line:90% And a couple of things. 00:02:52.145 --> 00:02:55.370 align:middle line:90% 00:02:55.370 --> 00:02:59.900 align:middle line:84% fj of x for all x is less than or equal to 4 to the minus j. 00:02:59.900 --> 00:03:01.700 align:middle line:84% Because cosine of whatever you put into 00:03:01.700 --> 00:03:03.080 align:middle line:90% it is always bounded by 1. 00:03:03.080 --> 00:03:07.520 align:middle line:90% 00:03:07.520 --> 00:03:20.880 align:middle line:84% And since this converges, this implies 00:03:20.880 --> 00:03:28.380 align:middle line:84% that this series, which I'm writing in this way, which you 00:03:28.380 --> 00:03:35.880 align:middle line:84% should think of as a limit of functions 00:03:35.880 --> 00:03:41.330 align:middle line:84% of the partial sums, which are functions, 00:03:41.330 --> 00:03:48.230 align:middle line:90% converges uniformly on R. 00:03:48.230 --> 00:03:50.390 align:middle line:84% Not only that, this function defined by-- 00:03:50.390 --> 00:03:52.400 align:middle line:90% I take x, I stick it into here-- 00:03:52.400 --> 00:03:53.300 align:middle line:90% is continuous. 00:03:53.300 --> 00:03:57.020 align:middle line:84% Because last time, we proved that the uniform limit 00:03:57.020 --> 00:04:01.220 align:middle line:84% of continuous functions is continuous. 00:04:01.220 --> 00:04:03.590 align:middle line:84% This function here is the uniform limit 00:04:03.590 --> 00:04:07.820 align:middle line:84% of the partial sums, which is just finitely many cosines 00:04:07.820 --> 00:04:08.730 align:middle line:90% of things. 00:04:08.730 --> 00:04:10.000 align:middle line:90% So that's continuous. 00:04:10.000 --> 00:04:14.600 align:middle line:84% So this gives another proof of one of the things 00:04:14.600 --> 00:04:18.140 align:middle line:84% we did by hand when we spoke about differentiability, namely 00:04:18.140 --> 00:04:25.610 align:middle line:84% that this function that we considered was continuous. 00:04:25.610 --> 00:04:29.610 align:middle line:90% 00:04:29.610 --> 00:04:31.950 align:middle line:84% So let's go back to power series, which 00:04:31.950 --> 00:04:36.110 align:middle line:90% was our original motivation. 00:04:36.110 --> 00:04:54.970 align:middle line:90% 00:04:54.970 --> 00:04:57.080 align:middle line:84% So I'm going to be saying that a series-- 00:04:57.080 --> 00:05:00.580 align:middle line:84% so let me, again, make this perfectly clear. 00:05:00.580 --> 00:05:06.190 align:middle line:84% When I say, if you like, make this a definition. 00:05:06.190 --> 00:05:07.900 align:middle line:90% Or really, this is just a-- 00:05:07.900 --> 00:05:10.684 align:middle line:90% 00:05:10.684 --> 00:05:14.450 align:middle line:84% I was using this terminology up there in that theorem 00:05:14.450 --> 00:05:17.510 align:middle line:84% and when I was talking about this example afterwards. 00:05:17.510 --> 00:05:19.850 align:middle line:84% But when I say a series involving 00:05:19.850 --> 00:05:23.480 align:middle line:84% functions converges uniformly, I mean the partial sums 00:05:23.480 --> 00:05:25.040 align:middle line:90% converge uniformly. 00:05:25.040 --> 00:05:46.360 align:middle line:84% So means there exist a function f such that-- 00:05:46.360 --> 00:05:53.970 align:middle line:90% 00:05:53.970 --> 00:05:57.050 align:middle line:84% and all of these are from S to R as well, such that-- 00:05:57.050 --> 00:06:02.700 align:middle line:90% 00:06:02.700 --> 00:06:05.358 align:middle line:84% just so that there's no ambiguity. 00:06:05.358 --> 00:06:06.900 align:middle line:84% So if I have a sequence of functions, 00:06:06.900 --> 00:06:09.960 align:middle line:84% f sub j, and I form their series, 00:06:09.960 --> 00:06:12.000 align:middle line:84% and I say that converges uniformly, that 00:06:12.000 --> 00:06:14.220 align:middle line:84% means there exists some function f, 00:06:14.220 --> 00:06:16.800 align:middle line:84% so that the sequence of partial sums-- 00:06:16.800 --> 00:06:18.810 align:middle line:90% these are now functions-- 00:06:18.810 --> 00:06:23.380 align:middle line:84% converges uniformly to the function f. 00:06:23.380 --> 00:06:24.840 align:middle line:90% That's what power series are. 00:06:24.840 --> 00:06:29.190 align:middle line:84% They're expressions involving a free variable x. 00:06:29.190 --> 00:06:30.690 align:middle line:84% And therefore, you can think of them 00:06:30.690 --> 00:06:35.680 align:middle line:90% as a series involving functions. 00:06:35.680 --> 00:06:40.740 align:middle line:84% So let me just state the following theorem about when 00:06:40.740 --> 00:06:43.660 align:middle line:90% we have a uniform convergence. 00:06:43.660 --> 00:07:05.080 align:middle line:84% So let's suppose we have a power series with radius 00:07:05.080 --> 00:07:14.270 align:middle line:84% of convergence rho, which I will recall 00:07:14.270 --> 00:07:20.750 align:middle line:84% is defined to be the limit as j goes to infinity, aj 00:07:20.750 --> 00:07:26.070 align:middle line:90% 1 over j inverse. 00:07:26.070 --> 00:07:30.990 align:middle line:84% If this limit is 0, we say that the radius of convergence 00:07:30.990 --> 00:07:31.950 align:middle line:90% is infinity. 00:07:31.950 --> 00:07:33.690 align:middle line:84% So if this is a finite number, this 00:07:33.690 --> 00:07:35.730 align:middle line:90% is a meaningful expression. 00:07:35.730 --> 00:07:38.100 align:middle line:84% If this is 0, then this is shorthand 00:07:38.100 --> 00:07:42.570 align:middle line:84% for saying the radius of convergence is infinity. 00:07:42.570 --> 00:07:47.970 align:middle line:90% 00:07:47.970 --> 00:07:50.130 align:middle line:90% So let's assume this is-- 00:07:50.130 --> 00:07:55.350 align:middle line:90% 00:07:55.350 --> 00:07:59.250 align:middle line:84% then, as long as I stay strictly inside the radius 00:07:59.250 --> 00:08:02.950 align:middle line:84% of convergence, I have uniformed convergence. 00:08:02.950 --> 00:08:05.230 align:middle line:84% This is the statement of the theorem. 00:08:05.230 --> 00:08:18.450 align:middle line:84% And for all R in 0 rho, this power series, now thought 00:08:18.450 --> 00:08:34.460 align:middle line:84% of as a series of functions of x, converges uniformly on x0 00:08:34.460 --> 00:08:41.060 align:middle line:90% minus R, x0 plus R. 00:08:41.060 --> 00:08:48.470 align:middle line:84% So the picture is, we have some radius of convergence, x0, x0 00:08:48.470 --> 00:08:51.800 align:middle line:90% plus rho, x0 minus rho. 00:08:51.800 --> 00:08:55.670 align:middle line:84% If we take any closed interval, basically, 00:08:55.670 --> 00:09:00.080 align:middle line:84% inside of that-- strictly inside of that interval, 00:09:00.080 --> 00:09:03.155 align:middle line:84% then we have uniform convergence of the power series. 00:09:03.155 --> 00:09:10.190 align:middle line:90% 00:09:10.190 --> 00:09:12.680 align:middle line:84% So we'll prove this using the Weierstrass M-test. 00:09:12.680 --> 00:09:18.920 align:middle line:90% 00:09:18.920 --> 00:09:21.300 align:middle line:90% So let R be in 0, rho. 00:09:21.300 --> 00:09:30.120 align:middle line:90% 00:09:30.120 --> 00:09:31.440 align:middle line:90% And for all j-- 00:09:31.440 --> 00:09:33.450 align:middle line:84% and OK, so the Weierstrass M-test 00:09:33.450 --> 00:09:36.930 align:middle line:84% is for j equals starting from 1 and going to infinity. 00:09:36.930 --> 00:09:39.270 align:middle line:84% But you don't have to, just like for series, we 00:09:39.270 --> 00:09:40.380 align:middle line:90% don't have to start at 1. 00:09:40.380 --> 00:09:42.570 align:middle line:90% It can start at 0. 00:09:42.570 --> 00:09:54.580 align:middle line:84% So for all j union 0, what do we have? 00:09:54.580 --> 00:09:55.660 align:middle line:90% What are we writing? 00:09:55.660 --> 00:10:02.410 align:middle line:84% And for all x in this interval where 00:10:02.410 --> 00:10:04.600 align:middle line:84% we want to show uniform convergence, 00:10:04.600 --> 00:10:08.580 align:middle line:84% we have that this function a sub j, 00:10:08.580 --> 00:10:13.960 align:middle line:84% x minus x0 to the j in absolute value. 00:10:13.960 --> 00:10:15.580 align:middle line:90% What is this bounded by? 00:10:15.580 --> 00:10:23.100 align:middle line:84% This is bounded by absolute value of aj, and then j. 00:10:23.100 --> 00:10:27.900 align:middle line:84% And because x is in this interval, its distance to x0 00:10:27.900 --> 00:10:33.350 align:middle line:84% is less than or equal to R. So R to the j. 00:10:33.350 --> 00:10:40.460 align:middle line:84% So this will be my Mj that I'll use for the Weierstrass M-test. 00:10:40.460 --> 00:10:45.390 align:middle line:90% 00:10:45.390 --> 00:10:48.710 align:middle line:84% Let me put that in parentheses, because I may not use that Mj. 00:10:48.710 --> 00:10:53.720 align:middle line:84% So these individual functions, just polynomials, 00:10:53.720 --> 00:10:59.600 align:middle line:84% are bounded by this number for each j. 00:10:59.600 --> 00:11:09.150 align:middle line:84% And now, I want to see if the series of involving 00:11:09.150 --> 00:11:11.830 align:middle line:90% these numbers converges. 00:11:11.830 --> 00:11:14.790 align:middle line:90% So let me apply the root test. 00:11:14.790 --> 00:11:21.900 align:middle line:90% 00:11:21.900 --> 00:11:31.370 align:middle line:90% We have this is equal to-- 00:11:31.370 --> 00:11:34.790 align:middle line:84% now, this R to the 1 over j just becomes R. 00:11:34.790 --> 00:11:37.970 align:middle line:84% And it can just come out of this limit. 00:11:37.970 --> 00:11:42.290 align:middle line:90% And I pick up a to the 1 over j. 00:11:42.290 --> 00:11:49.610 align:middle line:84% And this is equal to 1 over rho by-- 00:11:49.610 --> 00:11:57.230 align:middle line:84% so this is equal to, if you like, I can put R over rho, 00:11:57.230 --> 00:12:01.640 align:middle line:84% as long as if rho is less than infinity. 00:12:01.640 --> 00:12:03.200 align:middle line:84% Meaning if the radius of convergence 00:12:03.200 --> 00:12:04.535 align:middle line:90% is less than infinity and 0. 00:12:04.535 --> 00:12:09.320 align:middle line:90% 00:12:09.320 --> 00:12:10.280 align:middle line:90% And what do we notice? 00:12:10.280 --> 00:12:12.110 align:middle line:90% Now, R is less than rho. 00:12:12.110 --> 00:12:17.720 align:middle line:84% R is coming from the closed interval 0 up to the rho, 00:12:17.720 --> 00:12:18.890 align:middle line:90% not including rho. 00:12:18.890 --> 00:12:22.190 align:middle line:84% So this number-- so this is always less than 1. 00:12:22.190 --> 00:12:23.930 align:middle line:84% But this number is always less than 1 00:12:23.930 --> 00:12:26.270 align:middle line:84% too, as long as R is less than rho. 00:12:26.270 --> 00:12:30.950 align:middle line:84% Which implies that the series sum from j 00:12:30.950 --> 00:12:40.310 align:middle line:84% equals 0 to infinity of aj R to the j converges. 00:12:40.310 --> 00:12:43.580 align:middle line:84% So now we have the sequence of functions, 00:12:43.580 --> 00:12:47.180 align:middle line:84% the polynomials that we used to build up our power series, 00:12:47.180 --> 00:12:51.920 align:middle line:84% each of them bounded by these numbers, a sub j, 00:12:51.920 --> 00:12:54.080 align:middle line:90% absolute value times R to the j. 00:12:54.080 --> 00:12:57.590 align:middle line:90% And these numbers are summable. 00:12:57.590 --> 00:12:59.040 align:middle line:90% The series converges. 00:12:59.040 --> 00:13:10.850 align:middle line:84% So by the Weierstrass M-test, this 00:13:10.850 --> 00:13:21.400 align:middle line:84% implies that the power series converges 00:13:21.400 --> 00:13:22.825 align:middle line:90% uniformly on this interval. 00:13:22.825 --> 00:13:34.680 align:middle line:90% 00:13:34.680 --> 00:13:39.360 align:middle line:84% So as long as we stay strictly within inside the radius 00:13:39.360 --> 00:13:42.487 align:middle line:84% of convergence, we have uniform convergence 00:13:42.487 --> 00:13:43.320 align:middle line:90% of the power series. 00:13:43.320 --> 00:13:46.780 align:middle line:90% 00:13:46.780 --> 00:13:52.450 align:middle line:84% And therefore, using the theorems that we proved before, 00:13:52.450 --> 00:13:55.870 align:middle line:84% we can differentiate and integrate term by term 00:13:55.870 --> 00:13:56.605 align:middle line:90% the power series. 00:13:56.605 --> 00:14:01.240 align:middle line:90% 00:14:01.240 --> 00:14:05.460 align:middle line:84% So which follows immediately from this previous theorem 00:14:05.460 --> 00:14:09.300 align:middle line:84% and then what we proved last time, which I'll state now. 00:14:09.300 --> 00:14:25.960 align:middle line:90% 00:14:25.960 --> 00:14:39.490 align:middle line:84% The power series with radius of convergence rho positive. 00:14:39.490 --> 00:14:44.290 align:middle line:90% So let me write it this way. 00:14:44.290 --> 00:14:52.010 align:middle line:90% 00:14:52.010 --> 00:15:02.410 align:middle line:84% And the first is for all c inside x0 minus R, x0 plus R, 00:15:02.410 --> 00:15:09.190 align:middle line:84% the function given by the power series, which I'm not going 00:15:09.190 --> 00:15:10.730 align:middle line:90% to call this f or anything. 00:15:10.730 --> 00:15:13.420 align:middle line:84% I'm just going to refer the power series directly, 00:15:13.420 --> 00:15:21.400 align:middle line:90% this is differentiable at c. 00:15:21.400 --> 00:15:23.350 align:middle line:84% And to compute the derivative, you 00:15:23.350 --> 00:15:25.388 align:middle line:84% can just do it term by term, meaning I 00:15:25.388 --> 00:15:26.680 align:middle line:90% can take the derivative inside. 00:15:26.680 --> 00:15:29.840 align:middle line:90% 00:15:29.840 --> 00:15:44.850 align:middle line:84% By dx of j equals 0 to infinity, a sub j, 00:15:44.850 --> 00:15:48.900 align:middle line:90% evaluated at x equals c. 00:15:48.900 --> 00:15:53.760 align:middle line:84% This is equal to simply differentiating term by term. 00:15:53.760 --> 00:16:13.930 align:middle line:90% 00:16:13.930 --> 00:16:32.500 align:middle line:84% And number two, for all a less than be, well, with x0 00:16:32.500 --> 00:16:38.620 align:middle line:84% minus rho less than a, less than b, less than x0 plus rho-- 00:16:38.620 --> 00:16:42.250 align:middle line:84% so here I'm sticking to strictly with inside the radius 00:16:42.250 --> 00:16:43.780 align:middle line:90% of convergence. 00:16:43.780 --> 00:16:47.810 align:middle line:84% There's the x0 plus rho, x0 minus rho. 00:16:47.810 --> 00:16:50.480 align:middle line:84% And so, now I'm going to integrate over 00:16:50.480 --> 00:16:51.380 align:middle line:90% some interval inside. 00:16:51.380 --> 00:16:54.110 align:middle line:90% 00:16:54.110 --> 00:16:57.950 align:middle line:84% Then I can integrate term by term. 00:16:57.950 --> 00:17:01.040 align:middle line:84% The integral from a to b, sum from j 00:17:01.040 --> 00:17:08.164 align:middle line:84% equals 0 to infinity of a sub j, x minus x0 jdx equals-- 00:17:08.164 --> 00:17:10.849 align:middle line:90% 00:17:10.849 --> 00:17:12.650 align:middle line:84% and I'll just write it this way, sum from j 00:17:12.650 --> 00:17:14.750 align:middle line:90% equals 0 to infinity of-- 00:17:14.750 --> 00:17:23.500 align:middle line:90% 00:17:23.500 --> 00:17:24.000 align:middle line:90% OK. 00:17:24.000 --> 00:17:32.040 align:middle line:90% 00:17:32.040 --> 00:17:35.538 align:middle line:84% And in fact, let me just to really emphasize 00:17:35.538 --> 00:17:37.080 align:middle line:84% that we're interchanging limits here, 00:17:37.080 --> 00:17:43.110 align:middle line:84% let me instead write this equivalently as sum from j 00:17:43.110 --> 00:17:53.230 align:middle line:84% equals 0 to infinity of sub j d by dx of a sub j, 00:17:53.230 --> 00:17:59.970 align:middle line:84% x minus x0, j evaluated at x equals c. 00:17:59.970 --> 00:18:05.910 align:middle line:84% So for power series, I can interchange the limits. 00:18:05.910 --> 00:18:08.550 align:middle line:84% The limit being taking a derivative in the sum, 00:18:08.550 --> 00:18:09.720 align:middle line:90% I can interchange. 00:18:09.720 --> 00:18:15.990 align:middle line:84% And then, I can also interchange the integration and the sum 00:18:15.990 --> 00:18:19.680 align:middle line:84% as long as I stay within the radius of convergence. 00:18:19.680 --> 00:18:23.790 align:middle line:84% And this statement here is a point-wise. 00:18:23.790 --> 00:18:26.610 align:middle line:84% So let's see why one is the case. 00:18:26.610 --> 00:18:30.930 align:middle line:84% Two follows immediately from what we've proven, 00:18:30.930 --> 00:18:34.300 align:middle line:84% and so, this previous theorem and the theorem 00:18:34.300 --> 00:18:36.970 align:middle line:84% we have about integration and uniform convergence. 00:18:36.970 --> 00:18:38.340 align:middle line:90% So what about one? 00:18:38.340 --> 00:18:43.500 align:middle line:90% 00:18:43.500 --> 00:18:48.690 align:middle line:84% So first off, we know we have uniform convergence 00:18:48.690 --> 00:18:55.300 align:middle line:84% of the series strictly inside of the radius of convergence. 00:18:55.300 --> 00:18:58.590 align:middle line:84% So how about, we need to check to see 00:18:58.590 --> 00:19:03.300 align:middle line:84% if the formal derivative also has radius of convergence 00:19:03.300 --> 00:19:06.240 align:middle line:90% as well equal to rho. 00:19:06.240 --> 00:19:26.090 align:middle line:84% So I claim that the radius of convergence of the derivatives 00:19:26.090 --> 00:19:26.720 align:middle line:90% which equals-- 00:19:26.720 --> 00:19:29.330 align:middle line:90% 00:19:29.330 --> 00:19:31.085 align:middle line:90% I can write it this way. 00:19:31.085 --> 00:19:45.670 align:middle line:90% 00:19:45.670 --> 00:19:46.170 align:middle line:90% No. 00:19:46.170 --> 00:19:51.000 align:middle line:90% 00:19:51.000 --> 00:19:54.990 align:middle line:84% So the derivative of this guy is j times a sub j times 00:19:54.990 --> 00:19:57.360 align:middle line:90% x minus x0 to the j minus 1. 00:19:57.360 --> 00:20:02.580 align:middle line:84% So just shifting indices, this is equal to times j plus 1, 00:20:02.580 --> 00:20:07.890 align:middle line:90% times x minus x0 to the j. 00:20:07.890 --> 00:20:13.260 align:middle line:84% So our claim is that this power series here 00:20:13.260 --> 00:20:22.480 align:middle line:84% has radius of convergence equal to rho, the original radius 00:20:22.480 --> 00:20:23.270 align:middle line:90% of convergence. 00:20:23.270 --> 00:20:25.330 align:middle line:84% So rho is the radius of convergence 00:20:25.330 --> 00:20:26.620 align:middle line:90% of-- why did I come over here. 00:20:26.620 --> 00:20:27.850 align:middle line:90% I still had a board left. 00:20:27.850 --> 00:20:30.690 align:middle line:90% Sorry about that. 00:20:30.690 --> 00:20:34.920 align:middle line:84% The original power series has radius of convergence rho. 00:20:34.920 --> 00:20:37.260 align:middle line:84% What we need to show to prove part 00:20:37.260 --> 00:20:40.170 align:middle line:84% one is that the radius of convergence 00:20:40.170 --> 00:20:44.520 align:middle line:84% of the derivatives of taking the derivatives inside also 00:20:44.520 --> 00:20:47.490 align:middle line:90% has radius of convergence rho. 00:20:47.490 --> 00:20:51.630 align:middle line:84% Because then, this would imply that this power series, which 00:20:51.630 --> 00:20:53.490 align:middle line:84% is the derivative of that one, converges 00:20:53.490 --> 00:20:55.870 align:middle line:90% uniformly on the same set. 00:20:55.870 --> 00:20:57.630 align:middle line:84% So then we have uniform convergence 00:20:57.630 --> 00:21:02.070 align:middle line:84% of the power series and uniform convergence of the derivative 00:21:02.070 --> 00:21:03.070 align:middle line:90% of the power series. 00:21:03.070 --> 00:21:06.870 align:middle line:84% So by the theorem we proved in the last lecture, 00:21:06.870 --> 00:21:12.600 align:middle line:84% we would have that the derivative of the power series 00:21:12.600 --> 00:21:17.220 align:middle line:84% equals the power series of the derivatives. 00:21:17.220 --> 00:21:20.640 align:middle line:84% So I'm just going to focus on this claim. 00:21:20.640 --> 00:21:26.950 align:middle line:84% And this, again, just follows from what we know about limits. 00:21:26.950 --> 00:21:35.930 align:middle line:84% And so, we compute that if we take the limit as j 00:21:35.930 --> 00:21:39.560 align:middle line:84% goes to infinity of-- now, these are the coefficients 00:21:39.560 --> 00:21:41.070 align:middle line:90% for the new power series. 00:21:41.070 --> 00:21:51.640 align:middle line:84% So aj plus 1, j plus 1 raised to the 1 over j, 00:21:51.640 --> 00:21:58.210 align:middle line:84% this is equal to limit j goes to infinity 00:21:58.210 --> 00:22:10.370 align:middle line:84% of a sub j plus 1 now raised to the 1 over j plus 1, j plus 00:22:10.370 --> 00:22:15.980 align:middle line:90% 1, 1 over j plus 1. 00:22:15.980 --> 00:22:22.220 align:middle line:90% Now all raised j plus 1 over j. 00:22:22.220 --> 00:22:25.340 align:middle line:90% 00:22:25.340 --> 00:22:29.300 align:middle line:84% Now, this was a special limit that we looked at last time. 00:22:29.300 --> 00:22:31.860 align:middle line:90% 00:22:31.860 --> 00:22:34.910 align:middle line:90% This, the limit as j goes to-- 00:22:34.910 --> 00:22:36.680 align:middle line:84% not last time, but way back when we 00:22:36.680 --> 00:22:38.480 align:middle line:84% were looking at sequences-- the limit 00:22:38.480 --> 00:22:42.770 align:middle line:84% as j goes to infinity of this guy is 1. 00:22:42.770 --> 00:22:50.320 align:middle line:84% So this in fact equals limit as j goes to infinity, 00:22:50.320 --> 00:22:52.320 align:middle line:90% this goes to 1, this goes to 1. 00:22:52.320 --> 00:22:55.540 align:middle line:84% So you can actually say this converges to 1 as well. 00:22:55.540 --> 00:23:06.850 align:middle line:84% So this equals to the j plus 1 over j. 00:23:06.850 --> 00:23:11.870 align:middle line:90% 00:23:11.870 --> 00:23:17.460 align:middle line:84% Now, strictly speaking, you would have to-- 00:23:17.460 --> 00:23:22.140 align:middle line:84% so this thing also converges to-- what does it converge to? 00:23:22.140 --> 00:23:26.940 align:middle line:84% This remember, is-- so where do we have it? 00:23:26.940 --> 00:23:29.170 align:middle line:90% Do we still have it up here? 00:23:29.170 --> 00:23:30.920 align:middle line:84% So the radius-- well, it's way over there. 00:23:30.920 --> 00:23:35.660 align:middle line:84% So let me recall that the radius of convergence rho, 00:23:35.660 --> 00:23:37.190 align:middle line:84% let me put an inverse here, this is 00:23:37.190 --> 00:23:43.590 align:middle line:84% equal to the limit as let me put a k goes to infinity, of a sub 00:23:43.590 --> 00:23:44.670 align:middle line:90% k, 1 over k. 00:23:44.670 --> 00:23:48.220 align:middle line:90% 00:23:48.220 --> 00:23:50.830 align:middle line:84% So this is the exponents converging to 1. 00:23:50.830 --> 00:23:53.930 align:middle line:84% This is converging to rho inverse again. 00:23:53.930 --> 00:24:00.190 align:middle line:90% So I get rho inverse to the 1. 00:24:00.190 --> 00:24:05.320 align:middle line:84% So therefore, this is supposed to be the radius-- 00:24:05.320 --> 00:24:10.120 align:middle line:84% 1 over the radius of convergence to the differentiated power 00:24:10.120 --> 00:24:21.630 align:middle line:84% series, which implies the differentiated power series. 00:24:21.630 --> 00:24:33.400 align:middle line:90% 00:24:33.400 --> 00:24:41.590 align:middle line:84% j is 1 over this limit, 1 over that limit, 00:24:41.590 --> 00:24:46.330 align:middle line:84% which we computed is to be is 1 over rho, 00:24:46.330 --> 00:24:49.840 align:middle line:84% which is the original radius of convergence. 00:24:49.840 --> 00:24:53.440 align:middle line:84% So all this to say is that the radius of convergence 00:24:53.440 --> 00:24:55.750 align:middle line:84% of the differentiated power series, the formerly 00:24:55.750 --> 00:24:57.940 align:middle line:84% differentiated power series, is the same 00:24:57.940 --> 00:25:03.510 align:middle line:84% as the radius of convergence of the original power series. 00:25:03.510 --> 00:25:05.700 align:middle line:84% And therefore, you have uniform convergence 00:25:05.700 --> 00:25:08.880 align:middle line:84% of the derivatives wherever you have uniform convergence 00:25:08.880 --> 00:25:11.070 align:middle line:90% of the original power series. 00:25:11.070 --> 00:25:17.470 align:middle line:84% And therefore, by the theorem we proved last time, 00:25:17.470 --> 00:25:21.040 align:middle line:90% since both the derivatives and-- 00:25:21.040 --> 00:25:23.710 align:middle line:84% so since the power series and the derivative of the power 00:25:23.710 --> 00:25:28.480 align:middle line:84% series converges uniformly on the same set, 00:25:28.480 --> 00:25:29.410 align:middle line:90% the power series-- 00:25:29.410 --> 00:25:32.300 align:middle line:90% 00:25:32.300 --> 00:25:35.150 align:middle line:84% the infinite sum, so this guy is actually differentiable. 00:25:35.150 --> 00:25:37.550 align:middle line:84% And the derivative of the power series 00:25:37.550 --> 00:25:39.530 align:middle line:84% is the power series of the derivatives. 00:25:39.530 --> 00:25:42.200 align:middle line:90% 00:25:42.200 --> 00:25:47.960 align:middle line:84% But you can iterate this, because I now have this power 00:25:47.960 --> 00:25:49.970 align:middle line:84% series with radius of convergence 00:25:49.970 --> 00:25:53.062 align:middle line:84% equal to the radius of convergence of the original. 00:25:53.062 --> 00:25:54.770 align:middle line:84% And then, I can take a derivative of that 00:25:54.770 --> 00:25:58.550 align:middle line:84% and show that has the same radius of convergence 00:25:58.550 --> 00:26:00.330 align:middle line:90% as the original. 00:26:00.330 --> 00:26:05.175 align:middle line:84% So let me just leave this as a remark 00:26:05.175 --> 00:26:06.425 align:middle line:90% and not state it as a theorem. 00:26:06.425 --> 00:26:10.610 align:middle line:90% 00:26:10.610 --> 00:26:17.640 align:middle line:84% Iterating can prove that if I want the k-th derivative 00:26:17.640 --> 00:26:19.635 align:middle line:90% of the power series. 00:26:19.635 --> 00:26:24.180 align:middle line:90% 00:26:24.180 --> 00:26:38.830 align:middle line:84% So on-- so let me say, all x in here, the k-th derivative 00:26:38.830 --> 00:26:52.200 align:middle line:84% of the power series is equal the power series 00:26:52.200 --> 00:26:55.362 align:middle line:90% of differentiating term-by-term. 00:26:55.362 --> 00:27:01.760 align:middle line:90% 00:27:01.760 --> 00:27:06.230 align:middle line:84% So this holds for k equals 1, 2, and so on. 00:27:06.230 --> 00:27:10.210 align:middle line:84% And so in particular, if I evaluate at x0, 00:27:10.210 --> 00:27:17.800 align:middle line:84% this tells me that k factorial, a sub k 00:27:17.800 --> 00:27:25.170 align:middle line:84% is equal to the derivative of this function, which 00:27:25.170 --> 00:27:28.785 align:middle line:84% is what you get out when you stick x into this power series. 00:27:28.785 --> 00:27:42.030 align:middle line:90% 00:27:42.030 --> 00:27:45.090 align:middle line:84% And you can interpret this, although we never called 00:27:45.090 --> 00:27:50.790 align:middle line:84% them Taylor series, as a statement that every power 00:27:50.790 --> 00:27:52.785 align:middle line:84% series is the Taylor series of a function. 00:27:52.785 --> 00:27:55.762 align:middle line:90% 00:27:55.762 --> 00:27:57.720 align:middle line:84% At least in this setting that we're looking at. 00:27:57.720 --> 00:28:04.510 align:middle line:90% 00:28:04.510 --> 00:28:16.390 align:middle line:84% So we've answered pretty definitively, 00:28:16.390 --> 00:28:19.340 align:middle line:84% at least for the scope of this class, 00:28:19.340 --> 00:28:22.460 align:middle line:84% when we can interchange limits, we 00:28:22.460 --> 00:28:24.830 align:middle line:84% can do that as long as we have uniform convergence 00:28:24.830 --> 00:28:29.090 align:middle line:84% of the objects that we're interested in, 00:28:29.090 --> 00:28:34.250 align:middle line:84% be it the function, the continuous functions, 00:28:34.250 --> 00:28:38.060 align:middle line:90% or function and its derivative. 00:28:38.060 --> 00:28:41.782 align:middle line:84% But there are more powerful statements out there 00:28:41.782 --> 00:28:44.240 align:middle line:84% that one can make, especially when it comes to integration. 00:28:44.240 --> 00:28:54.140 align:middle line:84% This is why essentially why a different theory of integration 00:28:54.140 --> 00:28:56.330 align:middle line:90% was created, or one reason why. 00:28:56.330 --> 00:29:01.030 align:middle line:90% 00:29:01.030 --> 00:29:08.740 align:middle line:84% But hopefully, if I see some of you in 18 102, 00:29:08.740 --> 00:29:11.110 align:middle line:84% which is what I'm teaching next semester, 00:29:11.110 --> 00:29:13.000 align:middle line:84% we'll get into that further when we 00:29:13.000 --> 00:29:15.280 align:middle line:90% discuss Lebesgue integration. 00:29:15.280 --> 00:29:19.690 align:middle line:84% That was thought of because somehow Riemann integration is 00:29:19.690 --> 00:29:23.180 align:middle line:84% not complete the same way that the rational numbers are not 00:29:23.180 --> 00:29:23.680 align:middle line:90% complete. 00:29:23.680 --> 00:29:28.280 align:middle line:90% 00:29:28.280 --> 00:29:31.650 align:middle line:84% Riemann integration is not complete, 00:29:31.650 --> 00:29:34.275 align:middle line:84% Lebesgue integration is complete in a certain sense. 00:29:34.275 --> 00:29:35.900 align:middle line:84% I'm being very vague here for a reason. 00:29:35.900 --> 00:29:43.390 align:middle line:84% So I'm just giving you where you can go with this. 00:29:43.390 --> 00:29:52.110 align:middle line:84% Is what the next step is at least improving when can 00:29:52.110 --> 00:29:57.420 align:middle line:84% you interchange two limits is really a topic that's 00:29:57.420 --> 00:29:59.310 align:middle line:90% fundamental to the study of-- 00:29:59.310 --> 00:30:02.113 align:middle line:84% in the study of Lebesgue integration. 00:30:02.113 --> 00:30:03.780 align:middle line:84% And you have much more powerful theorems 00:30:03.780 --> 00:30:07.920 align:middle line:84% there than you do here, which allows 00:30:07.920 --> 00:30:12.930 align:middle line:84% you to prove interesting results, and especially about 00:30:12.930 --> 00:30:13.980 align:middle line:90% Fourier series. 00:30:13.980 --> 00:30:18.770 align:middle line:84% In fact, if I had more time, we would 00:30:18.770 --> 00:30:20.192 align:middle line:84% apply, in fact, some of the stuff 00:30:20.192 --> 00:30:21.650 align:middle line:84% that we've done here we could apply 00:30:21.650 --> 00:30:23.150 align:middle line:90% to the study of Fourier series. 00:30:23.150 --> 00:30:29.620 align:middle line:84% But maybe we'll do that in 18 102. 00:30:29.620 --> 00:30:40.450 align:middle line:84% So I want to now prove the last theorem of the class, which 00:30:40.450 --> 00:30:44.160 align:middle line:90% is also due to the godfather. 00:30:44.160 --> 00:30:48.930 align:middle line:84% So when we had these power series, 00:30:48.930 --> 00:30:58.260 align:middle line:84% so these are defining functions, very special types 00:30:58.260 --> 00:31:01.370 align:middle line:84% of functions, what are called analytic. 00:31:01.370 --> 00:31:04.100 align:middle line:90% 00:31:04.100 --> 00:31:08.150 align:middle line:84% They are by definition essentially the limit 00:31:08.150 --> 00:31:09.680 align:middle line:90% of polynomials. 00:31:09.680 --> 00:31:15.350 align:middle line:84% They're the limit of these finite sums of a sub j's times 00:31:15.350 --> 00:31:19.700 align:middle line:84% x minus x0 raised to the j power. 00:31:19.700 --> 00:31:20.700 align:middle line:90% That's a polynomial. 00:31:20.700 --> 00:31:24.730 align:middle line:84% So for analytic functions, which are 00:31:24.730 --> 00:31:27.670 align:middle line:84% these functions equal to power series, 00:31:27.670 --> 00:31:31.960 align:middle line:84% they are the limits of polynomials. 00:31:31.960 --> 00:31:37.280 align:middle line:84% But that's a pretty small class of functions, 00:31:37.280 --> 00:31:38.450 align:middle line:90% analytic functions. 00:31:38.450 --> 00:31:41.000 align:middle line:90% 00:31:41.000 --> 00:31:47.350 align:middle line:84% But Weierstrass proved this very interesting result 00:31:47.350 --> 00:31:52.190 align:middle line:84% that actually something like this 00:31:52.190 --> 00:31:54.840 align:middle line:84% is true for all continuous functions. 00:31:54.840 --> 00:31:58.850 align:middle line:84% So roughly speaking, Weierstrass proved 00:31:58.850 --> 00:32:04.700 align:middle line:84% that basically every continuous function is in some sense 00:32:04.700 --> 00:32:07.370 align:middle line:90% almost a polynomial. 00:32:07.370 --> 00:32:10.310 align:middle line:84% Just as we solve for these analytic functions, meaning 00:32:10.310 --> 00:32:15.170 align:middle line:84% defined by power series, they are very close to being, 00:32:15.170 --> 00:32:18.980 align:middle line:84% at least within their radius of convergence, a polynomial. 00:32:18.980 --> 00:32:22.070 align:middle line:84% In fact, this is true for all continuous functions, 00:32:22.070 --> 00:32:24.470 align:middle line:84% that not necessarily equal to a power series, 00:32:24.470 --> 00:32:30.440 align:middle line:84% but every continuous function is almost a polynomial. 00:32:30.440 --> 00:32:34.160 align:middle line:84% And in what sense do I mean that? 00:32:34.160 --> 00:32:47.230 align:middle line:84% So this is Weierstrass's approximation theorem, 00:32:47.230 --> 00:32:48.610 align:middle line:90% which states the following. 00:32:48.610 --> 00:32:52.220 align:middle line:90% 00:32:52.220 --> 00:32:58.170 align:middle line:84% If f is in continuous function and on the unit interval, 00:32:58.170 --> 00:33:01.950 align:middle line:84% say, you can make it a, b, just by rescaling the variables. 00:33:01.950 --> 00:33:04.660 align:middle line:84% But I'll just state it for continuous functions 00:33:04.660 --> 00:33:05.535 align:middle line:90% on the unit interval. 00:33:05.535 --> 00:33:08.710 align:middle line:90% 00:33:08.710 --> 00:33:23.190 align:middle line:84% Then there exists a sequence polynomials 00:33:23.190 --> 00:33:31.770 align:middle line:84% p sub n of x, such that p sub n converges 00:33:31.770 --> 00:33:37.440 align:middle line:90% to f uniformly on 0, 1. 00:33:37.440 --> 00:33:43.020 align:middle line:90% 00:33:43.020 --> 00:33:47.310 align:middle line:84% And so, in this sense, every continuous function 00:33:47.310 --> 00:33:49.540 align:middle line:84% is well approximated by polynomials. 00:33:49.540 --> 00:33:56.400 align:middle line:84% So every continuous function is close to being a polynomial. 00:33:56.400 --> 00:34:01.780 align:middle line:90% 00:34:01.780 --> 00:34:08.230 align:middle line:84% So I need to first prove a couple 00:34:08.230 --> 00:34:10.840 align:middle line:90% of things that will be needed. 00:34:10.840 --> 00:34:15.130 align:middle line:84% So first off, I'm only going to look at only a-- 00:34:15.130 --> 00:34:17.515 align:middle line:84% so let me make some remarks before we go to the proof. 00:34:17.515 --> 00:34:21.639 align:middle line:90% 00:34:21.639 --> 00:34:23.980 align:middle line:84% So we're putting the proof on hold now. 00:34:23.980 --> 00:34:27.610 align:middle line:90% So let me just make a remark. 00:34:27.610 --> 00:34:30.340 align:middle line:84% I don't necessarily need to prove this statement 00:34:30.340 --> 00:34:32.650 align:middle line:90% for every f, just for certain f. 00:34:32.650 --> 00:34:35.192 align:middle line:84% And then for every f will follow from this special class. 00:34:35.192 --> 00:34:44.530 align:middle line:90% 00:34:44.530 --> 00:34:46.570 align:middle line:90% So we only need to consider-- 00:34:46.570 --> 00:34:56.300 align:middle line:84% we'll only consider the case, f of 0 equals 0, and f of 1 00:34:56.300 --> 00:34:56.900 align:middle line:90% equals 0. 00:34:56.900 --> 00:34:58.903 align:middle line:90% So f is 0 at the end points. 00:34:58.903 --> 00:35:00.320 align:middle line:84% Why we're doing this is so that we 00:35:00.320 --> 00:35:03.200 align:middle line:84% can extend f to a continuous function outside of the unit 00:35:03.200 --> 00:35:05.800 align:middle line:84% interval by just setting it equal to 0. 00:35:05.800 --> 00:35:09.020 align:middle line:90% 00:35:09.020 --> 00:35:11.482 align:middle line:84% And so, let's suppose we've proven this special case 00:35:11.482 --> 00:35:12.815 align:middle line:90% and we look at the general case. 00:35:12.815 --> 00:35:25.240 align:middle line:90% 00:35:25.240 --> 00:35:33.010 align:middle line:84% Then if I take any continuous function now, 00:35:33.010 --> 00:35:37.420 align:middle line:84% what do we know that there exists sequence of polynomials, 00:35:37.420 --> 00:35:46.010 align:middle line:84% p sub n, such that these polynomials, 00:35:46.010 --> 00:35:50.210 align:middle line:84% p sub n converge uniformly to a small modification 00:35:50.210 --> 00:35:52.160 align:middle line:90% of f that results in-- 00:35:52.160 --> 00:36:06.180 align:middle line:90% 00:36:06.180 --> 00:36:10.450 align:middle line:84% so this function here, if I look at it, 00:36:10.450 --> 00:36:12.360 align:middle line:84% it's a continuous function on 0, 1, 00:36:12.360 --> 00:36:17.070 align:middle line:84% because I'm just modifying it by constants and then times x. 00:36:17.070 --> 00:36:22.470 align:middle line:84% And then at f of 1, I get f of 1 minus f of 0 minus f of 1 00:36:22.470 --> 00:36:23.370 align:middle line:90% minus f of 0. 00:36:23.370 --> 00:36:24.390 align:middle line:90% So I get 0. 00:36:24.390 --> 00:36:27.300 align:middle line:90% And at 0, I get f of 0 minus 0. 00:36:27.300 --> 00:36:28.590 align:middle line:90% f of 0 is 0. 00:36:28.590 --> 00:36:31.410 align:middle line:84% Minus 0 times something, I don't care, equals 0. 00:36:31.410 --> 00:36:35.430 align:middle line:84% So there exist polynomials converging to this function 00:36:35.430 --> 00:36:36.690 align:middle line:90% now. 00:36:36.690 --> 00:36:40.860 align:middle line:84% If we've been able to prove the case just for f of 0 equals 0 00:36:40.860 --> 00:36:42.120 align:middle line:90% and f of 1 equals 0. 00:36:42.120 --> 00:36:45.080 align:middle line:90% 00:36:45.080 --> 00:36:51.205 align:middle line:84% So uniformly, and therefore, the polynomials given by p 00:36:51.205 --> 00:37:05.100 align:middle line:84% sub n of x plus now x times f of 1 minus f of 0 plus-- 00:37:05.100 --> 00:37:07.680 align:middle line:84% so this is still, if this was a polynomial, 00:37:07.680 --> 00:37:09.270 align:middle line:90% so is adding this to it. 00:37:09.270 --> 00:37:16.430 align:middle line:84% That's still a polynomial, converges to f tilde 00:37:16.430 --> 00:37:17.480 align:middle line:90% of x uniformly. 00:37:17.480 --> 00:37:19.580 align:middle line:84% And this, again, this is still a polynomial. 00:37:19.580 --> 00:37:25.810 align:middle line:90% 00:37:25.810 --> 00:37:29.260 align:middle line:84% So the whole point of this remark 00:37:29.260 --> 00:37:32.470 align:middle line:84% is that we only need to consider the case f of 0 equals 0 00:37:32.470 --> 00:37:34.110 align:middle line:90% and f of 1 equals 0. 00:37:34.110 --> 00:37:35.860 align:middle line:84% And we're going to do that just so that we 00:37:35.860 --> 00:37:39.130 align:middle line:84% can extend f to a continuous function outside the interval. 00:37:39.130 --> 00:37:55.380 align:middle line:90% 00:37:55.380 --> 00:37:59.940 align:middle line:84% Now, the way we're going to build these polynomials that 00:37:59.940 --> 00:38:03.660 align:middle line:84% converge to f is by what's called an approximation 00:38:03.660 --> 00:38:04.620 align:middle line:90% to the identity. 00:38:04.620 --> 00:38:07.200 align:middle line:90% 00:38:07.200 --> 00:38:09.750 align:middle line:84% Really, I guess you could think of it as an approximation 00:38:09.750 --> 00:38:11.910 align:middle line:90% to the delta function. 00:38:11.910 --> 00:38:13.810 align:middle line:84% And I'll explain that in just a minute. 00:38:13.810 --> 00:38:23.790 align:middle line:84% So for all n natural number defined c sub n, 00:38:23.790 --> 00:38:27.900 align:middle line:84% this will be the integral from minus 1 to 1 of 1 00:38:27.900 --> 00:38:33.000 align:middle line:84% minus x squared raised to the n, dx, 1 over. 00:38:33.000 --> 00:38:37.350 align:middle line:84% And then, q sub n of x, this is going 00:38:37.350 --> 00:38:41.190 align:middle line:84% to be equal to c sub n times 1 minus x 00:38:41.190 --> 00:38:44.210 align:middle line:90% squared raised to the n. 00:38:44.210 --> 00:38:51.910 align:middle line:84% Then a few simple consequences are, so first off, this 00:38:51.910 --> 00:38:58.000 align:middle line:84% is the integral of a function which is non-negative, 00:38:58.000 --> 00:39:01.270 align:middle line:84% but it's positive at plenty of places between minus 1 and 1. 00:39:01.270 --> 00:39:03.190 align:middle line:84% And you proved in the homework that this 00:39:03.190 --> 00:39:06.610 align:middle line:84% means that the integral has to be positive. 00:39:06.610 --> 00:39:17.210 align:middle line:84% Then the first is, for all N natural number, for all x in 0, 00:39:17.210 --> 00:39:17.710 align:middle line:90% 1. 00:39:17.710 --> 00:39:21.175 align:middle line:84% So the first two of these observations are very clear. 00:39:21.175 --> 00:39:23.810 align:middle line:90% 00:39:23.810 --> 00:39:30.350 align:middle line:84% qn of x is greater than-- so minus 1 to 1. 00:39:30.350 --> 00:39:33.320 align:middle line:84% qn of x is bigger than or equal to 0. 00:39:33.320 --> 00:39:36.680 align:middle line:84% So it's just c sub n, which is a positive number times 1 00:39:36.680 --> 00:39:40.760 align:middle line:84% minus x squared. x squared, so x is between minus 1 and 1. 00:39:40.760 --> 00:39:45.970 align:middle line:84% And therefore, 1 minus x squared is always non-negative. 00:39:45.970 --> 00:39:52.840 align:middle line:84% Two is, this is also pretty clear. 00:39:52.840 --> 00:39:58.870 align:middle line:84% The integral from minus 1 to 1 of q sub n of x, dx equals 1. 00:39:58.870 --> 00:40:04.520 align:middle line:90% 00:40:04.520 --> 00:40:06.350 align:middle line:90% Now, why is this clear? 00:40:06.350 --> 00:40:08.690 align:middle line:84% Because this is equal to the integral of 1 00:40:08.690 --> 00:40:12.890 align:middle line:84% minus x squared raised to the n, times c to the n, or c sub n. 00:40:12.890 --> 00:40:16.860 align:middle line:84% But c sub n is the inverse of that integral. 00:40:16.860 --> 00:40:18.980 align:middle line:90% So we should just pick up 1. 00:40:18.980 --> 00:40:23.600 align:middle line:84% Now, the third and less trivial thing, which is important, 00:40:23.600 --> 00:40:35.330 align:middle line:84% is that for all delta in 0, 1, this function q sub n, 00:40:35.330 --> 00:40:40.490 align:middle line:84% I mean, this is just a polynomial, converges to 0 00:40:40.490 --> 00:40:52.770 align:middle line:84% uniformly on the set delta is less than-- so 00:40:52.770 --> 00:40:56.520 align:middle line:84% x such that delta is bigger than or equal to the absolute value 00:40:56.520 --> 00:40:58.320 align:middle line:90% of x. 00:40:58.320 --> 00:41:00.930 align:middle line:84% No, delta is less than or equal to the absolute value of x 00:41:00.930 --> 00:41:03.420 align:middle line:90% is less than or equal to 1. 00:41:03.420 --> 00:41:11.850 align:middle line:84% So in other words, here's minus 1, 1, 0, delta, minus delta. 00:41:11.850 --> 00:41:15.820 align:middle line:84% If I look in these regions, so, if you 00:41:15.820 --> 00:41:20.880 align:middle line:84% like, in the union of those two intervals, then 00:41:20.880 --> 00:41:24.180 align:middle line:84% q sub n, this polynomial, is converging to 0 00:41:24.180 --> 00:41:32.130 align:middle line:84% uniformly as n goes to infinity on the union of these two 00:41:32.130 --> 00:41:34.390 align:middle line:90% intervals. 00:41:34.390 --> 00:41:36.520 align:middle line:84% So what is the picture of what's going on? 00:41:36.520 --> 00:41:38.020 align:middle line:84% What do these q sub n's look like? 00:41:38.020 --> 00:41:44.280 align:middle line:90% 00:41:44.280 --> 00:41:45.590 align:middle line:90% So here's minus 1. 00:41:45.590 --> 00:41:46.580 align:middle line:90% Here's 1. 00:41:46.580 --> 00:41:51.980 align:middle line:84% So the first one it looks like some constant times 1 00:41:51.980 --> 00:41:53.070 align:middle line:90% minus x squared. 00:41:53.070 --> 00:41:58.590 align:middle line:90% So there's the first one. 00:41:58.590 --> 00:42:02.960 align:middle line:84% And then, as n keeps getting bigger and bigger, 00:42:02.960 --> 00:42:08.030 align:middle line:84% this is 0 at higher and higher order at 1 and minus 1, 00:42:08.030 --> 00:42:09.950 align:middle line:84% and getting pretty small near here. 00:42:09.950 --> 00:42:17.520 align:middle line:84% And in fact, according to three, if I take any interval around 0 00:42:17.520 --> 00:42:20.430 align:middle line:84% and look outside of it, as n goes to infinity, 00:42:20.430 --> 00:42:22.320 align:middle line:84% q sub n in is going to 0 uniformly. 00:42:22.320 --> 00:42:27.670 align:middle line:84% So what it should look like is, maybe the next one, 00:42:27.670 --> 00:42:28.860 align:middle line:90% it's like that. 00:42:28.860 --> 00:42:35.230 align:middle line:84% And then further still, like that. 00:42:35.230 --> 00:42:45.280 align:middle line:84% So that if I look inside over here, or over here, 00:42:45.280 --> 00:42:47.500 align:middle line:90% q sub n is going to 0 uniformly. 00:42:47.500 --> 00:42:51.250 align:middle line:84% So what you should think of is that the q sub 00:42:51.250 --> 00:42:54.370 align:middle line:84% n's as n goes to infinity is something like a direct delta 00:42:54.370 --> 00:42:56.635 align:middle line:84% function, which is not exactly a function. 00:42:56.635 --> 00:43:00.560 align:middle line:90% 00:43:00.560 --> 00:43:06.350 align:middle line:84% So again, so let me re-emphasize that-- 00:43:06.350 --> 00:43:08.920 align:middle line:90% and this is in quotes-- 00:43:08.920 --> 00:43:37.980 align:middle line:84% you should think of q sub n acts like delta function centered 00:43:37.980 --> 00:43:40.438 align:middle line:90% at x equals 0. 00:43:40.438 --> 00:43:42.480 align:middle line:84% So that's in quotes, because that is meaningless. 00:43:42.480 --> 00:43:44.250 align:middle line:84% But some of you who've taken physics 00:43:44.250 --> 00:43:48.360 align:middle line:84% know what the properties are of a delta function. 00:43:48.360 --> 00:43:53.100 align:middle line:84% And the integral is 1, it's somehow 0 everywhere away 00:43:53.100 --> 00:43:54.180 align:middle line:90% except for the origin. 00:43:54.180 --> 00:43:56.130 align:middle line:90% And it's infinite there. 00:43:56.130 --> 00:43:57.600 align:middle line:90% But somehow, it's integral 1. 00:43:57.600 --> 00:44:05.820 align:middle line:90% 00:44:05.820 --> 00:44:09.960 align:middle line:84% So these q sub n's will show a form, 00:44:09.960 --> 00:44:12.510 align:middle line:84% an approximation to the identity in a certain sense. 00:44:12.510 --> 00:44:14.235 align:middle line:90% But let me-- let's prove these. 00:44:14.235 --> 00:44:19.210 align:middle line:90% 00:44:19.210 --> 00:44:25.730 align:middle line:84% Let's prove the only non-trivial one, which is three. 00:44:25.730 --> 00:44:28.550 align:middle line:84% Number one and two are clear based on how they're defined. 00:44:28.550 --> 00:44:33.550 align:middle line:90% 00:44:33.550 --> 00:44:36.130 align:middle line:84% So let's first estimate how big is this. 00:44:36.130 --> 00:44:38.170 align:middle line:84% So this constant c sub n, we don't really 00:44:38.170 --> 00:44:40.270 align:middle line:90% know what it is explicitly. 00:44:40.270 --> 00:44:43.640 align:middle line:84% But let's compute at least a rough size of it. 00:44:43.640 --> 00:44:53.101 align:middle line:84% So and then, we'll use this to prove the third part. 00:44:53.101 --> 00:45:02.150 align:middle line:84% So we have for all N and natural number, 00:45:02.150 --> 00:45:10.850 align:middle line:84% the following inequality in xn minus 1, 1, 1 minus x squared 00:45:10.850 --> 00:45:13.130 align:middle line:84% raised to the n, this is greater than or equal 00:45:13.130 --> 00:45:18.050 align:middle line:90% to 1 minus nx squared. 00:45:18.050 --> 00:45:20.510 align:middle line:90% Now this, if it's not-- 00:45:20.510 --> 00:45:22.850 align:middle line:90% so it shouldn't be like-- 00:45:22.850 --> 00:45:24.680 align:middle line:84% hit you in the face clear why this is true. 00:45:24.680 --> 00:45:26.780 align:middle line:84% But one way you can prove it is that if you 00:45:26.780 --> 00:45:33.780 align:middle line:84% look at the function g of x equals 1 minus x squared n. 00:45:33.780 --> 00:45:37.140 align:middle line:84% So first off, this inequality is even in x. 00:45:37.140 --> 00:45:39.340 align:middle line:84% It doesn't matter if x is negative or positive. 00:45:39.340 --> 00:45:41.085 align:middle line:90% So let's look at this function. 00:45:41.085 --> 00:45:49.980 align:middle line:90% 00:45:49.980 --> 00:45:52.755 align:middle line:90% On 0, 1, then what's the point? 00:45:52.755 --> 00:45:56.120 align:middle line:90% 00:45:56.120 --> 00:45:58.700 align:middle line:90% I look at g of 0, this is 0. 00:45:58.700 --> 00:46:01.940 align:middle line:84% And if I compute g prime of x, this 00:46:01.940 --> 00:46:08.860 align:middle line:84% is equal to n times 2x times what? 00:46:08.860 --> 00:46:15.520 align:middle line:84% Times 1 minus 1 minus x squared to the n minus 1. 00:46:15.520 --> 00:46:19.810 align:middle line:84% And this is always less than or equal to 1, 00:46:19.810 --> 00:46:21.280 align:middle line:90% this thing in parentheses. 00:46:21.280 --> 00:46:28.160 align:middle line:84% So this thing is always bigger than or equal to 0 on 0, 1. 00:46:28.160 --> 00:46:31.340 align:middle line:84% So on 0, 1, this function is increasing. 00:46:31.340 --> 00:46:33.230 align:middle line:90% And its value at 0's 0. 00:46:33.230 --> 00:46:44.395 align:middle line:84% So we get that, which is exactly what we wanted to prove. 00:46:44.395 --> 00:46:51.120 align:middle line:90% 00:46:51.120 --> 00:46:55.670 align:middle line:84% So now we compute the size of c sub n. 00:46:55.670 --> 00:47:00.360 align:middle line:84% So to do that, let me look at 1 over c sub n. 00:47:00.360 --> 00:47:02.400 align:middle line:84% If I want an upper bound on c sub n, 00:47:02.400 --> 00:47:05.550 align:middle line:84% I need to prove a lower bound on 1 over c sub n. 00:47:05.550 --> 00:47:11.070 align:middle line:84% So let's take 1 over c sub n and find a lower bound for it. 00:47:11.070 --> 00:47:14.160 align:middle line:84% This is the integral from minus 1 to 1. 00:47:14.160 --> 00:47:18.510 align:middle line:84% So 1 minus x squared raised to the n, dx. 00:47:18.510 --> 00:47:22.440 align:middle line:84% Now, I can't remember if I made this a homework problem or not. 00:47:22.440 --> 00:47:28.440 align:middle line:84% But for even functions integrating over an interval 00:47:28.440 --> 00:47:31.030 align:middle line:90% even with respect to the origin. 00:47:31.030 --> 00:47:33.508 align:middle line:90% This is just twice-- 00:47:33.508 --> 00:47:35.550 align:middle line:84% I mean, I'm sure you remember this from calculus. 00:47:35.550 --> 00:47:38.810 align:middle line:84% It's not hard to prove with what we 00:47:38.810 --> 00:47:40.800 align:middle line:84% know of the change of variables formula 00:47:40.800 --> 00:47:44.210 align:middle line:84% and so on, that this is equal to twice times the integral from 0 00:47:44.210 --> 00:47:47.240 align:middle line:84% to 1 of 1 minus x squared raised at the n. 00:47:47.240 --> 00:47:50.570 align:middle line:90% 00:47:50.570 --> 00:47:53.690 align:middle line:84% Now, this is bigger than or equal to 2 times 00:47:53.690 --> 00:47:55.220 align:middle line:84% if I integrate to a certain point. 00:47:55.220 --> 00:47:58.280 align:middle line:84% This certain point is chosen so that I 00:47:58.280 --> 00:48:03.053 align:middle line:84% get a result in the end that's pretty, essentially. 00:48:03.053 --> 00:48:15.140 align:middle line:90% 00:48:15.140 --> 00:48:20.330 align:middle line:84% Now, this is where I replace this 00:48:20.330 --> 00:48:23.790 align:middle line:84% by the smaller thing, which is easy to integrate. 00:48:23.790 --> 00:48:25.520 align:middle line:84% So this is greater than or equal to 2, 00:48:25.520 --> 00:48:34.370 align:middle line:84% integral 0 1 over root n, 1 minus nx squared dx. 00:48:34.370 --> 00:48:38.750 align:middle line:84% And I leave it to you to verify that with this choice 00:48:38.750 --> 00:48:48.120 align:middle line:84% of the endpoint what I get is 4 over 3 root n, 1 over root n. 00:48:48.120 --> 00:48:54.690 align:middle line:90% 00:48:54.690 --> 00:48:59.690 align:middle line:84% And therefore, so and this is bigger than 1 over root n. 00:48:59.690 --> 00:49:02.030 align:middle line:84% So I started with 1 over c sub n. 00:49:02.030 --> 00:49:05.690 align:middle line:84% And I showed that it was bigger than 1 over root n. 00:49:05.690 --> 00:49:12.950 align:middle line:84% And therefore, c sub n is less than 1 over-- 00:49:12.950 --> 00:49:16.240 align:middle line:84% is c sub n is less than the square root of n. 00:49:16.240 --> 00:49:23.567 align:middle line:84% Now, we'll use this to compute what we want. 00:49:23.567 --> 00:49:25.150 align:middle line:84% In fact, I mean, we really just needed 00:49:25.150 --> 00:49:31.900 align:middle line:84% to show that c sub n is bounded by some polynomial in n. 00:49:31.900 --> 00:49:34.730 align:middle line:90% But this will suffice. 00:49:34.730 --> 00:49:37.110 align:middle line:90% So we now want to show-- 00:49:37.110 --> 00:49:43.760 align:middle line:90% 00:49:43.760 --> 00:49:47.240 align:middle line:90% so let delta be positive. 00:49:47.240 --> 00:49:50.650 align:middle line:84% We now want to show that Q7 q sub n converges uniformly to 0 00:49:50.650 --> 00:49:54.345 align:middle line:84% on that set where the absolute value of x 00:49:54.345 --> 00:49:56.845 align:middle line:84% is less than or equal to 1 is bigger than or equal to delta. 00:49:56.845 --> 00:49:59.610 align:middle line:90% 00:49:59.610 --> 00:50:09.050 align:middle line:84% Now, we note that the following sequence converges to 0, 00:50:09.050 --> 00:50:14.690 align:middle line:84% that square root of n times 1 minus delta squared 00:50:14.690 --> 00:50:15.590 align:middle line:90% raised to the n. 00:50:15.590 --> 00:50:18.095 align:middle line:84% So we should also put deltas in 0, 1. 00:50:18.095 --> 00:50:26.970 align:middle line:90% 00:50:26.970 --> 00:50:34.670 align:middle line:84% So this sequence here converges to 0 as n goes to infinity. 00:50:34.670 --> 00:50:36.530 align:middle line:90% Now intuitively, why is this? 00:50:36.530 --> 00:50:41.140 align:middle line:84% This is because this is some number less than 1 00:50:41.140 --> 00:50:43.990 align:middle line:90% raised to the n-th power. 00:50:43.990 --> 00:50:51.610 align:middle line:84% Exponential always beats just a polynomial or a power of n. 00:50:51.610 --> 00:50:54.460 align:middle line:84% If you want to see exactly why this is, 00:50:54.460 --> 00:50:57.790 align:middle line:84% we could compute the limit as n goes 00:50:57.790 --> 00:51:06.450 align:middle line:84% to infinity of this sequence raised to 1 over n power. 00:51:06.450 --> 00:51:09.300 align:middle line:84% Then this is equal to the limit as n 00:51:09.300 --> 00:51:15.480 align:middle line:84% goes to infinity of n to the 1 over n 1/2, 00:51:15.480 --> 00:51:18.180 align:middle line:90% 1 minus delta squared. 00:51:18.180 --> 00:51:19.590 align:middle line:90% And this converges to 1. 00:51:19.590 --> 00:51:20.440 align:middle line:90% We proved that. 00:51:20.440 --> 00:51:22.510 align:middle line:84% And so, this equals 1 minus delta squared, 00:51:22.510 --> 00:51:24.420 align:middle line:90% which is less than 1. 00:51:24.420 --> 00:51:31.200 align:middle line:84% And we have this theorem from our section on sequences 00:51:31.200 --> 00:51:34.530 align:middle line:84% that says, if this limit is less than 1, 00:51:34.530 --> 00:51:38.610 align:middle line:84% then the thing here converges to 0. 00:51:38.610 --> 00:51:40.320 align:middle line:84% Or you could interpret this as saying 00:51:40.320 --> 00:51:45.480 align:middle line:84% that the series with this as the individual terms converges. 00:51:45.480 --> 00:51:49.920 align:middle line:84% And therefore, the individual terms have to converge to 0. 00:51:49.920 --> 00:52:02.860 align:middle line:84% So which implies limit as n goes to infinity equals 0. 00:52:02.860 --> 00:52:04.100 align:middle line:90% So now we have this. 00:52:04.100 --> 00:52:08.870 align:middle line:84% And once we have this, we'll have what we want. 00:52:08.870 --> 00:52:16.340 align:middle line:84% So we want to prove uniform convergence on that set. 00:52:16.340 --> 00:52:17.800 align:middle line:90% So let epsilon be positive. 00:52:17.800 --> 00:52:20.710 align:middle line:90% 00:52:20.710 --> 00:52:24.610 align:middle line:84% Then since this sequence converges to 0, 00:52:24.610 --> 00:52:27.250 align:middle line:84% there exists in M a natural number, 00:52:27.250 --> 00:52:32.440 align:middle line:84% so that for all n bigger than or equal to M, square root of n 00:52:32.440 --> 00:52:36.160 align:middle line:84% times 1 minus delta squared raised to the n 00:52:36.160 --> 00:52:39.470 align:middle line:90% is less than epsilon. 00:52:39.470 --> 00:52:43.390 align:middle line:84% And for all n bigger than or equal to M, 00:52:43.390 --> 00:52:51.250 align:middle line:84% for all x so that delta is between less than 1, 00:52:51.250 --> 00:52:58.090 align:middle line:84% we get that q sub n, so it's non-negative 00:52:58.090 --> 00:52:59.470 align:middle line:90% minus 0 in absolute value. 00:52:59.470 --> 00:53:06.360 align:middle line:84% That's just q sub n of x raised to the n. 00:53:06.360 --> 00:53:09.570 align:middle line:84% This is less than or equal to the square root of n. 00:53:09.570 --> 00:53:10.740 align:middle line:90% That's for this guy. 00:53:10.740 --> 00:53:15.720 align:middle line:84% 1 minus x squared, this is getting smaller 00:53:15.720 --> 00:53:17.790 align:middle line:90% as I get closer to 1. 00:53:17.790 --> 00:53:19.710 align:middle line:90% So its biggest at delta. 00:53:19.710 --> 00:53:23.660 align:middle line:90% 00:53:23.660 --> 00:53:27.095 align:middle line:84% And how we chose M is, this is less than epsilon. 00:53:27.095 --> 00:53:36.080 align:middle line:90% 00:53:36.080 --> 00:53:41.740 align:middle line:84% So now, we're ready for the proof of the Weierstrass 00:53:41.740 --> 00:53:42.940 align:middle line:90% approximation theorem. 00:53:42.940 --> 00:53:55.930 align:middle line:90% 00:53:55.930 --> 00:54:01.780 align:middle line:84% So suppose f is continuous function on 0, 1. 00:54:01.780 --> 00:54:07.165 align:middle line:84% f of 0 equals 0, f of 1 equals 0. 00:54:07.165 --> 00:54:17.570 align:middle line:90% 00:54:17.570 --> 00:54:25.307 align:middle line:84% So this polynomial here is very concentrated at 0. 00:54:25.307 --> 00:54:28.650 align:middle line:90% 00:54:28.650 --> 00:54:32.430 align:middle line:90% So well first off, suppose f-- 00:54:32.430 --> 00:54:33.610 align:middle line:90% and it's continuous here. 00:54:33.610 --> 00:54:45.110 align:middle line:84% So you can check that if it's 0 there, 00:54:45.110 --> 00:54:48.800 align:middle line:84% and I extend it to be 0 outside of 0, 1, 00:54:48.800 --> 00:54:50.450 align:middle line:84% this is still a continuous function. 00:54:50.450 --> 00:54:52.670 align:middle line:84% I'm just defining it this way because I 00:54:52.670 --> 00:54:55.550 align:middle line:84% want to write certain symbols in a little bit 00:54:55.550 --> 00:54:59.010 align:middle line:84% without specifying exactly where the bounds of the integration 00:54:59.010 --> 00:54:59.510 align:middle line:90% are. 00:54:59.510 --> 00:55:02.350 align:middle line:90% 00:55:02.350 --> 00:55:15.610 align:middle line:84% So we extend f by 0 outside of 0, 1. 00:55:15.610 --> 00:55:19.990 align:middle line:84% And this function f is continuous function 00:55:19.990 --> 00:55:22.240 align:middle line:90% on the real number line. 00:55:22.240 --> 00:55:29.840 align:middle line:90% 00:55:29.840 --> 00:55:31.850 align:middle line:84% So we define-- now we're going to define 00:55:31.850 --> 00:55:37.350 align:middle line:84% this polynomial, sequence of polynomials, p sub n. 00:55:37.350 --> 00:55:40.370 align:middle line:90% 00:55:40.370 --> 00:55:45.760 align:middle line:84% This is going to be equal to the integral from 0 to 1 00:55:45.760 --> 00:55:52.690 align:middle line:84% f of t times q sub n of t minus x, dt. 00:55:52.690 --> 00:55:54.940 align:middle line:84% And just to remind you, this is equal to the integral 00:55:54.940 --> 00:55:58.930 align:middle line:84% from 0 to 1, of f of t times c sub 00:55:58.930 --> 00:56:07.140 align:middle line:84% n, 1 minus x minus t squared raised to the n, dt. 00:56:07.140 --> 00:56:13.560 align:middle line:84% So this is just c sub n times this thing. 00:56:13.560 --> 00:56:16.550 align:middle line:84% If you expand everything out, is just a-- 00:56:16.550 --> 00:56:21.350 align:middle line:90% 00:56:21.350 --> 00:56:24.290 align:middle line:84% using the binomial theorem, this is equal to j equals-- 00:56:24.290 --> 00:56:29.575 align:middle line:90% 00:56:29.575 --> 00:56:30.950 align:middle line:84% I'm just going to put a sum here, 00:56:30.950 --> 00:56:37.760 align:middle line:84% meaning a finite sum, of some numbers a sub j, n times 00:56:37.760 --> 00:56:43.050 align:middle line:90% x to the j, times t to the-- 00:56:43.050 --> 00:56:51.040 align:middle line:90% 00:56:51.040 --> 00:56:59.220 align:middle line:84% well, just some finite numbers a jk, times x the j, t to the k. 00:56:59.220 --> 00:57:00.810 align:middle line:84% All I'm saying is, if you expand this 00:57:00.810 --> 00:57:02.790 align:middle line:84% out using the binomial theorem, you 00:57:02.790 --> 00:57:06.275 align:middle line:84% get this theorem, this polynomial and x sub j, 00:57:06.275 --> 00:57:08.190 align:middle line:90% t sub k. 00:57:08.190 --> 00:57:16.180 align:middle line:84% And then, this is getting integrated against f of t dt. 00:57:16.180 --> 00:57:19.578 align:middle line:84% And so, this is in fact a polynomial. 00:57:19.578 --> 00:57:20.370 align:middle line:90% And I'll give you-- 00:57:20.370 --> 00:57:24.570 align:middle line:90% 00:57:24.570 --> 00:57:29.400 align:middle line:84% so let's write this out and be precise, 00:57:29.400 --> 00:57:32.670 align:middle line:84% just so that you're convinced that it's a polynomial. 00:57:32.670 --> 00:57:36.060 align:middle line:84% This is equal to the integral from 0 to 1 of f of t times 00:57:36.060 --> 00:57:37.080 align:middle line:90% c sub n. 00:57:37.080 --> 00:57:42.740 align:middle line:90% Now we use the binomial theorem. 00:57:42.740 --> 00:57:46.925 align:middle line:84% This is equal to j 0 to n, and choose j. 00:57:46.925 --> 00:57:51.320 align:middle line:90% 00:57:51.320 --> 00:57:59.000 align:middle line:84% And then, minus x minus t squared 00:57:59.000 --> 00:58:03.990 align:middle line:90% raised to the n minus j. 00:58:03.990 --> 00:58:07.120 align:middle line:90% Or I could put j, 2j. 00:58:07.120 --> 00:58:19.590 align:middle line:90% 00:58:19.590 --> 00:58:23.770 align:middle line:84% And then, det. I'm not sure if this 00:58:23.770 --> 00:58:25.360 align:middle line:90% is even helping or anything. 00:58:25.360 --> 00:58:29.470 align:middle line:84% But just so that you see this is actually a polynomial. 00:58:29.470 --> 00:58:34.300 align:middle line:84% f of t, c sub n, sum from j equals 0 to n. 00:58:34.300 --> 00:58:41.920 align:middle line:84% And now, sum from k equals 0 to 2j, n choose j minus 1 00:58:41.920 --> 00:58:42.490 align:middle line:90% to the j. 00:58:42.490 --> 00:58:45.870 align:middle line:90% 00:58:45.870 --> 00:58:56.670 align:middle line:84% Now, 2j choose k minus t to the k times x to the j minus k. 00:58:56.670 --> 00:59:01.680 align:middle line:90% 00:59:01.680 --> 00:59:03.480 align:middle line:90% And then, dt. 00:59:03.480 --> 00:59:07.230 align:middle line:84% So I have all this junk integrated dt, 00:59:07.230 --> 00:59:09.510 align:middle line:84% and then this is x to the 2j minus k. 00:59:09.510 --> 00:59:10.800 align:middle line:90% So that just pops out. 00:59:10.800 --> 00:59:12.780 align:middle line:90% So this is a polynomial. 00:59:12.780 --> 00:59:15.790 align:middle line:90% 00:59:15.790 --> 00:59:17.110 align:middle line:90% I mean, this is a polynomial. 00:59:17.110 --> 00:59:19.000 align:middle line:84% But then, when it integrates against f of t, 00:59:19.000 --> 00:59:22.120 align:middle line:84% this x to the 2j minus k comes out of the integral. 00:59:22.120 --> 00:59:25.930 align:middle line:90% And I get that times just all-- 00:59:25.930 --> 00:59:28.930 align:middle line:84% this becomes a sum of terms with x to the 2j minus 00:59:28.930 --> 00:59:33.640 align:middle line:84% k outside this integral, times f of t, times 00:59:33.640 --> 00:59:37.420 align:middle line:84% integrated against minus 2 to the k. 00:59:37.420 --> 00:59:39.350 align:middle line:84% I think I said more than I needed to there, 00:59:39.350 --> 00:59:46.150 align:middle line:84% but so the point is p sub n of x is a polynomial. 00:59:46.150 --> 00:59:49.240 align:middle line:84% Now, we can write this slightly differently. 00:59:49.240 --> 00:59:56.440 align:middle line:90% 00:59:56.440 --> 01:00:08.120 align:middle line:84% So piece sub n of x, this is equal to just as it was before. 01:00:08.120 --> 01:00:11.140 align:middle line:84% Now this, we change variables and is equal to-- 01:00:11.140 --> 01:00:13.750 align:middle line:90% 01:00:13.750 --> 01:00:18.190 align:middle line:84% so we do a u substitution now, where u is equal to x minus t. 01:00:18.190 --> 01:00:31.070 align:middle line:84% And then I'm going to change the dt. 01:00:31.070 --> 01:00:33.230 align:middle line:84% So what I did here was a change of variables. 01:00:33.230 --> 01:00:43.630 align:middle line:90% 01:00:43.630 --> 01:00:47.860 align:middle line:84% By setting u is equal to x minus t. 01:00:47.860 --> 01:00:50.620 align:middle line:84% Or I guess, if you'd like, t minus x. 01:00:50.620 --> 01:00:53.150 align:middle line:90% 01:00:53.150 --> 01:00:54.800 align:middle line:84% And then, I just recalled-- and then 01:00:54.800 --> 01:00:56.780 align:middle line:90% I just called ut then again. 01:00:56.780 --> 01:01:05.680 align:middle line:90% 01:01:05.680 --> 01:01:09.360 align:middle line:84% So I'm eventually looking at these polynomials only 01:01:09.360 --> 01:01:11.490 align:middle line:90% in the interval 0, 1. 01:01:11.490 --> 01:01:16.170 align:middle line:90% So this is what it looks like. 01:01:16.170 --> 01:01:24.760 align:middle line:84% But now, this f of x plus t, for t 01:01:24.760 --> 01:01:27.490 align:middle line:84% between minus x and 1 minus x, it's 01:01:27.490 --> 01:01:29.930 align:middle line:90% actually 0 outside of that. 01:01:29.930 --> 01:01:36.640 align:middle line:84% So I can extend the integration to minus 1 and 1. 01:01:36.640 --> 01:01:42.250 align:middle line:84% Again, with the understanding that I've extended 01:01:42.250 --> 01:01:46.220 align:middle line:90% f to be 0 outside of 0, 1. 01:01:46.220 --> 01:01:49.720 align:middle line:90% 01:01:49.720 --> 01:01:59.300 align:middle line:90% So since fy equals 0-- 01:01:59.300 --> 01:02:00.490 align:middle line:90% or let me write it this way. 01:02:00.490 --> 01:02:10.860 align:middle line:84% Since f of x plus t equals 0 for t not n minus x and 1 minus x. 01:02:10.860 --> 01:02:14.780 align:middle line:90% 01:02:14.780 --> 01:02:18.630 align:middle line:84% So that was a lot of explaining for some stuff. 01:02:18.630 --> 01:02:23.390 align:middle line:84% But here, let me get to the point. 01:02:23.390 --> 01:02:25.190 align:middle line:90% What is p sub n really? 01:02:25.190 --> 01:02:30.320 align:middle line:84% So why should you expect this to converge to f? 01:02:30.320 --> 01:02:32.510 align:middle line:90% So this is a little discussion. 01:02:32.510 --> 01:02:36.500 align:middle line:90% 01:02:36.500 --> 01:02:38.390 align:middle line:90% So I said that-- 01:02:38.390 --> 01:02:41.240 align:middle line:84% and this is for those who understood my comment 01:02:41.240 --> 01:02:42.680 align:middle line:90% about delta functions. 01:02:42.680 --> 01:02:45.420 align:middle line:90% 01:02:45.420 --> 01:02:53.740 align:middle line:84% In q sub n is very concentrated at 0, at t equals 0. 01:02:53.740 --> 01:03:04.370 align:middle line:84% So this, you should think of as being approximately 01:03:04.370 --> 01:03:07.478 align:middle line:90% like a delta function at t. 01:03:07.478 --> 01:03:09.770 align:middle line:84% If you don't know what delta function is or never heard 01:03:09.770 --> 01:03:12.140 align:middle line:84% of it, then forget the rest of this remark. 01:03:12.140 --> 01:03:15.830 align:middle line:90% But and then I'll say something. 01:03:15.830 --> 01:03:19.790 align:middle line:84% So q sub n kind of looks like a delta function. 01:03:19.790 --> 01:03:29.460 align:middle line:84% And therefore, minus 1 of x plus t qn of t dt 01:03:29.460 --> 01:03:37.080 align:middle line:84% should look more and more like f of x plus t delta of t dt. 01:03:37.080 --> 01:03:40.900 align:middle line:84% And what we know about direct delta functions is 01:03:40.900 --> 01:03:42.900 align:middle line:84% that when you integrate them against a function, 01:03:42.900 --> 01:03:48.555 align:middle line:84% you just pick up the function evaluated at 0, which is-- 01:03:48.555 --> 01:03:54.700 align:middle line:90% 01:03:54.700 --> 01:03:55.240 align:middle line:90% OK. 01:03:55.240 --> 01:03:59.170 align:middle line:84% So that's kind of why you expect it. 01:03:59.170 --> 01:04:02.020 align:middle line:84% I mean, if you look back at this picture of what the qn's 01:04:02.020 --> 01:04:06.040 align:middle line:84% are looking like, they are concentrating more and more 01:04:06.040 --> 01:04:07.120 align:middle line:90% at 0. 01:04:07.120 --> 01:04:12.490 align:middle line:84% So all of the contribution to this integral 01:04:12.490 --> 01:04:16.690 align:middle line:84% here, which defines these polynomials, is happening at t 01:04:16.690 --> 01:04:17.620 align:middle line:90% equals 0. 01:04:17.620 --> 01:04:20.335 align:middle line:84% At t equals 0, I just pick up f of x. 01:04:20.335 --> 01:04:23.040 align:middle line:90% 01:04:23.040 --> 01:04:25.070 align:middle line:84% So that's why you should kind of expect 01:04:25.070 --> 01:04:33.160 align:middle line:84% these polynomials to converge back to the function. 01:04:33.160 --> 01:04:37.663 align:middle line:90% 01:04:37.663 --> 01:04:39.380 align:middle line:84% So I think I can finish on this board. 01:04:39.380 --> 01:04:53.135 align:middle line:90% 01:04:53.135 --> 01:04:55.510 align:middle line:84% So this is a picture you should have seared in your mind. 01:04:55.510 --> 01:04:59.010 align:middle line:84% This is what the q sub n's look like, 01:04:59.010 --> 01:05:01.410 align:middle line:84% that they're concentrating all of their mass right 01:05:01.410 --> 01:05:02.460 align:middle line:90% at the origin. 01:05:02.460 --> 01:05:09.330 align:middle line:84% And therefore, I should just pick up f at the point x. 01:05:09.330 --> 01:05:11.910 align:middle line:84% since this is concentrating at the origin, t equals 0, 01:05:11.910 --> 01:05:14.730 align:middle line:84% I should just pick up f of x plus t at t 01:05:14.730 --> 01:05:16.440 align:middle line:90% equals 0, which is f of x. 01:05:16.440 --> 01:05:19.560 align:middle line:90% 01:05:19.560 --> 01:05:22.820 align:middle line:84% And why f of x plus t, not some constant times f of x plus t, 01:05:22.820 --> 01:05:26.540 align:middle line:84% that's because the integral of q sub n's is 1. 01:05:26.540 --> 01:05:30.040 align:middle line:90% 01:05:30.040 --> 01:05:34.390 align:middle line:84% So now, let's prove that the pn's converge to f uniformly. 01:05:34.390 --> 01:05:54.740 align:middle line:90% 01:05:54.740 --> 01:05:56.450 align:middle line:90% So let epsilon be positive. 01:05:56.450 --> 01:06:00.630 align:middle line:90% 01:06:00.630 --> 01:06:06.690 align:middle line:84% Since f is a continuous function on a closed and bounded 01:06:06.690 --> 01:06:09.240 align:middle line:84% interval, we know that it's uniformly continuous. 01:06:09.240 --> 01:06:16.620 align:middle line:90% 01:06:16.620 --> 01:06:20.460 align:middle line:84% And therefore, there exists delta positive such that-- 01:06:20.460 --> 01:06:25.215 align:middle line:90% 01:06:25.215 --> 01:06:26.340 align:middle line:90% how am I writing this here? 01:06:26.340 --> 01:06:32.050 align:middle line:90% 01:06:32.050 --> 01:06:38.200 align:middle line:84% Such that if-- let's write it this way-- for all y, 01:06:38.200 --> 01:06:41.680 align:middle line:84% so that z minus y less than delta, 01:06:41.680 --> 01:06:48.220 align:middle line:84% we get that f of z minus f of y is-- 01:06:48.220 --> 01:07:01.730 align:middle line:90% 01:07:01.730 --> 01:07:02.910 align:middle line:90% actually, we don't need-- 01:07:02.910 --> 01:07:03.410 align:middle line:90% OK. 01:07:03.410 --> 01:07:09.460 align:middle line:84% So I mean, we don't need so much of about f. 01:07:09.460 --> 01:07:11.110 align:middle line:90% But we'll go with this anyways. 01:07:11.110 --> 01:07:13.980 align:middle line:90% 01:07:13.980 --> 01:07:15.510 align:middle line:90% We really just know-- 01:07:15.510 --> 01:07:17.400 align:middle line:90% OK, never mind. 01:07:17.400 --> 01:07:18.090 align:middle line:90% I'll stop. 01:07:18.090 --> 01:07:21.500 align:middle line:84% So we know that f is uniformly continuous 01:07:21.500 --> 01:07:24.370 align:middle line:84% so there exists a delta, so that I have this. 01:07:24.370 --> 01:07:30.120 align:middle line:84% Yeah, so less than epsilon over 2. 01:07:30.120 --> 01:07:32.610 align:middle line:84% So z and y are within delta to each other. 01:07:32.610 --> 01:07:35.410 align:middle line:84% Then f of z and f of y, no matter what z and y are, 01:07:35.410 --> 01:07:39.330 align:middle line:84% are within epsilon over 2 of each other. 01:07:39.330 --> 01:07:53.410 align:middle line:84% Now, since f is continuous, it has a maximum on this interval. 01:07:53.410 --> 01:07:56.240 align:middle line:84% Well, I should say, since f is a continuous function, 01:07:56.240 --> 01:07:58.940 align:middle line:84% it has both a max and a min on this interval. 01:07:58.940 --> 01:08:16.899 align:middle line:84% And m there exist some number c such that f of x 01:08:16.899 --> 01:08:19.180 align:middle line:84% is bounded by c for all x in 0, 1. 01:08:19.180 --> 01:08:24.760 align:middle line:90% 01:08:24.760 --> 01:08:28.640 align:middle line:84% So now I have this delta coming from the uniform continuity 01:08:28.640 --> 01:08:29.140 align:middle line:90% of f. 01:08:29.140 --> 01:08:30.850 align:middle line:84% I have this c coming from the fact 01:08:30.850 --> 01:08:32.689 align:middle line:84% that it's bounded on this interval. 01:08:32.689 --> 01:08:36.069 align:middle line:84% And now, I'm going to choose my M for the uniform convergence 01:08:36.069 --> 01:08:38.020 align:middle line:84% of the polynomials just depending 01:08:38.020 --> 01:08:41.590 align:middle line:90% on these pieces of input. 01:08:41.590 --> 01:08:50.689 align:middle line:84% Then as we showed, since square root of n 1 minus delta 01:08:50.689 --> 01:08:55.130 align:middle line:84% squared and converges to 0, this implies 01:08:55.130 --> 01:08:58.939 align:middle line:84% there exists an M natural number so 01:08:58.939 --> 01:09:05.240 align:middle line:84% that for all n bigger than or equal to M, 01:09:05.240 --> 01:09:09.859 align:middle line:84% I get that square root of n 1 minus delta squared raised 01:09:09.859 --> 01:09:17.395 align:middle line:84% to the n is less than epsilon over 8c. 01:09:17.395 --> 01:09:24.109 align:middle line:90% 01:09:24.109 --> 01:09:25.790 align:middle line:90% I claim this M works. 01:09:25.790 --> 01:09:29.240 align:middle line:84% That for all n bigger than or equal to M, for all x in 0, 1, 01:09:29.240 --> 01:09:31.489 align:middle line:90% pn minus f is less than epsilon. 01:09:31.489 --> 01:09:36.040 align:middle line:90% 01:09:36.040 --> 01:09:42.180 align:middle line:84% And for all n bigger than or equal to M, for all x in 0, 01:09:42.180 --> 01:09:52.140 align:middle line:84% 1, if we look at p sub n of x minus f of x, 01:09:52.140 --> 01:09:54.720 align:middle line:84% we're going to use that this thing is 01:09:54.720 --> 01:09:57.370 align:middle line:84% an approximation to the identity, 01:09:57.370 --> 01:10:03.210 align:middle line:84% meaning it's essentially that it satisfies those three 01:10:03.210 --> 01:10:07.060 align:middle line:84% properties that I wrote a minute ago. 01:10:07.060 --> 01:10:11.250 align:middle line:84% So this is equal to integral from minus 1 to 1 01:10:11.250 --> 01:10:21.720 align:middle line:84% of f of x plus t times qn of t dt minus f of x. 01:10:21.720 --> 01:10:25.070 align:middle line:90% 01:10:25.070 --> 01:10:27.980 align:middle line:84% What's the integral from minus 1 to 1 of qn sub t? 01:10:27.980 --> 01:10:29.150 align:middle line:90% That's equal to 1. 01:10:29.150 --> 01:10:31.920 align:middle line:90% So f of x is at times-- 01:10:31.920 --> 01:10:34.220 align:middle line:84% so this is-- all of analysis is writing 01:10:34.220 --> 01:10:38.840 align:middle line:90% 1 or 0 in a certain way. 01:10:38.840 --> 01:10:46.670 align:middle line:84% Not all, but-- so I can write this as f of x plus t now 01:10:46.670 --> 01:10:53.660 align:middle line:90% minus f of x times qn of t, dt. 01:10:53.660 --> 01:10:58.460 align:middle line:84% Again, because the integral of q sub n of t equals 1. 01:10:58.460 --> 01:11:01.370 align:middle line:84% So again, if you just expand this out, this is f of x. 01:11:01.370 --> 01:11:03.260 align:middle line:84% I'm integrating with respect to t. 01:11:03.260 --> 01:11:04.790 align:middle line:84% So I just pick up what I had before. 01:11:04.790 --> 01:11:08.263 align:middle line:90% 01:11:08.263 --> 01:11:10.180 align:middle line:84% Now, by the triangle inequality for integrals, 01:11:10.180 --> 01:11:13.510 align:middle line:84% this is less than or equal to the integral for minus 1 to 1 01:11:13.510 --> 01:11:20.205 align:middle line:84% of f of x plus t minus f of x times the absolute value of q 01:11:20.205 --> 01:11:21.010 align:middle line:90% sub n of t. 01:11:21.010 --> 01:11:26.730 align:middle line:84% But since q sub n of t is not negative, 01:11:26.730 --> 01:11:29.710 align:middle line:90% that's just q sub n of t. 01:11:29.710 --> 01:11:33.085 align:middle line:84% And now, I'm going to split this integral up into two parts. 01:11:33.085 --> 01:11:35.650 align:middle line:90% 01:11:35.650 --> 01:11:41.040 align:middle line:84% This is equal to a part where t is 01:11:41.040 --> 01:11:45.420 align:middle line:84% less than or equal to-- so should 01:11:45.420 --> 01:11:50.880 align:middle line:84% minus delta, delta f of x plus t, 01:11:50.880 --> 01:11:56.160 align:middle line:90% minus f of x, q sub n of t dt. 01:11:56.160 --> 01:12:03.600 align:middle line:84% And then plus the other part, which I can write as-- 01:12:03.600 --> 01:12:10.260 align:middle line:84% so it's a union of the two intervals now away from delta 01:12:10.260 --> 01:12:13.240 align:middle line:90% out to 1. 01:12:13.240 --> 01:12:17.500 align:middle line:84% So it's the sum of the integral over these two intervals, which 01:12:17.500 --> 01:12:20.980 align:middle line:84% I'm going to write as the integral over 01:12:20.980 --> 01:12:25.300 align:middle line:84% delta less than or equal to t is less than or equal to 1. 01:12:25.300 --> 01:12:32.800 align:middle line:90% And times qn of t. 01:12:32.800 --> 01:12:38.970 align:middle line:90% 01:12:38.970 --> 01:12:43.890 align:middle line:84% Now, t is between minus delta and delta. 01:12:43.890 --> 01:12:48.300 align:middle line:84% And therefore, x plus t minus x in absolute value 01:12:48.300 --> 01:12:50.460 align:middle line:90% is less than delta. 01:12:50.460 --> 01:12:53.980 align:middle line:90% 01:12:53.980 --> 01:12:58.000 align:middle line:84% So in this integral here, I want to note 01:12:58.000 --> 01:13:07.680 align:middle line:84% that if t is in this interval here, x plus t minus x equals 01:13:07.680 --> 01:13:10.870 align:middle line:84% t is less than delta in this here. 01:13:10.870 --> 01:13:14.940 align:middle line:84% So since this guy minus this guy is less than delta 01:13:14.940 --> 01:13:21.630 align:middle line:84% in absolute value, I can use this uniform continuity part 01:13:21.630 --> 01:13:29.350 align:middle line:84% to say that this is less than epsilon over 2qn of t dt. 01:13:29.350 --> 01:13:32.620 align:middle line:90% 01:13:32.620 --> 01:13:36.490 align:middle line:84% Plus now the absolute value of this guy is less than-- 01:13:36.490 --> 01:13:39.040 align:middle line:84% by the triangle inequality-- is less than or equal to the sum 01:13:39.040 --> 01:13:43.750 align:middle line:84% of the absolute values, which is less than or equal to c-- 01:13:43.750 --> 01:13:44.380 align:middle line:90% 2c. 01:13:44.380 --> 01:13:50.630 align:middle line:90% So 1, and then 2c. 01:13:50.630 --> 01:13:54.430 align:middle line:90% 01:13:54.430 --> 01:13:56.975 align:middle line:84% Because that's bounded by c, that's bounded by c. 01:13:56.975 --> 01:13:58.600 align:middle line:84% So the absolute value of the difference 01:13:58.600 --> 01:14:01.600 align:middle line:84% is less than or equal to the sum of the absolute values, which 01:14:01.600 --> 01:14:04.210 align:middle line:84% is bounded by each of those bounded by c. 01:14:04.210 --> 01:14:08.560 align:middle line:90% 01:14:08.560 --> 01:14:12.490 align:middle line:84% Now, this is less than epsilon over 2 integral 01:14:12.490 --> 01:14:18.190 align:middle line:84% from, if I just integrate the whole thing, from minus 1 to 1, 01:14:18.190 --> 01:14:19.240 align:middle line:90% that just gives me 1. 01:14:19.240 --> 01:14:23.530 align:middle line:90% Plus 2c times qn of t. 01:14:23.530 --> 01:14:30.500 align:middle line:84% On this interval is, again, so I shouldn't put a delta there. 01:14:30.500 --> 01:14:39.500 align:middle line:84% 2c, c sub n, 1 minus delta squared raised to the n, dt. 01:14:39.500 --> 01:14:48.940 align:middle line:90% And so there's no t here. 01:14:48.940 --> 01:14:52.560 align:middle line:84% So this is equal to epsilon over 2, 01:14:52.560 --> 01:14:56.505 align:middle line:84% because this integral is 1 plus this-- 01:14:56.505 --> 01:15:00.740 align:middle line:90% 01:15:00.740 --> 01:15:04.430 align:middle line:84% c sub n, remember, is less than the square root of n times 01:15:04.430 --> 01:15:07.670 align:middle line:84% 1 minus delta squared and times the integral 01:15:07.670 --> 01:15:09.950 align:middle line:84% over this region dt, which I can make bigger 01:15:09.950 --> 01:15:11.750 align:middle line:90% by going from minus 1 to 1. 01:15:11.750 --> 01:15:12.890 align:middle line:90% It gives me 2. 01:15:12.890 --> 01:15:14.600 align:middle line:84% And this is equal to epsilon over 01:15:14.600 --> 01:15:20.420 align:middle line:84% 2 plus 4c square root of n, 1 minus delta squared n. 01:15:20.420 --> 01:15:26.570 align:middle line:84% And this is less than epsilon over 2, plus epsilon over 2, 01:15:26.570 --> 01:15:27.710 align:middle line:90% equals epsilon. 01:15:27.710 --> 01:15:29.210 align:middle line:84% I think I have one minute to spare. 01:15:29.210 --> 01:15:34.872 align:middle line:84% So this was quite an experience teaching to an empty room. 01:15:34.872 --> 01:15:36.830 align:middle line:84% I hope you did get something out of this class. 01:15:36.830 --> 01:15:39.390 align:middle line:90% 01:15:39.390 --> 01:15:41.530 align:middle line:84% Unfortunately, I wasn't able to meet a lot of you. 01:15:41.530 --> 01:15:44.670 align:middle line:84% And that's one of the best parts about teaching and being 01:15:44.670 --> 01:15:50.240 align:middle line:84% able to see you grasp in real time what I'm talking about. 01:15:50.240 --> 01:15:53.830 align:middle line:84% So hopefully this nightmare will end soon, 01:15:53.830 --> 01:15:57.690 align:middle line:84% and we'll get to see each other in the future. 01:15:57.690 --> 01:16:06.000 align:middle line:90%