WEBVTT 00:00:00.000 --> 00:00:01.992 align:middle line:90% [SQUEAKING] 00:00:01.992 --> 00:00:02.988 align:middle line:90% [RUSTLING] 00:00:02.988 --> 00:00:04.980 align:middle line:90% [CLICKING] 00:00:04.980 --> 00:00:14.320 align:middle line:90% 00:00:14.320 --> 00:00:17.410 align:middle line:84% CASEY RODRIQUEZ: So we're going to continue with our discussion 00:00:17.410 --> 00:00:21.060 align:middle line:90% of the derivative. 00:00:21.060 --> 00:00:27.900 align:middle line:84% So now, let me recall the definition 00:00:27.900 --> 00:00:30.520 align:middle line:84% we introduced at the end of last time of the derivative. 00:00:30.520 --> 00:00:34.170 align:middle line:84% So let I be an interval, meaning it could be open, 00:00:34.170 --> 00:00:37.450 align:middle line:84% closed, it could go out to plus infinity, 00:00:37.450 --> 00:00:40.320 align:middle line:84% it could go out to minus infinity. 00:00:40.320 --> 00:00:43.700 align:middle line:84% But you know what an interval is. 00:00:43.700 --> 00:00:46.700 align:middle line:84% And let's take a function from that interval to R. 00:00:46.700 --> 00:00:50.420 align:middle line:84% So we say, f is differentiable at a point c in I. 00:00:50.420 --> 00:00:54.410 align:middle line:84% If this limit of f of x minus f of c 00:00:54.410 --> 00:00:59.135 align:middle line:84% over x minus c, this difference quotient, if this limit exists. 00:00:59.135 --> 00:01:05.220 align:middle line:84% If this limit exists, we also denote it by f prime of c. 00:01:05.220 --> 00:01:08.950 align:middle line:90% And last time, we showed that-- 00:01:08.950 --> 00:01:16.250 align:middle line:84% we gave a simple example, that from last time, 00:01:16.250 --> 00:01:20.430 align:middle line:84% that if x equals alpha times x to the n, 00:01:20.430 --> 00:01:25.410 align:middle line:84% then this function is differentiable at every c, 00:01:25.410 --> 00:01:29.990 align:middle line:84% and its derivative is equal to n times alpha x 00:01:29.990 --> 00:01:31.580 align:middle line:90% c to the n minus 1. 00:01:31.580 --> 00:01:35.980 align:middle line:90% 00:01:35.980 --> 00:01:43.220 align:middle line:84% Now, let's state this in an equivalent way using sequences. 00:01:43.220 --> 00:01:45.460 align:middle line:84% Remember, we had this characterization 00:01:45.460 --> 00:01:48.490 align:middle line:90% of limits of functions. 00:01:48.490 --> 00:01:51.190 align:middle line:84% And the function we're looking at is f of x 00:01:51.190 --> 00:01:53.530 align:middle line:90% minus f of c over x minus c. 00:01:53.530 --> 00:02:01.670 align:middle line:84% We had this equivalence between limits and limits of sequences 00:02:01.670 --> 00:02:04.700 align:middle line:84% from last time, that the limit as, let's say, 00:02:04.700 --> 00:02:09.110 align:middle line:84% g of x as x goes to c equals L, if and only 00:02:09.110 --> 00:02:12.170 align:middle line:84% if for every sequence x sub n converging to c g 00:02:12.170 --> 00:02:17.560 align:middle line:84% if x sub n converges to L as n goes to infinity. 00:02:17.560 --> 00:02:22.500 align:middle line:84% So we can restate what it means for a function 00:02:22.500 --> 00:02:24.630 align:middle line:84% to be differentiable at a point and its derivative 00:02:24.630 --> 00:02:29.040 align:middle line:84% to be L, say, if and only if for every sequence 00:02:29.040 --> 00:02:35.450 align:middle line:84% x sub n, with x sub n not equal to c for all n, 00:02:35.450 --> 00:02:39.020 align:middle line:84% and converging to c, we have that L is equal to this limit 00:02:39.020 --> 00:02:43.280 align:middle line:84% as n goes to infinity of now this sequence of numbers, 00:02:43.280 --> 00:02:48.320 align:middle line:84% f of xn minus f of c over x of n minus c. 00:02:48.320 --> 00:02:53.760 align:middle line:90% 00:02:53.760 --> 00:03:03.540 align:middle line:84% So today, the theme of today is the connection 00:03:03.540 --> 00:03:07.350 align:middle line:84% between differentiability and continuity. 00:03:07.350 --> 00:03:12.040 align:middle line:84% And we have a very easy implication, 00:03:12.040 --> 00:03:16.750 align:middle line:84% which is that if a function is differentiable at a point, 00:03:16.750 --> 00:03:20.647 align:middle line:84% then it must be continuous at that point. 00:03:20.647 --> 00:03:22.230 align:middle line:84% So that's a statement of this theorem, 00:03:22.230 --> 00:03:26.190 align:middle line:84% if f going from an interval to R is differential at c, 00:03:26.190 --> 00:03:30.020 align:middle line:90% then f is also continuous at c. 00:03:30.020 --> 00:03:39.830 align:middle line:84% So let me also add in here that continuity at the point 00:03:39.830 --> 00:03:42.380 align:middle line:84% c in an interval is equivalent to saying 00:03:42.380 --> 00:03:48.170 align:middle line:84% that the limit as x goes to c of f of x equals f of c. 00:03:48.170 --> 00:03:50.690 align:middle line:90% 00:03:50.690 --> 00:03:54.260 align:middle line:84% Now, this is just a subtle point that I want to make, 00:03:54.260 --> 00:03:58.460 align:middle line:84% is that for an interval, open, closed, half open, half 00:03:58.460 --> 00:04:01.250 align:middle line:84% closed, whatever, any point in that interval 00:04:01.250 --> 00:04:05.720 align:middle line:84% is a cluster point of that interval. 00:04:05.720 --> 00:04:08.000 align:middle line:84% And therefore, this definition I made up there 00:04:08.000 --> 00:04:10.340 align:middle line:90% is actually meaningful. 00:04:10.340 --> 00:04:14.450 align:middle line:84% And for continuity, something being continuous at c 00:04:14.450 --> 00:04:16.279 align:middle line:90% is equivalent to-- 00:04:16.279 --> 00:04:19.430 align:middle line:84% if c is a cluster point, which I just said it is-- it always is, 00:04:19.430 --> 00:04:22.550 align:middle line:84% this limit equals the function evaluated at the point. 00:04:22.550 --> 00:04:29.080 align:middle line:84% So that's just kind of a subtle comment I want to make. 00:04:29.080 --> 00:04:38.030 align:middle line:90% 00:04:38.030 --> 00:04:40.810 align:middle line:90% So how do we prove this? 00:04:40.810 --> 00:04:47.180 align:middle line:84% We write the limit as x goes to the c of f of x as in this way. 00:04:47.180 --> 00:04:50.650 align:middle line:84% So essentially, what I do is, I add and subtract f of c. 00:04:50.650 --> 00:04:54.130 align:middle line:84% And then, the part that is f of x minus f of c, 00:04:54.130 --> 00:04:57.880 align:middle line:84% I multiply by x minus c and divide 00:04:57.880 --> 00:04:59.890 align:middle line:84% by x minus c, which is perfectly fine. 00:04:59.890 --> 00:05:01.930 align:middle line:84% Because remember, for a limit, I'm 00:05:01.930 --> 00:05:06.280 align:middle line:84% never actually looking at points where x equals c. 00:05:06.280 --> 00:05:10.310 align:middle line:90% 00:05:10.310 --> 00:05:11.960 align:middle line:90% So that's fine. 00:05:11.960 --> 00:05:15.770 align:middle line:84% And so, I get this expression here. 00:05:15.770 --> 00:05:21.260 align:middle line:84% Now, what we know about limits, is that the limit of the sum 00:05:21.260 --> 00:05:25.430 align:middle line:84% is the sum of the limits as long as all these limits exists. 00:05:25.430 --> 00:05:27.260 align:middle line:84% And the limit of the product is a product 00:05:27.260 --> 00:05:30.350 align:middle line:84% of the limits, again, assuming all of these limits exists. 00:05:30.350 --> 00:05:36.230 align:middle line:84% And the limit as x goes to c of everything you see here 00:05:36.230 --> 00:05:37.170 align:middle line:90% do exist. 00:05:37.170 --> 00:05:41.480 align:middle line:84% So as x goes to c, this thing in brackets, or in the-- 00:05:41.480 --> 00:05:44.660 align:middle line:90% I guess these are brackets. 00:05:44.660 --> 00:05:47.540 align:middle line:84% I don't know what those are called. 00:05:47.540 --> 00:05:50.360 align:middle line:90% This limit here is f prime of c. 00:05:50.360 --> 00:05:54.560 align:middle line:84% As x goes to c of x minus c, that's just 0. 00:05:54.560 --> 00:05:56.870 align:middle line:90% And f of c is just a constant. 00:05:56.870 --> 00:05:59.210 align:middle line:90% There's nothing changing with x. 00:05:59.210 --> 00:06:01.980 align:middle line:84% So the limit of a constant is that constant. 00:06:01.980 --> 00:06:03.800 align:middle line:90% So I just get f of c. 00:06:03.800 --> 00:06:05.900 align:middle line:84% So the limit of this whole expression 00:06:05.900 --> 00:06:10.550 align:middle line:84% is f prime of c times 0 plus f of c, which equals f of c. 00:06:10.550 --> 00:06:16.150 align:middle line:84% So we've just proven that the limit as x goes to c of f of x 00:06:16.150 --> 00:06:21.190 align:middle line:84% is equal to f of c, which is what we wish to prove. 00:06:21.190 --> 00:06:24.250 align:middle line:84% And therefore, a function which is differentiable at a point 00:06:24.250 --> 00:06:25.900 align:middle line:84% must be continuous at that point. 00:06:25.900 --> 00:06:29.770 align:middle line:90% 00:06:29.770 --> 00:06:33.340 align:middle line:84% So as beginning math students, we're learning proofs, 00:06:33.340 --> 00:06:35.200 align:middle line:84% but we're also learning questions, 00:06:35.200 --> 00:06:39.030 align:middle line:90% what types of questions to ask. 00:06:39.030 --> 00:06:41.370 align:middle line:84% And whenever you come across a new theorem that 00:06:41.370 --> 00:06:50.130 align:middle line:84% has one implication, one hypothesis, one conclusion, 00:06:50.130 --> 00:06:53.100 align:middle line:84% then you should ask yourself, does the converse hold? 00:06:53.100 --> 00:06:57.330 align:middle line:84% Does the conclusion also imply the hypothesis? 00:06:57.330 --> 00:07:00.830 align:middle line:84% So we've shown that f differentiable at c 00:07:00.830 --> 00:07:02.690 align:middle line:90% implies f is continuous at c. 00:07:02.690 --> 00:07:04.250 align:middle line:90% So does the converse hold? 00:07:04.250 --> 00:07:07.130 align:middle line:84% Namely, if f is continuous at a point, 00:07:07.130 --> 00:07:09.725 align:middle line:84% does this imply that f is differentiable at that point? 00:07:09.725 --> 00:07:12.290 align:middle line:90% 00:07:12.290 --> 00:07:15.123 align:middle line:84% And as you can see already there, the answer is, no. 00:07:15.123 --> 00:07:17.165 align:middle line:84% I think you've probably covered this in calculus. 00:07:17.165 --> 00:07:21.150 align:middle line:90% 00:07:21.150 --> 00:07:25.650 align:middle line:84% But what's the function that gives this counterexample that 00:07:25.650 --> 00:07:28.770 align:middle line:84% is continuous at c but not differentiable at c? 00:07:28.770 --> 00:07:31.410 align:middle line:84% Well, let's take, for example, c equals 0. 00:07:31.410 --> 00:07:33.120 align:middle line:84% And let's look at the function f of x 00:07:33.120 --> 00:07:36.660 align:middle line:90% equals the absolute value of x. 00:07:36.660 --> 00:07:39.790 align:middle line:84% In this function here, this is continuous at every point. 00:07:39.790 --> 00:07:41.850 align:middle line:90% So in particular, at c equals 0. 00:07:41.850 --> 00:07:46.880 align:middle line:84% But it's not differentiable at c equals 0. 00:07:46.880 --> 00:07:48.530 align:middle line:90% And how do you come up with-- 00:07:48.530 --> 00:07:50.250 align:middle line:90% and how do you do that? 00:07:50.250 --> 00:07:53.060 align:middle line:84% How do you show something's not differentiable? 00:07:53.060 --> 00:07:56.420 align:middle line:84% The simplest way is to use this remark up here 00:07:56.420 --> 00:07:59.180 align:middle line:84% by finding a sequence converging to c, 00:07:59.180 --> 00:08:01.460 align:middle line:84% so that the difference quotient does not have 00:08:01.460 --> 00:08:03.230 align:middle line:90% a limit as n goes to infinity. 00:08:03.230 --> 00:08:06.167 align:middle line:90% 00:08:06.167 --> 00:08:07.750 align:middle line:84% So that's what we'll do here is, we'll 00:08:07.750 --> 00:08:12.250 align:middle line:84% find a sequence x sub n, with x sub n not 00:08:12.250 --> 00:08:16.830 align:middle line:84% equal to 0 for all n converging to 0, 00:08:16.830 --> 00:08:21.360 align:middle line:84% such that this limit does not exist. 00:08:21.360 --> 00:08:23.950 align:middle line:84% Rather than write that, because the limit does not exist, 00:08:23.950 --> 00:08:28.170 align:middle line:84% it doesn't make really sense for me to even write that. 00:08:28.170 --> 00:08:38.630 align:middle line:84% Let me write this as saying, as the sequence is divergent. 00:08:38.630 --> 00:08:46.140 align:middle line:90% 00:08:46.140 --> 00:08:49.990 align:middle line:84% Now, how are we going to come up with this sequence, x of n, 00:08:49.990 --> 00:08:53.320 align:middle line:84% so that the difference quotient does not converge? 00:08:53.320 --> 00:08:55.130 align:middle line:90% Well, what's the logic? 00:08:55.130 --> 00:09:04.670 align:middle line:84% So for f of x equals the absolute value of x, 00:09:04.670 --> 00:09:06.380 align:middle line:90% it looks like this. 00:09:06.380 --> 00:09:07.540 align:middle line:90% It looks like x-- 00:09:07.540 --> 00:09:10.550 align:middle line:84% it's equal to x over here minus x over here. 00:09:10.550 --> 00:09:13.270 align:middle line:84% So if you were to formally differentiate-- 00:09:13.270 --> 00:09:18.190 align:middle line:84% well, I mean, you could actually differentiate to the left of 0, 00:09:18.190 --> 00:09:21.490 align:middle line:84% you get the derivative is equal to minus 1. 00:09:21.490 --> 00:09:23.410 align:middle line:84% To the right, the derivative is 1. 00:09:23.410 --> 00:09:25.870 align:middle line:84% So that kind of suggests maybe there's 00:09:25.870 --> 00:09:29.270 align:middle line:84% some funny business going on at 0. 00:09:29.270 --> 00:09:31.670 align:middle line:84% So let's look at a sequence which 00:09:31.670 --> 00:09:35.870 align:middle line:84% alternates between being negative and positive, 00:09:35.870 --> 00:09:37.700 align:middle line:84% but it's converging to 0, and see 00:09:37.700 --> 00:09:44.990 align:middle line:84% if that sequence will provide us with this desired 00:09:44.990 --> 00:09:48.140 align:middle line:84% sequence which results in the difference quotient being 00:09:48.140 --> 00:09:49.710 align:middle line:90% divergent. 00:09:49.710 --> 00:09:51.980 align:middle line:90% So let's just take a guess. 00:09:51.980 --> 00:09:54.047 align:middle line:84% Let xn be minus 1 to the n over n. 00:09:54.047 --> 00:09:56.630 align:middle line:84% It doesn't have to be this one, just something that alternates 00:09:56.630 --> 00:09:58.790 align:middle line:90% back and forth would be enough. 00:09:58.790 --> 00:10:02.390 align:middle line:84% So this could be n squared here, or 2n, 00:10:02.390 --> 00:10:07.180 align:middle line:84% or n to the 2020, whatever you like. 00:10:07.180 --> 00:10:10.210 align:middle line:84% Then, this is clearly always non-zero. 00:10:10.210 --> 00:10:12.040 align:middle line:90% And it converges to 0. 00:10:12.040 --> 00:10:16.760 align:middle line:90% 00:10:16.760 --> 00:10:18.470 align:middle line:84% And of course, why it converges to 0? 00:10:18.470 --> 00:10:21.890 align:middle line:84% I mean, you could prove this by epsilon m definition, 00:10:21.890 --> 00:10:29.940 align:middle line:84% but also if I take the absolute value, that's 00:10:29.940 --> 00:10:32.430 align:middle line:90% less than or equal to 1 over n. 00:10:32.430 --> 00:10:37.110 align:middle line:84% And this one we know converges to 0, 0. 00:10:37.110 --> 00:10:40.560 align:middle line:84% So by squeeze theorem, the thing in the middle 00:10:40.560 --> 00:10:43.590 align:middle line:90% must converge to 0. 00:10:43.590 --> 00:10:46.530 align:middle line:90% And we're done. 00:10:46.530 --> 00:10:49.410 align:middle line:84% So we have this sequence x sub n equals minus 1 00:10:49.410 --> 00:10:54.410 align:middle line:84% to the n over n, which is alternating back and forth. 00:10:54.410 --> 00:10:59.330 align:middle line:84% And we hope it's going to result in this difference quotient 00:10:59.330 --> 00:11:01.920 align:middle line:84% being divergent as n goes to infinity. 00:11:01.920 --> 00:11:03.278 align:middle line:90% And so, we compute that. 00:11:03.278 --> 00:11:04.820 align:middle line:84% Let's look at the difference quotient 00:11:04.820 --> 00:11:06.770 align:middle line:90% f of x sub n minus f of 0. 00:11:06.770 --> 00:11:08.540 align:middle line:90% F of 0, which is 0. 00:11:08.540 --> 00:11:10.580 align:middle line:84% So I just get f of x sub n, which 00:11:10.580 --> 00:11:13.940 align:middle line:84% is the absolute value of minus 1 to the n over n 00:11:13.940 --> 00:11:17.720 align:middle line:84% over x sub n, which is minus 1 to the n over n. 00:11:17.720 --> 00:11:19.055 align:middle line:90% This is just equal to-- 00:11:19.055 --> 00:11:27.250 align:middle line:90% 00:11:27.250 --> 00:11:30.970 align:middle line:84% so this is just equal to 1 over n over minus 1 00:11:30.970 --> 00:11:36.850 align:middle line:84% to the n over n, which equals minus 1 to the minus n. 00:11:36.850 --> 00:11:40.600 align:middle line:84% But minus 1 to the minus n, this is 00:11:40.600 --> 00:11:42.760 align:middle line:84% equal to the same thing as minus 1 to the n. 00:11:42.760 --> 00:11:45.820 align:middle line:90% 00:11:45.820 --> 00:11:50.780 align:middle line:84% So this sequence here of difference quotients, 00:11:50.780 --> 00:11:53.980 align:middle line:84% so f of xn minus f of 0 over xn minus 0, 00:11:53.980 --> 00:11:58.680 align:middle line:84% this is just equal to the sequence minus 1 00:11:58.680 --> 00:12:01.675 align:middle line:84% to the n, which we know is divergent, 00:12:01.675 --> 00:12:02.550 align:middle line:90% doesn't have a limit. 00:12:02.550 --> 00:12:07.198 align:middle line:84% That's one of the first examples of a sequence 00:12:07.198 --> 00:12:08.365 align:middle line:90% we proved does not converge. 00:12:08.365 --> 00:12:11.060 align:middle line:90% 00:12:11.060 --> 00:12:14.550 align:middle line:84% So in summary, this function f of x 00:12:14.550 --> 00:12:19.020 align:middle line:84% equals the absolute value of x is continuous at c equals 0, 00:12:19.020 --> 00:12:21.810 align:middle line:84% but not differentiable at c equals 0. 00:12:21.810 --> 00:12:29.760 align:middle line:84% So now, a natural question that people asked is, so a function 00:12:29.760 --> 00:12:32.400 align:middle line:84% being continuous at a point does not necessarily 00:12:32.400 --> 00:12:35.460 align:middle line:90% imply that it's differentiable. 00:12:35.460 --> 00:12:36.990 align:middle line:84% But let's take a function which is 00:12:36.990 --> 00:12:40.240 align:middle line:84% continuous on the real number line. 00:12:40.240 --> 00:12:43.870 align:middle line:84% Is there a point where it's differentiable? 00:12:43.870 --> 00:12:47.170 align:middle line:90% So it's already on the board. 00:12:47.170 --> 00:12:51.550 align:middle line:84% But this is not completely a crazy question to ask. 00:12:51.550 --> 00:12:53.350 align:middle line:84% For example, let's go back to f of x 00:12:53.350 --> 00:12:55.570 align:middle line:90% equals the absolute value of x. 00:12:55.570 --> 00:12:59.350 align:middle line:84% This function is differentiable everywhere except 0. 00:12:59.350 --> 00:13:01.845 align:middle line:84% So there's lots of points where this function is 00:13:01.845 --> 00:13:02.470 align:middle line:90% differentiable. 00:13:02.470 --> 00:13:05.300 align:middle line:84% It's differentiable for c positive and c negative. 00:13:05.300 --> 00:13:07.780 align:middle line:84% So for this continuous function, there 00:13:07.780 --> 00:13:10.480 align:middle line:84% does exist points where it's differentiable. 00:13:10.480 --> 00:13:13.390 align:middle line:84% And you can imagine trying to draw any kind of curve 00:13:13.390 --> 00:13:14.840 align:middle line:84% that you can on a piece of paper. 00:13:14.840 --> 00:13:16.000 align:middle line:84% And you can probably find a point 00:13:16.000 --> 00:13:17.425 align:middle line:84% on that curve that has a tangent. 00:13:17.425 --> 00:13:19.987 align:middle line:90% 00:13:19.987 --> 00:13:21.570 align:middle line:84% I mean, if you sit there with a pencil 00:13:21.570 --> 00:13:23.632 align:middle line:84% and then try to draw something very jagged, 00:13:23.632 --> 00:13:25.090 align:middle line:84% I mean, there are still going to be 00:13:25.090 --> 00:13:30.630 align:middle line:84% little sections of your jagged curve where it has a tangent. 00:13:30.630 --> 00:13:34.320 align:middle line:84% So I imagine this is why people thought 00:13:34.320 --> 00:13:37.380 align:middle line:84% this was the case, that a continuous function has 00:13:37.380 --> 00:13:40.600 align:middle line:84% to have at least one point where it's differentiable. 00:13:40.600 --> 00:13:44.380 align:middle line:84% And Weierstrass, the godfather of analysis, said, no. 00:13:44.380 --> 00:13:47.830 align:middle line:84% He came up with a whole class of examples of functions, which 00:13:47.830 --> 00:13:52.770 align:middle line:84% are continuous on the real number line, 00:13:52.770 --> 00:13:55.820 align:middle line:90% but are differentiable nowhere. 00:13:55.820 --> 00:14:03.530 align:middle line:84% And this was a really surprising set of examples and result 00:14:03.530 --> 00:14:04.640 align:middle line:90% to the community. 00:14:04.640 --> 00:14:17.200 align:middle line:84% And to me, it's also one of the few results in this class 00:14:17.200 --> 00:14:21.340 align:middle line:84% that you really didn't see in your calculus class. 00:14:21.340 --> 00:14:23.380 align:middle line:84% So we're going to go through this example. 00:14:23.380 --> 00:14:25.570 align:middle line:84% Although in most analysis classes, 00:14:25.570 --> 00:14:29.390 align:middle line:90% it's reserved for later. 00:14:29.390 --> 00:14:30.820 align:middle line:90% But I think we can do it now. 00:14:30.820 --> 00:14:33.920 align:middle line:90% 00:14:33.920 --> 00:14:34.460 align:middle line:90% And so. 00:14:34.460 --> 00:14:35.710 align:middle line:90% What we're going to prove is-- 00:14:35.710 --> 00:14:49.623 align:middle line:90% 00:14:49.623 --> 00:14:52.040 align:middle line:84% so we're going to construct a function which is continuous 00:14:52.040 --> 00:14:54.920 align:middle line:84% everywhere but is differential nowhere, 00:14:54.920 --> 00:14:57.170 align:middle line:84% so a continuous nowhere differentiable function. 00:14:57.170 --> 00:15:00.400 align:middle line:90% 00:15:00.400 --> 00:15:04.870 align:middle line:84% And I'll even write down the function for you. 00:15:04.870 --> 00:15:07.330 align:middle line:84% So what's the idea that Weierstrass had? 00:15:07.330 --> 00:15:10.067 align:middle line:90% It was, let me-- 00:15:10.067 --> 00:15:12.400 align:middle line:84% well, I'm not going to write down the function just yet. 00:15:12.400 --> 00:15:15.640 align:middle line:84% But the idea of the constructing such a function 00:15:15.640 --> 00:15:19.798 align:middle line:84% is that it should be highly oscillatory. 00:15:19.798 --> 00:15:21.340 align:middle line:84% Imagine, again, you're trying to draw 00:15:21.340 --> 00:15:23.712 align:middle line:84% a picture of the graph of a function that's 00:15:23.712 --> 00:15:24.670 align:middle line:90% nowhere differentiable. 00:15:24.670 --> 00:15:27.850 align:middle line:84% You would sit there and break your pencil trying 00:15:27.850 --> 00:15:33.440 align:middle line:84% to draw just a highly oscillating function so 00:15:33.440 --> 00:15:36.970 align:middle line:90% that it never has a tangent. 00:15:36.970 --> 00:15:42.000 align:middle line:84% And that's his idea, is to build a function which 00:15:42.000 --> 00:15:46.980 align:middle line:84% is highly oscillatory but too oscillatory so 00:15:46.980 --> 00:15:49.020 align:middle line:84% that it's still continuous at every point. 00:15:49.020 --> 00:16:17.500 align:middle line:90% 00:16:17.500 --> 00:16:21.080 align:middle line:84% So to start with the construction of this function, 00:16:21.080 --> 00:16:23.240 align:middle line:84% which is continuous but nowhere differentiable, 00:16:23.240 --> 00:16:28.030 align:middle line:84% we're going to need a few simple facts to start off with. 00:16:28.030 --> 00:16:33.800 align:middle line:90% So again, let me write out-- 00:16:33.800 --> 00:16:34.550 align:middle line:90% state our goal. 00:16:34.550 --> 00:16:45.490 align:middle line:84% We're going to construct a continuous function from R 00:16:45.490 --> 00:16:58.560 align:middle line:84% to R, which is nowhere differentiable 00:16:58.560 --> 00:17:02.920 align:middle line:84% Not differentiable, it's differentiable at no point. 00:17:02.920 --> 00:17:09.859 align:middle line:84% So let's start off with a few simple facts about-- 00:17:09.859 --> 00:17:12.470 align:middle line:84% so I said that we're going to build a function which 00:17:12.470 --> 00:17:15.109 align:middle line:90% is oscillating quite a lot. 00:17:15.109 --> 00:17:18.170 align:middle line:84% There's two functions you know that oscillate cosine and sine. 00:17:18.170 --> 00:17:20.970 align:middle line:90% 00:17:20.970 --> 00:17:37.540 align:middle line:84% So those are going to be our building blocks. 00:17:37.540 --> 00:17:39.190 align:middle line:90% We'll choose one of them. 00:17:39.190 --> 00:17:44.290 align:middle line:84% Let's choose cosine, say, as to be our building block. 00:17:44.290 --> 00:17:47.080 align:middle line:84% But anyway, so let's start off with some elementary facts 00:17:47.080 --> 00:17:48.310 align:middle line:90% about cosine. 00:17:48.310 --> 00:17:56.740 align:middle line:84% So first is that for all x, y, and R, cosine x minus cosine y 00:17:56.740 --> 00:17:59.410 align:middle line:84% is less than or equal to the absolute value of x times 00:17:59.410 --> 00:18:01.980 align:middle line:90% x minus y. 00:18:01.980 --> 00:18:13.390 align:middle line:84% The second is that for every real number c, 00:18:13.390 --> 00:18:26.000 align:middle line:84% and for all k natural number, there exist a y in c plus pi 00:18:26.000 --> 00:18:33.950 align:middle line:84% over k, c plus 3 pi over k, such that-- 00:18:33.950 --> 00:18:36.440 align:middle line:84% and let me label this as theorem one-- 00:18:36.440 --> 00:18:39.360 align:middle line:90% 00:18:39.360 --> 00:18:43.110 align:middle line:90% such that cosine of kc-- 00:18:43.110 --> 00:18:50.600 align:middle line:84% that's funny-- minus cosine of ky 00:18:50.600 --> 00:18:52.130 align:middle line:90% is greater than or equal to 1. 00:18:52.130 --> 00:19:00.810 align:middle line:90% 00:19:00.810 --> 00:19:10.180 align:middle line:84% So both of these are simple facts about cosine. 00:19:10.180 --> 00:19:14.140 align:middle line:84% Really, this one just follows from the angle sum formula. 00:19:14.140 --> 00:19:18.670 align:middle line:84% This one follows from the periodicity of cosine. 00:19:18.670 --> 00:19:21.710 align:middle line:90% But so, why are these true? 00:19:21.710 --> 00:19:23.950 align:middle line:84% So first off, we did prove-- and maybe I 00:19:23.950 --> 00:19:26.540 align:middle line:84% should have used sine instead of cosine in all of this. 00:19:26.540 --> 00:19:38.220 align:middle line:84% But if you recall from our continuity section, 00:19:38.220 --> 00:19:43.770 align:middle line:84% we proved that for all x and y, sine x minus sine y 00:19:43.770 --> 00:19:46.680 align:middle line:84% is less than or equal to the absolute value of x 00:19:46.680 --> 00:19:49.560 align:middle line:90% minus y for all x and y. 00:19:49.560 --> 00:19:55.110 align:middle line:84% And there was a simple relation between cosine x and sine 00:19:55.110 --> 00:19:55.740 align:middle line:90% of something. 00:19:55.740 --> 00:19:59.820 align:middle line:84% And that's just a shift by pi over 2. 00:19:59.820 --> 00:20:07.060 align:middle line:84% Then cosine of x minus cosine of y, 00:20:07.060 --> 00:20:11.740 align:middle line:84% this is equal to sine x plus pi over 2 00:20:11.740 --> 00:20:16.870 align:middle line:90% minus sine y plus pi over 2. 00:20:16.870 --> 00:20:20.860 align:middle line:84% And if you like, let me instead of using x and y here, 00:20:20.860 --> 00:20:22.180 align:middle line:90% let me use a and b. 00:20:22.180 --> 00:20:28.220 align:middle line:90% 00:20:28.220 --> 00:20:31.760 align:middle line:84% So then, this is less than or equal to x plus pi over 2 00:20:31.760 --> 00:20:38.850 align:middle line:84% minus y plus pi over 2 using this inequality. 00:20:38.850 --> 00:20:41.650 align:middle line:90% 00:20:41.650 --> 00:20:43.900 align:middle line:90% So that proves number one. 00:20:43.900 --> 00:20:51.100 align:middle line:84% For number two, it's just a simple fact about cosine 00:20:51.100 --> 00:20:54.700 align:middle line:90% being 2 pi periodic. 00:20:54.700 --> 00:21:05.350 align:middle line:84% So function g of x equals cosine k of x. 00:21:05.350 --> 00:21:06.640 align:middle line:90% This is also periodic. 00:21:06.640 --> 00:21:07.840 align:middle line:90% But now, what's the period? 00:21:07.840 --> 00:21:10.390 align:middle line:90% It's 2 pi over k. 00:21:10.390 --> 00:21:21.870 align:middle line:90% 00:21:21.870 --> 00:21:25.660 align:middle line:84% Now, this interval has link 2 pi, 00:21:25.660 --> 00:21:28.810 align:middle line:84% except it's missing the point c plus pi over 2 00:21:28.810 --> 00:21:31.210 align:middle line:90% and c plus 3 pi over k. 00:21:31.210 --> 00:21:44.670 align:middle line:90% 00:21:44.670 --> 00:21:56.710 align:middle line:84% Thus, if I look at the image of-- 00:21:56.710 --> 00:22:02.110 align:middle line:90% 00:22:02.110 --> 00:22:04.250 align:middle line:90% OK. 00:22:04.250 --> 00:22:07.340 align:middle line:84% Then, so this function g of x equals cosine k of x. 00:22:07.340 --> 00:22:22.070 align:middle line:84% If I look at the image of this interval, 00:22:22.070 --> 00:22:28.370 align:middle line:84% this will contain all of the real numbers between minus 1 00:22:28.370 --> 00:22:28.940 align:middle line:90% and 1. 00:22:28.940 --> 00:22:32.240 align:middle line:90% 00:22:32.240 --> 00:22:34.070 align:middle line:90% So this is a period. 00:22:34.070 --> 00:22:37.730 align:middle line:84% This is a interval of link 2 pi over k. 00:22:37.730 --> 00:22:40.460 align:middle line:84% Cosine kx is 2 pi over k periodic. 00:22:40.460 --> 00:22:43.280 align:middle line:84% So I should get everything between minus 1 00:22:43.280 --> 00:22:48.150 align:middle line:84% and one except possibly the value at the endpoints. 00:22:48.150 --> 00:22:54.600 align:middle line:84% So take away-- and the value at the endpoints 00:22:54.600 --> 00:22:59.490 align:middle line:90% is minus cosine ck. 00:22:59.490 --> 00:23:05.950 align:middle line:90% 00:23:05.950 --> 00:23:10.630 align:middle line:84% Because cosine of k times c plus pi is equal-- 00:23:10.630 --> 00:23:13.960 align:middle line:84% pi over k is equal to cosine of kc plus pi. 00:23:13.960 --> 00:23:16.890 align:middle line:90% 00:23:16.890 --> 00:23:18.680 align:middle line:84% And cosine of something plus pi is 00:23:18.680 --> 00:23:20.630 align:middle line:84% equal to minus cosine of that thing. 00:23:20.630 --> 00:23:23.430 align:middle line:90% 00:23:23.430 --> 00:23:28.800 align:middle line:84% So the image of the set by cosine kc 00:23:28.800 --> 00:23:32.040 align:middle line:84% contains everything between minus 1 and 1, 00:23:32.040 --> 00:23:35.520 align:middle line:84% except for possibly the endpoint. 00:23:35.520 --> 00:23:44.056 align:middle line:84% This really only happens at if cosine of k times E equals 1. 00:23:44.056 --> 00:23:48.550 align:middle line:90% 00:23:48.550 --> 00:23:53.080 align:middle line:90% So OK. 00:23:53.080 --> 00:24:01.200 align:middle line:90% 00:24:01.200 --> 00:24:19.570 align:middle line:84% Then, if cosine of kc is bigger than or equal to 0, 00:24:19.570 --> 00:24:32.260 align:middle line:84% then we choose y in this interval c plus pi over kc 00:24:32.260 --> 00:24:34.440 align:middle line:90% plus 3 pi over k. 00:24:34.440 --> 00:24:53.050 align:middle line:90% 00:24:53.050 --> 00:24:55.160 align:middle line:90% Let's write it this way. 00:24:55.160 --> 00:24:57.910 align:middle line:84% So if this thing is bigger than or equal to 0, 00:24:57.910 --> 00:24:59.040 align:middle line:90% then we choose y-- 00:24:59.040 --> 00:25:03.540 align:middle line:90% 00:25:03.540 --> 00:25:05.810 align:middle line:90% why am I confusing myself? 00:25:05.810 --> 00:25:08.160 align:middle line:90% So this is simple enough. 00:25:08.160 --> 00:25:14.750 align:middle line:84% We choose y in this interval so that cosine of ky equals 00:25:14.750 --> 00:25:15.440 align:middle line:90% minus 1. 00:25:15.440 --> 00:25:18.170 align:middle line:90% 00:25:18.170 --> 00:25:25.310 align:middle line:84% And if cosine kc is less than 0, then we 00:25:25.310 --> 00:25:38.430 align:middle line:84% choose y so that cosine of ky equals 1. 00:25:38.430 --> 00:25:44.170 align:middle line:90% 00:25:44.170 --> 00:25:48.220 align:middle line:90% So again, here's-- so forget-- 00:25:48.220 --> 00:25:49.570 align:middle line:90% just to get an idea. 00:25:49.570 --> 00:25:53.230 align:middle line:84% So I'm quibbling over a minor point, the fact 00:25:53.230 --> 00:25:56.515 align:middle line:84% that we have to take away possibly the point where-- 00:25:56.515 --> 00:25:59.090 align:middle line:90% 00:25:59.090 --> 00:26:02.000 align:middle line:90% possibly the value of cosine kc. 00:26:02.000 --> 00:26:05.540 align:middle line:84% Anyways, that's erase this for a minute. 00:26:05.540 --> 00:26:09.650 align:middle line:84% And let's imagine that the range of this guy-- 00:26:09.650 --> 00:26:13.620 align:middle line:84% I mean, it's a 2 pi over k periodic interval. 00:26:13.620 --> 00:26:15.380 align:middle line:84% So this should basically cover all values 00:26:15.380 --> 00:26:17.090 align:middle line:90% between minus 1 and 1. 00:26:17.090 --> 00:26:20.360 align:middle line:84% So cosine of kc will either be plus or minus. 00:26:20.360 --> 00:26:24.860 align:middle line:84% So if it's not negative, then I can find a y in this interval, 00:26:24.860 --> 00:26:26.690 align:middle line:90% since it's 2 pi over k periodic. 00:26:26.690 --> 00:26:29.330 align:middle line:84% So that cosine of ky equals minus 1. 00:26:29.330 --> 00:26:31.910 align:middle line:90% 00:26:31.910 --> 00:26:36.710 align:middle line:84% And since this is not negative and this is equal to minus 1, 00:26:36.710 --> 00:26:39.290 align:middle line:84% the difference between these two in absolute value 00:26:39.290 --> 00:26:43.440 align:middle line:84% will be greater than or equal to 1. 00:26:43.440 --> 00:26:47.160 align:middle line:84% Now, if cosine of kc is less than 0, 00:26:47.160 --> 00:26:51.030 align:middle line:84% then again, since this interval contains minus 1 to 1, 00:26:51.030 --> 00:26:54.630 align:middle line:84% it's 2 pi over k periodic, we choose a y so that cosine of ky 00:26:54.630 --> 00:26:55.740 align:middle line:90% equals 1. 00:26:55.740 --> 00:26:57.510 align:middle line:90% And then, this is negative. 00:26:57.510 --> 00:26:58.510 align:middle line:90% This is 1. 00:26:58.510 --> 00:27:01.500 align:middle line:84% So the difference between something negative and 1 00:27:01.500 --> 00:27:03.370 align:middle line:90% is greater than or equal to 1. 00:27:03.370 --> 00:27:05.910 align:middle line:90% 00:27:05.910 --> 00:27:08.950 align:middle line:90% And that's the proof. 00:27:08.950 --> 00:27:10.870 align:middle line:90% I made a kind of a mess of it. 00:27:10.870 --> 00:27:17.410 align:middle line:84% But this is very easy to understand if you just 00:27:17.410 --> 00:27:18.595 align:middle line:90% draw a picture really. 00:27:18.595 --> 00:27:22.710 align:middle line:90% 00:27:22.710 --> 00:27:25.770 align:middle line:90% So let's continue. 00:27:25.770 --> 00:27:29.910 align:middle line:90% 00:27:29.910 --> 00:27:34.350 align:middle line:84% And again, what these two parts of this theorem say 00:27:34.350 --> 00:27:39.660 align:middle line:84% is that cosine can be quite oscillatory 00:27:39.660 --> 00:27:42.900 align:middle line:90% if you insert a k here. 00:27:42.900 --> 00:27:47.460 align:middle line:84% Because the difference between this interval 00:27:47.460 --> 00:27:50.580 align:middle line:90% is actually quite small. 00:27:50.580 --> 00:27:52.650 align:middle line:84% The length of this interval is 2 pi over k. 00:27:52.650 --> 00:27:55.410 align:middle line:84% If k is very large, that's a very small interval. 00:27:55.410 --> 00:27:57.150 align:middle line:84% Yet somehow, we can find two points 00:27:57.150 --> 00:27:59.970 align:middle line:84% which differ by at least 1 in value 00:27:59.970 --> 00:28:02.880 align:middle line:84% if we plug them into the function. 00:28:02.880 --> 00:28:06.060 align:middle line:84% But cosine is not too wild, because it 00:28:06.060 --> 00:28:07.620 align:middle line:90% satisfies this bound. 00:28:07.620 --> 00:28:09.600 align:middle line:84% So these two are kind of-- they're 00:28:09.600 --> 00:28:15.040 align:middle line:84% going to be the ingredients in that idea I said at the start, 00:28:15.040 --> 00:28:18.700 align:middle line:84% that we're going to build a function which oscillates, 00:28:18.700 --> 00:28:23.020 align:middle line:84% which is quite oscillatory, but it's not too oscillatory 00:28:23.020 --> 00:28:26.440 align:middle line:90% that it's still continuous. 00:28:26.440 --> 00:28:34.252 align:middle line:84% And we're going to build our function out of cosine kx, 00:28:34.252 --> 00:28:35.585 align:middle line:90% where k is going to be changing. 00:28:35.585 --> 00:28:38.160 align:middle line:90% 00:28:38.160 --> 00:28:40.950 align:middle line:84% So I'm going to need one more very simple fact. 00:28:40.950 --> 00:28:45.960 align:middle line:84% So this one, I won't mess up too much. 00:28:45.960 --> 00:28:47.710 align:middle line:90% Which is the following. 00:28:47.710 --> 00:28:53.460 align:middle line:84% For all a, b, c, and R, the absolute value of a plus 00:28:53.460 --> 00:28:57.150 align:middle line:84% b plus c, this is bigger than or equal to the absolute value 00:28:57.150 --> 00:29:00.180 align:middle line:84% of a, minus the absolute value of b, 00:29:00.180 --> 00:29:04.160 align:middle line:90% minus the absolute value of c. 00:29:04.160 --> 00:29:06.710 align:middle line:84% And the proof of this is just to use the triangle inequality 00:29:06.710 --> 00:29:08.090 align:middle line:90% twice. 00:29:08.090 --> 00:29:09.470 align:middle line:90% We have the absolute value of a. 00:29:09.470 --> 00:29:19.310 align:middle line:84% This is equal to a plus b plus c, minus b plus c. 00:29:19.310 --> 00:29:24.530 align:middle line:84% And this is less than or equal to, by the triangle inequality, 00:29:24.530 --> 00:29:28.200 align:middle line:84% the absolute value of a, plus b, plus c, 00:29:28.200 --> 00:29:31.350 align:middle line:84% plus the absolute value of minus times b plus c. 00:29:31.350 --> 00:29:33.600 align:middle line:84% But that minus goes away with the absolute value. 00:29:33.600 --> 00:29:37.080 align:middle line:90% 00:29:37.080 --> 00:29:39.570 align:middle line:84% And then, I use a triangle inequality one more time here 00:29:39.570 --> 00:29:40.140 align:middle line:90% to get-- 00:29:40.140 --> 00:29:47.330 align:middle line:90% 00:29:47.330 --> 00:29:49.842 align:middle line:90% and then if I subtract-- 00:29:49.842 --> 00:29:51.050 align:middle line:90% oh, I didn't want to do that. 00:29:51.050 --> 00:29:58.080 align:middle line:90% 00:29:58.080 --> 00:30:04.610 align:middle line:84% And so, if I subtract this side over to the other side 00:30:04.610 --> 00:30:12.610 align:middle line:84% of this inequality, I get the statement of the theorem. 00:30:12.610 --> 00:30:19.280 align:middle line:90% 00:30:19.280 --> 00:30:21.590 align:middle line:84% So now, we're going to introduce-- 00:30:21.590 --> 00:30:22.640 align:middle line:90% call this theorem two-- 00:30:22.640 --> 00:30:25.250 align:middle line:90% 00:30:25.250 --> 00:30:31.970 align:middle line:84% now let me introduce the guest of honor. 00:30:31.970 --> 00:30:35.800 align:middle line:84% So first, I have the following claim. 00:30:35.800 --> 00:30:44.880 align:middle line:84% For all x in R, the series given by sum from n 00:30:44.880 --> 00:30:46.740 align:middle line:90% equals 0 to infinity-- 00:30:46.740 --> 00:30:52.350 align:middle line:84% and instead of n, I'm going to use k, cosine 160 kx. 00:30:52.350 --> 00:30:58.530 align:middle line:84% So here's our very oscillatory guy over 4 00:30:58.530 --> 00:31:04.690 align:middle line:84% to the k is absolutely convergent. 00:31:04.690 --> 00:31:08.700 align:middle line:90% 00:31:08.700 --> 00:31:11.900 align:middle line:84% So for each x in R, this series converges absolutely. 00:31:11.900 --> 00:31:13.530 align:middle line:90% So it spits out a real number. 00:31:13.530 --> 00:31:17.930 align:middle line:84% So I can define a function in terms of this series. 00:31:17.930 --> 00:31:22.260 align:middle line:90% 00:31:22.260 --> 00:31:30.700 align:middle line:84% Let f from R to R be defined by a prime-- 00:31:30.700 --> 00:31:39.920 align:middle line:84% f of x is just the number that I get when 00:31:39.920 --> 00:31:41.270 align:middle line:90% I stick in x to the series. 00:31:41.270 --> 00:31:48.180 align:middle line:90% 00:31:48.180 --> 00:31:50.100 align:middle line:84% Which is meaningful, because for all x, 00:31:50.100 --> 00:31:51.780 align:middle line:84% this series converges absolutely. 00:31:51.780 --> 00:31:53.670 align:middle line:90% So this is just a function. 00:31:53.670 --> 00:31:55.620 align:middle line:90% I put in x. 00:31:55.620 --> 00:31:57.090 align:middle line:90% I get out a real number. 00:31:57.090 --> 00:32:09.700 align:middle line:84% Then the claim is, f is bounded and continuous. 00:32:09.700 --> 00:32:12.160 align:middle line:90% 00:32:12.160 --> 00:32:13.910 align:middle line:84% So this is going to be our function, which 00:32:13.910 --> 00:32:17.630 align:middle line:84% is continuous but nowhere differentiable. 00:32:17.630 --> 00:32:21.200 align:middle line:84% As you can see, it's built out of a bunch 00:32:21.200 --> 00:32:25.850 align:middle line:84% of very oscillatory functions, cosine of 160 kx, 00:32:25.850 --> 00:32:29.160 align:middle line:84% this is just a really big number here. 00:32:29.160 --> 00:32:31.610 align:middle line:84% And as I said, when we were talking about this theorem 00:32:31.610 --> 00:32:34.670 align:middle line:84% here, each one of those pieces is very oscillatory. 00:32:34.670 --> 00:32:39.530 align:middle line:84% On a very small interval, it oscillates between two values 00:32:39.530 --> 00:32:40.895 align:middle line:90% that differ by at least 1. 00:32:40.895 --> 00:32:47.340 align:middle line:90% 00:32:47.340 --> 00:32:50.820 align:middle line:84% And so, all of these guys are oscillatory. 00:32:50.820 --> 00:32:54.390 align:middle line:84% And they're oscillatory on smaller and smaller intervals. 00:32:54.390 --> 00:32:56.640 align:middle line:84% And somehow, we're adding them all up 00:32:56.640 --> 00:33:00.000 align:middle line:84% in a way to get a function that's oscillatory 00:33:00.000 --> 00:33:01.890 align:middle line:90% on arbitrarily small intervals. 00:33:01.890 --> 00:33:04.265 align:middle line:84% And therefore, it will not be differentiable. 00:33:04.265 --> 00:33:07.930 align:middle line:90% 00:33:07.930 --> 00:33:20.110 align:middle line:84% So the proof of one is very easy. 00:33:20.110 --> 00:33:22.050 align:middle line:84% We just use the comparison principle. 00:33:22.050 --> 00:33:24.570 align:middle line:90% 00:33:24.570 --> 00:33:29.820 align:middle line:84% Cosine of 160 k times x over 4 to the k, 00:33:29.820 --> 00:33:32.130 align:middle line:84% this is less than or equal to-- because cosine 00:33:32.130 --> 00:33:35.070 align:middle line:84% of no matter what you plug in is bounded by 1. 00:33:35.070 --> 00:33:38.820 align:middle line:84% This is always bounded by 1 over 4 to the k. 00:33:38.820 --> 00:33:45.410 align:middle line:84% And therefore, by the comparison principle, 00:33:45.410 --> 00:33:58.190 align:middle line:84% we get that this series is convergent. 00:33:58.190 --> 00:34:00.890 align:middle line:84% And therefore, the original series 00:34:00.890 --> 00:34:04.790 align:middle line:90% is absolutely convergent. 00:34:04.790 --> 00:34:08.060 align:middle line:84% So each of these is bounded by 1 over 4 to the k. 00:34:08.060 --> 00:34:10.310 align:middle line:90% This series converges. 00:34:10.310 --> 00:34:12.320 align:middle line:90% This is a geometric series. 00:34:12.320 --> 00:34:15.500 align:middle line:84% Therefore, by the comparison principle, 00:34:15.500 --> 00:34:18.489 align:middle line:90% this series converges. 00:34:18.489 --> 00:34:24.270 align:middle line:84% So now, we have this function defined by this series. 00:34:24.270 --> 00:34:25.830 align:middle line:84% Note, this is not a power series, 00:34:25.830 --> 00:34:29.750 align:middle line:84% because a power series involves polynomials in x. 00:34:29.750 --> 00:34:33.449 align:middle line:84% This is cosine of x, or a number times x. 00:34:33.449 --> 00:34:35.400 align:middle line:90% So let's show it's bounded. 00:34:35.400 --> 00:34:38.775 align:middle line:84% The same proof actually that we gave here shows it's bounded. 00:34:38.775 --> 00:34:41.345 align:middle line:90% 00:34:41.345 --> 00:34:48.290 align:middle line:84% Let x be in R then, f of x equals-- 00:34:48.290 --> 00:34:56.260 align:middle line:90% 00:34:56.260 --> 00:35:16.360 align:middle line:84% this is equal to the limit as n goes to infinity of this sum. 00:35:16.360 --> 00:35:25.460 align:middle line:84% And now, for our absolute values, this limit pulls out. 00:35:25.460 --> 00:35:31.820 align:middle line:84% Whenever the limit exists, the limit of the absolute value 00:35:31.820 --> 00:35:34.340 align:middle line:84% equals the absolute value of the limit. 00:35:34.340 --> 00:35:43.290 align:middle line:90% 00:35:43.290 --> 00:35:46.440 align:middle line:84% And this is less than or equal to the limit 00:35:46.440 --> 00:35:50.880 align:middle line:84% as n goes to infinity of bringing the [INAUDIBLE].. 00:35:50.880 --> 00:35:55.240 align:middle line:84% So by the triangle inequality, this is just a finite sum. 00:35:55.240 --> 00:35:58.230 align:middle line:84% So I can bring the absolute values in. 00:35:58.230 --> 00:36:06.030 align:middle line:84% And I get sum from a equals 0 to m of cosine 160 k, x over 4 00:36:06.030 --> 00:36:08.140 align:middle line:90% to the k, absolute value. 00:36:08.140 --> 00:36:10.740 align:middle line:84% And this is less than or equal to limit 00:36:10.740 --> 00:36:16.080 align:middle line:84% as n goes to infinity of sum k equals 00:36:16.080 --> 00:36:20.920 align:middle line:84% 0 to m of 4 to the minus k, again, 00:36:20.920 --> 00:36:23.840 align:middle line:84% because cosine of anything is bounded by 1. 00:36:23.840 --> 00:36:26.600 align:middle line:90% 00:36:26.600 --> 00:36:30.950 align:middle line:90% And this equals 4/3. 00:36:30.950 --> 00:36:37.930 align:middle line:84% This is just a sum from k equals 0 to infinity 4 to the minus k. 00:36:37.930 --> 00:36:40.060 align:middle line:84% So this function is always bounded by 4/3. 00:36:40.060 --> 00:36:47.810 align:middle line:90% 00:36:47.810 --> 00:36:49.430 align:middle line:90% So the function is bounded. 00:36:49.430 --> 00:36:52.240 align:middle line:90% Let's now show it's continuous. 00:36:52.240 --> 00:37:04.950 align:middle line:84% So to show a function's continuous, 00:37:04.950 --> 00:37:09.510 align:middle line:84% remember, we have that other characterization of continuity 00:37:09.510 --> 00:37:11.760 align:middle line:84% that a function is continuous at a point 00:37:11.760 --> 00:37:13.680 align:middle line:84% if and only if for every sequence converging 00:37:13.680 --> 00:37:20.170 align:middle line:84% to that point f of xn converges to c. 00:37:20.170 --> 00:37:30.820 align:middle line:84% So before I start writing all this down, let c be in R, 00:37:30.820 --> 00:37:34.040 align:middle line:84% and let xn be a sequence converging to c. 00:37:34.040 --> 00:37:39.850 align:middle line:84% So what we want to show is that limit 00:37:39.850 --> 00:37:46.270 align:middle line:84% as n goes to infinity of f of xn minus f of c in absolute value 00:37:46.270 --> 00:37:47.620 align:middle line:90% equals 0. 00:37:47.620 --> 00:37:49.690 align:middle line:84% That's the same as saying the limit 00:37:49.690 --> 00:37:52.300 align:middle line:84% as n goes to infinity of f of xn equals f of c. 00:37:52.300 --> 00:37:56.440 align:middle line:90% 00:37:56.440 --> 00:38:00.560 align:middle line:90% Now, f as a bounded function. 00:38:00.560 --> 00:38:04.750 align:middle line:84% So this sequence, f of xn minus f of c in absolute 00:38:04.750 --> 00:38:09.880 align:middle line:90% is a bounded sequence. 00:38:09.880 --> 00:38:11.980 align:middle line:90% So it has a lim sup. 00:38:11.980 --> 00:38:14.680 align:middle line:84% And we did an exercise on the assignment 00:38:14.680 --> 00:38:17.770 align:middle line:84% that the limit of a sequence equals 0 if 00:38:17.770 --> 00:38:23.470 align:middle line:84% and only if the lim sup equals 0. 00:38:23.470 --> 00:38:42.420 align:middle line:84% So equivalently, we'll show that lim sup of f of x sub 00:38:42.420 --> 00:38:47.860 align:middle line:90% n minus f of c equals 0. 00:38:47.860 --> 00:38:51.520 align:middle line:84% This thing always exists for a bounded sequence. 00:38:51.520 --> 00:38:53.260 align:middle line:84% Which is one of the reasons which 00:38:53.260 --> 00:39:01.080 align:middle line:84% is what makes lim sup so useful, is that they do always exist. 00:39:01.080 --> 00:39:04.130 align:middle line:84% So if we show that this lim sup equals 0, 00:39:04.130 --> 00:39:10.103 align:middle line:84% then this is equivalent to showing this limit equals 0. 00:39:10.103 --> 00:39:11.520 align:middle line:84% Again, this was an exercise, where 00:39:11.520 --> 00:39:14.857 align:middle line:84% you can take it as the lim inf of something 00:39:14.857 --> 00:39:17.190 align:middle line:84% that's non-negative is always bigger than or equal to 0. 00:39:17.190 --> 00:39:20.525 align:middle line:84% So if we show this is equal to 0, we would have 0 is less than 00:39:20.525 --> 00:39:22.650 align:middle line:84% or equal to the lim inf of this thing, is less than 00:39:22.650 --> 00:39:24.870 align:middle line:84% or equal to the lim sup, which equals 0. 00:39:24.870 --> 00:39:29.160 align:middle line:84% And therefore, the lim inf equals the lim sup equals 0. 00:39:29.160 --> 00:39:31.470 align:middle line:84% So that's another way of saying that this 00:39:31.470 --> 00:39:34.200 align:middle line:90% is equivalent to this. 00:39:34.200 --> 00:39:37.680 align:middle line:84% So we're going to show this at the lim sup of f of x 00:39:37.680 --> 00:39:40.350 align:middle line:90% sub n minus f of c equals 0. 00:39:40.350 --> 00:39:46.280 align:middle line:84% But we don't-- so that might be a little tough. 00:39:46.280 --> 00:39:51.740 align:middle line:84% But what we can show, and what we will show, 00:39:51.740 --> 00:39:55.010 align:middle line:84% is we'll give ourself a little room. 00:39:55.010 --> 00:39:59.150 align:middle line:84% We'll show that for all epsilon positive, 00:39:59.150 --> 00:40:05.590 align:middle line:84% the lim sup of f of x sub n, minus f 00:40:05.590 --> 00:40:11.220 align:middle line:84% of c, which is a non-negative number, is less than epsilon. 00:40:11.220 --> 00:40:14.310 align:middle line:84% So this is a fixed number, non-negative number, 00:40:14.310 --> 00:40:17.430 align:middle line:90% which is always smaller than-- 00:40:17.430 --> 00:40:19.020 align:middle line:84% put less than or equal to there-- 00:40:19.020 --> 00:40:21.640 align:middle line:84% which is smaller than any number that I want. 00:40:21.640 --> 00:40:23.070 align:middle line:90% And therefore, it has to be 0. 00:40:23.070 --> 00:40:26.890 align:middle line:90% 00:40:26.890 --> 00:40:30.360 align:middle line:90% And proving thus, lim sup. 00:40:30.360 --> 00:40:38.210 align:middle line:90% 00:40:38.210 --> 00:40:41.930 align:middle line:84% So just to recap, we want to show-- 00:40:41.930 --> 00:40:44.210 align:middle line:84% we have a sequence converging to c. 00:40:44.210 --> 00:40:48.320 align:middle line:84% We want to show that f of xn converges to f of c. 00:40:48.320 --> 00:40:50.240 align:middle line:84% Another way of stating that is that this limit 00:40:50.240 --> 00:40:51.890 align:middle line:90% of the absolute value between-- 00:40:51.890 --> 00:40:54.080 align:middle line:84% of the difference between these two converges to 0. 00:40:54.080 --> 00:40:56.930 align:middle line:84% That's again a equivalent way of stating the limit. 00:40:56.930 --> 00:41:00.170 align:middle line:90% 00:41:00.170 --> 00:41:05.150 align:middle line:84% And in another assignment, we proved 00:41:05.150 --> 00:41:09.740 align:middle line:84% that for a sequence of non-negative numbers, 00:41:09.740 --> 00:41:13.530 align:middle line:84% or if you like, just for a sequence of numbers, 00:41:13.530 --> 00:41:17.650 align:middle line:84% it converges to 0 if and only if the lim sup 00:41:17.650 --> 00:41:19.510 align:middle line:84% of the absolute value converges to 0. 00:41:19.510 --> 00:41:21.760 align:middle line:84% But this is a sequence of non-negative numbers, 00:41:21.760 --> 00:41:24.670 align:middle line:84% this absolute value of f of xn minus f of c. 00:41:24.670 --> 00:41:29.000 align:middle line:90% So this is equivalent to this. 00:41:29.000 --> 00:41:34.250 align:middle line:84% So this thing we want to show is equivalent to this thing 00:41:34.250 --> 00:41:36.450 align:middle line:90% right below it. 00:41:36.450 --> 00:41:40.400 align:middle line:84% Now, rather than show directly that this limit lim sup equals 00:41:40.400 --> 00:41:42.995 align:middle line:84% 0, we're going to show that for all epsilon positive 00:41:42.995 --> 00:41:44.870 align:middle line:84% it's less than or equal to that small number. 00:41:44.870 --> 00:41:46.328 align:middle line:84% And therefore, proving that it's 0, 00:41:46.328 --> 00:41:48.530 align:middle line:84% because it's a non-negative number smaller 00:41:48.530 --> 00:41:54.020 align:middle line:90% than every positive number. 00:41:54.020 --> 00:41:57.500 align:middle line:84% So this is our goal, to show this lim 00:41:57.500 --> 00:41:58.595 align:middle line:90% sup is less than epsilon. 00:41:58.595 --> 00:42:01.850 align:middle line:90% 00:42:01.850 --> 00:42:08.530 align:middle line:90% So let epsilon be positive. 00:42:08.530 --> 00:42:11.090 align:middle line:90% 00:42:11.090 --> 00:42:12.770 align:middle line:84% We're now going to prove that that lim 00:42:12.770 --> 00:42:14.360 align:middle line:90% sup is less than epsilon. 00:42:14.360 --> 00:42:21.570 align:middle line:90% 00:42:21.570 --> 00:42:32.640 align:middle line:84% So first off, let m0 be a natural number such 00:42:32.640 --> 00:42:39.000 align:middle line:84% that the sum from k equals m0 plus 1 to infinity of 4 00:42:39.000 --> 00:42:43.980 align:middle line:84% to the minus k is less than epsilon over 2. 00:42:43.980 --> 00:42:48.750 align:middle line:84% So this series here, right this is a convergent series 00:42:48.750 --> 00:42:52.980 align:middle line:84% if I go from k equals 0 to infinity. 00:42:52.980 --> 00:42:58.360 align:middle line:84% And we have this Cauchy criterion 00:42:58.360 --> 00:43:02.050 align:middle line:84% for convergent series, which can be equivalently stated 00:43:02.050 --> 00:43:05.920 align:middle line:84% as for all m0 and natural number, there exists-- 00:43:05.920 --> 00:43:08.910 align:middle line:84% or for all epsilon, there exist-- 00:43:08.910 --> 00:43:11.690 align:middle line:84% so first off, rather than go through all 00:43:11.690 --> 00:43:13.370 align:middle line:84% that, we can actually just compute this. 00:43:13.370 --> 00:43:16.250 align:middle line:90% 00:43:16.250 --> 00:43:30.320 align:middle line:84% That the left-hand side equals 4 to the minus m0 minus 1, 00:43:30.320 --> 00:43:34.410 align:middle line:84% times sum from L equals 0 to infinity, 00:43:34.410 --> 00:43:40.380 align:middle line:84% 4 to the minus L, which equals 4/3. 00:43:40.380 --> 00:43:45.990 align:middle line:84% 4 to the minus m0 not minus 1, which equals one 1/2, 00:43:45.990 --> 00:43:47.190 align:middle line:90% 4 to the minus m0. 00:43:47.190 --> 00:43:49.940 align:middle line:90% 00:43:49.940 --> 00:43:53.750 align:middle line:90% So if m0 is chosen very large-- 00:43:53.750 --> 00:43:55.730 align:middle line:90% this is 4 to the minus m0. 00:43:55.730 --> 00:43:58.550 align:middle line:90% So this is-- 00:43:58.550 --> 00:44:00.920 align:middle line:84% I can write this as 1/2 over 1/4. 00:44:00.920 --> 00:44:03.926 align:middle line:90% 00:44:03.926 --> 00:44:06.830 align:middle line:84% And as long as m0 is chosen very large, 00:44:06.830 --> 00:44:09.787 align:middle line:84% I can always make this very small. 00:44:09.787 --> 00:44:11.120 align:middle line:90% That's the left side over there. 00:44:11.120 --> 00:44:14.540 align:middle line:84% So I can always find m0 so that this is the case. 00:44:14.540 --> 00:44:25.950 align:middle line:90% 00:44:25.950 --> 00:44:36.870 align:middle line:84% And now, we compute the lim sup of f of xn minus f of x. 00:44:36.870 --> 00:44:38.590 align:middle line:84% We split this up into two pieces. 00:44:38.590 --> 00:44:47.400 align:middle line:84% This is equal to lim sup then of two parts, sum from k 00:44:47.400 --> 00:44:56.640 align:middle line:84% equals 0 to m0 of cosine of 160 k xn minus-- 00:44:56.640 --> 00:45:01.490 align:middle line:84% so each of these is defined in terms of a sum. 00:45:01.490 --> 00:45:04.670 align:middle line:84% So I'm going to break the sum up to m0. 00:45:04.670 --> 00:45:06.380 align:middle line:84% And then, everything past m0, this 00:45:06.380 --> 00:45:13.760 align:middle line:84% is equal to 1 over 4k times cosine 160 k 00:45:13.760 --> 00:45:26.700 align:middle line:84% xn minus cosine 160 kx, plus sum from k equals 0 to m0 00:45:26.700 --> 00:45:46.150 align:middle line:84% plus 1 to infinity, 1 over 4 k, same thing, absolute value. 00:45:46.150 --> 00:45:52.310 align:middle line:84% And now, we use the triangle inequality and the fact 00:45:52.310 --> 00:45:55.980 align:middle line:84% that lim sups preserve inequality. 00:45:55.980 --> 00:45:58.070 align:middle line:84% So the absolute value of this thing 00:45:58.070 --> 00:46:00.650 align:middle line:84% is less than or equal to the sum of the absolute values. 00:46:00.650 --> 00:46:02.660 align:middle line:84% And the lim sup of those sequences 00:46:02.660 --> 00:46:04.970 align:middle line:84% is less than or equal to the sum of the lim sups. 00:46:04.970 --> 00:46:08.960 align:middle line:84% That was another exercise from an assignment, 00:46:08.960 --> 00:46:11.180 align:middle line:84% or you did something similar with the lim int. 00:46:11.180 --> 00:46:17.170 align:middle line:84% So this is less than or equal to lim sup of-- 00:46:17.170 --> 00:46:19.300 align:middle line:84% now I use the triangle inequality, 00:46:19.300 --> 00:46:32.210 align:middle line:84% sum from k equals 0 to m0, 4 minus k, cosine 160 k x of n 00:46:32.210 --> 00:46:35.150 align:middle line:90% minus cosine 160 k-- 00:46:35.150 --> 00:46:38.152 align:middle line:90% 00:46:38.152 --> 00:46:41.330 align:middle line:90% oh dear. 00:46:41.330 --> 00:46:45.270 align:middle line:90% 00:46:45.270 --> 00:46:46.630 align:middle line:90% So I've been writing x. 00:46:46.630 --> 00:46:48.550 align:middle line:90% I meant to write c. 00:46:48.550 --> 00:46:51.340 align:middle line:90% Sorry about that. 00:46:51.340 --> 00:47:00.390 align:middle line:90% c, c, c, c. 00:47:00.390 --> 00:47:00.890 align:middle line:90% OK. 00:47:00.890 --> 00:47:16.320 align:middle line:90% 00:47:16.320 --> 00:47:20.730 align:middle line:84% And then, same thing in these brackets here. 00:47:20.730 --> 00:47:25.410 align:middle line:90% 00:47:25.410 --> 00:47:29.300 align:middle line:84% Now, [INAUDIBLE] might keep that. 00:47:29.300 --> 00:47:32.220 align:middle line:90% 00:47:32.220 --> 00:47:36.870 align:middle line:84% So we have that this lim sup of f of x sub n minus f of c 00:47:36.870 --> 00:47:39.630 align:middle line:84% is less than or equal to the lim sup of this guy, 00:47:39.630 --> 00:47:41.670 align:middle line:90% plus the lim sup of this guy. 00:47:41.670 --> 00:47:45.480 align:middle line:84% And now, I'm going to use the triangle inequality, again, 00:47:45.480 --> 00:47:50.910 align:middle line:84% bringing the absolute values inside of the sums, which 00:47:50.910 --> 00:47:56.210 align:middle line:84% is perfectly valid, even for the infinite sum 00:47:56.210 --> 00:47:58.490 align:middle line:84% by the same argument I used basically over there. 00:47:58.490 --> 00:48:01.220 align:middle line:90% 00:48:01.220 --> 00:48:03.290 align:middle line:84% This is less than or equal to the lim sup 00:48:03.290 --> 00:48:23.470 align:middle line:84% as n goes to infinity of minus cosine of 160 kc, 00:48:23.470 --> 00:48:36.690 align:middle line:84% and sup plus lim sup of sum from k equals m0 plus 1 00:48:36.690 --> 00:48:40.200 align:middle line:84% to infinity of, now, the absolute value 00:48:40.200 --> 00:48:42.000 align:middle line:90% of the same thing. 00:48:42.000 --> 00:48:47.580 align:middle line:90% 00:48:47.580 --> 00:48:51.450 align:middle line:84% Now, I applied the triangle inequality to that. 00:48:51.450 --> 00:49:05.580 align:middle line:90% 00:49:05.580 --> 00:49:08.730 align:middle line:84% Now, cosine of anything is always bounded by 1. 00:49:08.730 --> 00:49:12.300 align:middle line:90% So this is bounded by-- 00:49:12.300 --> 00:49:18.250 align:middle line:84% first off, let me come back to this second one. m0 is fixed. 00:49:18.250 --> 00:49:20.010 align:middle line:84% Remember, this is not changing within. 00:49:20.010 --> 00:49:20.910 align:middle line:90% It's just fixed. 00:49:20.910 --> 00:49:24.360 align:middle line:90% It depended only on epsilon. 00:49:24.360 --> 00:49:26.580 align:middle line:90% So m0 is fixed. 00:49:26.580 --> 00:49:30.870 align:middle line:84% And this is less than or equal to lim sup of int. 00:49:30.870 --> 00:49:36.120 align:middle line:84% Now we use what we know about cosine, 00:49:36.120 --> 00:49:39.270 align:middle line:84% that cosine of something minus cosine of something else 00:49:39.270 --> 00:49:42.520 align:middle line:84% is bounded by the difference in the argument. 00:49:42.520 --> 00:49:45.480 align:middle line:84% So this is bounded by minus 4 to the minus 00:49:45.480 --> 00:49:59.280 align:middle line:84% k times 160 k xn minus 160 kc, plus lim soup 00:49:59.280 --> 00:50:11.850 align:middle line:84% of k equals m0 plus 1 to infinity, 4 minus k times 2. 00:50:11.850 --> 00:50:14.160 align:middle line:84% As this is bounded by 1, that's bounded by 1. 00:50:14.160 --> 00:50:17.740 align:middle line:90% Now, there's no n left here. 00:50:17.740 --> 00:50:22.570 align:middle line:84% So this is really just now equal to that. 00:50:22.570 --> 00:50:25.710 align:middle line:84% And we chose m0 so that this quantity here 00:50:25.710 --> 00:50:27.060 align:middle line:90% is less than epsilon over 2. 00:50:27.060 --> 00:50:29.680 align:middle line:84% So times 2, this is less than epsilon. 00:50:29.680 --> 00:50:37.770 align:middle line:84% So this is less than or equal to lim sup, sum from k 00:50:37.770 --> 00:50:42.780 align:middle line:90% equals 0 to m0 of 40 k. 00:50:42.780 --> 00:50:45.210 align:middle line:84% Now, again, this is just a fixed number. 00:50:45.210 --> 00:50:48.420 align:middle line:90% 00:50:48.420 --> 00:50:51.570 align:middle line:90% The lim sup is in n. 00:50:51.570 --> 00:50:57.510 align:middle line:84% This is just a fixed number times x sub n minus c. 00:50:57.510 --> 00:51:00.540 align:middle line:84% The difference in this is just equal to 160 k times 4 00:51:00.540 --> 00:51:02.550 align:middle line:90% to the minus k, that's 40 k. 00:51:02.550 --> 00:51:06.985 align:middle line:84% And then, we chose m0 so that this quantity over here 00:51:06.985 --> 00:51:07.860 align:middle line:90% is less than epsilon. 00:51:07.860 --> 00:51:12.040 align:middle line:90% 00:51:12.040 --> 00:51:16.170 align:middle line:84% And now, as n goes to infinity, x sub n is converging to c. 00:51:16.170 --> 00:51:19.590 align:middle line:84% So this thing here converges to 0, 00:51:19.590 --> 00:51:32.580 align:middle line:84% equals times 0 plus epsilon equals epsilon. 00:51:32.580 --> 00:51:39.170 align:middle line:84% So to me, this is a real first proof 00:51:39.170 --> 00:51:41.390 align:middle line:84% of analysis, where you're using everything 00:51:41.390 --> 00:51:45.110 align:middle line:84% that you've been exposed to up to this point 00:51:45.110 --> 00:51:48.170 align:middle line:90% to prove a really deep theorem. 00:51:48.170 --> 00:51:50.900 align:middle line:90% So go through this slowly. 00:51:50.900 --> 00:51:54.650 align:middle line:90% But there's not too much-- 00:51:54.650 --> 00:51:59.640 align:middle line:90% 00:51:59.640 --> 00:52:01.860 align:middle line:84% the estimates are not all that tricky 00:52:01.860 --> 00:52:04.170 align:middle line:84% once you have them in front of you. 00:52:04.170 --> 00:52:07.170 align:middle line:84% So we've shown that for all epsilon, 00:52:07.170 --> 00:52:09.570 align:middle line:84% this lim sup is less than or equal to epsilon. 00:52:09.570 --> 00:52:14.040 align:middle line:84% Because remember, we ended with epsilon here. 00:52:14.040 --> 00:52:19.060 align:middle line:84% But we started off estimating this lim sup here. 00:52:19.060 --> 00:52:24.270 align:middle line:90% 00:52:24.270 --> 00:52:29.743 align:middle line:84% Now we're in a position to prove the final theorem 00:52:29.743 --> 00:52:30.660 align:middle line:90% that we want to prove. 00:52:30.660 --> 00:52:32.310 align:middle line:90% And this is due to Weierstrass. 00:52:32.310 --> 00:52:45.100 align:middle line:90% 00:52:45.100 --> 00:52:48.460 align:middle line:84% The function which we've been studying, 00:52:48.460 --> 00:52:55.510 align:middle line:84% f of x equals sum from k equals 0 to infinity cosine 160 kx 00:52:55.510 --> 00:53:07.233 align:middle line:84% over 4 to the k is nowhere differentiable. 00:53:07.233 --> 00:53:12.890 align:middle line:90% 00:53:12.890 --> 00:53:16.320 align:middle line:84% So with everything we've done so far, 00:53:16.320 --> 00:53:18.680 align:middle line:84% so we this function is continuous. 00:53:18.680 --> 00:53:19.710 align:middle line:90% We've proven that. 00:53:19.710 --> 00:53:23.810 align:middle line:84% So this theorem provides you with an example 00:53:23.810 --> 00:53:25.850 align:middle line:84% of a function which is continuous but nowhere 00:53:25.850 --> 00:53:28.650 align:middle line:90% differentiable. 00:53:28.650 --> 00:53:33.960 align:middle line:84% And with what we have on the board, 00:53:33.960 --> 00:53:37.350 align:middle line:84% namely the key parts are going to be what's 00:53:37.350 --> 00:53:40.410 align:middle line:90% in that first theorem there. 00:53:40.410 --> 00:53:45.073 align:middle line:84% And with this triangle inequality we have over here, 00:53:45.073 --> 00:53:46.240 align:middle line:90% we'll be able to prove this. 00:53:46.240 --> 00:53:51.270 align:middle line:84% So in fact, let me give myself a little space. 00:53:51.270 --> 00:53:54.000 align:middle line:90% 00:53:54.000 --> 00:54:06.620 align:middle line:84% And I'm going to state theorem two from over there, 00:54:06.620 --> 00:54:08.810 align:middle line:84% because I need some space to right in a minute. 00:54:08.810 --> 00:54:14.330 align:middle line:84% For all a, b, c, the absolute value of a, plus b, plus c 00:54:14.330 --> 00:54:18.350 align:middle line:84% is bigger than or equal to a minus 00:54:18.350 --> 00:54:21.140 align:middle line:84% the absolute value of b minus the absolute value of c. 00:54:21.140 --> 00:54:32.750 align:middle line:90% 00:54:32.750 --> 00:54:34.054 align:middle line:90% So let me re-summarize. 00:54:34.054 --> 00:54:42.045 align:middle line:90% 00:54:42.045 --> 00:54:43.420 align:middle line:84% Since I've summarized it already. 00:54:43.420 --> 00:54:45.410 align:middle line:84% What we've done up to this point, 00:54:45.410 --> 00:54:47.860 align:middle line:84% we've shown that this function here is well-defined. 00:54:47.860 --> 00:54:51.010 align:middle line:84% Absolutely, the series is always absolutely convergent. 00:54:51.010 --> 00:54:53.860 align:middle line:84% And therefore, this function is bounded, 00:54:53.860 --> 00:54:57.910 align:middle line:84% and/or we also proved that it's bounded and continuous. 00:54:57.910 --> 00:55:00.600 align:middle line:90% That was the previous theorem. 00:55:00.600 --> 00:55:02.820 align:middle line:84% So this function is bounded and continuous. 00:55:02.820 --> 00:55:06.540 align:middle line:84% And we're going to prove it's nowhere differentiable. 00:55:06.540 --> 00:55:08.500 align:middle line:90% And again, what's the idea? 00:55:08.500 --> 00:55:10.920 align:middle line:84% The idea is that we've built it out of functions 00:55:10.920 --> 00:55:15.660 align:middle line:84% which are highly oscillatory at smaller and smaller scales. 00:55:15.660 --> 00:55:18.750 align:middle line:84% So somehow this function is highly 00:55:18.750 --> 00:55:21.290 align:middle line:90% oscillatory at every scale. 00:55:21.290 --> 00:55:23.040 align:middle line:84% And if you have a function which is highly 00:55:23.040 --> 00:55:25.080 align:middle line:84% oscillatory of every scale, if some of you 00:55:25.080 --> 00:55:32.090 align:middle line:84% have heard of Brownian motion, which is a function which is-- 00:55:32.090 --> 00:55:38.680 align:middle line:84% I mean, which is a path which is highly oscillatory, 00:55:38.680 --> 00:55:43.873 align:middle line:84% then that function will not be differentiable anywhere. 00:55:43.873 --> 00:55:46.550 align:middle line:90% 00:55:46.550 --> 00:55:58.100 align:middle line:84% So proof, let c be any real number. 00:55:58.100 --> 00:55:59.840 align:middle line:84% And what we're going to do, just as 00:55:59.840 --> 00:56:02.780 align:middle line:84% in when we looked at f of x equals the absolute value of x, 00:56:02.780 --> 00:56:15.600 align:middle line:84% we're going to construct or find a sequence x sub n, 00:56:15.600 --> 00:56:26.200 align:middle line:84% such that x sub n does not equal c for all n, xn converges to c. 00:56:26.200 --> 00:56:33.650 align:middle line:84% And the sequence, f of x of n, minus f of c, over x 00:56:33.650 --> 00:56:38.050 align:middle line:84% sub n minus c, in fact, is divergent. 00:56:38.050 --> 00:56:44.170 align:middle line:84% But we'll go even further, is unbounded. 00:56:44.170 --> 00:56:46.980 align:middle line:90% 00:56:46.980 --> 00:56:48.870 align:middle line:84% And therefore, the sequence cannot converge. 00:56:48.870 --> 00:56:51.120 align:middle line:84% And therefore the function is not differentiable at c. 00:56:51.120 --> 00:57:01.910 align:middle line:90% 00:57:01.910 --> 00:57:07.430 align:middle line:84% So we're going to use theorem one to find the sequence. 00:57:07.430 --> 00:57:14.780 align:middle line:90% 00:57:14.780 --> 00:57:22.260 align:middle line:84% By theorem one, namely part two, for all N, a natural number, 00:57:22.260 --> 00:57:28.440 align:middle line:84% there exists the x of n, such that what? 00:57:28.440 --> 00:57:31.330 align:middle line:84% x sub n is in this interval of the form. 00:57:31.330 --> 00:57:44.905 align:middle line:84% So c plus pi over 160 n is less than x, is less than c 00:57:44.905 --> 00:57:48.640 align:middle line:90% plus plus 3 pi over 160 n. 00:57:48.640 --> 00:57:53.520 align:middle line:90% 00:57:53.520 --> 00:58:07.522 align:middle line:84% And cosine of 160 n x of n minus cosine of 160 nc 00:58:07.522 --> 00:58:08.730 align:middle line:90% is bigger than or equal to 1. 00:58:08.730 --> 00:58:16.880 align:middle line:90% 00:58:16.880 --> 00:58:20.750 align:middle line:84% So let me call these two properties a and b. 00:58:20.750 --> 00:58:25.290 align:middle line:90% 00:58:25.290 --> 00:58:31.920 align:middle line:84% So another way of writing a is, we could have instead 00:58:31.920 --> 00:58:36.000 align:middle line:84% written, by subtracting c across the board, 00:58:36.000 --> 00:58:46.080 align:middle line:84% that x sub n minus c is between pi over 160 n and 3 pi 00:58:46.080 --> 00:58:48.890 align:middle line:90% over 160 n. 00:58:48.890 --> 00:58:55.700 align:middle line:84% So by a, for all n, x sub n does not 00:58:55.700 --> 00:58:58.190 align:middle line:84% equal to c, because their difference is bounded below 00:58:58.190 --> 00:58:59.090 align:middle line:90% by a positive number. 00:58:59.090 --> 00:59:02.753 align:middle line:90% 00:59:02.753 --> 00:59:04.670 align:middle line:84% I mean, their difference is actually positive, 00:59:04.670 --> 00:59:05.690 align:middle line:90% and it's non-zero. 00:59:05.690 --> 00:59:18.750 align:middle line:84% So-- and by the squeeze theorem, again, 00:59:18.750 --> 00:59:22.090 align:middle line:84% if you like putting the c on both sides, 00:59:22.090 --> 00:59:24.960 align:middle line:84% we get that x sub n converges to c. 00:59:24.960 --> 00:59:27.150 align:middle line:90% So this will be our sequence-- 00:59:27.150 --> 00:59:29.880 align:middle line:84% our bad sequence, for which the difference quotient 00:59:29.880 --> 00:59:30.720 align:middle line:90% is unbounded. 00:59:30.720 --> 00:59:37.760 align:middle line:90% 00:59:37.760 --> 00:59:43.000 align:middle line:84% So in fact, let me write this out a little bit more. 00:59:43.000 --> 00:59:50.650 align:middle line:84% And x sub n minus c, which is equal to x 00:59:50.650 --> 00:59:55.110 align:middle line:84% sub n minus c, because this is positive, 00:59:55.110 --> 01:00:00.730 align:middle line:90% is less than 3 pi over 160 n. 01:00:00.730 --> 01:00:02.340 align:middle line:90% And this goes to 0. 01:00:02.340 --> 01:00:06.220 align:middle line:84% Therefore, the absolute value of xn minus c converges to 0. 01:00:06.220 --> 01:00:08.110 align:middle line:84% And therefore, x sub n converges to c. 01:00:08.110 --> 01:00:19.430 align:middle line:90% 01:00:19.430 --> 01:00:21.230 align:middle line:84% Now, to lessen the number of times 01:00:21.230 --> 01:00:30.070 align:middle line:84% I have to write cosine 160 k, let fk of x be cosine 01:00:30.070 --> 01:00:34.660 align:middle line:90% 160 kx over 4 to the k. 01:00:34.660 --> 01:00:43.030 align:middle line:84% So f of x is equal to the sum of k equal to 0 of fk of x. 01:00:43.030 --> 01:00:52.962 align:middle line:90% 01:00:52.962 --> 01:00:54.420 align:middle line:84% And now, what we're going to do is, 01:00:54.420 --> 01:00:57.120 align:middle line:84% we're going to find a lower bound 01:00:57.120 --> 01:01:00.900 align:middle line:84% on the absolute value of f of xn minus f of c over x sub 01:01:00.900 --> 01:01:02.160 align:middle line:90% n minus c. 01:01:02.160 --> 01:01:16.840 align:middle line:90% So now, find a lower bound on-- 01:01:16.840 --> 01:01:29.480 align:middle line:90% 01:01:29.480 --> 01:01:32.000 align:middle line:84% if I can find a lower bound on this absolute value, which 01:01:32.000 --> 01:01:34.070 align:middle line:84% is getting large, as large as I wish, 01:01:34.070 --> 01:01:37.280 align:middle line:84% then that proves that this sequence is unbounded, 01:01:37.280 --> 01:01:38.240 align:middle line:90% and I'm done. 01:01:38.240 --> 01:01:54.430 align:middle line:90% 01:01:54.430 --> 01:02:03.501 align:middle line:84% So let's look at the absolute value of f of xn minus f of c. 01:02:03.501 --> 01:02:07.190 align:middle line:90% 01:02:07.190 --> 01:02:11.843 align:middle line:84% Let's just write this in a few different ways. 01:02:11.843 --> 01:02:13.010 align:middle line:90% Or not a few different ways. 01:02:13.010 --> 01:02:14.520 align:middle line:90% But let's split it up as a sum. 01:02:14.520 --> 01:02:19.190 align:middle line:84% So all of these guys are equal to a sum of f sub k. 01:02:19.190 --> 01:02:28.960 align:middle line:90% So I have the n'th one. 01:02:28.960 --> 01:02:31.848 align:middle line:84% So remember, f sub k, this is just one 01:02:31.848 --> 01:02:32.890 align:middle line:90% of these building blocks. 01:02:32.890 --> 01:02:36.300 align:middle line:90% 01:02:36.300 --> 01:03:05.480 align:middle line:84% Plus sum from k equals 0 to n minus 1, plus I put x here. 01:03:05.480 --> 01:03:06.950 align:middle line:90% I meant to write c. 01:03:06.950 --> 01:03:09.700 align:middle line:90% 01:03:09.700 --> 01:03:19.480 align:middle line:84% And so, I'm going to let a sub n be this first number, 01:03:19.480 --> 01:03:25.720 align:middle line:84% b sub n be the second number, including the sum. 01:03:25.720 --> 01:03:28.030 align:middle line:90% And this will be c sub n. 01:03:28.030 --> 01:03:36.140 align:middle line:84% So this is equal to a sub n plus b sub n plus c sub n. 01:03:36.140 --> 01:03:38.640 align:middle line:84% And now I'm, going to use that triangle inequality I proved. 01:03:38.640 --> 01:03:41.890 align:middle line:90% 01:03:41.890 --> 01:03:50.020 align:middle line:84% Then f of x sub n minus f of c is greater than-- which 01:03:50.020 --> 01:03:54.855 align:middle line:84% is equal to a sub n plus c sub n plus c 01:03:54.855 --> 01:04:03.040 align:middle line:84% sub n is bigger than or equal to a sub n, minus b sub n, 01:04:03.040 --> 01:04:06.040 align:middle line:90% minus c sub n. 01:04:06.040 --> 01:04:08.290 align:middle line:84% Now, a sub n, this is going to be-- 01:04:08.290 --> 01:04:10.360 align:middle line:84% this is just the difference between these guys. 01:04:10.360 --> 01:04:13.360 align:middle line:90% This is bounded below by 1. 01:04:13.360 --> 01:04:15.910 align:middle line:90% So it's kind of large. 01:04:15.910 --> 01:04:18.700 align:middle line:90% Not 1, but 1 over 4 to the n. 01:04:18.700 --> 01:04:20.770 align:middle line:84% These other guys we'll prove are very small 01:04:20.770 --> 01:04:22.900 align:middle line:90% compared to the a sub n. 01:04:22.900 --> 01:04:24.700 align:middle line:90% And then, this lower bound-- 01:04:24.700 --> 01:04:26.980 align:middle line:84% this upper bound we have on x sub n minus c 01:04:26.980 --> 01:04:29.610 align:middle line:84% will be the nail in the coffin, as we say. 01:04:29.610 --> 01:04:34.510 align:middle line:84% So our goal is now to estimate from below a sub n-- remember, 01:04:34.510 --> 01:04:36.160 align:middle line:84% we're trying to find a lower bound 01:04:36.160 --> 01:04:39.070 align:middle line:84% for this quantity over x sub n minus c. 01:04:39.070 --> 01:04:43.120 align:middle line:84% So we need to estimate these things from below, 01:04:43.120 --> 01:04:45.010 align:middle line:90% or this sum from below. 01:04:45.010 --> 01:04:47.290 align:middle line:84% That means, we need to estimate this from below, 01:04:47.290 --> 01:04:49.270 align:middle line:84% and then b sub n and c sub n, since they 01:04:49.270 --> 01:04:50.815 align:middle line:90% have minus signs, from above. 01:04:50.815 --> 01:04:57.586 align:middle line:90% 01:04:57.586 --> 01:05:07.930 align:middle line:84% Now, by b, a sub n, which is equal to 4 to minus n, 01:05:07.930 --> 01:05:15.010 align:middle line:84% cosine 160 nx of n, minus cosine 160 to the nc. 01:05:15.010 --> 01:05:18.460 align:middle line:90% 01:05:18.460 --> 01:05:24.850 align:middle line:84% This is-- remember how we chose these x sub n's. 01:05:24.850 --> 01:05:30.580 align:middle line:84% This is bounded above by 1, or bounded below by 1. 01:05:30.580 --> 01:05:39.650 align:middle line:84% So this is bigger than or equal to 4 to the minus n. 01:05:39.650 --> 01:05:42.170 align:middle line:90% So that's a sub n. 01:05:42.170 --> 01:05:44.270 align:middle line:84% Now, let's look at how big b sub n is. 01:05:44.270 --> 01:05:47.157 align:middle line:90% 01:05:47.157 --> 01:05:48.740 align:middle line:84% Now, we want to bound this from above. 01:05:48.740 --> 01:05:50.900 align:middle line:84% Because when it hits the minus sign, 01:05:50.900 --> 01:05:52.670 align:middle line:84% this will flip the inequality, and we'll 01:05:52.670 --> 01:05:56.390 align:middle line:84% have that this would be bounded from below by something. 01:05:56.390 --> 01:06:01.790 align:middle line:84% So the absolute value of B7, that's equal to sum from k 01:06:01.790 --> 01:06:16.610 align:middle line:84% equals 0 to n minus 1 of fk of x of n minus f of c, k of c. 01:06:16.610 --> 01:06:19.240 align:middle line:90% 01:06:19.240 --> 01:06:21.580 align:middle line:84% And bringing the absolute values inside 01:06:21.580 --> 01:06:23.860 align:middle line:84% by the triangle inequality, this is less than or equal 01:06:23.860 --> 01:06:27.460 align:middle line:84% to sum from k equals n minus 1 of f 01:06:27.460 --> 01:06:32.080 align:middle line:90% of fk x of n minus fk of c. 01:06:32.080 --> 01:06:35.000 align:middle line:90% 01:06:35.000 --> 01:06:41.880 align:middle line:84% And now, so this is equal to sum from k 01:06:41.880 --> 01:06:48.870 align:middle line:84% equals 0 to n minus 1, 4 to the minus k, cosine 160 k, 01:06:48.870 --> 01:06:54.240 align:middle line:90% x sub n minus cosine of 160 kc. 01:06:54.240 --> 01:07:01.470 align:middle line:84% And now we use theorem one, number two-- 01:07:01.470 --> 01:07:02.630 align:middle line:90% or number one. 01:07:02.630 --> 01:07:04.730 align:middle line:90% I'm sorry, theorem one part one. 01:07:04.730 --> 01:07:08.000 align:middle line:90% 01:07:08.000 --> 01:07:12.240 align:middle line:90% So this is theorem 1, 1. 01:07:12.240 --> 01:07:15.070 align:middle line:90% 01:07:15.070 --> 01:07:17.080 align:middle line:84% The difference in these is bounded 01:07:17.080 --> 01:07:20.140 align:middle line:84% by 160 k times the difference in those two. 01:07:20.140 --> 01:07:33.070 align:middle line:84% So that's-- and now, these things we can sum up in closed 01:07:33.070 --> 01:07:33.570 align:middle line:90% form. 01:07:33.570 --> 01:07:36.690 align:middle line:84% Well, actually, x sub n minus c, we proved 01:07:36.690 --> 01:07:40.260 align:middle line:90% is less than 3 pi over 160 n. 01:07:40.260 --> 01:07:44.020 align:middle line:90% 01:07:44.020 --> 01:07:55.480 align:middle line:84% So this is less than a sum from k equals 0, n minus 140 k. 01:07:55.480 --> 01:08:00.040 align:middle line:90% And this equals 3 pi over 160 n. 01:08:00.040 --> 01:08:04.210 align:middle line:84% Now, sum from k equals 0 to n minus 1 40k, 01:08:04.210 --> 01:08:10.450 align:middle line:84% we used this formula for summing a geometric sum. 01:08:10.450 --> 01:08:11.470 align:middle line:90% This is equal to-- 01:08:11.470 --> 01:08:14.240 align:middle line:90% 01:08:14.240 --> 01:08:15.310 align:middle line:90% what was this equal to? 01:08:15.310 --> 01:08:18.750 align:middle line:90% 01:08:18.750 --> 01:08:22.094 align:middle line:90% 40 to the n minus 1 over 39. 01:08:22.094 --> 01:08:25.880 align:middle line:90% 01:08:25.880 --> 01:08:29.630 align:middle line:90% And which is less than-- 01:08:29.630 --> 01:08:32.090 align:middle line:90% OK, so take away the one. 01:08:32.090 --> 01:08:40.670 align:middle line:84% This is 1 over 13, four to the minus n times pi. 01:08:40.670 --> 01:08:45.380 align:middle line:90% 01:08:45.380 --> 01:08:48.979 align:middle line:90% And that's less than-- 01:08:48.979 --> 01:08:50.240 align:middle line:90% pi is less than 4. 01:08:50.240 --> 01:08:54.319 align:middle line:90% So plus 1. 01:08:54.319 --> 01:08:59.000 align:middle line:90% So let me summarize this. 01:08:59.000 --> 01:09:05.970 align:middle line:90% 01:09:05.970 --> 01:09:13.260 align:middle line:84% It is less than or to 1 over 13 fours and minus n plus 1. 01:09:13.260 --> 01:09:23.390 align:middle line:90% 01:09:23.390 --> 01:09:27.500 align:middle line:84% Now, for the last box, this is less-- 01:09:27.500 --> 01:09:29.430 align:middle line:90% the absolute value of c sub n. 01:09:29.430 --> 01:09:31.534 align:middle line:84% And absolute value is less than or equal to-- 01:09:31.534 --> 01:09:36.060 align:middle line:90% 01:09:36.060 --> 01:09:39.020 align:middle line:84% now, we just kind of do a brutal estimate. 01:09:39.020 --> 01:09:47.520 align:middle line:90% 01:09:47.520 --> 01:09:50.700 align:middle line:84% We bring the absolute values inside. 01:09:50.700 --> 01:09:53.700 align:middle line:84% So this is equal-- so just as we had before, 01:09:53.700 --> 01:09:56.340 align:middle line:84% with the b sub n's, except now we're 01:09:56.340 --> 01:09:58.320 align:middle line:84% summing from k equals n plus 1 to infinity. 01:09:58.320 --> 01:10:00.240 align:middle line:84% We brought the absolute values inside. 01:10:00.240 --> 01:10:02.910 align:middle line:84% And then, let's use the triangle inequality on that. 01:10:02.910 --> 01:10:14.960 align:middle line:84% This is-- remember, fk is equal to cosine of 160 k 01:10:14.960 --> 01:10:16.850 align:middle line:90% times x over 4 to the k. 01:10:16.850 --> 01:10:18.860 align:middle line:84% So that's no matter what you plug-in, 01:10:18.860 --> 01:10:22.700 align:middle line:84% that's bounded by 1 over 4 to the k. 01:10:22.700 --> 01:10:36.337 align:middle line:90% 01:10:36.337 --> 01:10:38.270 align:middle line:84% And now this we can actually sum. 01:10:38.270 --> 01:10:48.840 align:middle line:84% This is equal to 4 to the minus n minus 1, times sum from L 01:10:48.840 --> 01:10:56.930 align:middle line:84% equals 0 to infinity of 4 to the minus L. What does this equal? 01:10:56.930 --> 01:11:03.410 align:middle line:84% This equals 2 times 4 to the minus int minus 1 times 4/3 01:11:03.410 --> 01:11:06.520 align:middle line:90% equals 2/3 4 to the minus n. 01:11:06.520 --> 01:11:13.740 align:middle line:90% 01:11:13.740 --> 01:11:18.720 align:middle line:84% So, my board work's getting a little shoddy, but it'll be OK. 01:11:18.720 --> 01:11:21.960 align:middle line:90% 01:11:21.960 --> 01:11:22.560 align:middle line:90% I.e. 01:11:22.560 --> 01:11:24.630 align:middle line:84% we've proven that the absolute value of c sub n 01:11:24.630 --> 01:11:29.070 align:middle line:84% is bounded by 2/3 4 to the minus n. 01:11:29.070 --> 01:11:38.520 align:middle line:90% 01:11:38.520 --> 01:11:42.720 align:middle line:84% So combining everything that we've done. 01:11:42.720 --> 01:11:49.530 align:middle line:84% We've shown that f of x sub n minus f of c. 01:11:49.530 --> 01:12:06.870 align:middle line:84% Which remember, is bounded below by a sub n-- remember, 01:12:06.870 --> 01:12:10.690 align:middle line:84% so the first box is bounded by 4 to the minus n. 01:12:10.690 --> 01:12:14.190 align:middle line:84% So first off, everything, the a sub n's, b sub n's, 01:12:14.190 --> 01:12:16.740 align:middle line:84% c sub n's, they have some forward 01:12:16.740 --> 01:12:21.966 align:middle line:90% to the minus n involved. 01:12:21.966 --> 01:12:24.290 align:middle line:84% What happened to my estimate for b? 01:12:24.290 --> 01:12:26.840 align:middle line:90% Oh, covering it up-- 01:12:26.840 --> 01:12:27.890 align:middle line:90% 4 to the minus n. 01:12:27.890 --> 01:12:34.730 align:middle line:84% So this is one coming from the a sub n's, and minus 4 over 13 01:12:34.730 --> 01:12:44.820 align:middle line:84% coming from the b sub n's, minus 2/3 coming from the c sub n's. 01:12:44.820 --> 01:12:49.880 align:middle line:90% 01:12:49.880 --> 01:12:55.100 align:middle line:84% And if you do the arithmetic, this 01:12:55.100 --> 01:12:59.285 align:middle line:90% is 4 to the minus n, 1 over 39. 01:12:59.285 --> 01:13:02.660 align:middle line:90% 01:13:02.660 --> 01:13:06.290 align:middle line:84% So f of x sub n minus c is bounded 01:13:06.290 --> 01:13:10.130 align:middle line:84% from below by 4 to the minus n times 1 over 39. 01:13:10.130 --> 01:13:15.320 align:middle line:84% Now I just divide by x sub n minus c, which is bounded 01:13:15.320 --> 01:13:18.200 align:middle line:90% above by 3 pi over 160 n. 01:13:18.200 --> 01:13:34.240 align:middle line:84% And therefore, when I take reciprocals, I get that. 01:13:34.240 --> 01:13:36.210 align:middle line:84% I get that the absolute value of f 01:13:36.210 --> 01:13:42.240 align:middle line:84% of x of n minus f of c over x of n minus c 01:13:42.240 --> 01:13:49.350 align:middle line:84% is bounded from above by 1 over x of n minus C times 4 01:13:49.350 --> 01:13:55.710 align:middle line:84% to the minus n, 1 over 39, which is bounded below by now 01:13:55.710 --> 01:13:57.810 align:middle line:90% inserting this estimate here. 01:13:57.810 --> 01:14:01.660 align:middle line:90% I get 40 n over-- 01:14:01.660 --> 01:14:05.630 align:middle line:90% I think that's 1017 pi. 01:14:05.630 --> 01:14:06.130 align:middle line:90% Yes. 01:14:06.130 --> 01:14:10.160 align:middle line:90% 01:14:10.160 --> 01:14:13.250 align:middle line:84% And therefore, the absolute value 01:14:13.250 --> 01:14:17.510 align:middle line:84% of the difference quotient is bounded below by 40 01:14:17.510 --> 01:14:20.480 align:middle line:90% to the n over some fixed number. 01:14:20.480 --> 01:14:24.170 align:middle line:84% And this thing on the right-hand side is unbounded, as in, 01:14:24.170 --> 01:14:28.020 align:middle line:84% I guess, as in varies because from 1 to infinity. 01:14:28.020 --> 01:14:30.260 align:middle line:84% And therefore, the absolute value 01:14:30.260 --> 01:14:49.220 align:middle line:84% of this difference quotient is unbounded, 01:14:49.220 --> 01:14:50.620 align:middle line:90% which finishes the proof. 01:14:50.620 --> 01:14:54.490 align:middle line:90% 01:14:54.490 --> 01:14:57.460 align:middle line:84% So we've done some things, maybe a few things 01:14:57.460 --> 01:14:59.710 align:middle line:84% that you haven't seen in calculus course. 01:14:59.710 --> 01:15:02.980 align:middle line:84% Definitely haven't seen this, most likely. 01:15:02.980 --> 01:15:04.480 align:middle line:84% And really, this is the first result 01:15:04.480 --> 01:15:09.280 align:middle line:84% that involves a lot of math that we've covered up to this point 01:15:09.280 --> 01:15:20.280 align:middle line:84% to prove a very non-trivial and deep theorem that 01:15:20.280 --> 01:15:25.110 align:middle line:84% differentiability is a bit of a miracle. 01:15:25.110 --> 01:15:26.460 align:middle line:90% It's a miracle when it happens. 01:15:26.460 --> 01:15:28.970 align:middle line:90% OK, we'll stop there. . 01:15:28.970 --> 01:15:35.438 align:middle line:90%