1 00:00:01,050 --> 00:00:04,090 Now we'll look at something called Taylor Series Expansions. 2 00:00:09,100 --> 00:00:12,670 So before I assumed that my function f was differentiatiable 3 00:00:12,670 --> 00:00:15,260 n times, so if I wanted to make order and 4 00:00:15,260 --> 00:00:19,180 Taylor polynomial so for a linear approximation I just needed 5 00:00:19,180 --> 00:00:22,780 to assume that it was differentialable at least one time. 6 00:00:24,012 --> 00:00:28,852 most of the functions were going to encounter are differentiable 7 00:00:28,852 --> 00:00:34,572 infinitely many times, and this means there so, the key thing to notice 8 00:00:34,572 --> 00:00:39,304 here is the function zero has a derivative that's also zero. 9 00:00:39,304 --> 00:00:41,449 So any polynomial is derivative, 10 00:00:41,449 --> 00:00:44,042 defferentiable an infinite number of times. 11 00:00:44,042 --> 00:00:48,688 So if it's say x squared, I take the derivative, I get 2x. 12 00:00:48,688 --> 00:00:52,427 I take the derivative, I get two. I take the derivative, I get zero. 13 00:00:52,427 --> 00:00:54,077 And then I can just continually take 14 00:00:54,077 --> 00:00:56,340 derivatives and I always have zero after that. 15 00:00:58,798 --> 00:01:00,870 so it's, it's difficult. 16 00:01:00,870 --> 00:01:04,740 I actually was trying to think up a function that I could put on a quiz. 17 00:01:04,740 --> 00:01:08,240 Or something that wouldn't, that would look like a normal 18 00:01:08,240 --> 00:01:12,960 function, and would not be differentiable after a certain point. 19 00:01:12,960 --> 00:01:15,379 And I couldn't really come up with anything that 20 00:01:15,379 --> 00:01:18,570 was, was pretty easy so there They do exist and 21 00:01:18,570 --> 00:01:20,550 they do show up when you start making, kind 22 00:01:20,550 --> 00:01:23,905 of, more complicated types of functions, for instance, these things 23 00:01:23,905 --> 00:01:27,140 that are defined as limits of other functions. 24 00:01:27,140 --> 00:01:30,612 But, for something you can write down, generally, they're going 25 00:01:30,612 --> 00:01:32,830 to have as many derivatives as you as you want. 26 00:01:35,580 --> 00:01:38,804 So I'm going to make something called a Taylor series expansion. 27 00:01:38,804 --> 00:01:41,269 So I'm going to use a notation t of x to 28 00:01:41,269 --> 00:01:45,519 mean a Taylor serious expansion, and that's just going to 29 00:01:45,519 --> 00:01:48,239 be the limit as n goes to infinity of the 30 00:01:48,239 --> 00:01:52,309 Taylor polynomial evaluated at x as n goes to infinity. 31 00:01:54,800 --> 00:01:58,328 And so what I mean by that is just, the limit as 32 00:01:58,328 --> 00:02:03,750 this, limit of this sum as the number of terms gets, infinite. 33 00:02:07,410 --> 00:02:09,298 And it turns out that a Taylor series is 34 00:02:09,298 --> 00:02:12,630 a special case of something called a Power series, so. 35 00:02:12,630 --> 00:02:16,290 For a Taylor series, we know exactly what the, the coefficients are. 36 00:02:16,290 --> 00:02:19,490 They're determined by the function f that I'm trying to approximate. 37 00:02:20,680 --> 00:02:25,633 So, a Power series is just a coefficient a sub k 38 00:02:25,633 --> 00:02:26,268 [SOUND] 39 00:02:26,268 --> 00:02:31,810 times the quantity x minus a to the kth power. 40 00:02:31,810 --> 00:02:34,569 And if I look at my definition of the, 41 00:02:34,569 --> 00:02:38,129 of the Taylor polynomial, it just tells me that 42 00:02:38,129 --> 00:02:44,390 a sub k has to equal the kth derivative of f evaluated at a, divided by k factorial. 43 00:02:46,460 --> 00:02:49,125 And then another little bit of notation that 44 00:02:49,125 --> 00:02:52,810 you're going to start seeing pretty often is. 45 00:02:52,810 --> 00:02:57,300 The limit as n goes to infinity of this finite sum. 46 00:02:57,300 --> 00:03:00,108 So from k equals 0 up to n is going 47 00:03:00,108 --> 00:03:04,750 to be written the sum from k equals 0 to infinity. 48 00:03:04,750 --> 00:03:08,290 So it's probably what you would have guessed it would be anyways, but 49 00:03:08,290 --> 00:03:11,840 the real definition of what's going on here is this means the limit. 50 00:03:12,900 --> 00:03:18,480 As n goes to infinity of the finite sum from k equals 1 to n of this series. 51 00:03:22,160 --> 00:03:24,528 So the Power series coefficients, a sub k, 52 00:03:24,528 --> 00:03:27,536 when I'm talking about a Taylor series expansion, are 53 00:03:27,536 --> 00:03:30,288 just the kth derivative of the function I'm trying 54 00:03:30,288 --> 00:03:33,610 to approximate, evaluated at a, divided by k factorial. 55 00:03:35,810 --> 00:03:37,695 And the reason I want to show that the 56 00:03:37,695 --> 00:03:40,100 Taylor series of expansion is a special case 57 00:03:40,100 --> 00:03:42,310 of Power series, is there a more general 58 00:03:42,310 --> 00:03:45,300 results for convergence of power series, so there's 59 00:03:45,300 --> 00:03:48,290 convergence results for power series in general, and 60 00:03:48,290 --> 00:03:51,280 those, because of Taylor series, there's a special 61 00:03:51,280 --> 00:03:53,165 case of a power series, we can also 62 00:03:53,165 --> 00:03:56,260 use those convergence results for a Taylor series. 63 00:04:00,780 --> 00:04:03,180 So the important one, the most important one, 64 00:04:03,180 --> 00:04:06,660 I suppose, is something called the radius of convergence. 65 00:04:06,660 --> 00:04:11,217 And this is going to be a number capital R greater than 0, and 66 00:04:11,217 --> 00:04:15,309 I want to look at my Power series, and if this sums up to 67 00:04:15,309 --> 00:04:20,982 a finite number so when I write less than, strictly less than infinity, 68 00:04:20,982 --> 00:04:26,004 what I mean is that whatever is on the less than side, so whatever 69 00:04:26,004 --> 00:04:29,120 is over here is a finite real number. 70 00:04:29,120 --> 00:04:33,760 And then I'm going to define S of x just to be this 71 00:04:33,760 --> 00:04:39,096 infinite sum and so if this is, if x, so x here is in this 72 00:04:39,096 --> 00:04:44,200 interval a minus R to a plus R, where R is the radius of 73 00:04:44,200 --> 00:04:49,830 the convergence, so it's giving me and interval. 74 00:04:49,830 --> 00:04:51,015 If I take any point 75 00:04:51,015 --> 00:04:53,464 in that interval then when I make this 76 00:04:53,464 --> 00:04:56,629 infinite sum I'm going to get a finite number. 77 00:04:58,280 --> 00:05:00,986 So clearly, If we don't get a finite number then 78 00:05:00,986 --> 00:05:04,022 we really can't, you know it's not going to be useful 79 00:05:04,022 --> 00:05:07,718 for approximation because that's telling me as I add more terms, 80 00:05:07,718 --> 00:05:12,410 my approximation's just getting bigger, and bigger, and bigger, and bigger. 81 00:05:12,410 --> 00:05:16,442 So unless I know that f of x is equal to infinity I, I don't 82 00:05:16,442 --> 00:05:19,046 want to be trying to approximate with 83 00:05:19,046 --> 00:05:22,660 something that's diverging as I add more terms. 84 00:05:22,660 --> 00:05:25,710 So, the, the whole idea is I want to be able to add more terms 85 00:05:25,710 --> 00:05:27,296 to my Taylor polynomial to get a 86 00:05:27,296 --> 00:05:30,440 better approximation, not to get a worse approximation. 87 00:05:36,920 --> 00:05:42,037 Let's see so, it also turns out that this function 88 00:05:42,037 --> 00:05:46,760 S of x, we can take derivatives of these too. 89 00:05:46,760 --> 00:05:51,460 So, I can take the derivative of this when it's finite. 90 00:05:57,450 --> 00:05:59,770 And then S of x we're going to say is not defined. 91 00:05:59,770 --> 00:06:02,350 So, if I'm outside of the radius of 92 00:06:02,350 --> 00:06:06,160 convergence, then I don't know what happens to S. 93 00:06:06,160 --> 00:06:07,600 It, it goes to a infinity. 94 00:06:07,600 --> 00:06:09,264 So, I'm not really going to make, be 95 00:06:09,264 --> 00:06:11,920 able to make any meaningful statement about its property. 96 00:06:11,920 --> 00:06:13,705 It's like I certainly couldn't define a 97 00:06:13,705 --> 00:06:16,910 derivative For something that's equal to infinity everywhere. 98 00:06:21,570 --> 00:06:25,560 And so this is the convergence property that comes from Power series 99 00:06:25,560 --> 00:06:29,260 that I want to also be able to use for Taylor series. 100 00:06:29,260 --> 00:06:36,830 So it turns out, I need, it's, it's going to depend on These coefficients a sub k. 101 00:06:37,980 --> 00:06:41,030 But if the limit as k goes to infinity. 102 00:06:42,700 --> 00:06:47,208 So I just need to look at these coefficients a sub k absolute 103 00:06:47,208 --> 00:06:49,950 value, to the one over kth power. 104 00:06:49,950 --> 00:06:57,090 So this is the same thing as taking the kth root Of the absolute value of a sub k. 105 00:06:57,090 --> 00:06:59,610 So if it was one half, that would be the square root. 106 00:06:59,610 --> 00:07:02,260 If it was one third, that would be the cube root. 107 00:07:02,260 --> 00:07:04,970 If it was one over four, that would be the fourth root. 108 00:07:04,970 --> 00:07:06,440 And so on. 109 00:07:06,440 --> 00:07:13,006 So if you can make sense of that limit, if that's a finite value, then I 110 00:07:13,006 --> 00:07:18,010 can define the radius of convergence to be one divided by that limit. 111 00:07:19,260 --> 00:07:24,125 And now for our Taylor series expansion, 112 00:07:24,125 --> 00:07:28,610 I know what these a sub k's have to be. 113 00:07:31,990 --> 00:07:37,723 And you can use a property so the sort of annoying thing is that a sub k has this 114 00:07:37,723 --> 00:07:41,545 k factorial in it and sometimes that's going to be 115 00:07:41,545 --> 00:07:45,420 a pain to try and deal with in the limit. 116 00:07:45,420 --> 00:07:49,389 So luckily somebody proved this additional theorem where 117 00:07:49,389 --> 00:07:52,100 instead of k factorial, I just have k. 118 00:07:53,180 --> 00:07:57,390 And so if this limit so it's k divided by. 119 00:07:57,390 --> 00:08:00,276 This is the kth derivative of f evaluated at a 120 00:08:00,276 --> 00:08:03,318 and then I take the absolute value of that to the 121 00:08:03,318 --> 00:08:05,970 one over kth power so it looks kind of like 122 00:08:05,970 --> 00:08:10,730 this, except I don't have this k factorial in there anymore. 123 00:08:10,730 --> 00:08:13,260 So, somehow the k factorial. 124 00:08:13,260 --> 00:08:16,588 To the 1 over k has become this k, up in 125 00:08:16,588 --> 00:08:22,660 the, just this k to with no factorial up in the numerator. 126 00:08:22,660 --> 00:08:25,108 So if this limit exists, then the radius of 127 00:08:25,108 --> 00:08:27,964 convergence, so this is for a Taylor series now, 128 00:08:27,964 --> 00:08:31,364 because I'm defining these coefficients in terms of the, 129 00:08:31,364 --> 00:08:35,170 the derivatives that I'm using to build my Taylor polynomials. 130 00:08:37,250 --> 00:08:41,710 The radius of convergence is 1 divided by e times the value of that limit. 131 00:08:48,930 --> 00:08:54,455 And so far, what we've been able to show then is that T of x is a finite number so 132 00:08:54,455 --> 00:09:00,070 that, that's a good first step but it's still not doing what we want to do. 133 00:09:00,070 --> 00:09:05,650 What we really want to be able to do is to use T of x, or, or a polynomial 134 00:09:05,650 --> 00:09:10,869 that we would build on the way to T of x to approximate our function f. 135 00:09:12,550 --> 00:09:13,942 So if T of x is at least a finite 136 00:09:13,942 --> 00:09:16,494 number, you know, we're doing a lot better than if 137 00:09:16,494 --> 00:09:20,010 it was infinite, there's still the possibility that it would work. 138 00:09:22,660 --> 00:09:27,600 But what we'd really like to know is if or where T of x is equal to 139 00:09:27,600 --> 00:09:33,130 f of x, because if T of x, so if this infinite sum is converging 140 00:09:33,130 --> 00:09:38,630 To f of x at a point x, then the idea would be that I can use some 141 00:09:38,630 --> 00:09:44,160 of those terms to make an approximation of f at that point x. 142 00:09:47,210 --> 00:09:48,810 So we have a, another theorem. 143 00:09:49,970 --> 00:09:53,020 And this is, this one's a little bit trickier. 144 00:09:53,020 --> 00:09:55,708 So we need to know what the radius of convergence is 145 00:09:55,708 --> 00:09:59,320 and we can get that from the formula on the previous slide. 146 00:09:59,320 --> 00:10:03,900 So you can use either the formula for the Power series in general. 147 00:10:03,900 --> 00:10:08,360 Or you can use the special case for the Taylor's expansion. 148 00:10:09,550 --> 00:10:12,559 And you, sometimes it's actually easier to use the general 149 00:10:12,559 --> 00:10:16,158 formula even when you're working with the Taylor expansion just to, it 150 00:10:16,158 --> 00:10:19,495 depends on whether when you want this k factorial in there or not. 151 00:10:19,495 --> 00:10:20,005 [COUGH] 152 00:10:20,005 --> 00:10:25,105 Then once you know what the radius of convergence is, so you 153 00:10:25,105 --> 00:10:30,220 know for what values of x is T of x going to at least be finite? 154 00:10:33,190 --> 00:10:34,600 We can look at this limit. 155 00:10:34,600 --> 00:10:38,031 So I need to take R to the k, so this is some number that's 156 00:10:38,031 --> 00:10:43,110 going to be inside the, it's going to be smaller than the radius of convergence. 157 00:10:44,870 --> 00:10:48,300 So I take R to the kth power, divided by k factorial. 158 00:10:50,070 --> 00:10:53,220 And then I have to look at the kth derivative of f. 159 00:10:55,890 --> 00:11:02,830 Evaluated at a point z, that's in between a minus r and a plus r. 160 00:11:02,830 --> 00:11:07,720 So I'm making an interval around a of width lower case r. 161 00:11:07,720 --> 00:11:12,060 And I need to find the maximum value of the kth derivative on that interval. 162 00:11:14,510 --> 00:11:18,590 So, once you're done doing this part, that should just be a number. 163 00:11:20,050 --> 00:11:22,970 So it's the maximum value of this derivative on 164 00:11:22,970 --> 00:11:26,370 this interval so, but it might involve this number r. 165 00:11:29,470 --> 00:11:35,110 And then I'm going to look at that, this limit as k goes to infinity. 166 00:11:35,110 --> 00:11:40,730 If that limit is equal to 0, then T of x is equal to f of x. 167 00:11:43,210 --> 00:11:45,859 And so what that's going to tell me is that every x. 168 00:11:46,950 --> 00:11:54,510 That's within r, of my point a, has the property that the Taylor series expansion. 169 00:11:54,510 --> 00:11:59,740 So if I just make this infinitely long Taylor polynomial. 170 00:11:59,740 --> 00:12:03,529 That's going to be exactly equal to the function that I'm trying to approximate. 171 00:12:07,070 --> 00:12:10,250 So let me work through a quick example of this. 172 00:12:10,250 --> 00:12:13,862 So the Taylor series expansion for the function f of 173 00:12:13,862 --> 00:12:16,634 x equals log of 1 plus x around the point 174 00:12:16,634 --> 00:12:18,986 a equals 0 is going to be, so I can 175 00:12:18,986 --> 00:12:22,900 either define this just with k equals 0 to infinity. 176 00:12:22,900 --> 00:12:27,820 Or I can think about making Taylor polynomials of order n and then taking 177 00:12:27,820 --> 00:12:32,248 the limit as n goes to infinity, either one of these they mean the 178 00:12:32,248 --> 00:12:36,676 same thing, they're just two different ways of communicating 179 00:12:36,676 --> 00:12:40,572 this same idea. But the one on the right hand side is sort 180 00:12:40,572 --> 00:12:47,410 of the more precise of the two. So I'm just going to take x minus a. 181 00:12:47,410 --> 00:12:49,970 But here, I've said I want to make the expansion around the point 0. 182 00:12:49,970 --> 00:12:57,425 So that's going to be x minus 0 to the kth power, divided by k factorial, times f k, 183 00:12:57,425 --> 00:13:03,097 so the kth derivative of my function f evaluated at the point 0. 184 00:13:03,097 --> 00:13:08,281 So it turns out, if I, if I have log of 1 plus x, the first 185 00:13:08,281 --> 00:13:13,465 derivative of that is just going to be 1 divided by 1 plus 186 00:13:13,465 --> 00:13:18,250 x, so I can write that as 1 plus x to the negative 1. 187 00:13:18,250 --> 00:13:22,468 And if I were to take another derivative of that, I'd have negative 188 00:13:22,468 --> 00:13:25,390 1 times 1 plus x to the negative 2. 189 00:13:25,390 --> 00:13:27,658 If I took another derivative of that I would 190 00:13:27,658 --> 00:13:31,190 have negative three times whatever coefficient I have out here. 191 00:13:31,190 --> 00:13:35,040 And because every time I take a derivative I'm getting a negative sign. 192 00:13:36,558 --> 00:13:39,880 the sign of each one of these derivatives is going to be alternating. 193 00:13:39,880 --> 00:13:45,200 So the way I'm going to capture that is just put a minus one To a power. 194 00:13:45,200 --> 00:13:47,510 And then I just choose this power so that 195 00:13:47,510 --> 00:13:51,860 it's even when I want a positive term and odd when I want a negative term. 196 00:13:54,380 --> 00:13:58,745 So the first derivative, when I have one prime here, 197 00:13:58,745 --> 00:14:03,260 those are the ones that I want to be positive. 198 00:14:03,260 --> 00:14:07,036 So when I have an odd derivative, so first derivative, third derivative, 199 00:14:07,036 --> 00:14:10,970 fifth derivative and so on, I want to have a positive expression. 200 00:14:10,970 --> 00:14:14,256 So I'm just going to put k plus 1 as my power on the negative 1, 201 00:14:14,256 --> 00:14:16,240 and that's going to take care of the flip 202 00:14:16,240 --> 00:14:18,640 flopping sign as the, as the sum goes along. 203 00:14:20,670 --> 00:14:24,420 And then, every time I take a derivative, I'm going to get 1 204 00:14:24,420 --> 00:14:27,270 and then a two and then a three and then a four. 205 00:14:28,910 --> 00:14:32,026 But, because the first derivative I got from 206 00:14:32,026 --> 00:14:35,440 log I'm actually one behind in my count. 207 00:14:35,440 --> 00:14:39,510 So I end up with k minus 1 factorial as my coeefiecient here. 208 00:14:40,560 --> 00:14:43,650 And then I have 1 plus x to the minus k. 209 00:14:43,650 --> 00:14:45,930 So when I take the first derivative, I have a minus 1. 210 00:14:45,930 --> 00:14:49,260 The second derivitive would have a minus 2 and so on. 211 00:14:51,550 --> 00:14:53,678 And so this is what I want to plug in 212 00:14:53,678 --> 00:14:57,110 to this term right here, and I'm going to evaluate. 213 00:14:57,110 --> 00:15:02,297 So this is the kth derivitive at x I want to evaluate the kth derivative 214 00:15:02,297 --> 00:15:06,210 at the point 0, so when I put in 0 in this part of the 215 00:15:06,210 --> 00:15:11,397 expression here, this is the only part that actually has an x in it, 216 00:15:11,397 --> 00:15:16,766 I'm going to get 1 plus 0 to the minus kth power, and that's just going to 217 00:15:16,766 --> 00:15:22,750 be 1 to the minus kth power. And 1 to any power is equal to 1. 218 00:15:22,750 --> 00:15:25,742 So when I evaluate this at 0, this final 219 00:15:25,742 --> 00:15:29,500 part of the expression is just going to drop away. 220 00:15:34,880 --> 00:15:37,370 So I end up with just this alternating sign. 221 00:15:38,440 --> 00:15:43,948 And then a k minus 1 factorial, for my fk of 0 derivatives, 222 00:15:43,948 --> 00:15:49,564 and now, when I plug this back into my Taylor expansion, what 223 00:15:49,564 --> 00:15:54,964 I end up getting is x minus 0 to the k, so that just going to 224 00:15:54,964 --> 00:16:00,148 be an x to the k in the numerator I have a k factorial and then 225 00:16:00,148 --> 00:16:05,548 this expression fk of 0 that we just worked out over here and 226 00:16:05,548 --> 00:16:10,538 this has a k minus one factorial in it. So the k minus one 227 00:16:10,538 --> 00:16:15,480 factorial time, divided by k factorial is just going to give me. 228 00:16:16,970 --> 00:16:18,170 Why can't I select you? 229 00:16:21,270 --> 00:16:23,520 Okay I'm not going to try that anymore. 230 00:16:23,520 --> 00:16:27,640 It's just going to give me this factor of k in the denominator. 231 00:16:27,640 --> 00:16:31,780 So that's k minus one factorial divided by k factorial. 232 00:16:31,780 --> 00:16:35,020 Everything cancels itself out except for the k. 233 00:16:35,020 --> 00:16:38,332 So I divide by k In the numerator, I have x to 234 00:16:38,332 --> 00:16:42,430 the k which came from this x minus 0 to the kth power. 235 00:16:44,350 --> 00:16:46,450 And then I have this flip flopping 236 00:16:46,450 --> 00:16:47,210 sine term. 237 00:16:48,540 --> 00:16:51,990 So it turns out that my Taylor series expansion is going to 238 00:16:51,990 --> 00:16:55,670 be x minus x squared over two plus x cubed over three. 239 00:16:55,670 --> 00:16:58,100 Minus x to the 4th over 4 and so on. 240 00:17:00,950 --> 00:17:03,960 And so now I want to find the radius of convergence of that. 241 00:17:07,430 --> 00:17:10,451 And so first I'll try, because this is a Taylor series expansion, 242 00:17:10,451 --> 00:17:13,540 I'll try using the formula we had for the Taylor series expansion. 243 00:17:13,540 --> 00:17:14,390 So I had the. 244 00:17:16,310 --> 00:17:18,500 Fk evaluated at the point a. 245 00:17:18,500 --> 00:17:22,400 So my point a was 0, so I have an expression for that. 246 00:17:22,400 --> 00:17:26,313 But now I have this k minus 1 factorial to the 1 over 247 00:17:26,313 --> 00:17:31,100 kth power in the denominator, and I have a k in the numerator. 248 00:17:32,420 --> 00:17:35,129 And so I don't really know how to make sense out of that limit. 249 00:17:36,470 --> 00:17:37,830 So, I said, hmh. 250 00:17:37,830 --> 00:17:40,000 But I have another possibility. 251 00:17:40,000 --> 00:17:42,925 Because this is a special case of a Power series, I can 252 00:17:42,925 --> 00:17:46,060 go back and try the formula for the power series as well. 253 00:17:47,970 --> 00:17:51,374 And so that said that if the limit of these coefficients 254 00:17:51,374 --> 00:17:56,820 existed, then the radius of convergence would be one over this limit. 255 00:17:56,820 --> 00:17:57,452 So let's 256 00:17:57,452 --> 00:18:00,060 see what happens if I just plug this in. 257 00:18:00,060 --> 00:18:02,710 So I have minus one to the k plus 258 00:18:02,710 --> 00:18:07,100 one divided by k as my sequence of coefficients. 259 00:18:07,100 --> 00:18:12,125 And when I take the absolute value of that, the whole numerator's just going to 260 00:18:12,125 --> 00:18:17,156 go away. So what I end up with is the limit As k 261 00:18:17,156 --> 00:18:22,796 goes to infinity of 1 over k to the one over kth power, and so notice that 262 00:18:22,796 --> 00:18:28,070 the power is exactly equal to the thing that I am taking the power. 263 00:18:28,070 --> 00:18:33,452 So I'm going to say this is equal to the limit as u of u to the 264 00:18:33,452 --> 00:18:38,717 uth power, and if one over k if k is going to infinity, 265 00:18:38,717 --> 00:18:43,820 then 1 over k is going to 0. So I can just say, this 266 00:18:43,820 --> 00:18:48,990 is the limit as u goes down to 0 of u to the uth power. 267 00:18:50,330 --> 00:18:53,159 And if you go back to our lectures on limits, this 268 00:18:53,159 --> 00:18:56,059 was one of the things that we said was equal to 1. 269 00:18:57,290 --> 00:19:02,590 So if this limit is equal to 1, the radius of convergence is 1 divided by that. 270 00:19:05,050 --> 00:19:08,920 So that tells me that the radius of convergence is equal to 1. 271 00:19:08,920 --> 00:19:14,128 And then the last thing we have to do is 272 00:19:14,128 --> 00:19:19,320 see if Or where t of x is equal to the log of one plus x. 273 00:19:19,320 --> 00:19:21,030 So this is my function f of x. 274 00:19:25,090 --> 00:19:28,030 So I know that the radius of convergence is equal to one. 275 00:19:28,030 --> 00:19:31,960 So I'm looking for some number r that's between 0. 276 00:19:31,960 --> 00:19:36,346 So some number little r that's in between 0 and one 277 00:19:36,346 --> 00:19:41,038 and so again, I'm just going to use the theorem that I had 278 00:19:41,038 --> 00:19:44,608 on a few slides ago, and that said that if I 279 00:19:44,608 --> 00:19:50,524 looked at this derivative, but now instead of being able to evaluate 280 00:19:50,524 --> 00:19:55,522 z it at the point 0, I have to evaluate it at the point, and 281 00:19:55,522 --> 00:20:00,590 find out where it's maximum on the interval minus r to r. 282 00:20:03,480 --> 00:20:07,340 And what I end up having here is minus 1 to the n plus 1. 283 00:20:07,340 --> 00:20:11,790 So my absolute value is going to take care of this flip-flopping sign for me. 284 00:20:11,790 --> 00:20:17,829 I have an n minus 1 factorial in the numerator and then a 1 plus z to the nth 285 00:20:17,829 --> 00:20:23,769 power in the denominator and I'm trying to make this as big as I possibly 286 00:20:23,769 --> 00:20:29,426 can. So I want to make the denominator 287 00:20:29,426 --> 00:20:33,500 as small as I possibly can. 288 00:20:33,500 --> 00:20:37,170 So the smallest I can make this is if I take one minus r 289 00:20:41,690 --> 00:20:47,748 And so that gives me is that 290 00:20:47,748 --> 00:20:53,161 right? I must have 291 00:20:53,161 --> 00:20:59,027 had a missing one 292 00:20:59,027 --> 00:21:04,490 line in here. 293 00:21:04,490 --> 00:21:07,056 For what I should do. but essentially 294 00:21:07,056 --> 00:21:10,752 because r, I can't actually make this denominator 0, 295 00:21:10,752 --> 00:21:13,530 because r has to be strictly less than 1. 296 00:21:13,530 --> 00:21:17,138 And so what I end up with is, I have something 297 00:21:17,138 --> 00:21:21,180 to the n down here, and this r to the n here. 298 00:21:21,180 --> 00:21:24,491 And that actually gives, I can think of that as then 299 00:21:24,491 --> 00:21:27,478 just r times 1, or, r divided by 1 minus r. 300 00:21:27,478 --> 00:21:31,824 To the nth power but I'm taking the limit as n goes to infinity, 301 00:21:31,824 --> 00:21:35,690 and so r divided by 1 minus r, that's just a constant. 302 00:21:37,340 --> 00:21:41,136 And I have, so I have some constant to the nth power divided by n 303 00:21:41,136 --> 00:21:46,820 factorial, and that limit is again one of the results we had from the first week. 304 00:21:46,820 --> 00:21:51,053 That any constant to a power divided by n factorial, if I take 305 00:21:51,053 --> 00:21:55,120 the limit of that as n goes to infinity, I'm going to get 0. 306 00:21:58,320 --> 00:22:01,682 And so that tells me, I can make r as close as I want 307 00:22:01,682 --> 00:22:06,540 to capital R but I can not actually make it equal to capital R. 308 00:22:06,540 --> 00:22:10,875 So it's just turned this sort of open interval, this radius 309 00:22:10,875 --> 00:22:15,720 of convergence of being 1 it goes, it's a closed interval its 310 00:22:15,720 --> 00:22:18,865 now an open interval so our little r can be any 311 00:22:18,865 --> 00:22:23,370 number if it's smaller than 1 but it can't actually be 1. 312 00:22:23,370 --> 00:22:26,770 And so that tells me that T of x is equal to f 313 00:22:26,770 --> 00:22:31,173 of x as long as the absolute value of x is less than 1.