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These are the user uploaded subtitles that are being translated: 1 00:00:00,420 --> 00:00:05,010 Today we're going to be talking about how to estimate the error or remainder of the series using the 2 00:00:05,010 --> 00:00:06,780 first 10 terms of the series. 3 00:00:06,780 --> 00:00:12,050 And in this particular problem we've been given the infinite sum from any equals 1 to infinity of 1 4 00:00:12,060 --> 00:00:16,570 divided by the square root of the quantity and to the fourth plus one. 5 00:00:16,620 --> 00:00:20,490 One thing that I quickly want to point out error versus remainder of the series. 6 00:00:20,490 --> 00:00:21,580 It's really the same thing. 7 00:00:21,600 --> 00:00:23,780 Some books and professors will call it the air. 8 00:00:23,790 --> 00:00:25,380 Others will call it the remainder. 9 00:00:25,380 --> 00:00:30,300 But it means the same thing and you can be asked either way whenever you're asked to find the air or 10 00:00:30,300 --> 00:00:32,480 remainder of the series you want to do. 11 00:00:32,580 --> 00:00:33,450 Three things really. 12 00:00:33,450 --> 00:00:36,540 First of all you want to identify how many terms you're going to use. 13 00:00:36,540 --> 00:00:40,620 And in this particular problem we've been given that figure exactly we've been told that we're going 14 00:00:40,620 --> 00:00:43,670 to use the first 10 terms of the series. 15 00:00:43,770 --> 00:00:49,770 What you want to do with the first 10 terms is find the some of the first 10 terms and then estimate 16 00:00:49,860 --> 00:00:55,500 the remainder of the first 10 terms and the remainder or the air is going to be different depending 17 00:00:55,500 --> 00:01:00,480 on how many terms you use so it's really important to know how many terms of the series we're using 18 00:01:00,480 --> 00:01:02,210 and in this case we're using 10. 19 00:01:02,220 --> 00:01:07,620 So the first thing we'll do is find the some of the first 10 terms then we'll use that to find error. 20 00:01:07,620 --> 00:01:11,670 So the first thing we're going to do is find the some of the first 10 terms and then after that we'll 21 00:01:11,670 --> 00:01:12,740 find air. 22 00:01:12,990 --> 00:01:19,140 So finding some is really easy the first thing we want to do is just identify that this value here one 23 00:01:19,140 --> 00:01:25,740 divided by the square root of end of the fourth plus one is equal to a sub and we'll just call a sub 24 00:01:25,740 --> 00:01:28,980 in our series that'll represent our series. 25 00:01:28,980 --> 00:01:35,040 So when we want to find the sum to the first 10 terms we'll pretend that the sum of the series is Subhan 26 00:01:35,310 --> 00:01:40,500 and therefore finding the sum of the first 10 terms is going to give us as sub 10. 27 00:01:40,620 --> 00:01:42,360 So we'll call it as subtend. 28 00:01:42,360 --> 00:01:47,250 And really this is just going to be plugging in the numbers 1 2 three four five six seven eight nine 29 00:01:47,280 --> 00:01:50,970 10 for N and adding all those values together. 30 00:01:50,970 --> 00:01:56,450 So we'll plug in one for and one to the fourth is just one plus one is two. 31 00:01:56,460 --> 00:02:02,400 So the first term is 1 over square root 2 then only plug into. 32 00:02:02,550 --> 00:02:06,650 We get 2 to the fourth which is 16 plus 1 which is 17. 33 00:02:06,750 --> 00:02:10,100 So that's 1 over square root 17. 34 00:02:10,170 --> 00:02:18,110 If we plug in 3 3 to the fourth is 81 plus one is 82 so we get 1 over the square root of eighty two. 35 00:02:18,300 --> 00:02:24,780 And we keep going like that all the way until 10 and when we get to 10 we can just say dot dot dot here 36 00:02:25,050 --> 00:02:26,820 when we get to 10. 37 00:02:26,850 --> 00:02:33,220 We'll have the square root 1 over the square root of the fourth is ten thousand plus one is 10000 ones 38 00:02:33,220 --> 00:02:37,290 will say 10000 and one we take the square root of that. 39 00:02:37,470 --> 00:02:41,670 So you need to find all the ones in the middle but then you add them together and you can do this on 40 00:02:41,670 --> 00:02:47,940 your calculator and it's perfectly acceptable to provide a decimal representation of the sum. 41 00:02:47,940 --> 00:02:53,460 So in our case the decimal value or the value of the sum of the first 10 terms rounded to four decimal 42 00:02:53,460 --> 00:02:58,580 places is 1 point 2 4 8 6. 43 00:02:58,590 --> 00:03:04,260 Now when it comes to estimating the error or remainder what we want to do is use the comparison test 44 00:03:04,260 --> 00:03:12,780 to verify that a 7 converges and then use the remainder of the comparison Series B 7 to more specifically 45 00:03:12,780 --> 00:03:16,490 define the remainder of our given series a 7. 46 00:03:16,530 --> 00:03:22,920 So when we're looking for a comparison series for our series we want to do is pick a series that is 47 00:03:23,010 --> 00:03:26,580 similar to ours for which we can easily determine convergence. 48 00:03:26,580 --> 00:03:31,290 And that takes a little bit of practice but one thing you can look for are values in your series that 49 00:03:31,290 --> 00:03:35,480 are kind of trivial to the sum and what I mean by that is for example. 50 00:03:35,590 --> 00:03:41,820 To the fourth is going to have a much greater effect on this series than this value here is this plus 51 00:03:41,820 --> 00:03:42,850 one value. 52 00:03:42,870 --> 00:03:47,940 For example if I plug in three for n I get three to the fourth which is eighty one. 53 00:03:48,090 --> 00:03:53,420 So this 81 value is a much more significant value than just this simple plus one right. 54 00:03:53,430 --> 00:03:59,700 I can have 81 the plus one changes it to 82 but 81 does most of the work here it's much greater than 55 00:03:59,700 --> 00:04:01,450 this value here of 1. 56 00:04:01,460 --> 00:04:06,820 So end of fourth is going to have a much greater effect on this whole series than just this plus one. 57 00:04:06,820 --> 00:04:13,760 So one thing we considered doing is getting rid of this plus one if we do let's talk about here B. 58 00:04:13,770 --> 00:04:16,180 Sub an R comparison series. 59 00:04:16,320 --> 00:04:22,440 If we do that what we're left with is 1 over the square root and to the Fourth and this is another clue 60 00:04:22,440 --> 00:04:23,280 for us. 61 00:04:23,370 --> 00:04:28,410 This is going to turn out to be a very convenient comparison series when we take the square root of 62 00:04:28,420 --> 00:04:30,510 end of the fourth we get and squared. 63 00:04:30,520 --> 00:04:33,060 We get one over and squared. 64 00:04:33,090 --> 00:04:39,360 Now what we know about this series is that it's a P series that converges because when you use the pre-series 65 00:04:39,360 --> 00:04:46,530 test you know that if this value here which is P this exponent when P is greater than 1 this series 66 00:04:46,530 --> 00:04:47,770 will converge. 67 00:04:47,790 --> 00:04:54,510 So this is a piece with p equals to P is greater than 1 and we know therefore that piece of and converges. 68 00:04:54,600 --> 00:04:57,630 So B-7 is a convergence sequence. 69 00:04:57,630 --> 00:05:04,690 Now we can say is that if Subban is less than B-7 then we know that A7 also must converge and you can 70 00:05:04,690 --> 00:05:10,680 think about it like this if you had a graph like this for example and would just do this really quickly 71 00:05:10,690 --> 00:05:16,480 these graphs don't represent a 7 or B 7 but if you just think about a convergence series and let's say 72 00:05:16,480 --> 00:05:24,280 that this series is be Subhan for example and it's converging down here toward the x axis if a submarine 73 00:05:24,640 --> 00:05:28,770 is always less than the series right it's always underneath it. 74 00:05:28,770 --> 00:05:32,370 If it's always underneath B-7 then it also must converge. 75 00:05:32,380 --> 00:05:38,670 It can't divert because then if it diverged at some point it would become greater than be Subhan. 76 00:05:38,680 --> 00:05:46,240 So if we can show that A-7 is always less than B-7 then we can prove that A7 also converges. 77 00:05:46,240 --> 00:05:50,980 So let's look at the values of A7 versus B-7. 78 00:05:51,040 --> 00:05:58,630 If we plug in values here for B 7 and we start with an equals 1 we'll say that when we plug in 1 we 79 00:05:58,630 --> 00:06:02,700 get 1 squared which is just 1 so the first term is 1. 80 00:06:02,830 --> 00:06:06,910 When we plug in 2 we get 2 square which is for us we get 1 fourth. 81 00:06:07,330 --> 00:06:14,180 When we plug in three we get 1 ninth and we kept going you'd see one 16 tier 1 25th etc.. 82 00:06:14,230 --> 00:06:20,230 What we need to do now is compare each term in our series so this was the turn that we got when we plugged 83 00:06:20,230 --> 00:06:22,690 in 1 to a 7. 84 00:06:22,690 --> 00:06:26,000 This is the term that we got when we plugged in 1 to be 7. 85 00:06:26,020 --> 00:06:28,960 This is the term that we got only plugged into to a 7. 86 00:06:29,050 --> 00:06:32,270 And this is the term that we got when we plugged into to be Subban. 87 00:06:32,320 --> 00:06:36,590 If we compare these if you plug these square roots into your calculator and you do one with squirted 88 00:06:36,600 --> 00:06:44,650 to 1 over the square to 17 etc. what we'll find is that the values in B-7 and the B-7 series will always 89 00:06:44,650 --> 00:06:47,020 be greater than the values in A-7. 90 00:06:47,020 --> 00:06:49,970 So one is greater than 1 over the square of two. 91 00:06:49,990 --> 00:06:55,960 One fourth is greater than 1 over the square to 17 when 9 is greater than 1 over the square of 82 et 92 00:06:55,990 --> 00:06:56,780 cetera. 93 00:06:56,890 --> 00:07:04,420 So we know that B-7 be Subhan is always greater than a Sabun. 94 00:07:04,420 --> 00:07:12,910 Therefore we know that a 7 converges since B 7 converges and is always greater than a 7. 95 00:07:12,910 --> 00:07:16,260 So by the comparison test we've proven that a 7 converges. 96 00:07:16,390 --> 00:07:22,060 Now we can use the remainder estimate for the integral test to get an estimate of the remainder for 97 00:07:22,150 --> 00:07:22,990 a 7. 98 00:07:23,140 --> 00:07:28,100 So we'll call the remainder for a sub n we'll call it our Subhan would be the remainder. 99 00:07:28,210 --> 00:07:30,370 In this case we're doing 10 terms right. 100 00:07:30,370 --> 00:07:31,990 The first 10 terms of the series. 101 00:07:32,140 --> 00:07:35,480 So we're going to call it our subtend by this remainder estimate. 102 00:07:35,470 --> 00:07:42,100 Theorem we know that our subtend is always going to be less than or equal to the integral from here 103 00:07:42,130 --> 00:07:50,140 10 because we're doing the first 10 terms to infinity of our comparison series B so so we'll say 1 over 104 00:07:50,140 --> 00:07:51,750 instead of using end squared. 105 00:07:51,760 --> 00:07:54,990 We'll just use x squared and then D x. 106 00:07:55,120 --> 00:08:00,520 Really what this theorem is telling us is that the remainder is going to be less than the value of this 107 00:08:00,520 --> 00:08:06,610 integral and what it's actually saying is that the remainder of the series B 7 is going to be less than 108 00:08:06,610 --> 00:08:07,420 this integral. 109 00:08:07,470 --> 00:08:13,720 However we've already proven that a Sabahan will always be less than be 7. 110 00:08:13,720 --> 00:08:19,410 So what we have for an inequality here is our subtend which is the remainder of the series. 111 00:08:19,440 --> 00:08:22,700 A-7 will always be less than or equal to. 112 00:08:22,870 --> 00:08:29,770 We'll call it T sub and or sub 10 which is going to be the remainder of the comparison series B 7 which 113 00:08:29,770 --> 00:08:37,630 is always going to be less than or equal to the integral of our comparison series B 7 from this value 114 00:08:37,630 --> 00:08:39,880 here 10 to infinity. 115 00:08:39,880 --> 00:08:46,330 So because we plugged in our comparison series B 7 here we know that our remainder for B Subban has 116 00:08:46,330 --> 00:08:49,580 to be less than or equal to that value that's t subtend. 117 00:08:49,780 --> 00:08:56,200 And because A7 is always underneath be subtend like we drew in that graph the remainder for Ace habban 118 00:08:56,440 --> 00:08:59,330 is going to have to be less than the remainder for B-7. 119 00:08:59,500 --> 00:09:04,450 So we can just go ahead and write this inequality here and then evaluate the integrals so we'll say 120 00:09:04,510 --> 00:09:08,340 our subtend is less than or equal to when we take the integral. 121 00:09:08,340 --> 00:09:15,010 Here we can consider this instead of 1 over x squared we'll call it X to the negative 2 we'll move this 122 00:09:15,100 --> 00:09:16,840 x squared to the numerator. 123 00:09:16,840 --> 00:09:23,020 Changing the exponent to a negative value then we just add 1 to our exponent negative two plus one is 124 00:09:23,020 --> 00:09:23,770 negative one. 125 00:09:23,770 --> 00:09:31,150 So we get X to the negative 1 and then we divide by our new exponent so we divide by negative 1 like 126 00:09:31,150 --> 00:09:36,220 this and we're going to be evaluating that on the interval 10 to infinity. 127 00:09:36,220 --> 00:09:38,770 Now of course we can simplify this function. 128 00:09:38,770 --> 00:09:41,900 Here we have X to the negative 1 in the numerator. 129 00:09:41,950 --> 00:09:46,490 We can move that to the denominator and change the exponent from a negative to a positive. 130 00:09:46,510 --> 00:09:48,780 So that just becomes X and the denominator. 131 00:09:48,970 --> 00:09:53,050 And we have this negative one here in the denominator which we can move to the numerator. 132 00:09:53,100 --> 00:10:00,070 What we really end up with here is just negative 1 over X that's our integrated value there and then 133 00:10:00,070 --> 00:10:02,590 we evaluate on this interval. 134 00:10:02,590 --> 00:10:08,350 So instead of evaluating for intend to infinity the technically correct way to write is we want to say 135 00:10:08,350 --> 00:10:17,520 that this is less than or equal to the limit as a goes to infinity of negative 1 over X evaluated and 136 00:10:17,530 --> 00:10:18,300 then the interval. 137 00:10:18,310 --> 00:10:19,340 Ten to a. 138 00:10:19,360 --> 00:10:21,790 So we've really just replaced the infinity with a. 139 00:10:21,790 --> 00:10:27,610 And we're saying we're going to evaluate the limit as a goes to infinity or approaches infinity. 140 00:10:27,640 --> 00:10:32,860 Of course here when we're taking this definite integral like this we want to plug in our upper limit 141 00:10:32,860 --> 00:10:34,260 here a first. 142 00:10:34,270 --> 00:10:41,320 So we're going to say the limit as a goes to infinity we'll plug in a and we'll get negative 1 overlay 143 00:10:41,770 --> 00:10:46,200 then we'll subtract whatever we get when we plug in our lower limit here 10. 144 00:10:46,210 --> 00:10:48,940 So we're going to get negative 1 over 10. 145 00:10:49,090 --> 00:10:57,190 So say negative 1 over 10 like this are negative signs here are going to cancel when we evaluate a at 146 00:10:57,190 --> 00:11:00,420 infinity and we get this infinite value here in the denominator. 147 00:11:00,430 --> 00:11:06,880 This whole fraction negative 1 over infinity is going to become 0 because any constant divided by an 148 00:11:06,940 --> 00:11:10,520 infinitely large value here is always going to tend toward zero. 149 00:11:10,720 --> 00:11:12,160 So that's going to go away. 150 00:11:12,310 --> 00:11:18,790 And what you see that we're left with is an estimate for the remainder are sub 10 as being less than 151 00:11:18,820 --> 00:11:21,190 or equal to one tenth. 152 00:11:21,670 --> 00:11:25,340 Or we can also write that zero point one either way. 153 00:11:25,510 --> 00:11:29,890 So that's our final answer for the remainder of the way that we want to write our complete final answer 154 00:11:29,890 --> 00:11:35,680 is we want to say that some will say a subtend is going to be approximately equal to because remember 155 00:11:35,680 --> 00:11:37,110 this was an approximation. 156 00:11:37,330 --> 00:11:50,890 1 point 2 4 8 6 with a remainder value with our sub 10 less than or equal to zero point 1 which we calculated 157 00:11:51,190 --> 00:11:57,340 using the comparison Series B 7 knowing that be Sabayon also had to be less than 0.1. 17782