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These are the user uploaded subtitles that are being translated: 1 00:00:00,660 --> 00:00:05,340 Today we're going to be talking about how to use the alternating series estimation theorem to approximate 2 00:00:05,340 --> 00:00:09,720 the sum of the series and to find the remainder of the approximation. 3 00:00:09,720 --> 00:00:14,520 And this particular problem we've been given the infinite sum from equals 1 to infinity of the quantity 4 00:00:14,520 --> 00:00:21,210 negative one raised to the end minus one power times and squared divided by 10 to the power. 5 00:00:21,420 --> 00:00:25,950 For reference here I've written the alternating series estimation theorem which we'll get to in a second. 6 00:00:25,950 --> 00:00:34,050 We really only need this to find the remainder term R7 for now to find the approximation of the series. 7 00:00:34,140 --> 00:00:39,750 What we need to do is list out the first several terms of this series so that we can find the sum correct 8 00:00:39,750 --> 00:00:41,000 to 4 decimal places. 9 00:00:41,010 --> 00:00:47,190 So if we plug n equals 1 into this series this exponent here in the negative 1 will become 1 minus 1 10 00:00:47,220 --> 00:00:53,880 which is zero negative 1 raise to the zero power is just one plugging in here for end squared we get 11 00:00:53,880 --> 00:00:57,810 1 squared or just 1 divided by 10 to the first power. 12 00:00:57,810 --> 00:01:02,130 So we just have one tenth so our value here is one tenth. 13 00:01:02,370 --> 00:01:08,670 When we plug in an equals two we'll get negative 1 race to the two minus one or negative 1 to the first 14 00:01:08,670 --> 00:01:15,960 power which will just give us negative one negative one at times 2 squared is negative 4 divided by 15 00:01:16,020 --> 00:01:22,460 10 squared so we get negative for over 100 and equals three. 16 00:01:22,500 --> 00:01:29,450 If we keep going here is going to give us positive 9 over 10 to the third or one thousand. 17 00:01:29,790 --> 00:01:35,160 And we could keep computing our terms but remember that the problem is asked us to find the sum of the 18 00:01:35,160 --> 00:01:37,500 series correct to 4 decimal places. 19 00:01:37,530 --> 00:01:41,200 So we're going to need to turn each of these terms into a decimal value. 20 00:01:41,250 --> 00:01:46,950 So let's look here at one tenth our decimal value is going to be zero point 1. 21 00:01:47,090 --> 00:01:55,770 Our decimal value for negative for over 100 will be negative zero point zero for our decimal value for 22 00:01:55,840 --> 00:01:56,390 9. 23 00:01:56,430 --> 00:02:00,480 Over 1000 will be zero point 0 0 9. 24 00:02:00,510 --> 00:02:07,350 And if we keep including the fourth fifth sixth and seventh terms what we get is negative zero point 25 00:02:07,570 --> 00:02:17,070 0 0 1 6 for the fourth term we get positive zero point zero zero zero to 5 for the fifth term we get 26 00:02:17,070 --> 00:02:30,420 negative zero point 0 0 0 0 3 6 for the sixth term and positive zero point 0 0 0 0 0 4 9 for the seventh 27 00:02:30,420 --> 00:02:31,110 term. 28 00:02:31,110 --> 00:02:34,700 Now remember we're looking to approximate correct to 4 decimal places. 29 00:02:34,740 --> 00:02:41,070 So if we basically take a series here of partial sums we start adding all of these decimals together 30 00:02:41,070 --> 00:02:41,610 here. 31 00:02:41,790 --> 00:02:49,980 If we take 0.1 and we subtract zero point zero for what we get here is zero point zero six. 32 00:02:50,040 --> 00:02:52,570 That's the scale of the first two terms. 33 00:02:52,680 --> 00:03:00,270 If we add to that point 0 0 9 some of the first three terms gives us zero point zero six nine. 34 00:03:00,630 --> 00:03:09,660 When we subtract from that point 0 0 1 6 we get zero point zero 6 7 4 which is the sum of the first 35 00:03:09,690 --> 00:03:10,770 four terms. 36 00:03:10,950 --> 00:03:19,560 And then when we add to that point 0 0 0 to 5 we get zero point zero six seven six five. 37 00:03:19,560 --> 00:03:25,320 Now we've already gotten 1 decimal place past the first four decimal places but we want to make sure 38 00:03:25,320 --> 00:03:31,560 that if we continue adding and subtracting terms in our series from this sum that we aren't affecting 39 00:03:31,590 --> 00:03:32,970 the fourth decimal place. 40 00:03:33,120 --> 00:03:38,090 So we go ahead and subtract point 0 0 0 0 3 6. 41 00:03:38,100 --> 00:03:40,590 And what we get is zero point zero. 42 00:03:40,590 --> 00:03:43,350 6 7 6 1 4. 43 00:03:43,370 --> 00:03:49,350 Now the reason that it's helpful to do that is because as you can see when we had this value here our 44 00:03:49,350 --> 00:03:55,350 five in this fifth decimal place would have rounded this fourth decimal place up to seven and we would 45 00:03:55,350 --> 00:03:59,420 have point 0 6 7 7 for our first four decimal places. 46 00:03:59,610 --> 00:04:05,550 But now that we've subtracted our sixth term here you can see that we're not going to be rounding up 47 00:04:05,790 --> 00:04:11,100 because the value in the fifth decimal place is a 1 leaving our first four decimal places as point 0 48 00:04:11,100 --> 00:04:12,540 6 7 6. 49 00:04:12,630 --> 00:04:20,910 If we add to this sum the seventh term in our series what we get is zero point zero 6 7 6 1 8 9. 50 00:04:21,060 --> 00:04:24,920 And what we can see is that that the decimal place is still holding is a 1. 51 00:04:24,930 --> 00:04:30,940 Even if we use the 8 to rounded up to a 2 it's not going to be rounding up our fourth decimal place. 52 00:04:31,080 --> 00:04:37,620 So we can conclude that the terms past this point are no longer going to be affecting our fourth decimal 53 00:04:37,620 --> 00:04:38,240 place. 54 00:04:38,250 --> 00:04:43,790 That's as far as we had to go to make sure that our fourth decimal place would stay intact at a 6. 55 00:04:43,800 --> 00:04:48,990 If you remember correctly this is the last term that affected our fourth decimal place. 56 00:04:48,990 --> 00:04:55,740 So what we want to say is that the sum of the series we'll call it as sub 6 because we use the first 57 00:04:55,770 --> 00:05:03,240 six terms to get the fourth decimal place accurately we'll say subsects is equal to approximately zero 58 00:05:03,240 --> 00:05:06,230 point zero six seven six. 59 00:05:06,240 --> 00:05:09,890 Now at this point we have our Some correct to four decimal places. 60 00:05:09,890 --> 00:05:17,070 Some of our series but we don't know how accurate this approximation is as an approximation of the exact 61 00:05:17,070 --> 00:05:22,080 sum of the series in order to find out how accurate this approximation is. 62 00:05:22,080 --> 00:05:25,640 We need to use the alternating series estimation theorem. 63 00:05:25,650 --> 00:05:31,500 What that tells us what the theorem tells us is that if we have a sum and we'll call that some S and 64 00:05:31,500 --> 00:05:38,070 that sum is the sum of an alternating series or we have negative 1 to the end minus 1 Power time some 65 00:05:38,220 --> 00:05:43,050 Series B 7 here if S is the sum of that alternating series. 66 00:05:43,200 --> 00:05:50,270 And if that term be sub and plus 1 is always less than or equal to the piece of term and the limit as 67 00:05:50,280 --> 00:05:57,330 end goes to infinity of the be seven series is equal to zero then our remainder term satisfies this 68 00:05:57,390 --> 00:05:58,290 inequality here. 69 00:05:58,290 --> 00:06:01,250 So it sounds a lot more complicated than it is. 70 00:06:01,260 --> 00:06:03,270 All we really need to do is this. 71 00:06:03,270 --> 00:06:08,450 Notice that this infinite sum here in the alternating series estimation theorem has this negative one 72 00:06:08,460 --> 00:06:13,540 raised to the end minus one power multiplied by some Series B 7. 73 00:06:13,710 --> 00:06:20,580 Well in our infinite sum we have this this value here negative one race to the minus one power which 74 00:06:20,580 --> 00:06:27,790 makes the series alternate but it's multiplied by and squared over 10 to the power. 75 00:06:27,960 --> 00:06:29,600 Right here it's multiplied by that. 76 00:06:29,760 --> 00:06:35,070 So we're going to carve that away from this alternating value right here and we're going to call that 77 00:06:35,070 --> 00:06:40,680 be saw then whatever we can carve away from this negative one race to the end minus one power is what 78 00:06:40,680 --> 00:06:46,500 we'll call B sevens we'll say this here is going to be b c then given that we have this SB So then we 79 00:06:46,500 --> 00:06:47,750 need to prove two things. 80 00:06:47,760 --> 00:06:52,540 We need to prove that B sub and plus 1 is always less than or equal to be sabahan. 81 00:06:52,770 --> 00:06:54,930 And we need to prove this limit here. 82 00:06:54,930 --> 00:06:57,380 So let's take it one piece at a time. 83 00:06:57,480 --> 00:06:59,050 We need to show that B. 84 00:06:59,100 --> 00:07:04,690 And plus one or in other words whatever we get when we plug and plus one into this be value here so 85 00:07:04,830 --> 00:07:05,610 it'll be. 86 00:07:05,610 --> 00:07:15,300 And plus 1 squared divided by 10 to the end plus one that that is always less than or equal to be Sabun 87 00:07:15,360 --> 00:07:19,250 which is just and squared over tend to the N. 88 00:07:19,290 --> 00:07:20,120 How do we show this. 89 00:07:20,130 --> 00:07:25,950 Well if we write out the first several terms of each of these series we'll be able to compare the two 90 00:07:26,250 --> 00:07:32,310 and see if the terms in this series are always less than the corresponding terms in this series. 91 00:07:32,310 --> 00:07:38,910 So if we plug in N equals 1 to our B sub and plus 1 series we'll get one plus one which is 2 squared 92 00:07:38,910 --> 00:07:42,230 which is four divided by 10. 93 00:07:42,230 --> 00:07:48,360 To the one plus one or 10 squared and we'll get 10 square there if we plug in N equals 2 We'll get two 94 00:07:48,360 --> 00:07:56,100 plus one is three squared is 9 divided by 10 to the two plus one or 10 cubed. 95 00:07:56,160 --> 00:08:02,820 And if we plug in an equals three we get three plus one is four squared is 16 divided by 10 to the three 96 00:08:02,820 --> 00:08:05,970 plus one or 10 to the fourth power. 97 00:08:05,970 --> 00:08:11,550 We need to make sure that each of these is either less than or equal to what we get when we plug in 98 00:08:11,610 --> 00:08:14,480 1 2 and 3 to this second series here. 99 00:08:14,520 --> 00:08:20,280 So we want to say less than or equal to we plug one into this series here we get 1 squared in the numerator 100 00:08:20,580 --> 00:08:23,260 we get tens of the first power in the denominator. 101 00:08:23,550 --> 00:08:31,720 If we plug in 2 we get 2 squared in our numerator which is four divided by 10 squared in our denominator. 102 00:08:31,920 --> 00:08:37,660 If we plug in three we get three squared in the numerator which is nine divided by ten. 103 00:08:37,650 --> 00:08:39,750 To the third in our denominator. 104 00:08:39,750 --> 00:08:44,700 Now if we just do the division for each of these terms on our calculator what we find is that these 105 00:08:44,760 --> 00:08:51,180 inequalities are in fact true the terms on the left are in fact always less than or equal to the terms 106 00:08:51,180 --> 00:08:51,750 on the right. 107 00:08:51,750 --> 00:08:59,500 So that shows us that be Saab and plus 1 will always be less than or equal to be Subhan. 108 00:08:59,520 --> 00:09:03,270 So thats the first part of our alternating series estimation theorem. 109 00:09:03,270 --> 00:09:09,480 The next part is proving that this limit that the limit of be Sabahan as and goes to infinity is equal 110 00:09:09,480 --> 00:09:10,200 to zero. 111 00:09:10,200 --> 00:09:16,970 So for that part we're just going to take the limit as and it goes to infinity of B 7. 112 00:09:17,010 --> 00:09:21,520 Or in our case and squared divided by 10 to the power. 113 00:09:21,690 --> 00:09:27,850 If we evaluate this at some very large number we plug in you know 1 million or 10 million on our calculator. 114 00:09:27,870 --> 00:09:34,350 What we find is that while and squared is increasing very quickly 10 raised to the end power is increasing 115 00:09:34,410 --> 00:09:35,860 much much faster. 116 00:09:35,880 --> 00:09:41,190 And therefore since the denominator will be significantly larger than the numerator we'll have some 117 00:09:41,340 --> 00:09:46,680 much smaller constant value in the numerator and some extremely large value in the denominator. 118 00:09:46,680 --> 00:09:51,110 This is just going to tend toward zero and the limit will be zero. 119 00:09:51,390 --> 00:09:56,170 So in fact we have proven that the limit as and goes to infinity of B-7 is equal to zero. 120 00:09:56,190 --> 00:09:57,940 Both of those parts are true. 121 00:09:58,170 --> 00:10:05,190 So given that we can use the alternating series estimation theorem to find this remainder term are seven. 122 00:10:05,330 --> 00:10:11,630 Now when it comes to this remainder inequality all we really need to look at is this value here that 123 00:10:11,630 --> 00:10:18,230 we found before remember we said that the sixth term of the series was the last term to affect our sum. 124 00:10:18,350 --> 00:10:24,140 Well this is basically an equals six right here and so what we want to do is we want to plug 6 in for 125 00:10:24,150 --> 00:10:30,090 and here we want to plug 6 in for and here and we want to plug 6 in for and right here. 126 00:10:30,230 --> 00:10:36,530 So what that leaves us with is the absolute value of our subsidies is going to be equal to the absolute 127 00:10:36,530 --> 00:10:45,410 value of S minus as subsects which is going to be less than or equal to be sub 6 plus one or in other 128 00:10:45,410 --> 00:10:47,560 words be sub 7. 129 00:10:47,600 --> 00:10:53,780 Now everything to the left of this less than or equal sign just says that the exact sum of the entire 130 00:10:53,780 --> 00:11:01,680 series minus the sum of the first six terms is equal to the remainder when we're approximating the sum. 131 00:11:01,740 --> 00:11:08,450 After six terms so we can use either one of these either the absolute value of X minus as subsects or 132 00:11:08,540 --> 00:11:13,580 our subsects and we really just want to use our subsects So we're going to say the absolute value of 133 00:11:13,670 --> 00:11:17,470 our subsects is less than or equal to be sub 7. 134 00:11:17,480 --> 00:11:20,980 Now remember we have a value for B subs 7. 135 00:11:21,080 --> 00:11:28,250 We got it in this column here that I erased when we plug 7 into our B sub and series up here. 136 00:11:28,250 --> 00:11:37,530 And what we got for we saw then was zero point and then 5 0 0 0 0 0 0 4 9. 137 00:11:37,610 --> 00:11:42,860 And so we just take that value for B sub 7 that we already found and we say that that's going to be 138 00:11:43,100 --> 00:11:46,990 greater than or equal to our sub 6. 139 00:11:47,000 --> 00:11:54,410 Now what these two things tell us together is that the some of the first six terms the approximation 140 00:11:54,440 --> 00:12:01,070 of the some of the first six terms is zero point zero six seven six when we round to four decimal places 141 00:12:01,460 --> 00:12:09,260 with a remainder on that approximation less than or equal 2 points 0 0 0 0 0 4 9. 142 00:12:09,260 --> 00:12:15,950 In other words this gives us an idea of how accurate this estimation of the sun is compared to the exact 143 00:12:15,950 --> 00:12:16,680 sum. 144 00:12:16,700 --> 00:12:22,640 In other words if we just sum up the first six terms of the series the difference between the sum of 145 00:12:22,640 --> 00:12:28,700 the first six terms and the sum of the entire infinite series every single term in the infinite series. 146 00:12:28,800 --> 00:12:32,290 They're only going to be different by this value here. 147 00:12:32,330 --> 00:12:39,260 So it goes to show you that summing up only the first few terms of an alternating series like this actually 148 00:12:39,260 --> 00:12:44,310 gets you very very close to the exact size of the entire infinite series. 149 00:12:44,310 --> 00:12:49,730 We're only off by this point 0 0 0 0 0 4 9 value. 150 00:12:49,790 --> 00:12:54,980 So that's how you use the alternating series estimation theorem to find the sum correct to a certain 151 00:12:54,980 --> 00:12:59,920 number of decimal places and get an idea of the remainder estimate. 152 00:12:59,930 --> 00:13:05,920 In other words how accurate your sum is compared to the sum of the entire infinite series. 17908