0:00:01.630,0:00:08.566 This video is all about Cigna[br]notation. This is a concise and 0:00:08.566,0:00:11.456 precise way of explaining long 0:00:11.456,0:00:17.468 sums. We study some examples and[br]some special cases and then will 0:00:17.468,0:00:19.100 drive some general results. 0:00:20.100,0:00:22.816 But first, let us look at these. 0:00:23.570,0:00:26.300 In math we have to, sometimes 0:00:26.300,0:00:30.970 some. A number of terms in the[br]sequence, like these two. 0:00:31.820,0:00:37.243 The first one, 1 + 2 + 3 +[br]4 + 5. It's the sum of the 0:00:37.243,0:00:38.519 first five whole numbers. 0:00:39.770,0:00:47.618 The second one, 1 + 4 + 9 +[br]16 + 25 + 36. It's the sum of 0:00:47.618,0:00:49.798 this first 6 square numbers. 0:00:51.740,0:00:56.670 Now, if we had a general[br]sequence of numbers such 0:00:56.670,0:00:57.656 as this. 0:00:59.040,0:01:06.420 You want you to you three, and[br]so on. We could write the sum as 0:01:06.420,0:01:13.308 Air Sequel you one, plus you 2[br]plus you three an so off. Now 0:01:13.308,0:01:21.180 this some goes on and on and on.[br]But if we want to some and terms 0:01:21.180,0:01:27.576 then we can write that the sum[br]for end terms equals you one 0:01:27.576,0:01:29.052 plus you too. 0:01:29.070,0:01:32.720 Bless you 3 up to 0:01:32.720,0:01:36.470 UN. And those are in terms. 0:01:37.250,0:01:41.540 Now we can use the Sigma[br]notation to write this more 0:01:41.540,0:01:46.744 concisely. And Sigma comes from[br]the Greek Capital Letter, which 0:01:46.744,0:01:52.096 corresponds to the S in some.[br]And it looks like this Sigma. 0:01:53.430,0:01:58.842 And if we want to express this[br]some using Sigma notation, we 0:01:58.842,0:02:01.548 take the general term you are. 0:02:02.530,0:02:08.928 And we say that we are summing[br]all terms like you are from R 0:02:08.928,0:02:10.756 equal 1 two N. 0:02:11.610,0:02:16.540 And that equals are some SN[br]which is written above. 0:02:17.200,0:02:22.997 More generally, we might take[br]values of our starting at any 0:02:22.997,0:02:29.848 point rather than just at one[br]and finishing it in so we can 0:02:29.848,0:02:37.226 write Sigma from our equal A to[br]be have you are and that means 0:02:37.226,0:02:40.388 the sum of all the terms. 0:02:40.930,0:02:48.080 Like you are starting with our[br]equal, a two are equal, BA is 0:02:48.080,0:02:49.730 the lower limit. 0:02:51.530,0:02:55.400 And B is the upper limit. 0:02:56.330,0:03:01.530 Now what I'd like to do[br]is explore some examples 0:03:01.530,0:03:03.610 using the Sigma Notation. 0:03:04.620,0:03:10.698 So take for[br]example this one. 0:03:11.340,0:03:13.680 It's Sigma of R cubed. 0:03:14.330,0:03:19.465 With our starting are equal 1[br]and ending with R equals 4. So 0:03:19.465,0:03:24.600 the lower limit is are equal 1[br]and the upper limit is are 0:03:24.600,0:03:30.130 equal. 4 will all we have to do[br]is just expanded out and write 0:03:30.130,0:03:34.870 it as along some using the[br]values are equal 123 and ending 0:03:34.870,0:03:36.450 at our equal 4. 0:03:37.220,0:03:44.288 When RS1R cubed is 1 cubed,[br]are equal 2 two cubed are 0:03:44.288,0:03:50.178 equal 3 three cubed and are[br]equal 4 four cubed. 0:03:51.250,0:03:52.898 And we can just. 0:03:53.520,0:03:59.052 Work the Zeit and add them[br]together. 1 cubed, 12 cubed is 0:03:59.052,0:04:05.967 it 3 cubed? Is 3 * 3 *[br]3 so that's 936-2074 cubed is 4 0:04:05.967,0:04:13.343 * 4 which is 16 times by 4[br]which is 64 and when we Add all 0:04:13.343,0:04:17.492 of those together you get a[br]lovely answer 100. 0:04:19.080,0:04:23.040 What about this one? It's[br]similar, but instead of using R 0:04:23.040,0:04:27.720 is the variable. You can use N.[br]In fact, you can use any 0:04:27.720,0:04:29.160 variable that you want. 0:04:29.810,0:04:34.874 And in this case were summing[br]terms like N squared for men 0:04:34.874,0:04:39.516 starting from 2 to 5, going up[br]in increments of one. 0:04:40.050,0:04:43.038 So starting with an equal 2. 0:04:43.800,0:04:45.960 And squared is 2 squared. 0:04:46.580,0:04:48.040 Plus 0:04:49.370,0:04:55.530 An equal 3 three squared. An[br]equal 4 four squared and 0:04:55.530,0:04:58.330 finishing with an equal 55 0:04:58.330,0:05:04.113 squared. And all we have to do[br]now is square these out and add 0:05:04.113,0:05:08.524 them together 2 squares for[br]three squared, 9 four squared 16 0:05:08.524,0:05:10.529 and 5 squared is 25. 0:05:11.120,0:05:14.574 And we can add these very[br]simply. Foreign 16 is 20. 0:05:15.110,0:05:21.545 9 and 25 is 3420[br]and 34 is 54. 0:05:22.490,0:05:23.998 What could be simpler? 0:05:24.630,0:05:30.520 Look at these two written using[br]the Sigma notation, just 0:05:30.520,0:05:35.821 slightly different from the[br]previous ones. Remember we look 0:05:35.821,0:05:38.177 at all the terms. 0:05:38.800,0:05:43.672 Such as two to the K and we[br]substitute in the values for K 0:05:43.672,0:05:47.500 in increments of one starting[br]with zero going up to five. 0:05:48.530,0:05:55.982 So the first one 2 to the K when[br]K is 0, is 2 to the North. Then 0:05:55.982,0:06:03.434 we add 2 to the one +2 to the[br]2 + 2 to 3 + 2 to the 0:06:03.434,0:06:05.918 4 + 2 to the five. 0:06:06.530,0:06:10.202 Notice we've got six terms and[br]the reason why we've got six 0:06:10.202,0:06:13.568 terms is we started K at zero[br]and not at one. 0:06:14.310,0:06:21.707 And then we evaluate those two[br]to the zero 1 + 2 plus 0:06:21.707,0:06:29.104 force plus three 2 cubed, which[br]is it +2 to 416. Let's do 0:06:29.104,0:06:36.670 the 5:32. And when we[br]add those up, 1 + 2 0:06:36.670,0:06:41.220 is 347-1531.[br]And then 63. 0:06:43.100,0:06:45.850 Now look at this one. 0:06:46.100,0:06:50.570 A little bit more involved, but[br]quite straightforward if you 0:06:50.570,0:06:55.934 follow the same steps that we've[br]done before, we take the general 0:06:55.934,0:07:00.851 term here and we substitute in[br]values looking at where it 0:07:00.851,0:07:06.662 starts from and where it ends.[br]So R equals one is the lower 0:07:06.662,0:07:12.473 limit, so we look at our general[br]term and substituted are equal 1 0:07:12.473,0:07:16.943 first and that will give us one[br]times by two. 0:07:16.950,0:07:20.846 Times by 1/2 added 0:07:20.846,0:07:28.576 two. Now we substitute are equal[br]to for the next one, so that's 0:07:28.576,0:07:35.332 two are plus one is 3 +[br]1/2. An hour equals 3. 0:07:35.750,0:07:38.040 3 plus One is 4. 0:07:38.590,0:07:41.380 Plus 1/2 an. 0:07:41.920,0:07:48.839 And I forgot my arm, but my are[br]in this case is 4 and we add 1 0:07:48.839,0:07:55.351 to it which is 5 and continue on[br]in the same then until we get to 0:07:55.351,0:08:01.049 our upper limit our equals 6. So[br]that's a half six times by 7. 0:08:01.760,0:08:05.881 And we can evaluate this side,[br]but we can do a lot of 0:08:05.881,0:08:09.685 canceling on the way before we[br]do. And if the adding two 0:08:09.685,0:08:13.489 cancers with the two here and[br]here, 2 cancers with the four 0:08:13.489,0:08:17.927 to give us two here 2 cancers[br]with this forward to give us 2 0:08:17.927,0:08:21.731 two counters with this six to[br]give us three and two counters 0:08:21.731,0:08:23.950 with this six to give us 3. 0:08:25.410,0:08:31.052 And then we evaluate eyes adding[br]together terms. So this one is 1 0:08:31.052,0:08:38.278 plus. And we're left with 1[br]* 1 * 3, which is 3 0:08:38.278,0:08:43.318 + 1 * 3 * 2, which[br]is 6 plus. 0:08:44.220,0:08:46.386 2 * 5 which is 10. 0:08:47.210,0:08:52.677 Plus 3 * 5 which is 15 and[br]the last 121. 0:08:53.250,0:08:58.020 Now if you look at these[br]numbers, these are quite 0:08:58.020,0:09:02.313 special numbers. There's a[br]name given to them, they 0:09:02.313,0:09:03.744 either triangular numbers. 0:09:04.840,0:09:08.788 And you should be able to[br]recognize these the same way as 0:09:08.788,0:09:11.749 you would recognize square[br]numbers or prime numbers, or 0:09:11.749,0:09:17.221 order even numbers. I'm finally[br]what would like to do is just 0:09:17.221,0:09:22.947 add these altogether. 1 + 3 + 6[br]is 10 + 10 is 20. 0:09:24.520,0:09:31.060 20 + 1535 thirty six and then[br]add the 20 is 56. 0:09:32.250,0:09:34.506 So we get to some 56. 0:09:35.350,0:09:38.020 Now, what 0:09:38.020,0:09:44.436 about this?[br]We've got our general term 2 to 0:09:44.436,0:09:50.804 the K and we want to start with[br]K equal 1 decay equal an if we 0:09:50.804,0:09:55.580 write it out substituting in our[br]values for K starting with K 0:09:55.580,0:10:03.116 equal 1. Our first time will be[br]2 to the one plus then case two 0:10:03.116,0:10:10.348 would be 2 to the 2 + 2[br]to 3 + 2 to the 4th and 0:10:10.348,0:10:14.274 so on. Until we get to[br]two to the N. 0:10:15.880,0:10:19.354 Now we can't evaluate[br]this to a finite number, 0:10:19.354,0:10:23.214 because at this stage we[br]don't know what Aaron is. 0:10:25.960,0:10:33.144 And what if we[br]have these two expressions? 0:10:33.990,0:10:38.401 But still using the Sigma[br]notation, but in this case are 0:10:38.401,0:10:42.010 general term involves a negative[br]sign in both cases. 0:10:42.610,0:10:46.029 So we'll just take a little bit[br]of time and carefully work the 0:10:46.029,0:10:50.832 might. Because it's dead easy to[br]go wrong in these sorts of ones, 0:10:50.832,0:10:55.326 so take a little bit of time and[br]you will hopefully work them out 0:10:55.326,0:10:57.252 OK using the same steps that 0:10:57.252,0:11:01.390 we've done before. In this[br]particular example, our general 0:11:01.390,0:11:07.620 term is negative, one raised to[br]the part of our and R equals 1 0:11:07.620,0:11:13.850 to 4. So the first term will be[br]negative. One reason the part of 0:11:13.850,0:11:19.773 one. Plus negative one race to[br]the power of 2 plus negative 0:11:19.773,0:11:24.646 one. Raise the pirate three and[br]finally negative one raised to 0:11:24.646,0:11:26.418 the power of 4. 0:11:27.260,0:11:32.525 And then all we have to do is[br]multiply these item, making sure 0:11:32.525,0:11:37.385 that we get our signs correct,[br]negative one raise the part of 0:11:37.385,0:11:42.245 one is negative, one negative[br]one raised to the part of two 0:11:42.245,0:11:44.270 that's squared is positive one. 0:11:44.950,0:11:47.038 Negative 1 cubed. 0:11:47.640,0:11:52.008 That will give negative one and[br]negative one is part of four. 0:11:52.008,0:11:53.464 That'll give positive one. 0:11:54.560,0:11:56.549 So in fact. 0:11:57.470,0:12:02.891 Are some work site to be O[br]negative 1 + 1 zero plus 0:12:02.891,0:12:04.976 negative 1 + 1 zero? 0:12:05.700,0:12:07.206 Quite a surprise, in fact, in 0:12:07.206,0:12:10.790 fact. Now look at this one. 0:12:11.440,0:12:16.632 Again, take care. Our general[br]term is negative one over K 0:12:16.632,0:12:22.768 squared and we have to use[br]values of K from one to three. 0:12:23.650,0:12:31.534 So we start with K equal[br]1. That's negative one over 1 0:12:31.534,0:12:39.418 squared plus negative one over 2[br]squared plus negative one over 3 0:12:39.418,0:12:45.659 squared. And then again, taking[br]care with the negative signs. 0:12:45.659,0:12:51.610 Square these values out negative[br]one over one is negative, 1 0:12:51.610,0:12:54.856 squared will give us one plus. 0:12:55.470,0:13:00.606 Negative one over 2 all squared[br]that would be positive and is 0:13:00.606,0:13:03.174 1/2 * 1/2, which is quarter. 0:13:03.940,0:13:07.603 Then added two negative one over[br]3 squared. Again, that's going 0:13:07.603,0:13:12.598 to be positive, and it's going[br]to be one over 3 * 1 over 3, 0:13:12.598,0:13:14.929 which is one of her and I. 0:13:15.690,0:13:20.019 We can add these together,[br]adding the two fractions 0:13:20.019,0:13:20.500 together. 0:13:21.560,0:13:23.768 The common denominators 36. 0:13:24.310,0:13:31.465 And it will be 9 + 4 and[br]you've got your one there. So my 0:13:31.465,0:13:37.666 final answer is one and 13 over[br]36. Quite a grotty answer, but. 0:13:38.180,0:13:42.860 We've been able to workout what[br]seemed to be quite a complicated 0:13:42.860,0:13:44.030 Sigma notation some. 0:13:44.660,0:13:46.646 And we finally got an answer. 0:13:47.660,0:13:53.344 But what if we had along some[br]and we wanted to write it in 0:13:53.344,0:13:55.856 Sigma Notation? How would we do 0:13:55.856,0:14:01.768 it? Well, why don't we first[br]look at the two sums that we 0:14:01.768,0:14:03.010 started off with? 0:14:03.020,0:14:10.050 The first one was 1[br]+ 2 + 3 + 0:14:10.050,0:14:12.159 4 + 5. 0:14:12.930,0:14:17.130 This one is quite an easy one to[br]start off with because it's the 0:14:17.130,0:14:21.147 sum. What type of terms? Well,[br]it's a dead easy term because 0:14:21.147,0:14:25.528 they are adding one on each[br]time. So if we take the general 0:14:25.528,0:14:26.539 term to BK. 0:14:28.040,0:14:34.449 K starts the lower limit one and[br]ends with the upper limit 5. 0:14:35.750,0:14:39.240 So there's no problem with this[br]one. This is very 0:14:39.240,0:14:41.884 straightforward. But look at the 0:14:41.884,0:14:46.720 second one. Again, it's not too[br]bad because we notice and we 0:14:46.720,0:14:48.650 said before that these are 0:14:48.650,0:14:53.972 square numbers. So we[br]could actually write these 0:14:53.972,0:15:00.642 as one squared +2 squared,[br]+3 squared, +4 squared, 5 0:15:00.642,0:15:03.310 squared, and six squared. 0:15:04.740,0:15:08.046 Once we've done this, are Sigma 0:15:08.046,0:15:15.377 notation some? Is dead easy[br]because we could easily see our 0:15:15.377,0:15:18.935 general term must be K squared. 0:15:20.450,0:15:23.411 And K must start with cake, will 0:15:23.411,0:15:26.628 one. Takei equals 6. 0:15:27.240,0:15:32.784 But what[br]about these 0:15:32.784,0:15:39.920 ones? This is[br]a long some. 0:15:41.370,0:15:46.290 What's different about this?[br]It's got fractions and it also 0:15:46.290,0:15:47.766 has alternating signs. 0:15:48.910,0:15:52.995 Negative. Positive negative[br]positive negative positive 0:15:52.995,0:15:56.005 silver. Now the trick in this 0:15:56.005,0:16:01.630 case. Is to rewrite all of these[br]in terms of fractions. 0:16:02.740,0:16:06.044 And then use negative 1 to a 0:16:06.044,0:16:12.194 power. So if we look at this, we[br]can rewrite this as negative one 0:16:12.194,0:16:15.150 over 1. Plus 1/2. 0:16:16.830,0:16:20.590 Minus one over 3 + 0:16:20.590,0:16:25.459 1/4 plus. And So what to[br]one over 100? 0:16:27.470,0:16:33.302 Now we have to deal with[br]the signs alternating. 0:16:34.620,0:16:38.734 And if we think back we did have[br]signs alternative before. 0:16:39.290,0:16:42.782 And that's to do with negative[br]numbers. Now we don't want to 0:16:42.782,0:16:45.983 change the value of the[br]fraction, we just want to change 0:16:45.983,0:16:51.934 the sign. So if I rewrite[br]this as negative one. 0:16:52.570,0:16:54.720 Times by one over 1. 0:16:55.760,0:16:59.710 Plus Negative one. 0:16:59.710,0:17:01.398 The power of 2. 0:17:02.750,0:17:04.019 One over 2. 0:17:05.010,0:17:10.246 Plus negative 1 to the power of[br]3 one over 3. 0:17:10.780,0:17:15.620 Plus negative 1 to the power of[br]4 one over 4. 0:17:17.230,0:17:23.500 Time 2 plus negative one to[br]the 101 over 100. 0:17:24.030,0:17:25.968 Have we got the same sum? 0:17:26.890,0:17:29.669 One yes, because[br]look at each term. 0:17:31.760,0:17:34.807 Negative 1 * 1 over one is[br]negative, one over 1. 0:17:36.070,0:17:40.965 Negative 1 squared will be[br]positive 1 * 1/2, which will 0:17:40.965,0:17:43.190 give us our plus 1/2. 0:17:43.850,0:17:48.734 Negative 1 cubed will be[br]negative one times by 1/3 which 0:17:48.734,0:17:54.062 is negative 1/3, so that's where[br]we get the negative 1/3 negative 0:17:54.062,0:17:56.726 1 to the power of 4. 0:17:57.340,0:18:01.825 Is positive one times by 1/4[br]gives us our plus one over 4. 0:18:02.940,0:18:06.570 So rewriting our sequence that[br]we started off with here. 0:18:07.360,0:18:08.809 Into this format. 0:18:09.620,0:18:13.820 That then helps us to write it[br]in Sigma Notation. 0:18:15.430,0:18:19.705 Because we can easily see that[br]this is Sigma. 0:18:21.120,0:18:24.950 We know it's negative one. 0:18:25.180,0:18:31.879 To the power and the power[br]is always the number underneath 0:18:31.879,0:18:37.570 the fraction. So we'll say that[br]our fraction is one over K 0:18:37.570,0:18:42.764 because we start with K equal 1[br]and we raise negative 1 to the 0:18:42.764,0:18:43.877 power of K. 0:18:44.570,0:18:49.180 And to complete our Sigma[br]notation for this long some. 0:18:50.050,0:18:53.600 We start with the lower limit of[br]K equal 1. 0:18:54.130,0:18:57.105 And the upper limit K equal 100. 0:18:57.930,0:19:00.926 So you can easily see that this 0:19:00.926,0:19:03.429 notation. Is very concise. 0:19:04.100,0:19:08.434 And I really lovely way of[br]writing that big long some. 0:19:09.600,0:19:16.950 Now I'd like to move[br]on some special cases using 0:19:16.950,0:19:21.744 Sigma notation. The first[br]special case is when you have to 0:19:21.744,0:19:24.290 sum a constant. Like this? 0:19:24.840,0:19:29.351 If we have Sigma of three[br]from K equal 1 to 5, what 0:19:29.351,0:19:30.739 does that actually mean? 0:19:31.950,0:19:33.940 Well, we work at night. 0:19:34.460,0:19:41.740 We take each term for cake with[br]one up to K equal 5. 0:19:42.660,0:19:49.764 Each time remains the same. The[br]matter whether cake was one K 0:19:49.764,0:19:56.868 equals 2 cakes, 3, four, or[br]five. So we actually just get 0:19:56.868,0:19:59.828 the sum of 5 threes. 0:20:02.430,0:20:04.380 Which is 15th. 0:20:06.080,0:20:12.262 So with five terms, five times[br]the constant to make 15. 0:20:13.170,0:20:18.994 So we can get a general result[br]from that if instead of three we 0:20:18.994,0:20:20.658 have a constant C. 0:20:22.290,0:20:26.520 And we take our[br]K from one to N. 0:20:27.690,0:20:32.790 What does this work? I to be?[br]Well, we do it in the same way. 0:20:32.800,0:20:35.062 We workout are some using cake 0:20:35.062,0:20:39.580 with one. 2, three, and so on up[br]to K equal an. 0:20:40.280,0:20:41.600 And evaluate it. 0:20:42.100,0:20:48.832 So can equal 1 is say K[br]Equal 2C and so on. 0:20:50.780,0:20:57.854 Up to. The last see[br]such that we have and terms. 0:20:58.780,0:21:05.176 Because we've got N lots of see[br]this, some can be simplified 0:21:05.176,0:21:08.374 dead easily to an times C. 0:21:10.620,0:21:15.360 So this is a general result[br]which is very useful when using 0:21:15.360,0:21:18.915 Sigma notation. If you have to[br]sum a constant. 0:21:19.700,0:21:26.434 From cable one to N, then the[br]value is and it is a finite 0:21:26.434,0:21:28.839 value and times by C. 0:21:29.420,0:21:30.988 What could be easier? 0:21:31.820,0:21:34.668 But what about this? 0:21:35.910,0:21:42.950 What if you had to sum[br]a variable like this 3K? 0:21:43.480,0:21:45.688 Can this be written more easily? 0:21:46.300,0:21:48.390 Well, we'll take this particular 0:21:48.390,0:21:53.924 example. 3K some 4K. Will[br]want to four. 0:21:54.540,0:22:01.400 So will sub in cake. Will 1[br]first, then K equal 2 then K 0:22:01.400,0:22:07.280 equals 3 and then K equals 4 and[br]we get that some. 0:22:07.820,0:22:12.318 We notice that three is a factor[br]if each one of these terms. 0:22:12.850,0:22:20.172 So we factorize IR 3 and we[br]get three bracket 1 + 2 + 0:22:20.172,0:22:21.741 3 + 4. 0:22:23.220,0:22:29.688 Now that is equal to three times[br]by. If you want to add these 0:22:29.688,0:22:34.308 altogether, that's three 610. So[br]the answer is actually 30. 0:22:35.070,0:22:39.102 Now I'm not so much interested[br]in the answer, but in the 0:22:39.102,0:22:43.990 process. At anyone stage, can[br]you spot something that we know 0:22:43.990,0:22:49.110 already? Well, look at this[br]second line. You've got 3 times 0:22:49.110,0:22:52.454 by 1 + 2 + 3 + 4. 0:22:53.770,0:23:00.166 That's the addition of K&K is[br]one 2, three and four, and we 0:23:00.166,0:23:01.642 can rewrite this. 0:23:02.410,0:23:06.174 Using Sigma notation that 0:23:06.174,0:23:08.056 is 3. 0:23:08.580,0:23:15.460 Sigma. Values[br]K from K equals 0:23:15.460,0:23:16.270 124. 0:23:17.500,0:23:23.429 So our expression that we[br]started off with Sigma of three 0:23:23.429,0:23:29.897 K equals 1 to 4K was one[br]to four equals 3 times. 0:23:30.470,0:23:36.482 Sigma of K from cable one to[br]four. So effectively we've taken 0:23:36.482,0:23:41.492 out with factorized at three[br]from our first expression, and 0:23:41.492,0:23:44.498 we brought it outside the Sigma 0:23:44.498,0:23:48.200 notation. So more generally. 0:23:48.820,0:23:56.506 We can take that if we've got[br]the Sigma of CK from K equals 0:23:56.506,0:23:58.153 1 to N. 0:23:58.940,0:24:00.160 What would that equal? 0:24:00.680,0:24:07.610 Wow, quite easy. All we have to[br]do is that we see that that is. 0:24:08.280,0:24:14.869 See times one plus C times[br]2 plus C times 3. 0:24:16.020,0:24:17.109 Right up to. 0:24:17.680,0:24:20.571 See times an. We've got M terms 0:24:20.571,0:24:23.665 here. We can factorize by the 0:24:23.665,0:24:30.980 sea. And we get 1[br]+ 2 + 3 plus up 0:24:30.980,0:24:37.190 to N and then we can[br]rewrite this using Sigma 0:24:37.190,0:24:44.021 notation, that is C Sigma[br]of care from K equals 1 0:24:44.021,0:24:45.263 to N. 0:24:46.440,0:24:47.548 And as I say. 0:24:48.420,0:24:53.337 Quite straightforward, when[br]you've got a Sigma of C times a 0:24:53.337,0:24:57.807 variable. See being a constant,[br]you can take the constant 0:24:57.807,0:25:03.171 outside the Sigma notation and[br]you left with C times by the 0:25:03.171,0:25:06.747 Sigma of K from K equal 1 to 0:25:06.747,0:25:11.733 N. That's another very useful[br]result that you should remember 0:25:11.733,0:25:13.401 when using Sigma notation. 0:25:14.190,0:25:20.185 But what if we had this[br]expression? The sum of the 0:25:20.185,0:25:26.725 variable K plus two where K[br]takes the lower limit one to 0:25:26.725,0:25:33.810 four? What does that work? I'd[br]be? Can we write it more concert 0:25:33.810,0:25:35.990 concisely and easily well? 0:25:36.690,0:25:40.988 We do it stage by stage, step by[br]step, putting in values for K 0:25:40.988,0:25:42.830 from K equal 1 to 4. 0:25:43.480,0:25:47.310 So the first time case one says[br]1 + 2. 0:25:48.720,0:25:53.780 Plus second, her K equals 2, so[br]that's 2 + 2. 0:25:54.470,0:26:01.370 The third term is 3 + 2 and[br]the fourth term is 4 + 2. 0:26:02.580,0:26:07.926 Now what I want to do next is[br]rearrange these values. 0:26:09.040,0:26:15.823 If I rearrange such that I have[br]the addition of 1 + 2 + 3 + 4 0:26:15.823,0:26:23.114 first. And then see what[br]we've got left. We've got 0:26:23.114,0:26:29.974 add 2 four times, so[br]that is 4 * 2. 0:26:31.420,0:26:37.630 And then we can easily evaluate[br]this expression 1 + 2 + 3 + 4 0:26:37.630,0:26:41.356 is 10 + 4 two 38. So the answer 0:26:41.356,0:26:46.898 is 18. But as I said before,[br]we're not so much concerned with 0:26:46.898,0:26:51.910 the answer, but in the process,[br]can we look and see at any line? 0:26:52.590,0:26:54.948 Can we use the Sigma notation 0:26:54.948,0:26:58.050 more easily? Well, if we look at 0:26:58.050,0:27:03.686 this line. We've seen part[br]of this before. 0:27:04.370,0:27:07.506 1 + 2 + 3 + 4. 0:27:08.640,0:27:14.305 Can be written more easily using[br]Sigma notation as Sigma. Care 0:27:14.305,0:27:17.395 from K equals 1 to 4. 0:27:18.080,0:27:23.337 Plus then we've got 4[br]* 2. 0:27:25.430,0:27:31.022 So looking back at what we[br]started with, we've got Sigma of 0:27:31.022,0:27:38.012 K plus two from K equals one to[br]four and it expanded I to this 0:27:38.012,0:27:44.536 Sigma notation Sigma of can at[br]all, from cable one to 4 + 4 0:27:44.536,0:27:50.594 * 2 or two is the constant that[br]was added to the care. 0:27:52.410,0:27:54.828 So we can generalize[br]this result. 0:27:56.950,0:28:01.290 If we had Sigma[br]of. 0:28:02.340,0:28:08.820 G of K function of K[br]plus a constant C. 0:28:09.320,0:28:16.509 As our variable and we want to[br]take care from one to N. 0:28:17.910,0:28:23.430 We can use the pattern that[br]we spotted here to write this 0:28:23.430,0:28:27.110 out using Sigma notation plus[br]some other value. 0:28:28.410,0:28:32.070 The first part is Sigma. 0:28:32.760,0:28:38.545 Off the function that we started[br]off with in our variable and the 0:28:38.545,0:28:40.325 function is GF care. 0:28:41.010,0:28:48.270 And we see that it should be[br]from K to the upper limit from K 0:28:48.270,0:28:54.548 equal 1. Takei equal the upper[br]limit in this case is for, but 0:28:54.548,0:28:57.334 in our general case it's K equal 0:28:57.334,0:29:00.720 N. Says K equal 1 to an. 0:29:01.320,0:29:08.384 Plus Now the next part, remember[br]it was the upper limit by the 0:29:08.384,0:29:13.688 number of terms times by the[br]constant that you've had in your 0:29:13.688,0:29:15.898 variable. Well, look here. Are 0:29:15.898,0:29:22.748 constantly see. And our[br]upper variable is an. 0:29:23.280,0:29:25.870 So it's End Times C. 0:29:26.780,0:29:30.255 So this Sigma notation can 0:29:30.255,0:29:37.540 be expanded. Two Sigma GF[br]K from K Equal 1M Plus M 0:29:37.540,0:29:38.590 Times C. 0:29:40.210,0:29:44.746 And we can even generalize[br]further using previous results 0:29:44.746,0:29:49.786 if we took Sigma from K equals[br]1 to N. 0:29:50.370,0:29:54.210 Of a GF K 0:29:54.210,0:29:57.820 Plus C. As our 0:29:57.820,0:30:01.368 variable. We can expand that 0:30:01.368,0:30:07.750 out. Looking at this time 1st[br]and then dealing with the 0:30:07.750,0:30:12.997 constant now, this term is a[br]multiple a constant times by 0:30:12.997,0:30:18.721 your GF K so we can bring that[br]outside the Sigma notation. 0:30:18.810,0:30:26.766 And say it's a Times by[br]Sigma GfK from K equals 1 0:30:26.766,0:30:28.092 to N. 0:30:29.560,0:30:30.630 Plus 0:30:32.670,0:30:38.065 Just what's left to do with the[br]sea and we know when that's 0:30:38.065,0:30:41.385 inside the bracket when we[br]expand it out. 0:30:42.070,0:30:44.416 We're left with end times C. 0:30:45.730,0:30:46.810 What could be easier? 0:30:47.560,0:30:51.260 All right? But 0:30:51.260,0:30:57.566 what if? We wanted to[br]use Sigma notation and the 0:30:57.566,0:31:03.350 variable we wanted to add[br]together is K Plus K squared IE 0:31:03.350,0:31:05.278 two functions of K. 0:31:07.060,0:31:11.053 What would we get when we write[br]out this long some? 0:31:11.660,0:31:14.480 Start simply by substituting in 0:31:14.480,0:31:20.683 for K. Case one to begin with.[br]So the first term is 1 + 1 0:31:20.683,0:31:28.043 squared. Second time is 2 +[br]2 squared. Third time is 3 + 3 0:31:28.043,0:31:33.224 squared, and that's our final[br]term because the upper limit for 0:31:33.224,0:31:34.637 K is 3. 0:31:35.300,0:31:41.528 Now we'll rearrange the way we[br]did before to make our work 0:31:41.528,0:31:46.718 easier. Will take the 1 + 2[br]+ 3 together. 0:31:49.040,0:31:53.473 And then we'll take the one[br]squared and the two squared 0:31:53.473,0:31:56.697 and the three squared bits[br]that are left. 0:31:57.860,0:32:02.684 And you can see why I've done[br]that, because here we're adding 0:32:02.684,0:32:05.096 our constant case and here we're 0:32:05.096,0:32:07.828 adding. Our case squareds. 0:32:08.750,0:32:14.282 When we add these together, we[br]just add them up and multiply 0:32:14.282,0:32:20.736 out using the square sign you[br]get 1 + 2 + 3 is 6, 0:32:20.736,0:32:26.729 one squared is 1, two squared is[br]4 three squared 9. Add back 0:32:26.729,0:32:31.339 together that gives you[br]fourteen. 6 + 14 is 20. 0:32:33.160,0:32:36.286 But look at this second line. 0:32:37.420,0:32:42.685 We can simplify this second line[br]using Sigma notation. 0:32:43.510,0:32:48.762 And it's using Sigma notation[br]that we've done before. 1 + 2 + 0:32:48.762,0:32:54.418 3 is the same as Sigma of care[br]from K equals 1 to 3. 0:32:55.350,0:32:56.100 Plus 0:32:57.480,0:33:02.560 1 squared, +2 squared, +3[br]squared, that's dead easy. It's 0:33:02.560,0:33:09.164 the sum of all variables, K[br]squared from K equals 1 to 3. 0:33:10.320,0:33:14.159 Now look at what would started[br]and look what we've got. 0:33:14.870,0:33:21.580 We have broken this Sigma[br]notation up into two sums to 0:33:21.580,0:33:28.540 Sigma sums. So here we've[br]got the Sigma offer variable 0:33:28.540,0:33:30.490 plus another variable. 0:33:31.420,0:33:34.000 And with split it up to. 0:33:34.720,0:33:39.364 Sigma of one variable added to[br]the Sigma of the other variable. 0:33:40.420,0:33:42.330 We've used the distributive law. 0:33:43.350,0:33:46.880 So to write this more 0:33:46.880,0:33:52.133 generally. We can say that the[br]sum of a function of care. 0:33:53.140,0:33:57.466 Added to another[br]function of K. 0:33:58.950,0:34:04.610 From K equals 1 to N must[br]equal the sum. 0:34:05.900,0:34:13.284 Two separate sums. The sum of[br]JFK from K equals 1 to M 0:34:13.284,0:34:19.755 Plus Sigma. FFK from K[br]equals 1 to add. 0:34:21.580,0:34:26.546 And in fact, we can continue[br]this on if we added on other 0:34:26.546,0:34:28.074 functions within our variable 0:34:28.074,0:34:34.606 here. Then all we have to do[br]is add on another Sigma of 0:34:34.606,0:34:35.965 that particular function. 0:34:37.980,0:34:43.047 Finally, here's a real life[br]example using Sigma notation. 0:34:46.440,0:34:50.388 Take for example if we want[br]to find the mean of a set of 0:34:50.388,0:34:52.926 numbers. That could be the[br]marks in a test. 0:34:54.230,0:34:58.790 If we want to find the main what[br]we want to do is we have to 0:34:58.790,0:34:59.930 workout the total sum. 0:35:01.500,0:35:04.530 Divided by the number of values. 0:35:05.580,0:35:10.816 And that's what the main is.[br]Well, that's actually what we've 0:35:10.816,0:35:15.576 been doing in the previous[br]examples. We have been finding 0:35:15.576,0:35:21.764 the total sum, and we have been[br]looking at the number of values. 0:35:21.764,0:35:27.952 So if I take this example, just[br]say we had the marks 23456. 0:35:29.210,0:35:34.150 And we want to find the mean of[br]those marks. The main will 0:35:34.150,0:35:41.924 equal. 2 + 3 + 4 +[br]5 + 6 and we'd have to divide 0:35:41.924,0:35:48.112 it by the number of values,[br]which is 5 when we work that 0:35:48.112,0:35:51.444 out. That is 5 + 510, another 0:35:51.444,0:35:57.056 10. 20 over 5 which works out to[br]be the value 4. 0:35:57.910,0:35:59.650 But more generally. 0:36:00.210,0:36:07.280 If you've got a set of marks,[br]say XI, we can write the main 0:36:07.280,0:36:13.845 in terms of Sigma Notation. It's[br]the total sum of all the marks. 0:36:14.380,0:36:15.829 Such as Zhao. 0:36:17.630,0:36:20.090 From I equals 1 to N. 0:36:20.830,0:36:25.940 And then we want to divide it by[br]the number of values. Now we 0:36:25.940,0:36:27.765 know there are N Marks. 0:36:28.570,0:36:35.738 So we want to pie by one over[br]N, so the main is equal to one 0:36:35.738,0:36:41.114 over an Sigma X of I from I[br]equal 1 to M.