Introduction to Summation

Introduction to Summation The purpose of this lecture is to understand what we mean by sigma notation, which we often use in statistics. Sigma is represented by this big Greek letter, which looks like a capital S with a long horizontal stroke through the middle. The goal of the lecture is not to bombard you with computations or to judge you on right or wrong answers, but simply to demystify notation that would otherwise be mystifying. For today's lecture, we'll consider three examples of sigma notation. The first example is the sum from i = 1 to 4 of i squared. The second example is the sum from i = 1 to 5 of 2i + 3. The last example is the sum from j = 3 to 7 of j over 2. These are all just different ways of writing numbers. Let's dive in to do the first one. We will compute the sum from i = 1 to 4 of i squared. The sum from i=1 to 4 of i squared is equal to 1 squared + 2 squared + 3 squared + 4 squared. If someone paid me to do this problem, I would consider myself done after computing this expression and translating it into a formula using sigma notation for summation. A stickler for details might ask that we continue, but we'd say you need to pay us more money. After our payment negotiations are finished, we'll continue by evaluating this expression: 30. The purpose of this lecture is to understand why the first equals sign is true. In order to do so, it is important to first realize that there are a number of things happening in this symbol. First, there is an index in here, which in this case is represented by the symbol i squared. If someone on the street asked us what i squared was, we would tell them that it is the same thing as i times i. But that isn't really fair because I know what i is; the person who asked me does not. There is a trick here though: at the bottom of this equation, there is a range specified for i . We know that we should begin with i equal to 1 and end with i equal to 4 . Let us do some scratch work and work out that formula on the side. Here we have i = 1, here I have i = 2, here I have i = 3 and here I have i = 4. So you will notice that the starting range is in fact 1, ending at 4. And actually there's something here that's a little unfair: nothing in the symbol tells us that we count by one as I proceed from the bottom of my range to its top. That's okay; it's sort of a cultural agreement that you start from low i and end at high i , and count by one as you proceed upward . To each one of these numbers, we apply what this equation tells us to do. In this case, the equation tells us to square the given number. So if i equals 1, then i squared equals 1 squared. If i equals 2, then i squared equals 2 squared. If i equals 3, then i squared equals 3 squared. We have completed our scratch work on the side. Next we add up all of these answers to get our final answer of 4i squared. For those of you with a business background, you can think of this as an example of parallel processing. One worker computes 1i = 1; another computes 1i = 2; another does 1i = 3; and so on. At the end of the day, they compare their answers. Here is another example in which we are asked to find the sum of (2i + 3) for i = 1 through 5. The only difference here is that we have changed the upper limit of our summation and we have changed the variable in the expression. As before, let us do this problem in detail then sum from i = 1 to 5. Here, we take 2i + 3 and multiply it by 2: (2i + 3)(2). Next, we add 3: 2i + 3 +3. Next, we multiply by 2 again: (2i + 3)(4). Now we add 3 again: 2i + 3 +3. Finally, we multiply by 2 again: (2i + 3)(6). Now add 3 again: 2i +3+3. Finally, add them all up: 6+9+12+15+18. Just as we did in the previous example, we can add up long strings of numbers using the sigma notation. So here we have 2 times 1, 3, 2 and 3 added together. This notation gives us a compact way of representing the work order. If you do choose to do this tedious task yourself, it will take you much longer than if you were to use our services. Then we have 5 and 3 added together, and again, we can say that this equals 45. Let's work through an example in which you don't have to start from 1 and use i. This time, we're going to take the sum from j = 3 to 7, j over 2. First, let's not do our scratch work. We think we get the idea. Let's just tell this to do something to j. What do we have to do to j? We divide it by 2. Okay, but what j do we want to divide it by 2? We want to divide it from j = 3 incrementing 1 up to j7—in other words, starting at 3 and counting up by 1 until we get to 7. Here, we'll be upfront and admit that we did this ourselves at home in the morning. It was actually 5 AM. This equals 25 divided by 2. Okay, so those were three easy examples. Now we`re going to give you the 100 gold coin problem. The sum of 3 to 7 over 2 times R. There are two ways we could've done this. One is to do it directly just as we did before, the other is to be a bit clever and say, nah, it's a trick question; it really means 25 over 2. The only difference between these two equations is one uses the letter j, and the other uses the letter R. Both use the same range of values for j and R, and both are solving for the same unknowns j and R. The letter j stands for a dummy variable, meaning it is not a real variable like in algebra when you solve 2x plus 1 equals a, where you have to solve for x to get points. Ordinary variables do not have an independent existence; they are symbols representing counters. They denote starting at 3, incrementing that something up to 7 and doing the following with that something. To drive this point home, note that the sum from a smiley face equals 3 to 7 of a smiley face over 2 is just 25 halves 2. There is nothing special about it. That said, let us not get too wild; there is generally a cultural agreement that when we use dummy indices we tend to use symbols like i, j or K sometimes L maybe R sometimes M sometimes N.