Introduction to Summation

Introduction to Summation The purpose of this lecture is to understand what we mean by sigma notation, whichwe often use in statistics. Sigma is represented by this big Greek letter, which lookslike a capital S with a long horizontal stroke through the middle. The goal of thelecture is not to bombard you with computations or to judge you on right or wronganswers, but simply to demystify notation that would otherwise be mystifying. For today's lecture, we'll consider three examples of sigma notation. The firstexample is the sum from i = 1 to 4 of i squared. The second example is the sum fromi = 1 to 5 of 2i + 3. The last example is the sum from j = 3 to 7 of j over 2. These areall 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 doneafter computing this expression and translating it into a formula using sigma notationfor summation. A stickler for details might ask that we continue, but we'd say youneed to pay us more money. After our payment negotiations are finished, we'llcontinue by evaluating this expression: 30.The purpose of this lecture is to understand why the first equals sign is true. In orderto do so, it is important to first realize that there are a number of things happening inthis symbol. First, there is an index in here, which in this case is represented by thesymbol i squared. If someone on the street asked us what i squared was, we wouldtell them that it is the same thing as i times i. But that isn't really fair because I knowwhat i is; the person who asked me does not. There is a trick here though: at thebottom of this equation, there is a range specified for i . We know that we shouldbegin 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 thestarting range is in fact 1, ending at 4. And actually there's something here that's alittle unfair: nothing in the symbol tells us that we count by one as I proceed from thebottom of my range to its top. That's okay; it's sort of a cultural agreement that youstart 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 thiscase, the equation tells us to square the given number. So if i equals 1, then isquared equals 1 squared. If i equals 2, then i squared equals 2 squared. If i equals3, then i squared equals 3 squared. We have completed our scratch work on theside. 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 ofparallel processing. One worker computes 1i = 1; another computes 1i = 2; anotherdoes 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 = 1through 5. The only difference here is that we have changed the upper limit of oursummation and we have changed the variable in the expression. As before, let us dothis problem in detail then sum from i = 1 to 5. Here, we take 2i + 3 and multiply it by2: (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 3again: 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 usingthe sigma notation. So here we have 2 times 1, 3, 2 and 3 added together. Thisnotation 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 ourservices. Then we have 5 and 3 added together, and again, we can say that thisequals 45. Let's work through an example in which you don't have to start from 1 and use i. Thistime, we're going to take the sum from j = 3 to 7, j over 2. First, let's not do ourscratch work. We think we get the idea. Let's just tell this to do something to j. Whatdo we have to do to j? We divide it by 2. Okay, but what j do we want to divide it by2? We want to divide it from j = 3 incrementing 1 up to j7—in other words, starting at3 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. Itwas actually 5 AM. This equals 25 divided by 2. Okay, so those were three easyexamples. Now we`re going to give you the 100 gold coin problem. The sum of 3 to 7over 2 times R. There are two ways we could've done this. One is to do it directly justas we did before, the other is to be a bit clever and say, nah, it's a trick question; itreally means 25 over 2. The only difference between these two equations is one uses the letter j, and theother uses the letter R. Both use the same range of values for j and R, and both aresolving 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 symbolsrepresenting counters. They denote starting at 3, incrementing that something up to7 and doing the following with that something. To drive this point home, note that thesum 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 generallya 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.