Summation and Product Notation

Summation notation uses the symbol S, which is the capital Greek letter Sigma (S) for "sum." It acts like a verb in a sentence. It's a shorthand way to write, "Add this and this and this and this and this and this and this and this." In order to see how it works, look at the following translations:

	= x1 + x2 + x3 + x4
	= x1 + x2 + . . . + xn-1 + xn
	= x3 + x4 + . . . + x56 + x57

(Remember, x could represent anything: the number of part-time employees who take a break in each of the four hours they work, the number of visits to a website via advertisements in each of n other sites, the number of people buying a good during 57 days of a sale -- not counting the first two days because those were for special customers -- , etc., etc. But to continue . . . .)

Looking at the first case above, we could try this translation: "Add x₁ and x₂ and x₃ and x₄." The second could be translated, "Add all the x_i." The third is, "Add all the x_is 3 through 57." (By the way, it's important to remember a further shorthand: if the summation sign stands alone, with no indication where i is to start or stop, it means that you should sum over all values available.)

The notation is new, but the operation isn't. The operation is just plain old addition. To see that, let's actually do the first operation above. Suppose that the x_i take the following values:

These add up to 22:

= x1 + x2 + x3 + x4

= 10 + 5 + 0 + 7 = 22

Of course, we usually want to do more than add values up; we usually want to do other operations on them as well. Suppose we have three different x values and we want (for some reason) to multiply each of them by 5. We can correctly write that two ways:

	= 5x1 + 5x2 + 5x3
	= 5(x1 + x2 + x3)

= 5 · (x1 + x2 + x3)

= 5 · (6 + 4 + 15) = 5 · 25 = 125

In a very common situation, we can use the above technique to describe what we do to take an average. Consider again the two expressions in the first table, where we had first 4 and then n values. This is how we would write the average of those values:

	= (x1 + x2 + x3 + x4)/4
	= (x1 + x2 + . . . + xn-1 + xn)/n

Using the x_i values we had above (10, 5, 0, 7), we can use summation notation to make the following calculation:

= (x1 + x2 + x3 + x4)/4 =

= (10 + 5 + 0 + 7)/4 = 22/4 = 5.5

Of course, there are many more complex examples. We can even add more than one type of variable: we can have, for example, n values of x and m values of y and do operations on them. For example, consider this:

	= p1q1 + p2q2 + p3q3 + p4q4 + p5q5
	= a1x1 + a1x2 + a2x1 + a2x2 + a3x1 + a3x2

(What might these mean? The first could be a statement that the revenue earned by a company is equal to the revenue -- price times quantity -- of each of the five goods it sells, or the revenue received on each of five workdays. The second could be the amount produced in each of three factories, where each factory has two different production lines.)

At any rate, what you can see from the above is that summation notation provides a powerful shorthand, enabling you to write very complex (or at least long) procedures down quite easily.