A.2 Sigma Notation; Increasing the Number of Subintervals

On this screen we review sigma notation. We will be using a lot of it on the upcoming screens — because the integral is defined using this notation — so we need to be sure you are completely comfortable with it. Once we've done some review, we'll revisit Scenario 1 from the preceding screen and continue our development now using sigma notation.

Sigma Notation and Sums

The key points:

• The left side of the equation is "sigma notation" because it uses the Greek upper-case capital letter Σ, "Sigma," to indicate that this is a Sum.
• The formula we're using for each term in the sum, (2𝑖 +1), is called the summand.
• The letter 𝑖 is the sum's index, meaning it's the variable we substitute into each term in the sum.
• The sum's lower limit 𝑖 =0 is shown underneath the Σ, and is the first value of i we use for the sum. The first term here is thus (2 ⋅0 +1).
• We then increment 𝑖 by 1, so for the second term 𝑖 =1 and the summand value is (2 ⋅1 +1). We then increment i by 1 again and continue, until. . .
• we reach the sum's upper limit, 3, shown at the top of the Σ. Our final term is thus (2 ⋅3 +1).

We could of course have a larger upper limit than 3: the slider below lets you choose whatever value you'd like, up to n = 25. (We could go larger still, but you get the idea.)

The key point is that as you increase the upper limit, the number of terms on the right-hand side of the equation becomes quite unwieldy. No one wants to write all of those out! The utility of the sigma notation is that we don't have to spend time with all that writing; instead we have this very compact way to represent them all.

But let's be clear: the sigma notation on the left-hand side and all of the terms on the right-hand side of the equation are simply two ways of representing the same thing. When you see such sigma notation, you must broadly envision the many terms on the right as if they were all written out. There's nothing magical happening with any of this; the Σ notation merely saves us a ton of writing.

Summation Index and Dummy Variables

Traditionally letters from the middle of the alphabet — specifically, i, j and k — are used as summation indicies. (By the way, in mathematics the plural of "index" is usually "indicies." "Indexes" is not wrong, just not typically used in this context.)

Notice that we can choose whatever letter we'd like for the index without affecting the actual sum:

The index is a dummy variable

The index is merely a placeholder that lets us track where we are between the lower and upper limits, and as such we can call it whatever we want. For this reason the index is known as a dummy variable: you can change it within the sigma notation without consequence. You do have to be consistent and use the same name in the lower limit and in the summand itself. For example, mixing the variables i and k in 3∑𝑖=0(2𝑘+1) does not get us the same sum as we have above, and is actually nonsensical.

Quick note for those familiar with computer programming: The index is a variable local to the sigma summation, the same as it is within a for-loop: it doesn't exist outside of the loop. The usage is exactly the same here.

Shifting the Index

Let's introduce how to "shift the index." We'll do some index shifting on upcoming screens, and it'll be easier to handle then if you already understand the approach.

Sticking with our original 4-term example from above,

Let's say that, for some reason, you don't want your sigma-summation 3∑𝑖=0(2𝑖+1) to start with 𝑖 =0, and instead you want your lower limit to be 1. You can handle this easily by making the substitution 𝑗=𝑖+1 You must make this substitution everywhere the index appears, which is three places: (1) the lower limit, (2) the upper limit, and (3) in the summand. Since we have 𝑗 =𝑖 +1:

1. The lower limit was 𝑖 =0, and so the new lower limit is 𝑗 =0 +1 =1.
2. The upper limit was 𝑖 =3, and so the new upper limit is 𝑗 =3 +1 =4.
3. For the summand, we can rewrite 𝑗 =𝑖 +1 as 𝑖 =𝑗 −1.
   Then our original summand (2𝑖 +1) becomes [2(𝑗 −1) +1] =(2𝑗 −1).

The original sum 3∑𝑖=0(2𝑖+1) rewritten with the shifted index is thus

Notice that our new sigma notation using j produces exactly the same sum as we had above. (Of course: math works!) Remember, the sigma-notation is merely a compact way of writing the sum that appears on the right-hand side, and how exactly you write it — using i, j or k as the index, or shifting the index up or down — doesn't matter, as long as the compact representation correctly represents the (expanded) sum. If you'd like, try shifting the index by 99, or -63, or whatever number you'd like to verify this for yourself.

And notice that since j is a dummy variable, we're free to switch the name from j back to i if we'd like:

The following problem lets you practice shifting an index for yourself. This question illustrates an occasional occurance: you end up with a summand that just doesn't "look right" given the context, and so you want to alter it by shifting.

We typically start our indexing
with i = 1.

Our standard practice will be use 𝑖 =1 as our starting index. You may encounter other resource materials that start with 𝑖 =0 instead. Those are equally valid, and merely represent an index-shift as you've just seen.

Revisiting Draining Container

Let's use sigma notation to revisit, and extend, our examination of how much liquid drains from the container on the previous screen. As a reminder, in this scenario we're pretending that the liquid flows out at the rate 𝐹(𝑡) =(6−90211𝑡) gal/min, and we want to know how much liquid flows out from 𝑡 =0 to 𝑡 =3 minutes.

Overestimate

To begin, we'll again consider an overestimate for the actual amount of water that flows out over the first three minutes after we open the drain. Recall that for this overestimate, we use the flowrate value at the start of each subinterval, at its leftmost point, and thereby compute the left sum, 𝑆𝐿.

We're going to extend a bit and use twice as many rectangles as we did before, so 6 total, as shown in the figure below. As you see, we've also assigned an index to each subinterval: i = 1, 2, 3, 4, 5, and 6. While we happen to be using 6 subintervals to build up our sum, we'll need to generalize to any integer number n of them; you'll see below how this number n of subintervals is the same as the n we'll use as the upper-limit in our sigma notation.

Also recall the two Big Steps to our approximation approach for the overestimate:

1. Determine the overestimate for the amount accumulated in each subinterval:

2. Sum the overestimated amounts accumluated for all of the subintervals.

We know for the 6 subintervals the overestimate of the total accumulated amount is

Now we want to generalize that sum, using sigma notation, for any number intervals n. What we're aiming for is this:

Overestimate for total amount accumulated=𝑛∑𝑖=1𝐹(something in terms of 𝑖)Δ𝑡𝑖

So we need to focus on the summand, which is to say Step 1 above. Since this is our first time determining how to form the summand in terms of i, we're going to proceed rather slowly through this part. We'll be developing a lot of intergrals this way going forward, and want to be sure you're with us line-by-line so you can do this on your own.

Let's consider first the duration of the subintervals, which we see from the figure are Δ𝑡1 =Δ𝑡2 =... =Δ𝑡6 =0.5 min. Since they're all the same, let's leave off the subscript and simply write Δ𝑡 =0.5. And so we can generalize, notice that the intervals are this length because we have 𝑛 =6 and that means Δ𝑡 =3 min6 =0.5 min. Generalizing, for any number n subintervals the duration is Δ𝑡=3 min𝑛◂

Next, remember that our flowrate is a function of time t: 𝐹(𝑡) =(6−90211𝑡) gal/min, as shown in the figure above. And we want the maximum flowrate (since we're currently computing the overestimate), which occurs at the first instant of each subinterval i for this decreasing function. Here's the important part: we need to write that first instant in terms of i. A good way to approach this is to write out the times for a few values of i:For 𝑖=1, the max flowrate occurs at 𝑡1=0 min:𝐹(0)For 𝑖=2, the max flowrate occurs at 𝑡2=0.5 min:𝐹(0.5)For 𝑖=3, the max flowrate occurs at 𝑡3=1 min:𝐹(1)⋮For 𝑖=6, the max flowrate occurs at 𝑡6=2.5 min:𝐹(2.5) We see immediately that each 𝑡𝑖 advances by Δ𝑡 =0.5 min from the preceding one, and the first one occurs at 𝑡1 =0. Hence the equation that does what we need is 𝑡𝑖=(𝑖−1)(0.5 min) You can check that the equation works for each of the i values we have above.

The result of this step will always look something like that; the only trick is deciding whether the i-term is (𝑖 −1), 𝑖, or (1 +𝑖). The correct choice depends on how you choose your indicies (for instance, whether you start with 𝑖 =0 or 𝑖 =1) and whether you're after the maximum or minimum on the subinterval. A quick check of the first and last intervals will tell you whether you've gotten this bit right. If not, try one of the other forms and check again. It's super-easy to be "off by one" here, so don't feel bad if you don't get the correct generalized equation on your first try.

Generalizing that equation for 𝑡𝑖 so it applies for any subinterval duration Δ𝑡 gives us 𝑡𝑖=(𝑖−1)Δ𝑡◂ where Δ𝑡 =3 min𝑛. Finally, then ▸For each subinterval 𝑖, the max flowrate occurs at 𝑡𝑖=(𝑖−1)Δ𝑡 min, and has the value𝐹((𝑖−1)Δ𝑡)◂ We know your eyes may be glazing over at this point, so please stop and check to make sure that last statement makes sense to you — it's everything. You can quickly verify, for instance, For subinterval 𝑖=2, the max flowrate occurs at 𝑡2=(2−1)Δ𝑡=0.5 min:𝐹((2−1)Δ𝑡)=𝐹(0.5) which matches the statement we wrote above. ✓

We can now complete Step 1 for the overestimate (whew!), which recall is (1) Determine the overestimate for the amount accumulated in each subinterval: Overestimate\ for amount accumluatedduring each subinterval=(max flowrate on this subinterval)⋅(subinterval duration)=𝐹((𝑖−1)Δ𝑡)⋅Δ𝑡◂ where Δ𝑡 =3 min𝑛.

Step 2 is then straightforward: we sum the amount accumulated for subintervals 1 through n, for however-many subintervals n we happen to be using: Overestimate for=𝑛∑𝑖=1max rate on subint i⏞¯¯¯⏞¯¯¯⏞𝐹((𝑖−1)Δ𝑡)⋅subint duration⏞Δ𝑡,where Δ𝑡=3 min𝑛◂total amount accumulated

Now, ready? Let's put all of the pieces together and tie the sum to the graph for any number n of subintervals that you can set:

Underestimate

Let's consider now the underestimate for the amount of water that flows out during the first three minutes, again using six subintervals. Having laid so much groundwork above, we'll go a bit faster here.

To determine the underestimate for the total amount we:

1. Determine the underestimate for the amount accumulated in each subinterval: Underestimate\ for amount accumluatedduring each subinterval=(min rate on this subinterval)⋅(subinterval duration)

2. Sum the underestimated amounts accumluated for all of the subintervals.

As above, the duration of each of the 6 subintervals is Δ𝑡 =0.5 minutes, and more generally for n subintervals Δ𝑡 =3 min𝑛.

Indeed, the only difference from the overestimate above is that in Step 1 we now want the minimum flowrate instead of the maximum flowrate. As you recall from the previous screen, that minimum flowrate occurs at the end of each subinterval as shown in the figure. We thus use the rightmost point, and are computing the right sum, 𝑆𝑅.For Step 1, let's focus on one subinterval to get the details right; we'll arbitrarily choose 𝑖 =2. From the figure we see that

because the minimum flowrate for subinterval #2 is 𝐹(1). Since Δ𝑡 =0.5 min, we see we can rewrite that statement in terms of i as "the minimum flowrate for subinterval 𝑖 =2 is 𝐹(𝑖Δ𝑡)." And in general, ▸For each subinterval 𝑖, the min flowrate occurs at 𝑡𝑖=𝑖Δ𝑡 min:𝐹(𝑖Δ𝑡)◂ As we wrote above, you know the function's input will always be 𝑖Δ𝑡, (𝑖 −1)Δ𝑡, or (1 +𝑖)Δ𝑡; in this case we recognize that the correct choice is the first option there.

Hence completing Step 1, we have

Then immediately moving on to Step 2, the sum of the accumulated amount for n subintervals is

Let's check by writing out the terms of that sum for n = 6:

If you compare those terms to the figure above, you'll see that our sigma notation does indeed correctly match the sum that we're after.

And with that, we'll let you again vary the value of n for yourself:

We've illustrated the key points we needed to for now, so we're actually to leave Context 1 here for the time being. On the next screen we'll introduce a new scenario, one that requires a bit more thought to develop the accumulation that we're after.

The Upshot

1. Sigma notation lets us compactly write many terms in a sum no matter how many terms there are.
2. We can write our over- and under-approximations for the amount of liquid that flows out of the container in Context 1 as 𝑛∑𝑖=1𝐹(𝑡𝑖)Δ𝑡 where n is the number of subintervals we are using, Δ𝑡 =3 min𝑛, and 𝑡𝑖 represents either the first instant in the subinterval (for the overestimate) or the last instant in the subinterval (for the underestimate).

We know the part above about determining what 𝐹(𝑡𝑖) to use can be a bit intimidating. What questions do you have, or how do you think we can make this all clearer for future students? Please post on the Forum and let us know! We would deeply appreciate it, as will future students.