What do temporal and spatial summations have in common?

Question about summations?

• (sorry in advance if the equation looks confusing, I will try to make it as clear as possible) In a solution to a homework problem it stated: (n/2) * [Summation from i=0 to i=(k-1)] of (1) minus [summation from i=0 to i=(k-1)] of (i) plus 1 equals (n/2)k - (k-1)k/2 + 1 I am a bit confused at how the two summations were solved... Question 1: I thought that the (n/2) times the summation of 1 from i=0 to i=(k-1) should be (n/2) * [1*(k-1)] so it should be (n/2)(k-1), but it appears that in the solution it is equal to (n/2)*k. Why is this so? I understand that the summation of i from i=0 to i=(k-1) should adhere to the formula for the summation of an arithmetic series, making it (1/2)(k-1)((k-1)+1) = (k-1)k/2 While I was trying to solve my first question I thought that perhaps the reason it was (n/2)k instead of (n/2)(k-1) was because since i starts at 0 not at 1, it actually summed 1 k times instead of k-1 times, which would give the answer (n/2)(k). If this is the case however, then why does the second summation equal (1/2)(k-1)((k-1)+1) instead of (1/2)(k)(k+1)?? I know that this is really confusing without being able to write out the equations properly, but if someone can understand what I am saying can you try to shed a little light on this problem? Thanks so much!

• Answer:

  You are right, the sum from i = 0 to i = k - 1 of 1 is k for the reason that this sum is k copies of 1. It's k copies of 1 since there are exactly k indices i that satisfy the inequality 0 <= i <= k - 1, namely i = 0, and then i = 1, 2, ..., k - 1. You are also right that the sum from i = 0 to i = k - 1 of i is (k-1)k/2. It doesn't matter that there are exactly k indices i that satisfy the inequality 0 <= i <= k - 1. Because the formula we're summing isn't constant, the number of indices isn't so important anymore. When i = 0 the thing being added is 0, so in fact this value of i doesn't contribute to the sum at all. So the sum from i = 0 to k - 1 of i, is equal to the sum from i = 1 to k - 1 of i. And hopefully it makes sense why this should be obtained by putting n = k - 1 (and not n = k) into the formula for the sum of the first n positive integers. Really, it's only when you're summing a constant that the *number* of indices you are summing over is an explicit consideration (because the sum in that case is [the number of incides] * [the constant]). If you are summing a genuine function i, then it's the values of this function that matter, and not the number of indices. This does lead to situations where a sum from i = m to n of [something] is equal to the sum from i = [something besides m] to [something besides n] of [the same thing], for the reason that the function of i that you are summing is sometimes zero, meaning that some values of the index don't contribute to the sum. It sounds like you actually understand most of what is going on, so you should not be too discouraged by how confusing it seems. Even if you are extremely experienced with sigma notation, tiny things like the fact that the sum from i = n to i = m of k is (m - n + 1)*k and not (m - n)*k will still trip you up from time to time. It is like making sign errors like (-2)(-3) = -6, or silly arithmetic errors like 2*3 = 5. Most people who understand what they are doing still make errors like that from time to time.