A Summary Of Some Summery Summation

"Counting is hard. This is what I keep saying."

- @realityminus3

It all stemmed from an arithmetic series problem with a known sum, but an unknown number of terms. As these things are prone to do, it led to a quadratic equation; as those things are prone to do, that led to two possible values for $n$, one of which was negative.

This is fine; there is no problem here; you discard the negative root as impossible due to the context, collect three marks and move on to the next question. Unless, of course, you're @realityminus3, in which case you ask "what does the negative root represent?"

That's not a simple question. I found a simpler, related one: what is $\sum_{k=100}^{0} k$?

It's a simpler question. However, it's most definitely not a simple answer; I think there are reasonable arguments to be made for four distinct answers. Let's start with the boring ones.

Undefined

"It doesn't make any sense to have the lower limit larger than the upper one" is, to me, a perfectly valid (if dull) way to read the notation. I don't think anyone I asked made this argument, but it'd be a bit off to ignore it.

0

If you're going to say something about the empty set, I will fight you.

-@christianp

I'm afraid Christian is going to have to take this one up with Wolfram|Alpha and Wikipedia, if they think they're hard enough.

Wikipedia, in particular, explicitly defines this as 0 - the logic being that this notation means $\sum_{100\le k \le 0} k$, which is indeed the empty set.

5050

This is by far the preferred answer among the Twitterers who responded to a poll on the subject (as if maths is decided by popular opinion!):

[Aside: You'll note that I erroneously put -5050 in the poll instead of -4950. This was not an attempt to invalidate the poll and force a rerun, honestly.]

Fully 82% of those 41 people chose the 'obvious' answer that summation doesn't care about order.

Some of the arguments include:

  • @adamcreen: "Substitute and add. Discrete so not area so can't be negative."
  • @nextlevelmaths: "This is silly, really, because with addition being commutative you'd just never have reason to write that, other than to confuse."
  • @mathteacher1729: "Worth noting that when I see it in AP Calculus, the smaller number is typically the subscript:
    $ \sum_{r=1}^{100} r $."
  • @graememcrae: "The integral from 100.5 to 0.5 of r dr is -5050, but the sum from 100 to 1 step -1 of r is 5050. Reason: integral $\approx$ sum × step."
  • @topometallo (on Mathstodon): "Summation doesn't care about order. If you think -4950 because of the 'step -1 reason' why should the first '100' become negative?"

Now, I have a dilemma: do I, @peterrowlett-like, stay out of the argument altogether, or get into commentary on these suggestions? I think, for once, I'll leave it. For now, at least.

-4950

The analysts on Mathstodon - notably @alephthought, @jeffgerickson and @byorgey1 - note that summation has, for $a<b<c$, a really useful property: $\sum_{r=a}^b f(r) + \sum_{r=b+1}^c f(r) = \sum_{r=a}^c f(r)$. That would be a nice property to have without the restriction, in which case $\sum_{r=0}^{99} r + \sum_{r=100}^{0} r = \sum_{r=0}^{0} r = 0$. The first term on the left is 4950, so the second must be -4950.

This was spiritually the same as my (piratical) argument: when you're integrating, flipping the limits negates the integral, so something similar ought to work for summation.

But if you think that's piratical, wait until you hear what Barney Maunder-Taylor has to say. Inspired by a Year 6 question, he pondered 'what happens if the numbers join up around the back?' This idea suggests that $\sum_{r=100}^{0} r = \sum_{r=100}^\infty r + \sum_{r=-\infty}^{0} r$. Of course, neither of those sums converge, but many of their individual terms correspond each other. In particular, you'd expect $\sum_{r=100}^\infty r + \sum_{r=-\infty}^{-100} r$ to be 0. All you're left with is $\sum_{r=-99}^{0} r$, which does indeed give -4950.

A final argument, based on this particular summation, is that $\sum_{r=a}^{b} r = \int_{r=a-\frac {1}{2}}^{b+\frac{1}{2}} r \d r$, if $a \le b$. If we remove the restriction, $\sum_{r=100}^{0} r = \int_{99.5}^{0.5} r \d r = \left[ \frac{1}{2}r^2\right]_{99.5}^{0.5} = -4950$.

Summing up

Personally, I'm much more convinced by the arguments for -4950 (for convenience) and 0 (by definition) than for 5050 (by simplicity) or for undefined (by dodging the question). I'll leave the last word to @xander (on Mathstodon), who says: "Regard the sum as a Lebesgue integral with respect to counting measure on the integers, and adopt the convention that orientation matters (that is, $\int_b^a f =−\int_a^b f$). Of course, any of the other three answers could be justified, as well. It depends on what you mean by $\sum$."

I'd be interested to hear different arguments and/or explanations in the comments below!

* Thanks to everyone who took part in the discussion!
* Edited 2017-06-05 to correct a couple of typos. I wondered who'd be the first to spot them, and it turned out to be @realityminus3. Thanks!
* Edited 2017-06-05 to give credit to Jeff Erickson, whose toot had got lost among the favourites.
* Edited 2017-06-05 to clarify Brent Yorgey's field of expertise.

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. Brent considers himself a combinatorialist rather than an analyst, but he probably also says po-tah-to. []

Share

One comment on “A Summary Of Some Summery Summation

  • Barney Maunder-Taylor

    Delighted to add to the confusion with my “numbers loop round from infinity back to the negatives” argument – and especially to be described as “piratical”. I will do my best to turn up at next month’s Maths Jam complete with parrot and eyepatch.

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Sign up for the Sum Comfort newsletter and get a free e-book of mathematical quotations.

No spam ever, obviously.

Where do you teach?

I teach in my home in Abbotsbury Road, Weymouth.

It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby.

On twitter