Summation Limits: Understanding When a>b

[EngWiPy]

Hello,

If we get a summation $$\sum_{r=a}^{b}$$, where a > b, how to treat this summation?

Regards

 

[CompuChip]

I don't know if it is official notation, but I think it is usually interpret it as just

$$\sum_{r = b}^a$$

 

[CRGreathouse]

Really? I'd interpret it as 0. That is, I take
$$\sum_{n=a}^bf(n)$$
as shorthand for
$$\sum_{n\in\mathbb{Z},a\le n\le b}f(n)$$

 

[EngWiPy]

Really? I'd interpret it as 0. That is, I take
$$\sum_{n=a}^bf(n)$$
as shorthand for
$$\sum_{n\in\mathbb{Z},a\le n\le b}f(n)$$

What is zero, the whole summation, or the index r?

 

[DrGreg]

What is zero, the whole summation, or the index r?

The whole summation.

To perform a sum, you start with zero, then for each value of n that satisfies the condition specified, you add f(n). In this case, there are no such values of n, so the answer remains zero.

That's my interpretation at least. But there may be other opinions.

Incidentally, some computer languages such as C work similarly. A loop of the form for(n=10; n<0; n++) is never executed.

 

[Moo Of Doom]

Most of the time, it's best to assume the summation is zero, but it really depends on your application. If a sum like this pops up somewhere for some reason, you should interpret it in context to see if it makes sense at all.

In some cases, for instance in finite calculus, it might be more useful to interpret it slightly differently. It is well known that

$$\int_a^b f(x) dx = -\!\!\!\int_b^a f(x) dx,$$

and since integrals are sort of like infinite sums, perhaps sums should work similarly. If $$a < b < c$$, we have

$$\sum_{k = a}^b f(k) + \sum_{k = b + 1}^c f(k) = \sum_{k = a}^c f(k).$$

It might be nice to extend this to cases other than $$a < b < c$$. For example, we would have

$$\sum_{k = a}^{b - 1} f(k) + \sum_{k = b}^a f(k) = \sum_{k = a}^a f(k) = f(a),$$

which would imply that

$$\sum_{k = b}^a = f(a) - \sum_{k = a}^{b - 1} = -\!\!\!\sum_{k = a + 1}^{b - 1}.$$

We would then have

$$\sum_{k = a}^{a - 1} f(k) = 0,$$

but

$$\sum_{k = a}^{a - 2} f(k) = -f(a - 1).$$

Which interpretation you should choose is up to the application.