Summation Limits: Understanding When a>b

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Discussion Overview

The discussion revolves around the interpretation of summation limits when the lower limit is greater than the upper limit, specifically in the case of \(\sum_{r=a}^{b}\) where \(a > b\). Participants explore various interpretations and implications of this scenario, considering both mathematical notation and practical applications.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests interpreting \(\sum_{r=a}^{b}\) as \(\sum_{r=b}^{a}\), implying a reversal of limits.
  • Another participant argues that the summation should be interpreted as zero, stating that there are no values of \(n\) satisfying the condition when \(a > b\).
  • A follow-up question seeks clarification on whether it is the summation or the index \(r\) that is considered zero.
  • It is clarified that the whole summation is interpreted as zero due to the absence of valid indices.
  • Some participants mention that the interpretation may depend on context, suggesting that in finite calculus, there might be alternative interpretations that could be useful.
  • One participant draws a parallel between summations and integrals, proposing that similar rules might apply, and discusses how this could lead to different interpretations based on the application.

Areas of Agreement / Disagreement

Participants express differing views on how to interpret the summation limits when \(a > b\). There is no consensus on a single interpretation, as various perspectives and conditions are presented.

Contextual Notes

Participants note that interpretations may vary based on the specific application or context in which the summation arises. The discussion highlights the potential for different mathematical frameworks to influence the understanding of summation limits.

EngWiPy
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Hello,

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

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

\sum_{r = b}^a
 
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)
 
CRGreathouse said:
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?
 
saeddawoud said:
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.
 
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 &lt; b &lt; 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 &lt; b &lt; 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.
 
Last edited:

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