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In this answer the summation index is indicated as $0\leq k < 0$.

I read this as,

k is less than zero
k equals zero
k is greater than zero

What is the value of $k$ ? Can $k$ be both less than and greater than zero? It seems that there is a simple convention that I'm not aware of. Can you please explain?

Also, in the same answer, there is the word "anything" inside the summation sign. Is "anything" a mathematical term? Probably it is meant "any function of $k$ goes here"?

zeynel
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    Do you think any number satisfies the summation index? So it means that the set of numbers over which you are going to do summation is actually the null set, so you don’t need to sum over anything. And yes you got ‘anything’ right. – Soham Saha Dec 16 '23 at 09:10
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    Does anyone know why this is being downvoted? I think there are very good answers and the question asks something about inequalities which is not widely known. – zeynel Dec 16 '23 at 11:51
  • @SohamSaha "so you don’t need to sum over anything" Can you express this in mathematical language? To me, non-existence does not imply 0: :$\neg \exists k \rlap{\quad\not}\implies 0$ What are the mathematical steps that allow you to set the summation that did not happen to 0? – zeynel Dec 19 '23 at 21:24
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    Added my thoughts on your previous comment as an answer. – Soham Saha Dec 20 '23 at 11:55

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Your point about "can $k$ be both gretaer than and less or equal to zero" hits right on the mark here: the point is there there is no such $k$! This means that we have an empty sum: as written in that answer, $$\sum_{0 \le k < 0} f(k) = 0.$$ "anything" stands, as you say, for any arbitrary function of $k$.

kipf
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Johannes Kloos
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  • Can we instead use any other empty set criteria as the index? Like ‘$x>5$ and $x<3$’? – Soham Saha Dec 16 '23 at 09:32
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    @SohamSaha The point of notation is to clearly communicate expressions. The notation $0\le k < 0$ is not direct, and the sole reason it was brought up in the original question was that it coincides with the general form $0\le k < n$. We could get away with using any pair of incompatible propositions such as yours, or something like $k=0, k=1$, but there would be no benefit to this. We may as well explicitly notate that $k$ does not exist, e.g. "DNE $k$" or $\not \exists k$. – Jam Dec 16 '23 at 09:48
  • @Jam got it now. Thanks – Soham Saha Dec 16 '23 at 09:49
  • There is something not clear about $f(k)$ in the summation. $k$ is the index, it does not represent the summands, $a_k$. To me, it would be more correct to write $f(a_k)$. And in that case, eg, $a=2$ we would get $2^0$ for $k=0$ and the summation will not be 0. What’s wrong with this argument? – zeynel Dec 18 '23 at 08:55
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    @zeynel you can write $f(k)$ or $a_k$, they're really two ways of representing the same thing. The notation $a_k$ is usually understood to be a function from the index set to the value set of a sequence, written differently to stress that this is a sequence we're dealing with. But "under the hood", it's really the same thing. – Johannes Kloos Dec 18 '23 at 09:21
  • Well, maybe I got the notation wrong but we have two distinct quantities in the summation, the index $k$ and the constant $a$. I need to write the function in the summation to make this clear. In the example in the linked question, $a=2$ and we sum $2^{k_i}$. Maybe I should write $f(a\cdot k_i)$. Because if $k=0$, in this case, the summation is not 0 and we don’t have an empty sum. – zeynel Dec 18 '23 at 09:40
  • @zeynel I think you may be overcomplicating things here. There are several entirely equivalent ways to write a sum: $\sum_{1 \le k \le n} a_k$, $\sum_{k \in { 1, \ldots, n}} a_k$ and $\sum_1 \le k \le n f(k)$, where $f(k) := a_k$ all mean exactly the same thing, for instance. – Johannes Kloos Dec 18 '23 at 10:19
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In my opinion, summation over an empty set can’t be defined using the standard meaning for summation, as you aptly describe ‘the summation that did not happen’ in one of your comments. However, due to its use in several mathematical formulae, it satisfies many situations to set such a summation to be equal to the additive identity ($0$). Therefore, I think it is a case of expansion of the originally intended definition to better suit many situations.

To put it another way, for two disjoint sets $A$ and $B$, if you define summation over an empty set to be $0$, the relation $$\sum_{x\in A\cup B}f(x)=\sum_{x \in A}f(x)+\sum_{x \in B}f(x)$$will still hold even when $A$ or $B$ or both is/are the empty set $\phi$

So defining summation over an empty set to be equal to $0$ is merely a choice, albeit a very good one, which is accepted by most; as it extends many already present formulae to a higher domain.

Soham Saha
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The sum $\sum_{0\le k<0}2^k$ asks us to add up terms $2^k$ for integers $k$ such that $0\le k < 0$.

Since there are no such $k$, $\sum_{0\le k<0}2^k=0$.

Observe $\{k \in \mathbb Z : 0\le k < 0\}$ is the empty set. So, we call $\sum_{0\le k<0}2^k$ an empty sum.

user182601
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  • I write the same as $k\in E = \emptyset =0$. $E$ is the set that contains $k_i$. If $E=\emptyset$ and $k$ is not in it then $k$ disappears from the summation sign and we end up with $\sum_{0\leq < 0} 2$. This sum is not empty, 2 is still there. How do you eliminate the given constant 2? – zeynel Dec 18 '23 at 09:22
  • @zeynel: then $k$ disappears from the summation sign and we end up with $\sum_{0≤<0}2$ This claim is wrong. (I'm not sure if you're just trolling or have some very deep misunderstandings.) Given a set $S$, the expression $\sum_{k\in S} 2^k$ asks us to add up $2^k$ in exactly those cases where $k\in S$. If $S$ happens to be $\varnothing$, then we add up nothing, so that $\sum_{k\in \varnothing} 2^k= 0$. (We do not claim that $\sum_{k\in \varnothing} 2^k$ equals $\sum_{\in \varnothing} 2$ which is an undefined/meaningless expression.) – user182601 Dec 18 '23 at 10:39