Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Often you need a notation for a finite sequence with one element is removed; i.e. $$(x_1,\dots,x_{i-1},x_{i+1}\dots, x_n).$$ I know one notation $$(x_1,\dots,\hat x_i,\dots, x_n)$$ and I hate it. It is too long and it has no sense; i.e., unless you know the meaning you will never guess what is it.

Question: Did you see any other?

share|improve this question

closed as too localized by Andreas Thom, Andres Caicedo, Felipe Voloch, S. Carnahan Dec 14 '10 at 9:34

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

12  
That's a very standard notation... why introduce another? "I hate it" is not a great reason to do that to your readers! –  Mariano Suárez-Alvarez Dec 13 '10 at 17:57
2  
Yes, actually I would have tried to convince the OP to give up his\her hate. The \hat notation is nice. It recalls me the \phantom command in TeX! –  Pietro Majer Dec 13 '10 at 18:39
1  
I love that notation! Is there a context in which it would be used but is problematic? –  Dr Shello Dec 13 '10 at 22:26
3  
Why I hate it: (1) it is too long; (2) it has no sense (unless you know the meaning you will never guess what is it). –  ε-δ Dec 14 '10 at 1:15
2  
The notation is a bit annoying when applied twice, if we have to skip the j- term of the sequence where the i-term has been skipped. –  Pietro Majer Dec 15 '10 at 10:52
show 2 more comments

4 Answers 4

up vote 8 down vote accepted

In game theory, such sequences are needed all the time, and the notation $x_{-i}$ has become so common that it is often not even defined in papers.

The reason is that much of game theory is concerned with situations where each player $j$ has a presupposed strategy $x_j$ and we think of one player $i$ deviating from his given strategy to some other strategy $y_i$, while the other players do not deviate. This new outcome is often denoted by $(y_i,x_{-i})$ or $(y_i; x_{-i})$ or some such abuse of notation, instead of the cumbersome $(x_1,\ldots, x_{i-1},y_i,x_{i+1},\ldots,x_n)$. Despite the fact that "the indices are out of order," it is very convenient notation for game theory once you get used to it.

In particular it allows one to write conditions like $u_i(x)\geq u_i(y_i,x_{-i})$ for all players $i$ and all $y_i$ to define what it means for $x$ to be a Nash equilibrium. Other solution concepts can also be defined compactly with this notation.

share|improve this answer
    
That's fine if you're replacing $x_i$ with something else, but OP just wants to delete $x_i$. –  Gerry Myerson Dec 13 '10 at 21:54
1  
@Gerry: In that case we just use $x_{-i}$. –  Noah Stein Dec 13 '10 at 22:07
2  
I like it, but it would be better to write as $x_{i-}$ (in pronciple one might have negative indexes). –  ε-δ Dec 14 '10 at 1:16
1  
I would do $x_{{\neq}i}$ –  Anton Petrunin Oct 16 '11 at 1:33
add comment

How about $x|_{[n]\setminus\{i\}}$ ?

share|improve this answer
1  
Thank you, I also thought about $x|_{\complement\{i\}}$, but i did not see it used... –  ε-δ Dec 14 '10 at 1:22
add comment

Borrowing from simplicial sets/complexes, $d_i: \mathbf{n-1} \to \mathbf{n}$ is the map that skips $i$, so your sequences would be $x\circ d_i$.

share|improve this answer
    
(but why $d_i$ is better than $\tau_i$, so as to require a new answer?) –  Pietro Majer Dec 15 '10 at 11:02
    
I didn't claim it was better; I simply offered other existing notation, as requested by the OP. –  Jeff Strom Dec 15 '10 at 14:48
add comment

For any $n\in\mathbb{N}$ and for any $i\in[n]:=\{1,\dots,n\}$, you may consider the maps $\tau_i:[n-1] \to [n]$ defined in Iverson notation by $$\tau_i(x):=x+[x\ge i]\, .$$ That is, $\tau_i(x)=x$ unless $x\ge i$, in which case it is $x+1$). It induces by composition the map $\tau_i^*:\mathbb{N}^n\to\mathbb{N}^{n-1}$ that takes the element $x\in \mathbb{N}^n$ to $x\circ\tau_i$, which is what you want. In case of need, to recall the domain we may write $\tau_{i,n}$ instead of $\tau_i$; also, to simplify the notation, $\tau_i\cdot x$ instead of $\tau_i^*(x)$.

This and similar notations are somehow useful e.g. in treating technicalities with the constructions in singular homology. You may write down a list of simple identities relating e.g. compositions of these simple maps, and the analogous identities obtained by counter-functoriality on the $^*$-maps.

share|improve this answer
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.