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I have a series of terms $\{t_n : t_n = a_n x_n\}$, and I want to talk about the product of each term except $t_j$.

Would any of these be an appropriate way to say that? I like this:

$$\prod_{i \ne j} t_i$$

or maybe this

$$\prod_{i=1,i \ne j}^n t_i$$

Is it a normal thing to do, to put qualifiers on an index like that, or is there a better way to do it?

Or would it be better to make a subset, $\{b_n : a_n, n \ne j \}$ and use that:

$$\prod_{i=1}^n b_i$$

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5  
I think that either of the first two suggestions is fine (but in the case of the second, I would typeset $i \neq j$ below the line $i = 1$, rather than separating them with a comma); the third adds a complicating layer of notation that I don't think is necessary. –  Matt E Sep 23 '10 at 19:10
1  
In addition to the fact that it is complicated (as Matt E pointed out), I do not think that the third suggestion is even correct. It still denotes the product of the n numbers $b_1,\ldots,b_n$, which is not defined unless you specify the value of $b_j$. –  Tsuyoshi Ito Sep 23 '10 at 19:21

2 Answers 2

up vote 4 down vote accepted

It is normal to write

$$\prod_{\begin{matrix}i=1 \\ i \ne j \end{matrix}}^n t_i$$

but another good notation is

$$\prod_{t \in T \setminus \{t_j\}} t$$

(with $T = \{t_n : t_n = a_n x_n\}$, also note that it is only defined up to permutation)

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Thanks, that's helpful. What does "only defined up to permutation" mean? –  Unkz Sep 23 '10 at 19:42
1  
When the first one defines $t_1 \cdot t_2 \cdot t_4$ but it's not clear which one of $\{ t_1 \cdot t_2 \cdot t_4,\;\; t_1 \cdot t_4 \cdot t_2,\;\; t_2 \cdot t_1 \cdot t_4,\;\; t_2 \cdot t_4 \cdot t_1,\;\; t_4 \cdot t_1 \cdot t_2,\;\; t_4 \cdot t_2 \cdot t_1 \}$ that the second defines - so unless they are all equal (i.e. multiplication is commutative) it is meaningless. –  anon Sep 23 '10 at 20:01
    
@muad: People write $\prod_{i=1}^{3} t_i$ for $t_1\cdot t_2\cdot t_3$ when multiplication is noncommutative? That's interesting. How do they write $t_3\cdot t_2\cdot t_1$? –  Rahul Sep 23 '10 at 20:30
    
@Rahul: Well they could just write t_(n-i) instead of t_i –  Cam Sep 23 '10 at 20:37

$\large{\text{Annex to Tsuyoshi Ito's Comment :}}$

I thought to provide some more detail. User Unkz defined: $b_n:= a_n, \forall \; n \neq j$. So $\color{red}{b_j}$ doesn't exist.

However, by definition of the Pi notation: $\prod_{i=1}^n b_i := b_1b_2...b_{j-1}\color{red}{b_j}b_{j+1}...b_n$.
Since $\color{red}{b_j}$ doesn't exist but this product contains $\color{red}{b_j}$, therefore this product cannot exist either.

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