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We write $\sum_{j=0}^n a_j =a_0+a_1+\cdots+a_n$ or $\prod_{j=0}^n a_j=a_0\cdot a_1 \cdots a_n$.

How to abbreviate $a_0=a_1=\cdots=a_n$?

Maybe ${\Large =}| _{j=0}^n a_j$? Or $\{\forall j,k| a_j=a_k\}$...

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    $\begingroup$ Not an abbreviation, but: $$\forall k\in\{1,2,\ldots,n\},\qquad a_k=a_0.$$ $\endgroup$
    – Did
    Nov 13, 2015 at 15:27
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    $\begingroup$ IMHO $a_0=a_1=\ldots =a_n$ is pretty short already. Don't see how you could abbriviate it anymore than that. $\endgroup$
    – gebruiker
    Nov 13, 2015 at 15:35
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    $\begingroup$ I really like your first suggested notation, that will also work with equality replaced by many other relation or operator symbols, it is simple and intuitive and preferable to set based hickery pokery IMHO. $\endgroup$
    – jimjim
    Nov 13, 2015 at 16:06

4 Answers 4

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I think leaving it as $a_0 = a_1 = \dots = a_n$ is the clearest and you don't need to abbreviate it.

You can write it like $\{a_i\}_{i=0}^n = \{a_0\}$ or $\forall i\, 0 \leq i \leq n\; a_i = a_0$ but I don't think it gets any better.

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You can always use words and be crystal clear:

All the $a_i$ are equal.

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I agree that $a_0=a_1=\cdots =a_n$ is optimal, but if you want an alternative, consider $$\max_{i,j\in [1,n]}(a_i-a_j) = 0$$

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  • $\begingroup$ In similar vein one can say that the set of values has zero variance; which could potentially be abbreviated to $\sigma^2 = 0$. $\endgroup$ Nov 13, 2015 at 16:13
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As there is already:

  • $\displaystyle\sum_i x_i$, $\displaystyle\prod_i x_i$, $\displaystyle\coprod_i x_i$, $\dots$
  • $\displaystyle\bigcup_i X_i$, $\displaystyle\bigcap_i X_i$, $\dots$

And one can also define things like:

  • $\underset{i}{+}\; x_i$, $\underset{i}{-}\; x_i$, $\underset{i}{\times}\; x_i$, $\dots$

for pretty much any operator, I suggest doing the same for comparation operators:

  • $\underset{i}{\Large =} x_i$, $\underset{i} > x_i$, $\underset{i} < x_i$, $\dots$
  • $\underset{i} \neq x_i$, $\underset{i} \ge x_i$, $\underset{i} \le x_i$, $\dots$
  • $\underset{i} \approx x_i$, $\underset{i} \equiv x_i$, $\underset{i} \cong x_i$, $\dots$

Therefore, are valid solutions the following:

  • $\overset{i = n}{\underset{i = 0}{\Large =}} a_i$
  • $a_0 \overset{i = n}{\underset{i = 1}{\Large =}} a_i$
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