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

Question

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 $\{\forall j,k| a_j=a_k\}$...

Answer 1

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.

Answer 2

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

Answer 3

You can always use words and be crystal clear:

All the $a_i$ are equal.

Answer 4

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:

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