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For addition, there is $\sum$

For multiplication, there is $\prod$

What about for division?

Something like ${/}_{i=1}^n a_i = a_1 \div a_2\ \div\ ...\ \div\ a_n$?

Nick
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    Like subtraction, division is not an associative operation. This is probably why these operators are not used, as their meaning might be ambiguous. –  Jul 14 '18 at 16:36
  • @YvesDaoust: $\sum_{i=1}^n a_0 - a_i$ is not ambiguous. – Nick Jul 14 '18 at 16:38
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    Do you mean $\sum_{i=1}^n(a_0-a_i)$ ? This has no connection with the definitions of repeated operators ! Btw, you bring a new interpretation I would never have thought of. Even more ambiguous. –  Jul 14 '18 at 16:58

2 Answers2

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I don't believe there is such a symbol, and you can likely get what you want pretty simply with $\prod$ alone.

For instance, if $N=4$, you probably intended one of:

$$a_1/a_2/a_3/a_4=\dfrac{a_1}{ {\displaystyle\prod_{i=2}^{4}} a_i}$$

or $$a_1/(a_2/(a_3/a_4))=\dfrac{{\displaystyle\prod_{i=1}^2a_{2i-1}}}{{\displaystyle\prod_{i=1}^2a_{2i}}}\text{.}$$

Mark S.
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There can be a justification for a repeated subtraction, provided it is defined with right to left associativity.

Because $a_0-(a_1-(a_2-(a_3-a_4)))=a_0-a_1+a_2-a_3+a_4$ is an alternating series, which is a very frequent pattern, whereas $(((a_0-a_1)-a_2)-a_3)-a_4=a_0-a_1-a_1-a_2-a_2$ is a mere summation in disguise, and we don't need a specific notation for this very rare case.

The case of repeated division is clearer:

$$\frac{a_0}{a_1\cdot a_2\cdot a_3\cdot a_4\cdots}$$ is well replaced by a product and

$$\frac{a_0\cdot a_2\cdot a_4\cdots}{a_1\cdot a_3\cdots}$$ is of very limited use because products with the same factor expressions at the numerator and denominator are exceptional.

By natural selection, these operators have died out.