Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We have $n$-ary sums ($\sum$) and products ($\prod$). Is there an $n$-ary exponentiation operator?

$$\underset{i=1}{\overset{n}{\LARGE{\text{E}}}}\, x_i = x_1 \text{^} (x_2 \text{^} (\cdots \text{^} (x_{n-1} \text{^} x_n)))$$

How about an $n$-ary tetration operator?

$$\underset{i=1}{\overset{n}{\boxed{\underset{\leftarrow}{\LARGE{\text{4}}}}}}\, x_i = x_1 \uparrow\uparrow (x_2 \uparrow\uparrow (\cdots \uparrow\uparrow (x_{n-1} \uparrow\uparrow x_n)))$$ $$\underset{i=1}{\overset{n}{\boxed{\underset{\rightarrow}{\LARGE{\text{4}}}}}}\, x_i = (((x_1 \uparrow\uparrow x_2) \uparrow\uparrow \cdots) \uparrow\uparrow x_{n-1}) \uparrow\uparrow x_n$$

What are the correct symbols for these operations, if they exist?

share|cite|improve this question
How about $\Delta$ and $\Phi$, by analogy with atomic orbitals S, P, D, F? – Dan Brumleve Nov 24 '12 at 8:44
@DanBrumleve Interesting thought. I picked $\text{E}$ for exponent and $4$ for tetration, but it wouldn't render them as accumulation symbols (which is why they are inside product symbols). This way $5$ would work for pentation, etc. – ctype.h Nov 26 '12 at 1:00
I think I finally figured out how to get them to look more like accumulation symbols, but they are higher than sum and product symbols. – ctype.h Nov 26 '12 at 3:06
I think the reason that such symbols don't exist is that exponentiation and tetration, unlike addition and multiplication, aren't associative (or for that matter, commutative), so in addition to specifying terms in an $n$-ary exponentiation, you would also need to specify how to associate them if you want the same versatility as the $\Sigma$ and $\Pi$ notations. For instance, $\underset{i=1}{\overset{3}{\LARGE{\text{E}}}}\, x_i$ could be $x_1^{(x_2^{x_3})}$ or $(x_1^{x_2})^{x_3}$, and the number of possibilities grows rapidly with the number of terms. – rayradjr Nov 26 '12 at 3:15
@rayradjr Since $((({x_1}^{x_2})^{\cdots})^{x_{n-1}})^{x_n} = x_1 \text{^}\Big(\prod\limits_{i=2}^{n} x_i \Big)$, we do not need a symbol for left-associative exponentiation. However, nothing like this would work for right-associative exponentiation, hence the question about notation. – ctype.h Nov 26 '12 at 3:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.