Is there a notation that is after pi product?

Question

I have been wondering that if there is a notation that works with powers. What I mean is:

How there is addition (+) and after it; is multiplication (x) and after it; is exponentiation (^) and after it; is tetration (^^) and goes so on...

And then there is summation ($\sum$) and after it; is product ($\prod$) and after it; ??? (is there something after it?)

So, summation works like that:

$\sum_{k=1}^{4}{k^2} = 1^2 + 2^2 + 3^2 + 4^2 = 1+4+9+16 = 30$,

And, product works like that:

$\prod_{i=2}^{4}{i^i} = 2^2 \cdot 3^3 \cdot 4^4 = 4 \cdot 9 \cdot 256 = 9216$,

Then, is there something that works with exponents? Like it would work with something like that:

(assume A is the notation in this case) $A_{n=2}^{4}{n} = 2^{3^4} = 2^{81} =$ (too lazy to calculate it, put it in your calculator if you want)

Thanks for answers.

Answer 1

The special case where the concerned operation is applied to the same variable multiple times is often represented using Knuth's up-arrow notation:

$x \uparrow y = x \cdot x \dots \text{(y times)} \dots x \cdot x = x^y \\ x \uparrow \uparrow y = x \uparrow ^2 y = x^{x^{ \text{(y times)}^{x^x}}}$

And so forth. Such repeated operations, originating from the basic operator addition, are known as hyperoperations and are conventionally represented by the letter $H$ :

$H_1 \left( x, y \right) = x+y \\ H_2 \left( x, y \right) = xy \\ H_3 \left( x, y \right) = x \uparrow y = x^y \\ H_4 \left( x, y \right) = x \uparrow^2 y \\ \vdots \\ H_n \left( x, y \right) = x \uparrow^{n-2} y \\ $

Note that for more than two variables, a repeated hyperoperation $H_n$ is no longer commutative after $n=2$, and we must consider right-associativity,

$x^{y^z} \equiv x^{\left( y^z \right)} \neq \left( x^y \right) ^z$

Keeping this in mind, one may devise a general notation for the repeated hyperoperation $H_{n}$ in the following manner. Let us start with three variables $x,y,z$,

$H_1 \left( x,y,z \right) = x + \left( y + z \right) = H_1 \left( x, H_1 \left( y, z \right) \right) \\ H_2 \left( x,y,z \right) = x \left( yz \right) = H_2 \left( x, H_2 \left( y, z \right) \right) \\ H_3 \left( x,y,z \right) = x ^ \left( y^z \right) = H_3 \left( x, H_3 \left( y, z \right) \right) \\ \vdots$

Moving on to four variables $x,y,z,t$,

$H_1 \left( x,y,z,t \right) = x + \left( y + \left( z+t \right) \right) = H_1 \left( x, H_1 \left( y, H_1 \left( z,t \right) \right) \right) \\ H_2 \left( x,y,z,t \right) = x \left( y \left( zt \right) \right) = H_2 \left( x, H_2 \left( y, H_2 \left( z,t \right) \right) \right) \\ H_3 \left( x,y,z,t \right) = x ^ \left( y^ {\left( z^t \right)} \right) = H_3 \left( x, H_3 \left( y, H_3 \left( z,t \right) \right) \right) \\ \vdots$

Therefore, for a general number of arguments $n$,

$$H_k \left( x_1, x_2, \dots , x_n \right) = H_k \left( x_1, H_k \left( x_2, x_3, \dots , x_n \right) \right) = H_k \left( x_1, H_k \left( x_2, H_k \left( x_3, x_4, \dots , x_n \right) \right) \right) = \dots$$

Or,

$$\boxed{H_k \left( x_1, x_2, \dots , x_n \right) = H_k \left( x_1, H_k \left( x_2, \dots \left( x_{n-2}, H_k \left( x_{n-1} , x_n \right) \right) \right) \right)}$$

Using the iterative function composition notation that I have described here, the same equation becomes more structured,

$$\boxed{H_k \left( x_1, x_2, \dots , x_n \right) = \underset{i=1}{\overset{n-2}{\Huge{\kappa}}} \: H_k \left( x_{n-i-1}, H_k \left( x_{n-1}, x_{n} \right) \right)}$$

From this, we learn that $\prod$ is just $H_2$ in the infinite-membered family of hyperoperations.