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.
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.