6
$\begingroup$

$$ \newcommand{\pow}{\mathop{\vcenter{\huge{\text{E}}}}\limits} $$

The $\sum$ operator can be defined recursively as $$ \sum_{i = a}^b f(i) = f(a) + \sum_{i = a + 1}^b f(i). $$

Likewise, the $\prod$ operator can be defined as $$ \prod_{i = a}^b f(i) = f(a) \prod_{i = a + 1}^b f(i). $$

One can define an operator $\pow$ via $$ \pow_{i = a}^b f(i) = f(a)^{\pow_{i = a + 1}^b f(i)}. $$

Logarithms have a well-known property of being able to turn products into sums $$ \log \prod = \sum \log $$ which is actually what majorly prompted their study in the first place.

Is there analogous function $\psi$ that turns powers into products $$ \psi \pow = \prod \psi? $$

I imagine that $\phi$ is either the super-logarithm or at the very least closely related to it, but I have not been able to verify this property for myself (I find it hard to wrap my mind around tetration).

$\endgroup$

1 Answer 1

8
$\begingroup$

The only such functions are constants, essentially because multiplication is commutative while exponentiation is not: for any $x$, $$ \psi(x) = \psi(x^1) = \psi(x)\psi(1) = \psi(1)\psi(x) = \psi(1^x) = \psi(1). $$ Indeed this shows that the only two possible constant functions are $\psi(x)=1$ and $\psi(x)=0$.

$\endgroup$
1
  • 1
    $\begingroup$ Oh, yeah. I just realized that for the most part, you cannot turn a noncommutative function into a commutative one with an operator because $\phi(a^b) = \phi(a)\phi(b) = \phi(b^a)$ in the case of this function. $\endgroup$ Commented Dec 14, 2022 at 5:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .