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Comments on Would sesquation have identity element 0, 1, or something else?

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Would sesquation have identity element 0, 1, or something else?

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If we allow a schematic of hyperoperations $\uparrow^a$ to take values for $a \not\in \mathbb{N}$, then setting $a = 0 := Succ(n)$, sesquation occurs for $\uparrow^{\frac{3}{2}}$. This is intended to be, as literally as can be, intermediary between $+$ and $\times$.

Now what would its identity element be, if it has one? Would it be $0$, like with addition, or $1$, like with multiplication? Or would it somehow be $\frac{1}{2}$? Would it "oscillate" between $0$ and $1$, or be "superpositioned" in both states? Might sesquation not even admit of an identity element? The nLab entry on identity elements reads at one point:

... a unit law is the statement that a given operation has an identity element.

... so I don't know that I should automatically guess that sesquation "should have" an identity element anyway?

Motivation: I'm experimenting with a form of "passing to the limit" for $\frac{2xh + h^2}{h}$ where $+$ is replaced by a variable hyperoperator $\uparrow^q$ such that $h$ covaries with $q$ so that $h \rightarrow 0$ when $q \rightarrow 1$ ("treat $h$ like $0$ when adding it to $2x$") and then $h \rightarrow 1$ when $q \rightarrow 2$ ("treat $h$ like $1$ when multiplying it by $2x$"). The "reason for this" is that when $x = 2$, the formula for the slope can be "toggled between" $2x, 2 + x$. (I don't have much of anything like a very clean explanation for why "toggling between" addition and multiplication should be generalized over the formula in its consolidated form; for now, my intuition is just screaming at me that we can simulate the old practice of "neglecting" $h$ by this means.)

But what would this "passing to the limit" for $q$ be? Would we superpose the values $1, 2$ for $q$ or would we "pass through" hyperoperations purportedly in-between $\uparrow^1$ and $\uparrow^2$? In the latter event, I want to have a more definite sense for sesquation, esp. in terms of whether it has a unique identity element or not. (I'm imagining an "abstract algebraic" or "category-theoretic" argument where $h$ is "able to be neglected" because it is a term grounded in identity elements, "being $0$" relative to when $0$ is an identity element and "being non-$0$" when e.g. $1$ is an identity element.)

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This question confuses me (7 comments)
This question confuses me
wizzwizz4‭ wrote 6 months ago

Exponentiation has a different left and right identity. The successor function is a unary operation with no notion of "identity" at all. There are many ways we can interpolate between the hyperoperations (e.g. piecewise linear), and this question doesn't give enough constraints to choose one: so as it stands, the answer could be "whatever you like!".

I think we need more motivation, but the question might not be salvageable.

Ripheus24‭ wrote 6 months ago

wizzwizz4‭ unfortunately, it seemed like most of the "settled" information about sesquation I could find focused on how it would still conform to the normal "2 by 2 is 4" parameter. The rest was (to my memory) all relatively old (e.g. a 2006 report) guesstimation/graph work. Should I look less for direct uses of the word "sesquation" and study the Ackermann function and related items for some pointers? I haven't found much about the general concept of identity elements that suggested a non-arbitrary direction to explore...

clemens‭ wrote 6 months ago

@Ripheus24 Even tetration for non-integral exponents has various definitions none of which (IIRC) is really elegant, merely interpolation between known values and analytic continuation of the kind that could be done for any operator. Of course, you could do that with non-integral numbers of ↑'s (finding analytic continuations of values of $x↑^ny$ for integral values of $n$ and $y$) but it would lack mathematical interest.

I don't think that sesquation is any more likely to have an identity than nullation/the successor function. The reason that the additive right identity is 0 and the multiplicative/exponentiative right identity is 1 is that the base case of + is defined differently from the base case of the other hyperoperations. The Ackermann function is sometimes defined in a way that smooths over this inconsistency, but at the cost of considerable definitional obscurity.

Ripheus24‭ wrote 6 months ago

clemens‭ if I try to read addition off union and multiplication off iterated union, might I adapt something like fuzzy/rough/qua-set/etc. theories to extensions of the concept of union that would allow for "fuzzy addition" and "fuzzy multi-addition" and the like? I suppose it would be painstakingly tedious but it's work I'm willing to spend my time on...

clemens‭ wrote 6 months ago

@Ripheus Not tedious, just mathematically uninteresting without some concept of fuzzy iteration for the hyperoperations. As I said, to my knowledge even the far simpler case of iterated exponentiation (i.e. tetration) has not been successfully made "fuzzy".

There are some functions that at first appeared to have only discrete iterates but later turned out (surprisingly) to be iterable an arbitrary non-integral number of times. As a paradigmatic example of this look up the fractional integral—a fuzzily iterated integral, as its name suggests.

wizzwizz4‭ wrote 6 months ago · edited 6 months ago

Fractal dimension is another place where you can give meaning to a non-integer value… although there are a lot of different notions of fractal dimension, which all only coincide on the integers, so it's not really a counterexample. However, if you can parametrise a shape which is an $a+b$ line for $q=1$, and a space-filling $a \times b$ rectangle for $q=2$, such that $q$ corresponds to some kind of fractal dimension, there may be a measure of the "size" of that curve which gives you a natural definition of sesquation. (The only way I can see you solving this problem in a remotely satisfying way is if you can find a natural definition of sesquation cropping up somewhere; but it doesn't seem like a very natural concept to me.)

clemens‭ wrote 6 months ago · edited 6 months ago

@Ripheus24 @wizzwizz4 That might work, but it'd also somehow have to give rise to $a^b$ when the "modified fractal dimension" equals 3.

The general problem with things of this kind is that there are just too many degrees of freedom with which one can define one's function. See e.g. Wikipedia on the Half-exponential function:

"In particular, for every A in the open interval ( 0 , 1 ) and for every continuous strictly increasing function g from [ 0 , A ] onto [ A , 1 ], there is an extension of this function to a continuous strictly increasing function f on the real numbers such that f ( f ( x ) ) = exp ⁡ x."

And there's no especially elegant way to choose either the $A$ or the $g$, just many different inelegant ways.