It’s a derivation of the Y combinator from ruby lambdas
More on topic:
> No comparable primitive has been known for continuous mathematics: computing elementary functions such as sin, cos, sqrt, and log has always required multiple distinct operations.
I was taught that these were all hypergeometric functions. What distinction is being drawn here?
It's one of those facts that tends to blow minds when it's first encountered, I can see why one would name a company after it.
Although you also need to encode where to put the input.
The real question is what emoji to use for eml when written out.
Some Emil or another, I suppose. Maybe the one from Ratatouille, or maybe this one: https://en.wikipedia.org/wiki/Emil_i_L%C3%B6nneberga
I'm kidding, of course. You can encode anything in bits this way.
Posts like these are the reason i check HN every day
Exp and ln, isn't the operation its own inverse depending on the parameter? What a neat find.
This is a function from ℝ² to ℝ. It can't be its own inverse; what would that mean?
But no...
This is about continuous math, not ones and zeroes. Assuming peer review proves it out, this is outstanding.
Elementary functions such as exponentiation, logarithms and trigonometric functions are the standard vocabulary of STEM education. Each comes with its own rules and a dedicated button on a scientific calculator;
What?
and No comparable primitive has been known for continuous mathematics: computing elementary functions such as sin, cos, √ , and log has always required multiple distinct operations. Here we show that a single binary operator
Yeah, this is done by using tables and series. His method does not actually facilitate the computation of these functions.
There is no such things as "continuous mathematics". Maybe he meant to say continuous function?
The whole thing comes off a gibberish or rediscovering something that already exists.
I read the paper. Is there a table covering all other math operations translated to eml(x,y) form?
e^ix = cos x + i sin x
which means: e^-ix = cos -x + i sin -x
= cos x - i sin x
so adding them together: e^ix + e^-ix = 2 cos x
cos x = (e*ix - e^-ix) / 2
So I guess the real part of that.Multiplication, division, addition and subtraction are all straightforward. So are hyperbolic trig functions. All other trig functions can be derived as per above.
You can find this link on the right side of the arxiv page:
https://arxiv.org/src/2603.21852v2/anc/SupplementaryInformat...
Quick google seach brings up https://github.com/pr701/nor_vm_core, which has a basic idea
A stack of zeros and ones can be encoded in a single number by keeping with bit-shifting and incrementing.
Pushing a 0 onto the stack is equivalent to doubling the number.
Pushing a 1 is equivalent to doubling and adding 1.
Popping is equivalent to dividing by 2, where the remainder is the number.
I use something not too far off for my daily a programming based on a similar idea:Rejoice is a concatenative programming language in which data is encoded as multisets that compose by multiplication. Think Fractran, without the rule-searching, or Forth without a stack.
That's awesome. I always wondered if there is some way to do this.
Why use splines or polynomials or haphazardly chosen basis functions if you can just fit (gradient descent) your data or wave functions to the proper computational EML tree?
Got a multidimensional and multivariate function to model (with random samples or a full map)? Just do gradient descent and convert it to approximant EML trees.
Perform gradient descent on EML function tree "phi" so that the derivatives in the Schroedinger equation match.
But as I said, still reading, this sounds too good to be true, but I have witnessed such things before :)
Reminds me of the Iota combinator, one of the smallest formal systems that can be combined to produce a universal Turing machine, meaning it can express all of computation.
BobbyTables2•2h ago
bzax•1h ago
nick238•42m ago
xy = eml(eml(1, eml(eml(eml(eml(1, eml(eml(1, eml(1, x)), 1)), eml(1, eml(eml(1, eml(y, 1)), 1))), 1), 1)), 1)
From Table 4, I think addition is slightly more complicated?
Charon77•15m ago
exp(a) = eml(a, 1) ln(a)=eml(1,eml(eml(1,a),1))
Plugging those in is an excercise to the reader
simplesighman•4m ago