31–40 of 317 posts
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.
> For example, exp(x)=eml(x,1), ln(x)=eml(1,eml(eml(1,x),1)), and likewise for all other operations I read the paper. Is there a table covering all other math operations translated to eml(x,y) form?
You can find this link on the right side of the arxiv page:
https://arxiv.org/src/2603.21852v2/anc/SupplementaryInformat...
> For example, exp(x)=eml(x,1), ln(x)=eml(1,eml(eml(1,x),1)), and likewise for all other operations I read the paper. Is there a table covering all other math operations translated to eml(x,y) form?
Judging by the title, I thought I would have a good laugh, like when the doctor discovered numerical integration and published a paper. 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?
Looking at page 14, it looks like he reinvented the concept of the vector valued function or something. The whole thing is rediscovering something that already exists.
How does one actually add with this?
Don't know adding, but multiplication has diagram on the last page of the PDF. 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?
const eml = (x,y) => Math.exp(x) - Math.log(y);
const mul = (x,y) => eml(eml(1,eml(eml(eml(1,eml(eml(1,eml(1,x)),1)),eml(1,eml(eml(1,eml(y,1)),1))),1)),1);
console.log(mul(5,7));
For larger or negative inputs you get a NaN because ECMAScript has limited precision and doesn't handle imaginary numbers.
> eml(x,y)=exp(x)-ln(y) Exp and ln, isn't the operation its own inverse depending on the parameter? What a neat find.
> isn't the operation its own inverse depending on the parameter? This is a function from ℝ² to ℝ. It can't be its own inverse; what would that mean?
EDIT: please change the article link to the most recent version (as of now still v2), it is currently pointing to the v1 version which misses the figures. I'm still reading this, but if this checks out, this is one of the most significant discoveries in years. 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 computati…
Same reason all boolean logic isn't performed with combinations of NAND – it's computationally inefficient. Polynomials are (for their expressivity) very quick to compute.
Earlier quoted context omitted.
Don't know adding, but multiplication has diagram on the last page of the PDF. 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?
x+y = ln(exp(x) * exp(y)) exp(a) = eml(a, 1) ln(a)=eml(1,eml(eml(1,a),1)) Plugging those in is an excercise to the reader
How does one actually add with this?
Table 4 shows the "size" of the operators when fully expanded to "eml" applications, which is quite large for +, -, ×, and /.
Here's one approach which agrees with the minimum sizes they present:
eml(x, y ) = exp(x) − ln(y) # 1 + x + y
eml(x, 1 ) = exp(x) # 2 + x
eml(1, y ) = e - ln(y) # 2 + y
eml(1, exp(e - ln(y))) = ln(y) # 6 + y; construction from eq (5)
ln(1) = 0 # 7
eml(ln x, exp y) = x - y # 9 + x + y
x - (0 - y) = x + y # 25 + {x} + {y}