4 ms·
It reminds me of those 'surprising' rational approximations of π where xyz/abc≈π with 7 digits precision and abcd/efgh≈π with 9 digits precision. It's not that
by singularity2001 4y ago
It reminds me of those 'surprising' rational approximations of π where xyz/abc≈π with 7 digits precision and abcd/efgh≈π with 9 digits precision. It's not that surprising when you take into account that abcd/efgh does not save you many digits compared to just writing π out.
Somewhat similarly the ±23 chars of
"∫ 0 ∞ cos(2x)∏ n 1 ∞ cos(xn)dx"
giving 43 digits of another number is not _that_ surprising and in the realm of what you can expect with some likelihood.
I expect that one could theoretically find some double integral with less then 25 signs which approximates e^π or ANY number to 50 digits.
Now if you find some 20 char expression which approximates another 20 char expression up to 100000 digits and THEN suddenly takes a different turn, that would be really curious. Like those properties of natural numbers which are true for some orders of magnitude before someone found a counter example.
- version_five 4y agoFrom the article: Jaded nonmathematicians told us it’s just a coincidence, so what is there to explain? The actual reason they get into is interesting and much deeper and orthogonal to the information content of the representation
- Jabbles 4y ago(1+9^-4(6*7))^3^2^85 is an expression that uses all digits 1-9 once and approximates e to 10^25 digits. https://math.stackexchange.com/questions/1945026/an-amazing-approximation-of-e https://math.stackexchange.com/questions/1945026/an-amazing-...
- cozzyd 4y agoyeah but that's silly since lim n->\infty (1 + \frac{1}{n})^ n is even easier to remember
- wyager 4y agoThis one isn't so interesting, as it's super easy to generate `e` in lots of different ways (such as the limit the expression is approximating). `e` is a very low-kolmogorov-complexity constant. The error term in the integral, on the other hand, has no apparent reason to be.
- wyager 4y agoThe kolmogorov complexity of the provided integral is vastly lower than the >140 bits needed to naively represent the error term. Something else is going on here.
- hgsgm 4y agoWhat are 140bits of the error term? The approximation has 140 bits.
- mananaysiempre 4y agoHaving good “small” rational approximations is actually a characterization of transcendental numbers: there are upper bounds for how good rational approximations to an algebraic number can be as the maximum allowed denominator grows, so by proving that this bound is violated for your number you can prove it’s not algebraic; that’s how people initially went about constructing transcendental numbers and—later—proving π and e were such. (Nowadays, it’s easy to construct a transcendental number because it’s easy to construct a noncomputable one; constructing a transcendental computable number still requires additional ideas such as those bounds—I don’t really know of a simple way to do it.)
- actually_a_dog 4y agoHere's a computable transcendental number for you: https://en.wikipedia.org/wiki/Liouville_number#The_existence_of_Liouville_numbers_(Liouville's_constant https://en.wikipedia.org/wiki/Liouville_number#The_existence...)
- mananaysiempre 4y agoProving it’s transcendental still requires the bound I was talking about, which to my admittedly very ignorant eyes looks like it doesn’t lead to anything interesting outside number theory. This is unlike the undecidability of halting needed to construct a noncomputable number, or Cantor’s theorem used to show they exist; both are also quite simple conceptually, containing a grand total of one idea each and basically no computations. (The Lindemann–Weierstrass theorem you’d usually use for proving e or π is transcendental at least looks like interesting algebra.)
- Someone 4y ago> Now if you find some 20 char expression which approximates another 20 char expression up to 100000 digits and THEN suddenly takes a different turn, that would be really curious. Not necessarily. For example: 0 10^-100000