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Ugh. In the title: "Ancient" means 1941. "Simplify" means approximate. "Numbers like pi" means irrational numbers. "How to" means "how accurately you can". Ba
by gjm11 7y ago
Ugh. In the title:
"Ancient" means 1941. "Simplify" means approximate. "Numbers like pi" means irrational numbers. "How to" means "how accurately you can".
Background to the actual conjecture: if you take any irrational number and ask how well you can approximate it by rationals p/q, then it turns out (the proof isn't trivial but e.g. plenty of undergraduate mathematicians would be able to follow it without too much trouble) you can always find infinitely many approximations where the error is no bigger than of order 1/q^2.
The actual conjecture: suppose you want it to be true that for _almost all_ irrational numbers, you can find infinitely many p/q with an approximation error no more than f(q). Then (so conjectured Duffin and Schaeffer, and so apparently Maynard and Koukoulopoulos have proved):
you can do it if and only if the sum (over all positive integers q) of f(q) phi(q) is infinite, where phi is the "Euler totient function": phi(q) = number of things <= q and coprime to q.
Very crudely, how fast does this allow f to decrease with q? Well, the sum of 1/q is infinite but "only just"[1], so we want f(q) phi(q) not to decrease much faster than 1/q. phi(q) varies fairly wildly but crudely is "not usually much smaller than q", so we want f(q) q not to decrease much faster than 1/q; that is, we want f(q) not to decrease much faster than 1/q^2. In other words, the "classical" result I started with is kinda-sorta about the best you can do. (I think that at this level of handwaviness this has been well known for a while; the Duffin-Schaeffer conjecture is about making it extra-precise.)
[1] The sum of 1/n is infinite but that of 1/n^(1+h) is finite if h>0. The sum of 1/(n log n) is infinite but that of 1/(n (log n)^(1+h)) is finite if h>0. The sum of 1/(n log n log log n) is infinite but that of 1/(n log n (log log n)^(1+h)) is finite if h>0. Etc. So in some sense the boundary between finite and infinite is very close to the (kinda-nonsensical) 1 / (n log n log log n log log log n ...).
- svat 7y agoTo be clear, in the statement of the conjecture, we do not need f(q) to be a polynomial or “nice” function of q like 1/q or 1/q^2 — it may not even be an increasing function of q (so statements like “we want f(q) not to decrease much faster than 1/q^2” can be a bit misleading) — one can have arbitrary functions like, say, specifying that f(12)=1/4, f(20)=1/8, and for other q, f(q) = {0 if q is a multiple of 7 (so basically such denominators are never allowed), 1/3 if q does not contain the digit 9, 1/(q log q) otherwise}, etc. I imagine the fact that the proof has to work with arbitrary functions was part of the difficulty.
- gjm11 7y agoYes, f can be any function at all here, and indeed this allows for the sort of more complicated things you mention, and indeed this is where a lot of the difficulty lies. Duffin and Schaeffer, for whom the conjecture is named, had already proved it subject to a restriction on f that kinda-sorta requires it not to vary too wildly. (One other bit of precision that I glossed over: the conjecture is actually about approximations p/q for which p and q have no common factor and that genuinely makes a difference here.)
- warent 7y agoThank you. That title was completely nonsensical to me and it's bizarre that it would be worded that way.
- voldacar 7y agoI feel this way about every quanta article. It's like they just take a bunch of mathy words that have meaning on their own and assemble them in ways that don't actually say anything about anything
- gjm11 7y agoI've felt that way about other Quanta articles, but this one seems substantially worse than usual to me.
- dang 7y agoWe've switched to the article title now. It was submitted as "New Proof Settles Ancient Question of How to Simplify Numbers Like Pi". Perhaps the publication got the message and changed it?