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New Proof Settles How to Approximate Numbers Like Pi
- saagarjha 7y agoI understand that the article is trying to simplify the topic for a general audience, but there are some things that are just plain counterproductive (even if the article clarifies what they mean later): > New Proof Settles Ancient Question of How to Simplify Numbers Like Pi The title, obviously. This should say "approximate", since we're not simplifying pi. > Under what circumstances is it possible to represent irrational numbers that go on forever — like pi — with simple fractions, like 22/7? Never, unless you're approximating. > He proved that for every irrational number, there exist infinitely many fractions that approximate the number evermore closely. Specifically, the error of each fraction is no more than 1 divided by the square of the denominator. The first part is obviously true and should have been merged with the second part.
- jerf 7y agoTo the article's credit, I was able to back its simplifications into a reasonable understanding of the underlying problem. I think the author does genuinely understand the underlying mathematics of at least the problem statement, because when they don't the attempt to simplify for a high-school-level audience usually renders the results total gibberish, as we've seen many times on HN.
- inlined 7y agoMathematically speaking, Pi can’t be “simplified”, but I’m personally curious what ratio is Pi to the highest precision necessary to calculate anything tangible (the circumference of the universe to the Plank length). Practically speaking one could say that is the simplification of Pi.
- saagarjha 7y agoYou'd need something along the lines of 60ish digits: https://www.wolframalpha.com/input/?i=diameter+of+the+observable+universe%2Fplank+length https://www.wolframalpha.com/input/?i=diameter+of+the+observ...
- rrauenza 7y agohttps://www.jpl.nasa.gov/edu/news/2016/3/16/how-many-decimals-of-pi-do-we-really-need/ https://www.jpl.nasa.gov/edu/news/2016/3/16/how-many-decimal... Let's go to the largest size there is: the visible universe. The radius of the universe is about 46 billion light years. Now let me ask a different question: How many digits of pi would we need to calculate the circumference of a circle with a radius of 46 billion light years to an accuracy equal to the diameter of a hydrogen atom (the simplest atom)? The answer is that you would need 39 or 40 decimal places. If you think about how fantastically vast the universe is — truly far beyond what we can conceive, and certainly far, far, far beyond what you can see with your eyes even on the darkest, most beautiful, star-filled night — and think about how incredibly tiny a single atom is, you can see that we would not need to use many digits of pi to cover the entire range.
- hanche 7y agoIn all fairness, though, when you read on, the word “approximate” is seen more frequently, and the word “represent” less so.
- svat 7y agoHere's a demonstration I made recently, of best rational approximations: https://shreevatsa.net/fraction/best https://shreevatsa.net/fraction/best — much of this Quanta magazine article sets up background on such approximations, so playing with the numbers on this web page may help build some intuition about the problem. In particular, the Duffin-Schaeffer conjecture that has just been proved involves asking about the set of irrational numbers that are approximated to within an error of f(q) for each denominator q, while the webpage above simply finds the best possible approximations for a single fixed irrational number (well, rational…) from the (semi-)convergents of its continued fraction — so any denominator is allowed and the error (for convergents) turns out to be the Dirichlet bound 1/q^2. (For best results, choose a really large bound on the denominator and only ask for the “bestest” rational approximations. The source code is just a single HTML file and a bunch of plain JavaScript files to do the computation, so if you can improve it, please do so and let me know: https://gitlab.com/svat/web-fraction/tree/master/website https://gitlab.com/svat/web-fraction/tree/master/website .)
- papln 7y agoIn your calculator, what does "bestest" mean?
- svat 7y agoBest rational approximation of the second kind (as defined in the text higher up on the page). Partly a joke, as there's already a natural idea of “best rational approximations” (the first kind), and these (the second kind) are even better than that.
- rini17 7y agoI'd like to see rational numbers much more used in computers instead of floating point. Not just for improved representability of widespread numbers such as 1/3. Also what the article hints towards: accuracy of pi in half-precision 16-bit floating point is around 1/(2^11) ~ 0.0005, while 355/113 is more precise 1/(113^2) ~ 0.00008 and the fraction is still possible to be stored in 16-bit.
- inlined 7y agoI like Go’s strategy. Consts have no type and are the only thing with implicit conversion. math.Pi is defined explicitly with a whole lot of digits and the type system can somehow do inline math on all those digits with full precision before downcasting the final result to an int, float32, float64, etc.
- rini17 7y agoI prefer Common Lisp's way, where fractions are first class citizens. Pity it is not supported directly in hardware.
- msla 7y agoI'm not convinced supporting it in hardware would make it any faster, compared to doing what we do now with fast integer arithmetic and reasonable software algorithms. Hardware only makes things faster if there's some good parallelism you can't express in software, like doing integer addition using modern hardware versus doing it the long way bit-by-bit. Here's a whole post on the issue by an actual hardware designer: https://yosefk.com/blog/its-done-in-hardware-so-its-cheap.html https://yosefk.com/blog/its-done-in-hardware-so-its-cheap.ht... Quote: To see the limitations of hardware support, let's first look at what hardware can do to speed things up. Roughly, you can really only do two things: 1. Specialization - save dispatching costs in speed and energy. 2. Parallelization - save time, but not energy, by throwing more hardware at the job.
- saagarjha 7y agoRational numbers are an absolute pain to work with once you perform more than a couple of additions with denominators that are relatively prime.
- gjm11 7y agoUgh. 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.
- leiroigh 7y agoI hate how the article waffles between "almost none", "virtually none" and "negligible". There is a word with precise mathematical meaning that matches naive non-mathematical intuition: "almost all" / "almost none" (aka full measure set). This concept is easy to explain to lay people: count up lengths of intervals, do some limiting process trickery to deal with the fact that the set cannot be written as a finite union of intervals. Alternatively, choose a real number uniformly at random from some interval, probability one will hit the "almost all" set and probability zero will hit one of the others. "negligible set" is often used to describe sets that have zero measure and small Baire category. "Virtually" also has different mathematical connotations. Upside is that it was well explained that the contents of the new result is "don't worry about overlaps between different denominators".
- oluckyman 7y agoWow. Maynard FTW yet again. A Fields medal down the track would be no surprise at this rate. He has several more years to qualify.