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October 9[edit]

Number theoretic function -- growth, properties[edit]

Define $g(n)$ to be the maximum value of k in Z such that k is divisible by every positive number less than or equal to its own nth root. I've calculated that $g(1)=2,g(2)=24$ , but I'm not sure about good ways to bound g in order to calculate higher values. Does anyone have insight about the properties and growth of this function? Cheers. —Anonymous Dissident^Talk 00:56, 9 October 2012 (UTC)[reply]

2, 24, 420, 27720, 720720, 36756720, 5354228880, 481880599200, 25619985190800, 10685862914126400, 876240758958364800, 113035057905629059200, 24792356033967973651200, 9690712164777231700912800, 2364533768205644535022723200, 396059406174445459616306136000

oeis series A003102, the definition given is Largest number divisible by all numbers < its n-th root, but there's obviously no difference between the series for less than and the one for less than or equal to since a number is always divisible by it's root. See the page for references. Ssscienccce (talk) 01:39, 9 October 2012 (UTC)[reply]

Probability question[edit]

Hi,

If I am given a random number, r, that has been chosen uniformly between 1 and n, though n is unknown to me, it seems I ought to be able to say something about n (in a probabilistic sense). For example, if r = 2, then it seems intuitively more likely that n = 10 than n = 10^100000000000. However, I cannot seem to grasp a way to fully explain or quantify this. Is it just because there are "more big numbers than small ones"? E.g. I am unsure, in the same scenario, whether it is more likely that n = 10 than n > 10^100000000000. Can anyone shed any light on this problem? 86.160.83.247 (talk) 03:02, 9 October 2012 (UTC)[reply]

Problems like this are usually thought of in terms of Bayesian inference. You know P(r|n); you want a formula for P(n|r); Bayes's formula tells you how to get one. The tricky thing, though, is that the formula requires a prior probability distribution for n, and it isn't obvious what you should use. It can't be a uniform distribution, because there is no such thing as a uniform distribution over the full set of natural numbers. The most reasonable choice is an exponential distribution, and if that's what you use, then small values of n come out more likely than large values. How much more likely, though, depends on the specific form of exponential you choose. Looie496 (talk) 05:02, 9 October 2012 (UTC)[reply]

See German tank problem for a way to deal with problems like this. In fact that analysis doesn't work for just a single sample, but for very small numbers what is often more important is what the impact of any decision will be - you've got to push through the Bayesian prior distribution into the decision making rather than just coming up with some single figure that is good enough for most purposes. If you are jut interested in the most likely number rather than having an expectation then that is normally simply the number you got. If you go into a town and the first bus you see is a number 79 then if you have to bet on a single number as the highest you should choose 79. Dmcq (talk) 09:10, 9 October 2012 (UTC)[reply]

Thanks, so, to simplify things, suppose n can be anything from 1 to 10, all initially equally likely. If r = 2 then the probability of each value of n >= 2 is (1/n)/Σ, where Σ = 1/2 + 1/3 + ... + 1/10. Is that reasoning correct? 86.160.213.6 (talk) 13:39, 9 October 2012 (UTC)[reply]

Yes Dmcq (talk) 16:39, 9 October 2012 (UTC)[reply]

Good, thanks! 86.181.205.180 (talk) 19:06, 9 October 2012 (UTC)[reply]

Stabilizer of cosets[edit]

What is the stabilizer of the coset $[aH]$ for the operation $G$ on $G/H$ ? My answer is simply the set of all $g\in G$ such that $gaH=aH\implies a^{-1}gaH=H$ , so if g conjugated with a is in H then it is part of the stabilizer. Is there a more elementary way of expressing this? Widener (talk) 04:04, 9 October 2012 (UTC)[reply]

Which operation? Left multiplication? Conjugation? It depends. Based on what you've written, it looks like left multiplication. In that case, the orbit of aH in G/H has cardinality |G:H|, so by the orbit-stabiliser theorem (assuming G is finite) you get that the stabiliser of aH has size H. It is in fact H, which you can check by noting that

a^{-1}gaH=H\implies a^{-1}ga\in H\implies g\in H

,

since H is normal by assumption. —Anonymous Dissident^Talk 07:10, 9 October 2012 (UTC)[reply]

Dividing a circle question. (Peace sign)[edit]

This is from a symbol that I saw on a bumper sticker, sort of... Points A, B, Z and C are on a circle so that Arc AB = Arc AC = 150 degrees and Arc BZ = Arc ZC = 30 degrees (AZ is a diameter). Point D is on AZ and dist(AD)= 2* dist(BD) = 2*dist(CD). Assuming at dist(AZ) = 2, what are the areas each of the circle slices? (So this looks like a peace sign with the upper branch exactly twice the distance of the side branches. )Naraht (talk) 09:36, 9 October 2012 (UTC)[reply]

Like this symbol ? Those would be circular sectors if point D was in the center, but it sounds like it's not. StuRat (talk) 10:12, 9 October 2012 (UTC)[reply]

I assume you mean the angles to be relative to the centre of the circle. The points a constant ratio of distances to two other points form a circle so can you find two points on the line AC which satisfy the ratio and so would form the diameter of such a circle? That's a geometric way of getting the point D but you might be better doing it algebraically to get the area. I don't think there is any clean way of getting the areas except as sectors of the circle plus the little triangle from C to the centre of the main circle to D. Dmcq (talk) 12:00, 9 October 2012 (UTC)[reply]

You declare some specific ratio of distances of point D to two given points, so it is a point of some specific Apollonian circle – see Circles of Apollonius#Apollonius' definition of a circle and Apollonian circles#Definition.

Once you find the point D, you can compute the area of each part as a sum or a difference of areas of respective circular sector (which is either 1/12 or 5/12 of the circle's area) and the ODB triangle (where O denotes the circle's center). --CiaPan (talk) 13:23, 9 October 2012 (UTC)[reply]

And for that you can use Triangle: Area using trigonometry: knowing SAS. ~~End results should be (π-1)/12 and (5π+1)/12 imo.~~ Ssscienccce (talk) 18:53, 9 October 2012 (UTC)[reply]

I didn't get that. For the larger slice I got 1/12 * (5π + 2*sqrt(3) + 1 - 2*sqrt(1 + sqrt(3))). 86.181.205.180 (talk) 20:36, 9 October 2012 (UTC)[reply]

You may be right, I misread, took AD=2*DZ, instead of 2*DB. Ssscienccce (talk) 06:02, 10 October 2012 (UTC)[reply]

Formula for Poulet numbers[edit]

Is it possible to generate Poulet numbers (ie base 2 pseudoprimes) via a formula? By formula I mean something like F_n = 2^2ⁿ + 1 for Fermat numbers, where inserting all positive integers in order as value for n will generate all Fermat numbers. Is something like that possible for Poulet numbers as well? Is there any known result in that direction? -- Toshio Yamaguchi (tlk−ctb) 12:44, 9 October 2012 (UTC)[reply]

Looking at the OEIS page and the links there, I'd say it's pretty unlikely. When it comes to prime numbers and related properties, there never is one it seems. The formula for the Fermat numbers can hardly be called a generating function, "Fermat number" is simply the name we give to numbers of the form 2^2ⁿ + 1. There's no "essential" property that they all have in common, first five are prime, the rest, as far as we know, aren't. But you probably knew all that. All I can say is that at least in 2008 it looks like there wasn't one, otherwise building a database with Poulet numbers would be a waste of time. Don't know what happened in 2009 and why the database hasn't been updated after that. Ssscienccce (talk) 19:25, 11 October 2012 (UTC)[reply]

Thanks. I also don't really expect that such a formula exists (at least none that could be written down in a convenient way). -- Toshio Yamaguchi (tlk−ctb) 10:33, 12 October 2012 (UTC)[reply]