Wikipedia:Reference desk/Archives/Mathematics/2018 July 18

= July 18 =

Center of population
When calculating the center of population for a jurisdiction (specifically, the mean center), must one choose a location inside the jurisdiction's boundaries, or can it be outside the jurisdiction entirely? Depending on the distribution of the population, significantly concave jurisdictions like Croatia and Florida might have no place inside their boundaries on which a rigid, weightless map would balance perfectly were the population members represented as points of equal mass. [I assume that the geometric median can be outside the borders, but again I'm not clear]. And should a jurisdiction have no location within its borders that would balance in such a situation, would we say that a location outside the borders is the mean center, or would we say that there is no mean center? I suspect that this second question may overlap with the first, but I don't know enough about topology to understand whether I'm repeating myself. Nyttend backup (talk) 17:25, 18 July 2018 (UTC)
 * I see no reason why the centre of population can’t be outside the borders of a territory. In fact it’s pretty easy to imagine such a case. Suppose a country was in the form of a annulus with population evenly distributed. Then the centre would be at the centre of the annulus, and so outside the area of the territory. It’s not an especially interesting point. It’s not fixed like the geographical centre so you can’t put down a marker – it would be hard and contentious to work out exactly where it was anyway. It does not have any special significance in any other sense.-- JohnBlackburne wordsdeeds 17:36, 18 July 2018 (UTC)
 * For a real-life example, before East Pakistan seceded from Pakistan to become Bangladesh, it had 55% of Pakistan's population; therefore it should be obvious that the population center of Pakistan at that time was somewhere in either India, China, or Nepal. (I think it was in India but I'm not going to attempt to check that.)  Countries like Kiribati that contain widely separated islands would also likely be examples, but it depends on what areas of sea they claim possession of. --76.69.47.228 (talk) 02:46, 19 July 2018 (UTC)


 * By the way, "center of population" is not a topological concept. It depends on distances and topology does not concern itself with distances. --76.69.47.228 (talk) 02:48, 19 July 2018 (UTC)
 * A good example. My thinking was countries can have holes in – not internal voids like lakes but bits that are parts of other countries, often the whole of another country, like Lesotho which is entirely surrounded by South Africa. Or something like Somalia which wraps around Ethiopia-- JohnBlackburne wordsdeeds 00:51, 20 July 2018 (UTC)

Prove this equation is true for odd n
While trying to reverse engineer a formula used in a game, I was able to discover that for odd positive integers, the following equation is true:

$$13.75n^2 + 165n + 41.25 = 220n + 110 * \lfloor n/4 \rfloor * (2*(\lfloor n/4 \rfloor -1) + (n \bmod 4)) $$

But how would one go about proving this? (Obviously I can simplify a bit by subtracting 165n from both sides but that doesn't get me very far) Iffy★Chat -- 18:16, 18 July 2018 (UTC)


 * I don't have a complete answer, but I'll note that you can multiply through by 4/55 to simplify some of the constants and make them all integers. That gets you to
 * $$n^2 - 4n + 3 = 8 * \lfloor n/4 \rfloor * (2*(\lfloor n/4 \rfloor -1) + (n \bmod 4)) $$
 * Then you can multiply out the right hand side and get
 * $$n^2 - 4n + 3 = 16 * (\lfloor n/4 \rfloor)^2 - 16 * \lfloor n/4 \rfloor + 8 * \lfloor n/4 \rfloor * (n \bmod 4)) $$
 * The left side and right side are looking more similar now, but I'm not sure how to finish. Knuth presented some techniques in Concrete Mathematics for dealing with floors and mods, but I'm a bit rusty. CodeTalker (talk) 21:08, 18 July 2018 (UTC)

There are two cases: Both cases are OK. Bo Jacoby (talk) 21:47, 18 July 2018 (UTC).
 * 1) $$n=4x+1, \lfloor n/4 \rfloor=x, n \bmod 4=1$$ $$(4x+1)^2-4(4x+1)+3=16 x^2-16 x+8x$$
 * 2) $$n=4x+3, \lfloor n/4 \rfloor=x, n \bmod 4=3$$ $$(4x+3)^2-4(4x+3)+3=16 x^2-16 x+24x$$