Wikipedia:Reference desk/Archives/Mathematics/2016 July 15

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July 15

Quadrant equals four, __???__ equals two?

If something of four equal parts is a quadrant, what is something of two equal parts called (like quadralateral is four and bilateral is two)?166.186.168.37 (talk) 01:59, 15 July 2016 (UTC)

If you are talking about a circle, it is made up of four quadrants or two hemispheres. :) 64.94.31.206 (talk) 02:54, 15 July 2016 (UTC)

Semicircles, actually. Rojomoke (talk) 04:12, 15 July 2016 (UTC)

Half? Also see Orthant, 'ray' may be an option. Gap9551 (talk) 04:20, 15 July 2016 (UTC)

Analogous to your example that a four-sided polygon is a quadrilateral, a two-sided polygon (which has zero area) is a digon. Loraof (talk) 14:11, 15 July 2016 (UTC)

When the Cartesian plane or the complex plane divided into four parts by the axes, the four parts are called quadrants. When the Cartesian or complex plane is divided into just two parts by just one of the axes, the two parts are called half-planes. Loraof (talk) 14:37, 15 July 2016 (UTC)

I say "half-line", as in the "right half-line" and "left half-line". Sławomir Biały (talk) 17:05, 15 July 2016 (UTC)

If "quadra-nt" means four, wouldn't "bint" or "dint" mean two?166.186.169.121 (talk) 12:56, 16 July 2016 (UTC)

Since your IP geolocates to North America, I assume you're not aware that the first of your suggestions already has a rather different meaning... Double sharp (talk) 16:20, 16 July 2016 (UTC)

The derivation is actually from the Latin quadrans, "quarter". A quadrant is simply a quarter of a circle. Similarly derived are "sextant" from sextans, one sixth, and "octant" from octans, one eighth. A third and a half are triens and semis, which using this analogy would give "trient" and "semiss". 86.176.84.174 (talk) 14:46, 16 July 2016 (UTC)

Although much more idiomatic English would be "semiline", which does enjoy wide use as well. Sławomir Biały (talk) 18:12, 16 July 2016 (UTC)

There was a gentleman who in Britain was popularly referred to as Osama bin Liner. 86.176.84.174 (talk) 19:31, 16 July 2016 (UTC)

Since we are adding rays and lines to the mix, how bout a straight angle? 50.79.172.89 (talk) 02:13, 18 July 2016 (UTC)

An axis? 173.13.86.113 (talk) —Preceding undated comment added 11:56, 19 July 2016 (UTC)

Integer solutions to an equation

Given $b$, how many integer solutions $(n,k)$ are there for the equation

$n^{k}-\left((n^{k})\operatorname {mod} b^{k}\right)=nb^{k}$?

For $b=10$ we have 8 solutions: $(11,26),(12,14),(13,10),(14,8),(16,6),(18,5),(32,3),(100,2)$, which is the largest number of solutions for $b\leq 50$. Is the number of solutions always less than or equal to 8? 24.255.17.182 (talk) 07:09, 15 July 2016 (UTC)

I get 10 solutions for b=54. There doesn't appear to be a simple formula since the values bounce around. But you can rewrite the equation as logn/(logn-logb) ≤ k < log(n+1)/(logn-logb). Also, n has to be between b+1 and b², so you can write the number of solutions as a summation involving logs and the floor function.--RDBury (talk) 13:11, 15 July 2016 (UTC)

PS, according to my calculations the next record breakers are b=68 with 12 solutions and b=147 with 13 solutions. --RDBury (talk) 13:37, 15 July 2016 (UTC)

Here is the number of solutions for b = 2 .. 725:

{2, 2, 3, 3, 4, 4, 4, 7, 8, 3, 2, 6, 4, 5, 5, 5, 3, 5, 5, 6, 4, 5, 6, 6, 7, 5, 
6, 5, 6, 7, 5, 5, 5, 6, 5, 5, 8, 4, 8, 6, 7, 7, 5, 8, 8, 5, 6, 4, 8, 8, 5, 6, 
10, 6, 4, 5, 5, 5, 8, 5, 5, 6, 8, 7, 3, 5, 12, 4, 6, 8, 6, 7, 11, 6, 9, 8, 7, 
5, 7, 6, 10, 7, 5, 5, 10, 7, 5, 6, 7, 7, 7, 9, 7, 6, 8, 6, 5, 12, 9, 6, 7, 10, 
7, 6, 9, 10, 4, 7, 7, 8, 8, 5, 5, 6, 6, 8, 10, 7, 6, 9, 10, 7, 11, 7, 8, 5, 7, 
9, 9, 10, 8, 6, 8, 7, 6, 5, 9, 6, 6, 10, 4, 8, 9, 8, 5, 13, 6, 8, 7, 7, 7, 10, 
5, 9, 9, 8, 5, 6, 8, 9, 10, 8, 8, 9, 9, 11, 8, 7, 9, 6, 7, 12, 6, 9, 7, 12, 7, 
11, 6, 12, 7, 8, 4, 5, 10, 5, 8, 6, 8, 9, 6, 7, 8, 8, 7, 6, 6, 9, 7, 11, 5, 5, 
9, 5, 8, 7, 9, 9, 10, 8, 9, 7, 9, 6, 13, 11, 7, 9, 7, 9, 7, 8, 10, 9, 6, 7, 6, 
9, 7, 7, 11, 9, 7, 6, 9, 9, 9, 8, 7, 8, 7, 7, 6, 9, 7, 9, 4, 5, 10, 10, 6, 5, 
9, 8, 9, 6, 8, 9, 8, 11, 9, 8, 7, 9, 8, 11, 8, 5, 7, 11, 8, 6, 12, 9, 9, 8, 5, 
5, 7, 8, 11, 7, 11, 6, 9, 8, 7, 12, 9, 6, 7, 5, 5, 9, 10, 8, 6, 11, 5, 6, 7, 
6, 7, 10, 9, 8, 7, 13, 11, 10, 6, 5, 12, 13, 4, 10, 11, 8, 8, 6, 6, 9, 10, 12, 
5, 8, 7, 5, 9, 13, 8, 10, 8, 9, 7, 9, 8, 10, 8, 8, 8, 9, 6, 14, 4, 13, 7, 9, 
9, 9, 10, 8, 6, 10, 13, 7, 8, 14, 10, 7, 10, 6, 9, 9, 15, 9, 5, 5, 6, 7, 8, 
10, 9, 8, 9, 6, 7, 7, 8, 10, 11, 6, 11, 12, 9, 8, 12, 11, 5, 9, 10, 9, 13, 9, 
7, 18, 3, 6, 8, 10, 7, 8, 5, 11, 10, 6, 10, 14, 7, 6, 9, 8, 7, 13, 8, 6, 9, 
13, 3, 8, 9, 8, 8, 8, 16, 9, 9, 11, 9, 7, 13, 6, 9, 5, 7, 7, 6, 9, 6, 10, 10, 
22, 6, 6, 6, 8, 20, 7, 12, 7, 9, 14, 8, 7, 10, 19, 8, 11, 10, 9, 8, 7, 18, 3, 
11, 8, 3, 13, 8, 11, 6, 8, 8, 7, 7, 11, 7, 13, 8, 7, 7, 6, 5, 11, 10, 15, 9, 
4, 10, 7, 8, 13, 10, 11, 11, 11, 9, 10, 11, 4, 10, 11, 4, 8, 9, 15, 13, 6, 6, 
7, 16, 5, 9, 9, 8, 11, 10, 13, 10, 12, 6, 15, 12, 8, 12, 8, 17, 12, 11, 9, 13, 
12, 7, 7, 7, 10, 15, 6, 12, 4, 4, 10, 9, 7, 9, 7, 4, 10, 12, 12, 9, 7, 6, 13, 
7, 10, 10, 11, 12, 9, 12, 9, 7, 10, 13, 14, 9, 11, 11, 8, 10, 8, 7, 7, 12, 10, 
9, 11, 9, 8, 6, 11, 11, 5, 10, 11, 8, 7, 12, 10, 10, 7, 7, 8, 11, 10, 8, 7, 9, 
12, 9, 8, 8, 12, 11, 10, 12, 7, 9, 9, 11, 8, 9, 8, 13, 11, 7, 10, 15, 8, 12, 
9, 12, 11, 8, 6, 12, 11, 8, 7, 14, 11, 11, 7, 9, 12, 7, 11, 11, 8, 9, 13, 8, 
12, 7, 9, 11, 8, 9, 7, 9, 10, 10, 11, 11, 11, 11, 10, 9, 7, 14, 7, 9, 8, 8, 9, 
5, 8, 8, 9, 8, 9, 6, 10, 11, 4, 8, 5, 9, 12, 10, 9, 11, 9, 7, 9, 11, 8, 13, 
11, 9, 12, 16, 10, 11, 11, 15, 11, 8, 13, 9, 11, 10, 10, 10, 9, 10, 9, 10, 11, 
12, 10, 7, 10, 11, 8, 10, 8, 11, 12, 4, 7, 12, 5}

JBL (talk) 16:01, 20 July 2016 (UTC)

Average length of all subintervals in an interval of length 1

We are given an interval of length 1. What is the easiest way to calculate the average length of all subintervals in that interval? Gil_mo (talk) 21:02, 15 July 2016 (UTC)

It depends on how the intervals are selected, but assuming both endpoints are independently uniformly distributed on the interval the average length is 1/3. --RDBury (talk) 21:28, 15 July 2016 (UTC)

Thanks, but how did you calculate it? Gil_mo (talk) 21:29, 15 July 2016 (UTC)

$\int _{0}^{1}\int _{0}^{1}|x-y|\ dy\ dx=2\int _{0}^{1}\int _{0}^{x}x-y\ dy\ dx=1/3$.--2406:E006:1898:1:B4C0:D261:8235:552B (talk) 23:47, 15 July 2016 (UTC)