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April 5[edit]

What's the name for this?[edit]

I came up with a curious mathematical structure but can't find anything about it, but I guess it has already been studied by others, and there must be a name for it.

Let $(X,d)$ be a metric space that also has a probability measure $(X,{\mathfrak {B}},\mu )$ . Define a kind of distance on measurable sets as $\delta (A):=\int _{x\in A,y\in A}d(x,y)d\mu$ ., and $\beta (A):=\inf {\big \lbrace }\sum \delta (A_{i})|\bigcup _{i=1}^{N}A_{i}=A{\big \rbrace }$ .

The function $\beta$ is intended to be low when the points of A are strongly clustered.

Whats the name for this structure? 95.113.84.102 (talk) 10:48, 5 April 2014 (UTC)[reply]

Zero? Sławomir Biały (talk) 11:40, 6 April 2014 (UTC)[reply]

Yes for countable sets. For countable sets define

\beta _{N}(A):=\inf {\big \lbrace }\sum \delta (A_{i})|\bigcup _{i=1}^{N}A_{i}=A{\big \rbrace }

with fixed N. 93.132.27.1 (talk) 11:44, 6 April 2014 (UTC)[reply]

I'm not sure what role countability has. For the unit interval with the uniform measure, it's zero. For the real line with a normal distribution, it's zero. For a delta measure, it's zero. I think you would need a very pathological metric space for this to be non-zero. (Here I am assuming that the above "inf" is taken both with respect to coverings of A and the number N of sets in the cover. If you only cover with a fixed number of sets, then I have no idea.) Sławomir Biały (talk) 12:25, 6 April 2014 (UTC)[reply]

OK, so the above definition is probably not what I wanted it to be. What I wanted was some function that tells me how strong a set of points is clustered, and possibly also a way to tell the number of clusters. Standard deviation tells me how strong points are concentrated about the center which is gained by averaging the positions. but in case my points are the superposition of 2 or more normal distributions I can't see that from standard deviation. 93.132.27.1 (talk) 12:29, 6 April 2014 (UTC)[reply]

It probably depends on the specific application. My first idea would be to consider something from electrostatics: think of the probability distribution as charge. The total energy is greater for more concentrated charges. In good cases, this is given as an explicit integral of a potential (which, in turn, is also given as an explicit integral of the charge distribution). But precise formulas might not exist at the specified level of generality though. Sławomir Biały (talk) 23:06, 6 April 2014 (UTC)[reply]

Numbers whose prime factors include only 2 and 3[edit]

How we call this ensemble? What's the best algorithm that test if a given number is in that ensemble or not? Thanks for your answers. Hunsu (talk) 11:57, 5 April 2014 (UTC)[reply]

That would be {-6,0,6} . Unless you mean something else? Like numbers whose prime factors include only 2 and 3? Staecker (talk) 12:03, 5 April 2014 (UTC)[reply]

It's all numbers of the form

2^{a}3^{b}

for a,b non-negative integers. The best computer algorithm would probably exploit the binary representation of the number, which presumably is already available on the computer hardware. Truncate off all trailing zeros and if the remaining number is a power of 3, then the number you started with must be that power of three times a power of two (the power of two corresponding to the number of digits that you truncated on the first step). Sławomir Biały (talk) 12:07, 5 April 2014 (UTC)[reply]

I changed my question, it was a mistake. Hunsu (talk) 12:11, 5 April 2014 (UTC)[reply]

It has the snappy name of A003586. See also Smooth number. AndrewWTaylor (talk) 19:39, 5 April 2014 (UTC)[reply]