I was looking at List of probability distributions. Which is the proper distribution for a sine or cosine wave? Specifically, I am looking for one cycle from one min value to the next min value. -- kainaw™ 04:31, 20 March 2011 (UTC)[reply]

If I understand you correctly, that's the Raised cosine distribution. -- Meni Rosenfeld (talk) 10:42, 20 March 2011 (UTC)[reply]

Or is what you mean the distribution of ${\displaystyle \sin(X)}$ or ${\displaystyle \cos(X)}$, given that ${\displaystyle X}$ is uniformly distributed in ${\displaystyle [0,2\pi ]}$? That would be (an affine stretch of) the arcsine distribution. –Henning Makholm (talk) 14:10, 20 March 2011 (UTC)[reply]

Thanks. Unfortunately, I was using vector cosine, which I thought would yield a distribution similar to using cosine on random angles. Nope. Using vector cosine on random data produces a distribution that looks like nothing on that page. It is like the raised cosine distribution with -1 and 1 raised to about half the frequency of zero. I don't think I'll be able to describe what I'm getting from this and many other functions (such as Levenshtein) on random data vs. real data. I'll just show the actual plots. -- kainaw™ 22:39, 20 March 2011 (UTC)[reply]

If your data points are isotropically distributed around the origin, then the distribution will depend on the dimensionality of the data. The distribution will always peak at zero, and the higher the dimensionality, the stronger the peak. (You're basically asking what fraction of the surface of a hypersphere lies at each possible angle θ from the pole. For high-dimensional hyperspheres, most of the surface lies near the equator.) If the data is not isotropic, it is hard to make any generalizations about what can happen. Looie496 (talk) 23:31, 20 March 2011 (UTC)[reply]

(ec) The distribution would depend on the dimension of your vector space. For dimension 3 I'd expect a uniform distribution on [-1,1] -- cf Lambert cylindrical equal-area projection -- and increasingly narrow unimodal distribution for dimension 4 and above. My best bet (supported by those two data points and a bit of geometric intuition) would be a beta distribution with parameters ${\displaystyle \alpha =\beta ={\tfrac {D-1}{2}}}$. –Henning Makholm (talk) 23:37, 20 March 2011 (UTC)[reply]

Is there a list of the longest straight lines that can be drawn across each state? --CGPGrey (talk) 12:11, 20 March 2011 (UTC)[reply]

Well, maybe. No straight line can be drawn across any state, as the states are superimposed on a sphere (or an oblate spheroid, or an even more complex shape, depending on what level of precision we use). If you want to draw straight lines, you need to define the map projection to bring the state onto a flat plane. Alternatively, if you use the sphere abstraction, you can look at great circle segments. I don't know if there is any such list, though. --Stephan Schulz (talk) 12:36, 20 March 2011 (UTC)[reply]

Is this US states or state as in country? If you have to work it out yourself, of course, then you'll know. Grandiose (me, talk, contribs) 12:38, 20 March 2011 (UTC)[reply]

if you do do it yourself, here might eb useful. Grandiose (me, talk, contribs) 12:41, 20 March 2011 (UTC)[reply]

One question you'd need to resolve is whether your lines are allowed to leave the state, since most states are not convex. Is the "longest line across Oklahoma" allowed to cross into Texas? And how about coastlines -- can a line "across" Florida go out across the Gulf of Mexico? If not, then what about freshwater bodies (Michigan!); if yes, then do borders in water count? –Henning Makholm (talk) 13:05, 20 March 2011 (UTC)[reply]

I mean state as in US state, not country. I'm not really concerned if the line travels outside the state or not, I just want to know if there are any estimates using any methodology. --CGPGrey (talk) 13:46, 20 March 2011 (UTC)[reply]

Certainly allowing it to travel outside the state makes the problem simpler. Michael Hardy (talk) 03:36, 21 March 2011 (UTC)[reply]

The convex hull of the state is a polygon having the same longest straight line as the state itself. Bo Jacoby (talk) 05:53, 21 March 2011 (UTC).[reply]

Except that it's not generally a polygon. When a state is bounded by a river, as often happens, then a curved part of the river in some cases forms part of the boundary of the convex hull. Michael Hardy (talk) 17:47, 21 March 2011 (UTC)[reply]

OK, take two parallel lines passing opposite sides of the state on a map and move them toward each other until they bump into the state. Look at how far apart they are then. The rotate the direction in which the two lines go. For each direction, look at how far apart they are when they're as close together as they can get while just touching the state but not passing through it. Where that distance is largest, you have the answer. Michael Hardy (talk) 17:50, 21 March 2011 (UTC)[reply]

The convex hull is the intersection of all half-planes containing the state. It is approximated to sufficient precision by a convex polygon. Bo Jacoby (talk) 07:35, 22 March 2011 (UTC).[reply]

Hello. When we want to consider the limiting value of √x as x is near 0, we usually have to write

${\displaystyle \lim _{x\to 0^{+}}{\sqrt {x}}=0\quad {\text{not}}\quad \lim _{x\to 0}{\sqrt {x}}=0,}$

because if we approach 0 from the left, √x is undefined. However, this is only a problem if x can be any real number. The more general definition given here, "for functions defined on subsets of the real line", allows us to restrict x a bit more. Specifically, √x is defined on the non-negative reals, so if we adopt this more general definition and restrict x to the non-negative reals, could we not discard the left, negative approach and simply say

${\displaystyle \lim _{x\to 0}{\sqrt {x}}=0?}$

But looking more closely at the article, it says the limit point c must be within some open interval (a,b) on which the function is defined (though perhaps not at x = c). In the case of √x, there is no such (a, b) for c = 0. My questions: Does this mean we can never assign a value to

${\displaystyle \lim _{x\to c}f(x)}$

when c is an endpoint of f's domain, but must always write c⁺ or c⁻? If so, why do we make this exclusion even in the general definition? The open interval construction is confusing in general to me – why not replace (a, b) with dom(f)? This is a very long-winded post, so please ask me to clarify if I've been unclear at any point. Thanks in advance. —Anonymous Dissident^Talk 12:55, 20 March 2011 (UTC)[reply]

I think this is one of the areas where different authors make different choices and nobody cares very much because the different definitions agree in the cases that matter. It is not as if the confusion makes it easy to write down something that could have two different valid meanings depending on which definition one uses. Among mathematicians who trust each other to understand how limits work, the real rule is that anything goes, as long as the meaning is clear. However, a teacher who needs to deal with the possibility that the student has misunderstood something will often demand more rigor from students, in order to demonstrate that they can do the disambiguation that working mathematicians leave implicit.

(One reason not to bother with the c⁺ or c⁻ is that this notation is in any case insufficient to deal with similar situations in more abstract topological spaces. So a more abstract definition would likely just require that the domain of f contains points different from c in any neighborhood of c).

The downside of this situation, from a Wikipedia standpoint, is that our articles easily become internally inconsistent because different editors work from sources that use different conventions. –Henning Makholm (talk) 13:49, 20 March 2011 (UTC)[reply]

Also of course [0, ∞) is open when the domain is the non-negative reals. Dmcq (talk) 13:54, 20 March 2011 (UTC)[reply]