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April 27

choice function

In set theory, is it the case that the " choice function " itself is independent of any value in its range that is selected by that choice function ? In other words, is it that the choice function cannot be changed as a result of its own application? Intuitively, it would seem to be independent. But, what is the proof? 206.74.74.42 00:25, 27 April 2007 (UTC) Willie[reply]

I'm not sure what you mean here. A choice function

f

for a collection

X

of nonempty sets is simply a function

f\colon X\to \bigcup X

such that for all

S\in X

,

f(S)\in S

. What do you mean when you ask if it can be changed as a result of its own application? Algebraist 13:53, 27 April 2007 (UTC)[reply]

I should have asked: is it possible for a choice function over a set of sets, to be a function of an element within the domain ? Thank you . 206.74.74.42 03:07, 28 April 2007 (UTC) Willie[reply]

Could you clarify this further? Using the notation introduced by Algebraist above, if S₀ is some element of the domain X, what would it mean, specifically, for f to be "a function of" S₀? For a concrete example, take X = {{0, 1}, {0, 2}, {1, 2}}, f(S) = (ΣS)–1, and S₀ = {0, 2}. So then f(S₀) = 0+2–1 = 1. Now in this example, is f a function of S₀? If not, why not, or inasmuch as what not? --Lambiam ^Talk 04:37, 28 April 2007 (UTC)[reply]

Sums of cubes

Hi - I was mucking around with some figures and I noticed something odd - I was wondering what the proof of it was (if it's simple...):

(∑ positive integers to n)^2 = (∑ cubes of positive integers to n)

For instance: (1+2+3)^2 = 36, as does 1^3 + 2^3 + 3^3.

Thanks in advance, Grutness...wha? 01:19, 27 April 2007 (UTC)[reply]

See squared triangular number. —David Eppstein 01:50, 27 April 2007 (UTC)[reply]

More average than average

Here's a vaguely statistical, vaguely philisophical question. Is it possible to be more average than the average person (in terms of that person's average-ness)?--The Fat Man Who Never Came Back 01:57, 27 April 2007 (UTC)[reply]

In a manner of speaking, I suppose. There are very few people who are the average height, average age, average intelligence, have the average income and the average wealth, etc. So, if that happens to be you, you might say you are "more average than average". This wouldn't be mathematically correct, but that's OK. This vaguely reminds me of the town where "all the children are above average". :-) StuRat 03:55, 27 April 2007 (UTC)[reply]

You could say that if you were above 75% of the population you were above the average who are above the average (the first average being 50% the next at 75%.--Dacium 04:22, 27 April 2007 (UTC)[reply]

Yes, but my question relates more to whether there can be an average level of average-ness... perhaps this is synonymous with the concept of standard deviation from the mean, within a group of people... but I'm not sure.--The Fat Man Who Never Came Back 05:18, 27 April 2007 (UTC)[reply]

(edit conflict)

I think it implies some rather different meanings of the word "average"

usage of some type of measure or comparison of "averageness".
an average over that measure (does that make sense?)
the standard notion of average.

The question is how to define the first two notions.

I think one way to do it would be.

Find an average point X (standard notion of average)
Calculate a distance B of a certain point Y of interest from the average point
Find the average distance C of all points from the average point. Is B less than C? Then Y is more average than average.

But we didn't quite specify what the distance metric was or how we found average distances. That's more up to you I think. It would seem they ought to be related to the way the original average was calculated, though. BTW, it seems this concept is related to the concept of standard deviation. (as I see you've also noted). Root⁴(one) 05:28, 27 April 2007 (UTC)[reply]

You need to be more specific about your definition of "average". Firstly, there are several different types of "average", many of which are discussed at the article of the same name, such as the (arithmetic) mean, median, mode, etc. Then you might be able to argue that, for example, while the "average" man is found by the mean, you are in fact in the median position which is a "better" measure of central tendency (for a given value of "better"), or that your statistic occupies both the midrange position and the mode of the data, and so is the "average" value by twice as many methods as the "average" man. Both arguments appear pretty faulty to me though so I wouldn't recommend using them to call yourself the "average man" though. Confusing Manifestation 06:50, 27 April 2007 (UTC)[reply]

It might work better if you consider a sample of people from a population, rather than just a single person. Say your poulation is all the people in a country and your sample is the people in a particular town. Say your population has mean M and standard deviation σ. Now work out the sample mean and population m and s say. It could happen that the town has a strange demographic favoring normal sorts, in which case s will be less than σ. You can use the F-test (I think) to compare the deviations to see if its significant. --Salix alba (talk) 14:19, 27 April 2007 (UTC)[reply]

Whimsically, consider two examples. In a class of 51 students, 25 are female and 25 are male; the one remaining student has a medical problem. Among the mothers of those students, each has an average of 1.5 children; one exceptional mother is understandably distraught.

Too extreme? OK, suppose the student heights partition according to sex, with no overlap between the 25 males and 25 females mentioned; one remaining student has exactly the average height. (See bimodal distribution.) Or suppose we roll three dice; the average outcome is 10.5, but that would be a startling roll. (See discrete probability distribution.) --KSmrq^T 15:15, 27 April 2007 (UTC)[reply]

On a random note, that reminds of that quote from "Animal Farm": "All animals are equal, but some animals are more equal than others." - Akamad 08:40, 28 April 2007 (UTC)[reply]

If the number of toes people have is of interest, the mean (arithmetic, geometric, harmonic, ...) for any large group will not be exactly 10, as some will have been born with extra toes and some will have been lost. But 10 will overwhelmingly be the mode, so all those with this number will be more average than the average, if a suitable choice of "average" is defined on each occasion. → Semiable 10:57, 29 April 2007 (UTC)[reply]

Quarternions as applied to computer graphics

The entry for Quaternions mentions its use in 3D graphics, but fails to go into any detail. I've seen the term mentioned on a wishlist for the next version of some 3D software. Why is it good thing and how is it different from how 3D software works now? --24.249.108.133 21:46, 27 April 2007 (UTC)[reply]

Did you look at Quaternions and spatial rotation#Quaternions versus other representations of rotations? —David Eppstein 22:07, 27 April 2007 (UTC)[reply]

Computer graphics has used quaternions for rotations over the last two decades, robotics was using them before that, and the space program began using them even earlier. For each discipline their behavior and benefits have been defined and debated. If you want a computer graphics perspective, try the quaternion entry of the news:comp.graphics.algorithms FAQ. The 1948 NASA technical report by Stuelpnagel is also well worth reading.

For a mathematician, the relationship between quaternions and rotations in 3D is simple and compelling. Unfortunately, it draws on advanced topics, including Lie groups, universal coverings, topology, the Atiyah-Singer index theorem, and Riemannian geometry. Briefly, 3D rotations form a Lie group called SO(3). We can compose two rotations to get another (giving a group structure); and we can measure distances between rotations (giving geometric and topological structures). Quaternions efficiently give us both, with unique advantages. Unit quaternions form a sphere, S³, in the 4D space of all quaternions. Each pair of opposite quaternions maps to a 3D rotation. The distance between two quaternions corresponds to the distance between their rotations; the product of quaternions corresponds to the composition of rotations.

We must use at least three numbers to describe a 3D rotation, but we can prove that using only three numbers is guaranteed to have problems. Note especially that 3D rotations are not a 3D vector space — a 1+1+1 split, nor do they admit a 2+1 split like aiming and rolling. Quaternions use four numbers, in the manner of homogeneous coordinates, which is the minimum required to be trouble-free. But it is really the special nature of the correspondence that makes them so appealing.

You may enjoy searching the web; it is full of information (and, alas, misinformation) about quaternions. --KSmrq^T 04:57, 28 April 2007 (UTC)[reply]

As said above, quaternions are a very effective way of representing 3D rotations. The benefit of a quaternion is that you can rotate something about any axis. A lot of time in software, the programmer is quick to implement a scheme like Euler angles or Spherical coordinates or Tait-Bryan angles (yaw, pitch, roll), and very often these are suitable for specific purposes. For some situations though, these schemes become awkward (see Gimbal lock) or are limited to certain views. The quaternion, however, is very general and can cover every possible view of the object without these problems. (The implementation of quaternions is not too complicated, but it's usually a little more work than quickly throwing in some Euler angle controls.) - Rainwarrior 17:11, 28 April 2007 (UTC)[reply]

Mea culpa, I forgot to say that physics used quaternions earlier still. Two pivotal instances are Maxwell's creation of his famous equations for electricity and magnetism, and Pauli's spin matrices for quantum mechanics. In fact, quaternions are the simplest interesting example of a spin group, Spin(3).

As for viewing controls, Arcball is easy and appealing. Treat screen points as vectors to (the front half of) a unit sphere. A mouse drag from v₀ to v₁ produces the rotation quaternion q = v₁v₀, whose scalar component is the negated dot product −v₁·v₀ and whose vector components are the cross product v₁×v₀. (This is essentially a double reflection; see here for a planar illustration.) To play with a graphically captivating example, visit the Stanford Real-Time Programmable Shading Project and download an executable. The armadillo is a good object to play with; use the right-side mouse button to select an object and a surface. Be sure to press "c" to allow free rotation, then later try holding down SHIFT or CTRL to impose object-space or screen-space constraint. --KSmrq^T 18:30, 28 April 2007 (UTC)[reply]