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May 6

pearsons r and F test

When computing an F test and Person's r, if significance testing is conducted for both tests using the same set of data, why is it possible to achieve contrary results? please give the answer

Information theory and infinities

(Note that all of this is from a completely lay perspective): where would I go to read about the parts of information theory that deal with "infinite amounts" of information? Is that even a thing? For example, if I have some real number a (unspecified beyond the fact that it is real), then revealing that it is, say, a trillion has narrowed down my a-space from being uncountably large to being a single number. Using the traditional definition of information (base-2 log of the size of the old possibility space over the size of the new one), then I have transmitted infinite information (of course you couldn't actually construct a real-life function that outputs any real number with uniform probability, but I think the idea is clear). --122.150.246.242 (talk) 07:49, 6 May 2012 (UTC)[reply]

I think you're confusing information that a number is a real with knowing the number in "full detail", i.e. all digits. The former is not "infinite amount of information" anymore than linking to pi would make this page to never end loading :-) 86.104.57.242 (talk) 11:11, 6 May 2012 (UTC)[reply]

I do know all the digits. There's a 1, then 12 zeroes, then a decimal point, then infinitely many zeroes. A trillion, like I said. I narrowed down "any real number" to a particular real number, not "one of ten digits" to a particular digit. Put it this way: if I say, "a can be any real number, with uniform probability", then what is the entropy of a? --122.150.246.242 (talk) 11:45, 6 May 2012 (UTC)[reply]

Differential entropy - the entropy of a continuous distribution - is relevant. If X and Y are uniform and normal variables of the same mean and variance, then transmitting either of them exactly requires infinite information, but transmitting Y still requires a few bits more than X. So it's the difference in information content that is studied.

This can also be considered in terms not of transmitting the value of X exactly, but rather of the limiting process of transmitting it with arbitrarily high accuracy. The amount of information depends on the accuracy, but for any given accuracy, different variables will require different amounts of information, and this is the difference in their entropy. -- Meni Rosenfeld (talk) 12:09, 6 May 2012 (UTC)[reply]

If you can transmit an arbitrary real number, then you can indeed transmit infinitely much information. In some sense it's true that every real number (not just random-looking ones) has infinite information content; if you choose a real number from a uniform distribution on [0,1] and it turns out by random chance that it comes out to be exactly zero point five zero repeating, then yes, that event encodes infinitely many bits of information. Of course the probability of the number coming out to be exactly that is zero (same as the probability of it being any other particular number). --Trovatore (talk) 00:45, 8 May 2012 (UTC)[reply]

Is there an "abelian version" of the free product?

I'm thinking of an operation that would glue together two structures (say two groups G1, G2) such that any pair with one element from G1 and one from G2 would commute, e.g. the free abelian group of rank 2 would be the result of this operation applied to two free abelian groups of rank 1. If not for groups, is it possible to define it for categories/semigroups etc.? Thank you for your time. 86.104.57.242 (talk) 10:20, 6 May 2012 (UTC)[reply]

Maybe I'm misunderstanding, but doesn't the direct product do this? Staecker (talk) 12:58, 6 May 2012 (UTC)[reply]

It sounds like you want the result to be abelian, even if the components aren't. Then the best you'll find for groups will be to first abelianize, then take the direct product. But the abelianization of a group might be trivial, so you'll probably find this unsatisfying.--130.195.2.100 (talk) 00:17, 7 May 2012 (UTC)[reply]

You can take the free product of G1 and G2 and then force elements of G1 to commute with elements of G2 without abelianizing G1 and G2 themselves, but like Staecker said this is exactly the same as the direct product. Rckrone (talk) 00:42, 8 May 2012 (UTC)[reply]

Yeah, this is what I had in mind. Thanks. 86.104.57.242 (talk) 09:29, 8 May 2012 (UTC)[reply]

This notion extends to a class of groups called graph products (an unfortunately vague name). They are neatly defined by choosing a (finitely generated) group for each vertex of a (finite) simplicial graph, taking their free product, and quotienting out to get the commuting relations you want, whenever the groups' vertices are joined by an edge in the graph. If your graph is complete, you just get a direct product as mentioned above. However, you can get much more interesting groups. For instance, the class of graph products contains all right-angled Artin groups, which are interesting in and of themselves. Icthyos (talk) 13:18, 8 May 2012 (UTC)[reply]

Umm, unless you have more relations than commutations between some generators (those connected by the graph edges to be precise), what you describe are exactly the right-angled Artin groups. Those are known by many other names indeed; I've added some to the article. 86.104.57.242 (talk) 19:08, 8 May 2012 (UTC)[reply]

Ah, I see you're talking about an external version of that (which makes the RAAGs constructed by extending with only a single-vertex group at a time). [1] I'll add this too. 86.104.57.242 (talk) 19:12, 8 May 2012 (UTC)[reply]

You only get a right-angled Artin group if every vertex group is a copy of the integers (i.e. free of rank one), yes. You get right-angled Coxeter groups if every vertex group is of order two. You can use any finitely generated for any of the vertices, though, and get many different graph products. In my original statement of the definition, the relations from the vertex groups were inherited by the free product, to which the desired commutators were added. (Glad to see someone adding to the RAAG article!) Icthyos (talk) 20:20, 8 May 2012 (UTC)[reply]

Well, I had misunderstood that def. I've corrected the statement in the RAAG article now. Thanks. 86.104.57.242 (talk) 22:51, 8 May 2012 (UTC)[reply]

Meaning of a symbol in a specific context

In [2] on page 1, theorem 2 says

If p is an odd prime number, then $\left(-1\right)^{\tfrac {p-1}{2}}.{\mathit {C}}_{\tfrac {p-1}{2}}\equiv 2{\pmod {p}}$ .

What does the dot before the ${\mathit {C}}$ stand for? -- Toshio Yamaguchi (tlk−ctb) 12:14, 6 May 2012 (UTC)[reply]

Looks to me like multiplication, probably a typo. "Proof 1 of Theorem 2" proves it as if it were multiplication. Staecker (talk) 12:47, 6 May 2012 (UTC)[reply]

Okay, thanks. Looking at some other formulas in the paper it seems this is a result of the typeface used in the paper. -- Toshio Yamaguchi (tlk−ctb) 12:58, 6 May 2012 (UTC)[reply]

It's a typical LaTeX document. The author seems to have used "." instead of "\cdot". Staecker (talk) 12:59, 6 May 2012 (UTC)[reply]

The author ment to write:

If 2n+1 is a prime number, then $\left(-1\right)^{n}{\mathit {C}}_{n}\equiv 2{\pmod {2n+1}}$ .

Bo Jacoby (talk) 15:18, 6 May 2012 (UTC).[reply]

vector identities

Use identities to show that $curl(r\times (a\times r))=3r\times a$ where $r=xi+yj+zk$ and $a$ is a constant vector? Which identities do I use? Widener (talk) 17:55, 6 May 2012 (UTC)[reply]

Green's theorem

Consider the integral $\int _{C}{\frac {-y^{3}dx+xy^{2}dy}{(x^{2}+y^{2})^{2}}}$ where $C$ is the ellipse $x^{2}+4y^{2}=4$ . Explain why Green's theorem cannot be applied to the region interior to C to evaluate the above integral, and show that a blind application of this theorem would lead to the incorrect result $\int _{C}=0$ . I simply don't know why Green's theorem doesn't work. It looks like it should. As for applying Green's theorem, I get $\int \int {\frac {(x^{2}+y^{2})^{2}y^{2}-4x^{2}y^{2}(x^{2}+y^{2})+3y^{2}(x^{2}+y^{2})^{2}+2y^{4}(x^{2}+y^{2})}{(x^{2}+y^{2})^{4}}}dxdy=\int \int {\frac {-4x^{2}y^{2}+4y^{2}(x^{2}+y^{2})+2y^{4}}{(x^{2}+y^{2})^{3}}}dxdy$ which is a monstrosity. Can someone evaluate this (and is what I've done so far correct)? Widener (talk) 18:28, 6 May 2012 (UTC)[reply]

The problem is that the functions are not defined in (0,0) which is in the region bounded by the curve. The way to handle this problem is to draw a circle around (0,0), and since this is a conservative field in the region between the curves, the integral on the outer curve is equal to the one on the inner curve (which is simpler).

You have a missing negation in evaluating the integral, the two derivatives need to cancel each other out. The integral you wrote is not only unwieldy, it doesn't converge. -- Meni Rosenfeld (talk) 20:20, 6 May 2012 (UTC)[reply]

Do the derivatives cancel? Now I get

\int \int {\frac {(x^{2}+y^{2})^{2}y^{2}-4x^{2}y^{2}(x^{2}+y^{2})+3y^{2}(x^{2}+y^{2})^{2}-2y^{4}(x^{2}+y^{2})}{(x^{2}+y^{2})^{4}}}dxdy=\int \int {\frac {-4x^{2}y^{2}+4y^{2}(x^{2}+y^{2})-2y^{4}}{(x^{2}+y^{2})^{3}}}dxdy=\int \int {\frac {2y^{4}}{(x^{2}+y^{2})^{3}}}

. The derivatives don't have to cancel in order for the integral to be zero. Widener (talk) 20:51, 6 May 2012 (UTC)[reply]

The derivatives don't have to cancel, but in this case they do. This kind of problems is usually given with a conservative field. Also note that your latest integrand is nonnegative, so the integral can't be 0.

You have another derivative mistake in addition to the sign; it should be

\int \int {\frac {(x^{2}+y^{2})^{2}y^{2}-4x^{2}y^{2}(x^{2}+y^{2})+3y^{2}(x^{2}+y^{2})^{2}-4y^{4}(x^{2}+y^{2})}{(x^{2}+y^{2})^{4}}}dxdy

. -- Meni Rosenfeld (talk) 04:14, 7 May 2012 (UTC)[reply]

Second order ODEs

Hi folks. Can anyone supply some nice (applied) problem sheets for second order, linear, homogeneous ODEs? I'm trying to write a homework sheet, but I want to make the questions seem relevant. There's only so many times I can mention springs (simple and damped harming motion) before it wears thin. Can anyone suggest a link to some homework problems? — Fly by Night (talk) 22:18, 6 May 2012 (UTC)[reply]

Have you seen the examples in the article on oscillation? Note that the second order linear homogenous ODE with constant coefficients also describe the onset of explosively increasing oscillations. Bo Jacoby (talk) 06:07, 8 May 2012 (UTC).[reply]

Yes I have, have you? There is an extensive list of subject-specific examples which are inappropriate for a mathematics course. None of the ones I looked at contained any ODEs whatsoever. (Perhaps some do, but I couldn't find any.) — Fly by Night (talk) 01:52, 13 May 2012 (UTC)[reply]

All the oscillations are solutions to the ODE, even if it is not mentioned. Bo Jacoby (talk) 07:35, 13 May 2012 (UTC).[reply]