Wikipedia:Reference desk/Archives/Mathematics/2013 September 10

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September 10

Twisted Rack Cohomology

I saw a reference, in a paper on entropic algebras, about "twisted rack cohomology" with a reference to a paper I don't have access to, "Twisted quandle homology theory and cocycle knot invariants". Our article Racks and quandles mentions a quandle cohomology, but doesn't go into any details- does anyone have, or know where I can find, details on quandle cohomology and what "twisted" refers to? Does it have any relation to Twisted K-theory?Phoenixia1177 (talk) 05:25, 10 September 2013 (UTC)

Never mind, the paper is freely available! I don't know how I missed that, it's the first link that comes up in Google.Phoenixia1177 (talk) 05:36, 10 September 2013 (UTC)

If someone wants to remove this, please feel free. I'm not sure how to.Phoenixia1177 (talk) 05:36, 10 September 2013 (UTC)

nLab also has a nice discussion of twisted cohomology. --Mark viking (talk) 06:16, 10 September 2013 (UTC)

Thanks for the link:-) That's an interesting site, by the way:-)Phoenixia1177 (talk) 06:20, 10 September 2013 (UTC)

Power rule in calculus

Can someone explain to me the differences in the layout of the power rule? I have two versions of it but I don't know how or why they differ.

They are:

${\frac {d(x^{n})}{dx}}=n\cdot x^{n-1}$

And:

${\frac {d(u^{n})}{dx}}=n\cdot u^{n-1}{\frac {du}{dx}}$

I guess I don't understand the significance of the letters and what they represent because, before the equal sign, the first looks identical to the second. The only difference being the letters that are being used. Thanks, Dismas|^(talk) 05:46, 10 September 2013 (UTC)

No, if you changed the letter from x to u, you would get ${\frac {d(u^{n})}{du}}=n\cdot u^{n-1}$. Note that the left hand side has a u in the bottom, not an x as in the left hand side of your second expression. Looie496 (talk) 05:51, 10 September 2013 (UTC)

Usually, in textbooks I've seen, u denotes any differentiable function of x. The second case includes provisions for the chain rule.--Jasper Deng (talk) 05:53, 10 September 2013 (UTC)

There is a difference between the two left hand sides, both have a dx on the bottom, it's x -> u in the numerator only. In that case, it tells us that we are taking the deriv. of u with respects to x- so the first case is the second with u(x) = x.Phoenixia1177 (talk) 05:59, 10 September 2013 (UTC)

Jasper, I understood your response the best. I see what is meant now. Thank you! Dismas|^(talk) 06:21, 10 September 2013 (UTC)

Resolved

Triangular Prism Die.

I asked a more general question that got bogged down a few days ago, and I thought I'd ask it here. If a triangular prism of sides 1 (the side of the triangle) and x (the sides connecting the triangles) is thrown like a die, what is x so that all five sides are equally likely to come up?Naraht (talk) 14:41, 10 September 2013 (UTC)

I'm not certain how to go round this, but if you just want a die that gives 1 in 5 chances then you can use an icosahedron which has the advantage of being visibly fair. Dmcq (talk) 15:07, 10 September 2013 (UTC)

Yes, or a pentagonal or dodecagonal trapezohedron or pentagonal or dodecagonal bipyramid or even a Zocchihedron. I know there are other options for a die for a 1 in five chance (or even reroll a d6 if it comes up a 6). That's why I tried to ask it *very* specifically.Naraht (talk) 15:49, 10 September 2013 (UTC)

Here is a patent for such a die, with precise measurements included. Whether is it a fair die is another matter. Some say that the faces should have equal area for fairness, others say equal solid angle. But the position of the center of mass surely plays a role in the physics, too. --Mark viking (talk) 16:33, 10 September 2013 (UTC)

Yes, the fairness is claimed but not proven. Also, even if 1-5 are equiprobable, they cannot sum to 1, because the die could also land on the beveled edges labeled 22, 24. It would be unlikely, but the chance is still not zero :) SemanticMantis (talk) 18:54, 10 September 2013 (UTC)

The patent says that fairness was determined experimentally, in other words it wasn't derived using math. There is no guarantee that if the die is rolled on a a different surface or made of different material then the probabilities would not change. --RDBury (talk) 20:09, 10 September 2013 (UTC)

My own intuition is this could only be determined experimentally. Having done simulations of such problems even such a simple case like this is chaotic and so non-deterministic. It's simple enough though that it could be simulated on a computer, letting you do millions of trials and so estimate x with a high degree of precision.--JohnBlackburne^words_deeds 20:22, 10 September 2013 (UTC)

It can't land on the beveled edge "22" (or the beveled edges between the "22" sides) since it would not be stable (it would fall on the triangular side), OTOH, *theoretically*, the die could stand on the "24" side. But I wonder if what required the testing was the beveled sides.Naraht (talk) 21:26, 10 September 2013 (UTC)

From reading the spec, it appears that the bevelled edges are just to "make the shape of the present invention 10 interesting and ... capture and radiate illumination which enhances the lustrous and sparkling gem-like appearance", rather than to enhance the rolling (though it's entirely possible that this is an attempt by the patent author to mislead those trying to replicate the design and inadvertently make his own patent easier to invalidate for insufficiency). For those worrying about making their own, the patent has expired due to failure to pay renewal fees 4 years ago, so it's well beyond ressurection (in the US at least, I've not checked for related patents in other jusridictions). I could rant about those claims all day... the main claim is OK, the dependents are awful... MChesterMC (talk) 08:34, 11 September 2013 (UTC)

Neither fairness nor unfairness can be proven experimentally. Fairness and probability are matters of faith. I believe that if the five sides have equal solid angle as seen from the center of mass, then the die is fair. I may change my mind when confronted with theoretical arguments or experimental facts, but even that does not prove unfairness. Bo Jacoby (talk) 08:52, 11 September 2013 (UTC).

I'm pretty certain that's wrong. If you have a hexagon and roll it along it has 1/6 chance of coming up on each side. If you cut it down to a rectangle of 1 x sqrt(3) then by that argument it should end up on the short ends one third of the time. I'm pretty certain it will now end up on one of the long sides more than 2/3 of the time even though they take up only 2/3 of the angle and less than 2/3 of the perimeter. Dmcq (talk) 13:41, 11 September 2013 (UTC)

Do you have any argument besides being 'pretty certain'? By the way, I did not say that double solid angle means double probability, but only that equal solid angle means equal probability. Are you pretty certain that's wrong? Do you have suggestions for improvement? Bo Jacoby (talk) 19:33, 11 September 2013 (UTC).

(Assuming you meant any n and not just 5:) I don't think that there is a separate "equal solid angles" rule because I can subdivide all the faces of a die into many (almost) equal-solid-angle pieces. It certainly can't just be proportional to solid angle, because a face can be impossible (by having the center of gravity project outside it) and yet still subtend solid angle from the CG. Even if we reject such dice, I think the answer requires a precise definition of the "throwing the die" process in order to define probabilities and thence fairness. Certainly, throwing a die in a very viscous fluid will tend to follow the solid angles (because there is no rotation, so all that matters is the face through which the weight vector points). --Tardis (talk) 13:02, 12 September 2013 (UTC)

Let's stick to the problem at hand, the triangular prism of sides 1 (the side of the triangle) and x (the sides connecting the triangles). Do you have a better suggestion for constructing a fair die, or are you merely providing negative criticism? Bo Jacoby (talk) 13:42, 12 September 2013 (UTC).

It seems that even if you manage to compute x such that the die is actually fair, you can't convince people that this is the case, and so it is pointless after all. Bo Jacoby (talk) 16:42, 16 September 2013 (UTC).

Question about notation for partial derivatives and integrals

If ⌈x⌉ means the ceiling of x, i.e. the smallest integer greater than or equal to x, then why does ∂f/∂xi mean the partial derivative of f with respect to xi, where f is a function on (x1, …, xn)? Also, does ∫ab f(x) dx means the signed area between the x-axis and the graph of the function f between x = a and x = b? --Marticleeë (talk) 17:58, 10 September 2013 (UTC)

I don't understand the "if ... then" part of your question, but if you mean $\int _{a}^{b}\!f(x)\,dx\,$ for the "also" part then yes, see Integral. Dbfirs 21:56, 10 September 2013 (UTC)