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September 28

Polynomials and symmetries

For each of the point groups in three dimensions, can I assume that there exists a polynomial in x,y,z with that symmetry (and none higher)? How about three independent ones? —Tamfang (talk) 02:12, 28 September 2010 (UTC)[reply]

On second thought, my purpose would be satisfied by two or three functions each of which has higher symmetry, but whose combination as a 'vector' does not. —Tamfang (talk) 02:16, 28 September 2010 (UTC)[reply]

There are many interesting algebraic surfaces which have the same symmetries as polyhedra see for example [1]. A few more can be found at y own site [2].--Salix (talk): 09:37, 28 September 2010 (UTC)[reply]

Thank you, but I'm more interested in the chiral groups and S_n. —Tamfang (talk) 23:48, 4 October 2010 (UTC)[reply]

Generalization of cut and choose

What should be the rules of a game involving n people being allowed to cut and/or choose pieces of a cake, such that the optimal strategy for each person leads to the cake being divided equally? Count Iblis (talk) 03:51, 28 September 2010 (UTC)[reply]

Good question. I don't know the answer, but here is a thought-starter. All participants sit in a circle with the cake in the middle. In turn, each person cuts a segment out of the cake; then hands the segment to the person on the left; and then hands the knife to the person on the right. What are the pros and cons of that set of rules? Dolphin (t) 04:03, 28 September 2010 (UTC)[reply]

If one person does all the cutting and then gets to choose last, their optimal strategy is to cut the cake as evenly as possible regardless of the number of people. Rckrone (talk) 04:15, 28 September 2010 (UTC)[reply]

But my notion of equal shares might be wildly different from yours, if the cake is not homogeneous; so only the cutter and the first chooser are assured of fair shares. —Tamfang (talk) 06:38, 28 September 2010 (UTC)[reply]

The Divide and choose article doesn't mention it, but I seem to recall reading something over a decade(?) ago (in Scientific American, perhaps? - or maybe it was Discover Magazine) about a generalization of the process. All I recall is that with more than two people, the algorithm becomes unsuitable for practice, necessitating many cuts and recuts of the slices. - Also note that the Wikipedia article mentions "The divide and choose method does not guarantee each person gets exactly half the cake by their own valuations", simply that no one will get the "worst" piece by their own evaluation. (this is for the general case where all pieces of the cake are not identical - half chocolate/vanilla, frosting flowers, etc.) -- 174.31.192.131 (talk) 04:22, 28 September 2010 (UTC)[reply]

There are some methods described at Fair division#Many players. In particular, perhaps something described at Proportional (fair division)#Many players is what you're looking for. —Bkell (talk) 04:59, 28 September 2010 (UTC)[reply]

Possibly relevant: http://www.math.hmc.edu/~su/fairdivision/ —Tamfang (talk) 06:36, 28 September 2010 (UTC)[reply]

Cutting cake strategies is serious business, see e.g. [3]. Most of the papers seem to cite Evans and Paz [4] for the best upper bound.—Emil J. 11:54, 28 September 2010 (UTC)[reply]

Thanks everyone! I had heard about a complicated solution when there are more than 2 players a long time ago, but I was unable to find more about this (apparantly because you need to search for "fair division"). Count Iblis (talk) 18:29, 29 September 2010 (UTC)[reply]

Evaluating the sum of the sums of geometric progression

$\sum _{k=1}^{n}\sum _{r=1}^{k}z^{r}$ ?
$\sum _{k=1}^{n}{\frac {z(1-z^{k})}{1-z}}$ ?--Wikinv (talk) 05:08, 28 September 2010 (UTC)[reply]

Straightforward:

\sum _{k=1}^{n}{\frac {z(1-z^{k})}{1-z}}=\sum _{k=1}^{n}{\frac {z}{1-z}}-\sum _{k=1}^{n}{\frac {z^{k+1}}{1-z}}={\frac {nz}{1-z}}-{\frac {1}{1-z}}\sum _{k=1}^{n}z^{k+1}=\cdots

Bo Jacoby (talk) 06:55, 28 September 2010 (UTC).[reply]

Decomposing a standard normal CDF

Hi, a simple question (not homework): I have a standard normal CDF evaluated at some point: $\Phi ({\frac {a+b}{c}})$ . Is there any way to decompose this into $\Phi (\cdot )+\Phi (\cdot )$ , presumably where the first part contains only some transformation of a and c and the second part only b and c? a, b and c are just numbers. (Maybe I should rather formulate this as "are there any functions f and g so that $\Phi (a+b)=\Phi (f(a))+\Phi (g(b))$ .) I feel this should be possible (but it probably isn't, couldn't figure it out by myself). Jørgen (talk) 07:43, 28 September 2010 (UTC)[reply]

This is impossible. Differentiate both sides of

\Phi (a+b)=\Phi (f(a))+\Phi (g(b))

wrt a. You get

\phi (a+b)=f'(a)\phi (f(a))\;\!

. Thus

\phi (a+b)

does not depend on b, which is a contradiction. -- Meni Rosenfeld (talk) 10:00, 28 September 2010 (UTC)[reply]

Good point. Thanks! (What about multiplicative decomposition? Is that possible?) Jørgen (talk) 10:47, 28 September 2010 (UTC)[reply]

No, by the same argument, with first taking logs. -- Meni Rosenfeld (talk) 11:37, 28 September 2010 (UTC)[reply]

N throws of an unbalanced coin

I have an unbalanced coin that has a probability p of coming up heads when thrown. How do I calculate the probability of getting m heads when throwing it n times? Would it just be 1-(p^(n-m)) ? Not a homework question. Thanks. 92.28.249.130 (talk) 13:22, 28 September 2010 (UTC)[reply]

The answer is

{\binom {n}{m}}p^{m}(1-p)^{n-m}

. See Binomial distribution. -- Meni Rosenfeld (talk) 13:31, 28 September 2010 (UTC)[reply]

Simple check of your original idea: the probability of getting 1 head with 1 throw (i.e. m = n = 1) is, by definition, p. But 1 − (p^(1−1)) = 1 − 1 = 0. Gandalf61 (talk) 13:53, 28 September 2010 (UTC)[reply]

Thanks. Is there an online calculator for this anywhere please? 92.29.114.118 (talk) 20:33, 4 October 2010 (UTC)[reply]

I imagine you can use Wolfram Alpha if you know the relevant Mathematica syntax. (I don't.) —Tamfang (talk) 23:38, 4 October 2010 (UTC)[reply]

Topology

SHOW THAT THE COLLICTION AQ={R,Ø}U{(Q,∞):Q IS RATIONAL NUMBER} IS NOT ATOPOLOGICAL SPACE. —Preceding unsigned comment added by 86.108.51.220 (talk) 16:49, 28 September 2010 (UTC)[reply]

Note that the Reference Desk will not do your homework for you. Check the axioms of topological spaces: does your collection contain the empty set and the full set? It obviously does. Is it closed under finite intersections? Is it closed under arbitrary unions?—Emil J. 17:08, 28 September 2010 (UTC)[reply]

An arbitrary union of open sets in a topological space is open. So look at the union of all intervals (Q, ∞) such that Q > √2. That's not a set of the form (Q, ∞) where Q is rational. Michael Hardy (talk) 03:26, 29 September 2010 (UTC)[reply]

Horizontal acceleration

Okay a child pulls an 11 kg wagon with a horizontal handle whose mass is 1.8 kg, giving the wagon and handle an acceleration of 2.3 m/s^2. What is the magnitude of the other horizontal forces acting on the child, besides the tension in the handle? Assume that the child moves along with the wagon.

I calculated the tension on both ends of the handle to have magnitudes of 29.4 N (the end at his hand) and 25.3 N (the end attached to the wagon. What other horizontal forces should I be considering?209.6.54.248 (talk) 22:51, 28 September 2010 (UTC)[reply]

I agree that your figures of 25.3 N and 29.4 N are correct. To calculate the resultant force on the child you need to know the mass of the child. When you know the resultant force on the child you can determine the horizontal force between the child's shoes and the ground. Dolphin (t) 22:58, 28 September 2010 (UTC)[reply]

OP here, the mass of the child isn't given in the problem--I'm tempted to think that I can simply add 25.3 N and 29.4 N and divide by the acceleration 2.3 m/s^2 but that can't be right because there are additional forces which would be on the other side of the equation. Is there something that I'm missing?209.6.54.248 (talk) 00:49, 29 September 2010 (UTC)[reply]

Without knowing the mass of the child your only option is to give a qualitative answer to the question What is the other horizontal force acting on the child? There is a horizontal force between the ground and the child's shoes. Seeing the child is accelerating at 2.3 m.s^-2, the resultant horizontal force F on the child is 2.3 times the mass of the child. So there is a horizontal force between the child and the ground of F plus 29.4 N. Dolphin (t) 01:31, 29 September 2010 (UTC)[reply]