Wikipedia:Reference desk/Archives/Mathematics/2011 May 18

= May 18 =

Rubik's Cube Group Theory!
If you take a sequence of turns on a Rubik's cube and repeat them $$n$$ times without changing orientation of the cube between applications, then you are effectively computing $$g^n$$, where $$g$$ is the element of the Rubik's Cube group representing that sequence. It is a fact of group theory, then, that you are guaranteed to return to solved position eventually by repeating the sequence.

Question: is there a nontrivial example of a sequence of turns which will reach a solved position in the middle of one of the applications of the sequence of turns? More formally, if $$g = t_1 \circ \cdots \circ t_i$$, where $$t_i$$ are single turns of a face, then we know $$g^n = e$$ for some $$n$$, but is there the possibility that $$g^j \circ t_1 \cdots \circ t_k = e$$ where $$1<k<i$$?

Examples of trivial cases would be if the sequence of turns is, say, just 3 90° rotations of the same face clockwise; here it takes 4 applications to get back to identity, but it really only takes 1 application and then the turn of the next application. Another example would be if the sequence of turns itself returns to identity before reaching something else. I haven't made a precise formulation of what 'trivial' means here, and reserve the right to expand on what should be considered trivial if shown some further counterexamples. --72.179.51.84 (talk) 02:47, 18 May 2011 (UTC)
 * One possible first pass at making my first trivial case more precise would be to require that the sequence of turns not itself be a repetition of a sub-sequence more than one time. --72.179.51.84 (talk) 02:54, 18 May 2011 (UTC)


 * You are guaranteed to return to the original position eventually by repeating the sequence. You are not guaranteed to return to the solved position eventually by repeating the sequence. Bo Jacoby (talk) 06:03, 18 May 2011 (UTC).
 * Above is right. If you take a really jumbled up cube and g is a single face turn, then doing gn will never give a solved cube. It will just cycle between 4 different cube states. Staecker (talk) 11:33, 18 May 2011 (UTC)
 * I think the OP meant that he starts with the solved position... -- Meni Rosenfeld (talk) 12:26, 18 May 2011 (UTC)
 * I think Meni Rosenfeld is right, but who am I to judge what the OP meant? --72.179.51.84 (talk) 12:33, 18 May 2011 (UTC)
 * If you do not usurp yourself, you are. -- Meni Rosenfeld (talk) 14:06, 18 May 2011 (UTC)
 * Pick any group element $$g$$. Let n be its order. Pick any integer $$j$$ where $$0 < j < n$$. Let $$t_1 \cdots \circ t_k$$ be a solution to $$g^{-j}$$ found by Optimal_solutions_for_Rubik's_Cube (so $$k \le 20$$). Let $$t_{k+1} \cdots \circ t_i$$ be a solution to $$g^{j+1}$$ found by the optimal algorithm. Then $$g = t_1 \circ \cdots \circ t_i$$ satisfies your conditions where $$i \le 40$$. The example should be non-trivial for most choices of $$g$$ and $$j$$. 98.248.42.252 (talk) 14:57, 18 May 2011 (UTC)
 * Thanks, that makes sense. --72.179.51.84 (talk) 13:18, 19 May 2011 (UTC)

sum
∑$$\sum_{v=∈V}^{dv} i$$


 * Arfle barfle gloop? Dmcq (talk) 11:50, 18 May 2011 (UTC)

financial accounting 1
Mr mubaiwa a sole trader drew up his trial balance on 31/02/08. As it would not balance, a suspense account was opened and the difference debited to it. The trial balance is shown below: $           $ Premises                   20 000 Buildings                 30 000 Motor vehicles            10 000 Purchases                 20 000 Stock opening              1 000 Debtors                    6 700 Cash at bank               6 140 Cash in hand                 760 Discount allowed             900 General expenses           1 500 Rent                       1 400 Wages                      1 208 Capital                                61 000 Sales                                  35 000 Discount received                       4 000 Suspense account             392 100 000     100 000

Closing stock as at 31/01/08 is $7 000: Subsequently the following errors were discovered a)The wages had been overcast by $120 b)Sales had been overcast by $100 c)The purchase of machinery was posted to purchases account $2 000 d)Discount received $143 had been posted to the account as $134 e)Discount allowed $200 had been posted to the wrong side of the account f)Bank charges $21 had been posted to the ledger. g)Wages of $3 000 had not been entered into the books.

Required: a)Prepare journal entries to correct the above errors. (8 marks) b)Write up the suspense account, clearing the balance in the account and correcting the errors. (6 marks) —Preceding unsigned comment added by 209.88.92.29 (talk) 09:08, 18 May 2011 (UTC)


 * We're not just going to do your homework for you. First you have to try it, then show us what you did and ask for help where needed. StuRat (talk) 09:35, 18 May 2011 (UTC)

an "elementary calculation"
Hi. I'm reading a paper, and stuck on the following:


 * If $$x$$, $$y$$ are coprime integers and $$n > 1$$, an elementary calculation shows that:
 * $$\left( \frac{x^n\pm y^n}{x \pm y}, x \pm y \right) \mid n$$.
 * One writes $$x = (x \pm y) \mp y$$ and then expands.

How does that work? I tried writing down what it says and raising it to the power $$n$$, but that doesn't seem to help. Any ideas? Thanks in advance if anyone can shed some light here. -GTBacchus(talk) 16:15, 18 May 2011 (UTC)
 * Actually, I've got it worked out now, except in the case where $$n$$ is even and we're looking at $$\frac{x^n + y^n}{x+y}$$. The $$\frac{x^n - y^n}{x-y}$$ case was easy, and so was the plus case when $$n$$ was odd. -GTBacchus(talk) 16:33, 18 May 2011 (UTC)

Simplifying a polynomial fraction
How do I turn (3-3x)/[(3x-2)(x-1)] into -3/(3x-2)? When I distribute the denominator I get 3x^2-5x+2, so I guess that's wrong since I somehow need 3x-2 for the denominator. —Preceding unsigned comment added by 142.132.6.81 (talk) 17:38, 18 May 2011 (UTC)
 * $$3-3x = 3(1-x) = -3(x-1)$$ Readro (talk) 17:42, 18 May 2011 (UTC)


 * (ec) I added a more useful (sub)title. Ok, let me rewrite it first:

(3-3x) - (3x-2)(x-1)


 * Now take a 3 out of the numerator:

3(1-x) - (3x-2)(x-1)


 * Make that a negative 3:

-3(-1+x) - (3x-2)(x-1)


 * Reorder the portion of the numerator in parens:

-3(x-1) - (3x-2)(x-1)


 * The (x-1) cancels:

-3 (3x-2)


 * We're done. StuRat (talk) 17:48, 18 May 2011 (UTC)

I was trying so hard to change the denominator when it was just a thing with the numerator all along? Thanks Readro and thanks so much for taking it all apart StuRat!142.132.6.81 —Preceding unsigned comment added by 142.132.6.81 (talk) 19:02, 18 May 2011 (UTC)


 * You're welcome. (I've marked this Q resolved.) StuRat (talk) 20:22, 19 May 2011 (UTC)

Diophantine equations
Hello. WHat are some methods to solve simple Diophantine equations? I saw your article and I don't think I understood it very well, to a large part I suspect because it was being more rigorous than I need, and accounting for possibilities or special cases that I personally am unlikely to encounter. I would like a brief walk-through or a link to a site with a brief walk through, if possible. Thanks. 72.128.95.0 (talk) 23:55, 18 May 2011 (UTC)


 * I wrote a brief page about using the extended Euclidean algorithm to solve two-variable linear Diophantine equations for a class I taught last year, which includes a few examples. Perhaps you would find it helpful: . —Bkell (talk) 13:20, 19 May 2011 (UTC)