Wikipedia:Reference desk/Archives/Mathematics/2010 January 25

= January 25 =

Cryptography
I can't figure this out. Can someone else decrypt it?

9 ka 1f 2c 1t g1 sj g3 ni aa hm a2 ub dh g2 y6 1d8 ug 18q m3 dq 1i5 m6 18x 1fz 1bc ep 1e 1e g1 4r 1i kd uc g5 qq 43 p7 9b 5e f3 ce qt km 18n 16s

--J4\/4 &lt;talk&gt; 18:24, 25 January 2010 (UTC)

It is not possible to decrypt without knowing the key and encryption/decryption algorithm. More info is needed. 78.101.208.18 (talk) 20:48, 25 January 2010 (UTC)
 * I think he wants someone to codebreak (you know, like the NSA). --75.50.49.11 (talk) 22:57, 25 January 2010 (UTC)
 * Breaking a small sample of code like this is almost impossible unless you can guess at what it might say, or you have seen something similar before, or you happen to make a very lucky guess. A computer could spend a thousand years of processing without getting the true meaning, though it would come up with very many possible but wrong meanings.    D b f i r s   01:20, 31 January 2010 (UTC)

Differential equation
I was given a differential equation to solve: 3y'=5y^(2/5) or something, over the domain -infinity to infinity. We're also given than y(-1)=2 and y(4)=32...but how can this be? Solutions are of the form y=(x+C)^(3/5), so there would have to be two different C values for x=-1 and x=4. Now I know that, if say x=0 is not allowed, the domain is split in two an each part of the domain gets a seperate C value...but why would this apply to this case here? —Preceding unsigned comment added by 173.179.59.66 (talk) 19:09, 25 January 2010 (UTC)
 * Your solution isn't one for that DE, and you should only need one initial condition for a first-order equation. Are you stating the problem correctly?  (It's worth noting that your DE is autonomous; that's the source of the $$x+C$$ form.)  --Tardis (talk) 20:21, 25 January 2010 (UTC)
 * (ec) First note that the solution you wrote is not correct (have you checked it?). To find a solution, try the ansatz y(x):=c|x|αsgn(x) (which simply writes cxα in case α is a negative integer) and determine c and α plugging y(x) it into the equation. But first go and check the equation you were given: I bet it was 3y'=5|y|2/5, with the absolute value. Since the equation is autonomous (no x appears in it), y(x-a) is also a solution, for any real a. Not only: you may take y(x)=0 in some interval [a,b] and attach to it solutions of the above form for x>b and x<a vanishing at a and respectively b (check that such a function satisfies the ODE at any real x). For your problem, take a=-1 and b=4. You may be confused because the initial value problem for this equation does not have unicity, but note that the Cauchy-Lipschitz-Lindelöf-Picard-&c.. unicity theorem does not apply, because the RHS lacks the Lipschitz condition. pma.--84.220.118.69 (talk) 20:22, 25 January 2010 (UTC)

Step by step:
 * $$3\frac {dy}{dx}=5y^{\frac 2 5}$$
 * $$dx=\frac 3 5 y^{-\frac 2 5} dy=d(y^{\frac 3 5})$$
 * $$y^{\frac 3 5}=x-C$$
 * $$y=(x-C)^{\frac 5 3}$$

Another solution is
 * $$y=0\,$$

so the general solutions can be written
 * $$y=-[x< C_1](C_1-x)^{\frac 5 3}+[C_2< x](x-C_2)^{\frac 5 3}$$

using the Iverson bracket for notation, and assuming that
 * $$-\infty \le C_1 \le C_2 \le \infty$$.

Bo Jacoby (talk) 18:05, 26 January 2010 (UTC).

Topological Group Question
I'm having trouble figuring out this exercise: If G is a compact topological group, and g is in G, let A = {g0, g1, g2,... }.  Show that the closure of A is a subgroup of G.  This is problem 3 in section 15 of Bredon's Topology and Geometry (p. 55).

I'm trying to show that g-1 is in the closure of A and then I think the rest should be pretty straight forward. Since G is compact, the sequence g0, g1, g2,... has to have a subsequence that converges, but I can't figure out how to prove that there's one that converges to g-1. Rckrone (talk) 23:57, 25 January 2010 (UTC)


 * Maybe just show the closure K satisfies gK=K. gK ⊆ K since a subsequence times g is still a subsequence. K ⊆ gK, because a convergent subsequence doesn't change its limit when you remove g^0, and so a convergent subsequence (minus at most one term) divided by g is still a convergent subsequence. JackSchmidt (talk) 01:18, 26 January 2010 (UTC)


 * For showing K ⊆ gK, wouldn't you also need to show that g0 is in gK (which is equivalent to showing g-1 is in K)? I think what you did is not dependent on G being compact, and the property doesn't necessarily hold when compactness is removed. Rckrone (talk) 01:32, 26 January 2010 (UTC)


 * As far as equivalence goes, yes of course, that's how logic works. I show something equivalent to what you want is true, so what you want is true.  What I did is very much dependent on K being compact, otherwise why would K have any points besides A in it?  In other words, where do all the convergent subsequences come from? JackSchmidt (talk) 01:49, 26 January 2010 (UTC)
 * I think you misunderstood my objection. You did not show that g0 is in gK, which is the crux of the problem. Rckrone (talk) 04:40, 26 January 2010 (UTC)


 * (ec)The proposition is false if you don't assume compactness, so it must be used somehow. First, eliminate the case where A is finite since then g is finite order. Then by compactness, let x be a limit point. Let gni be a subsequence that converges to x. It shouldn't be hard to show that the set gni-nj (i>j) has 1 as a limit point. So 1 is a limit point of A and from there is trivial to show that g-1 is a limit point of A. I think that will work as an outline at least.--RDBury (talk) 01:46, 26 January 2010 (UTC)
 * The problem I have is that if gni and gnj are "close" I don't know that implies that gni-nj and 1 are "close". For example suppose in R that gn = 1/n.  Obviously that's not a group, but I'm not sure what about a topological group makes something like that not happen. Rckrone (talk) 02:28, 26 January 2010 (UTC)
 * Multiplying by g−nj is a homeomorphism. Algebraist 02:33, 26 January 2010 (UTC)
 * Note that a topological group is a uniform space, and that "x and y are close" exactly means that xy-1 is close to 1. --pm a 09:15, 26 January 2010 (UTC)
 * Sorry if this is just me being dense, but I'm not sure what to do with that. Say I have some open neighborhood U of 1.  I want to show that there's some gm in U (m>0).  So I try to show that there's some i and j, i<j such that gi(U) contains gj.  One way that came to mind was to argue that the gi(U)'s covered the closure of A.  But I don't how to show that. Rckrone (talk) 03:16, 26 January 2010 (UTC)
 * Ok I think I got it. For any neighborhood U of 1 you can find a symmetric open subset of U, call it V.  For any point x not A but in the closure of A, there is a subsequence that converges to x, so there's a gn in x(V) so y = x-1gn is in V.  gn = xy so gny-1 = x.  y-1 is also in V, so x is in gn(V).  That shows that the gi(V)'s and therefore the gi(U)'s cover the closure of A, which is compact, so there's a finite subcover.  Then for all open U around 1, there's some gj in gi(U) with j>i, so gj-i is in U.  So there's a subsequence that converges to 1, and from that a subsequence that converges to gn for each n<0. Thanks for the nudges. Rckrone (talk) 04:19, 26 January 2010 (UTC)
 * Just a small detail: since G is not assumed to be first countable, it may fail to be sequentially compact. I'd like more your proof rephrased this way: for any nbd U of the identity there exists n≥0, such that ggn &isin;U. This can be written as gn &isin; g-1 U, and reads exactly : g-1 belongs to the closure of A. (any nbd of g-1 meets A) pma. --84.220.118.69 (talk) 08:05, 26 January 2010 (UTC)
 * That makes sense but I'm confused about sequential compactness. If a space is compact, then any net has subnet that converges.  Isn't a sequence a net?  So wouldn't any sequence in a compact space have a convergent subsequence?  Obviously I'm doing something wrong. Rckrone (talk) 06:27, 27 January 2010 (UTC)
 * A sequence is a net, yes. But a subnet of a sequence is not necessarily a sequence. — Emil J. 12:35, 27 January 2010 (UTC)
 * Ok, I'll have to look into that more carefully. Thanks.  Rckrone (talk) 17:46, 28 January 2010 (UTC)